6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A) 173 m	B) 200 m
C) 273 m	D) 300 m

Answer & Explanation Answer: C) 273 m

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, $∠ A C B = 30 ° a n d ∠ A D B = 45 °$

$\frac{A B}{A C} = \tan 30 ° = \frac{1}{\sqrt{3}} = > A C = A B * \sqrt{3} = 100 \sqrt{3 m}$

$\frac{A B}{A D} = \tan 45 ° = 1 = > A D = A B = 100 m$

CD=(AC+AD)= $(100 \sqrt{3} + 100) m = 100 (\sqrt{3} + 1) = 100 * 2.73 = 273 m$

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Filed Under: Height and Distance

17 8465

Q:

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

A) 3/5	B) 1/3
C) 2/5	D) 3/4

Answer & Explanation Answer: A) 3/5

Explanation:

Let A be the event of A will tell truth. B be the event of B tell truth

$P (A) = \frac{2}{3}, P (A^{c}) = 1 - P (A) = \frac{1}{3}$

$P (B) = \frac{4}{5}, P (B^{c}) = 1 - P (B) = \frac{1}{5}$

When both agree then they say true or they say false together, that is

$A \cap B o r A^{c} \cap B^{c}$

Also these events will be mutually exclusive :

$P (A \cap B) + P (A^{c} \cap B^{c}) = P (A) P (B) P (C) = \frac{2}{3} * \frac{4}{5} + \frac{1}{3} * \frac{1}{5} = \frac{3}{5}$

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Filed Under: Probability

6 8460

Q:

A bill for Rs. 6000 is drawn on July 14 at 5 months. It is discounted on 5th October at 10%. Find the banker's discount, true discount, banker's gain and the money that the holder of the bill receives.

A) 4390	B) 6580
C) 5880	D) 5350

Answer & Explanation Answer: C) 5880

Explanation:

Face value of the bill = Rs. 6000.

Date on which the bill was drawn = July 14 at 5 months. Nominally due date = December 14.

Legally due date = December 17.

Date on which the bill was discounted = October 5.

Unexpired time : Oct. Nov. Dec.

26 + 30 + 17 = 73 days =1/ 5Years

B.D. = S.I. on Rs. 6000 for 1/5 year

= Rs. (6000 x 10 x1/5 x1/100)= Rs. 120.

T.D. = Rs.[(6000 x 10 x1/5)/(100+(10*1/5))]

=Rs.(12000/102)=Rs. 117.64.

B.G. = (B.D.) - (T.D.) = Rs. (120 - 117.64) = Rs. 2.36.

Money received by the holder of the bill = Rs. (6000 - 120) = Rs. 5880.

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Filed Under: Banker's Discount

5 8453

Q:

A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

A) 6/145	B) 36/145
C) 18/95	D) 12/145

Answer & Explanation Answer: B) 36/145

Explanation:

The probability that first ball is white:
=C112C130=1230=25

Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence the probability the second ball is black:
=C118C129=1829

Required probability =(25)×(1829)

=36/145

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Filed Under: Quantitative Aptitude - Arithmetic Ability

10 8453

Q:

1.12.91 is the first Tuesday. Which is the fourth Thursday of December 91 ?

A) 22.12.91	B) 25.12.91
C) 24.12.91	D) 23.12.91

Answer & Explanation Answer: B) 25.12.91

Explanation:

Given that 2.12.91 is the first Tuesday
Hence we can assume that 4.12.91 is the first Thursday

If we add 7 days to 4.12.91, we will get second Thursday
If we add 14 days to 4.12.91, we will get third Thursday
If we add 21 days to 4.12.91, we will get fourth Thursday

=> fourth Thursday = (4.12.91 + 21 days) = 25.12.91.

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Filed Under: Calendar
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

32 8448

Q:

(0.2 * 0.2 + 0.01) ( 0.1 * 0.1 + 0.02)^(-1) is equal to :

A) 5/3	B) 9/3
C) 11/3	D) 13/3

Answer & Explanation Answer: A) 5/3

Explanation:

Given = (0.2 * 0.2 + 0.01) / ( 0.1 * 0.1 + 0.02) = (0.04 + 0.01) / (0.01 + 0.02) = 0.05 / 0.03 = 5/3...

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Filed Under: Decimal Fractions

11 8438

Q:

If x = 1 - q and y = 2q + 1, then for what value of q, x is equal to y ?

A) 0	B) 1
C) -1	D) -2

Answer & Explanation Answer: A) 0

Explanation:

x = y <=> 1 - q = 2q + 1 <=> 3q = 0 <=> q = 0.

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Filed Under: Simplification

8 8438

Q:

A rectangular block has the dimensions 5x6x7 cm it is dropped into a cylindrical vessel of radius 6cm and height 10 cm. If the level of the fluid in the cylinder rises by 4 cm, What portion of the block is immersed in the fluid ?

A) 22/7 x 24/35	B) 22/7 x 36 x 4
C) 22/7 x 36/5	D) 22/7 x 37/21

Answer & Explanation Answer: A) 22/7 x 24/35

Explanation:

Since level of water increased in cylinder by height 4.

This is because of the rectangular block .

Therefore , area of rectangular block immersed in water is

= $\frac{22}{7} \times 6 \times 6 \times 4$

Thats why portion of block immersed in water is

= (22/7 * 36 * 4) / total vol. of rectangle

= (22/7 * 36 * 4)/ (7*5*6)

= (22/7 * 24)/ 35.

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Filed Under: Volume and Surface Area
Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

7 8436

Quantitative Aptitude - Arithmetic Ability Questions

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