CAT 2018 Slot 1 QA Questions with Answers & Solutions

Q67MCQLogarithms

If

x

is a positive quantity such that

2^x = 3^{\log_5 2}

, then

x

is equal to

$\log_5 8$
$1 + \log_3 \frac{5}{3}$
$\log_5 9$
$1 + \log_5 \frac{3}{5}$

Answer and solution

Answer: (D) $1 + \log_5 \frac{3}{5}$

Use the identity

a^{\log_b c} = c^{\log_b a}

, which holds because both sides have the same logarithm to base

b

, namely

\log_b a \cdot \log_b c

. So

3^{\log_5 2} = 2^{\log_5 3}

, and the equation becomes

2^x = 2^{\log_5 3}

, giving

x = \log_5 3

. No option is written this way, so rewrite it:

\log_5 3 = \log_5 \left(5 \times \frac{3}{5}\right) = \log_5 5 + \log_5 \frac{3}{5} = 1 + \log_5 \frac{3}{5}

. Option B looks similar, but

1 + \log_3 \frac{5}{3} = \log_3 3 + \log_3 \frac{5}{3} = \log_3 5

, which is the reciprocal of

\log_5 3

, not the same number. Option C is

\log_5 9 = 2\log_5 3

, twice the answer, and option A,

\log_5 8 = 3\log_5 2

, is a different number. Hence, option D (

1 + \log_5 \frac{3}{5}

).

Q68MCQPolygons & Circles

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

$\sqrt{13}$
$\sqrt{14}$
$\sqrt{11}$
$\sqrt{12}$

Answer and solution

Answer: (A) $\sqrt{13}$

Let the chords be AB = 6 cm and CD = 4 cm, with midpoints M and N. The perpendicular from the centre O to a chord bisects it, so AM = 3 cm and CN = 2 cm, and O, M, N lie on one line. Solution figure for question 68, CAT 2018 Slot 1

Solution figure for question 68, CAT 2018 Slot 1

The longer chord is nearer the centre. Both chords are on the same side of the centre, so if OM

= x

, then ON

= x + 1

. OA and OC are both radii, so

3^2 + x^2 = 2^2 + (x + 1)^2

, i.e.

9 + x^2 = x^2 + 2x + 5

, which gives

x = 2

. Radius

= \sqrt{3^2 + 2^2} = \sqrt{13}

cm. Check: the chords are then 2 cm and 3 cm from the centre, 1 cm apart, as required. Option D,

\sqrt{12}

, is close, but it would put the chords

\sqrt{12 - 9} = \sqrt{3}

cm and

\sqrt{12 - 4} = \sqrt{8}

cm from the centre, about 1.1 cm apart instead of 1 cm. Hence, option A (

\sqrt{13}

).

Q69MCQTime & Work

Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

45
36
32
40

Answer and solution

Answer: (C) 32

Let one human do

h

units of work a day and one robot

r

units. The job is the same in both cases, so

30(15h + 5r) = 60(5h + 15r)

. Dividing by 30 gives

15h + 5r = 10h + 30r

, so

5h = 25r

and

h = 5r

: one human does the work of 5 robots. Total work

= 30(15h + 5r) = 30(15h + h) = 480h

. Fifteen humans do

15h

a day, so they need

\frac{480h}{15h} = 32

days. This makes sense: the 5 robots in the first team do only as much as 1 human, so without them the team's daily work falls from

16h

to

15h

and 30 days stretch to

30 \times \frac{16}{15} = 32

. Option B, 36 days, would mean the 5 robots did the work of 3 humans. Hence, option C (32).

Q70MCQPolygons & Circles

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

24, 10
25, 9
25, 10
24, 12

Answer and solution

Answer: (A) 24, 10

We know that AC is the diameter and

\angle ABC = 90°

. So

AC = 2 \times 13 = 26

cm. In right angle triangle ABC:

AC^2 = AB^2 + BC^2

\Rightarrow AB^2 + BC^2 = 26^2 = 676

Checking the options: - Option A:

24^2 + 10^2 = 576 + 100 = 676

✓ - Option B:

25^2 + 9^2 = 625 + 81 = 706 \neq 676

- Option C:

25^2 + 10^2 = 625 + 100 = 725 \neq 676

- Option D:

24^2 + 12^2 = 576 + 144 = 720 \neq 676

Therefore, option A is the correct answer. Solution figure for question 70, CAT 2018 Slot 1

Q71MCQPolygons & Circles

Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

3 : 8
2 : 5
4 : 9
1 : 3

Answer and solution

Answer: (D) 1 : 3

Let AE : EB

= x : 1

. For EFGH to be a square, the four corner triangles must be congruent, so BF, CG and DH also equal

x

, while FC, GD and HA equal 1. The side of ABCD is

x + 1

, with CG

= x

and EB

= 1

.

Solution figure for question 71, CAT 2018 Slot 1

In right triangle EBF,

EF^2 = EB^2 + BF^2 = 1 + x^2

. This is the area of EFGH, which is 62.5%, i.e.

\frac{5}{8}

, of

(x + 1)^2

. So

8(x^2 + 1) = 5(x + 1)^2

, which gives

3x^2 - 10x + 3 = 0

, i.e.

(3x - 1)(x - 3) = 0

. Then

x = 3

or

x = \frac{1}{3}

. CG is longer than EB, so

x = 3

and EB : CG

= 1 : 3

. Check option B: EB : CG

= 2 : 5

would give EFGH an area of

\frac{2^2 + 5^2}{7^2} = \frac{29}{49} \approx 59\%

of ABCD, not 62.5%. Hence, option D (1 : 3).

Q72TITATime, Speed & Distance

Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 12

Let AP

= x

, so BP

= 3x

. Let car 1's speed be

s

; car 2's speed is then

\frac{s}{2}

. Both cars start together. Car 1 reaches P after

\frac{x}{s}

, and car 2 after

\frac{3x}{s/2} = \frac{6x}{s}

. Car 2 arrives 60 minutes later, so

\frac{6x}{s} - \frac{x}{s} = 60

. This gives

\frac{5x}{s} = 60

, so

\frac{x}{s} = 12

minutes. So car 1 takes 12 minutes to go from A to P, and car 2 takes 72 minutes to reach P. The answer is 12.

Q73TITAFunctions & Graphs

Let

f(x) = \min(2x^2, 52 - 5x)

, where

x

is any positive real number. Then the maximum possible value of

f(x)

is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 32

For

x > 0

,

2x^2

increases as

x

grows, while

52 - 5x

decreases. So

f(x)

, the smaller of the two, follows

2x^2

up to the point where they meet and then follows

52 - 5x

down. Its largest value is where they are equal.

2x^2 = 52 - 5x

gives

2x^2 + 5x - 52 = 0

, i.e.

(2x + 13)(x - 4) = 0

. Since

x > 0

,

x = 4

.

f(4) = \min(2 \times 16, 52 - 20) = \min(32, 32) = 32

. For smaller

x

,

f(x) = 2x^2 < 32

; for larger

x

,

f(x) = 52 - 5x < 32

. The answer is 32.

Q74TITAInequalities & Modulus

While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 40

Let the product of the other two numbers be

p

. Taking 73 instead of 37 raises the product by

73p - 37p = 36p

, so

36p = 720

and

p = 20

. Call the two numbers

a

and

b

, with

ab = 20

. For any real numbers,

(a - b)^2 \ge 0

, so

a^2 + b^2 \ge 2ab = 40

. Equality needs

a = b

, i.e.

a = b = \sqrt{20}

(or both

-\sqrt{20}

), which is allowed because the numbers are real. So the minimum of

a^2 + b^2

is

20 + 20 = 40

. The answer is 40.

Q75TITATime, Speed & Distance

Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 15

Let T's speed be

t

, so S's speed is

0.75t

. Let the trains meet

x

hours after 3 pm; S, which started at 4 pm, has then been running for

x - 1

hours. Z is three-fifths of the way from X to Y, so at the meeting T has covered

\frac{3}{5}

of the distance and S the other

\frac{2}{5}

:

\frac{xt}{0.75t(x - 1)} = \frac{3}{2} \Rightarrow 2x = 2.25(x - 1) \Rightarrow 0.25x = 2.25 \Rightarrow x = 9

. So T covers

\frac{3}{5}

of the journey in 9 hours, and the whole journey takes

9 \times \frac{5}{3} = 15

hours. Check: in its 8 hours S covers

8 \times 0.75t = 6t

, which T covers in 6 hours, and

\frac{2}{5}

of T's 15-hour journey is indeed 6 hours. The answer is 15.

Q76MCQAverages, Mixtures & Alligations

Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

17 : 25
18 : 25
19 : 24
21 : 25

Answer and solution

Answer: (C) 19 : 24

Let the cost prices of A and B be Rs.

a

and Rs.

b

per kg. The mixture sells at Rs. 40 per kg. Mixed 3 : 2 at 10% profit, the mixture costs

\frac{40}{1.1}

per kg, so

\frac{3a + 2b}{5} = \frac{40}{1.1}

, i.e.

3.3a + 2.2b = 200

. Mixed 2 : 3 at 5% profit,

\frac{2a + 3b}{5} = \frac{40}{1.05}

, i.e.

2.1a + 3.15b = 200

. Equating the left sides:

3.3a + 2.2b = 2.1a + 3.15b

, so

1.2a = 0.95b

and

\frac{a}{b} = \frac{0.95}{1.2} = \frac{95}{120} = \frac{19}{24}

. Option D (21 : 25) is the closest value, but it fails

1.2a = 0.95b

:

1.2 \times 21 = 25.2

, while

0.95 \times 25 = 23.75

. Hence, option C (19 : 24).

Q77MCQSequences & Series

Given an equilateral triangle

T_1

with side 24 cm, a second triangle

T_2

is formed by joining the midpoints of the sides of

T_1

. Then a third triangle

T_3

is formed by joining the midpoints of the sides of

T_2

. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles

T_1, T_2, T_3, \dots

will be

$188\sqrt{3}$
$248\sqrt{3}$
$164\sqrt{3}$
$192\sqrt{3}$

Answer and solution

Answer: (D) $192\sqrt{3}$

Joining the midpoints of the sides gives a triangle with half the side length. The three joins split

T_1

into four congruent equilateral triangles, and

T_2

is the middle one, so each new triangle has

\frac{1}{4}

of the area of the one before. Solution figure for question 77, CAT 2018 Slot 1

Area of

T_1 = \frac{\sqrt{3}}{4} \times 24^2 = \frac{\sqrt{3}}{4} \times 576 = 144\sqrt{3}

sq cm. The areas form an infinite geometric series with first term

144\sqrt{3}

and common ratio

\frac{1}{4}

:

\text{Sum} = \frac{144\sqrt{3}}{1 - \frac{1}{4}} = \frac{4}{3} \times 144\sqrt{3} = 192\sqrt{3}

sq cm. Option A (

188\sqrt{3}

) is the tempting near miss, but it cannot be right, and neither can C: the first three triangles alone give

144\sqrt{3} + 36\sqrt{3} + 9\sqrt{3} = 189\sqrt{3}

, and every further triangle adds more. Hence, option D (

192\sqrt{3}

).

Q78MCQLogarithms

If

\log_{12} 81 = p

, then

3\left(\frac{4-p}{4+p}\right)

is equal to

$\log_4 16$
$\log_6 16$
$\log_2 8$
$\log_6 8$

Answer and solution

Answer: (D) $\log_6 8$

Since

\log_{12} 81 = p

, we have

\log_{81} 12 = \frac{1}{p}

. As

81 = 3^4

, this says

\frac{1}{4}\log_3 12 = \frac{1}{p}

, so

\log_3 12 = \frac{4}{p}

. Write

k = \log_3 4

. Then

\log_3 12 = \log_3 3 + \log_3 4 = 1 + k

, so

\frac{1 + k}{1} = \frac{4}{p}

. By componendo and dividendo,

\frac{(1 + k) - 1}{(1 + k) + 1} = \frac{4 - p}{4 + p}

. So

\frac{4 - p}{4 + p} = \frac{k}{2 + k} = \frac{\log_3 4}{\log_3 9 + \log_3 4} = \frac{\log_3 4}{\log_3 36}

. Therefore

3\left(\frac{4 - p}{4 + p}\right) = \frac{3\log_3 4}{\log_3 36} = \frac{\log_3 64}{\log_3 36} = \log_{36} 64 = \log_{6^2} 8^2 = \log_6 8

. Option B (

\log_6 16

) has the right base but equals

\log_{36} 256

, not

\log_{36} 64

: it is about 1.55, while the expression is about 1.16. Hence, option D (

\log_6 8

).

Q79TITASimple & Compound Interest

John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 121000

Let each instalment be Rs.

x

. Left unpaid, the loan would grow to

2{,}10{,}000 \times 1.1 \times 1.1 = 2{,}54{,}100

rupees by the end of Year 2. The first instalment is paid at the end of Year 1, so by the end of Year 2 it is worth

1.1x

; the second is worth

x

. Together they must settle the loan:

1.1x + x = 2{,}54{,}100 \Rightarrow 2.1x = 2{,}54{,}100 \Rightarrow x = 1{,}21{,}000

. Check: Year 1 interest is Rs. 21,000, so the first instalment repays Rs. 1,00,000 of principal and leaves Rs. 1,10,000. Year 2 interest on that is Rs. 11,000, and

1{,}10{,}000 + 11{,}000 = 1{,}21{,}000

, the same amount. The answer is 121000.

Q80MCQTime & Work

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

20
22
16
18

Answer and solution

Answer: (A) 20

Take the whole job as 100 units, and let A and B do

a

and

b

units per day. B needs 25% more time than A, so

\frac{100}{b} = 1.25 \times \frac{100}{a}

, which gives

a = 1.25b

. A alone does the first 50 units and B alone the last 5, so the 4 days of joint work cover

100 - 50 - 5 = 45

units:

4a + 4b = 45 \Rightarrow 4(1.25b) + 4b = 45 \Rightarrow 9b = 45 \Rightarrow b = 5

. So B alone needs

\frac{100}{5} = 20

days. Check with the 13 days: A, at 6.25 units a day, takes 8 days for the first half, then come 4 days together, and B needs 1 day for the last 5 units:

8 + 4 + 1 = 13

. Option C (16) is the trap: it is A's time alone (

\frac{100}{6.25} = 16

), not B's. Hence, option A (20).

Q81TITAPermutations & Combinations

How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 502

Because the digits must appear in ascending order, each set of chosen digits gives exactly one number: once the digits are picked, their order is fixed. So we only need to count the ways to choose 2 or more of the 9 digits. That count is

\binom{9}{2} + \binom{9}{3} + \cdots + \binom{9}{9}

. All subsets of the 9 digits number

2^9 = 512 = \binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9}

. Removing the empty choice (

\binom{9}{0} = 1

) and the nine one-digit choices (

\binom{9}{1} = 9

):

512 - 1 - 9 = 502

. The answer is 502.

Q82TITAMensuration

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With

\pi = 22/7

, the volume, in cubic ft, of the remaining part of the cone is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 198

The cone has radius 4 ft (half the 8 ft diameter) and height 12 ft. In the figure, O is the tip, AB is the cut and CD is the 4 ft base radius. Solution figure for question 82, CAT 2018 Slot 1

Solution figure for question 82, CAT 2018 Slot 1

The cut is 9 ft above the base, so the piece removed is a small cone of height

12 - 9 = 3

ft at the tip. By similar triangles its radius is

\frac{3}{12} \times 4 = 1

ft. Remaining volume = full cone minus small cone:

\frac{1}{3}\pi (4^2)(12) - \frac{1}{3}\pi (1^2)(3) = \frac{1}{3}\pi (192 - 3) = \frac{189\pi}{3} = 63\pi

. With

\pi = \frac{22}{7}

, this is

63 \times \frac{22}{7} = 9 \times 22 = 198

cubic ft. The answer is 198.

Q83MCQRatios, Proportions & Partnership

Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

$\frac{1}{5}$
$\frac{6}{19}$
$\frac{1}{4}$
$\frac{7}{33}$

Answer and solution

Answer: (D) $\frac{7}{33}$

Let Raju and Lalitha start with

4x

and

9x

marbles, and let Lalitha give

a

marbles to Raju. Then

\frac{4x + a}{9x - a} = \frac{5}{6} \Rightarrow 24x + 6a = 45x - 5a \Rightarrow 11a = 21x

, so

a = \frac{21x}{11}

. The fraction of her original marbles that she gave away is

\frac{a}{9x} = \frac{21}{11 \times 9} = \frac{21}{99} = \frac{7}{33}

. Check with

x = 11

: they start with 44 and 99; she gives 21, leaving 65 and 78, which is

5 : 6

. Option A (

\frac{1}{5}

) is the closest value, but giving

\frac{1}{5}

of

9x

, i.e.

1.8x

, leaves

5.8x

and

7.2x

, a ratio of

29 : 36

, not

5 : 6

. Hence, option D (

\frac{7}{33}

).

Q84TITASet Theory

Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 52

Every student studies at least one subject, and anyone who studies P also studies H or E, so no one studies P alone. Known regions: 10 study all three, 20 study H and E but not P, and 0 study P only. Solution figure for question 84, CAT 2018 Slot 1

Solution figure for question 84, CAT 2018 Slot 1

The remaining

74 - 10 - 20 = 44

students fall into four regions: only H, H and P but not E, only E, and E and P but not H. Let

h

= (only H) + (H and P but not E) and

e

= (only E) + (E and P but not H). Then

h + e = 44

. The students of H number

h + 20 + 10

and those of E number

e + 20 + 10

. These are equal, so

h = e = 22

. Number studying H

= 22 + 20 + 10 = 52

. The answer is 52.

Q85MCQProfit, Loss & Discount

A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

96
98
86
84

Answer and solution

Answer: (A) 96

Let the wholesaler's cost of peanuts be Rs.

100x

per kg, so walnuts cost Rs.

300x

per kg. The shopkeeper pays 10% more for peanuts (

110x

per kg) and 20% more for walnuts (

360x

per kg). His total cost is

8 \times 110x + 16 \times 360x = 880x + 5760x = 6640x

. After losing 3 kg of peanuts and 5 kg of walnuts he has

5 + 11 = 16

kg, which he sells for

16 \times 166 = 2656

rupees at a 25% profit. So his cost was

\frac{2656}{1.25} = 2124.8

rupees.

6640x = 2124.8 \Rightarrow x = 0.32

, so the wholesaler bought walnuts at

300 \times 0.32 = 96

rupees per kg. Option B (98), the nearest value, would make the shopkeeper's cost

6640 \times \frac{98}{300} \approx 2169

rupees, a profit of only about 22%, not 25%. Hence, option A (96).

Q86MCQSet Theory

If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be

23
26
96
93

Answer and solution

Answer: (D) 93

Let

k

students like both. Then the number who like only burger is

134 - k

. At most all 105 pizza lovers also like burger, so

k \le 105

and only burger is at least

134 - 105 = 29

. The students who like at least one item number

105 + 134 - k = 239 - k

, which cannot exceed 200, so

k \ge 39

and only burger is at most

134 - 39 = 95

. Put simply, only-burger students are among the

200 - 105 = 95

who do not like pizza. So the number who like only burger lies between 29 and 95. Options A (23) and B (26) are below 29. Option C (96) is the closest trap, but it would need 96 students outside the pizza group, and there are only 95. Option D (93) lies in the range. Hence, option D (93).

Q87TITASequences & Series

If

f(x+2) = f(x) + f(x+1)

for all positive integers

x

, and

f(11) = 91, f(15) = 617

, then

f(10)

equals

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 54

Each term is the sum of the two terms before it. Let

f(10) = b

; we know

f(11) = 91

. Then

f(12) = b + 91

and

f(13) = 91 + (b + 91) = b + 182

. Next,

f(14) = (b + 91) + (b + 182) = 2b + 273

, and

f(15) = (b + 182) + (2b + 273) = 3b + 455

. Since

f(15) = 617

, we get

3b + 455 = 617

, so

3b = 162

and

b = 54

. Check: the terms from

f(10)

to

f(15)

are

54, 91, 145, 236, 381, 617

, and each is the sum of the two before it. The answer is 54.

Q88MCQAverages, Mixtures & Alligations

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

27
25
26
28

Answer and solution

Answer: (D) 28

The overall average is 38. Each of the 30 people aged 51 or above is at least

51 - 38 = 13

years above it, so together they are at least

30 \times 13 = 390

years above the average. The people below 51 must be below the average by the same total. If there are

n

of them, their average is at most

38 - \frac{390}{n}

, reached when all 30 older people are exactly 51. This bound grows with

n

, so it is largest at the maximum,

n = 39

:

38 - \frac{390}{39} = 38 - 10 = 28

. Check:

\frac{30 \times 51 + 39 \times 28}{69} = \frac{1530 + 1092}{69} = \frac{2622}{69} = 38

. Option B (25) is the trap of taking only 30 people below 51:

38 - \frac{390}{30} = 25

. Fewer people must share the same shortfall, so their average is lower. Hence, option D (28).

Q89MCQAverages, Mixtures & Alligations

A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

16
26
20
22

Answer and solution

Answer: (C) 20

The mixture sells for Rs. 264 at 10% profit, so it cost

\frac{264}{1.1} = 240

rupees for 10 litres. Let B cost Rs.

x

per litre, so A costs Rs.

(x + 8)

. If the mixture has

q

litres of A and

10 - q

litres of B:

q(x + 8) + (10 - q)x = 240 \Rightarrow 10x + 8q = 240 \Rightarrow x = 24 - 0.8q

. So B's price is highest when the mixture has as little A as possible. B does not exceed A, so

q \ge 5

, and

q = 5

gives

x = 24 - 4 = 20

. Option D (22) would need

q = 2.5

, i.e. 2.5 litres of A and 7.5 litres of B, which breaks the condition that B does not exceed A. Option B (26) is impossible, since the cheaper paint cannot cost more than the average of 24. Hence, option C (20).

Q90MCQPolygons & Circles

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

$32\sqrt{3}$
$18\sqrt{3}$
$24\sqrt{3}$
$12\sqrt{3}$

Answer and solution

Answer: (A) $32\sqrt{3}$

The area of a parallelogram is base times height. With CD as the base,

72 = 9 \times AP

, so the perpendicular from A to line CD is

AP = 8

cm.

Solution figure for question 90, CAT 2018 Slot 1

In the figure, AB = CD = 9 cm and BC = AD = 16 cm. Triangle APD is right-angled at P with hypotenuse

AD = 16

cm, so

DP = \sqrt{AD^2 - AP^2} = \sqrt{256 - 64} = \sqrt{192} = 8\sqrt{3}

cm. Area of triangle APD

= \frac{1}{2} \times AP \times DP = \frac{1}{2} \times 8 \times 8\sqrt{3} = 32\sqrt{3}

sq cm. Since

8\sqrt{3} \approx 13.9

is more than

CD = 9

, the foot P lies on CD extended beyond C, as drawn; the area is unaffected. With

AP

fixed at 8, the area is

4 \times DP

, and

DP

is fixed by

AD = 16

, so no other option is possible. For example, option C (

24\sqrt{3}

) would need

DP = 6\sqrt{3}

, which would make

AD = \sqrt{64 + 108} = \sqrt{172}

, not 16. Hence, option A (

32\sqrt{3}

).

Q91MCQQuadratic & Polynomial Equations

If

U^2 + (U - 2V - 1)^2 = -4V(U + V)

, then what is the value of

U + 3V

?

$0$
$\frac{1}{2}$
$-\frac{1}{4}$
$\frac{1}{4}$

Answer and solution

Answer: (C) $-\frac{1}{4}$

Move everything to one side:

U^2 + (U - 2V - 1)^2 + 4V(U + V) = 0

. Expanding,

(U - 2V - 1)^2 = U^2 + 4V^2 + 1 - 4UV - 2U + 4V

and

4V(U + V) = 4UV + 4V^2

. The

UV

terms cancel, leaving

2U^2 - 2U + 8V^2 + 4V + 1 = 0

. Completing the squares:

2\left(U - \frac{1}{2}\right)^2 + 8\left(V + \frac{1}{4}\right)^2 = 0

, since

2 \times \frac{1}{4} + 8 \times \frac{1}{16} = 1

. A sum of two non-negative terms is zero only if both are zero, so

U = \frac{1}{2}

and

V = -\frac{1}{4}

. Then

U + 3V = \frac{1}{2} - \frac{3}{4} = -\frac{1}{4}

. Option D (

\frac{1}{4}

) is the sign-slip trap: it is the value of

U + V

, not

U + 3V

. Hence, option C (

-\frac{1}{4}

).

Q92MCQPolygons & Circles

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

$(\frac{\pi}{4\sqrt{3}})^{\frac{1}{2}}$
$(\frac{\pi}{6})^{\frac{1}{2}}$
$(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$
$(\frac{\pi}{4})^{\frac{1}{2}}$

Answer and solution

Answer: (C) $(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$

R is a

60^\circ

sector of a circle of radius 1, so its area is

\frac{60}{360} \times \pi \times 1^2 = \frac{\pi}{6}

.

Solution figure for question 92, CAT 2018 Slot 1

Let

OC = OD = x

. Triangle OCD has the

60^\circ

angle at O, so its area is

\frac{1}{2}x^2 \sin 60^\circ = \frac{\sqrt{3}}{4}x^2

. This is half the area of R:

\frac{\sqrt{3}}{4}x^2 = \frac{\pi}{12} \Rightarrow x^2 = \frac{\pi}{12} \times \frac{4}{\sqrt{3}} = \frac{\pi}{3\sqrt{3}}

, so

OC = \left(\frac{\pi}{3\sqrt{3}}\right)^{\frac{1}{2}}

. Option B,

(\frac{\pi}{6})^{\frac{1}{2}}

, is the trap of dropping the

\sin 60^\circ

factor:

\frac{1}{2}x^2 = \frac{\pi}{12}

gives

x^2 = \frac{\pi}{6}

, but that formula needs a right angle at O. Hence, option C (

(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}

).

Q93MCQTime, Speed & Distance

The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is

6
4
3
5

Answer and solution

Answer: (D) 5

Let Partha take

x

hours for the 60 km, so Narayan takes

x - 4

hours. Partha reaches the midpoint, half the distance, at time

\frac{x}{2}

, and this is 2 hours before Narayan reaches B:

\frac{x}{2} + 2 = x - 4 \Rightarrow \frac{x}{2} = 6 \Rightarrow x = 12

. So Partha's speed is

\frac{60}{12} = 5

km/h. Check: Narayan takes 8 hours, and Partha is at the midpoint after 6 hours, 2 hours earlier. Option A (6 km/h) would give Partha 10 hours and Narayan 6 hours; Partha would reach the midpoint at 5 hours, only 1 hour before Narayan reaches B, not 2. Hence, option D (5).

Q94MCQSequences & Series

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

$\frac{3}{6}$
$\frac{1}{6}$
$\frac{5}{2}$
$\frac{3}{2}$

Answer and solution

Answer: (C) $\frac{5}{2}$

Let the terms be

x = a

,

y = ar

and

z = ar^2

, with

a > 0

. Since

5x

,

16y

and

12z

are in AP, twice the middle term equals the sum of the other two:

2 \times 16y = 5x + 12z

.

32ar = 5a + 12ar^2 \Rightarrow 12r^2 - 32r + 5 = 0 \Rightarrow (6r - 1)(2r - 5) = 0

, so

r = \frac{1}{6}

or

r = \frac{5}{2}

. The terms are positive and increasing (

x < y < z

), which needs

r > 1

. With

r = \frac{1}{6}

each term would be smaller than the one before, so option B (

\frac{1}{6}

), the other root, is ruled out. With

r = \frac{5}{2}

the terms increase as required. Hence, option C (

\frac{5}{2}

).

Q95MCQIndices & Surds

Given that

x^{2018}y^{2017} = 1/2

, and

x^{2016}y^{2019} = 8

, then the value of

x^2 + y^3

is

$\frac{31}{4}$
$\frac{35}{4}$
$\frac{37}{4}$
$\frac{33}{4}$

Answer and solution

Answer: (D) $\frac{33}{4}$

Dividing the second equation by the first:

\frac{x^{2016}y^{2019}}{x^{2018}y^{2017}} = \frac{y^2}{x^2} = \frac{8}{1/2} = 16

, so

y = 4x

or

y = -4x

. If

y = 4x

, the first equation gives

4^{2017}x^{4035} = \frac{1}{2}

. Since

4^{2017} = 2^{4034}

,

x^{4035} = \frac{1}{2^{4035}}

, so

x = \frac{1}{2}

and

y = 2

. If

y = -4x

, it gives

-4^{2017}x^{4035} = \frac{1}{2}

, so

x = -\frac{1}{2}

and again

y = 2

. In both cases

x^2 = \frac{1}{4}

and

y^3 = 8

, so

x^2 + y^3 = \frac{1}{4} + 8 = \frac{33}{4}

. Option A (

\frac{31}{4}

) is the value of

y^3 - x^2

, the trap of subtracting instead of adding. Hence, option D (

\frac{33}{4}

).

Q96TITATime & Work

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 10

Let each filling pipe fill

x

units per hour and each draining pipe drain

y

units per hour, and let the tank hold

W

units. 6 filling and 5 draining pipes take 6 hours:

W = 6(6x - 5y)

. 5 filling and 6 draining pipes take 60 hours:

W = 60(5x - 6y)

. Equating:

6x - 5y = 10(5x - 6y) = 50x - 60y

, so

44x = 55y

and

x = \frac{5y}{4}

. Then

W = 6\left(\frac{30y}{4} - 5y\right) = 6 \times \frac{10y}{4} = 15y

. With 2 filling pipes and 1 draining pipe the net rate is

2x - y = \frac{10y}{4} - y = \frac{3y}{2}

, so the time is

\frac{15y}{3y/2} = 10

hours. The answer is 10.

Q97MCQLogarithms

If

\log_2 (5 + \log_3 a) = 3

and

\log_5 (4a + 12 + \log_2 b) = 3

, then

a + b

is equal to

59
40
32
67

Answer and solution

Answer: (A) 59

From the first equation,

5 + \log_3 a = 2^3 = 8

, so

\log_3 a = 3

and

a = 3^3 = 27

. From the second,

4a + 12 + \log_2 b = 5^3 = 125

. Substituting

a = 27

:

108 + 12 + \log_2 b = 125

, so

\log_2 b = 5

and

b = 2^5 = 32

. Check:

\log_2(5 + 3) = \log_2 8 = 3

and

\log_5(108 + 12 + 5) = \log_5 125 = 3

. Therefore

a + b = 27 + 32 = 59

. Option C (32) is the trap of stopping at

b

; the question asks for

a + b

. Hence, option A (59).

Q98TITAAverages, Mixtures & Alligations

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 60

Let the number of tests be

n

and the overall average be

A

, so the total score is

nA

. The first 10 tests total

10 \times 20 = 200

. Leaving them out raises the average by 1:

(n - 10)(A + 1) = nA - 200 \Rightarrow n - 10A - 10 = -200 \Rightarrow 10A - n = 190

. The last 10 tests total

10 \times 30 = 300

. Leaving them out lowers the average by 1:

(n - 10)(A - 1) = nA - 300 \Rightarrow -n - 10A + 10 = -300 \Rightarrow 10A + n = 310

. Adding the two equations:

20A = 500

, so

A = 25

, and then

n = 250 - 190 = 60

. The answer is 60.

Q99TITAProperties of Numbers

The number of integers

x

such that

0.25 \le 2^x \le 200

and

2^x + 2

is perfectly divisible by either 3 or 4, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 5

0.25 \le 2^x \le 200

means

2^{-2} \le 2^x \le 200

. Since

2^7 = 128 \le 200 < 256 = 2^8

, the integers are

x = -2, -1, 0, 1, \dots, 7

. For

x = -2

and

x = -1

,

2^x + 2

is 2.25 or 2.5, not an integer, so it cannot be divisible by 3 or 4. For the rest,

2^x + 2

is:

x = 0

: 3 (divisible by 3);

x = 1

: 4 (by 4);

x = 2

: 6 (by 3);

x = 3

: 10 (neither);

x = 4

: 18 (by 3);

x = 5

: 34 (neither);

x = 6

: 66 (by 3);

x = 7

: 130 (neither). The valid values are

x = 0, 1, 2, 4, 6

, which is 5 values. The answer is 5.

Q100MCQPercentages

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

$N \le 200$
$243 \le N \le 252$
$201 \le N \le 242$
$N \ge 253$

Answer and solution

Answer: (B) $243 \le N \le 252$

The pass mark is

0.45N

. Falling short of it by 68% means the candidate's 36 marks are 68% below the pass mark, that is, 32% of it:

36 = 0.32 \times 0.45N = 0.144N \Rightarrow N = \frac{36}{0.144} = 250

. Check: the pass mark is

0.45 \times 250 = 112.5

, and the shortfall

112.5 - 36 = 76.5

is 68% of 112.5. Since

243 \le 250 \le 252

, option B is correct. Option A (

N \le 200

) is the trap of reading the shortfall as scoring 68% of the pass mark:

0.68 \times 0.45N = 36

gives

N \approx 118

. Hence, option B (

243 \le N \le 252

).

CAT 2018 Slot 1 — QA questions with answers

Quantitative Ability

Q67MCQLogarithms

Q68MCQPolygons & Circles

Q69MCQTime & Work

Q70MCQPolygons & Circles

Q71MCQPolygons & Circles

Q72TITATime, Speed & Distance

Q73TITAFunctions & Graphs

Q74TITAInequalities & Modulus

Q75TITATime, Speed & Distance

Q76MCQAverages, Mixtures & Alligations

Q77MCQSequences & Series

Q78MCQLogarithms

Q79TITASimple & Compound Interest

Q80MCQTime & Work

Q81TITAPermutations & Combinations

Q82TITAMensuration

Q83MCQRatios, Proportions & Partnership

Q84TITASet Theory

Q85MCQProfit, Loss & Discount

Q86MCQSet Theory

Q87TITASequences & Series

Q88MCQAverages, Mixtures & Alligations

Q89MCQAverages, Mixtures & Alligations

Q90MCQPolygons & Circles

Q91MCQQuadratic & Polynomial Equations

Q92MCQPolygons & Circles

Q93MCQTime, Speed & Distance

Q94MCQSequences & Series

Q95MCQIndices & Surds

Q96TITATime & Work

Q97MCQLogarithms

Q98TITAAverages, Mixtures & Alligations

Q99TITAProperties of Numbers

Q100MCQPercentages