CAT 2019 Slot 2 QA Questions with Answers & Solutions

Q67MCQAverages, Mixtures & Alligations

The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

3.5
5
4.5
4

Answer and solution

Answer: (C) 4.5

The 30 integers add up to

30 \times 5 = 150

. The other 10 integers exceed 5, so each is at least 6 and together they are at least

10 \times 6 = 60

. The 20 integers that do not exceed 5 therefore add up to at most

150 - 60 = 90

, and their average is at most

\frac{90}{20} = 4.5

. This is achievable: take ten 5s and ten 4s (sum 90) together with ten 6s (sum 60). Option B (5) would need all 20 to equal 5, a sum of 100, which leaves only 50 for ten integers that must each be at least 6. So 5 is impossible. Hence, option C (4.5).

Q68TITASimple & Compound Interest

Amal invests Rs 12000 at 8% interest, compounded annually, and Rs 10000 at 6% interest, compounded semi-annually, both investments being for one year. Bimal invests his money at 7.5% simple interest for one year. If Amal and Bimal get the same amount of interest, then the amount, in Rupees, invested by Bimal is

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

Answer and solution

Answer: 20920

Let's calculate the amount with Amal at the end of 1 year.
Amal invests Rs 12,000 at 8% compounded annually:
Amount $_1 = 12000 \times (1 + \frac{8}{100}) = 12000 \times 1.08 = 12960$ .

Amal also invests Rs 10,000 at 6% compounded semi-annually (so the rate is 3% per half-year):
Amount $_2 = 10000 \times (1 + \frac{3}{100})^2 = 10000 \times 1.03^2 = 10000 \times 1.0609 = 10609$ .

Total amount Amal receives = $12960 + 10609 = 23569$ .
Total principal invested by Amal = $12000 + 10000 = 22000$ .
Total interest received by Amal = $23569 - 22000 = 1569$ .

Bimal invests an amount $P$ at 7.5% simple interest for one year.
Interest received by Bimal = $\frac{P \times 7.5 \times 1}{100} = 0.075P$ .

We are given that Amal and Bimal get the same amount of interest:
$0.075P = 1569$
$P = \frac{1569}{0.075} = \frac{1569 \times 1000}{75} = \frac{1569000}{75} = 20920$ .

The amount invested by Bimal is Rs 20,920.

Q69MCQProperties of Numbers

What is the largest positive integer

n

such that

\frac{n^2+7n+12}{n^2-n-12}

is also a positive integer?

6
16
8
12

Answer and solution

Answer: (D) 12

Factorise:

n^2 + 7n + 12 = (n + 3)(n + 4)

and

n^2 - n - 12 = (n - 4)(n + 3)

. For a positive integer

n

,

n + 3 \neq 0

, so the expression equals

\frac{n + 4}{n - 4} = 1 + \frac{8}{n - 4}

. This is an integer only when

n - 4

divides 8, so

n - 4 \le 8

and

n \le 12

. At

n = 12

the value is

1 + \frac{8}{8} = 2

, a positive integer. Option B (16) is the largest choice, but it fails:

1 + \frac{8}{12} = \frac{5}{3}

is not an integer. Hence, option D (12).

Q70TITAProperties of Numbers

How many pairs

(m, n)

of positive integers satisfy the equation

m^2 + 105 = n^2

?

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

Answer and solution

Answer: 4

We are given the equation $m^2 + 105 = n^2$ .
Rearranging the terms, we get:
$n^2 - m^2 = 105$
$(n - m)(n + m) = 105$

Since $m$ and $n$ are positive integers, both $(n - m)$ and $(n + m)$ must be integers.
Also, $n + m > n - m$ , and since their product is positive and $n + m > 0$ , $(n - m)$ must also be positive.
Both factors must have the same parity (either both odd or both even) because their sum is $(n - m) + (n + m) = 2n$ , which is even. Since their product is 105 (odd), both factors must be odd.

Let's find the factor pairs of 105:
$105 = 1 \times 105$
$105 = 3 \times 35$
$105 = 5 \times 21$
$105 = 7 \times 15$

Each pair gives a valid system of equations:
1) $n - m = 1$ and $n + m = 105 \implies 2n = 106 \implies n = 53, m = 52$ .
2) $n - m = 3$ and $n + m = 35 \implies 2n = 38 \implies n = 19, m = 16$ .
3) $n - m = 5$ and $n + m = 21 \implies 2n = 26 \implies n = 13, m = 8$ .
4) $n - m = 7$ and $n + m = 15 \implies 2n = 22 \implies n = 11, m = 4$ .

All 4 pairs result in positive integers for $m$ and $n$ .
Thus, there are 4 such pairs.

Q71MCQTime, Speed & Distance

Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at

10:25 am
10:45 am
10:18 am
10:27 am

Answer and solution

Answer: (D) 10:27 am

Moving in opposite directions, A and B together cover one full lap by their first meeting. A has covered 60%, so B has covered 40%. After the meeting, A covers its remaining 40% in 12 minutes (10:00 to 10:12). So A took

\frac{60}{40} \times 12 = 18

minutes for its first 60%, and both started at 9:42. B covered 40% in those 18 minutes, so its remaining 60% takes

\frac{60}{40} \times 18 = 27

minutes. B returns to P at 10:27 am. Option B (10:45 am) adds B's full-lap time of 45 minutes to 10:00, forgetting that B had already covered 40% of the lap by then. Hence, option D (10:27 am).

Q72MCQSequences & Series

Let

a_1, a_2, \ldots

be integers such that

a_1 - a_2 + a_3 - a_4 + \ldots + (-1)^{n-1} a_n = n

, for all

n \ge 1

. Then

a_{51} + a_{52} + \ldots + a_{1023}

equals

0
1
10
-1

Answer and solution

Answer: (B) 1

Let

S_n = a_1 - a_2 + \dots + (-1)^{n-1}a_n = n

. Then

a_1 = S_1 = 1

, and for

n \ge 2

,

S_n - S_{n-1} = 1

gives

(-1)^{n-1}a_n = 1

. So

a_n = 1

when

n

is odd and

a_n = -1

when

n

is even. The sum

a_{51} + a_{52} + \dots + a_{1023}

has

1023 - 51 + 1 = 973

terms. Pair them from the start:

(a_{51} + a_{52}) + \dots + (a_{1021} + a_{1022})

is 486 pairs, each equal to

1 + (-1) = 0

. The unpaired last term is

a_{1023} = 1

, since 1023 is odd. So the sum is

0 + 1 = 1

. Option A (0) would be right only if the number of terms were even; with 973 terms, one odd-indexed term is left over. Hence, option B (1).

Q73MCQPolygons & Circles

Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

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

Answer and solution

Answer: (D) 1

Let the two circles of radius 4 cm have centres A and B, and let the third circle have centre C and radius

r

. All three touch the common tangent below them. Solution figure for question 73, CAT 2019 Slot 2

The big circles touch, so

AB = 8

cm, and both centres are 4 cm above the tangent. The third circle touches both equal circles, so by symmetry C lies directly below E, the midpoint of AB, and

AE = 4

cm. C is

r

above the tangent, so

EC = 4 - r

. The small circle touches the big one externally, so

AC = 4 + r

.

Triangle AEC has a right angle at E, so

(4 + r)^2 = 4^2 + (4 - r)^2

. Expanding,

16 + 8r + r^2 = 32 - 8r + r^2

, so

16r = 16

and

r = 1

cm. Option C (

\sqrt{2}

) fails: with

r = \sqrt{2}

,

AC^2 = (4 + \sqrt{2})^2 \approx 29.3

, but

AE^2 + EC^2 = 16 + (4 - \sqrt{2})^2 \approx 22.7

. Hence, option D (1).

Q74MCQQuadratic & Polynomial Equations

Let

A

be a real number. Then the roots of the equation

x^2 - 4x - \log_2 A = 0

are real and distinct if and only if

$A > \frac{1}{16}$
$A < \frac{1}{16}$
$A < \frac{1}{8}$
$A > \frac{1}{8}$

Answer and solution

Answer: (A) $A > \frac{1}{16}$

The roots of

x^2 - 4x - \log_2 A = 0

are real and distinct exactly when the discriminant is positive. Here

a = 1

,

b = -4

and

c = -\log_2 A

, so

(-4)^2 - 4(1)(-\log_2 A) > 0 \iff 16 + 4\log_2 A > 0 \iff \log_2 A > -4

. Since the base 2 is greater than 1, this gives

A > 2^{-4} = \frac{1}{16}

. This also keeps

A

positive, as the logarithm requires. Option B (

A < \frac{1}{16}

) reverses the inequality. For example,

A = \frac{1}{32}

gives

\log_2 A = -5

and a discriminant of

16 - 20 = -4 < 0

, so there are no real roots. Hence, option A (

A > \frac{1}{16}

).

Q75MCQQuadratic & Polynomial Equations

The quadratic equation

x^2 + bx + c = 0

has two roots

4a

and

3a

, where

a

is an integer. Which of the following is a possible value of

b^2 + c

?

3721
361
427
549

Answer and solution

Answer: (D) 549

The roots are

4a

and

3a

, so

b = -(4a + 3a) = -7a

and

c = 4a \times 3a = 12a^2

. Then

b^2 + c = 49a^2 + 12a^2 = 61a^2

. With

a

an integer,

b^2 + c

must be

61

times a perfect square. Check each option:

3721 = 61 \times 61

would need

a^2 = 61

, which is not a perfect square. This is the strongest trap, as it is a multiple of

61

.

361

is not a multiple of

61

(

61 \times 5 = 305

and

61 \times 6 = 366

).

427 = 61 \times 7

would need

a^2 = 7

, which is not a perfect square.

549 = 61 \times 9

gives

a^2 = 9

, so

a = \pm 3

works: the roots are

12

and

9

(or

-12

and

-9

), with

b^2 + c = 441 + 108 = 549

. Hence, option D (549).

Q76MCQMensuration

The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is

12
10√2
8√3
5√5

Answer and solution

Answer: (B) 10√2

Each face is an equilateral triangle of side 20 cm, so the square base has side 20 cm. In the figure, A is the apex, O is the centre of the square base and B is the midpoint of a base edge. The apex lies directly above O, so AO is the vertical height

h

.

Solution figure for question 76, CAT 2019 Slot 2

AB is the height of an equilateral face:

AB = \frac{\sqrt{3}}{2} \times 20 = 10\sqrt{3}

cm. OB is half the side of the square:

OB = 10

cm. Triangle AOB has a right angle at O, so

h^2 = AB^2 - OB^2 = 300 - 100 = 200

, giving

h = 10\sqrt{2}

cm. Option C (

8\sqrt{3}

) is close in size, but

(8\sqrt{3})^2 = 192

, not 200. Hence, option B (10√2).

Q77MCQTriangles & Lines

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

10
5
8√2
6√2

Answer and solution

Answer: (A) 10

Let the right angle be at A. In the figure,

BP = x

, so

PC = 20 - x

, and

AP = p

.

Solution figure for question 77, CAT 2019 Slot 2

Triangles ABP and CAP are similar, since each has a right angle at P and

\angle ABP = \angle CAP = 90^\circ - \angle C

. So

\frac{BP}{AP} = \frac{AP}{PC}

, which gives

p^2 = x(20 - x)

. Now

x(20 - x) = 100 - (x - 10)^2

, which is largest when

x = 10

. Then

p^2 = 100

and

p = 10

cm. This is the isosceles right triangle with

AB = AC

. Option C (

8\sqrt{2} \approx 11.3

) is impossible: A lies on the circle with diameter BC, whose radius is 10, so A is never more than 10 cm from BC. Hence, option A (10).

Q78MCQLogarithms

If

x

is a real number, then

\sqrt{\log_e \left(\frac{4x - x^2}{3}\right)}

is a real number if and only if

$1 \le x \le 3$
$1 \le x \le 2$
$-1 \le x \le 3$
$-3 \le x \le 3$

Answer and solution

Answer: (A) $1 \le x \le 3$

The square root is real only when

\log_e\left(\frac{4x - x^2}{3}\right) \ge 0

, that is, when

\frac{4x - x^2}{3} \ge 1

. This also makes the logarithm's argument positive. So

4x - x^2 \ge 3

, which is

x^2 - 4x + 3 \le 0

, or

(x - 1)(x - 3) \le 0

. This holds exactly when

1 \le x \le 3

. Option B (

1 \le x \le 2

) contains only values that work, but it is not 'if and only if':

x = 3

gives

\frac{12 - 9}{3} = 1

and

\sqrt{\log_e 1} = 0

, which is real, yet 3 lies outside that range. Hence, option A (

1 \le x \le 3

).

Q79TITAIndices & Surds

If

5^x - 3^y = 13438

and

5^{x-1} + 3^{y+1} = 9686

, then

x + y

equals

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

Answer and solution

Answer: 13

We are given two equations:
1) $5^x - 3^y = 13438$
2) $5^{x-1} + 3^{y+1} = 9686$

Let $A = 5^x$ and $B = 3^y$ . The equations can be written as:
1) $A - B = 13438 \implies A = 13438 + B$
2) $\frac{A}{5} + 3B = 9686$

Multiply equation 2 by 5 to eliminate the fraction:
$A + 15B = 48430$

Substitute $A = 13438 + B$ into the modified equation 2:
$(13438 + B) + 15B = 48430$
$13438 + 16B = 48430$
$16B = 48430 - 13438$
$16B = 35000 - 8 = 34992$
$B = \frac{34992}{16} = 2187$

Since $B = 3^y$ , we have:
$3^y = 2187$
$3^7 = 2187 \implies y = 7$ .

Now find $A$ :
$A = 13438 + B = 13438 + 2187 = 15625$
Since $A = 5^x$ , we have:
$5^x = 15625$
$5^6 = 15625 \implies x = 6$ .

We need to find $x + y$ :
$x + y = 6 + 7 = 13$ .

Q80MCQAverages, Mixtures & Alligations

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is

15
13
12
14

Answer and solution

Answer: (D) 14

A p% solution has p grams of salt in every 100 ml. Initially A has

0.10 \times 500 = 50

g, B has

0.22 \times 500 = 110

g and C has

0.32 \times 500 = 160

g of salt. A to B: 100 ml is

\frac{1}{5}

of A, so it carries 10 g. A is left with 40 g in 400 ml, and B has 120 g in 600 ml. B to C: 100 ml is

\frac{1}{6}

of B, so it carries 20 g. B keeps 100 g in 500 ml, and C has 180 g in 600 ml. C to A: 100 ml is

\frac{1}{6}

of C, so it carries 30 g. C keeps 150 g in 500 ml, and A has

40 + 30 = 70

g in 500 ml. The table tracks the volume, salt and water in each vessel after every transfer. Solution figure for question 80, CAT 2019 Slot 2

Strength of A

= \frac{70}{500} \times 100 = 14\%

. Option A (15) would need 75 g in A's 500 ml, but A keeps only 40 g after the first transfer and gains exactly 30 g from C. Hence, option D (14).

Q81MCQTime, Speed & Distance

A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?

22
23
15
20

Answer and solution

Answer: (C) 15

The cyclist takes 60 minutes. Motorcycles leave A at 10:01, 10:02 and so on, so the 45th leaves at 10:45 and the 46th at 10:46. Exactly 45 reach B by 11:00, so a motorcycle's travel time

t

satisfies

14 < t \le 15

minutes. At double speed the cyclist takes 30 minutes and reaches B at 10:30. A motorcycle leaving

m

minutes after 10:00 reaches B

m + t

minutes after 10:00, so it arrives by 10:30 when

m + t \le 30

. For

m = 15

,

15 + t \le 30

, so it arrives; for

m = 16

,

16 + t > 30

, so it does not. The motorcycles leaving from 10:01 to 10:15 arrive: 15 of them. Options A (22) and B (23) come from halving 45, but the count is not proportional to the cyclist's time: each motorcycle needs about 15 minutes, so only those leaving in the first 15 minutes arrive by 10:30. Hence, option C (15).

Q82MCQMensuration

A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is

1026(1+π)
8464π
928π
1044(4+π)

Answer and solution

Answer: (A) 1026(1+π)

All cylinders have the same volume

V

and every metal is used up, so

405 \div V

,

783 \div V

and

351 \div V

must be whole numbers. The fewest cylinders come from the largest such

V

, the HCF.

405 = 3^4 \times 5

,

783 = 3^3 \times 29

and

351 = 3^3 \times 13

, so

V = 3^3 = 27

cc. Number of cylinders

= 15 + 29 + 13 = 57

. With radius 3 cm:

\pi \times 3^2 \times h = 27

, so

h = \frac{3}{\pi}

cm. Surface area of one cylinder

= 2\pi r^2 + 2\pi r h = 18\pi + 2\pi \times 3 \times \frac{3}{\pi} = 18\pi + 18 = 18(1 + \pi)

sq cm. Total

= 57 \times 18(1 + \pi) = 1026(1 + \pi)

sq cm. Options B and C are pure multiples of

\pi

, but each curved surface,

2\pi r h = 18

, adds a part without

\pi

. Option D's

\pi

-part,

1044\pi

, would need 58 cylinders, not 57. Hence, option A (1026(1+π)).

Q83MCQIndices & Surds

The real root of the equation

2^{6x} + 2^{3x+2} - 21 = 0

is

$\log_2 9$
$\frac{\log_2 3}{3}$
$\log_2 27$
$\frac{\log_2 7}{3}$

Answer and solution

Answer: (B) $\frac{\log_2 3}{3}$

Write

2^{6x} = (2^{3x})^2

and

2^{3x+2} = 4 \cdot 2^{3x}

, and let

v = 2^{3x}

. The equation becomes

v^2 + 4v - 21 = 0

, which factorises as

(v + 7)(v - 3) = 0

. Since

2^{3x}

is always positive,

v = -7

is rejected and

v = 3

. Then

2^{3x} = 3

, so

3x = \log_2 3

and

x = \frac{\log_2 3}{3}

. Option D (

\frac{\log_2 7}{3}

) would need

v = 7

, but

7^2 + 4 \times 7 - 21 = 56 \neq 0

; the 7 comes from the rejected root

-7

with its sign dropped. Hence, option B (

\frac{\log_2 3}{3}

).

Q84TITAProperties of Numbers

How many factors of

2^4 \times 3^5 \times 10^4

are perfect squares which are greater than 1?

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

Answer and solution

Answer: 44

Let the given number be $N = 2^4 \times 3^5 \times 10^4$ .
First, find the prime factorization of $N$ :
$N = 2^4 \times 3^5 \times (2 \times 5)^4 = 2^4 \times 3^5 \times 2^4 \times 5^4 = 2^8 \times 3^5 \times 5^4$ .

Any factor of $N$ will be of the form $2^a \times 3^b \times 5^c$ , where $0 \le a \le 8$ , $0 \le b \le 5$ , and $0 \le c \le 4$ .
For the factor to be a perfect square, all the exponents $a, b, c$ must be even numbers.

Possible values for $a$ (even numbers from 0 to 8): 0, 2, 4, 6, 8 (5 options).
Possible values for $b$ (even numbers from 0 to 5): 0, 2, 4 (3 options).
Possible values for $c$ (even numbers from 0 to 4): 0, 2, 4 (3 options).

Total number of perfect square factors = $5 \times 3 \times 3 = 45$ .

The question asks for the number of perfect squares which are strictly greater than 1.
The value 1 corresponds to $a=0, b=0, c=0$ , which is included in the 45 factors.
So we must subtract 1 from the total.

Number of perfect squares greater than 1 = $45 - 1 = 44$ .

Q85TITADigits & Base Systems

In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

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

Answer and solution

Answer: 7

Let the six-digit number be $ABCDEF$ , where $A$ is the first digit (leftmost) and $F$ is the sixth digit (rightmost).
We are given the following relationships:
$F = A + B + C$
$E = A + B$
$C = A$
$B = 2A$
$D = E + F$

Let's express all digits in terms of $A$ :
$C = A$
$B = 2A$
$E = A + 2A = 3A$
$F = A + 2A + A = 4A$
$D = 3A + 4A = 7A$

Since $ABCDEF$ is a six-digit number, the leading digit $A$ cannot be 0, so $A \ge 1$ .
Also, every letter represents a single digit, so they must be between 0 and 9.
In particular, $D$ must be a single digit, so $D = 7A \le 9$ .
Since $A \ge 1$ and $A$ is an integer, the only possible value for $A$ is 1.
If $A = 2$ , then $D = 14$ , which is not a single digit.

So, $A = 1$ .
Then $D = 7 \times 1 = 7$ .

The largest possible value of the fourth digit ( $D$ ) is 7.

Q86TITATime, Speed & Distance

John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?

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

Answer and solution

Answer: 48

Let the length of track A be $L_A$ and the length of track B be $L_B$ .
Total length $L_A + L_B = 325$ m.
Speed of John = $6$ kmph.
Speed of Mary = $7.5$ kmph.

While John makes 9 rounds of track A, Mary makes 5 rounds of track B in the same amount of time.
Time taken by John = $\frac{\text{Distance}}{\text{Speed}} = \frac{9L_A}{6}$ .
Time taken by Mary = $\frac{5L_B}{7.5}$ .

Equating the times:
$\frac{9L_A}{6} = \frac{5L_B}{7.5}$
$1.5 L_A = \frac{2}{3} L_B$
$\frac{3}{2} L_A = \frac{2}{3} L_B \implies L_B = \frac{9}{4} L_A = 2.25 L_A$

Substitute $L_B$ into the total length equation:
$L_A + 2.25 L_A = 325$
$3.25 L_A = 325$
$L_A = 100$ m.

We need to find the time it takes Mary to make one round of track A (100 m).
Speed of Mary = $7.5$ kmph = $7.5 \times \frac{5}{18}$ m/s = $\frac{37.5}{18}$ m/s = $\frac{75}{36}$ m/s = $\frac{25}{12}$ m/s.

Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{100}{25/12} = \frac{100 \times 12}{25} = 4 \times 12 = 48$ seconds.

Q87MCQPercentages

In 2010, a library contained a total of 11500 books in two categories - fiction and non-fiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

6160
6600
6000
5500

Answer and solution

Answer: (B) 6600

Let the 2010 fiction count be

F

; non-fiction is then

11500 - F

. After the increases,

1.1F + 1.12(11500 - F) = 12760

, so

12880 - 0.02F = 12760

, giving

0.02F = 120

and

F = 6000

. That is the 2010 figure. In 2015, fiction

= 1.1 \times 6000 = 6600

. As a check, non-fiction

= 1.12 \times 5500 = 6160

, and

6600 + 6160 = 12760

. Option C (6000) is the 2010 fiction count, and option A (6160) is the 2015 non-fiction count; neither answers the question. Hence, option B (6600).

Q88TITALinear Equations

John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

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

Answer and solution

Answer: 12

Let John work for $R$ hours of regular work and $O$ hours of overtime work.
Total hours = $R + O = 172$ .
His income from regular work is $57R$ .
His income from overtime work is $114O$ .

We are given that his income from overtime is 15% of his income from regular hours:
$114O = 0.15 \times 57R$

Notice that $114 = 2 \times 57$ , so we can divide both sides by 57:
$2O = 0.15R$
$R = \frac{2O}{0.15} = \frac{200O}{15} = \frac{40O}{3}$

Substitute $R$ into the total hours equation:
$\frac{40O}{3} + O = 172$
$\frac{40O + 3O}{3} = 172$
$\frac{43O}{3} = 172$
$43O = 516$
$O = \frac{516}{43} = 12$

John worked 12 hours overtime.

Q89TITASequences & Series

If

(2n+1) + (2n+3) + (2n+5) + \ldots + (2n+47) = 5280

, then what is the value of

1 + 2 + 3 + \ldots + n

?

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

Answer and solution

Answer: 4851

The given series is $(2n+1) + (2n+3) + (2n+5) + \dots + (2n+47) = 5280$ .
This is an arithmetic progression where the first term $a = 2n+1$ and the common difference $d = 2$ .
The last term is $l = 2n+47$ .

Let the number of terms be $k$ .
$l = a + (k-1)d$
$2n+47 = 2n+1 + (k-1)2$
$46 = (k-1)2$
$k-1 = 23 \implies k = 24$ .

The sum of an arithmetic progression is given by $S = \frac{k}{2} (a + l)$ .
$5280 = \frac{24}{2} (2n+1 + 2n+47)$
$5280 = 12(4n + 48)$
$\frac{5280}{12} = 4n + 48$
$440 = 4n + 48$
$4n = 440 - 48 = 392$
$n = \frac{392}{4} = 98$ .

We need to find the value of $1 + 2 + 3 + \dots + n$ .
This is the sum of the first $n$ natural numbers, given by $\frac{n(n+1)}{2}$ .
Sum = $\frac{98(98+1)}{2} = 49 \times 99$
$49 \times 99 = 49(100 - 1) = 4900 - 49 = 4851$ .

Q90MCQProfit, Loss & Discount

A shopkeeper sells two tables, each procured at cost price p, to Amal and Asim at a profit of 20% and at a loss of 20%, respectively. Amal sells his table to Bimal at a profit of 30%, while Asim sells his table to Barun at a loss of 30%. If the amounts paid by Bimal and Barun are x and y, respectively, then (x − y) / p equals

1
1.2
0.50
0.7

Answer and solution

Answer: (A) 1

The shopkeeper sells to Amal at

1.2p

(20% profit) and to Asim at

0.8p

(20% loss). Amal sells at a 30% profit, so Bimal pays

x = 1.3 \times 1.2p = 1.56p

. Asim sells at a 30% loss, so Barun pays

y = 0.7 \times 0.8p = 0.56p

. So

\frac{x - y}{p} = \frac{1.56p - 0.56p}{p} = 1

. Options B (1.2) and D (0.7) are single price factors from the working, Amal's purchase multiplier and Asim's 30% loss multiplier, not the ratio asked for. Hence, option A (1).

Q91MCQTriangles & Lines

In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

78
80
72
68

Answer and solution

Answer: (C) 72

The medians meet at the centroid G, which divides each median in the ratio

2 : 1

. So

AG = \frac{2}{3} \times 12 = 8

cm,

GD = 4

cm,

BG = \frac{2}{3} \times 9 = 6

cm and

GE = 3

cm.

Solution figure for question 91, CAT 2019 Slot 2

The medians are perpendicular, so

\triangle AGB

has a right angle at G and area

\frac{1}{2} \times 8 \times 6 = 24

sq cm.

\triangle ABD

is half of

\triangle ABC

, since D is the midpoint of BC.

\triangle AGB

is

\frac{2}{3}

of

\triangle ABD

, since

AG = \frac{2}{3} AD

and both triangles have the same height from B. So

\triangle AGB

is

\frac{1}{3}

of

\triangle ABC

. Area of

\triangle ABC = 3 \times 24 = 72

sq cm. Option B (80) would need

\triangle AGB

to have area

\frac{80}{3}

, but it is exactly

\frac{1}{2} \times 8 \times 6 = 24

. Hence, option C (72).

Q92MCQSequences & Series

The number of common terms in the two sequences: 15, 19, 23, 27, . . . . , 415 and 14, 19, 24, 29, . . . , 464 is

21
20
18
19

Answer and solution

Answer: (B) 20

The first sequence rises in steps of 4 from 15, and the second in steps of 5 from 14. Their first common term is 19, and after that common terms recur every

\text{LCM}(4, 5) = 20

: 19, 39, 59, and so on. A common term cannot exceed 415, where the first sequence ends. So

19 + 20(k - 1) \le 415

, giving

k - 1 \le 19.8

and

k \le 20

. The last common term is

19 + 20 \times 19 = 399

, which also lies within the second sequence. Option A (21) would add 419, which is beyond 415 and so not in the first sequence. Hence, option B (20).

Q93MCQInequalities & Modulus

Let

a, b, x, y

be real numbers such that

a^2 + b^2 = 25

,

x^2 + y^2 = 169

, and

ax + by = 65

. If

k = ay - bx

, then

$0 < k \le \frac{5}{13}$
$k > \frac{5}{13}$
$k = \frac{5}{13}$
$k = 0$

Answer and solution

Answer: (D) $k = 0$

We are given the following equations:
$a^2 + b^2 = 25$
$x^2 + y^2 = 169$
$ax + by = 65$
$k = ay - bx$

We can use the algebraic identity (Brahmagupta-Fibonacci identity) relating the products of sums of squares:
$(a^2 + b^2)(x^2 + y^2) = (ax + by)^2 + (ay - bx)^2$

Substitute the given values into the identity:
$(25)(169) = (65)^2 + k^2$
$4225 = 4225 + k^2$
$k^2 = 0 \implies k = 0$

Hence, Option D is the correct answer.

Q94MCQProfit, Loss & Discount

Mukesh purchased 10 bicycles in 2017, all at the same price. He sold six of these at a profit of 25% and the remaining four at a loss of 25%. If he made a total profit of Rs. 2000, then his purchase price of a bicycle, in Rupees, was

6000
8000
4000
2000

Answer and solution

Answer: (C) 4000

Let the purchase price (CP) of one bicycle be $P$ .
Total cost for 10 bicycles = $10P$ .

Mukesh sold 6 bicycles at a profit of 25%:
Selling price of these 6 bicycles = $6 \times 1.25P = 7.5P$ .

He sold the remaining 4 bicycles at a loss of 25%:
Selling price of these 4 bicycles = $4 \times 0.75P = 3.0P$ .

Total selling price of all 10 bicycles = $7.5P + 3.0P = 10.5P$ .

Total profit = Total Selling Price - Total Cost Price
$2000 = 10.5P - 10P$
$2000 = 0.5P$
$P = \frac{2000}{0.5} = 4000$

The purchase price of a bicycle was Rs 4000.

Hence, Option C is correct.

Q95TITAPercentages

In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was

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

Answer and solution

Answer: 80

Let the score of D be $d$ .
The score of C is 20% less than that of D:
$C = d - 0.20d = 0.8d$ .

The score of B is 25% more than that of C:
$B = C + 0.25C = 1.25C$
$B = 1.25(0.8d) = 1.0d = d$ .

The score of A is 10% less than that of B:
$A = B - 0.10B = 0.9B$
$A = 0.9d$ .

We are given that A scored 72:
$0.9d = 72$
$d = \frac{72}{0.9} = \frac{720}{9} = 80$ .

The score of D was 80.

Q96MCQRatios, Proportions & Partnership

The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Ramesh’s salary increased by 25% during 2010- 2015, then the percentage increase in Rajesh’s salary during this period is closest to

10
7
9
8

Answer and solution

Answer: (B) 7

Let the salaries of Ramesh, Ganesh, and Rajesh in 2010 be $6x$ , $5x$ , and $7x$ respectively.
Let their salaries in 2015 be $3y$ , $4y$ , and $3y$ respectively.

We are given that Ramesh's salary increased by 25% during 2010-2015.
So, Ramesh's 2015 salary = 125% of his 2010 salary.
$3y = 1.25(6x)$
$3y = 7.5x$
$y = 2.5x$

Now, let's look at Rajesh's salary.
Rajesh's salary in 2010 = $7x$ .
Rajesh's salary in 2015 = $3y$ .
Substitute $y = 2.5x$ into Rajesh's 2015 salary:
$3(2.5x) = 7.5x$ .

The percentage increase in Rajesh's salary is:
$\frac{\text{Increase}}{\text{Initial Salary}} \times 100$
$= \frac{7.5x - 7x}{7x} \times 100 = \frac{0.5x}{7x} \times 100 = \frac{50}{7}\%$
$\frac{50}{7} \approx 7.14\%$ .

The closest integer percentage is 7%.

Hence, Option B is correct.

Q97TITAPolygons & Circles

Let A and B be two regular polygons having

a

and

b

sides, respectively. If

b = 2a

and each interior angle of B is

3/2

times each interior angle of A, then each interior angle, in degrees, of a regular polygon with

a + b

sides is

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

Answer and solution

Answer: 150

The formula for each interior angle of a regular polygon with $n$ sides is $\frac{(n-2) \times 180^\circ}{n}$ .

For polygon A (sides = $a$ ):
Interior angle of A = $\frac{(a-2)180}{a}$

For polygon B (sides = $b$ ):
Interior angle of B = $\frac{(b-2)180}{b}$

We are given two conditions:
1) $b = 2a$
2) Interior angle of B = $\frac{3}{2} \times$ Interior angle of A

Substitute these into the equation:
$\frac{(2a-2)180}{2a} = \frac{3}{2} \times \frac{(a-2)180}{a}$
$\frac{2(a-1)}{2a} = \frac{3(a-2)}{2a}$
$\frac{a-1}{a} = \frac{3a-6}{2a}$

Multiply both sides by $2a$ (since $a > 0$ ):
$2(a-1) = 3a - 6$
$2a - 2 = 3a - 6$
$3a - 2a = 6 - 2$
$a = 4$

So, $a = 4$ and $b = 2(4) = 8$ .
We need to find the interior angle of a regular polygon with $a + b$ sides.
$a + b = 4 + 8 = 12$ sides.

Interior angle = $\frac{(12-2) \times 180}{12} = \frac{10 \times 180}{12} = 10 \times 15 = 150^\circ$ .

Q98TITAFunctions & Graphs

Let

f

be a function such that

f(mn) = f(m) f(n)

for every positive integers

m

and

n

. If

f(1), f(2)

and

f(3)

are positive integers,

f(1) < f(2)

, and

f(24) = 54

, then

f(18)

equals

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

Answer and solution

Answer: 12

Put

m = n = 1

:

f(1) = f(1)^2

. As

f(1)

is a positive integer,

f(1) = 1

. Since

24 = 2^3 \times 3

, we get

f(24) = f(2)^3 \times f(3) = 54

. So

f(2)^3

is the cube of a positive integer that divides

54 = 2 \times 3^3

. The only such cubes are

1

and

27

, so

f(2) = 1

or

f(2) = 3

. The condition

f(1) < f(2)

rules out

f(2) = 1

. Hence

f(2) = 3

and

f(3) = \frac{54}{27} = 2

. Now

18 = 2 \times 3^2

, so

f(18) = f(2) \times f(3)^2 = 3 \times 4 = 12

. The answer is 12.

Q99MCQTime & Work

Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

12
13
15
14

Answer and solution

Answer: (B) 13

Let the total amount of work be the LCM of 20 and 40, which is 40 units.

Anil's rate = $\frac{40}{20} = 2$ units/day.
Sunil's rate = $\frac{40}{40} = 1$ unit/day.

Anil starts the job and works alone for 3 days.
Work done by Anil in these 3 days = $3 \times 2 = 6$ units.

We are given that Bimal does 10% of the total job.
Work done by Bimal = 10% of 40 = 4 units.

The rest of the work must have been completed by Anil and Sunil.
Total work done by Anil and Sunil = Total work - Work done by Bimal = $40 - 4 = 36$ units.

Out of this, 6 units were done by Anil in the first 3 days.
So, the remaining work done by Anil and Sunil working together (and possibly alongside Bimal, but we only care about their contribution) = $36 - 6 = 30$ units.

The combined rate of Anil and Sunil is $2 + 1 = 3$ units/day.
Time taken by Anil and Sunil to complete these 30 units = $\frac{30}{3} = 10$ days.

So, Anil worked alone for 3 days, and then Anil and Sunil worked for 10 more days to finish their portion of the work (and Bimal finished his 4 units during this time too).
Total time taken for the job = $3 + 10 = 13$ days.

Hence, Option B is correct.

Q100MCQRatios, Proportions & Partnership

In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali's score exceeded Rama's score by

26
32
35
24

Answer and solution

Answer: (B) 32

Let the initial scores of Rama, Anjali, and Mohan be $r$ , $a$ , and $m$ respectively.
We are given that Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali:
$r = \frac{1}{12}(m + a)$
$12r = m + a$ --- (1)

After a review, the score of each of them increased by 6.
Their new scores are $r+6$ , $a+6$ , and $m+6$ .
The revised scores of Anjali, Mohan, and Rama were in the ratio 11 : 10 : 3.
Let $a+6 = 11x$ , $m+6 = 10x$ , and $r+6 = 3x$ .
From this, we get:
$a = 11x - 6$
$m = 10x - 6$
$r = 3x - 6$

Substitute these expressions into equation (1):
$12(3x - 6) = (10x - 6) + (11x - 6)$
$36x - 72 = 21x - 12$
$36x - 21x = 72 - 12$
$15x = 60 \implies x = 4$ .

We need to find by how much Anjali's score exceeded Rama's score:
$a - r = (11x - 6) - (3x - 6) = 11x - 3x = 8x$
Substitute $x = 4$ :
$a - r = 8(4) = 32$ .

Hence, Option B is correct.

CAT 2019 Slot 2 — QA questions with answers

Quantitative Ability

Q67MCQAverages, Mixtures & Alligations

Q68TITASimple & Compound Interest

Q69MCQProperties of Numbers

Q70TITAProperties of Numbers

Q71MCQTime, Speed & Distance

Q72MCQSequences & Series

Q73MCQPolygons & Circles

Q74MCQQuadratic & Polynomial Equations

Q75MCQQuadratic & Polynomial Equations

Q76MCQMensuration

Q77MCQTriangles & Lines

Q78MCQLogarithms

Q79TITAIndices & Surds

Q80MCQAverages, Mixtures & Alligations

Q81MCQTime, Speed & Distance

Q82MCQMensuration

Q83MCQIndices & Surds

Q84TITAProperties of Numbers

Q85TITADigits & Base Systems

Q86TITATime, Speed & Distance

Q87MCQPercentages

Q88TITALinear Equations

Q89TITASequences & Series

Q90MCQProfit, Loss & Discount

Q91MCQTriangles & Lines

Q92MCQSequences & Series

Q93MCQInequalities & Modulus

Q94MCQProfit, Loss & Discount

Q95TITAPercentages

Q96MCQRatios, Proportions & Partnership

Q97TITAPolygons & Circles

Q98TITAFunctions & Graphs

Q99MCQTime & Work

Q100MCQRatios, Proportions & Partnership