6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

Read the data given below and answer the questions based on it.

In the CBSE Board exams last year, 53% passed in Biology, 61% percent passed in English, 60% in Social studies, 24% in Biology and English, 35% in English and social studies, 27% in Biology and social studies and 5% in none.

If the number of students in the class is 200, how many passed in only one subject?

A) 48	B) 46
C) More than 50	D) Less than 40

Answer & Explanation Answer: B) 46

Explanation:

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Filed Under: Percentage
Exam Prep: Bank Exams

3 29143

Q:

Identify the odd one from the list below.

A) Stream	B) Bridge
C) Canal	D) River

Answer & Explanation Answer: B) Bridge

Explanation:

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Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 29127

Q:

The sale price of an article including the sales tax is Rs. 616. The rate of sales tax is 10%. If the shopkeeper has made a profit of 12%, then the cost price of the article is :

A) Rs. 560	B) Rs. 530
C) Rs. 500	D) Rs. 514

Answer & Explanation Answer: C) Rs. 500

Explanation:

110% of S.P. = Rs. 616

S.P. = (616 x 100)/110 = Rs. 560

C.P = (110 x 560)/112 = Rs. 500

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Filed Under: Profit and Loss
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

65 29060

Q:

Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :

A) 20 %	B) 40 %
C) 60 %	D) 80 %

Answer & Explanation Answer: C) 60 %

Explanation:

Let the total original sale be Rs. 100. Then, original number of visitors = 100.

New number of visitors = 120/0.75 = 160.

Increase % = 60 %.

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Filed Under: Percentage

95 29048

Q:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A) B=900;H=300	B) B=300;H=900
C) B=600;H=700	D) B=500;H=900

Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
$13.5 * 10000 m^{2} = 135000 m^{2}$
Let altitude = x metres and base = 3x metres.
Then, $\frac{1}{2} * 3 x * x = 135000 \Rightarrow x^{2} = 90000 \Rightarrow x = 300$

Base = 900 m and Altitude = 300 m.

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

62 29040

Q:

The average temperature of the town in the first four days of a month was 58 degrees. The average for the second, third, fourth and fifth days was 60 degrees. If the temperatures of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ?

A) 62 degrees	B) 64 degrees
C) 65 degrees	D) 66 degrees

Answer & Explanation Answer: B) 64 degrees

Explanation:

Sum of temperatures on 1st, 2nd, 3rd and 4th days = (58 * 4) = 232 degrees ... (1)

Sum of temperatures on 2nd, 3rd, 4th and 5th days - (60 * 4) = 240 degrees ....(2)

Subtracting (1) From (2), we get :

Temp, on 5th day - Temp on 1st day = 8 degrees.

Let the temperatures on 1st and 5th days be 7x and 8x degrees respectively.

Then, 8x - 7x = 8 or x = 8.

Temperature on the 5th day = 8x = 64 degrees.

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Filed Under: Average

56 29039

Q:

A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Answer

(A+B)'s two days work = $\frac{1}{40} + \frac{1}{50} = \frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22 \times \frac{9}{200} = \frac{198}{200}$

Remaining work = $1 - \frac{198}{200}$ = 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40 \times \frac{1}{100} = \frac{2}{5}$

Therefore, Total time taken = 44 $\frac{2}{5}$ daya

Workspace

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Subject: Time and Work
Job Role: Bank PO

72 28989

Q:

Which of the following represents the sides of an acute angled triangle?

A) 6, 9, 16	B) 7, 8, 10
C) 5, 12, 13	D) None of these

Quantitative Aptitude - Arithmetic Ability Questions

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