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Test: CAT Quantitative Aptitude- 7 - CAT MCQ


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22 Questions MCQ Test CAT Mock Test Series 2024 - Test: CAT Quantitative Aptitude- 7

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Test: CAT Quantitative Aptitude- 7 - Question 1

The question is followed by two statements, I and II. Answer each question using the following instructions: Mark (1) if the question can be answered by using statement I alone but not by using statement II alone.
Mark (2) if the question can be answered by using statement II alone but not by using statement I alone.
Mark (3) if the question can be answered by using both the statements together.
Mark (4) if the question can be answered by using either of the statements.
Is xy < 5?
I. x < 3 andy < 3 II. 1/2 < x < 2/3 and y2< 16

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 1

Statement I alone is not sufficient as xy may or may not be less than 5.
Statement II restricts x to 1/2 < x < 2/3 and y to the interval - 4 < y< 4 .

Thus, largest possible value of xy would be less than 8/3, which is less than 5.

Thus, Statement II is sufficient to answer.
Hence, option 2.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 2

Find the sum of all the natural numbers between 1 and 100, which are divisible by 3.


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 2

Numbers between 1 and 100 divisible by 3 are 3, 6, 9, …, 99.
Sum = 3 + 6 + 9 + 12 + ... + 99
Sum = 3(1 + 2 + 3 + 4 + ... + 33)
Now, using the formula,
Sum of first n natural numbers = 

1 + 2 + 3 + 4 + ... + 33 =  = 561

Sum = 3(561) = 1683

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Test: CAT Quantitative Aptitude- 7 - Question 3

Krunal played 11 matches in Indian Cricket T-20 championship. Average runs scored by him in 11 matches was 16. His runs scored in nth match was 2 less than that of the (n - l)th match. Find the average runs scored by him in third and ninth match.

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 3

Runs scored by him forms A.R with common difference = 2.
He played 11 matches.

∴  Runs scored by him in 6th match = Average runs scored in 11 matches = 1 6

For an A.R with n = 11 (1st term + 11th term)/2 = average Similarly,
(2nd term + 10th term) / 2 = average and so on...
∴ Average runs socred in the 3rd match and the 9th match = Average runs scored in 11 matches = 16
Hence, option 3.

Test: CAT Quantitative Aptitude- 7 - Question 4

A boat takes 5 hours to row downstream along the entire length of the river. If the speed of the stream increases by 1 km/hr and the speed of the boat remains the same, the boat will take 40 hours to row upstream along the entire length of the river. If the speed of boat in still water is 5 km/hr, find the length of the river (in km).

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 4

Let the speed of stream be x km/hr.

∴ Length of the river = (5 + x) x 5 = (5x + 25) km

Also, [5 - (x + 1)] x 40 = 5x + 25

Solving this we get, x = 3 km/hr

∴ Length of the river = (5 x 3 + 25) = 40 km

Answer: 40

Hence Option 2

Test: CAT Quantitative Aptitude- 7 - Question 5

A leap year starts on a Sunday. What is the probability that if a day is chosen from that year, it is a Tuesday?

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

A leap year has 366 days, i.e., 52 weeks and 2 days.
Therefore the year will have 52 Tuesdays.

∴ Required probabilty = 52/366 = 26/183

Hence, option 3.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 6

A certain shipping company has a certain free baggage allowance for each voyager. It charges for excess baggage at a fixed rate per kg. 2 voyagers, Mr. P and Mr. Q, have a total of 50 kg of baggage between them. They were charged Rs. 2,800 and Rs. 1,400, respectively for excess baggage. If the free baggage allowance were halved and the entire baggage belonged to one of them, then the excess baggage charge would have been Rs. 6,300. Find the weight (in kg) of Mr. P's baggage.


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 6

Let the free baggage allowed on board = f kg
It is seen that Mr. P was charged double the amount of Mr. Q.
Thus, as the rate per kg is the same, the excess baggage of Mr. P must be 2 times that of Mr. Q.
Let the excess baggage of Mr. Q be q kg.
Thus, the excess baggage of Mr. P = 2q kg
Let the rate charged for the excess baggage be Rs. r/kg
So, the quantity of baggage carried by Mr. P = (f + 2q) kg and that carried by Mr. Q = (f + q) kg
Now, as the total baggage carried is 50 kg, we have f + 2q + f + q = 50
Or 2f + 3q = 50 .... (i)
As Mr. Q was charged Rs. 1,400, we have
qr = 1,400 .... (ii)
Now, when 1 of them has all the baggage, the total baggage becomes (2f + 3q) kg.
As the free baggage allowed is halved, the charge becomes Rs. (1.5r + 3qr) = Rs. 6,300
Or 1.5r + 3qr = 6,300 ....(iii)
Plugging in the value of qr from equation (ii) into equation (iii), we have
1.5r + 4,200 = 6,300 .... (iv)
Or fr = 1,400 .... (v)
Now, from equations (ii) and (v), we get that
f = q .... (vi)
Plugging in the value of f as q from equation (vi) into equation (i) and solving for q, we get
q = 10
Thus, f = 10 as well.
Now, total weight of Mr. P's baggage = (f + 2q) kg
= (10 + 2 × 10) kg
= (10 + 20) kg
= 30 kg

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

Calculate the area(in sq. m) of the shaded region in the following figure:(Type your answer to the nearest integer)


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

The bigger triangle is a right triangle.
So, area = [1/2] × 120 × 22 = 1320 m[1]
For the triangle having the lengths of sides as 22 m, 24 m and 26 m,
s =  [1/2]
Area of the triangle =  [1/2]


= 245.92 m[1]
Therefore, area of the shaded region
= (1320 – 245.92) m2
 1074 m(Approx.) [1]
1074 is the nearest integer answer.

Test: CAT Quantitative Aptitude- 7 - Question 8

Forth and fifth terms of a series in geometric progression are 378 and 1134 respectively. What is the first term of the series?

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

Let the first term and common ratio be a and r respectively.
Forth term = ar3 = 378

Similarly, Fifth term = ar4 =1134 Now,

r = ar4lar3 = 1134/378 = 3

∴ a =1134/81 = 14

Answer: 14

Hence Option 2

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 9

In an examination, Jack scored 55 marks and Jill scored 30 more marks than Jack in the same examination. For every correct answer, a student gets 1 mark, but the negative marks per wrong answer for the first 20 wrong answers is different from that for the remaining wrong answers. Jack and Jill attempted 160 and 150 questions, respectively. If Jack and Jill correctly answered half and two third of the questions that they attended, respectively, then find the negative mark for each wrong answer beyond the first 20 wrong answers.
(Enter your answer rounded to two decimal places)


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 9

Let the negative marking per mistake for the first 20 mistakes be n1 and for all the subsequent mistakes, let it be n2.
Jack attempted 160 questions and got only 80 correct and Jill attempted 150 questions and got only 100 correct.
So, 80(1) - 20(n1) - 60(n2) = 55
100(1) - 20(n1) - 30(n2) = 85
20 + 30n= 30
n2 = 1/3 = 0.33

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 10

Each angle of a regular r-sided polygon is  times each angle of a regular s-sided polygon. What is the largest possible value of s?


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 10

The angles are 180°and 180°.
So, 58(r – 2)s = 59r(s – 2)
Or r = 116s/(118 – s)
But r is positive; so s  117.
s = 117 (Maximum)

Test: CAT Quantitative Aptitude- 7 - Question 11

The ratio of the cost price of cookie A to that of cookie B is 2 : 5. The ratio of profits made on selling the cookies is in the ratio 4 : 5. If the selling price of cookie B is Rs. 100, find the selling price of cookie A.

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 11

Let the cost price for cookies A and B be 2x and 5x respectively and the profits for cookies A and B be 4y and 5y respectively.

Thus, for cookie B, SP = 5x + 5y = 100

∴ x + y = 20

For cookie A, SP = 2x + 4y But we do not know the value of x andy hence the selling price of cookie A cannot be determined.
Hence, option 4.

Test: CAT Quantitative Aptitude- 7 - Question 12

A man standing at the bottom of a hill wants to reach the top, which is at a distance of 10√13 km from him. He starts climbing at an angle of 30°. After sometime he finds that he is 10 km above ground level and decides to climb the remaining distance at an angle of 60°. What is the height of the hill?

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 12

DE = BF = 10 km
tan 60° = √3 = y / EF
∴ EF = y / √3 = DB
Let CF be y km.
tan 60° = √3 = y / EF
∴ EF = y / √3 = DB
AB = 10√3 + y / √3
BC = 10 +y
AC = 10√13 km
AB2+BC2 = AC2  ...(Pythagoras theorem)
(10√3 + y / √3)2 + (10 + y)2 = (10√13)2
Solving, y = 15 km

∴ Height = 10 + 15 = 25 km

Hence, option 3.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 13

Between 1 and 31, 'm' arithmetic means have been inserted in such a way that the ratio of the 7th and (m - 1)th means is 5 : 9, respectively. Find the value of m


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 13

Let A1, A2, A3, A4, …, Am be 'm' A.M.s between 1 and 31.
Therefore, 1, A1, A2, …, Am, 31 are in A.P.
Let d be the common difference of the A.P.
Here, the total number of terms is m + 2 and Tm + 2 = 31.
1 + (m + 2 - 1)d = 31
(m + 1)d = 30
d = 
A7 = T8 = a + 7d = 1 + 7  =  = 
…………………
…………………
Am - 1 = Tm = 1 + (m - 1)d
= 1 + (m - 1)  = 

 =  =  =  [Given]

 = 

9m + 1899 = 155m - 145
146m = 2044
m =  = 14
Hence, m = 14

Test: CAT Quantitative Aptitude- 7 - Question 14

Slant height of the cone is 25 cm. Diameter of the cone is 3/2 times the height of the cone. Find the volume of the cone (in cm3).

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 14

Let x be the height of the cone.
∴ Diameter = 3x/2

∴ Radius = 3x/4

So we get,
252 = x2 + (3x/4)2

∴ x = 20 cm

∴ Radius of the cone = (3 x 20)/4 = 15 cm

Volume of the cone = (1/3) x π x 152 x 20 = 1500π cm3 
Hence, option 1.

Test: CAT Quantitative Aptitude- 7 - Question 15

Each question is followed by two statements, I and II. Answer each question 3 using the following instructions: Marks Mark (1) if the question can be answered by using statement I alone but not by using statement II alone.
Mark (2) if the question can be answered by using statement II alone but not by using statement I alone.
Mark (3) if the question can be answered by using either of the statements alone.
Mark (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark (5) if the question cannot be answered by using any of the statements. Find the area of the triangle. (Enter the correct answer option.)

I. The measures of the sides of the triangle are in arithmetic progression.

II. The ratio of the sides is 2 : 3 : 4.

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 15

The question cannot be answered using either of the statements alone. We need not combine the statements as statement II implies statement I.
Answer: 5

Hence Option 5

Test: CAT Quantitative Aptitude- 7 - Question 16

For which of the following functions f is f(x) = - f (1 +x) satisfies for all x?

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 16

Let x = 4.
So we get,

f(4) = -f(5)
Now we will try the options.
Option 1: f(x) = 1 - x ⇒ f(4) = 1 - 4 = - 3 and - f(5) = -[1 - 5] = 4 As, f(4) ≠ - f(5)
Hence, option 1 is incorrect.
Option 2: f(x) = 1 - x2 ⇒ f(4) = 1 - (4)2 = 1 - 16 = -15 and -f(5) = -[1 - (5)2] = 24 As, f (4)  -f(5)
Hence, option 2 is incorrect.
Option 3: f(x) =x2(1 -x)2
f(4) = 1 - (4)2 = 1 - 4)2 = 16 x 9 and -f(5) = -(5)2 x [1 - (5)]2 = 25 x 16
∴ f(4) ≠ -f(5)
Hence, option 3 is incorrect.
Hence, option 4.

Test: CAT Quantitative Aptitude- 7 - Question 17

A solid cylinder is immersed in a hemispherical bowl filled with water. What is the maximum volume of the liquid (in cubic units) that can be displaced, if the height of the cylinder is equal to the radius of the hemispherical bowl (given by r) and the radius of the cylinder is one third that of the bowl if cylinder is placed vertically?

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 17

Consider the figure,

The radius of hemispherical bowl = r Radius of cylinder below the surface of water = Radius of cylinder PQRS = r/3
Height of cylinder = r

Volume of the water displaced = Volume of the cylinder immersed.

Maximum height of the cylinder that can be immersed = AB - AC = BC

Using Pythagoras theorem in ΔPOT, we get,
∴ BC = 2√2r / 3
∴ Height of cylinder below the surface of water 
= 2√2π / 3 
∴ Volume of Liquid displaced = Volume of the cylinder immersed
= π x r2 / 9 x 2√2π / 3 = 2√2πr3 / 27
Hence, option 2.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 18

If  and  are different complex numbers with  = 1, find the value of .


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 18


Test: CAT Quantitative Aptitude- 7 - Question 19

A shopkeeper buys certain goods at 20% discount from the wholesaler. The shopkeeper then marks a price on the goods that would fetch him 50% profit. What percentage discount should he offer on this marked price so that he sells the goods at the marked price of the wholesaler (i.e., the price at which the wholesaler usually sells his goods) ?

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 19

Let the marked price of the wholesaler be Rs. x.
After 20% discount, shopkeeper buys the goods at Rs. 0.8x To fetch 50% profit,
Shopkeeper marks the goods at 0.8x + 0.4x = Rs. 1.2x
To sell the goods at the marked price of wholesaler, shopkeeper should offer
∴ discount of 0.2x Discount percentage = (0.2x/1.2x) x 100 = 16.67%
Hence, option 2.

Test: CAT Quantitative Aptitude- 7 - Question 20

In a class of 120 students, for every 2 students taking up Physics there are 4 students taking up Chemistry. Further for every student taking Physics there are 3 students taking up Maths as well as 1 student taking up Biology. Then the maximum number of students taking up all the four subjects will be:

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 20

Please refer to the Venn diagram. Here P stands for Physics, C stands for Chemistry, M stands for Maths and B stands for Biology.

We can write the given statements in form of ratios as following:
P C M B
2x 4x
1x 2x 3x 1x

∴The ratio of students taking up P:C:M:B will be 1:2:3:1.

Since we are supposed to maximize the number of students taking up all subjects, place lx the minimum of them in the intersection area of all four subjects. After placing the remaining students we get the values for different areas as given in the Venn diagram.

∴The total number of students will be 3x = 120

∴ x = 40
Hence, option 2.

Test: CAT Quantitative Aptitude- 7 - Question 21

Everday, Murugan starts at 4:00 pm from his home to pick up his son from school. They reach house at 6:00 pm. One day, school was over at 4:00 pm and Murugan, not aware of this, started from home as usual. He met his son on the way and they reached home 20 minutes earlier than usual. If the speed of the car is 30 km/hr, find his son's speed (in km/hr).

Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 21

Murugan takes two hours to complete the journey in which he travels from home to school and back.

∴ Car takes 1 hour to travel from home to school and vice versa.
When son starts walking towards home, 20 minutes saved in whole journey i.e., 10 minutes journey from home to school and vice versa.

∴ Distance travelled by son = Distance travelled by car in 10 minutes.

As car takes 1 hour to reach from home to school of which 10 minutes distance is covered by son, car travels for 50 minutes from home to school.

∴ Son walks for 50 minutes.

Distance from school to car = 5 km

Speed of the boy = 6 km/hr

Hence, option 2.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 7 - Question 22

A lattice point (x, y) in the rectangular coordinate plane is a point in which both x and y are integers. How many lattice points are there in the interior of the circle x2 + y2 = 11?


Detailed Solution for Test: CAT Quantitative Aptitude- 7 - Question 22

If a point (p, q) lies inside the circle x2 + y2 = 11, then p2 + q2 < 11.
If p = 0, q can be –3, –2, –1, 0, 1, 2 or 3.
If p = 1 or –1, q can be –3, –2, –1, 0, 1, 2 or 3.
If p = 2 or –2, q can be –2, –1, 0, 1 or 2.
If p = 3 or –3, q can be –1, 0 or 1.
So, total = 7 + 14 + 10 + 6 = 37

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