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Test: CAT Quantitative Aptitude- 8 - CAT MCQ


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22 Questions MCQ Test CAT Mock Test Series 2024 - Test: CAT Quantitative Aptitude- 8

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Test: CAT Quantitative Aptitude- 8 - Question 1

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. 

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 1

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
Test: CAT Quantitative Aptitude- 8 - Question 2

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?


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 2

Let's 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. The 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 = 10

=>  x = 5/32

ie. The filling capacity of  tap B per hour

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

=> capacity of tank/ filling capacity of B per hour  =  32/5 hours 

=>  32/5 * 60  

=> 384 min

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*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 3

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


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 3

(35a246772)9

= (3 x 98) + (5 x 97) + (a x 96) + ... + (2 x 90)

= [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

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 4

Harry forgot his 7 digit driving licence number which he wanted to write. However, he remembered that the digit 1 appeared exactly once in the number and the remaining 6 digits were 3 non-zero digits, each appearing exactly twice. What is the minimum number of different numbers that Harry would have to write, to be certain of writing the right number?


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 4

The number of ways of choosing the 3 non-zero and non-unity digits out of 8 digits = 8C3 = 56
These can be arranged in (6!)/(2! × 2! × 2!) = 90 ways and the digit 1 can be placed in any of the 7 possible locations with respect to the 6 digits (7 × 56 × 90 = 35,280).
There are 35,280 trials that are needed.

Test: CAT Quantitative Aptitude- 8 - Question 5

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

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 5

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.

Test: CAT Quantitative Aptitude- 8 - Question 6

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?

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 6

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

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 7

In the figure, ABCDEF is a regular hexagon and ABPQR is a regular pentagon. The side AB is common. Calculate AFR (in degrees).


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 7

Interior angle of an n-sided regular polygon = [(n - 2) × 180]/n

BAF = (6 - 2) × 180° ×  = 120°
∠BAR = (5 - 2) × 180° ×  = 108°
∠FAR + ∠BAF + ∠BAR = 360° (angles around the point)

 ∠FAR = 360° - 120° - 108° = 132°

As m(AF) and m(AR) are equal, angle opposite to the equal side is equal.

 ∠AFR =  = 24°
Therefore, 24 is the correct answer.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 8

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?


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 8

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

Test: CAT Quantitative Aptitude- 8 - Question 9

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

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 9

For a triangle to be formed, 7 < x < 17 
Case I: x ≥ 12 i.e. x is the largest side
x2 < 132, 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 
x2 > 122 - 52, 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.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 10

In a 100 km car race, Car-1 can beat Car-2 by 10 km and Car-3 by 20 km. In a certain 500 km race, if Car-1 starts from a point 50 km behind the starting point, Car-2 starts from the starting point and Car-3 starts 50 km ahead of the starting point, then which of the following cars shall win the race?
(Enter your answer as 1 for Car-1, 2 for Car-2, and so on.)


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 10

According to the first race, in the time Car-1 covers a distance of 100 km, Car-2 covers a distance of 90 km and Car-3 covers a distance of 80 km.
So, in the time Car-1 covers a distance of 1 km, Car-2 covers a distance of 0.9 km and Car-3 covers a distance of 0.80 km.
Now, for the second race, Car-1 has to cover a distance of 550 km, Car-2 has to cover a distance of 500 km and Car-3 has to cover a distance of 450 km.
In the time Car-1 covers a distance of 550 km, Car-2 covers a distance of 0.9 × 550 km = 495 km and Car-3 covers a distance of 0.80 × 550 km = 440 km
Thus, in the second race, it is Car-1 that wins the race.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 11

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.


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 11

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

Test: CAT Quantitative Aptitude- 8 - Question 12

Mitali has 2 watches one is Polex and the other is Tastfrack. At 7:00 am on Friday, 1st 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?

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 12

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 2nd January 2016.
Hence, option 1.

Test: CAT Quantitative Aptitude- 8 - Question 13

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

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 13

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.

Test: CAT Quantitative Aptitude- 8 - Question 14

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). 

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 14

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.

Test: CAT Quantitative Aptitude- 8 - Question 15

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

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 15

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.

Test: CAT Quantitative Aptitude- 8 - Question 16

Meena invested Rs. 5,000 in a scheme offering 8% interest p.a. compounded annually for two years while Seema invested some amount in another scheme offering 12% simple interest for two years. If the interest received by Seema after 2 years was 30.77% more than that received by Meena, then find the approximate amount invested by Seema in the scheme.

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 16

Interest received by Meena after 2 years = 5,000(1.08)2 - 5,000 = Rs. 832
Suppose Seema invested Rs. y in the investment scheme.
According to the given condition:
0.24y = 1.3077 × 832
y = Rs. 4,533 (approx.)

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 17

How many solutions does the inequalty x2 < - 16 hold, if x is a real number?


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 17

For a real x, x2 is greater than zero.

x2 < -16 has no solution.
Answer: 0

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 18

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


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 18

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

Test: CAT Quantitative Aptitude- 8 - Question 19

Area of an equilateral triangle is 8√3 cm2 . 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. 

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 19

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
r2 = 32/3
πr2 = 32π/3 
Hence, option 2.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 8 - Question 20

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.


Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 20

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

Test: CAT Quantitative Aptitude- 8 - Question 21

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?

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 21

As we do not know in which direction the other train is running.
Hence cannot determine the time taken.
Hence, option 4.

Test: CAT Quantitative Aptitude- 8 - Question 22

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.

Detailed Solution for Test: CAT Quantitative Aptitude- 8 - Question 22

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 = 16C2 =120 
Hence, option 3

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