CAT 2020 Slot 2 QA Questions with Answers & Solutions

Q51MCQTime, Speed & Distance

The distance from B to C is thrice that from A to B. Two trains travel from A to C via B. The speed of train 2 is double that of train 1 while traveling from A to B and their speeds are interchanged while traveling from B to C. The ratio of the time taken by train 1 to that taken by train 2 in travelling from A to C is

5:7
4:1
1:4
7:5

Answer and solution

Answer: (A) 5:7

Let the distance from A to B be $x$ . Then the distance from B to C is $3x$ .

From A to B:
Speed of Train 1 = $v$
Speed of Train 2 is double that of Train 1 = $2v$
Time taken by Train 1 to travel from A to B = $\frac{x}{v}$
Time taken by Train 2 to travel from A to B = $\frac{x}{2v}$

From B to C:
The speeds are interchanged.
Speed of Train 1 = $2v$
Speed of Train 2 = $v$
Time taken by Train 1 to travel from B to C = $\frac{3x}{2v}$
Time taken by Train 2 to travel from B to C = $\frac{3x}{v}$

Total Time from A to C:
Total time taken by Train 1 = $\frac{x}{v} + \frac{3x}{2v} = \frac{2x + 3x}{2v} = \frac{5x}{2v}$
Total time taken by Train 2 = $\frac{x}{2v} + \frac{3x}{v} = \frac{x + 6x}{2v} = \frac{7x}{2v}$

Ratio of Time Taken:
$\text{Ratio} = \frac{\text{Total time of Train 1}}{\text{Total time of Train 2}} = \frac{\frac{5x}{2v}}{\frac{7x}{2v}} = \frac{5}{7}$

Hence, the required ratio is $5:7$ . Option A is correct.

Q52TITATime & Work

John takes twice as much time as Jack to finish a job. Jack and Jim together take one-third of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

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

Answer and solution

Answer: 4

Let Jack take $t$ days to complete the work alone. Then John takes $2t$ days to complete the work alone.
Work done by Jack in 1 day = $\frac{1}{t}$ .
Work done by John in 1 day = $\frac{1}{2t}$ .

Let Jim take $m$ days to complete the work alone. Work done by Jim in 1 day = $\frac{1}{m}$ .

Jack and Jim together take one-third of the time John takes alone.
Time taken by John alone = $2t$ .
Time taken by Jack and Jim together = $\frac{2t}{3}$ .
Therefore, the combined work done by Jack and Jim in 1 day is $\frac{3}{2t}$ .

$\frac{1}{t} + \frac{1}{m} = \frac{3}{2t}$
$\frac{1}{m} = \frac{3}{2t} - \frac{1}{t} = \frac{3 - 2}{2t} = \frac{1}{2t}$
$\implies m = 2t$

So Jim takes $2t$ days to complete the work alone.

Now, John takes 3 days more than the time taken by all three working together.
Let the three of them complete the work together in $p$ days.
Work done by all three in 1 day = $\frac{1}{t} + \frac{1}{2t} + \frac{1}{2t} = \frac{2 + 1 + 1}{2t} = \frac{4}{2t} = \frac{2}{t}$ .
So, time taken by all three together, $p = \frac{t}{2}$ days.

We are given that John takes 3 days more than all three together:
$2t = p + 3$
$2t = \frac{t}{2} + 3$
$2t - \frac{t}{2} = 3$
$\frac{3t}{2} = 3 \implies t = 2$

Since Jim takes $2t$ days to finish the job alone, the time taken by Jim is $2(2) = 4$ days.

Q53MCQQuadratic & Polynomial Equations

Let

f(x) = x^2+ax+b

and

g(x) = f(x+1) - f(x-1)

. If

f(x) \ge 0

for all real

x

, and

g(20) = 72

, then the smallest possible value of

b

is

16
4
1
0

Answer and solution

Answer: (B) 4

Expanding,

f(x+1)=x^2+2x+1+ax+a+b

and

f(x-1)=x^2-2x+1+ax-a+b

, so

g(x)=f(x+1)-f(x-1)=4x+2a

. Then

g(20)=80+2a=72

gives

a=-4

, so

f(x)=x^2-4x+b

. For

f(x)\ge0

for all real

x

, the discriminant must not be positive:

16-4b\le0

, so

b\ge4

. Equivalently,

f(x)=(x-2)^2+b-4

, whose minimum value

b-4

must be at least

0

. So the smallest possible value of

b

is

4

. Options C (1) and D (0) are below 4 and would make

f(2)

negative, while option A (16) is allowed but not the smallest. Hence, option B (4).

Q54TITASimple & Compound Interest

For the same principal amount, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by Rs 1125. Then the principal amount in rupees is

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

Answer and solution

Answer: 90000

Let the principal amount be $P$ .

The compound interest for 2 years at 5% per annum is:
$CI = P \left(1 + \frac{5}{100}\right)^2 - P = P(1.05)^2 - P = P(1.1025) - P = 0.1025P$

The simple interest for 3 years at 3% per annum is:
$SI = \frac{P \times 3 \times 3}{100} = \frac{9P}{100} = 0.09P$

The difference between the compound interest and the simple interest is Rs 1125.
$CI - SI = 1125$
$0.1025P - 0.09P = 1125$
$0.0125P = 1125$

$P = \frac{1125}{0.0125} = \frac{11250000}{125} = 90000$

The principal amount is Rs 90000.

Q55MCQPolygons & Circles

Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

8.8
7.8
6.6
7.2

Answer and solution

Answer: (A) 8.8

Place

O

at the origin with north along the

y

-axis. Then

A=(0,4)

and

B=(3,0)

, and the chord lies on the line

\frac{x}{3}+\frac{y}{4}=1

, i.e.

4x+3y=12

.

Solution figure for question 55, CAT 2020 Slot 2

Let

C

be the foot of the perpendicular from

O

to the chord. Then

OC=\frac{12}{\sqrt{4^2+3^2}}=\frac{12}{5}=2.4

m. Equivalently, in the right triangle

AOB

with

AB=5

, the altitude to

AB

is

\frac{3\times4}{5}=2.4

. The perpendicular from the centre bisects the chord, so

PC=\sqrt{5^2-2.4^2}=\sqrt{19.24}\approx4.39

m, and

PQ=2\times PC\approx8.77

m. Option B (7.8) is the next closest but is about a metre short. Note that

AB=5

m is only part of the chord, since

A

and

B

lie inside the circle. Hence, option A (8.8).

Q56MCQTime, Speed & Distance

In a car race, car A beats car B by 45 km. car B beats car C by 50 km. and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is

475
450
500
550

Answer and solution

Answer: (B) 450

Let the race be

D

km. When A finishes, B has run

D-45

and C has run

D-90

; when B finishes, C has run

D-50

. So

\frac{S_A}{S_B}=\frac{D}{D-45}

,

\frac{S_B}{S_C}=\frac{D}{D-50}

and

\frac{S_A}{S_C}=\frac{D}{D-90}

. Since

\frac{S_A}{S_C}=\frac{S_A}{S_B}\cdot\frac{S_B}{S_C}

, we get

\frac{D}{D-90}=\frac{D^2}{(D-45)(D-50)}

, so

(D-45)(D-50)=D(D-90)

. This gives

D^2-95D+2250=D^2-90D

, so

5D=2250

and

D=450

. Check: over 450 km, B runs 405 while A runs 450, and C runs

405\times\frac{400}{450}=360

, so A beats C by 90. Option C (500) fails: B would run 455 and C

455\times\frac{450}{500}=409.5

, a gap of 90.5, not 90. Hence, option B (450).

Q57TITAPermutations & Combinations

How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

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

Answer and solution

Answer: 315

We need to form 4-digit numbers greater than 1000 with all four digits distinct such that 7 appears before 3.

First, we select 4 distinct digits from the 10 available digits (0 to 9). Since 7 and 3 must be among the digits, we need to choose 2 more digits from the remaining 8 digits (which include 0). There are two cases depending on whether 0 is selected or not.

Case 1: 0 is not among the selected digits.
We choose 2 digits from the remaining 7 non-zero digits: $\binom{7}{2} = 21$ ways.
We now have 4 non-zero digits (including 7 and 3).
Total permutations of these 4 digits is $4! = 24$ .
By symmetry, in exactly half of these permutations, 7 will appear before 3.
So, $\frac{24}{2} = 12$ valid arrangements per selection.
Total numbers for Case 1 = $21 \times 12 = 252$ .

Case 2: 0 is one of the selected digits.
We need 1 more digit from the remaining 7 non-zero digits: $\binom{7}{1} = 7$ ways.
We now have 4 digits: 7, 3, 0, and one other non-zero digit.
Total permutations of these 4 digits is $4! = 24$ .
However, numbers cannot start with 0. We must subtract the cases where 0 is the first digit.
If 0 is the first digit, the remaining 3 digits can be arranged in $3! = 6$ ways.
Out of these 6 arrangements, 7 appears before 3 in exactly half of them: $\frac{6}{2} = 3$ ways.
Total valid permutations for these 4 digits = (Total where 7 is before 3) - (Those starting with 0 where 7 is before 3)
$= \frac{24}{2} - 3 = 12 - 3 = 9$ valid arrangements.
Total numbers for Case 2 = $7 \times 9 = 63$ .

Total 4-digit numbers = $252 + 63 = 315$ .

Q58MCQMensuration

The sum of the perimeters of an equilateral triangle and a rectangle is 90cm. The area,

T

, of the triangle and the area,

R

, of the rectangle, both in sq cm, satisfying the relationship

R = T^2

. If the sides of the rectangle are in the ratio 1:3, then the length, in cm, of the longer side of the rectangle, is

27
21
24
18

Answer and solution

Answer: (A) 27

Let the triangle's side be

a

and the rectangle's sides be

x

and

3x

. The perimeters give

3a+8x=90

. The areas are

T=\frac{\sqrt3}{4}a^2

and

R=3x^2

. From

R=T^2

:

3x^2=\frac{3}{16}a^4

, so

x^2=\frac{a^4}{16}

and

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

. Substituting:

3a+2a^2=90

, i.e.

2a^2+3a-90=0

, which factorises as

(a-6)(2a+15)=0

. So

a=6

and

x=\frac{36}{4}=9

. The rectangle is 9 cm by 27 cm. Check:

18+72=90

,

T=9\sqrt3

and

T^2=243=9\times27=R

. Option B (21) would give

x=7

and

3a=34

, but then

\frac{a^2}{4}\approx32

, not 7, so

R=T^2

fails. Hence, option A (27).

Q59TITARatios, Proportions & Partnership

A sum of money is split among Amal, Sunil and Mita so that the ratio of the shares of Amal and Sunil is 3:2, while the ratio of the shares of Sunil and Mita is 4:5. If the difference between the largest and the smallest of these three shares is Rs.400, then Sunil’s share, in rupees, is

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

Answer and solution

Answer: 800

The ratio of shares of Amal and Sunil is $3:2$ .
The ratio of shares of Sunil and Mita is $4:5$ .

We can equalize the ratio for Sunil to combine them.
Amal : Sunil = $(3:2) \times 2 = 6:4$
Sunil : Mita = $4:5$

Combining the ratios, we get Amal : Sunil : Mita = $6:4:5$ .
Let their shares be $6x$ , $4x$ , and $5x$ respectively.

The largest share is Amal's ( $6x$ ) and the smallest share is Sunil's ( $4x$ ).
The difference between the largest and smallest shares is Rs 400.
$6x - 4x = 400$
$2x = 400 \implies x = 200$ .

Sunil's share is $4x = 4(200) = 800$ rupees.

Q60MCQLogarithms

The value of

\log_a \frac{a}{b} + \log_b \frac{b}{a}

, for

1 < a \le b

cannot be equal to

0
-1
1
-0.5

Answer and solution

Answer: (C) 1

Expand:

\log_a\frac{a}{b}+\log_b\frac{b}{a}=(1-\log_a b)+(1-\log_b a)=2-\left(x+\frac1x\right)

, where

x=\log_a b

. Since

1<a\le b

,

x=\log_a b\ge1>0

, so by AM-GM

x+\frac1x\ge2

and the expression is at most

0

. It equals

0

when

a=b

, and as

x

grows it takes every negative value:

x=2

gives

2-2.5=-0.5

, and

x+\frac1x=3

gives

-1

. So 0,

-1

and

-0.5

are all possible; option A (0) is attained when

a=b

, so it is not excluded. A positive value such as 1 is impossible. Hence, option C (1).

Q61MCQSet Theory

Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is

22
21
20
19

Answer and solution

Answer: (C) 20

Every student takes at least two subjects, so each student is in exactly one group: Maths and Physics only (

x

), Physics and Chemistry only (

y

), Maths and Chemistry only (

z

), or all three (18). Maths gives

x+z+18=23

, so

x+z=5

. Physics gives

x+y+18=25

, so

x+y=7

. Chemistry is

y+z+18=(7-x)+(5-x)+18=30-2x

. To make this as small as possible, make

x

as large as possible. Since

z=5-x\ge0

,

x\le5

. With

x=5

:

z=0

,

y=2

, and Chemistry is

2+0+18=20

. Option D (19) would need

x=5.5

, which is not a whole number and exceeds

x\le5

; option B (21) would need

x=4.5

, also not a whole number, so it is impossible; option A (22) is possible (with

x=4

,

y=3

,

z=1

) but not the smallest. Hence, option C (20).

Q62MCQAverages, Mixtures & Alligations

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

5
4
3
6

Answer and solution

Answer: (B) 4

Sort the scores

x_1\le x_2\le\dots\le x_{10}

. The lowest nine sum to

9\times42=378

and the highest nine to

9\times47=423

, so

x_{10}-x_1=45

and the total is

378+x_{10}

. Maximum mean: make

x_{10}

, and so

x_1

, as large as possible. As

x_1

is the smallest of nine scores averaging 42,

x_1\le42

. Take

x_1=\dots=x_9=42

and

x_{10}=87

; the mean is

\frac{378+87}{10}=46.5

. Minimum mean: make

x_{10}

as small as possible. As

x_{10}

is the largest of nine scores averaging 47,

x_{10}\ge47

. Take

x_2=\dots=x_{10}=47

and

x_1=2

; the mean is

\frac{378+47}{10}=42.5

. The difference is

46.5-42.5=4

. Option A (5) is just

47-42

, the gap between the two nine-score means, not between possible means of all ten. Hence, option B (4).

Q63MCQTime, Speed & Distance

A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other, Ram and Rahim reach their destination in one minute and four minutes, respectively. If they start at the same time, then the ratio of Ram's speed to Rahim's speed is

$\frac{1}{2}$
$\sqrt{2}$
2
$2\sqrt{2}$

Answer and solution

Answer: (C) 2

Let Ram's and Rahim's speeds be

u

and

v

, and let them meet after

t

minutes. The stretch Ram covers before meeting,

ut

, is what Rahim covers after meeting, in 4 minutes:

ut=4v

. The stretch Rahim covers before meeting,

vt

, is what Ram covers after meeting, in 1 minute:

vt=u

. Dividing,

\frac{u}{v}=\frac{4v}{u}

, so

\left(\frac{u}{v}\right)^2=4

and

\frac{u}{v}=2

. (They meet after

t=2

minutes.) Option A inverts the ratio: Ram is the faster runner, since he needs only 1 minute to finish the remaining distance while Rahim needs 4. Hence, option C (2).

Q64MCQTime, Speed & Distance

Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track

T_1

and track

T_2

at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is

5
3
2
4

Answer and solution

Answer: (B) 3

The two tracks touch only at A, so Ram and Rahim can meet only at A. That happens when each has completed a whole number of rounds. Solution figure for question 64, CAT 2020 Slot 2

Solution figure for question 64, CAT 2020 Slot 2

Ram: track length

2\pi \times 100 = 200\pi

m, speed

15

km/hr

= 250

m/min, so one round takes

\dfrac{200\pi}{250} = \dfrac{4\pi}{5}

min. Rahim: track length

2\pi \times 20 = 40\pi

m, speed

5

km/hr

= \dfrac{250}{3}

m/min, so one round takes

\dfrac{40\pi \times 3}{250} = \dfrac{12\pi}{25}

min. The round times are in the ratio

\dfrac{4\pi}{5} : \dfrac{12\pi}{25} = 20 : 12 = 5 : 3

. So both are first back at A together after

3

of Ram's rounds, which take as long as

5

of Rahim's:

3 \times \dfrac{4\pi}{5} = 5 \times \dfrac{12\pi}{25} = \dfrac{12\pi}{5}

min. Option A (5) is the number of rounds Rahim makes in that time, not Ram. Hence, option B (3).

Q65TITAPolygons & Circles

Let C1 and C2 be concentric circles such that the diameter of C1 is 2cm longer than that of C2. If a chord of C1 has length 6cm and is a tangent to C2, then the diameter, in cm, of C1 is

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

Answer and solution

Answer: 10

Let the radius of the smaller circle

C_2

be

r

. The diameter of

C_1

is 2 cm more than that of

C_2

, so the radius of

C_1

is

r+1

. Let

O

be the common centre and

AB

the 6 cm chord of

C_1

, touching

C_2

at

M

. The radius

OM

is perpendicular to the tangent

AB

, so

M

is the midpoint of

AB

and

AM=3

cm.

Solution figure for question 65, CAT 2020 Slot 2

In the right triangle

OMA

,

OA=r+1

,

OM=r

and

AM=3

. By Pythagoras,

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

, so

r^2+2r+1=r^2+9

, giving

2r=8

and

r=4

cm. The radius of

C_1

is

4+1=5

cm, so its diameter is

10

cm. The answer is 10.

Q66MCQProfit, Loss & Discount

Anil buys 12 toys and labels each with the same selling price. He sells 8 toys initially at 20% discount on the labeled price. Then he sells the remaining 4 toys at an additional 25% discount on the discounted price. Thus, he gets a total of Rs 2112, and makes a 10% profit. With no discounts, his percentage of profit would have been

50
60
54
55

Answer and solution

Answer: (A) 50

Let the labelled price be

M

and the cost be

C

per toy. Eight toys sell at

0.8M

each and four at

0.75\times0.8M=0.6M

each, so the revenue is

6.4M+2.4M=8.8M=2112

, giving

M=240

. A 10% profit means

2112=1.1\times12C

, so

C=\frac{2112}{13.2}=160

. Without discounts, each toy sells for 240, a profit of 80 on a cost of 160, which is

\frac{80}{160}\times100=50

per cent. Check: all 12 toys at 240 bring 2880 against a cost of 1920, a profit of 960, exactly half the cost, so option D (55) is too high. Hence, option A (50).

Q67MCQIndices & Surds

The number of integers that satisfy the equality

(x^2-5x+7)^{x+1} = 1

is

3
2
4
5

Answer and solution

Answer: (A) 3

A power

a^b

with an integer exponent equals 1 in three ways. Exponent zero with a non-zero base:

x+1=0

gives

x=-1

, and the base is

1+5+7=13\ne0

, so

x=-1

works. Base equal to 1:

x^2-5x+7=1

gives

x^2-5x+6=0

, so

x=2

or

x=3

. Base equal to

-1

with an even exponent:

x^2-5x+8=0

has discriminant

25-32=-7<0

, so there is no real

x

. In fact

x^2-5x+7=(x-2.5)^2+0.75>0

always, so the base is never negative. The solutions are

x=-1

,

2

and

3

. Option B (2) misses the zero-exponent case

x=-1

, and option C (4) would wrongly count a base of

-1

. Hence, option A (3).

Q68TITALinear Equations

The number of pairs of integers

(x,y)

satisfying

x \ge y \ge -20

and

2x+5y = 99

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

Answer and solution

Answer: 17

We are given the equation $2x + 5y = 99$ and the inequality $x \ge y \ge -20$ .
We need to find the number of integer pairs $(x, y)$ that satisfy these conditions.

From the equation, we can express $x$ in terms of $y$ :
$2x = 99 - 5y \implies x = \frac{99 - 5y}{2}$ .

Since $x$ must be an integer, $99 - 5y$ must be even.
This means $5y$ must be odd, so $y$ must be an odd integer.

We are given $x \ge y$ :
$\frac{99 - 5y}{2} \ge y$
$99 - 5y \ge 2y$
$99 \ge 7y \implies y \le \frac{99}{7} = 14.14$

Since $y$ is an integer, $y \le 14$ .
We are also given $y \ge -20$ .
So the range for $y$ is $-20 \le y \le 14$ .

Within this range, $y$ must be an odd integer.
The odd integers from $-20$ to $14$ are:
$-19, -17, -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13$ .

There are exactly 17 such odd values for $y$ .
For each odd value of $y$ , there is a corresponding unique integer value for $x$ .

Therefore, there are 17 pairs of integers $(x, y)$ satisfying the conditions.

Q69MCQTriangles & Lines

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is

s

. Then the area of the triangle is

$\frac{\sqrt{3}s^2}{2}$
$\frac{2s^2}{\sqrt{3}}$
$\frac{s^2}{2\sqrt{3}}$
$\frac{s^2}{\sqrt{3}}$

Answer and solution

Answer: (D) $\frac{s^2}{\sqrt{3}}$

Let the side be

a

, and let the perpendiculars from the interior point

G

to the three sides be

h_1, h_2, h_3

, so

h_1+h_2+h_3=s

.

Solution figure for question 69, CAT 2020 Slot 2

Joining

G

to the three vertices splits the triangle into three smaller triangles. Each has base

a

and one of the perpendiculars as its height, so

\text{Area}=\frac{1}{2}ah_1+\frac{1}{2}ah_2+\frac{1}{2}ah_3=\frac{1}{2}as

. The area is also

\frac{\sqrt{3}}{4}a^2

. Equating,

\frac{1}{2}as=\frac{\sqrt{3}}{4}a^2

, so

a=\frac{2s}{\sqrt{3}}

. In other words,

s

equals the altitude of the triangle. The figure puts the point at the centre, but this holds for any interior point. Then

\text{Area}=\frac{1}{2}\cdot\frac{2s}{\sqrt{3}}\cdot s=\frac{s^2}{\sqrt{3}}

. Option B,

\frac{2s^2}{\sqrt{3}}

, is exactly twice this. It comes from dropping the

\frac{1}{2}

in this last step and taking the area as base

\times

altitude. Hence, option D (

\frac{s^2}{\sqrt{3}}

).

Q70MCQSequences & Series

Let the m-th and n-th terms of a geometric progression be

\frac{3}{4}

and

12

, respectively, where

m < n

. If the common ratio of the progression is an integer

r

, then the smallest possible value of

r + n - m

is

6
2
-4
-2

Answer and solution

Answer: (D) -2

Let the first term be

a

. Then

ar^{m-1}=\frac{3}{4}

and

ar^{n-1}=12

. Dividing the second by the first,

r^{n-m}=12\times\frac{4}{3}=16

. Since

m<n

,

k=n-m

is a positive integer, and

r

is an integer with

r^k=16

. The possible pairs are

r=16, k=1

(sum

17

);

r=4, k=2

(sum

6

);

r=-4, k=2

(sum

-2

);

r=2, k=4

(sum

6

); and

r=-2, k=4

(sum

2

).

r=-16

does not work, since

(-16)^1=-16

. So the smallest value of

r+n-m

is

-2

, from

r=-4

and

n-m=2

. Option C,

-4

, cannot be reached. With

k\ge 1

it would need

r\le -5

, but the only negative ratios allowed are

-4

and

-2

. Hence, option D (-2).

Q71MCQPercentages

In May, John bought the same amount of rice and the same amount of wheat as he had bought in April, but spent ₹ 150 more due to price increase of rice and wheat by 20% and 12%, respectively. If John had spent ₹ 450 on rice in April, then how much did he spend on wheat in May?

Rs.560
Rs.570
Rs.590
Rs.580

Answer and solution

Answer: (A) Rs.560

Let John's April spending on wheat be

W

. He bought the same quantities in both months, so each May amount is the April amount raised by that item's price increase. Rice: April

450

, May

1.2\times 450=540

, an increase of

90

. The total increase is

150

, so the increase on wheat is

150-90=60

. Wheat's price rose by 12%, so

0.12W=60

, which gives

W=500

in April. Wheat in May

=1.12\times 500=560

. Option D (Rs.580) fails this check. It would mean April wheat spending of

\frac{580}{1.12}\approx 518

and a wheat increase of about

62

, so the total increase would be about

152

, not

150

. Hence, option A (Rs.560).

Q72TITALinear Equations

If

x

and

y

are non-negative integers such that

x+9=z

,

y+1=z

and

x+y < z+5

, then the maximum possible value of

2x+y

equals

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

Answer and solution

Answer: 23

From

x+9=z

and

y+1=z

, we get

x=z-9

and

y=z-1

. Substitute into

x+y<z+5

:

(z-9)+(z-1)<z+5 \implies 2z-10<z+5 \implies z<15

. Since

x

is an integer,

z

is an integer, so

z\le 14

. Also

x\ge 0

needs

z\ge 9

. Now

2x+y=2(z-9)+(z-1)=3z-19

, which grows as

z

grows. So take the largest value,

z=14

. This gives

x=5

and

y=13

, so

2x+y=10+13=23

. Check:

x+y=18

and

z+5=19

, so the inequality holds. The answer is 23.

Q73MCQLinear Equations

Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

33
27
30
36

Answer and solution

Answer: (A) 33

Let a pencil cost

p

, so a sharpener costs

p+2

. Aron buys

x

pencils and

y

sharpeners. Aditya buys

2x

pencils and

y-10

sharpeners. Equal spending gives

px+(p+2)y=2px+(p+2)(y-10)

. Cancelling

(p+2)y

from both sides,

px=2px-10(p+2) \implies p(x-10)=20

. Since

p>0

,

x-10=\frac{20}{p}>0

, so

x>10

. Pencils are counted in whole numbers, so the least possible

x

is

11

. That needs

p=20

(pencil ₹20, sharpener ₹22), which works for any

y\ge 10

. Together they bought

x+2x=3x

pencils, so the minimum is

3\times 11=33

. Option C (30) would need

x=10

, but then

p(x-10)=0

, not

20

. So Aron must buy more than 10 pencils. Hence, option A (33).

Q74MCQInequalities & Modulus

For real

x

, the maximum possible value of

\frac{x}{\sqrt{1+x^4}}

is

$\frac{1}{2}$
1
$\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{2}}$

Answer and solution

Answer: (D) $\frac{1}{\sqrt{2}}$

Let

f(x)=\frac{x}{\sqrt{1+x^4}}

. For

x\le 0

,

f(x)\le 0

, so the maximum occurs at some

x>0

. For

x>0

, divide the numerator and the denominator by

x=\sqrt{x^2}

:

f(x)=\frac{1}{\sqrt{\frac{1}{x^2}+x^2}}

. So

f(x)

is largest when

\frac{1}{x^2}+x^2

is smallest. By AM-GM,

\frac{1}{x^2}+x^2\ge 2\sqrt{\frac{1}{x^2}\cdot x^2}=2

, with equality at

x=1

. The maximum is therefore

f(1)=\frac{1}{\sqrt{2}}

. Option A,

\frac{1}{2}

, is a value

f

does take, but it is not the largest, since

f(1)\approx 0.707>0.5

. Option B,

1

, would need

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

, which is impossible because that sum is at least

2

. Hence, option D (

\frac{1}{\sqrt{2}}

).

Q75MCQInequalities & Modulus

In how many ways can a pair of integers

(x, a)

be chosen such that

x^2 - 2|x| + |a-2| = 0

?

6
5
4
7

Answer and solution

Answer: (D) 7

Since

x^2=|x|^2

, complete the square in

|x|

:

|x|^2-2|x|+1+|a-2|=1 \implies (|x|-1)^2+|a-2|=1

. Both terms are non-negative integers, so one of them is

1

and the other is

0

. Case 1:

(|x|-1)^2=1

and

|a-2|=0

. Then

a=2

, and

|x|-1=\pm 1

gives

|x|=2

or

|x|=0

, so

x=2,-2,0

. That is 3 pairs:

(2,2)

,

(-2,2)

,

(0,2)

. Case 2:

(|x|-1)^2=0

and

|a-2|=1

. Then

x=\pm 1

and

a=3

or

a=1

. That is

2\times 2=4

pairs:

(1,1)

,

(1,3)

,

(-1,1)

,

(-1,3)

. Total:

3+4=7

pairs. Option A (6) is what you get if you miss

x=0

in Case 1, that is, if you forget that

|x|-1=-1

also satisfies

(|x|-1)^2=1

. Hence, option D (7).

Q76TITAInequalities & Modulus

If

x

and

y

are positive real numbers satisfying

x+y=102

, then the minimum possible value of

2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)

is

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

Answer and solution

Answer: 2704

We are given $x, y > 0$ and $x + y = 102$ .
We need to find the minimum value of $E = 2601 \left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right)$ .

Let's expand the expression in the parentheses:
$\left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right) = 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} = 1 + \frac{x + y}{xy} + \frac{1}{xy}$

Substitute $x + y = 102$ :
$1 + \frac{102}{xy} + \frac{1}{xy} = 1 + \frac{103}{xy}$

To minimize $E = 2601 \left(1 + \frac{103}{xy}\right)$ , we need to maximize $xy$ .
Using the AM-GM inequality for positive numbers $x$ and $y$ :
$\frac{x + y}{2} \ge \sqrt{xy}$
$\frac{102}{2} \ge \sqrt{xy}$
$51 \ge \sqrt{xy} \implies xy \le 51^2$
$51^2 = 2601$ . So the maximum value of $xy$ is $2601$ .

Substitute $xy = 2601$ into the expression for $E$ :
Minimum $E = 2601 \left(1 + \frac{103}{2601}\right) = 2601 + 103 = 2704$ .

The minimum possible value is 2704.

CAT 2020 Slot 2 — QA questions with answers

Quantitative Ability

Q51MCQTime, Speed & Distance

Q52TITATime & Work

Q53MCQQuadratic & Polynomial Equations

Q54TITASimple & Compound Interest

Q55MCQPolygons & Circles

Q56MCQTime, Speed & Distance

Q57TITAPermutations & Combinations

Q58MCQMensuration

Q59TITARatios, Proportions & Partnership

Q60MCQLogarithms

Q61MCQSet Theory

Q62MCQAverages, Mixtures & Alligations

Q63MCQTime, Speed & Distance

Q64MCQTime, Speed & Distance

Q65TITAPolygons & Circles

Q66MCQProfit, Loss & Discount

Q67MCQIndices & Surds

Q68TITALinear Equations

Q69MCQTriangles & Lines

Q70MCQSequences & Series

Q71MCQPercentages

Q72TITALinear Equations

Q73MCQLinear Equations

Q74MCQInequalities & Modulus

Q75MCQInequalities & Modulus

Q76TITAInequalities & Modulus