2. Disclaimer
This session is meant for those who are
not good in quant.
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3. SESSION OBJECTIVE
2. Solve CAT level questions
– Without long calculations
– Using simple logic
– Spotting wrong options…
• To score marks sufficient to clear the cut off in quantitative
section
• To demonstrate that a relaxed state of mind can help us
significantly while attempting quant
• To demonstrate that all the above is possible without really
knowing quant in detail
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4. Point To Ponder
The psychology of a quant paper setter
She/he spends 3-4 hours to make a good question
and 1-2 minutes to make 5 options
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6. Commandment 1
All questions are equally tough for me, hence I will read
all of them
Reading means:
Reading the questions and all the choices
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7. Commandment 2
If I am able to
eliminate 3 out of 5 options,
I will take a chance and mark an answer.
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8. Commandment 3
Geometry is easy !
4. If no figure is given I will draw it roughly to scale.
5. Wherever possible I will assume a particular case for a
general case
– Hexagon implies a regular hexagon
– Quadrilateral/parallelogram implies a square
– Right angle triangle with sides 3,4,5 or its multiples
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9. Commandment 4
Algebra is my friend !
4. In all ‘series’ questions, I will add first 2-3 terms and
eliminate options.
5. In quadratic equations, I will assume a simple equation
satisfying the options & eliminate options.
6. In all a, b, c or questions where n terms are given I will
always assume them to be easy numbers like 0,1,-1, 2
etc.
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10. Commandment 5
Probability (Hey bhagwan ! aaj subah kiska mooh
dekha tha …)
is not as bad as it looks.
4. In questions of probability involving ‘n’ trials I
will check for n= 1,2 so on
5. Probability of any event lies between the best
and the worst case scenarios. Thus, I will take
the best and worst case scenarios for anything
to happen or not to happen.
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11. Commandment 6
Graph questions are not as ugly as they look, in
fact in CAT, like in life, looks are deceiving.
The more intimidating they look,
the easier they are!
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12. Commandment 7
Bring your relaxed brain with you for the exam,
there are some really easy questions, we just
need to be patient with.
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13. Commandment 8
I will use all ‘illegitimate methods’ at my
disposal to find the answer.
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14. Commandment 9
I will not be afraid of any question in quant,
however lengthy or scary it looks.
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15. Commandment 10
CAT is like a festival. It occurs every year. So
enjoy the festival and its preparation.
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17. Application 1 - Question
If ab X cd = 1073 and ba X cd = 2117, find the value of (ab
+ cd) given that ab, ba and cd are all two digit positive
integers.
1.66 2.65 3.63 4.67 5.69
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18. Application 1 - Solution
If ab X cd = 1073 and ba X cd = 2117, find the value of (ab
+ cd) given that ab, ba and cd are all two digit positive
integers.
1.66 2.65 3.63 4.67 5.69
ab and cd are odd because ab X cd = odd,
Thus ab + cd = even,
Hence, choice 2,3,4,5 are incorrect.
Answer: Choice 1 (37 & 29)
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19. Application 2 - Question
If the points (1, 2), (4, 5), (8, 9) and (x, y) are vertices of
a parallelogram, then find (x, y).
1.(1, 5) 2.(5, 12) 3.(7, 13)
4.(5, 7) 5. Data Insufficient
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20. Application 2 - Solution
If the points (1, 2), (4, 5), (8, 9) and (x, y) are vertices of
a parallelogram, then find (x, y).
1.(1, 5) 2.(5, 12) 3. (7, 13)
4. (5,7) 5. Data Insufficient
3 of the 4 given points lie on the same line y = x + 1.
Hence, they are Collinear and a parallelogram cannot be
formed.
Answer: Choice 5
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21. Application 3 – Question – CAT 2002
The nth element of a series is represented as
Xn = (-1)n Xn-1
If X0 = x and x > 0, then which of the following is always
true.
1. Xn is positive if n is even 2. Xn is positive if n is odd
3. Xn is negative if n is even 4. None of these
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22. Application 3 – Solution – CAT 2002
The nth element of a series is represented as
Xn = (-1)n Xn-1
If X0 = x and x > 0, then which of the following is always
true.
1. Xn is positive if n is even 2. Xn is positive if n is odd
3. Xn is negative if n is even 4. None of these
Substitute for n = 1,2,3 …
X0 = X, X1 = -X, X2 = -X … We find there is no sequence
Answer: Choice 4
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23. Application 4 – Question – CAT 2003
Given that -1 ≤ v ≤ 1, -2 ≤ u ≤ -0.5 and -2 ≤ z ≤ -0.5 and
w = vz/u, then which of the following is necessarily true
1. -0.5 ≤ w ≤ 2 2. -4 ≤ w ≤ 4
3. -4 ≤ w ≤ 2 4. -2 ≤ w ≤ -0.5
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24. Application 4 – Solution – CAT 2003
Given that -1 ≤ v ≤ 1, -2 ≤ u ≤ -0.5 and -2 ≤ z ≤ -0.5 and
w = vz/u, then which of the following is necessarily true
1. -0.5 ≤ w ≤ 2 2. -4 ≤ w ≤ 4
3. -4 ≤ w ≤ 2 4. -2 ≤ w ≤ -0.5
U is always negative. For minimum value of w, vz should
always be maximum positive. This means vz is 2.
For maximum value of w, vz should be the smallest
negative. This means vz is -2. Consider decimal value of
W, to find the extreme values.
Answer: Choice 2
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25. Application 5 – Question – CAT 2004
If f(x) = x3 -4x + p, and f(0) and f(1) are of opposite signs,
then which of the following is necessarily true
1. -1 < p < 2 2. 0 < p < 3
3. -2 < p < 1 4. -3 < p < 0
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26. Application 5 – Solution – CAT 2004
If f(x) = x3 -4x + p, and f(0) and f(1) are of opposite signs,
then which of the following is necessarily true
1. -1 < p < 2 2. 0 < p < 3
3. -2 < p < 1 4. -3 < p < 0
Substitute x = 0,1
f(0) = p, f(1) = -3 + p
From the above we can infer that p cannot be negative.
From -3 we know that p cannot be greater than 3.
Answer: Choice 2
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27. Application 6 – Question
Bag A contains 6 white balls and 4 black balls and bag B
contains 3 white balls and 2 black balls. A white ball is
picked from bag A and put into bag B. Then, 3 balls are
picked from bag B and put into bag A. Find the probability
that a ball picked now from bag A is black.
1. 1/4 2. 1/3
3. 7/12 4. 5/12
5. 11/24
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28. Application 6 – Solution
Probability will lie between the best and the worst case
scenario.
White Black
Bag A 6-1=5 4
Bag B 3+1=4 2
3 balls are transferred from bag B to bag A.
Best Case 2 black and 1 white ball from B to A
Worst Case 3 white balls from B to A
White Black P(Black Ball)
Best Case 5+1=6 4+2=6 6/12
Worst Case 5+3=8 4+0=4 4/12
The answer should lie between the best and the worst case. This
eliminates option 1,2 and 3.
Answer: Choice 4
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30. Application 7 – Solution – CAT 2005
Then x equals. (2 Marks)
1. 3 2. (√13 - 1) / 2
3. (√13 + 1) / 2 4. √13
The above can be written as X2 = 4 + √(4-x)
The first term is √4 which is greater than 2.
The second term is √(4 + 2) which means the value of x
must be less than 3. √13 = around 3.6
1. 3 2. 1.3
3. 2.3 4. 3.6
Answer: Choice 3
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31. Application 8 – Question – CAT 2006
Let f(x) = max(2x + 1, 3 – 4x), where x is any real
number. Then the minimum possible value of f(x) is.
1. 1/3 2. 1/2
3. 2/3 4. 4/3
5. 5/3
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32. Application 8 – Solution – CAT 2006
Let f(x) = max(2x + 1, 3 – 4x), where x is any real
number. Then the minimum possible value of f(x) is.
1. 1/3 2. 1/2
3. 2/3 4. 4/3
5. 5/3
We know that f(x) would be minimum at the point of
intersection of these curves.
2x + 1 = 3 – 4x 6x = 2 x = 1/3
Hence, min f(x) = 5/3
Answer: Choice 5
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33. Application 9 – Question
MTNL has 1500 subscribers in ‘Paradise’ and they earn a
revenue of Rs.300 from each subscriber. They wish to
maximize their revenue but for every increase in revenue
of Rs.1 per subscriber, one subscriber drops out. What will
be the maximum revenue of MTNL.
1. Rs.810,000 2. Rs.640,000
3. Rs.570,000 4. Rs.720,000
5. None of these
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34. Application 9 – Solution
MTNL has 1500 subscribers in ‘Paradise’ and they earn a
revenue of Rs.300 from each subscriber. They wish to
maximize their revenue but for every increase in revenue
of Rs.1 per subscriber, one subscriber drops out. What will
be the maximum revenue of MTNL.
1. Rs.810,000 2. Rs.640,000
3. Rs.570,000 4. Rs.720,000
5. None of these
For every increase in revenue the subscriber base falls by same
amount. Therefore (1500 – x) + (300 + x) = 1800 is a constant.
When the sum of 2 numbers is constant, the product of the 2 is
maximum when they are equal. (1500 – x) = (300 + x) = 900
Therefore, f(x)max = 900 * 900
Answer: Choice 1
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35. Application 10 – Question – CAT 2005
If x = (163 + 173 + 183 + 193), then x divided by 70 leaves a
remainder of
1. 0 2. 1
3. 69 4. 35
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36. Application 10 – Solution – CAT 2005
If x = (163 + 173 + 183 + 193), then x divided by 70 leaves a
remainder of
1. 0 2. 1
3. 69 4. 35
X is a even number. Why ??
(a + b) always divides a3 + b3
Therefore,
(163 + 193) and (173 + 183)are divisible by 35.
Hence, x is divisible by 70.
Answer: Choice 1
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