# IIM CAT Preparation Tips

IIM CAT Preparation Tips: June 2013

## Jun 21, 2013

### CAT Pipes and Cisterns - Do-able but atypical

Question
A drain pipe can drain a tank in 12 hours, and a fill pipe can fill the same tank in 6 hours. A total of n pipes – which include a few fill pipes and the remaining drain pipes – can fill the entire tank in 2 hours. How many of the following values could ‘n’ take?
a)      24
b)      16
c)       33
d)      13
e)      9
f)       8

A.      3
B.      4
C.      2
D.      1

Two drain pipes can drain the same volume that one fill pipe fills. This means that a D-D-F combination has to have a net volume effect of 0.
In spite of this, the tank still gets filled. Only the fill pipes can manage to fill the tank. In addition to all the net zero effect pipes, we need three more fill pipes in order to fill the tank in 2 hours.
So, we can have as many D-D-Fs as we want, but we need one F-F-F at the end to ensure that the tank gets filled in 2 hours.
So the number of pipes will be → (D – D - F).......(D – D - F) + (F – F - F)
The number of pipes has to be a multiple of 3. Only options A, C and E fit the description.

### CAT - Profit and Loss Simple Question

Question

Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of x%. If both make the same profit, find x
A.      25%
B.      12.5%
C.      37.5%
D.      40%

SP of trader A = 1000 (1 + x)
Profit of trader A = 1000 (1 + x) - 1000
MP of trader B = 2000 (1 + 2x)
SP of trader B = 2000 (1 + 2x) (1 - x)
Profit of trader B = 2000(1 + 2x) (1 - x)- 2000
Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 - x)- 2000
1000x = 2000 – 4000x2 + 4000x – 2000x - 2000
4000x2 -1000x = 0
1000x (4x - 1) = 0
ð  x = 25%

## Jun 17, 2013

### CAT Linear Equations

Question
(|x| -3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?
1. 4
2. 10
3. 6
4. 8

Product of two integers is 12. Further |y| + 4 > 4 . So, we know that one of the numbers has to be greater than 4.
So, we can have
1 x 12 => x = +4, y = +8
2 x 6 => x = +5, y = +2
3 x 4 => x = +6, y = 0

4 + 4 + 2 = 10 possible solutions

### CAT Calendar Questions

Question

How many of the following statements have to be true?
1. No year can have 5 Sundays in the month of May and 5 Thursdays in the month of June
2. If Feb 14th of a certain year is a Friday, May 14th of the same year cannot be a Thursday
3. If a year has 53 Sundays, it can have 5 Mondays in the month of May
A.      0
B.      1
C.      2
D.      3

Explanation

I.            No year can have 5 Sundays in the month of May and 5 Thursdays in the month of June
A year has 5 Sundays in the month of May => it can have 5 each of Sundays, Mondays and Tuesdays, or 5 each of Saturdays, Sundays and Mondays, or 5 each of Fridays, Saturdays and Sundays. Or, the last day of the Month can be Sunday, Monday or Tuesday.
Or, the 1st of June could be Monday, Tuesday or Wednesday. If the first of June were a Wednesday, June would have 5 Wednesdays and 5 Thursdays. So, statement I need not be true.

II.            If Feb 14th of a certain year is a Friday, May 14th of the same year cannot be a Thursday
From Feb 14 to Mar 14, there are 28 or 29 days, 0 or 1 odd day
Mar 14 to Apr 14, there are 31 days, or 3 odd datys
Apr 14 to May 14 there are 30 days or 2 odd days
So, Feb 14 to May 14, there are either 5 or 6 odd days
So, if Feb 14 is Friday, May 14 can be either Thursday or Wednesday. So, statement 2 need not be true.

III.            If a year has 53 Sundays, it can have 5 Mondays in the month of May
Year has 53 Sundays => It is either a non-leap year that starts on Sunday, or leap year that starts on Sunday or Saturday.
Non-leap year starting on Sunday: Jan 1st = Sunday, jan 29th = Sunday. Feb 5th is Sunday. Mar 5th is Sunday, Mar 26 is Sunday. Apr 2nd is Sunday. Apr 30th is Sunday, May 1st is Monday. May will have 5 Mondays.
So, statement C can be true.

Only one of the three statements needs to be true. Answer Choice (B)

## Jun 13, 2013

### CAT Geometry

Question
x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.

A.    2
B.    3
C.    1
D.    More than one possible value of z exists

xy = 4
xy could be 2 x 2 or 4 x 1

221
222            These are the possible triangles
223
441

22x will be a triangle if x is 1, 2 or 3 (trial and error)
44x is a triangle only if x is 1.

Ø  221 is acute.  12 + 22> 22
Ø  222 is equilateral. So acute.
Ø  223 is obtuse.  22 + 22< 32
Ø  144 is acute.  12 + 42> 42

Only triangle 223 is obtuse.  Hence, the third side has to be 3.

## Jun 12, 2013

### CAT Geometry

Question
Sides of a triangle are 6, 10 and x for what value of x is the area of the D the maximum?
A.    8 cms
B.    9 cms
C.    12 cms
D.    None of these

Side of a triangle are. 6, 10, x.
Area = 1/2 x 6 x 10 sin ÐBAC.
Area is maximum, when ÐBAC = 90o
x = sqrt(100 + 36) = sqrt(136)

There is a more algebraic method using hero’s formula. Try that also.

## Jun 11, 2013

### CAT Averages

Question

Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum possible average weight of the class now?
1. 40.18 kgs
2. 40.56 kgs
3. 40.67 kgs
4. 40.49 kgs

Explanation:
If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40.
So, n > 40
Consequently, m has to be <10 (as n + m = 50)
Working with the “differences” approach, we know that the total additional weight added by “m” students would be (n - 40) each, above the already existing average of 40. m(n - 40) is the total extra additional weight added, which is shared amongst 40+m students. So, m (n – 40)/ (m + 40) has to be maximum for the overall average to be maximum.
At this point, use the trial and error approach (or else, go with the answer options) to arrive at the answer.
The maximum average occurs when m = 5,and n = 45
And the average is 40 + (45 – 5) * 5/45 = 40 + 5/9 = 40.56

## Jun 10, 2013

### CAT Averages,

Question
1.       5. Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90 and 91-100. The number of students who have scored in each range is given below
51 -60 - 3 students
61 -70 - 8 students
71 -80 - 7 students
81 -90 - 4 students
91 -100 - 3 students
Furthermore, we know that at least as many students scored 76 or more as those who scored below 75. What is the minimum possible average overall of this class?
1. 72
2. 71.2
3. 70.6
4. 69.2

Let's employ the idea of a total of 25 students (all of the same weight) sitting on a see-saw, which has numbers from 51 to 100 marked on it. At least as many students are sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But you can't do that, as you have other constraints as well.

First of all, you have to seat 3 students from 51 to 60, and 8 students from 61 to 70. Secondly, you also have to make sure that the average is the least. This means that the see-saw should be tilting as much to the left as possible, which in turn means that the number of people sitting to the left of 75 should be the highest possible.

This makes it 12 students to the left of 75, and the remaining 13 students on 76 or to its right.

Next, how do you ensure that the average is least, i.e. how do you ensure that the balance tilts as much as possible to the left? Make each student score as little as possible given the constraints.

So, the first 3 students only score 51 each. The next 8 students score only 61 each. 11 Students are now fixed. The 12th student has to be below 75, so seat him on 71. The remaining 6 students (who are in the 71 to 80 range) have to score 76. The next 4 score 81 and the next 3 score 91.

This would give you the least average.
The lowest possible average would be:
= [(3 * 51) + (8 * 61) + 71 + (6 * 76) + (4* 81) +  (3 * 91)]/25
= 70.6

### CAT Functions, Onto functions

Question
How many onto functions can be defined from the set A = {1, 2, 3, 4} to {a, b, c}?
1. 81
2. 79
3. 36
4. 45

First let us think of the number of potential functions possible. Each element in A has three options in the co-domain. So, the number of possible functions = 34 = 81.

Now, within these, let us think about functions that are not onto. These can be under two scenarios

Scenario 1: Elements in A being mapped on to exactly two of the elements in B (There will be one element in the co-domain without a pre-image).

Ø  Let us assume that elements are mapped into A and B. Number of ways in which this can be done = 24 – 2 = 14
o   24 because the number of options for each element is 2. Each can be mapped on to either A or B
o   -2 because these 24 selections would include the possibility that all elements are mapped on to A or all elements being mapped on to B. These two need to be deducted
Ø  The elements could be mapped on B & C only or C & A only. So, total number of possible outcomes = 14 * 3 = 42.

Scenario 2: Elements in A being mapped to exactly one of the elements in B. (Two elements in B without pre-image). There are three possible functions under this scenario. All elements mapped to a, or all elements mapped to b or all elements mapped to c.

Total number of onto functions = Total number of functions – Number of functions where one element from the co-doamin remains without a pre-image - Number of functions where 2 elements from the co-doamin remain without a pre-image

ð  Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36

## Jun 7, 2013

### CAT Linear Equations

Another question from the book Quantitative Aptitude for CAT .

Question
1.       x + |y| = 8, |x| + y = 6.How many pairs of x, y satisfy these two equations?
A.      2
B.      4
C.      0
D.      1

We start with the knowledge that the modulus of a number can never be negative, though the number itself may be negative.

The first equation is a pair of lines defined by the equations
y = 8 – x        ------- (i) {when y is positive}
y = x – 8        ------- (ii) {when y is negative}
With the condition that x 8 (because if x becomes more than 8, |y| will be forced to be negative, which is not allowed)

The second equation is a pair of lines defined by the equations:
y = 6 – x        ------- (iii) {when x is positive}
y = 6 + x        ------- (iv) {when x is negative}
with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be negative.

On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and (iv) are parallel (same slope).

Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv) holds good only when x is negative

Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This satisfies both equations.

Only one solution is possible for this system of equations.

## Jun 6, 2013

### CAT Functions

Question
Qn. f(x + y) = f(x)f(y) for all x, y, f(4) = +3 what is f(-8)?
A.      1/3
B.      1/9
C.      9
D.      6