Test: CAT Quantitative Aptitude- 5 - CAT MCQ

# Test: CAT Quantitative Aptitude- 5 - CAT MCQ

Test Description

## 22 Questions MCQ Test Quantitative Aptitude (Quant) - Test: CAT Quantitative Aptitude- 5

Test: CAT Quantitative Aptitude- 5 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Test: CAT Quantitative Aptitude- 5 questions and answers have been prepared according to the CAT exam syllabus.The Test: CAT Quantitative Aptitude- 5 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: CAT Quantitative Aptitude- 5 below.
Solutions of Test: CAT Quantitative Aptitude- 5 questions in English are available as part of our Quantitative Aptitude (Quant) for CAT & Test: CAT Quantitative Aptitude- 5 solutions in Hindi for Quantitative Aptitude (Quant) course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt Test: CAT Quantitative Aptitude- 5 | 22 questions in 40 minutes | Mock test for CAT preparation | Free important questions MCQ to study Quantitative Aptitude (Quant) for CAT Exam | Download free PDF with solutions
Test: CAT Quantitative Aptitude- 5 - Question 1

### If the interest compounded annually on a certain sum at a certain rate of interest for 2 years is equal to 40% of the simple interest on the same amount at same interest rate for thrice the time period, find the rate percentage.

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 1

Let P be the sum and r be the rate of interest (expressed as a decimal fraction).
We have,
P (1 + r)2 - P = 0.40(6Pr)
P (1 + 2r + r2 - 1) = 2.4Pr
2r + r2 = 2.4r
r2 = 0.4r
r = 0.4 i.e., 40%
Hence, option 3.

Test: CAT Quantitative Aptitude- 5 - Question 2

### Jagan, the video dealer, lends at most 10 DVDs at the cost of Rs. 300 per day and 12 CDs at the cost of Rs. 200 per day. Direct costs are Rs. 50 per day for DVDs and Rs. 30 per day for CDs. Also because some of the customers mishandled the products he had to spend Rs. 5000 on maintenance. In the month of March due to exams DVDs and CDs lending rate was 30% and 40% respectively, which kept on increasing gradually by 5% and 8% every month (i.e. rate became 35% and 48% respectively the next month). What is the profit for the month of May if his fixed monthly cost is Rs. 40,000?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 2

Fixed monthly cost = Rs. 40,000
First we need to compute the Total Revenue.
Revenue made by lending 10 DVDs in May = 31 x 10 x 300 = Rs. 93,000
Revenue made by lending 12 CDs in May = 31 x 12 x 200 = Rs. 74,400
But then the lending rate in May is 40% (it is gradually increasing by 5%)

= Rs. 37,200
Similarly, the lending rate in May is 56% (it is gradually increasing by 8%)

= Rs. 41,664
Total direct cost of a DVD = 50x31 x 10 = Rs. 15,500
Total direct cost of a CD = 30 x 31 x 12 = Rs. 11,160
Maintenance cost = Rs.5,000
The revenue earned from all these products = (Actual money made by lending DVDs and CDs) - (direct cost of DVDs and CDs) - (Maintenance)
= Rs. 37,200 + Rs. 41,664 - Rs. 15,500 - Rs. 11,160 - Rs. 5,000
= Rs. 47,204
Profit or loss = revenue - expenses
= Rs. 47 ,204 - Rs . 40,000
= Rs. 7,204
This is positive that means his revenue is more than expense.
We have a profit of Rs. 7,204.
Hence, option 2.

 1 Crore+ students have signed up on EduRev. Have you?
Test: CAT Quantitative Aptitude- 5 - Question 3

### A group of 10 students started playing pass-the-bucket. They stood in a row and the 1st student passed a bucket with 10 litres of milk. While passing it to the 2nd student in the row he spilled 5/3 litres. The 2nd student compensated for the lost volume by adding same amount of water and spilled 5/3 litres while passing the bucket to the 3rd student. This continued till the bucket was passed to the 10th student who replaced the volume of the spilled solution with an equal volume of water. What was the percentage of milk in the bucket when it was in the hands of the 5th student just before he handed it over to the 6th student?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 3

When the bucket reached the 5th student, 5/3rd solution in the bucket has been replaced by water 4 times.
Also, when the bucket reaches the 5th student 5/3rd of the solution in the bucket has been spilled.
After 4 replacements, amount of milk in solution

When the bucket is passed to the 5th student 5/3 litres of the solution is spilled.
1/6th of the solution is spilled.
1/6th of the milk is spilled.
As equal percentages of milk and water are spilled, the percentage of milk in the bucket is
4.8/10 x 100 = 48%
Percentage of milk in the bucket when it was passed to the 5th student = 48%
Hence, option 3.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 4

Robin has a silver ball and a platinum ball each of radii 4 cm. The worth of each ball is Rs. 50,000. The worth of silver doubles every 2 years, while that of platinum remains the same. At the end of 6 years, Robin sells the silver ball and invests the money in a new platinum ball. The worth of a platinum ball is directly proportional to its volume. How much should the radius (in cm) of the new platinum ball be?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 4

The initial worth of the ball was Rs. 50,000. The worth of silver doubles every 2 years.
Worth of the silver ball at the end of 6 years = Rs. (50,000 x 8) = Rs. 4,00,000
Robin invests Rs. 4,00,000 in a new platinum ball and the worth of a platinum ball is directly proportional to its volume.

r2 = 8 cm

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 5

Ten years ago, the Edwards had 8 members and the average of their ages then was 33 years. Four years later, Mrs. Edward died at the age of 64 years and William Edward was born. After three more years, Mr. Edward died, at the age of 72 years and Liz Edward was born. Find the present average age of the family (in years).

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

Sum of the ages of the family 10 years ago = 33 × 8 = 264 years
After another 4 years, sum of their ages = 264 + (4 × 8) = 296 years
As Mrs. Edwards then died at the age of 64 years and William was born, sum of their ages would be 232 years but the number of members would still be 8.
Similarly, after 3 more years, the sum of the ages of the family members would be 184 years (232 +( 3 × 8) - 72 + 0) and the number of members would still be 8.
Presently i.e. after 3 more years, the sum of the ages of the members would be 208 years and the number of family members would still be 8.
Thus, the present average age of the Edwards = 208/8 = 26 years

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 6

Samrat covers the total journey of 810 km to his village partly by train and then by bus. The train is 20% faster than the bus. If the train were to travel for the number of hours, which normally would be hours travelled by bus, and the bus were to travel the remaining distance, then the time for the journey would reduce by 24 minutes. The normal travelling hours of bus and train are in the ratio 3:2. What is the speed(in km/hr) of the bus?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 6

Let the speed of the bus be b km/hr
The speed of the train = 12b km/hr
Let the bus normally travel for 3x hours and let the train normally travel for 2x hours.
3xb + 2.4xb = 810
xb = 150
If the train travels for the number of hours that the bus travels,

5.6xb - 810 = 0.4b
5.6(150) -810 = 0.4b
b = 75
The speed of the bus = 75 km/hr

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 7

If find the value of

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 7

Thus, f(x) + f(1 - x) =
Therefore,

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 8

What is the number of solutions of the equation

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

Only one solution is possible.

Test: CAT Quantitative Aptitude- 5 - Question 9

If x3 - x - 1 = 0, then (x - 1) is equal to

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 9

Consider the equation, x3 - x - 1 = 0,
x3 = x + 1 ...(i)
On multiplying the equation by x and x2 to get,
x4 = x2 + x . . . (ii)
x5 = x3 + x2 . ..(iii)
x5 = x3 + x2 = (x + 1) + x2 ... [Using (i)]
= (x + x2) + 1
= x4 + 1 ... [Using (ii)]

x5 - x4 = 1

Hence, option 4.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 10

How many three digit numbers exist such that their value is decreased by 99% when their digits are reversed?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 10

The value of the three digit number decreases by 99% when its digits are reversed, we can see that the number with reversed digits must be 0.01 times the original three digit number.
The number formed on reversal of the digits must be a single digit number
(The number thus formed would become more than three digits on multiplication by 100).
The last two digits of the original three digit number must be zero, and the first digit must be a nonzero digit.
There are 9 such numbers: 100, 200, 300,..., 900.

Test: CAT Quantitative Aptitude- 5 - Question 11

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 11

We have 1 + 2x > 0

Also we have

Consider the given inequality

4(1 + 2x) < 49

Hence, option 3.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 12

Given a cube, I pick three random points from among its vertices. Let N be the number of ways in which three points can be selected such that they form a right angled triangle. What is N?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 12

The number of ways in which three points can be selected out of the eight vertices of a cube is 8C3 = 56
We can see that in most of these 56 ways, the three points selected form a right angled triangle. So, we look at possible selections of three points which would not give a right angled triangle.
We can see that if two of these three points are the endpoints of one of the edges of the cube, then irrespective of which of the remaining vertices is the third point, we always get a right angle. If two of the three points are endpoints of a body diagonal of the cube, then the third point will always be such that it will form an edge of the cube with one of the two points, so we again get a right angle.
So, the sides of the triangle formed by the three points should neither be edges nor the body diagonals of the cube, i.e. all three sides should be face diagonals.
To count the number of such triangles, consider the 'top' face ABCD of a cube ABCDWXYZ (the points corresponding to A, B, C and D on the bottom face are W, X, Y and Z respectively). It has two face diagonals, AC and BD. With AC as one of the sides, we can have two acute triangles, ΔACX and ΔACZ. With BD as one of the sides, we can have two acute triangles, ΔBDY and ΔBDW.
Similarly, WXYZ has two face diagonals, WY and XZ, from which we get four acute triangles, ΔBWY, ΔDWY, ΔAXZ and ΔCXZ. Thus, 8 triangles out of 56 are not right angled triangles.
56 - 8 = 48

Test: CAT Quantitative Aptitude- 5 - Question 13

A circle with centre O has radius 12 cm. AD is the diameter and AB is the tangent to the circle at point A.

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 13

Let BC = x, BD = 4x
CD = BD - BC = 3x
Also AB2 = BC x BD = 4x2
AB = 2x
Now , in right angled  ΔABD
(4x)2 = (2x)2 + (24)2
16x2 = 4x2 + (24)2
12x2 = (24)2
x2 = 48 = 16 x 3

CD = 3x

Hence, option 3.

Test: CAT Quantitative Aptitude- 5 - Question 14

If a, b and c are the sides of a triangle, then which of the following is true?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 14

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
In the above identity if we replace c by -c, we get,
a3 + b3 + c3 - 3abc = (a + b - c)(a2 + b2 + c2 - ab + bc + ca) ... (i)
By triangle inequality we know that
a + b> c or (a + b - c) > 0 ...(ii)
Also we can write,

Combining (ii) and (iii) and replacing in (i), we get,

Hence, option 4.

Test: CAT Quantitative Aptitude- 5 - Question 15

The difference in the areas of circumscribing and inscribing circles of the region bounded by the curves y = -|x| + 12 andy = |x| + 4 will be:

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 15

As shown in the graph, we get the intersection points (-4, 8), (4, 8), (0, 12) and (0, 4).
The enclosed region would be a square with each side of length of 4√2
The radius of the inscribed circle = 2√2
It's area = 8π
The radius of the circle circumscribing the square will be 4.
It’s area = 16π
The difference between these two areas = 16π - 8π = 8π
Hence, option 2.

Test: CAT Quantitative Aptitude- 5 - Question 16

If we draw graphs of the following equations using standard conventions, for how many of the graphs will a line parallel to the y-axis intersect more than once?

i. xy = 4
ii. y = tan(x)
iii. x2 +y2 = 8
iv. x = y2
v. |x| + |y| = 2

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 16

i.xy = 4

∴ A line parallel to the y-axis can intersect the graph o f xy = 4 at only one point.
ii. y = tan(x)

∴ A line parallel to the y -axis can intersect the graph o f y = tan(x) at only one point.
iii. x2 + y2 = 4

∴ A line parallel to the y-axis can intersect the graph of x2 + y2 = 4 twice.
iv. x = y2

∴ A line parallel to the y -a x is can intersect the graph o f x = y2 twice.

v. |x|+ |y| = 2

∴ A line parallel to the 7-axis can intersect the graph of |x| + |y| = 2 twice.
Hence, option 3.
Alternatively,
A line parallel to the y-axis intersects the graph of an equation twice if for a given value of x, there can be two values of y.
Consider the given equations:
i. xy = 4
∴ y = 4/x
For a given value of x, y can take only one value.
A line parallel to the y-axis intersects the graph of xy = 4 only once.
ii. y = tan x
For a given value of x, y can take only one value.
iii. x2 + y2 = 8 .
∴ y2 = 8 - x2
Clearly, for a given value of x, y can take two values.
∴ A line parallel to the y-axis intersects the graph of x2 + y2 = 8 twice.
iv. x = y2
Again, here it is clear that for a given value of x, y can take two values.
v. |x| + [y| = 2
∴ |y| = 2 - |x|
For every value of |x|, y can take a positive and a negative value.
Thus we can see that a line parallel to the y-axis intersects the graphs of (iii), (iv) and (v) twice.
Hence, option 3.

Test: CAT Quantitative Aptitude- 5 - Question 17

People who like chocolate and ice-cream is 25% of people who like only chocolate. People who like ice-cream are 66.66% of people who don’t like chocolate. People who like only ice-cream is 90% of people who don’t like ice-cream. What is the ratio of people who like both to none?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 17

Let people who:

like ice-cream only = Do not like chocolate = x
like chocolate only = Do not like ice-cream = y
like neither chocolate nor ice-cream = z
From conditions given in the question,
People who like both ice-cream and chocolate = 0.25y
x + 0.25y = 0.66(x + z) ...(i)
x = 0.9 (y + z) ... (ii)
Solving (i) and (ii),
y/z = 2/3
The ratio of people who like both to none = 0.25y : z = 1 : 6
Hence, option 2.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 18

A student has to attempt exactly 10 questions out of 15 in an exam ensuring that at least 3 out of the first 5 questions are attempted. In how many ways can the questions be selected?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 18

Three questions are attempted from the first five.
These can be selected in 5C3 = 10 ways
Also the remaining 7 questions can be attempted from the next 10 in 10C7 = 120 ways
Therefore the total number of ways in which questions can be selected in the exam, if three questions are attempted from the first five questions, is 10 x 120 = 1200

Case II:

Four questions are attempted from the first five.
These can be attempted in 5C4 = 5 ways
Also the remaining 6 questions can be attempted from the next 10 in 10C6 = 210 ways
Therefore the total number of ways in which questions can be selected in the exam, if four questions are attempted from the first five questions, is 5 x 210 = 1050

Case III:

All five questions are attempted from the first five.
These can be attempted in 5C5 = 1 way
Also the remaining 5 questions can be attempted from the next 10 in 10C5 = 252 ways
Therefore the total number of ways in which questions can be selected in the exam, if five questions are attempted from the first five questions is 1 x 252 = 252
Therefore the total number of ways of selecting questions in the exam is 1200+ 1050 + 252 = 2502

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 19

A potter makes more than 120 but less than 300 pots and arranges them in rows, with each row consisting of the same number of pots, for drying. He finds that if he places 6 pots more per row, he can arrange the pots in 10 less rows. How many pots does the potter make? Key in the value.

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 19

Suppose the potter places m pots in 1 row and makes a total of n rows, then the total number of pots he makes is mn, where: 120 < mn < 300

According to the question, mn = (m + 6) ( n - 10)
Solving m =
Since m cannot be a non-integral value, n must be a multiple of 5.
Also, n > 10 because if n ≤ 10, then m is negative or 0 (at n = 10).
Putting n = 15, we get m = 3 and mn = 45 which is less than 120. Hence, it is incorrect.
Putting n = 20, we get m = 6 and mn = 120 which is equal to 120. Hence, it is incorrect.
Putting n = 25, we get m = 9 and mn = 225 which is between 120 and 300. Hence, it is correct.
Putting n = 30, we get m = 12 and mn = 360 which is beyond 300. Hence, it is incorrect.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 20

The arithmetic mean of a set of 20 observations is X. When three of the twenty observations, 12, 19 and 74 are discarded, the mean of the new set remains unchanged. What is the arithmetic mean, X?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 20

Sum of the original set of 20 observations = 20 x X

Sum of the new set of 17 observations = 17 x X
Now, the new set is formed by discarding three of the elements of the original set. 20X- (12 +19 + 74) = 17X
3X= 105
X = 35

Alternatively,

As the mean of the original set does not change when the three numbers are discarded, hence, the mean of the original set is same as the mean of the three discarded numbers.
The mean of the three discarded numbers is (19 + 74 + 12)/3 = 35
X= 35

Test: CAT Quantitative Aptitude- 5 - Question 21

In a company three projects were assigned to all the members of a company. Following information is also known: 12 members were selected for the entire three projects.
The ratio for number of members selected for only project A : only project B : only project C is 3 : 4 : 5.
The ratio for number of members selected for project B and C only : project C and A only : project A and B only is 2 : 3 : 4.

Q. If the ratio of sum of number of members for only A, only B, only C to the sum of members for A and B only, B and C only, A and C only is 20/7 and the ratio of sum of number of members of A and B only, B and C only, A  and C only to the members selected for all the three projects is 21/4, then find the number of members selected for only project C.

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 21

According to the diagram in the previous question,

∴ Number of members selected for only C = 5x = 15 x 5 = 75

Hence, option 4.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 5 - Question 22

Group Question

Answer the following question based on the information given below.

A and B together can complete some work in 24 days. If A, B and C work together, they can complete the same work in 18 days. B and C both have the same efficiency.

Q. , B and C start working together from the very first day. If A works for 8 days, B works for 16 days and C works till the end of the given work, how many days are required to complete the given work?

Detailed Solution for Test: CAT Quantitative Aptitude- 5 - Question 22

Let A require ‘a’ days, B require ‘b’ days and C require ‘c’ days to complete the work separately.
A and B together both can complete a work in 24 days

A, B and C all three can complete a work in 18 days.

Solving equation (i) and (ii), we get,
∴ c = 72 days
∴ C requires 72 days to complete the work separately.
So, B requires 72 days to complete the work separately.
Putting this value of b = 72 in equation (i), we get,
a = 36 days

∴ The number of days required to complete the work = 16 + 24 = 40 days
Alternatively,
Let the total amount of work to be done be the L.C.M. of 24 and 18 i.e. 72 units.

Let A, B and C complete a, b and c units of work per day respectively when working alone.
∴ a + b = 72/24 = 3 units/day
and a + b + c = 72/18 = 4 units/day
∴ c = 4 - 3 = 1 unit/day
∴  b =1 unit/day and a = 2 units/day
Now, A, B and C work together for the first 8 days.
In this time frame, they complete 8(a + b + c) = 8(1 + 1 + 2) = 32 units of work.
Then, B and C work together for the next 8 days.
So, in this period, they complete 8(b + c) = 8(1 + 1) = 16 units o f work.
Thus, by the time, C starts working alone, 32 + 16 = 48 units of work has been completed.
Now, C completes the remaining 24 units of work alone by completing 1 unit per day.
Thus, C takes 24 days to complete the remaining work alone.
So, total time taken = 8 + 8 + 24 = 40 days.

## Quantitative Aptitude (Quant)

183 videos|150 docs|113 tests
Information about Test: CAT Quantitative Aptitude- 5 Page
In this test you can find the Exam questions for Test: CAT Quantitative Aptitude- 5 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: CAT Quantitative Aptitude- 5, EduRev gives you an ample number of Online tests for practice

## Quantitative Aptitude (Quant)

183 videos|150 docs|113 tests