6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

Two stations A and B are 200 km apart on a straight track. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

A) 12 a.m.	B) 1 p.m.
C) 11 a.m.	D) 12 p.m.

Answer & Explanation Answer: D) 12 p.m.

Explanation:

Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.

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Filed Under: Time and Distance
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4 7184

Q:

The Value of $[\log (\tan (1^{0})) + \log (\tan (2^{0})) + \dots \dots + \log (\tan (89^{0}))]$ is

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

Answer & Explanation Answer: B) 0

Explanation:

$= [\log \tan (1^{0}) + \log \tan (89^{0})] + [\log \tan (2^{0}) + \log \tan (88^{0})] + \dots \dots + [\log \tan (45^{0})]$

$= \log [\tan (1^{0}) \times \tan (89^{0})] + \log [\tan (2^{0}) \times \tan (88^{0})] + \dots \dots + \log (1)$

$[∵ \tan (90 - θ) = c o t θ a n d \tan (45^{0}) = 1]$

= log 1 + log 1 +.....+log 1

= 0.

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Filed Under: Logarithms
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27 7181

Q:

Rice worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, then the price of third variety of rice per kg?

A) Rs. 571.25	B) Rs. 175.5
C) Rs. 751.75	D) Rs. 850

Answer & Explanation Answer: B) Rs. 175.5

Explanation:

Let the price of required variety = Rs. P/kg

Then, respective amounts were m kg, m kg and 2m kg

= 126m + 135m + 2pm = 153 x 4m

=> 2p = 351

p = 175.5 / kg

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Filed Under: Ratios and Proportions
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37 7180

Q:

If the radius of a circle is decreased by 50% , find the percentage decrease in its area.

A) 75%	B) 65%
C) 35%	D) 25%

Answer & Explanation Answer: A) 75%

Explanation:

let original radius = r and new radius = (50/100) r = r/2

original area = ${πr}^{2}$ and new area = $π {(r / 2)}^{2}$

decrease in area = $3 {πr}^{2} / 4$ * $\frac{1}{{πr}^{2}}$ *100 = 75%

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Filed Under: Area
Exam Prep: Bank Exams
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0 7167

Q:

A tap can fill a tank in 6 hours. After half the tank is filled three more similar taps are opened. What is the total time taken to fill the tank completely ?

A) 4 hrs 15 min	B) 3 hrs 24 min
C) 4 hrs 51 min	D) 3 hrs 45 min

Answer & Explanation Answer: D) 3 hrs 45 min

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.

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Filed Under: Pipes and Cistern
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13 7167

Q:

Which value is equal to 5% of 1,500?

A) 35	B) 55
C) 75	D) 45

Answer & Explanation Answer: C) 75

Explanation:

In the question it is asked for 5% of 1500

5 x 1500/100 = 15 x 5 = 75.

Hence, 5% of 1,500 is equal to 75.

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Filed Under: Percentage
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10 7154

Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is neither a spade nor a king.

A) 0	B) 9/13
C) 1/2	D) 4/13

Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus n(E) = 13 + 3 = 16

Hence $n (\bar{E}) = 52 - 16 = 36$

Therefore, $P (E) = \frac{36}{52} = \frac{9}{13}$

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Filed Under: Probability

6 7147

Q:

Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?

A) 3	B) 6
C) 4	D) 2

Answer & Explanation Answer: A) 3

Explanation:

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4

b = 225 - 19a/4----------(1)

b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

0 < 225 - 19a/4 < 99

225 < -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.

Hence possible values of a are (28,32,36,40,44)

For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40

Quantitative Aptitude - Arithmetic Ability Questions

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