Quantitative Aptitude MCQ Quiz - Objective Question with Answer for Quantitative Aptitude - Download Free PDF

Last updated on Apr 20, 2025

Testbook brings Quantitative Aptitude MCQs with answers and explanations. Some tricks and tips to solve Quantitative Aptitude objective questions are also listed to help you easily crack this section. Quant is an important section of competitive exams such as Bank PO, SBI PO, RBI Assistant, UPSC, NET, Defense exams, etc. Many job recruiters include Quant in Aptitude Tests or Interviews to test a candidate's logical approach. Read this article and practice these Quantitative Aptitude Question Answers to ace your examination.

Latest Quantitative Aptitude MCQ Objective Questions

Quantitative Aptitude Question 1:

അടുത്ത സംഖ്യ ഏത്?
125, 135, 120, 130, 115, 125

  1. 110
  2. 115
  3. 120
  4. 105

Answer (Detailed Solution Below)

Option 1 : 110

Quantitative Aptitude Question 1 Detailed Solution

Solution:

Looking at the pattern of changes:

  • 135 - 125 = 10

  • 120 - 135 = -15

  • 130 - 120 = 10

  • 115 - 130 = -15

  • 125 - 115 = 10

The pattern alternates between adding 10 and subtracting 15. Therefore, the next change should be -15.

So, the next number will be:

12515=110

Answer: (A) 110

Quantitative Aptitude Question 2:

X = -, /= +, += / - = x ആയാൽ താഴെ തന്നിരിക്കുന്നവയിൽ ശരി ഏത്?

  1. 6 * 2 + 3 / 12 - 3 = 15
  2.  3 / 7 - 5 * 10 + 3 = 10
  3. 15 - 5 / 5 * 20 + 16 = 6
  4.  8 / 10 - 3 + 5 * 6 = 8

Answer (Detailed Solution Below)

Option 3 : 15 - 5 / 5 * 20 + 16 = 6

Quantitative Aptitude Question 2 Detailed Solution

Solution:

We will solve each expression by replacing the operators according to the given rule.

  1. Option (A):

    62+3/123=12+0.253=9.2515
  2. Option (B):

    3/7510+3=0.4285750+3=46.5714310
  3. Option (C):

    155/520+16=15120+16=1520+16=6
  4. Option (D):

    8/103+56=0.83+30=27.88

So, the correct answer is (C).

Answer: (C) 15 - 5 / 5 * 20 + 16 = 6

Quantitative Aptitude Question 3:

ഒരു ദിവസത്തിൽ എത്ര തവണ ഒരു ക്ലോക്കിൻ്റെ മണിക്കൂർ സൂചിയും മിനിറ്റ് സൂചിയും നേർരേഖയിൽ വരും?

  1.  44
  2. 24
  3. 22
  4. 48

Answer (Detailed Solution Below)

Option 3 : 22

Quantitative Aptitude Question 3 Detailed Solution

Solution:

The hour hand and minute hand are aligned 22 times a day. This happens because the two hands overlap every 12 hours, which gives us 11 overlaps in 12 hours, so 22 overlaps in a 24-hour period.

Answer: (C) 22

Quantitative Aptitude Question 4:

a, b, c യുടെ ശരാശരി m ആണ്. കൂടാതെ ab + bc + ca = 0 ആയാൽ a², b², c² യുടെ ശരാശരി എത്ര?

 

  1. 3m²
  2. 6m²
  3. 9m²

Answer (Detailed Solution Below)

Option 2 : 3m²

Quantitative Aptitude Question 4 Detailed Solution

Solution:

Given that the average of ab, and c is m, we have:

a+b+c3=msoa+b+c=3m

Also, given that ab+bc+ca=0, we can use the identity:

a2+b2+c2=(a+b+c)22(ab+bc+ca)

Substituting a+b+c=3m and ab+bc+ca=0:

a2+b2+c2=(3m)22(0)=9m2

Thus, the average of a2b2, and c2 is:

a2+b2+c23=9m23=3m2

Answer: (B) 3m²

Quantitative Aptitude Question 5:

ഒറ്റയാൻ ഏത്?
56, 72, 90, 110, 132, 150

  1.  72
  2.  110
  3. 132
  4. 150

Answer (Detailed Solution Below)

Option 4 : 150

Quantitative Aptitude Question 5 Detailed Solution

Solution: Looking at the pattern:

The difference between consecutive numbers is increasing by 2:

72 - 56 = 16

90 - 72 = 18

110 - 90 = 20

132 - 110 = 22

150 - 132 = 18 (which is different)

The odd number here is 150, which breaks the pattern. Hence, the missing number is 150.

Answer: (D) 150

Top Quantitative Aptitude MCQ Objective Questions

If x − \(\rm\frac{1}{x}\) = 3, the value of x3 − \(\rm\frac{1}{x^3}\) is

  1. 36
  2. 63
  3. 99
  4. none of these

Answer (Detailed Solution Below)

Option 1 : 36

Quantitative Aptitude Question 6 Detailed Solution

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Given:

x - 1/x = 3

Concept used:

a3 - b3 = (a - b)3 + 3ab(a - b)

Calculation:

x3 - 1/x3 = (x - 1/x)3 + 3 × x × 1/x × (x - 1/x)

⇒ (x - 1/x)3 + 3(x - 1/x)

⇒ (3)3 + 3 × (3)

⇒ 27 + 9 = 36

∴ The value of x3 - 1/x3 is 36.

Alternate Method If x - 1/x = a, then x3 - 1/x3 = a3 + 3a

Here a = 3

x - 1/x3 = 33 + 3 × 3

= 27 + 9

= 36

A shopkeeper earns a profit of 25 percent on selling a radio at 15 percent discount on the Printed price. Finds the ratio of the Printed price and the cost price of the radio.

  1. 17 : 25
  2. 25 : 27
  3. 27 : 25
  4. 25 : 17
  5. None

Answer (Detailed Solution Below)

Option 4 : 25 : 17

Quantitative Aptitude Question 7 Detailed Solution

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Given:

Profit = 25 Percent

Discount = 15 Percent

Formula:

MP/CP = (100 + Profit %)/(100 – Discount %)

MP = Printed Price

CP = Cost Price

Calculation:

We know that –

MP/CP = (100 + Profit %)/(100 – Discount %)   ………. (1)

Put all given values in equation (1) then we gets

MP/CP = (100 + 25)/(100 – 15)

⇒ 125/85

⇒ 25/17

∴ The Ratio of the Printed price and cost price of radio will be 25 ∶ 17

Six chords of equal lengths are drawn inside a semicircle of diameter 14√2 cm. Find the area of the shaded region?

F4 Aashish S 21-12-2020 Swati D7

  1. 7
  2. 5
  3. 9
  4. 8

Answer (Detailed Solution Below)

Option 1 : 7

Quantitative Aptitude Question 8 Detailed Solution

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Given:

Diameter of semicircle = 14√2 cm

Radius = 14√2/2 = 7√2 cm

Total no. of chords = 6

Concept:

Since the chords are equal in length, they will subtend equal angles at the centre. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.

Formula used:

Area of sector = (θ/360°) × πr2

Area of triangle = 1/2 × a × b × Sin θ

Calculation:

F4 Aashish S 21-12-2020 Swati D8

The angle subtended by each chord = 180°/no. of chord

⇒ 180°/6

⇒ 30°

Area of sector AOB = (30°/360°) × (22/7) × 7√2 × 7√2

⇒ (1/12) × 22 × 7 × 2

⇒ (77/3) cm2

Area of triangle AOB = 1/2 × a × b × Sin θ

⇒ 1/2 × 7√2 × 7√2 × Sin 30°

⇒ 1/2 × 7√2 × 7√2 × 1/2

⇒ 49/2 cm2

∴ Area of shaded region = 6 × (Area of sector AOB - Area of triangle AOB)

⇒ 6 × [(77/3) – (49/2)]

⇒ 6 × [(154 – 147)/6]

⇒ 7 cm2

Area of shaded region is 7 cm2

There is a rectangular garden of 220 metres × 70 metres. A path of width 4 metres is built around the garden. What is the area of the path?

  1. 2472 metre2
  2. 2162 metre2
  3. 1836 metre2
  4. 2384 metre2

Answer (Detailed Solution Below)

Option 4 : 2384 metre2

Quantitative Aptitude Question 9 Detailed Solution

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Formula used

Area = length × breath

Calculation

8-July-2012 Morning 1 1 Hindi Images Q7

The garden EFGH is shown in the figure. Where EF = 220 meters & EH = 70 meters.

The width of the path is 4 meters.

Now the area of the path leaving the four colored corners

= [2 × (220 × 4)] + [2 × (70 × 4)]

= (1760 + 560) square meter

= 2320 square meters

Now, the area of 4 square colored corners:

4 × (4 × 4)

{∵ Side of each square = 4 meter}

= 64 square meter

The total area of the path = the area of the path leaving the four colored corners + square colored corners

⇒ Total area of the path = 2320 + 64 = 2384 square meter

∴ Option 4 is the correct answer.

In an election between two candidates, the winning candidate got 70 percent votes of the valid votes and he won by a majority of 3630 votes. If out of total votes polled 75 percent votes are valid, then what is the total number of votes polled?

  1. 15200
  2. 13000
  3. 16350
  4. 12100

Answer (Detailed Solution Below)

Option 4 : 12100

Quantitative Aptitude Question 10 Detailed Solution

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Given:

Valid votes = 75% of total votes

Winning Candidate = 70% of Valid votes

He won by a majority of 3630 votes

Losing Candidate = 30% of Valid votes

Calculation:

Let 100x be the total number of votes polled

Valid votes = 75% of total votes

= 0.75 × 100x

= 75x

Majority of the Winning Candidate is 3630

Then, Difference between Winning and Losing Candidate = (70 % - 30 %) of valid votes

= 40% of the valid votes

Valid Votes = 75x

Then,

= 0.40 × 75x

= 30x

Hence, 30x is Majority of winning candidate

30x = 3630

x = 121

Total number of votes is 100x

= 100 × 121

= 12100

Answer is 12100.

Which of the following number is largest among all?

\(0.7,\;0.\bar 7,\;0.0\bar 7,0.\overline {07}\)

  1. \(0.\overline {07} \)
  2. \(0.0\bar 7\)
  3. 0.7
  4. \(0.\bar 7\)

Answer (Detailed Solution Below)

Option 4 : \(0.\bar 7\)

Quantitative Aptitude Question 11 Detailed Solution

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Concebt used

a.b̅ = a.bbbbbb

a.0b̅ = a.0bbbb

Calculation

0.7 = 0.700000 ̇....

\(0.\bar7 = 0.77777 \ldots\)

\(0.0\bar7 = 0.077777 \ldots\)

\(0.\overline {07} = 0.070707 \ldots\)

Now, 0.7777…  or \(0.\bar7\) is largest among all.

A train of length 400 m takes 15 seconds to cross a train of length 300 m traveling at 60 km per hour from the opposite direction along a parallel track. What is the speed of the longer train, in km per hour? 

  1. 108
  2. 102
  3. 98
  4. 96

Answer (Detailed Solution Below)

Option 1 : 108

Quantitative Aptitude Question 12 Detailed Solution

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Given

Length of first train (L1) = 400 m

Length of second train (L2) = 300 m

Speed of second train (S2) = 60 km/hr

Time taken to cross each other (T) = 15 s

Concept:

Relative speed when two objects move in opposite directions is the sum of their speeds.

Calculations:

Let the speed of the first train = x km/hr

Total length = 300 + 400

Time = 15 sec

According to the question:

700/15 = (60 + x) × 5/18

28 × 6 = 60 + x

x = 108 km/hr.

Therefore, the speed of the longer train is 108 km per hour.

u : v = 4 : 7 and v : w = 9 : 7. If u = 72, then what is the value of w?

  1. 98
  2. 77
  3. 63
  4. 49

Answer (Detailed Solution Below)

Option 1 : 98

Quantitative Aptitude Question 13 Detailed Solution

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Given:

u : v = 4 : 7 and v : w = 9 : 7

Concept Used: In this type of question, number can be calculated by using the below formulae

Calculation:

u : v = 4 : 7 and v : w = 9 : 7

To make ratio v equal in both cases

We have to multiply the 1st ratio by 9 and 2nd ratio by 7

u : v = 9 × 4 : 9 × 7 = 36 : 63 ----(i)

v : w = 9 × 7 : 7 × 7 = 63 : 49 ----(ii)

Form (i) and (ii), we can see that the ratio v is equal in both cases

So, Equating the ratios we get,

u v w = 36 63 49

u w = 36 49

When u = 72,

w = 49 × 72/36 = 98

Value of w is 98

If the price of petrol has increased from Rs. 40 per litre to Rs. 60 per litre, by how much percent a person has to decrease his consumption so that his expenditure remains same.

  1. 66.67%
  2. 40%
  3. 33.33%
  4. 45%
  5. None of these 

Answer (Detailed Solution Below)

Option 3 : 33.33%

Quantitative Aptitude Question 14 Detailed Solution

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GIVEN :

If the price of petrol has increased from Rs. 40 per litre to Rs. 60 per litre

CALCULATION :

Let the consumption be 100 litres.

When price is Rs. 40 per litres, then, the expenditure = 100 × 40

⇒ Rs. 4,000.

At Rs. 60 per litre, the 60 × consumption = 4000

Consumption = 4,000/60 = 66.67 litres.

∴ Required decreased % = 100 - 66.67 = 33.33%

What is the value of \(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}?\)

  1. 36
  2. 37
  3. 39
  4. 38

Answer (Detailed Solution Below)

Option 2 : 37

Quantitative Aptitude Question 15 Detailed Solution

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Solution:

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 25/2 + 37/3 + 73/6

= (75 + 74 + 73)/6

= 222/6

= 37


Shortcut Trick

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 12 + 12 + 12 + (1/2 + 1/3 + 1/6)

= 36 + 1 = 37

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