How to Prepare for Quantitative Aptitude for CAT (2014)

MOCK TEST PAPERS

MODEL TEST PAPER-1

Solutions and Shortcuts

Solution 1: Level of Difficulty (2)

Let p be the amount of work Priyanka can do per day and s be amount of work Sanjana can do per day. According to the given statement ... ((w1/p) – (w1/(p + s)))/(w1/(p + s)) = 0.8 this gives s = 0.8p, what we were asked is, ((w2/s) – (w2/(p + s)))/(w2/(p + s)) substituting s = 0.8p, we get the required value as 1.25 i.e. 125%. Hence, choice (1) is the right answer.

Solution 2: Level of Difficulty (3)

The expression needs to be evaluated at different values of x and we can easily see that at x = 0, the value of the function becomes –1. Further at x = 0.5 we can find that the value is –3/2. So we can understand that the value of the function is reducing when we move to the right of 0. It can also be seen that to the left of 0 also there will be a drop in the value of the function. For instance at x = –0.1 also the value of the function will be less than –1. So obviously the function is reaching a kind of a maximum at –1 and is not going beyond that when the range of values are in this range.

It can be observed that after x = 1, the function will become positive. At x = 1.1 it can be seen that the value of the function would become around 10-11. As you would increase the value of x beyond 1, 1 the function would reduce in value. Also, it can be seen that after x = 1, the funtion would achieve its minimum value at x = 2 Æ where its value would be 3. After 2 the value would start increasing. Hence, the value of the function cannot be between –1 to +3. Hence, option 3 is the correct answer.

Solution 3: Level of Difficulty (1)

Equate 12 – x = 8 + x to give you the intersection point between the two lines 12 – x and 8 + x. The intersection occurs at a value of x as 2. It can be visualized by plotting both these lines that the maximum value of the given function would occur at x = 2. Hence, the correct answer would be 10.

Solution 4: Level of Difficulty (2)

Solutions would exist for the following structures of making the value of 5: 0 + 5 Æ This would happen if we take the value of x as 4 and y can take the values of 7 or –3. Hence, there would be 2 sets of integral (x, y) values giving us 0 + 5 = 5

1 + 4 Æ (5, 6), (5, –2), (3, 6), (3, –2) Æ four solutions

2 + 3 Æ four possibilities again

3 + 2 Æ four possibilities again

4 + 1 Æ four possibilities again

5 + 0 Æ 2 possibilities

Solution to Question 5: Level of Difficulty (2)

There would be 10 chapters with even number of pages. Place them to start with—each of them would start on an odd numbered page. After that start to place the chapters with an odd number of pages—the first one would start on an odd numbered page, the second on an even numbered page, the third on an odd numbered page and so on. Thus there would be 10 + 5 = 15 chapters out of 20 which can at the maximum start on an odd numbered page. Hence, option 2 is correct.

Solution to Question 6: Level of Difficulty (2)

For this question, you would have to count the actual number of numbers. In the hundreds, the first numbers you would find would be in the 120s. The first numbers are 124, 126, 128, 134, 136, 138, 146, 148, 156, 158, 168, 178.

In the 200s, the values would be 234, 236, 238, 246, 248, 256, 258, 268, 278

In the 300s the values would be 346, 348, 356, 358, 368, 378

In the 400s the values would be 456, 458, 468, 478

In the 500s there would be only 2 values.

1 value in the 600s and no value after that. Hence 34 values. But in all there are 450 even three digit numbers starting from 100, 102, 104 …998. Hence, the required answer is 450 – 34 = 416.

Solution 7: Level of Difficulty (2)

Each of the 3 places can take 3 letters fi 27. But we don’t want the combination (1, 2, 3) fi 3! = 6 are out fi 27 – 6 = 21.

Solution 8: Level of Difficulty (3)

It can be seen by plotting the graph of this expression that the function y = 3^x – 2x – 1 would cut the x axis twice. Hence, the equation would have 2 real solutions.

Solution 9: Level of Difficulty (3)

log(2x – 3/4) > 2log x solving we get 2 cases:

Case 1: When x > 1 fi (2x – 3/4) > x².

Case 2: When x < 1 fi (2x – 3/4) < x²

Solving these inequalities we get: x lies in (3/8, 1/2) » (1, 3/2) viz. (0.375, 0.5) » (1, 1.5)

In these ranges we have two independent values which could be expressed as p/5 viz. 0.4 = 2/5 and 1.2 = 6/5. Since in both the cases, p is co prime with 5, we can say that both these cases satisfy the conditional requirements. Hence, choice (2) is the right answer.

Solution 10: Level of Difficulty (2)

Total matches being played = 8C2*8C2 = 28² = 784. Thus, a total of 784 × 4 = 3136 people are part of these 784 matches. Each of the 16 players would play in the same number of matches = 3136/16 = 196.

Hence, choice (1) is the right answer.

Solution 11: Level of Difficulty (2)

In this case you would get the sets like: {1, 2, 3, 4, 5, 6}, {8, 9, 10, 11, 12, 13}, {15, 16, 17, 18, 19, 20}…. As we can see the starting digits for each of these sets consists of an Arithmetic Progression as 1, 8, 15…. The next step is to find the number of terms in this series before 999. This can be done as: Series is 1, 8, 15, 22 … 995 …. And this series would have [(995 – 1)/7] + 1 = 143 terms.

Solution 12: Level of Difficulty (1)

Join AC fi < ACB = 90° fi < CAB = 55°. But < BDC = < CAB as they are subscribed by the same arc. Hence, choice (3) is the right answer.

Solution 13: Level of Difficulty (2)

Use t_n = a + (n – 1)d

6000 = a₁ + a₂ + (n – 1) (d₁ + d₂)

or, n = 429

so, we have two A.P. series

3, 9, 15, 21, ———, 2571 (Calculate t_n)

5, 13, 21, ———, 3429 (Calculate t_n)

so, common series is 21, 45 ———, x where x £ 2571

On solving, x comes for n = 107. Hence, choice (3) is the right answer.

Solution 14:

Let the distance between Delhi and Lucknow be 300 kms. The total time taken would be 3 + 6 + 10 + 5 = 24 hours. The total distance would be 1200 kms. Hence, the average speed is given by

Average speed = (total distance/total time)

= 1200/24 = 50 kmph

Hence, choice (4) is the right answer.

Solution 15: Level of Difficulty (2)

Divya needs to travel 100 km @ 80 km/h fi Time taken = 5/4 hours. Hence, choice (2) is the right answer.

Solution 16: Level of Difficulty (3)

Go through trial and error for both situations.

We can see that x + y + z = 60 is satisfied for z = 44, 47 and 50. Hence, A = 3

We can see that x + y + z = 84 is satisfied for z = 73, 70 and 67. Hence, B = 3.

Hence, choice (4) is the right answer.

Solution 17: Level of Difficulty (2)

The letters available to us are: A, E H, S and T in their alphabetical order. When we form 5 lettter words out of these, the following order of appearing in the dictionary would hold…

Words starting with A___4! = 24

Words starting with E___4! = 24

Words starting with HA___3! = 6

Words starting with HE … complete description:

First word: HEAST, Second word = HEATS…

Hence, the 56^th word in the list would be HEATS. Option 3 is correct.

Solution 18: Level of Difficulty (2)

Let a solve just A, b solve just B, c solve just C, and d solve B and C but not A. Then 25 – a – b – c – d solve A and at least one of B or C. The conditions give: b + d = 2(c + d); a = 1 + 25 – a – b – c – d; a = b + c. Eliminating a and d, we get: 4b + c = 26. But d = b – 2c >= 0, so b = 6, c = 2. Hence, choice (1) is the right answer.

Solution 19: Level of Difficulty (2)

If we take x as 2, we will get 2 integers,

With x as 3 we will get 2 integers,

With x as 4 we have 3 integers,

With x as 5 we have 2 integers,

With x as 6 we have 4 integers,

With x as 7, we have 2 integers,

With x as 8 we have 4 integers,

With x as 9 we have 3 integers,

With x as 10 we have 4 integers,

With x as 11 we have 2 integers,

With x as 12 we have 6 integers,

With x as 13 we have 2 integers

With x as 14 we have 4 integers,

With x as 15 we have 4 integers.

Thus there would be a total of 44 such integral ratios. Hence, the option 4 is correct.

Solution 20: Level of Difficulty (1)

From the table, it is clear that the difference between Arun and Aishwarya’s lottery ticket value is constant upto the 3^rd exchange and equal to 2. After the 4^th exchange also it equals 2.