tag:blogger.com,1999:blog-80443742024-03-14T02:41:23.667+05:30IIM CAT Preparation Tips2IIM's blog on cracking the quant section of the CAT.
Frequent posts on interesting questions, tips, shortcuts in math.Anonymoushttp://www.blogger.com/profile/01307897270318146959noreply@blogger.comBlogger157125tag:blogger.com,1999:blog-8044374.post-44909792673049741012015-09-22T18:01:00.002+05:302015-09-30T12:20:05.722+05:30Wilson's Theorem for CAT<div dir="ltr" style="text-align: left;" trbidi="on">
Wilson's theorem states that for any prime number 'p', (p-1)! divided by p leaves a remainder of p - 1.<br />
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In other words,<br />
16! divided by 17, remainder is 16.<br />
12! divided by 13, remainder is 12<br />
10! by 11, remainder is 10<br />
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and so on.<br />
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This has an interesting extrapolation. For any prime number 'p', (p-2)! divided by p leaves a remainder of 1.<br />
15! divided by 17, remainder is 1.<br />
11! divided by 13, remainder is 1<br />
9! by 11, remainder is 1<br />
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and so on.<br />
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The proof for this theorem is interesting. If we take any prime 'p', we can have two interesting observations about it<br />
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1. For any number that leaves a remainder 'a' on division by p, we can find a unique remainder 'b' on division by p such that a*b leaves a remainder of 1 on division by p. (An extrapolation of Linear Congruence Theorem)<br />
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2. For any prime p, of there is a number n such that n^2 leaves a remainder 1 on division by p, then n has to leave a remainder either 1 or -1 on division by p. The proof for this is fairly simple.<br />
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If n^2 leaves a remainder of 1 on division by p, then n^2 = kp + 1. or (n -1) (n +1) should be a multiple of p. If p > 2, then either n -1 or n + 1 has to be a multiple of p. Or, n divided by p leaves a remainder of 1 or -1 on division by p.<br />
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So, if we take (p-1)!, this is nothing but 1 * 2 * 3 * 4 * ....(p-1). Of these let us remove 1 and (p-1), the remaining p - 3 numbers can be paired up as a * b such that a*b leaves a remainder of 1 on division by p. They will be ( p-3)/2 such pairs. The product of all these pairs multiplied together will leave a remainder of 1 on division by p.<br />
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So, 1 * 2 * 3 * ......(p-1), will effectively leave a remainder 1 * 1 * (p-1) = (p -1).<br />
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Fabulous theorem. But one can almost take it for granted that the CAT will not have a question based on this theorem in the exam. This will get slotted under the category of 'too tough to be tested in CAT'.<br />
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Similarly, <a href="http://iimcat.blogspot.in/2010/11/fermats-little-theorem-and-eulers-phi.html" target="_blank">Fermat's Little Theorem and Euler's Phi function</a> would also be considered too tough for CAT.<br />
<br /></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-89270810767814001022015-08-24T20:18:00.004+05:302015-09-30T12:21:08.938+05:30Data Interpretation - Average growth rate vs. CAGR<div dir="ltr" style="text-align: left;" trbidi="on">
Growth rate, average growth rate, CAGR - these are three terms that appear frequently in exams. I am going to examine each of this with an example.<br />
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Let us say company XYZ has revenues in 5 consecutive years as follows<br />
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2011 - Rs. 1200 crores<br />
2012 - Rs. 1250 crores<br />
2013 - Rs. 1310 crores<br />
2012 - Rs. 1380 crores<br />
2012 - Rs. 1500 crores<br />
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For the period from 2011 to 2015, we can calculate all three - growth rate, average annual growth and CAGR. Let us do each of this.<br />
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<b>Growth rate from 2011 to 2015.</b><br />
Revenues have grown from Rs. 1200 crores to Rs. 1500 crores, a growth of Rs. 300 crores. Growth rate = 300/1200 expressed as a percentage = 25%.<br />
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Average annual growth rate from 2011 to 2015<br />
We need to calculate growth rate in each year and then compute the average of those growth rates<br />
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Year Revenues growth rate<br />
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2011 - Rs. 1200 crores<br />
2012 - Rs. 1250 crores 4.2%<br />
2013 - Rs. 1310 crores 4.8%<br />
2012 - Rs. 1380 crores 5.3%<br />
2012 - Rs. 1500 crores 8.7%<br />
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Average of 4.2%, 4.8%, 5.3% and 8.7% = 5.75%<br />
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<b>CAGR of revenues from 2011 to 2015</b><br />
The CAGR is an interesting idea. Conceptually, it is that equal growth rate, if it had been present for all 4 years that would have taken us from Rs. 1200 crores to 1500 crores. Or, if we assume some growth rate 'r' in each year that takes us from the starting point to the end point, that r is the CAGR.<br />
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So, in our question 1200( 1 + r)^4 = 1500.<br />
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How so, each year's revenues would be (1 + r) times the previous year's revenue. So, the final year's revenue would be (1 + r) ^4 of the first year's revenue.<br />
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(1 + r)^4 = 1.25<br />
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Or, CAGR = 5.73%.<br />
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Usually, CAGR and average growth rate will be reasonably close to each other. Both give a reasonable sense of how things have been. If the growth rates are wildly variant, then CAGR and average growth rate will be far apart. Otherwise they will be close to each other.<br />
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<b>Why CAGR is often better</b><br />
Imagine a stock price that jumps by 100% in year 1 and falls by 50% in the second year. Growth rates are 100% and -50%. Average growth rate would be 25%. Sounds like fantastic returns for an investor. But this is wrong. Dramatically wrong. If the stock's original price had been Rs. 20. At the end of year 1, it would have been Rs. 40, at the end of year 2 it would have been back to Rs. 20. So, the overall growth is zero. Zlich. Nada. CAGR is 0. Remember that percentages are calculated on the starting point. So, for the same change, going up will you will see a higher percentage growth than coming down. If a stock goes from 100 to Rs. 1000 it is a 900% growth rate. If it goes the other way around, it is only a 90% decline. Oops!<br />
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So, if any mutual fund manager pitches his fund to you based on average growth, ask him for the CAGR and then politely ask him to leave. :-)<br />
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<b>How to compute CAGR using a calculator.</b><br />
Let us do this with an example. If revenues from $300m to $ 369m in 4 years, what is the CAGR? 300( 1+r)^4 = 369.<br />
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(1+r)^4 = 369/300 = 1.23<br />
(1+r) = 1.23^0.25 = 1.053.<br />
r = 5.3%<br />
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If we go from Rs. x to Rs. y in 'n' years. x(1+r)^n = y.<br />
(1 + r) = (y/x) ^ (1/n)<br />
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r = (y/x)^(1/n) - 1.<br />
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Now, remember, this formula will give 'r' as 0.625 or something like that. We need to think of it as 6.25%.<br />
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Let us finish with another example.<br />
Revenues grow from Rs. 6000 crores to Rs. 8250 crores in 5 years, what is the CAGR?<br />
8250/6000 = 1.375<br />
1.375^0.2 = 1.06576.<br />
Or, r = 0.06576 or 6.58%.<br />
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<br /></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-31709049518782532762015-07-20T22:20:00.000+05:302015-09-30T12:22:01.240+05:30CAT Coaching Onlne - Links across topics in Math<div dir="ltr" style="text-align: left;" trbidi="on">
Some simple interesting patterns emerge when we look across topics. Very often, thinking about these patterns helps us wind our heads around one or the other topic. In this post, we discuss a few of those links. Nothing profound, but just an outline to help remember simple ideas<br />
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<b>Simple Interest Compound Interest and Progressions.</b><br />
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The amounts at the end of each year form an AP if interest is calculated on a simple interest basis, and form a GP if interest is calculated on a compound interest basis.<br />
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For instance if amount invested is Rs. 1000 and interest rate were 10%, then amount at the end of yr 1, yr 2, yr 3,.. would be Rs. 1100, 1200, 1300,...Each year's amount is previous year's amount + Rs. 100<br />
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The same amount of Rs. 1000 invested at same 10% but on a compound interest basis would give us amounts at the end of yr 1, yr 2 of Rs. 1100, 1210, 1331...Each year's amount is previous year's amount * 1.1. This is the basis for the formula for amount on principal invested in Compound Interest basis<br />
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<b>Combinatorics - This can be linked with almost any topic</b><br />
See if you can spot some idea in some other topic that is 'sitting inside' the following combinatorics questions<br />
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1. Consider a shelf that has 4 copies of a book and 3 copies of a painting. In how many ways can we select at least one article from this shelf?<br />
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2. Consider a shelf that has 4 different books and three different paintings. In how many ways can we select at least one article from this shelf?<br />
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3. In how many ways can be place 5 different toys into 3 different boxes such that all 5 toys are allotted and no box is empty?<br />
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How else these can be interpreted is given below.<br />
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1. Consider a shelf that has 4 copies of a book and 3 copies of a painting. In how many ways can we select at least one article from this shelf?<br />
<span style="color: blue;">How many factors greater than 1 does the number 2^4 & 3^3 have?</span><br />
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2. Consider a shelf that has 4 different books and three different paintings. In how many ways can we select at least one article from this shelf?<br />
<span style="color: blue;">How many non-empty subsets does the set {1, 2, 3, 4, 5, 6 7} have?</span><br />
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3. In how many ways can be place 5 different toys into 3 different boxes such that all 5 toys are allotted and no box is empty?<br />
<span style="color: blue;">How many onto fucntions can be defined from the set {1, 2, 3, 4, 5} to the set {a, b, c}?</span><br />
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<b>Weighted averages and Mixtures.</b><br />
Class A has 40 students whose average mark is 40, class B has 60 students whose average mark is 50. What is the overall average? Class A has an average mark of 40 and class B has a an average mark of 50. If the overall average is 46, what is the ratio of students in class A and class B? These two questions are two sides of the same coin. Sometimes we teach one in weighted averages and the other in mixtures.<br />
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Mathematicians are supposed to love connections. There are far more powerful connections across topics than the ones I have mentioned above. It is instructive to observe even these simple connections.<br />
<br /></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-87764750781212870072015-06-02T23:07:00.000+05:302015-09-30T12:22:43.642+05:30Online CAT Coaching: A few interesting True/False questions from Geometry<div dir="ltr" style="text-align: left;" trbidi="on">
State whether the following statements are true or false<br />
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1. A parallelogram that circumscribes a circle has to be a square<br />
2. A trapezium inscribed in a circle has to be an isosceles trapezium<br />
3. Orthocenter of a triangle can lie outside the triangle<br />
4. Triangle with sides a, b and c has the relationship a^2 + b^2 > c^2, the triangle has to be acute-angled.<br />
5. Diagonals of a parallelogram are angle bisectors of the angles of a parallelogram.<br />
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Scroll down for answers and explanation<br />
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1. A parallelogram that circumscribes a circle has to be a square: FALSE<br />
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In a parallelogram, opposite sides are equal. In a quadrilateral, the sums of pairs of opposite sides are equal. So, a parallelogram that circumscribes a circle should have all 4 of its sides equal. Or, it should be a Rhombus; it need not be a square.<br />
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2. A trapezium inscribed in a circle has to be an isosceles trapezium: TRUE<br />
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An isosceles trapezium is a symmetric diagram. The two base angles should be equal and the two top angles should be equal. So, a trapezeium where the base angles were equal would be an isosceles trapezium.<br />
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In any cyclic quadrilateral, opposite angles would be supplementary. In a trapezium, co-interior angles between the parallel lines would be supplementary. So, if we took a trapezium ABCD with AB parallel to CD inscribed in a circle. Angle A and Angle D would be supplementary (co-interior angles). And Angle A and Angle C would be supplementary (opposite angles of a cyclic quadrilateral). Or angle B would be equal to angle C. Ergo, isosceles trapezium.<br />
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3. Orthocenter of a triangle can lie outside the triangle: TRUE<br />
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For any obtuse-angled triangle, two of the altitudes would lie outside the triangle, and would intersect at a point outside the triangle. So, the orthocenter can lie outside the triangle.<br />
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4. Triangle with sides a, b and c has the relationship a^2 + b^2 > c^2, the triangle has to be acute-angled: FALSE<br />
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Let us take triangle with sides 2, 3 and 4. 4^2 + 3^2 > 2^2. But as 2^2 + 3^3 < 4^2, the triangle is obtuse-angled. Is a^2 + b^2 > c^2, we can say angle C is acute-angled. We cannot say all three angles are acute-angled. One can use cosine rule also for having a go at this question (though it should be considered inelegant)<br />
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5. Diagonals of a parallelogram are angle bisectors of the angles of a parallelogram: FALSE<br />
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Diagonals of a parallelogram bisect each other. They need not bisect the angles of the parallelogram. Imagine this, if we took a rectangle and studied its diagonals. if the diagonals bisected each other, the angle between diagonal and a side would be 45 degrees. Or, we would end up having a square. So, any rectangle that was not a square would have diagonals that were not angle bisectors. So, diagonals of a parallelogram NEED NOT be angle bisectors of the angles of a parallelogram.</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-79714846020366530832015-04-29T15:19:00.000+05:302015-09-30T12:25:31.729+05:30CAT Online Coaching - Permutation and Combination, Fixing the Errors III<div dir="ltr" style="text-align: left;" trbidi="on">
<div class="MsoNormal" style="margin-bottom: 0.0001pt;">
<span lang="EN-US" style="font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";"><a href="http://iimcat.blogspot.in/2015/04/cat-online-preparation-permutation-and.html" target="_blank">This post</a> had given a series of
questions with incorrect solutions. Given below are the "debugged"
solutions to questions 7, 8 and 9.
<div class="separator" style="clear: both; text-align: center;">
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<div class="MsoNormal">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">7. What is the
probability of selecting 3 cards from a card pack such that all three are face
cards? what is the probability that none of the three is a face card? <o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution</span></b><span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Number of cards in a
card pack = 52<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Numbert of face cards
in a card pack = 12<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Number of ways of
selecting 3 cards from a card pack = 52C3<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Number of ways of
selecting 3 face cards from a card pack = 12C3<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Probability of
selecting three cards such that all three are face cards = 12C3/52C3<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Probability of
selecting three cards such that none of the three are face cards = 1 -
12C3/52C3<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution:</span></b><span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Other possibilities
exist. As in, if we select three cards from a card pack, all three could be
face cards, all three could be non-face cards, one could be a face card with 2
non-face cards or we could have two face cards and one non-face card. This is
why we cannot use the 1 minus idea.<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">As a rule we can say
P(A) = 1 – P(B) – P(C) if A, B and C are mutually exclusive and collectively
exhaustive events. As in among them they should account for all possible events.
And there should be no overlap. Creating a group of MECE equiprobable events is
the most fundamentally brilliant idea in all of probability. Go on, look it up.<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:<o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">This is simple. Probability
of selecting three cards such that none of the three are face cards = 40C3/52C3<o:p></o:p></span></div>
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<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">8. A die is rolled
thrice. In how many outcomes will two throws be same and the third one
different?<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution</span></b><span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Let the three
outcomes be ABC.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">'A' can take all
values from 1 to 6<o:p></o:p></span></div>
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<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">'B' can also take all
values from 1 to 6 <o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">'C' can take all
values except A - so it has 5 possibilities<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Total number of
outcomes = 6 * 6 * 5 = 180.<o:p></o:p></span></div>
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<br /></div>
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<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution:</span></b><span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">The given solution is
actually absurd. If two throws are to be same, then if A can take values from 1
to 6 and if B were equal to A, then b can take only one value. There can be no
6 * 6 * 5
<div class="separator" style="clear: both; text-align: center;">
<a href="http://www.chennai.2iim.com/democlass.shtml" style="margin-left: 1em; margin-right: 1em;"><img src="http://2iim.com/picts/DC.png" height="15" width="200" /></a></div><o:p></o:p></span></div>
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<br /></div>
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<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:<o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">'A' can take all
values from 1 to 6<o:p></o:p></span></div>
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<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">'B' should be equal
to A --- One possibility <o:p></o:p></span></div>
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<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">'C' can take all
values except A - so it has 5 possibilities<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">6 * 1 * 5 = 30
outcomes.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">There are 30 possible
outcomes when A = B but not equal to C. Likewise, we could have A = C but not
equal to B and B = C but not equal to A. So, there are totally 90 different
possibilities.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">9. How many 7 letter
words can we have in English that have two distinct vowels and 5 distinct
consonants.<o:p></o:p></span></div>
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<br /></div>
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<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution</span></b><span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0cm; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Now, we know there
are 5 vowels and 21 onsonants. So, we need to select 2 from these 5 and 5 from
the remaining 21. Since order is important, we need to select keeping order in
mind.<o:p></o:p></span></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
<span lang="EN-US" style="background: white; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt;">So, we
have 5P2 * 21P5.</span><span lang="EN-US" style="font-size: 16.0pt; line-height: 107%; mso-bidi-font-size: 12.0pt;"><o:p></o:p></span></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution:<o:p></o:p></span></b></div>
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">In this solution we
do not account for intermingling of vowels and consonants. As in, we do not
count words such as BACEDFG. We account for order within vowels and order
within consonants, but we do not account for order across both categories.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><br /></span></div>
<div class="MsoNormal">
<b><span lang="EN-US" style="color: #002060; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><span style="font-size: 12.0pt; line-height: 107%; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;">Select without
accounting for order, then arrange everything put together<o:p></o:p></span></div>
<br />
<div class="MsoNormal">
<span lang="EN-US" style="background: white; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt;">So, we
have 5C2 * 21C5 * 7!.</span><span style="font-size: 12.0pt; line-height: 107%; mso-ansi-language: EN-IN; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-IN;"><o:p></o:p></span></div>
<div class="MsoNormal">
<span lang="EN-US" style="background: white; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial; mso-bidi-font-size: 10.0pt;"><br /></span></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-88442451102319948422015-04-28T18:29:00.003+05:302015-09-30T12:28:40.586+05:30CAT Online Coaching - Permutation and Combination, Fixing the Errors II<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";"><a href="http://iimcat.blogspot.in/2015/04/cat-online-preparation-permutation-and.html" target="_blank">This post</a> had given a series of questions with incorrect solutions. Given below are
the "debugged" solutions to questions 4, 5 and 6. The questions and the incorrect solutions <a href="http://iimcat.blogspot.in/2015/04/cat-online-preparation-permutation-and.html" target="_blank">here</a></span></div>
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<div class="separator" style="clear: both; text-align: center;">
<a href="http://www.chennai.2iim.com/democlass.shtml" style="margin-left: 1em; margin-right: 1em;"><img src="http://2iim.com/picts/DC.png" height="15" width="200" /></a></div>
<br /></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">4. In how
many ways can we select 2 cards from a card pack such that both belong to the
same suit?<o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution<o:p></o:p></span></b></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">The first
card can be selected in 52 ways, as it can be any card in the pack. Now, there
are 12 different ways of selecting the second card. Depending on the suit of
the first card, the choice set we have for the second card gets limited. If the
first card were a club, then the second card must be from among the 12
remaining clubs and so on.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">So, total
number of outcomes = 52 * 12.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution: <o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">We are
double-counting when we do this. If we select 2H as the first card and 5D as
the second card; this is same as selecting 5D as the first card and 2H as the
second card. However, in our current approach, we will end up counting both of
these.<o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><b><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></b></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">The
correct approach would be to select one of the four suits first and then select
2 cards from 13 in that suit. Or, the number of ways of doing this = 4 * <sup>13</sup>C<sub>2</sub>
= 52 * 6.<o:p></o:p></span></div>
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<br /></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">5. When a
coin is tossed 10 times, what is the probability of getting more heads than
tails?<o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution<o:p></o:p></span></b></div>
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<div class="separator" style="clear: both; text-align: center;">
<a href="http://online.2iim.com/" style="margin-left: 1em; margin-right: 1em;"><img src="http://2iim.com/picts/onlinebanner1.jpg" height="12" width="400" /></a></div>
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">When a
coin is tossed 'n' times, the probability of getting more heads than tails =
Probability of getting more tails than heads. So, we do not need to really
compute this. Probability = 1/2.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution: <o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">We are
not counting the scenarios where there are an equal number of heads and tails.
If we threw the coin an odd number of times, the probability of getting more
heads than tails will be ½. <o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><b><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></b></div>
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%;">Number of
heads could be 6, 7, 8, 9, or 10. Probability = ( <sup>10</sup>C<sub>6</sub> + <sup>10</sup>C<sub>7</sub>
+ <sup>10</sup>C<sub>8</sub> + <sup>10</sup>C<sub>9</sub> + <sup>10</sup>C<sub>10</sub>)/
2<sup>10</sup><o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%;">The other
way of thinking about this would be ½ (1 – <sup>10</sup>C<sub>5</sub>)/ 2<sup>10</sup>.
If we removed the scenario where the number of heads and tails are equal, the
probability of more heads than tails should be equal to probability of more
tails than heads. <o:p></o:p></span></div>
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<br /></div>
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<span style="font-size: 12.0pt; line-height: 107%;">6. In how
many ways can we arrange 3 boys and 3 girls on a straight line such that no two
boys stand next to each othe</span><span style="font-size: 12pt; line-height: 107%;">r</span></div>
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<span style="font-size: 12pt; line-height: 107%;"><br /></span></div>
<div class="MsoNormal">
<b><span style="color: #002060; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution</span></b><span style="font-size: 12.0pt; line-height: 107%;"><o:p></o:p></span></div>
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%;">We can
arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we
would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total
number of possibilities = 2 * 3! * 3! = 72 ways.<o:p></o:p></span></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><br /></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution: <o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">We are
not accounting for arrangement such as BGBGGB and BGGBGB. Note that the
question says that no two boys should stand next to each other. Two girls can
stand next to each other.<o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><b><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></b></div>
<br />
<div class="MsoNormal">
<span style="font-size: 12.0pt; line-height: 107%;">There are 4
possible outlines – BGBGBG, GBGBGB, BGBGGB or BGGBGB. So, total number of
possibilities = 3! * 3! * 4 = 144. <o:p></o:p></span></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-87885184435297501372015-04-25T11:56:00.002+05:302015-09-30T12:30:29.454+05:30CAT Online Coaching - Permutation and Combination, Fixing the Errors<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";"><a href="http://iimcat.blogspot.in/2015/04/cat-online-preparation-permutation-and.html" target="_blank">This post</a> had given a series of
questions with incorrect solutions. Given below are the "debugged"
solutions to the first three questions.
<div class="separator" style="clear: both; text-align: center;">
<a href="http://online.2iim.com/" style="margin-left: 1em; margin-right: 1em;"><img src="http://2iim.com/picts/onlinebanner1.jpg" height="12" width="400" /></a></div><o:p></o:p></span></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: "Times New Roman"; mso-fareast-font-family: "Times New Roman";"><br /></span></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">1. Five
boys need to be allotted to 4 different rooms such that each boy is allotted a
room and no room is empty. In how many ways can this be done?<o:p></o:p></span></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution<o:p></o:p></span></b></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Let the
boys be A, B, C, D and E. Let the rooms be 101, 102, 103 and 104. Now, we know
that exactly one room will have two occupants. First, let us try to send 4 boys
to 4 rooms, we can worry about the fifth occupant later on. Let us select 4 out
of the 5 boys first. This can be done in 5C4 ways. Now, these 4 can be allotted
to 4 different rooms in 4! ways. So, 4 boys in 4 rooms can be done in 5C4 * 4!
= 5 * 24 = 120 ways.<o:p></o:p></span></div>
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<br /></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Now, the
fifth boy has to go into one of the rooms. He can do this in 4 ways as there
are 4 different rooms available. So, total number of outcomes = 120 * 4 = 480.<o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution:</span></b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"> </span><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">We end up
double-counting here.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">A, B, C
and D could be allotted rooms 101, 102, 103 and 104 in that order. Post this, E
could “double-up” with A. So, we would have A and E in 101, B in 102, C in 103
and D in 104.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">In
another scenario, E, B, C, D could be allotted 101, 102, 103 and 104 in that
order. Post this, A could “double-up” with E. So, we would have A and E in 101,
B in 102, C in 103 and D in 104.<o:p></o:p></span></div>
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<br /></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">The two
scenarios mentioned above are identical. However, we end up counting both. This
is the reason for the double count. <o:p></o:p></span></div>
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<br /></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Note that
in our method, we end up having a ‘first’ occupant for a room and a ‘second’
occupant. We say, A goes into room number 101 and then E ‘joins’ him. The moment
you do that, ‘order’ creeps in. We end up factoring in order when we shouldn’t.
<o:p></o:p></span></div>
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<br /></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Awesome, isn’t
it.
<div class="separator" style="clear: both; text-align: center;">
<a href="http://www.chennai.2iim.com/democlass.shtml" style="margin-left: 1em; margin-right: 1em;"><img src="http://2iim.com/picts/DC.png" height="15" width="200" /></a></div><o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><b><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">The boys
should be allotted into rooms as 2 + 1 + 1 + 1. As in, exactly one room has 2
occupants. Some two guys have to be room-mates. Let us first select these two
room-mates. This can be done in <sup>5</sup>C<sub>2</sub> ways. After we have
selected these two room-mates, we practically need to just allot these 4 ‘groups’
into 4 rooms. This can be done in 4! Ways.<o:p></o:p></span></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Total
number of options = <sup>5</sup>C<sub>2 </sub>* 4! = 10 * 24 = 240 ways.<o:p></o:p></span></div>
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<br /></div>
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<br /></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">2. How
many 4-digit numbers with 4 distinct digits are possible?<o:p></o:p></span></div>
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<br /></div>
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<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution<o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Let the
4-digit number be 'abcd'. Now,<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'d' can
take 10 possible values - 0 to 9. <o:p></o:p></span></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'c' can
take 9 possible values - 0 to 9 except d<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'b' can
take 8 possible values - 0 to 9 except d and c<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'a' can
take 6 possible values - 1 to 9 except d, b and c<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">So, total
number of possibilities = 6 * 8 * 9 * 10 = 4320.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution: <o:p></o:p></span></b></div>
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<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">The simple
way of stating the bug is as follows “But one of b, c, d might have been zero”.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">When we
say a can take 6 possible values. We eliminate 0, b, c, and d from the list 0
to 9. If b, c or d were zero, then that in that case, a will still have 7
possible options. <o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">In this
question, the leading digit has a constraint. So, start with that. No point factoring
in the constraint in the last step. Get that out of the way first.<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><b><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Let the
4-digit number be 'abcd'. Now,<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'a' can
take 9 possible values - 1 to 9. <o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'b' can
take 9 possible values - 0 to 9 except a<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'c' can
take 8 possible values - 0 to 9 except a and b<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">'d' can
take 7 possible values - 1 to 9 except a, b and c<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Total
number of options = 9 * 9 * 8 * 7.<o:p></o:p></span></div>
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<br /></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">3. In how
many ways can we rearrange the letters of the word TWOIIM such that the vowels
appear together?<o:p></o:p></span></div>
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<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Given solution<o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Let us
take the three vowels together and call it as X. Now, we need to rearrange the
letters of the word TWMX. This can be done in 4! ways. Or, there are 24 ways of
rearranging the letters of the word TWOIIM with vowels appearing together.<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Bug in the solution: <o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">We do not
account for the different variations that ‘X’ can come in. Now, the three
vowels have been replaced with X. Or, X is nothing but OII in some order. The
key point here is that X could be OIO, IOO or OOI. X can take three different
forms. The solution has overlooked this. <o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<b><span style="color: #002060; font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Correct solution:</span></b><b><span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";"><o:p></o:p></span></b></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Let us
take the three vowels together and call it as X. Now, we need to rearrange the
letters of the word TWMX. This can be done in 4! ways. <o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">Now, the
three vowels have been replaced with X. Or, X is nothing but OII in some order.
The key point here is that X could be OIO, IOO or OOI. X can take three
different forms.<o:p></o:p></span></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<br /></div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<span style="font-size: 12.0pt; mso-bidi-font-family: Arial; mso-fareast-font-family: "Times New Roman";">So total
number of possibilities = 4! * 3 = 72 ways. <o:p></o:p></span></div>
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</div>
<div class="MsoNormal" style="background: white; line-height: 15.75pt; margin-bottom: .0001pt; margin-bottom: 0in; vertical-align: baseline;">
<br /></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-52350902701121763312015-04-12T17:57:00.003+05:302015-09-30T12:31:28.913+05:30CAT Online Preparation - Permutation and Combination<div dir="ltr" style="text-align: left;" trbidi="on">
<div>
Permutation and Combination is a classcic topic where getting something wrong is often more instructive than getting it right the first time, provided we have another look at what exatly went wrong with the incorrect solution.</div>
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<br /></div>
<div>
To reinforce this idea of "debugging" a wrong solution in order to understand this topic better, I have given a bunch of questions with incorrect solutions that need to be fixed. So, have a go at this. Before you try, let me give a few ground rules.</div>
<div>
<br /></div>
<div>
1. You should try to figure out what exactly is wrong with this solution. It is not enough to say 240 is wrong as 120 is the correct answer. Try to articulate which step of the given solution is wrong. </div>
<div>
<br /></div>
<div>
2. There might be 1-2 questions where the answer + solution are indeed correct. Otherwise, there is no thrill in this.</div>
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<br /></div>
<div>
3. The best counter-solutions will get published on the blog with due credit to whoever contributed this. There might be some thrill in getting an admit to join IIM A or B or some such, but I am sure it will pale in comparison to getting a badge of honour from this blog. So, go for it.</div>
<div>
<br /></div>
<div>
Now, on to the questions and solutions.</div>
<div>
<br /></div>
<div>
1. Five boys need to be allotted to 4 different rooms such that each boy is allotted a room and no room is empty. In how many ways can this be done?</div>
<div>
<br /></div>
<div>
Solution: Let the boys be A, B, C, D and E. Let the rooms be 101, 102, 103 and 104. Now, we know that exactly one room will have two occupants. First, let us try to send 4 boys to 4 rooms, we can worry about the fifth occupant later on. Let us select 4 out of the 5 boys first. This can be done in 5C4 ways. Now, these 4 can be allotted to 4 different rooms in 4! ways. So, 4 boys in 4 rooms can be done in 5C4 * 4! = 5 * 24 = 120 ways.</div>
<div>
<br /></div>
<div>
Now, the fifth boy has to go into one of the rooms. He can do this in 4 ways as there are 4 different rooms available. So, total number of outcomes = 120 * 4 = 480.<br />
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and.html" target="_blank">here</a></div>
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2. How many 4-digit numbers with 4 distinct digits are possible?</div>
<div>
<br /></div>
<div>
Let the 4-digit number be 'abcd'. Now,</div>
<div>
'd' can take 10 possible values - 0 to 9. </div>
<div>
'c' can take 9 possible values - 0 to 9 except d</div>
<div>
'b' can take 8 possible values - 0 to 9 except d and c</div>
<div>
'a' can take 6 possible values - 1 to 9 except d, b and c</div>
<div>
<br /></div>
<div>
So, total number of possibilities = 6 * 8 * 9 * 10 = 4320.<br />
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and.html" target="_blank">here</a></div>
<div>
<br /></div>
<div>
3. In how many ways can we rearrange the letters of the word TWOIIM such that the vowels appear together.</div>
<div>
<br /></div>
<div>
Let us take the three vowels together and call it as X. Now, we need to rearrange the letters of the word TWMX. This can be done in 4! ways. Or, there are 24 ways of rearranging the letters of the word TWOIIM with vowels appearing together.<br />
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and.html" target="_blank">here</a></div>
<div>
<br /></div>
<div>
4. In how many ways can we select 2 cards from a card pack such that both belong to the same suit?</div>
<div>
<br /></div>
<div>
The first card can be selected in 52 ways, as it can be any card in the pack. Now, there are 12 different ways of selecting the second card. Depending on the suit of the first card, the choice set we have for the second card gets limited. If the first card were a club, then the second card must be from among the 12 remaining clubs and so on.</div>
<div>
<br /></div>
<div>
So, total number of outcomes = 52 * 12.<br />
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and_28.html" target="_blank">here</a></div>
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<br /></div>
<div>
5. When a coin is tossed 10 times, what is the probability of getting more heads than tails?</div>
<div>
<br />
When a coin is tossed 'n' times, the probability of getting more heads than tails = Probability of getting more tails than heads. So, we do not need to really compute this. Probability = 1/2.</div>
<div>
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and_28.html" target="_blank">here</a><br />
<br /></div>
<div>
6. In how many ways can we arrange 3 boys and 3 girls on a straight line such that no two boys stand next to each other</div>
<div>
<br /></div>
<div>
We can arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total number of possibilities = 2 * 3! * 3! = 72 ways.<br />
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and_28.html" target="_blank">here</a></div>
<div>
<br /></div>
<div>
7. What is the probability of selecting 3 cards from a card pack such that all three are face cards? what is the probability that none of the three is a face card? </div>
<div>
<br /></div>
<div>
Number of cards in a card pack = 52</div>
<div>
Numbert of face cards in a card pack = 12</div>
<div>
Number of ways of selecting 3 cards from a card pack = 52C3</div>
<div>
Number of ways of selecting 3 face cards from a card [ack = 12C3</div>
<div>
<br /></div>
<div>
Probability of selecting three cards such that all three are face cards = 12C3/52C3</div>
<div>
Probability of selecting three cards such that none of the three are face cards = 1 - 12C3/52C3</div>
<div>
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and_29.html" target="_blank">here</a><br />
<br /></div>
<div>
8. A die is rolled thrice. In how many outcomes will two throws be same and the third one different?</div>
<div>
<br /></div>
<div>
Let the three outcomes be ABC.</div>
<div>
<br /></div>
<div>
'A' can take all values from 1 to 6</div>
<div>
'B' can also take all values from 1 to 6 </div>
<div>
'C' can take all values except A - so it has 5 possibilities</div>
<div>
<br /></div>
<div>
Total number of outcomes = 6 * 6 * 5 = 180.</div>
<div>
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and_29.html" target="_blank">here</a><br />
<br /></div>
<div>
9. How many 7 letter words can we have in English that have two distinct vowels and 5 distinct consonants.</div>
<div>
<br /></div>
<div>
Now, we know there are 5 vowels and 21 onsonants. So, we need to select 2 from these 5 and 5 from the remaining 21. Since order is important, we need to select keeping order in mind.</div>
<div>
<br /></div>
<div>
So, we have 5P2 * 21P5.</div>
<div>
<br />
This solution has been debugged <a href="http://iimcat.blogspot.in/2015/04/cat-online-coaching-permutation-and_29.html" target="_blank">here</a><br />
<br /></div>
<div>
10. This is an interesting question - Of the 10 questions given here, how many answers are wrong? </div>
<div>
<br /></div>
<div>
Solution: Can the 10th question have two possible answers?</div>
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<br /></div>
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<br /></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-55271738244883338832015-03-09T18:59:00.001+05:302015-09-30T12:33:05.601+05:30CAT Online Class - Interesting one from Number Systems<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: red;"><b>Question</b></span><br />
Actually, this is not a CAT preparation question. This is way tougher than what one can expect to see in CAT. I saw it on some Olympiad paper. But this is a wonderful question to think about one very interesting property. Anyway, let us get to the question -<br />
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A six-digit number has digits 'abcabd', where a, b, c, d take values from 0 to 9. Further we know that d = c + 1 and that this number is a perfect square. Find the number. If there is more than one number possible, find all such numbers.<br />
<br />
<span style="color: red;"><b>Answer</b></span><br />
183184, 328329, 528529, 715716<br />
<br />
<b><span style="color: red;">Explanation</span></b><br />
This is a fabulous question. It bears repeating that this is far far tougher than what we are likely to see in CAT. Nevertheless we will look at this question in order to discuss one key idea.<br />
<br />
Now, 'abcabd' is a perfect square. So, say, 'abcabd' = x^2<br />
Now, 'abcabd' = 'abcabc' + 1. So, 'abcabc' = x^2 - 1<br />
<br />
After this, we are halfway there to the solution<br />
<br />
'abcabc' = (x +1) (x - 1)<br />
<br />
Now, 'abcabc' = 'abc' * 1001. Or, 'abcabc' = 'abc' * 1001. This is the key idea that is very vital in a bunch of questions.<br />
<br />
'abc' * 1001 = 'abc * 7 * 11 * 13 = (x - 1) (x + 1)<br />
<br />
So, between x -1 and x + 1, we need to account for 7, 11 and 13 as factors. Wherever this works, we are through. Now, x - 1 or x + 1, either number alone cannot be a multiple of 7, 11 and 13. So, one of these two should be a multiple of one of the three primes and the other should be a multiple of the other two.<br />
<br />
So, of the two numbers one can be a multiple of 7 and the other 143. or,<br />
one can be a multiple of 11 and the other 91. or,<br />
one can be a multiple of 13 and the other 77,<br />
<br />
After this we are down to trial and error; but a very scientific form of trial and error.<br />
<br />
Let us say, we are picking two numbers such that one is a multiple of 7 and the other of 143. Since 143 is the far larger number, it helps to look for multiples of 143 and see if we can spot the scenario where the other number is a multiple of 7.<br />
<br />
If one number were 143, the other would have to be 145 or 141. Neither is a multiple of 7. This does not work.<br />
<br />
If one number were 286, the other would have to be 284 or 288. Neither is a multiple of 7. This does not work either.<br />
<br />
If one number were 429 (143 * 3), the other would have to be 427 or 429. 427 is a multiple of 7. Houston, we have an answer!<br />
<br />
(x -1) * (x + 1) = 427 * 429 works. 427 * 429 = 183183. <span style="color: blue;"><b>Or, 183184 is a perfect square.</b></span> 428 * 428.<br />
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<br />
Now, all we need to do is check out all other variants. But having said that, even this can be a very time-consuming process. So, let us see if we can fine-tune this a touch. If one number is a multiple of 143, the other number should be a multiple of 7. So, the first number + 2 or first number -2 should be a multiple of 7. If we say the first number were k, then either k + 2 or k - 2 should be a multiple of 7. In that case, we are through. or, k divided by 7 should give us a remainder of 2 or 5.<br />
<br />
Now, 143 divided by 7 gives us a remainder of 3. This is why this does not work<br />
<br />
2 * 143 divided by 7 will give us a remainder of 6. This does not work either<br />
3 * 143 divided by 7 will give us a remainder of 9, which is same as 2. This works. This is what gave us the solution 183184<br />
4 * 193 gives us a remainder of 12, which is same as 5. This should also work. Let us check this out. 4 * 143 = 572. 574 is a multiple of 7. Or, 572 * 574 will be a multiple of 1001.<br />
572 * 574 = 328328. or, 328329 = 573 *573. We have got our second possibility!!<br />
<br />
<span style="color: blue;"><b>183184 and 328329 are both perfect squares.</b></span><br />
<br />
Now, let us move to 5 * 143. This divided by 7 gives us a remainder of 15, which is same as 1. This does not work either<br />
<br />
Let us move to 6 * 143. This divided by 7 gives us a remainder of 18, which is same as 4. This does not work either<br />
<br />
7 * 143 = 1001. That is a 4-digit number, so we do not have to worry about this.<br />
<br />
So, we have got two solutions so far. With the numbers being divided as product of 143 and of 7. Let us move to the combination of the numbers being broken as product of 91 and 11.<br />
<br />
Now, let us go directly to the remainder approach. 91 divided by 11 gives us a remainder of 3. We need to find some multiple of 91 that on division by 11 gives us a remainder of either 2 or 9. (Why? scroll up to see this. We are looking for numbers that differ by 2 which between them account for all factors of 1001)<br />
<br />
1 * 91 divided by 11 will give us a remainder of 3. This does not work.<br />
2 * 91 divided by 11 will give us a remainder of 6. This does not work.<br />
3 * 91 divided by 11 will give us a remainder of 9. This should work. Let us check this out. 273 is a multiple of 91, 275 is a multiple of 11. 273 * 275 should be a multiple of 1001. 273 * 275 = 75075. 75076 is 274 * 274. But this does not count because it has only 5 digits. Close, but no cigar. Let us take this further<br />
<br />
4 * 91 divided by 11 will give us a remainder of 12, which is same as 1 This does not work.<br />
5 * 91 divided by 11 will give us a remainder of 15, which is same as 4 This does not work either<br />
6 * 91 divided by 11 will give us a remainder of 18, which is same as 7 This does not work either<br />
7 * 91 divided by 11 will give us a remainder of 21, which is same as 10 This does not work either<br />
8 * 91 divided by 11 will give us a remainder of 24, which is same as 2. Hello, do we have our third number here?!<br />
<br />
8 * 91 = 728. 726 is a multiple of 11. 728 * 726 should be a multiple of 1001. 726 * 728 = 528528. 528529 = 727*727. <span style="color: blue;"><b>So, our third number is 528529.</b></span><br />
<br />
Let us go further, but quicker from now on.<br />
<br />
9 * 91 divided by 11 will give us a remainder of 27, which is same as 5 This does not work.<br />
10 * 91 divided by 11 will give us a remainder of 30, which is same as 8 This does not work either.<br />
<br />
We do not need to worry about 11 * 91 as that is 1001.<br />
<br />
Now, let us move on to numbers that break as 77p * 13q.<br />
<br />
We need a multiple of 77 that when divided by 13 gives a remainder of either +2 or + 11. 77 divided by 13 gives a remainder of 12, or -1.<br />
<br />
77 * 2, on division by 13 will give us a remainder of -2. This should work. But this will be a waste of time as it will result in a 5-digit number.<br />
<br />
So, let us evaluate 77 * 3. This gives a remainder of -3. Not good enough.<br />
Straight away, we can sense that the next number that will work for us is 77 * 11, where we get a remainder of -11, which is nothing but + 2.<br />
<br />
77 * 11 = 847. 845 is a multiple of 13. 845 *847 is a multiple of 1001. 847 * 845 = 715715. 846 *846 <span style="color: blue;"><b>= 715716 is our fourth number. </b></span><br />
<br />
No other possibility exists.<br />
<br />
<span style="color: blue;"><b>There are 4 possible numbers - 183184, 328329, 528529 and 715716. Fantabulous question to nail down the fact that 1001 = 7 * 11 * 13. </b></span><br />
<br />
Since you have been brilliant and diligent enough to get through this far to the solution, I am going to leave you with another nugget. 'abcdabcd' = 'abcd' * 10001.<br />
<br />
This 10001 is not prime. It is a product of two primes. Which two? It is not for nothing that we at 2iim say CAT preparation can be fun.<br />
<br />
Again, for the nth time, let me reiterate that the above question is way tougher than what we will find in CAT. If you want questions (lots of them) at or around CAT level, visit the <span style="color: red;"><b><a href="http://iim-cat-questions-answers.2iim.com/" target="_blank">questionbank</a>.</b></span><br />
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<br /></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-85509028861315226022014-12-20T07:22:00.001+05:302015-09-30T12:33:36.615+05:30CAT Linear Equations<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="color: red;"><b>Question</b></span><br />
A system of equations has 3 equations<br />
3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.<br />
If the system of equations has no solution, find k.<br />
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<span style="color: red;"><b>Explanation</b></span><br />
This is a very interesting question. If we can generate one of the equations from the other two, we can then say that the system has infinite solutions. If we can generate one of the equations from the other two, but with the constant part alone being different, this would be akin to having parallel lines when we are dealing with 2 variables and that would result in the system having no solutions. So, lets look for that.<br />
<br />
So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) = 2x + 3y + kz.<br />
We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y ) = 2x + 3y . Then said a, b would give us k.<br />
<br />
Or, we are effectively solving for<br />
3a + 4b = 2<br />
4a + 5b = 3<br />
<br />
Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.<br />
<br />
Now, k = 5a + 3b = 10 -3 = 7.<br />
<br />
If k = 7, this system of equations would result in no solutions.<br />
<br />
If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + 7z = 6.<br />
First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.<br />
<br />
Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has no solution.<br />
<br />
Yeah, the more mechanical, rather deceptively straight-forward-looking determinant method is also there. But where is the joy in that.<br />
<br />
<br />
<br />
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<span style="color: red;"><b><br /></b></span></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-13040207653968419352014-10-31T12:10:00.000+05:302015-09-30T12:34:26.691+05:30CAT Preparation Online - Permutation and Combination<div dir="ltr" style="text-align: left;" trbidi="on">
<div class="MsoNormal">
<b><span style="color: red;">Question<o:p></o:p></span></b></div>
<div class="MsoNormal">
Product of the distinct digits of a natural number is 60.
How many such numbers are possible?
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<b><span style="color: red;">Explanation<o:p></o:p></span></b></div>
<div class="MsoNormal">
60 = 2^2 * 3 * 5<o:p></o:p></div>
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<br /></div>
<div class="MsoNormal">
60 cannot be written as a product of two single digit
numbers. SO, the number in question should either have 3 or more digits. <o:p></o:p></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
Three-digit numbers<o:p></o:p></div>
<div class="MsoNormal">
The digits could be 345 or 265<o:p></o:p></div>
<div class="MsoNormal">
Digits being 345 - there are 3! such numbers<o:p></o:p></div>
<div class="MsoNormal">
There are six numbers for each of these outlines. So, there
are 3! + 3! = 12 three-digit numbers<o:p></o:p></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
Four-digit numbers<o:p></o:p></div>
<div class="MsoNormal">
The digits could be 1345, 1265 or 2235. <o:p></o:p></div>
<div class="MsoNormal">
Digits being 1345 - there are 4! such numbers<o:p></o:p></div>
<div class="MsoNormal">
Digits being 1265 - there are 4! such numbers<o:p></o:p></div>
<div class="MsoNormal">
Digits being 2235 - this is not possible as digits have to be distinct.<o:p></o:p></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
So, there are totally 24 + 24 four-digit numbers
possible. 48 four-digit numbers. <o:p></o:p></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
Total number of numbers = 12 + 48 = 60.<o:p></o:p><br />
<br />
(this post has been modified to plug error)</div>
<br />
<div class="MsoNormal">
<br /></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-81816598989254752572014-10-23T22:50:00.000+05:302015-09-30T12:40:49.738+05:30CAT preparation - Ratio and Proportion (Tough)<div dir="ltr" style="text-align: left;" trbidi="on">
<div class="MsoNormal">
<span style="background: white; font-size: 13pt; line-height: 115%;"><span style="color: red;"><b>Question</b></span><span style="color: #141823;"><o:p></o:p></span></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">John
has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates
of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the
ratio 3 : 5. If there are more chocolates of type C than of type B, and more of
type B than of type A, what is the minimum possible number of chocolates
overall?
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<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; font-size: 13pt; line-height: 115%;"><span style="color: red;"><b>Explanation</b></span><span style="color: #141823;"><o:p></o:p></span></span></div>
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<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Again,
big thanks to Mukund Sukumar for excellent solution.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Let
john have 3x chocolates of type A and 7x of type B <o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Let
Mike have 5y chocolates of type B and 4y of type C<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Let
john have 3z chocolates of type C and 5z of type A<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">So
in total A=3x+5z ; B=7x+5y ; C=4y+3z<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Since
C>B we get solving y<3z-7x --->(1)<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Since
B>A we get solving 5y>5z-4x ---> (2)<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">What
gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only
y=1 as choice, for which second condition doesnt satisfy. <o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">So,
when x=1,z=4, we get y<5 from first condition and when y > 3.2 from
second condition. so which gives choice the only y=4.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Hence
x=1,y=4 and z=4 works and is the best possible answer.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">For
these values, we get A=23,B=27,C=28.<o:p></o:p></span></div>
<br />
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Minimum
possible number of chocolates overall is 76. <o:p></o:p></span></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-88753794149282046402014-10-23T22:40:00.002+05:302015-09-30T12:44:10.629+05:30CAT Preparation Online - Tough one from Permutation and Combination<div dir="ltr" style="text-align: left;" trbidi="on">
<div class="MsoNormal">
<span style="font-size: 13pt; line-height: 115%;"><span style="color: red;"><b>Question</b></span></span></div>
<div class="MsoNormal">
<span style="font-size: 13pt; line-height: 115%;"><br /></span></div>
<div class="MsoNormal">
<span style="font-size: 13pt; line-height: 115%;">A wonderful, but very tough question from
Permutation and Combinations.</span><o:p></o:p></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">In
how many ways, can we rearrange the word MONSOON such that no two adjacent
positions are taken by the same letter? (Tough one. Tougher than what we will
see in CAT)
<div class="separator" style="clear: both; text-align: center;">
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<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
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<span style="background: white; font-size: 13pt; line-height: 115%;"><span style="color: red;"><b>Explanation</b></span></span></div>
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<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
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<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">First
up, lets get the facts out of the way – The three O’s need to be kept apart,
then the 2 N’s. </span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Let us focus on the three O’s. <o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">We
can place the three O’s in some blanks around the other letters. Or, three O’s
can be placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in <sup>5</sup>C<sub>3</sub>,
or 10 ways.<o:p></o:p></span></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Or,
the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1,
4, 7} or {1, 5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7} - Whew.<o:p></o:p></span></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Now,
for each of these arrangements, there are 4!/2! = 12 arrangements for the other
4 letters. But the one thing we need to keep in mind now is the fact that 2 N’s
could be adjacent in these arrangements. We will need to eliminate these.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are
2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are
2! = 2 </span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">words like this. So, number of words that we need to count = 12 – 2 = 10<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}.
There are totally 4 words that we need to subtract. So, number of words that we
need to count = 12 – 4 = 8<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {1, 4, 6} or O__ __O__O__ - Ns
could be in the slots {2, 3}. There are 2! = 2 words like this. So, number of
words that we need to count = 12 – 2 = 10<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There
are totally 4 words that we need to subtract. So, number of words that we need
to count = 12 – 4 = 8
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<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}.
There are totally 4 words that we need to subtract. So, number of words that we
need to count = 12 – 4 = 8<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {2, 4, 6} or __O__O__O__ - Tehre
are no possible slots for N. So, we count all 12 words on this list. <o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2!
= 2 words like this. So, number of words that we need to count = 12 – 2 = 10<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2!
= 2 words like this. So, number of words that we need to count = 12 – 2 = 10.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">O’s
in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There
are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 =
10<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Total
number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Phew!.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">There
is a far more elegant method for accounting for the words where the 2 N’s
appear together. This one came from Mukund Sukumar.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">We
need to account for the number of possibilities of N,N being together. So to
subtract that part, consider 'NN' being together as one letter and place O's. In
3 out of the 4 slots in _M_NN_S_<o:p></o:p></span></div>
<div class="MsoNormal">
<br /></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">The
O’s can be selected in <sup>4</sup>C<sub>3</sub> ways. The MNNS can be
rearranged in 3! Ways if the N’s have to appear together.<o:p></o:p></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #141823; font-size: 13.0pt; line-height: 115%; mso-bidi-font-family: Helvetica; mso-bidi-font-size: 12.0pt;">Or,
we get <sup>4</sup>C<sub>3</sub> * 3! = 24 ways. So, we have 120-24 = 96 ways
totally. <o:p></o:p></span></div>
<div class="MsoNormal">
<br /></div>
<br />
<div class="MsoNormal">
<br /></div>
<br />
<br /></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-38895333332199732092014-10-15T20:13:00.000+05:302015-09-30T12:46:04.992+05:30CAT Preparation Online - A simple question from Quadratic Equations<div dir="ltr" style="text-align: left;" trbidi="on">
Questions<br />
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<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">x^2 - 17x + |p| = 0 has integer solutions. How many values can p take?
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<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">Explanation</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">To begin with, if we assume roots to be a and b, sum of the roots is 17 and product of the roots is |p|. Product of the roots is positive and so is the sum of the two roots. </span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">So, both roots need to be positive.</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">So, we are effectively solving for </span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">Number of positive integer solutions for a + b = 17.</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">We could have (1,16), (2, 15), (3, 14)......(8, 9) - There are 8 sets of pairs of roots. Each of these will yield a different product.</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">So, |p| can take 8 different values. Or, p can take 16 different values.</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">is that it? Or, are we missing something? Can p be zero? What are the roots of x^2 - 17x = 0. This equation also has integer solutions. </span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">So, p can also be zero.</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">So, number of possible values of p = 16 + 1 = 17.
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<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">Wonderful question - chiefly because there are two really good wrong answers one can get. 8 and 16. So, pay attention to detail. No point telling yourself "Just missed, I just didnt think of zero. I deserve this mark". Being just wrong, will earn us a -1 instead of the honourable 0.</span><br />
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>
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<b><span style="background: white; color: red; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">Question</span></b><br />
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<span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">
<br />
</span><span style="background: white; color: #141823; font-family: "Helvetica","sans-serif"; font-size: 10.5pt; line-height: 107%;">How many integer solutions
exist for the equation x<sup>2</sup> - 8|x| - 48 =0?</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">
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</span><span style="background: white; font-size: 12pt; line-height: 107%;"><br />
<b><span style="color: red;"><span style="font-size: small;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial; font-size: 12pt; line-height: 107%;"></span></span>Explanation</span></b></span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
One approach is to solve when x > 0 and then solve for x
< 0. However, there is a slightly simpler method. <o:p></o:p></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">Note that x<sup>2</sup> is the same as
|x|<sup>2</sup>, so we can treat the equation as a quadratic in |x|.<o:p></o:p></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br /></span></div>
<div class="MsoNormal">
<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">Or, |x|<sup>2 </sup>– 8|x| - 48 = 0<o:p></o:p></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">(|x| - 12|) (|x| + 4) = 0<o:p></o:p></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">|x| could be 12. |x| cannot be -4. <o:p></o:p></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br /></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">If |x| could be 12, x can be -12 and 12.<o:p></o:p></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br /></span></div>
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<span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">Two possible solutions exist. </span><span style="font-size: 12.0pt; line-height: 107%;"><o:p></o:p></span></div>
</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-43915830738698053122014-10-14T11:48:00.001+05:302015-09-30T12:48:11.892+05:30CAT - Question from Pipes and Cisterns<div dir="ltr" style="text-align: left;" trbidi="on">
<div class="MsoNormal">
<b><span style="background: white; color: red; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">Question</span></b><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
<br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">There are n pipes that fill a
tank. Pipe 1 can fill the tank in 2 hours, Pipe 2 in 3 hours, Pipe 3 in 4 hours
and so on. Pipe 1 is kept open for 1 hour, pipe 2 for 1 hour, then pipe 3 and
so on. In how many hours will the tank get filled completely?</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">
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</span><span style="background: white; font-size: 12pt; line-height: 107%;"><br />
<b><span style="color: red;"><span style="font-size: small;"><span style="background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial; font-size: 12pt; line-height: 107%;"></span></span>Explanation</span></b></span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
Almost all questions can be approached well by asking the
question "What happens in 1 hour" (or 1 day, or 1 minutes)</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
So, let us start with that</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first
hour, the tank will not be filled</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">In 1 hour, pipe 2 fills 1/3 of
the tank. So, in two hours we would have filled 1/2 + 1/3 of the tank, or
5/6 of the tank. So, by the end of the second hour, the tank would still not be
filled.</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;">Let us move to the third hour.
In 1 hour, pipe 3 fills 1/4 of the tank. So, by the end of the third hour, we
should have filled 5/6 + 1/4 = 13/12 of the tank.</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
Oops, one cannot fill 13/12 of a tank. What this tells us is
that the tank gets filled in the 3rd hour. </span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
When exactly during the third hour?</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
At the beginning of the third hour, we still have 1/6 of the
tank still to fill. Pipe 3 can fill at the rate of 1/4 of the tank per hour.
Or, pipe 3 will take 2/3 hours to fill the tank.</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
Or, the tank will be filled in 2 hours and 40 minutes.</span><span style="color: #333333; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
</span><span style="background: white; color: #222222; font-size: 12.0pt; line-height: 107%; mso-bidi-font-family: Arial;"><br />
This is an example of a sequence that is a harmonic
progression. The formulae for HP are needlessly confusing. So, simple
step-by-step approach works best. </span><span style="font-size: 12.0pt; line-height: 107%;"><o:p></o:p></span></div>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-31032119537163034082014-10-10T13:06:00.000+05:302015-09-30T12:48:55.593+05:30CAT Permutation Combination and Functions<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><b><span style="color: red;">Question</span></b></span><br />
<br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">How many functions can be defined from Set A -- {1, 2, 3, 4} to Set B = {a, b, c, d} that are neither one-one nor onto? </span>
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><b><span style="color: red;">Explanation</span></b></span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">To start with, if you do not know the meaning of one-one or onto, look these up. For good measure know the meanings of the terms surjective, injective etc also. CAT tests these terms.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Let us start by answering a far simpler question. How many functions are possible from Set A to Set B. This is equal to 4^4 = 256. </span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Note that any function from Set A to Set B that is one-one will also be onto and vice versa. How? Why? - Think about this. Remember, tutors can only take the horse to the pond. :-)</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">So, we need to subtract only those functions that are one-one AND onto. Or, effectively we have to eliminate those functions where {1, 2, 3, 4} are mapped to {a, b, c, d} such that each is mapped to a different element. This is effectively same as the number of ways of rearranging 4 elements. Or, number of ways of doing this is 4! .</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">So, total number of functions that are neither one-one nor onto = 256 - 24 = 232. </span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-58335762771887126252014-10-10T12:59:00.001+05:302015-09-30T12:49:55.400+05:30CAT Permutation and Combinations<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><span style="color: red;"><b>Question</b></span></span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">If we list all the words that can be formed by rearranging the letters of the word SLEEPLESS in alphabetical order, what would be the rank of SLEEPLESS?</span>
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><b><span style="color: red;">Explanation</span></b></span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">First let us think about the number of possible rearrangements. SLEEPLESS can be rearranged in 9!/ (2! 3! 3!) = 5040. </span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Now, if rearrange these, we would have words starting with E, L, P and S.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Now, SLEEPLESS starts with S. So, let us think about how many words start with S. Number of words starting with S = 8</span><span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">!/ (2! 2! 3!) = 1680.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">So, there would have been 5040 - 1680 words that have gone by before the first word starting with S. Or, 3360 words start with E, L or P.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">The first word with S is SEEELLPSS. This would have a rank of 3361.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><br /></span>
<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Now, let us go step by step. </span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Words starting with SE____ - Number of such words = 7!/(2!*2!*2!) = 630 words</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Next we have words starting with SL, but our word also starts with SL, so let us go deeper</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Words starting with SLE would be the next step, but our word starts with SLE as well.SO, let us go one further step deeper</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Words starting with SLEE. The first such word would be SLEEEPLSS</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">So, let us think about words starting with SLEEE - there would be 4!/(2!) words like this = 12 words like this</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">words starting with SLEEL - </span><span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">there would be 4!/(2!) words like this = 12 words like this</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">So, we have accounted for 3360 + 630 + 12 + 12 = 4014 words thus far.
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Now, on to words starting with SLEEP - there are again </span><span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">4!/(2!) words like this = 12 words like this</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Our word is within these 12 words</span><br />
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><span style="background-color: white;">Words starting with SLEEPE - there are 3!/2! words like this, or 3 words like this</span></span><br />
<span style="color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><span style="background-color: white;">Our word comes in the next bunch.</span></span><br />
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><span style="background-color: white;">Words starting with </span></span><span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;">SLEEPL - SLEEPLESS is the first such word. Or, the rank fo SLEEPLESS = 4014 + 3 + 1 = 4018.</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;">A very good question to get lots of practice on letters rearrangement. However, it is unlikely that you will face a question this time-consuming in CAT. Wonderful question to practice though.</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><br /></span></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-14234483352008978252014-10-08T14:24:00.001+05:302015-09-30T12:50:32.192+05:30CAT Number Systems - Fun question based on Armstrong numbers<div dir="ltr" style="text-align: left;" trbidi="on">
<span style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: 13px;"><span style="color: red;"><b>Question</b></span></span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">A 3-digit Armstrong number is a three-digit number where the number is equal to the sum of the cubes of the three digits. Give a few Armstrong numbers.</span>
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<span style="color: red; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><span style="background-color: white;"><b>Explanation</b></span></span><br />
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><span style="background-color: white;">To start with, be clear that CAT does not ask questions like these. You do not need to know what Armstrong number are; nor do you need to know how to get these to crack these exam. This is just a fun question to do some trial and error with.</span></span><br />
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: x-small;"><span style="background-color: white;">The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and error (although a lil scientifically)</span></span></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-46990568481415323472014-10-08T14:21:00.000+05:302015-09-30T12:52:04.933+05:30CAT Number Theory - Interesting Question from Factorial<div dir="ltr" style="text-align: left;" trbidi="on">
Question<br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">How many trailing zeroes will be present in the base 12 representation of 55!?</span>
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Explanation</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">The question can be restated as follows - "What is the highest power of 12 that divides 55!"</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.</span><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: 13px;">So, in order to find the highest power of 12 that divides 55!, we need to look at the highest powers of 2 and 3 that divide 55!. </span><br />
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Successive division by 2 will give us the highest power of 2 that divides 55!:</div>
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55/2 = <b>27</b>, 27/2 = <b>13</b>, 13/2 = <b>6</b>, 6/2 = <b>3</b>, 3/2 = <b>1.</b></div>
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Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 = 50</div>
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Successive division by 3 will give us the highest power of 3 that divides 55!:</div>
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55/3 = <b>18</b>, 18/3 = <b>6</b>, 6/3 = <b>2.</b></div>
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Highest power of 3 that divides 55! = 18 + 6 + 2 = 26. </div>
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So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide 55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!</div>
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We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's. </div>
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Thus 12^25 <b>will</b> divide 55! - there are 25 trailing zeros in base 12 representation of 55!.</div>
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</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-11335528070971585382014-10-08T14:12:00.004+05:302015-09-30T12:53:17.725+05:30Question on Number Theory<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="color: red;"><b>Question</b></span><br />
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A bus service runs from Chennai to Blr every third day with the first bus starting on Jan 1st. Another bus service runs from Chennai to Mumbai every 5th day starting from Jan 2nd. A third bus service from Chennai to Cochin runs on every 7th day starting from Jan 4th. In that year (which is not a leap year), on how many different days will a bus run from Chennai to all three cities?</div>
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<span style="color: red;"><b>Explanation</b></span><br />
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Let us call Jan 1st as day 1, jan 2nd as day 2 and so on.</div>
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So, </div>
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The Chennai-Blr services runs on days 1, 4, 7, 10, 13, 16, 19,......</div>
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The Chennai-Mumbai services runs on days 2, 7, 12, 17, 22, 27,......</div>
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The Chennai-Cochin services runs on days 4, 11, 18, 25, ......</div>
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First up, let us see if we can find one day where all three buses ply.</div>
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Chennai Blr runs on days of the form 3a + 1, Chennai_Mumbai on 5b + 2, and Chennai-Cochin on 7c + 4. If N can be written as 3a + 1, 5b + 2 and 7c + 4, then we should be able to get n as 105d + ___ . (105 is the LCM of 3, 5 and 7)</div>
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This is a very important idea. Get lots of practice on this idea. Wrap your head around this idea very clearly.</div>
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First let us combine 3a + 1 and 5b + 2.</div>
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A number of the form 3a + 1, on division by 15 can have one of the following remainders { 1, 4, 7, 10, 13}</div>
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A number of the form 5b + 2, on division by 15 can have one of the following remainders { 2, 7, 12}</div>
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Or, if a number is of the form 3a + 1 and 5b + 2, it has to be of the form 15k + 7.</div>
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Now, N = 15K + 7 and 7c + 4.</div>
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Number 15k + 7, on division by 105 can have one of the following remainders { 7, 22, 37, 52, 67, 82, 97}}</div>
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Number 7c + 4, on division by 105 can have one of the following remainders { 4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95, 102, 109}</div>
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Or, the number is of the form 105d + 67.
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So, all three services will run on days 67, 67 + 105, 67 + 105*2 and so on.</div>
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Or, on days 67, 172 and 277. (Any other day would go beyond 365)</div>
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So, all three services would run together 3 days of the year.</div>
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Wonderful question testing concepts of LCM and Remainders. </div>
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</div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-63974928125001327672014-08-13T12:49:00.000+05:302015-09-30T12:54:09.922+05:30CAT Coordinate Geometry Question and Solution<div dir="ltr" style="text-align: left;" trbidi="on">
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<b style="font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 15px; line-height: 1.4;"><span style="color: red; font-family: Verdana, sans-serif;">Question</span></b></h3>
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<span style="color: #404040; font-family: Roboto, arial, sans-serif; font-size: 13px; line-height: 18.200000762939453px;">What is the area enclosed by the region defined by y = |x -1| + 2, the line x = 1; X-axis and Y-axis?
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<span style="color: red; font-family: Verdana, sans-serif; line-height: 18px;"><b>Explanation</b></span></div>
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<span style="color: #404040; font-family: Roboto, arial, sans-serif; font-size: 12.727272033691406px; line-height: 18.200000762939453px;">Solution is available on the video given below. </span><br />
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<iframe allowfullscreen="" frameborder="0" height="315" src="//www.youtube.com/embed/PdXZ7ZK9RbE" width="560"></iframe></div>Rajesh Balasubramanianhttp://www.blogger.com/profile/06092210868780120343noreply@blogger.com0tag:blogger.com,1999:blog-8044374.post-63928182231478979712014-02-27T11:02:00.000+05:302015-09-30T12:55:37.105+05:30Inequalities Question and Solution<div dir="ltr" style="text-align: left;" trbidi="on">
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<b><span style="color: red; font-family: Verdana, sans-serif;">Question</span></b></div>
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<span style="background-color: white; color: #333333; font-family: Verdana, sans-serif; line-height: 18px;"><b>How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0?
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<span style="color: red; font-family: Verdana, sans-serif;"><b>Question</b></span><br />
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<span style="background-color: white; color: #333333; line-height: 18px;"><span style="font-family: Verdana, sans-serif;">|x + 2| + | x - 3| + | x - 5| < 15, find the range of x that satisfies this inequality.</span></span>
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<span style="background-color: white; font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;"><span style="color: red;">Question</span></span><br />
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<span style="background-color: white; color: #333333; font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">Consider set A with 'a' elements, set B with 'b' elements, set C with 'c' elements. We can define a function that is one-one but not onto from set A to set B, a function that is onto but not one-one from set B to set C; and a function that is injective but not surjective from C to A. Arrange a, b, c in ascending order.</span><br />
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