Online CAT Coaching: A few interesting True/False questions from Geometry

State whether the following statements are true or false

1. A parallelogram that circumscribes a circle has to be a square
2. A trapezium inscribed in a circle has to be an isosceles trapezium
3. Orthocenter of a triangle can lie outside the triangle
4. Triangle with sides a, b and c has the relationship a^2 + b^2 > c^2, the triangle has to be acute-angled.
5. Diagonals of a parallelogram are angle bisectors of the angles of a parallelogram.

Scroll down for answers and explanation
























1. A parallelogram that circumscribes a circle has to be a square: FALSE

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.

2. A trapezium inscribed in a circle has to be an isosceles trapezium: TRUE

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.

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.

3. Orthocenter of a triangle can lie outside the triangle: TRUE

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.

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

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)


5. Diagonals of a parallelogram are angle bisectors of the angles of a parallelogram: FALSE

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.

CAT Online Coaching - Permutation and Combination, Fixing the Errors

This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to the first three questions.

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?

Given 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.

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.

Bug in the solution:

We end up double-counting here.

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.

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.

The two scenarios mentioned above are identical. However, we end up counting both. This is the reason for the double count.

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.

Awesome, isn’t it.

Correct solution:

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 ⁵C₂ 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.

Total number of options = ⁵C₂ * 4! = 10 * 24 = 240 ways.

2. How many 4-digit numbers with 4 distinct digits are possible?

Given solution

Let the 4-digit number be 'abcd'. Now,

'd' can take 10 possible values - 0 to 9. 

'c' can take 9 possible values - 0 to 9 except d

'b' can take 8 possible values - 0 to 9 except d and c

'a' can take 6 possible values - 1 to 9 except d, b and c

So, total number of possibilities = 6 * 8 * 9 * 10 = 4320.

Bug in the solution:

The simple way of stating the bug is as follows “But one of b, c, d might have been zero”.

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.

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.

Correct solution:

Let the 4-digit number be 'abcd'. Now,

'a' can take 9 possible values - 1 to 9. 

'b' can take 9 possible values - 0 to 9 except a

'c' can take 8 possible values - 0 to 9 except a and b

'd' can take 7 possible values - 1 to 9 except a, b and c

Total number of options = 9 * 9 * 8 * 7.

3. In how many ways can we rearrange the letters of the word TWOIIM such that the vowels appear together?

Given solution

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.

Bug in the solution:

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.

Correct solution:

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.

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.

So total number of possibilities = 4! * 3 = 72 ways.

CAT Linear Equations

Question
A system of equations has 3 equations
3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.
If the system of equations has no solution, find k.

Explanation
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.

So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) =  2x + 3y + kz.
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.

Or, we are effectively solving for
3a + 4b = 2
4a + 5b = 3

Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.

Now, k = 5a + 3b = 10 -3 = 7.

If k = 7, this system of equations would result in no solutions.

If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + 7z = 6.
First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.

Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has no solution.

Yeah, the more mechanical, rather deceptively straight-forward-looking determinant method is also there. But where is the joy in that.

CAT preparation - Ratio and Proportion (Tough)

Question

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?

Explanation

Again, big thanks to Mukund Sukumar for excellent solution.

Let john have 3x chocolates of type A and 7x of type B

Let Mike have 5y chocolates of type B and 4y of type C

Let john have 3z chocolates of type C and 5z of type A

So in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since C>B we get solving y<3z-7x --->(1)

Since B>A we get solving 5y>5z-4x ---> (2)

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.

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.

Hence x=1,y=4 and z=4 works and is the best possible answer.

For these values, we get A=23,B=27,C=28.

Minimum possible number of chocolates overall is 76. 

CAT Preparation Online - Tough one from Permutation and Combination

Question

A wonderful, but very tough question from Permutation and Combinations.

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)

Explanation

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2 N’s. 

Let us focus on the three O’s.

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 ⁵C₃, or 10 ways.

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.

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’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’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2 

words like this. So, number of words that we need to count = 12 – 2 = 10

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’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’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

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’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’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’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’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

Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.

Phew!.

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.

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_

The O’s can be selected in ⁴C₃ ways. The MNNS can be rearranged in 3! Ways if the N’s have to appear together.

Or, we get ⁴C₃ * 3! = 24 ways. So, we have 120-24 = 96 ways totally.

CAT Preparation Online - A simple question from Quadratic Equations

Questions

x^2 - 17x + |p| = 0 has integer solutions. How many values can p take?

Explanation

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. 

So, both roots need to be positive.

So, we are effectively solving for 

Number of positive integer solutions for a + b = 17.

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.

So, |p| can take 8 different values. Or, p can take 16 different values.

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. 

So, p can also be zero.

So, number of possible values of p = 16 + 1 = 17.

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.

CAT Preparation Online - Simple one from Quadratic Equations

Question

How many integer solutions exist for the equation x² - 8|x| - 48 =0?

Explanation

One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly simpler method.

Note that x² is the same as |x|², so we can treat the equation as a quadratic in |x|.

Or, |x|² – 8|x| - 48 = 0

(|x| - 12|) (|x| + 4) = 0

|x| could be 12. |x| cannot be -4.

If |x| could be 12, x can be -12 and 12.

Two possible solutions exist. 

CAT - Question from Pipes and Cisterns

Question

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?

Explanation

Almost all questions can be approached well by asking the question "What happens in 1 hour" (or 1 day, or 1 minutes)

So, let us start with that

In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first hour, the tank will not be filled
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.
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.

Oops, one cannot fill 13/12 of a tank. What this tells us is that the tank gets filled in the 3rd hour. 

When exactly during the third hour?

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.

Or, the tank will be filled in 2 hours and 40 minutes.

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. 

CAT Permutation Combination and Functions

Question

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? 

Explanation

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.

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. 

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. :-)

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! .

So, total number of functions that are neither one-one nor onto = 256 - 24 = 232. 

CAT Permutation and Combinations

Question

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?

Explanation

First let us think about the number of possible rearrangements. SLEEPLESS can be rearranged in 9!/ (2! 3! 3!) = 5040. 

Now, if rearrange these, we would have words starting with E, L, P and S.

Now, SLEEPLESS starts with S. So, let us think about how many words start with S. Number of words starting with S = 8!/ (2! 2! 3!) = 1680.

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.

The first word with S is SEEELLPSS. This would have a rank of 3361.

Now, let us go step by step. 

Words starting with SE____ - Number of such words = 7!/(2!*2!*2!) = 630 words

Next we have words starting with SL, but our word also starts with SL, so let us go deeper

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

Words starting with SLEE. The first such word would be SLEEEPLSS

So, let us think about words starting with SLEEE - there would be 4!/(2!) words like this = 12 words like this

words starting with SLEEL - there would be 4!/(2!) words like this = 12 words like this

So, we have accounted for 3360 + 630 + 12 + 12 = 4014 words thus far.

Now, on to words starting with SLEEP - there are again 4!/(2!) words like this = 12 words like this

Our word is within these 12 words

Words starting with SLEEPE - there are 3!/2! words like this, or 3 words like this
Our word comes in the next bunch.

Words starting with SLEEPL -  SLEEPLESS is the first such word. Or, the rank fo SLEEPLESS = 4014 + 3 + 1 = 4018.

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.

CAT Number Systems - Fun question based on Armstrong numbers

Question

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.

Explanation

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.

The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and error (although a lil scientifically)

CAT Number Theory - Interesting Question from Factorial

Question

How many trailing zeroes will be present in the base 12 representation of 55!?

Explanation

The question can be restated as follows - "What is the highest power of 12 that divides 55!"

Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.

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!. 

Successive division by 2 will give us the highest power of 2 that divides 55!:

55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.

Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 =  50

Successive division by 3 will give us the highest power of  3 that divides 55!:

55/3 = 18, 18/3 = 6, 6/3 = 2.

Highest power of 3 that divides 55! = 18 + 6 + 2 = 26. 

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!

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. 

Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.

CAT Coordinate Geometry Question and Solution

Question

What is the area enclosed by the region defined by y = |x -1| + 2, the line x = 1; X-axis and Y-axis?

Explanation

Solution is available on the video given below.