CAT 2017 Slot 1 QA Questions with Answers & Solutions

Q67TITALinear Equations

Arun's present age in years is 40% of Barun's. In another few years, Arun's age will be half of Barun's. By what percentage will Barun's age increase during this period?

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

Answer and solution

Answer: 20

Let Arun's current age be A. Hence, Barun's current age is 2.5A Let Arun's age be half of Barun's age after X years. Therefore, 2*(X+A) = 2.5A + X Or, X = 0.5A Hence, Barun's age increased by 0.5A/2.5A = 20%

Q68TITATime & Work

A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

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

Answer and solution

Answer: 15

Let one person do

x

units of work per day, so the whole job is

120x

units. On day 1 one person works, on day 2 two people work, on day 3 three, and so on. So the work done on days 1, 2, 3, … is

x, 2x, 3x, \ldots

If the job ends on day

n

, the total work is

x(1 + 2 + \cdots + n) = \frac{n(n+1)}{2}x

. Setting this equal to

120x

gives

n(n + 1) = 240

. Since

15 \times 16 = 240

,

n = 15

; the other root of

n^2 + n - 240 = 0

is

-16

, which is impossible. The job is finished exactly at the end of day 15. The answer is 15.

Q69TITAInequalities & Modulus

An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

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

Answer and solution

Answer: 11

The group has one person of 57 kg and one of 53 kg. To fit the most people, everyone else should be as light as allowed, that is 53 kg each. Let there be

n

other people. The total weight must not exceed the limit:

57 + 53 + 53n \le 630

, so

53n \le 520

and

n \le \frac{520}{53} \approx 9.8

. So there are at most 9 other people, and the group has at most

9 + 2 = 11

people. Check: 11 people can weigh as little as

57 + 10 \times 53 = 587

kg, within 630 kg, while 12 people would weigh at least

57 + 11 \times 53 = 640

kg, over the limit. The answer is 11.

Q70TITATime, Speed & Distance

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is

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

Answer and solution

Answer: 20

We see that the man saves 20 minutes by changing his speed from 12 km/hr to 15 km/hr. Let

d

be the distance. Then:

\frac{d}{12} - \frac{d}{15} = \frac{20}{60} = \frac{1}{3}

d \cdot \frac{5 - 4}{60} = \frac{1}{3}

\frac{d}{60} = \frac{1}{3}

d = 20 \text{ km}

Q71TITAPercentages

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500, then Ravi's total monthly savings (in Rs) is

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

Answer and solution

Answer: 70000

Let Ravi's total monthly savings be

100x

. He puts 50%, that is

50x

, in fixed deposits. The rest is

50x

; 30% of it,

15x

, goes into stocks, and the remaining

50x - 15x = 35x

goes into the savings bank account. Total deposited in the bank (fixed deposits and savings account)

= 50x + 35x = 85x

. Given

85x = 59500

, we get

x = 700

, so the total savings are

100x = 70000

. Check: fixed deposits get 35000; of the other 35000, stocks take 10500 and the savings account gets 24500; and

35000 + 24500 = 59500

. The answer is 70000.

Q72MCQProfit, Loss & Discount

If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures that she makes a profit of 20%?

Give a discount of 5% on retail price.
Give a discount of 2% on retail price.
Increase the retail price by 2%.
Sell at retail price.

Answer and solution

Answer: (D) Sell at retail price.

Let the retail price be

M

and the cost price be

C

. A 15% discount still gives a 2% profit, so

0.85M = 1.02C

, which gives

M = \frac{1.02}{0.85}C = 1.2C

. A 20% profit means selling at

1.2C

, which is exactly the retail price

M

. So she should sell at the retail price, with no discount. The other options miss 20%. A 2% discount (B) gives

0.98 \times 1.2C = 1.176C

, a 17.6% profit; a 5% discount (A) gives

0.95 \times 1.2C = 1.14C

, a 14% profit; and raising the retail price by 2% (C) gives

1.02 \times 1.2C = 1.224C

, a 22.4% profit. Hence, option D (Sell at retail price.).

Q73MCQTime, Speed & Distance

A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is

$\sqrt{6}:\sqrt{2}$
$\sqrt{7}:2$
$2\sqrt{5}:3$
3:2

Answer and solution

Answer: (B) $\sqrt{7}:2$

Let the boat's speed be

u

, the river's speed

x

and the one-way distance

d

. Original time:

t = \frac{d}{u - x} + \frac{d}{u + x} = \frac{2ud}{u^2 - x^2}

. Doubling the boat's speed cuts the time by 75%, to

\frac{t}{4}

:

\frac{d}{2u - x} + \frac{d}{2u + x} = \frac{4ud}{4u^2 - x^2} = \frac{t}{4}

. So

\frac{4ud}{4u^2 - x^2} = \frac{2ud}{4(u^2 - x^2)}

, which gives

8(u^2 - x^2) = 4u^2 - x^2

, that is

4u^2 = 7x^2

. Hence

\frac{u}{x} = \frac{\sqrt{7}}{2}

. The ratio 3:2 (D) does not work: with

u = 3

and

x = 2

the time goes from

\frac{6d}{5}

to

\frac{3d}{8}

, a cut of only 68.75%, not 75%. Hence, option B (

\sqrt{7}:2

).

Q74MCQRatios, Proportions & Partnership

Suppose, C1, C2, C3, C4, and C5 are five companies. The profits made by C1, C2, and C3 are in the ratio 9 : 10 : 8 while the profits made by C2, C4, and C5 are in the ratio 18 : 19 : 20. If C5 has made a profit of Rs 19 crore more than C1, then the total profit (in Rs) made by all five companies is

438 crore
435 crore
348 crore
345 crore

Answer and solution

Answer: (A) 438 crore

First make the share of

C_2

the same in both ratios.

C_1 : C_2 : C_3 = 9 : 10 : 8 = 81 : 90 : 72

(multiplying by 9), and

C_2 : C_4 : C_5 = 18 : 19 : 20 = 90 : 95 : 100

(multiplying by 5). So

C_1 : C_2 : C_3 : C_4 : C_5 = 81 : 90 : 72 : 95 : 100

. Let the profits be

81k, 90k, 72k, 95k

and

100k

crore.

C_5

exceeds

C_1

by 19 crore:

100k - 81k = 19k = 19

, so

k = 1

. Total profit

= (81 + 90 + 72 + 95 + 100)k = 438

crore. The total must be

438k

with

k = 1

, so 435 crore (B) and the other options cannot occur; 348 crore (C) only swaps the digits of 438. Hence, option A (438 crore).

Q75MCQPercentages

The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is

35
50
60
65

Answer and solution

Answer: (D) 65

Let the number of boys be

x

, so the number of girls is

2x

and there are

3x

candidates. Girls admitted: 30% of

2x

is

0.6x

. Boys admitted: 45% of

x

is

0.45x

. Total admitted

= 0.6x + 0.45x = 1.05x

, so the number not admitted is

3x - 1.05x = 1.95x

. Percentage not admitted

= \frac{1.95x}{3x} \times 100 = 65

. Option A (35) is the percentage who do get admission,

\frac{1.05x}{3x} \times 100 = 35

; the question asks for those who do not. Hence, option D (65).

Q76MCQRatios, Proportions & Partnership

A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

1:1
8:7
4:3
6:5

Answer and solution

Answer: (A) 1:1

Popcorn packets (large, super, jumbo) are

7x, 17x, 16x

, a total of

40x

. Chips packets are

6y, 15y, 14y

, a total of

35y

. The totals are equal:

40x = 35y

, so

x = \frac{7y}{8}

. Jumbo popcorn

= 16x = 16 \times \frac{7y}{8} = 14y

, and jumbo chips

= 14y

. So the ratio is

14y : 14y = 1 : 1

. Option B (8:7) compares 16 with 14 directly, as if

x = y

. That ignores the condition that the two stocks have equal totals, which makes

x

smaller than

y

. Hence, option A (1:1).

Q77MCQProfit, Loss & Discount

In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is

6%
8%
10%
12%

Answer and solution

Answer: (B) 8%

Let medium mangoes cost

x

per kg, so good mangoes cost

2x

per kg. Cost: good mangoes

80 \times 2x = 160x

and medium mangoes

40 \times x = 40x

, a total of

200x

. All 120 kg are sold at 10% below the good-mango price:

0.9 \times 2x = 1.8x

per kg. Revenue

= 120 \times 1.8x = 216x

. Profit

= 216x - 200x = 16x

, so the profit percentage is

\frac{16x}{200x} \times 100 = 8

. Option C (10%) comes from dividing the profit by the cost of the good mangoes alone,

\frac{16x}{160x}

; the profit must be measured on the whole cost,

200x

. Hence, option B (8%).

Q78MCQProfit, Loss & Discount

If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?

30%
25%
24%
28%

Answer and solution

Answer: (D) 28%

Let the cost price of a toy be

C

and the printed price be

M

. Selling at a 40% discount gives a 20% profit:

0.6M = 1.2C

, so

M = 2C

. All 60 toys were paid for, so the cost is

60C

, and the original profit is 20% of

60C

, that is

12C

. To make the same profit, the revenue must be

60C + 12C = 72C

. Only 50 toys are left, each sold at

M(1 - d) = 2C(1 - d)

, where

d

is the discount:

50 \times 2C(1 - d) = 72C

, so

1 - d = 0.72

and

d = 0.28

, a 28% discount. The trap is using the cost of the 50 remaining toys only; the burnt toys were paid for too. A 30% discount (A) would bring in only

100C \times 0.7 = 70C

, short of

72C

. Hence, option D (28%).

Q79MCQLinear Equations

If

a

and

b

are integers of opposite signs such that

(a+3)^2 : b^2 = 9 : 1

and

(a-1)^2 : (b-1)^2 = 4 : 1

, then the ratio

a^2 : b^2

is

9:4
81:4
1:4
25:4

Answer and solution

Answer: (D) 25:4

Since the square root can be positive or negative, we get two cases for each equation. For the first equation:

(a+3)^2 : b^2 = 9 : 1

, so: -

a + 3 = 3b

... (i) -

a + 3 = -3b

... (ii) For the second equation:

(a-1)^2 : (b-1)^2 = 4 : 1

, so: -

a - 1 = 2(b-1)

... (iii) -

a - 1 = -2(b-1)

... (iv) Solving (i) and (iii):

a + 3 = 3b

and

a - 1 = 2b - 2 \Rightarrow a = 2b - 1

. Substituting:

2b - 1 + 3 = 3b \Rightarrow b = 2, a = 3

. But

a

and

b

must have opposite signs. ✗ Solving (i) and (iv):

a + 3 = 3b

and

a - 1 = -2(b-1) \Rightarrow a = -2b + 3

. So

-2b + 3 + 3 = 3b \Rightarrow b = 6/5

. Not an integer. ✗ Solving (ii) and (iii):

a + 3 = -3b

and

a = 2b - 1

. So

2b - 1 + 3 = -3b \Rightarrow 5b = -2 \Rightarrow b = -2/5

. Not an integer. ✗ Solving (ii) and (iv):

a + 3 = -3b

and

a - 1 = -2b + 2 \Rightarrow a = -2b + 3

. So

-2b + 3 + 3 = -3b \Rightarrow b = -6, a = 15

. Opposite signs ✓ Therefore

a^2 : b^2 = 225 : 36 = 25 : 4

Q80MCQAverages, Mixtures & Alligations

A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

9.5
10
4.5
6

Answer and solution

Answer: (A) 9.5

Work with total scores. The class has 50 students and its average rose by 2, so the class total rose by

50 \times 2 = 100

. The 30 girls' average fell by 3, so the girls' total fell by

30 \times 3 = 90

. The boys' total must therefore have risen by

100 + 90 = 190

, and the average of the 20 boys rose by

\frac{190}{20} = 9.5

. The mid-semester gap of 5 between girls and boys does not affect these changes. The table below does the same with the boys' mid-term average as

A

: the girls go from

A + 5

to

A + 2

, the class average from

A + 3

to

A + 5

, and the boys from

A

to

A + 9.5

.

Solution figure for question 80, CAT 2017 Slot 1

Check option B (10): the boys' total would rise by 200, the class total by

200 - 90 = 110

, and the class average by

2.2

, not 2. Hence, option A (9.5).

Q81MCQCoordinate Geometry

The area of the closed region bounded by the equation

| x | + | y | = 2

in the two-dimensional plane is

$4π$ sq. units
$4$ sq. units
$8$ sq. units
$2π$ sq. units

Answer and solution

Answer: (C) $8$ sq. units

In the first quadrant,

|x| + |y| = 2

is the line

x + y = 2

, from

(2, 0)

to

(0, 2)

. By symmetry, the other quadrants give the lines joining

(0, 2)

,

(-2, 0)

,

(0, -2)

and back to

(2, 0)

.

Solution figure for question 81, CAT 2017 Slot 1

So the region is a square with vertices

(2, 0)

,

(0, 2)

,

(-2, 0)

and

(0, -2)

. Its diagonals lie along the axes and each is 4 long. Area

= \frac{d_1 \times d_2}{2} = \frac{4 \times 4}{2} = 8

square units. Check: each side is

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

, and

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

. Option B (4) takes the side as 2, the distance from the origin to a vertex; the side is actually

2\sqrt{2}

. The options with

\pi

would need a circle, but the boundary is made of straight lines. Hence, option C (

8

sq. units).

Q82MCQTriangles & Lines

From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is

$225\sqrt{3}$
$\frac{500}{\sqrt{3}}$
$\frac{275}{\sqrt{3}}$
$\frac{250}{\sqrt{3}}$

Answer and solution

Answer: (B) $\frac{500}{\sqrt{3}}$

The sides are 40, 25 and 35 ft, so the semi-perimeter is

s = \frac{40 + 25 + 35}{2} = 50

. By Heron's formula, the area of

ABC

is

\sqrt{50 \times 10 \times 25 \times 15} = \sqrt{187500} = 250\sqrt{3}

sq ft.

Solution figure for question 82, CAT 2017 Slot 1

The three medians split a triangle into six small triangles of equal area. Triangle

GBC

is made of two of them, so it is one third of

ABC

:

\frac{250\sqrt{3}}{3} = \frac{250}{\sqrt{3}}

. The remaining portion, triangles

GAB

and

GCA

together, is two thirds of

ABC

:

\frac{2}{3} \times 250\sqrt{3} = \frac{500\sqrt{3}}{3} = \frac{500}{\sqrt{3}}

sq ft. Option D (

\frac{250}{\sqrt{3}}

) is the area of the piece

GBC

that is cut off, not of what remains. Hence, option B (

\frac{500}{\sqrt{3}}

).

Q83MCQPolygons & Circles

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is

$9π−18$
$18$
$9π$
$9$

Answer and solution

Answer: (B) $18$

AB = AC = 6

cm and the right angle is at

A

, so

BC = \sqrt{6^2 + 6^2} = 6\sqrt{2}

cm.

Solution figure for question 83, CAT 2017 Slot 1

Semicircle

BQC

has radius

3\sqrt{2}

, so its area is

\frac{\pi (3\sqrt{2})^2}{2} = 9\pi

. Arc

BPC

has centre

A

and radius 6, and

\angle BAC = 90^\circ

, so it is a quarter circle. The region between chord

BC

and arc

BPC

is the quarter circle minus triangle

ABC

:

\frac{\pi \times 6^2}{4} - \frac{1}{2} \times 6 \times 6 = 9\pi - 18

. The required region lies between arc

BPC

and arc

BQC

, so it is the semicircle minus that segment:

9\pi - (9\pi - 18) = 18

sq cm. Option A (

9\pi - 18

) is the area of the segment between

BC

and arc

BPC

, not the region between the two arcs. Hence, option B (

18

).

Q84MCQMensuration

A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to

10
50
60
20

Answer and solution

Answer: (B) 50

Let the five volumes be

a^3, a^3, 8a^3, 27a^3, 27a^3

. Then the sides are

a, a, 2a, 3a, 3a

. The original cube had volume

a^3 + a^3 + 8a^3 + 27a^3 + 27a^3 = 64a^3

, so its side was

4a

and its surface area

6(4a)^2 = 96a^2

. The five new cubes have total surface area

6(a^2 + a^2 + 4a^2 + 9a^2 + 9a^2) = 6 \times 24a^2 = 144a^2

. Excess

= \frac{144a^2 - 96a^2}{96a^2} \times 100 = 50

percent. The excess is exactly 50%, so 60 (C) is too large: a 60% excess would need a new total of

1.6 \times 96a^2 = 153.6a^2

. Hence, option B (50).

Q85TITAMensuration

A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9π cubic centimeters. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is

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

Answer and solution

Answer: 6

The cylinder's volume is

\pi r^2 \times 3 = 9\pi

, so

r^2 = 3

and

r = \sqrt{3}

cm. The ball's radius, 2 cm, is more than

\sqrt{3}

cm, so the ball cannot drop inside; it rests on the circular rim at the top of the cylinder, with its centre directly above the centre of the rim. In a vertical cross-section through the centre, shown below, the line from the ball's centre to a point where it touches the rim is a radius of the ball, 2 cm, and its horizontal part is the rim's radius,

\sqrt{3}

cm. So the ball's centre is

\sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4 - 3} = 1

cm above the rim. Solution figure for question 85, CAT 2017 Slot 1

The topmost point of the ball is one radius, 2 cm, above its centre. Its height above the base is therefore

3 + 1 + 2 = 6

cm. The answer is 6.

Q86TITATriangles & Lines

Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is

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

Answer and solution

Answer: 24

The shortest path from A to the hypotenuse BC is the altitude from A. Given:

AB = 15

km,

AC = 20

km (right angle at A). Hypotenuse:

BC = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25

km Length of altitude

h

from A to BC (using area method):

\frac{1}{2} \times BC \times h = \frac{1}{2} \times AB \times AC

25h = 15 \times 20 = 300

h = 12 \text{ km}

Time taken at 30 km/hr:

= \dfrac{12}{30} \times 60 = \mathbf{24}

minutes

Q87MCQLogarithms

Suppose

\log_3 x = \log_{12} y = a

, where

x, y

are positive numbers. If

G

is the geometric mean of

x

and

y

, then

\log_6 G

is equal to

$\sqrt{a}$
$2a$
$\frac{a}{2}$
$a$

Answer and solution

Answer: (D) $a$

From

\log_3 x = a

and

\log_{12} y = a

, we get

x = 3^a

and

y = 12^a

. The geometric mean is

G = \sqrt{xy} = \sqrt{3^a \times 12^a} = \sqrt{36^a} = 6^a

. So

\log_6 G = \log_6 6^a = a

. Option B (

2a

) is

\log_6(xy) = \log_6 36^a

, the value you get if you forget the square root in the geometric mean. Taking the square root halves the logarithm, which gives

a

. Hence, option D (

a

).

Q88MCQQuadratic & Polynomial Equations

If

x + 1 = x^2

and

x > 0

, then

2x^4

is

$6 + 4\sqrt{5}$
$3 + 3\sqrt{5}$
$5 + 3\sqrt{5}$
$7 + 3\sqrt{5}$

Answer and solution

Answer: (D) $7 + 3\sqrt{5}$

From

x + 1 = x^2

we have

x^2 = x + 1

. Then

x^4 = (x^2)^2 = (x + 1)^2 = x^2 + 2x + 1 = (x + 1) + 2x + 1 = 3x + 2

, so

2x^4 = 6x + 4

. Since

x > 0

, solving

x^2 - x - 1 = 0

gives

x = \frac{1 + \sqrt{5}}{2}

. So

2x^4 = 6 \times \frac{1 + \sqrt{5}}{2} + 4 = 3 + 3\sqrt{5} + 4 = 7 + 3\sqrt{5}

. Option B (

3 + 3\sqrt{5}

) is just

6x

; it leaves out the

+4

that comes from

x^4 = 3x + 2

. Hence, option D (

7 + 3\sqrt{5}

).

Q89MCQLogarithms

The value of

\log_{0.008} \sqrt{5} + \log_{\sqrt{3}} 81 - 7

is equal to

1/3
2/3
5/6
7/6

Answer and solution

Answer: (C) 5/6

Write each base and argument as a power of a prime.

0.008 = \frac{8}{1000} = \frac{1}{125} = 5^{-3}

and

\sqrt{5} = 5^{1/2}

, so

\log_{0.008} \sqrt{5} = \frac{1/2}{-3} = -\frac{1}{6}

.

\sqrt{3} = 3^{1/2}

and

81 = 3^4

, so

\log_{\sqrt{3}} 81 = \frac{4}{1/2} = 8

. The expression is

-\frac{1}{6} + 8 - 7 = 1 - \frac{1}{6} = \frac{5}{6}

. Option D (

\frac{7}{6}

) comes from taking the first logarithm as

+\frac{1}{6}

. But the base

0.008

is less than 1 while

\sqrt{5}

is more than 1, so that logarithm must be negative. Hence, option C (5/6).

Q90MCQIndices & Surds

If

9^{2x−1}−81^x−1=1944,

then x is

3
9/4
4/9
1/3

Answer and solution

Answer: (B) 9/4

The equation is

9^{2x-1} - 81^{x-1} = 1944

. Write both terms using

81^x

:

9^{2x-1} = \frac{9^{2x}}{9} = \frac{81^x}{9}

and

81^{x-1} = \frac{81^x}{81}

. So

81^x\left(\frac{1}{9} - \frac{1}{81}\right) = 1944

, that is

81^x \times \frac{8}{81} = 1944

. Then

81^x = \frac{1944 \times 81}{8} = 243 \times 81 = 19683 = 3^9

. Since

81^x = 3^{4x}

, we get

4x = 9

and

x = \frac{9}{4}

. Option C (

\frac{4}{9}

) inverts this result, and

x = 3

(A) would make the left side

9^5 - 81^2 = 52488

, far more than 1944. Hence, option B (9/4).

Q91MCQPermutations & Combinations

The number of solutions

(x,y,z)

to the equation

x−y−z=25

, where

x

,

y

, and

z

are positive integers such that

x≤40,y≤12

, and

z≤12

is

101
99
87
105

Answer and solution

Answer: (B) 99

Since

y, z \ge 1

, we have

x = 25 + y + z \ge 27

, and also

x \le 40

. Let

s = y + z = x - 25

, which runs from 2 to 15. For each

s

, count the pairs with

1 \le y \le 12

and

1 \le z \le 12

. For

s = 2

to

13

,

y

can be

1, 2, \ldots, s - 1

, giving

s - 1

pairs; together

1 + 2 + \cdots + 12 = 78

. For

s = 14

,

y

runs from 2 to 12 (11 pairs); for

s = 15

,

y

runs from 3 to 12 (10 pairs). Total

= 78 + 11 + 10 = 99

. Option D (105) forgets the cap of 12: counting

s - 1

pairs for every

s

gives

1 + 2 + \cdots + 14 = 105

, which includes pairs such as

y = 13, z = 1

. Hence, option B (99).

Q92TITAInequalities & Modulus

For how many integers n, will the inequality

(n−5)(n−10)−3(n−2)≤0

be satisfied?

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

Answer and solution

Answer: 11

(n−5)(n−10)−3(n−2)≤0

=>

n^2−15n+50−3n+6≤0

=>

n^2−18n+56≤0

=>

(n−4)(n−14)≤0

=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 - 4 + 1 = 11.

Q93TITAQuadratic & Polynomial Equations

If

f_1(x) = x^2 + 11x + n

and

f_2(x) = x

, then the largest positive integer

n

for which the equation

f_1(x) = f_2(x)

has two distinct real roots is

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

Answer and solution

Answer: 24

f_1(x) = f_2(x)

\Rightarrow x^2 + 11x + n = x

\Rightarrow x^2 + 10x + n = 0

For two distinct real roots:

b^2 - 4ac > 0

10^2 > 4n

100 > 4n

n < 25

The largest positive integer

n

can take is

\mathbf{24}

.

Q94TITAInequalities & Modulus

If

a, b, c

, and

d

are integers such that

a + b + c + d = 30

, then the minimum possible value of

(a-b)^2 + (a-c)^2 + (a-d)^2

is

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

Answer and solution

Answer: 2

Each term is the square of an integer, so the expression is a sum of three non-negative integers. It cannot be

0

: that needs

b = c = d = a

, so

4a = 30

, which has no integer solution. It cannot be

1

: the only way to get 1 from three squares is

1 + 0 + 0

, so two of

b, c, d

equal

a

and the third is

a \pm 1

. Then

4a \pm 1 = 30

, which again has no integer solution. A total of

2

is possible. Take

a = 8

,

b = 8

,

c = 7

,

d = 7

. The sum is

30

, and

(a-b)^2 + (a-c)^2 + (a-d)^2 = 0 + 1 + 1 = 2

. So the minimum value is 2. The answer is 2.

Q95TITAPermutations & Combinations

Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

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

Answer and solution

Answer: 160

There are 11 points: the 10 ends of the diameters, which lie on the circle, and the centre

O

. Any 3 of them can be chosen in

\binom{11}{3} = 165

ways. A choice fails to form a triangle only if the three points are collinear. No three points of a circle are collinear, so a collinear choice must include

O

.

O

is collinear with two points of the circle only when they are the two ends of the same diameter. That gives exactly 5 bad choices:

\{A, O, B\}

,

\{C, O, D\}

,

\{E, O, F\}

,

\{G, O, H\}

and

\{J, O, K\}

. So the number of triangles is

165 - 5 = 160

. The answer is 160.

Q96MCQFunctions & Graphs

The shortest distance of the point

\left(\frac{1}{2}, 1\right)

from the curve

y = |x - 1| + |x + 1|

is

1
0
$\sqrt{2}$
$\sqrt\frac{3}{2}$

Answer and solution

Answer: (A) 1

Write

y = |x - 1| + |x + 1|

piece by piece: for

x \le -1

,

y = -2x

; for

-1 \le x \le 1

,

y = 2

; for

x \ge 1

,

y = 2x

. So the curve is a flat segment at height 2 between

x = -1

and

x = 1

, with arms rising on both sides. Every point on it has

y \ge 2

.

Solution figure for question 96, CAT 2017 Slot 1

The point

\left(\frac{1}{2}, 1\right)

has

y = 1

, so every point of the curve is at least 1 unit above it, and hence at least 1 unit away. The point

\left(\frac{1}{2}, 2\right)

on the flat segment, directly above it, is exactly 1 unit away. So the shortest distance is 1. Option B (0) fails:

\left(\frac{1}{2}, 1\right)

lies on the line

y = 2x

, but that line is part of the curve only for

x \ge 1

; at

x = \frac{1}{2}

the curve is at height 2. Hence, option A (1).

Q97MCQSequences & Series

If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

2:3
3:2
3:4
4:3

Answer and solution

Answer: (A) 2:3

Let the first term be

a

and the common difference

d > 0

. The 7th, 3rd and 17th terms are

a + 6d

,

a + 2d

and

a + 16d

. We are given

(a + 6d)^2 = (a + 2d)(a + 16d)

. Expanding:

a^2 + 12ad + 36d^2 = a^2 + 18ad + 32d^2

, so

4d^2 = 6ad

. Since

d > 0

, divide by

2d

:

2d = 3a

, so

\frac{a}{d} = \frac{2}{3}

. Option B (3:2) is

d : a

, the ratio the other way round; the question asks for the first term to the common difference. Hence, option A (2:3).

Q98MCQPermutations & Combinations

In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

16
20
14
15

Answer and solution

Answer: (A) 16

Let the kids get

a, b, c, d

erasers, with

a + b + c + d = 7

and each at least 1. The number of such distributions is

\binom{7 - 1}{4 - 1} = \binom{6}{3} = 20

. Now remove those where a kid gets more than 3. The other three kids get at least 1 each, so no kid can get more than

7 - 3 = 4

. A kid gets exactly 4 only when the others get 1 each, and this can be any of the 4 kids: 4 cases. So the number of ways is

20 - 4 = 16

. Option B (20) counts every distribution with at least one eraser each, including the 4 where a kid gets 4 erasers. Hence, option A (16).

Q99MCQFunctions & Graphs

If

f(x) = \dfrac{5x + 2}{3x - 5}

and

g(x) = x^2 - 2x - 1

, then the value of

g(f(f(3)))

is

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

Answer and solution

Answer: (A) 2

First,

f(3) = \frac{5 \times 3 + 2}{3 \times 3 - 5} = \frac{17}{4}

. Next,

f\left(\frac{17}{4}\right) = \frac{\frac{85}{4} + 2}{\frac{51}{4} - 5} = \frac{93/4}{31/4} = \frac{93}{31} = 3

. So

f(f(3)) = 3

, and

g(3) = 3^2 - 2 \times 3 - 1 = 9 - 6 - 1 = 2

. A common slip is to stop after one application of

f

:

g\left(\frac{17}{4}\right) = \frac{137}{16}

, which is not an option. Once

f(f(3)) = 3

is found, the answer is

g(3) = 2

, not 6 (C) or a fraction such as

\frac{2}{3}

(D). Hence, option A (2).

Q100MCQSequences & Series

Let

a_1, a_2, \ldots, a_{3n}

be an arithmetic progression with

a_1 = 3

and

a_2 = 7

. If

a_1 + a_2 + \cdots + a_{3n} = 1830

, then what is the smallest positive integer

m

such that

m(a_1 + a_2 + \cdots + a_n) > 1830

?

8
9
10
11

Answer and solution

Answer: (B) 9

Here

a_1 = 3

and

a_2 = 7

, so the common difference is 4. Let

k = 3n

. The sum of the first

k

terms is

\frac{k}{2}[2 \times 3 + (k - 1) \times 4] = \frac{k}{2}(4k + 2) = k(2k + 1)

. So

k(2k + 1) = 1830

, that is

2k^2 + k - 1830 = 0

, which factors as

(k - 30)(2k + 61) = 0

. Hence

k = 30

and

n = 10

. Sum of the first 10 terms

= \frac{10}{2}[2 \times 3 + 9 \times 4] = 5 \times 42 = 210

. We need

210m > 1830

, that is

m > 8.71\ldots

, so the smallest integer is

m = 9

:

9 \times 210 = 1890 > 1830

. Option A (8) fails because

8 \times 210 = 1680

, which is less than 1830. Hence, option B (9).

CAT 2017 Slot 1 — QA questions with answers

Quantitative Ability

Q67TITALinear Equations

Q68TITATime & Work

Q69TITAInequalities & Modulus

Q70TITATime, Speed & Distance

Q71TITAPercentages

Q72MCQProfit, Loss & Discount

Q73MCQTime, Speed & Distance

Q74MCQRatios, Proportions & Partnership

Q75MCQPercentages

Q76MCQRatios, Proportions & Partnership

Q77MCQProfit, Loss & Discount

Q78MCQProfit, Loss & Discount

Q79MCQLinear Equations

Q80MCQAverages, Mixtures & Alligations

Q81MCQCoordinate Geometry

Q82MCQTriangles & Lines

Q83MCQPolygons & Circles

Q84MCQMensuration

Q85TITAMensuration

Q86TITATriangles & Lines

Q87MCQLogarithms

Q88MCQQuadratic & Polynomial Equations

Q89MCQLogarithms

Q90MCQIndices & Surds

Q91MCQPermutations & Combinations

Q92TITAInequalities & Modulus

Q93TITAQuadratic & Polynomial Equations

Q94TITAInequalities & Modulus

Q95TITAPermutations & Combinations

Q96MCQFunctions & Graphs

Q97MCQSequences & Series

Q98MCQPermutations & Combinations

Q99MCQFunctions & Graphs

Q100MCQSequences & Series