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This mock test of Sample CAT Quantitative Aptitude MCQ for CAT helps you for every CAT entrance exam.
This contains 34 Multiple Choice Questions for CAT Sample CAT Quantitative Aptitude MCQ (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Student was supposed to calculate the average of 10 numbers. He calculated the average as 261 but mistakenly took 34 instead of 43 and counted 12 twice. What is the actual average of 10 numbers?

Solution:

Sum of the 10 numbers = 261 x 10 = 2610

Difference in the sum = (43 - 34) + (-12) = -3

Actual sum = 2610 - 3 = 2607

Average = 2607/10 = 260.7

Hence, option 3.

QUESTION: 2

A cylindrical well of depth 15 m and radius 6 m is dug at a place to fetch water from it. The well is lined from inside with a layer of bricks of thickness 1 m. In rainy season the well is filled to the brim. Find the volume of water in the well at that time.

Solution:

The effective depth and radius after brick layering would be 14 m and 5 m respectively.

The volume of water in it would be = π x 5 x 5 x l 4 = 1100 cu. m Hence, option 1.

*Answer can only contain numeric values

QUESTION: 3

Tap A can fill a tank in 16 hours. Another tap B is opened 2 hours after opening of tap A and it takes 4 more hours to fill the tank. How many minutes will tap B alone require to fill the entire tank?

Solution:

Let assume the capacity of the tank is 1.

As, tap A fills the tank in 16 hours

So, tap A filling capacity per hour, ie. Volume of water tap A can fill in one hour.

=> capacity of tank / hours tap A take to fill tank = 1 / 16

Let tap B filling capacity per hour = x

As, given;

2 * ( tap A filling capacity per hour) + 4*

( tap A and tap B together filling capacity per hour) = capacity of tank

=> 2* 1/ 16 +4 ( 1/16 +x) = 1

=> 2 +4 + 64 x = 16

=> 64 x = 8

=> x = 1/ 8

ie. The filling capacity of tap B per hour

So, time tap B will take to fill the tank ;-

=> capacity of tank/ filling capacity of B per hour = 8 hours

=> 8 * 60

=> 480 min

*Answer can only contain numeric values

QUESTION: 4

The number 35a246772 is in base 9. This number when written in base 10 is divisible by 8. Find the value of digit a.

Solution:

(35a246772)_{9}

= (3 x 9^{8}) + (5 x 9^{7}) + (a x 9^{6}) + ... + (2 x 9^{0})

= [3 x (8 + 1)^{8}] + [5 x (8 + 1)^{7}] + [a x (8 + 1)^{6}] + ... + [2 x (8 + 1)^{6}]

When the above expression is expanded, only the last term of each binomial expression is not divisible by 8.

Thus, the number will be divisible when sum of its digits is divisible by 8. Sum of digits = 3 + 5 + <z + 2 + 4 + 6 + 7 + 7 + 2 = 36 + a The sum will be divisible by 8 when a = 4.

Answer: 4

QUESTION: 5

Find the value of x in the expression log_{x} 9 + log_{x} 8 + log_{x} 32 = 2.

Solution:

log_{x} 9 + log_{x} 8 + log_{x} 32 = 2

log_{x} (9 x 8 x 32) = 2

9 x 8 x 32 = x^{2}

x = 48

Hence, option 1.

QUESTION: 6

How many points with integer co-ordinates do not lie outside x^{2} + 4y^{2} = 36?

Solution:

We can see that the value ofy cannot exceed ±3 while x cannot exceed ±6.

When y = ±3; x^{2} ≤ 0. Thus, x = 0. Thus, there are (2 x 1 = 2) points.

When y = ±2; x^{2} ≤ 20. Thus, x is between -4 and 4. Thus, there are 2 x 9 = 18 points.

When y = ±1; x^{2} ≤ 32. Thus, x is between -5 and 5. Thus, there are 2 x 11 = 22 points.

When y = ±0; x^{2} ≤ 36. Thus, x is between -6 and 6. Thus, there are 2 x 6 + 1 = 13 points.

∴ Total points = 2 + 18 + 22 + 13 = 55 points

Hence, option 3.

QUESTION: 7

Which of the following is true about the absolute difference between a two- digit number and the number obtained by reversing its digits?

Solution:

Let the units digit and tens digit be 'a' and 'b' respectively.

The number = 10b + a

Required difference = |(10b + a) - (10a +b)|= |9b - 9a| = 9 x |b - a|

The difference will always be divisible by 9.

Hence, option 3.

QUESTION: 8

If team India has played (4/5)th of total 40 matches and wins 29 matches, then the maximum number of matches it can afford to lose from the remaining matches to maintain a more than 75% win record?

Solution:

4/5 of 40 = 32 matches played out of which 29 won.

8 more matches are to be played.

75% of 40 = 30

To maintain more than 75% win record, it has to win at least 31 matches.

Team India has to win at least two out of the remaining eight matches. Hence, team India can afford to lose at most six matches.

Answer: 6

QUESTION: 9

How many values of 'b' does the equation x^{2} + bx + 4b = 0 satisfy such that the roots are equal and integral?

Solution:

Let'a' be the root o f the equation x^{2} + bx + 4b = 0

2a = —b and a^{2} = 4b

Solving this we get, b= {0, 16}

There are 2 values of b which satisfies the given equation.

Hence, option 3.

QUESTION: 10

There are two people, one on each side of a tower making an angle of elevation of 45° and 60° respectively. Find the distance between them if the height of the tower is 10√3 m.

Solution:

The distance between the tower and the person making 45° with the tower is 10a/3 m and the distance between the tower and the person making 60° with the tower is 10 m.

The distance between them is

(10 + 10√3) m = 10(1 + √3) m

Hence, option 1.

*Answer can only contain numeric values

QUESTION: 11

An alloy of gold and silver weighs 50 gm. It contains 80% gold. How many grams of gold should be added to the alloy so that percentage of gold increases to 90?

Solution:

In 50 gm alloy 40 gm is gold.

Consider x gm gold to be added in the alloy to make gold 90%. (40 + x)/(50 + x) = 90/100

Solving the equation we get, x = 50

50 gm of gold should be added.

Answer: 50

QUESTION: 12

How many acute-angled triangles can be formed with sides 5 cm, 12 cm and x cm, if x is an integer.

Solution:

For a triangle to be formed, 7 < x < 17

**Case I:** x ≥ 12 i.e. x is the largest side

x^{2} < 13^{2}, for an acute-angled triangle.

Thus, the only possible integer value of x is 12.

**Case II:** x <12 i.e. 12 is the largest side

x^{2} > 12^{2} - 5^{2}, for an acute-angled triangle.

Thus, the only possible integer value of x is 11.

1 + 1 = 2 acute-angled triangles can be formed.

Hence, option 1.

QUESTION: 13

A group of 630 children is arranged in a row for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible ?

Solution:

Let no of children be x, x+3,x+6,x+9,x+12,x+15...........

Present in Rows,.......... respectively

Putting R=3,4,5 and 6

We see that when R=6

x+x+3+x+6+x+9+x+12+x+15=630

6x+45=630

6x=585

x=97.5

then x is not an integer.

So, R=6 does not satisfies.

QUESTION: 14

A man walks 4 steps forward and 1 step backward everytime on an escalator (each step in 1 second). He takes exactly 118 seconds to climb down a stationary escalator. When the escalator is moving, he takes 40 seconds. What is the speed of the escalator in steps per second?

Solution:

In a span of 5 seconds, the man moves forward 3 steps.

Now, 118 = 2 3 x 5 + 3F where F denotes a forward step.

Thus, the stationary escalator has 72 steps.

There are such 8 five second intervals in 40 seconds. Thus, the man takes 24 steps before he reaches down. Thus, the rest 48 steps are taken by the escalator in 40 seconds.

Speed of escalator = 48/40 = 1.2 steps/second Hence, option 2.

Alternatively,

We can see that the time required when the escalator is stationary is roughly three times the time required when the escalator is moving.

Since the speed of the man remains constant in both cases, we can say that the speed of the escalator is twice that of the person. Now, the man takes 3 steps every 5 seconds i.e. 0.8 steps/second. Thus, the speed of the escalator will be close to but less than 1.6 steps/second.

Hence, option 2.

QUESTION: 15

The marked price of a watch is Rs. 800. A shopkeeper gives two successive discounts and sells the watch at Rs. 612. If the first discount is 10%, the second discount is

Solution:

First discount is 10%, so, price becomes (100 - 10)% of 800 = Rs. 720

Let second discount be 'm'%, so, price becomes(100 - m)% of 720 = Rs. 612

∴ 100 - m / 100 x 720 = 612

m = 15 = Second discount

QUESTION: 16

If L = a^{2} — 16a + 48 and —2 ≤ a ≤ 2, then find the maximum value of L.

Solution:

L = a^{2} - 16^{a} + 48 = (a - 8)^{2} - 16

It will have maximum value at a = -2. Substituting a = - 2 , w e get ,

L = 84

Hence, option 3.

*Answer can only contain numeric values

QUESTION: 17

A certain number of chocolates are equally distributed among a certain number of students in such a way that each student gets 14 chocolates. However, if 4 of those students decide to distribute their chocolates equally amongst themselves, then the students who have not distributed their share get 4 extra chocolates. Find the total number of chocolates.

Solution:

Let the total number of students be x.

The total number of chocolates = 14x

Four students amongst them will now distribute 4 x 14 = 56 chocolates

This is equal to the number of chocolates which are recieved by all of them in this distribution = 4x

4x = 56 i.e. x = 14 Thus, there are 14 x 14 = 196 chocolates.

Answer: 196

QUESTION: 18

Mitali has 2 watches one is Polex and the other is Tastfrack. At 7:00 am on Friday, 1^{st} January 2016 (which is the correct time) Polex shows 6:22 am while Tastfrack shows 7:57 am. Polex gains 2 minutes and Tastfrack loses 3 minutes every hour. At what time and date will both watches show same time for first instance?

Solution:

Polex is set 38 minutes behind the original time and Tastfrack is set 57 minutes ahead of original time.

Thus, the total time difference = 95 minutes Polex gains 2 minutes in every hour and Tastfrack loses 3 minutes in every hour.

Thus, the relative speed of the watches = 5 minutes/hour They will cover 95 minutes in 95/5 = 19 hours i.e. at 2:00 am on 2^{nd }January 2016.

Hence, option 1.

*Answer can only contain numeric values

QUESTION: 19

An investment bank offers different return percentages for different amount of investments. If the rate of simple interest on Rs. 1,00,000 is 10% per annum then find the rate of simple interest it must offer for investment of Rs. 1,25,000 such that both investments amount to the same value after a tenure of five years.

Solution:

Let the rate of simple interest for Rs. 1,25,000 be x.

Rs. 1,00,000 will amount to Rs. 1,50,000 in 5 years.

125000(1 + 5x/100) = 150000

x = 4 Answer: 4

QUESTION: 20

What is the number of sides of the regular polygon whose interior angle is four times of the exterior angle?

Solution:

Exterior angle of a polygon = 360/n

Since, the sum of interior and exterior angle = 180°

(360/n) + 4 x (360/n)= 180

n = 10

Hence, option 3.

QUESTION: 21

If A is the set of all natural numbers divisible by 3 upto 100, B is the set of all even natural numbers upto 100, C is the set of all natural numbers divisible by 7 upto 100 and D is the set of all the perfect squares from 1 to 100, then find the sum of all the numbers in (A ∩ C) ∪ (B ∩ D).

Solution:

A = { 3 , 6 , 9 ....... 99}

B = {2, 4 , 6 ........100}

C = {7, 1 4 ,2 1 ..... 98}

D = {1,4, 9, 16, 2 5 ..... 100}

∴ (A ∩ C) = {21, 42, 63, 84} and (B ∩ D) = {4, 16, 21, 36, 64, 100}

∴ (A ∩ C) ∪ (B ∩ D) = {4, 16, 21, 36, 42, 63, 64, 84, 100}

The sum of the numbers of (A ∩ C) ∪ (B ∩ D) = 430

Hence, option 3.

QUESTION: 22

Seven men can complete a work in 12 days. They started the work and after 5 days two men left. In how many days will the work be completed by the remaining men?

Solution:

Let total work is 12 x 7 = 84 man days.

Now,Work done in 5 days = 7 persons worked for 5 days = 35 man days Remaining work is 49 man days and this work is to be done by 5 people Hence, days required = 49/5 = 9.8 days Hence, option 4.

*Answer can only contain numeric values

QUESTION: 23

If in the equation ax^{2} + bx + c = 0, a, b and c are in an A.R and sum of the roots is 5 greater than product of the roots, find the modulus of harmonic mean of the roots.

Solution:

Since, a, b and c are in an A.R

2 b = c + a

∴ 2b / a = c / a + 1...(i)

Also,

∴ -b / a - c / a = 5 ...(ii)

Solving (i) and (ii) we get,

—bla = 4/3 and da = - 11/3

The harmonic mean of roots = 2 x product of roots / Sum of the roots = -11 / 2 = -5.5

Answer: 5.5

QUESTION: 24

Sum of 10 consecutive even numbers is 330. Find the sum of the two middle terms.

Solution:

Sum of the 10 consecutive even numbers is 330.

Average = 33

This average will lie between the 5^{th} and 6^{th }term of the series.

As they are consecutive even numbers, they must be 32 and 34.

Required sum = 32 + 34 = 66

Hence, option 3.

QUESTION: 25

A number 242a27b is divisible by 18, what is the value of (a, b)?

Solution:

As the number is divisible by 18, it is a even number divisible by 9.

For a number to be divisible by 9, sum of the digits should be a multiple of 9.

Sum of digits = 2 + 4 + 2 + a + 2 + 7 + b=17+ a + b

a + b can either be 1 or 10.

Considering the options only valid option is (4, 6).

Hence, option 2.

QUESTION: 26

What is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

Solution:

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.

Length of largest tile can be obtained from HCF of length and breadth.

So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)

= 41 cm

Required number of tiles =

Area of floor / Area of tile

= (1517 x 902 / 41 x 41)

= 814

*Answer can only contain numeric values

QUESTION: 27

How many solutions does the inequalty x^{2} < - 16 hold, if x is a real number?

Solution:

For a real x, x^{2} is greater than zero.

x^{2} < -16 has no solution.

Answer: 0

*Answer can only contain numeric values

QUESTION: 28

Find the selling price (in Rs.) of a sari after successive discounts of 20%, 25% and 30% on the marked price of Rs. 700.

Solution:

Selling price after successive discount = 700 x 0.8 x 0.75 x 0.7 = Rs. 294

Answer: 294

*Answer can only contain numeric values

QUESTION: 29

The length and the b readth of a rectangle are in the ratio 7 : 5 . W hen the length of the rectangle is increased by 5 units and the breadth is decreased by 3 units the area decreases by 4 sq. units. Find the perimeter of the original rectangle.

Solution:

Let the length and the breadth be lx and 5x respectively.

The new area of the rectangle = (7x + 5) x (5x - 3) = 35x^{2} + 4x - 15

From the given condition, 35x^{2} + 4x - 15 = 35x^{2 }- 4

x = 2.75 units

The p erim eter o f the rectangle =12x2x2.75 units = 66 units

Answer: 66

*Answer can only contain numeric values

QUESTION: 30

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

Solution:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

QUESTION: 31

Area of an equilateral triangle is 8√3 cm^{2} . A circle passes through the incentre of the triangle such that one side of the triangle is chord to the circle such that the other two sides of the triangle are tangent to the circle. Find the area of the circle.

Solution:

I is the incentre => mzBIC = 120°

m(arc BDC) = 240° ... (Inscribed angle theorem)

So, m(arc BIC) = 120° => mzBOC = 120° Therefore, mZlOC = mzOIC = 60°

Thus, ΔIOC is an equilateral triangle.

ΔIEC is 30°-60°-90° triangle.

Let IO = IC = OC = r

∴ EC = (r√3) / 2

∴ AC = (r√3)

∴ (√3 / 4) x (r√3)^{2} = 8√3

r^{2} = 32/3

πr^{2} = 32π/3

Hence, option 2.

*Answer can only contain numeric values

QUESTION: 32

A chocolate selling firm manufactures one chocolate for Rs. 20 and sells them in a box of chocolates at Rs. 960 per box such that each box contains 40 chocolates. If the firm now increases the quantity of chocolates per box by 15%, find the increase (in Rs.) in the selling price per box, so that the percent profit increases by 5 percent points.

Solution:

The cost price per box = 40 x 20 = Rs. 800 The profit percentage = (160/800) x 100 = 20%

The required profit percentage = 25%

The number of chocolates now = 1.15 x 40 = 46

The cost price per box = 46 x 20 = Rs. 920 Selling price per box = 920 x 1.25 = Rs. 1150

Hence, the increase in the selling price per box = 1150- 960 = Rs. 190

Answer: 190

QUESTION: 33

A train 440 metres long passes a railway platform 876 metres long in 10 seconds. How many seconds will it take to pass a running train 990 metres long?

Solution:

As we do not know in which direction the other train is running.

Hence cannot determine the time taken.

Hence, option 4.

QUESTION: 34

If the angle formed by the diagonal (passing through the centre of the polygon) with one of the sides of a regular polygon is 7π/ 16, find the number of distinct lines which can be drawn between all the vertices of that polygon.

Solution:

Since, the polygon is regular the diagonal passing through the centre of the polygon will bisect the internal angle.

The angle formed at the centre of the triangle = π-(2 x (7π/16)) = π/8

Hence, the given polygon has 16 sides.

The number of distinct lines between sixteen vertices = ^{16}C_{2} =120

Hence, option 3

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