6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

When I was married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me.The average age of my family during my marriage is same as today. What is the age of Father when he died ?

A) 50 yrs	B) 60 yrs
C) 70 yrs	D) 65 yrs

Answer & Explanation Answer: B) 60 yrs

Explanation:

Let the Father be x years when he died

Average Age 10 years ago be A

Total Age 10 years ago = 6*A

Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60

Father Died and Baby was born => the Total number of people in the family is Same (6)
Baby born today so age of baby = 0

(6A +60 - x)/6 = 6A/6
=> A + 10 -(x/6) = A
=> x/6 = 10
=> x = 60

Therefore we can conclude that the father was 60 years old when he died.

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Filed Under: Problems on Ages
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
Job Role: Bank Clerk , Bank PO

57 22876

Q:

Find the next number in the series 14, 916, 2536,......

A) 4981	B) 6481
C) 4964	D) 6449

Answer & Explanation Answer: C) 4964

Explanation:

The pattern is $1^{2} 2^{2}, 3^{2} 4^{2}, 5^{2} 6^{2}$ ,

$∴$ The next number= $7^{2} 8^{2}$ = 4964

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Filed Under: Numbers
Exam Prep: CAT , Bank Exams
Job Role: Bank PO , Bank Clerk

45 22827

Q:

In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?

A) 4000	B) 8000
C) 12000	D) 16000

Answer & Explanation Answer: B) 8000

Explanation:

Let the number of candidates appeared from each state be x.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

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Filed Under: Percentage

54 22768

Q:

The area of a circle of radius 5 is numerically what percent its circumference?

A) 150%	B) 250%
C) 350%	D) 450%

Answer & Explanation Answer: B) 250%

Explanation:

required percentage = $\frac{{πr}^{2}}{2 πr} \times 100 = 250$ %

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

26 22764

Q:

At what time, between 3 o’clock and 4 o’clock, both the hour hand and minute hand coincide each other ?

A) 3:16 7/11	B) 3:16 11/4
C) 3:30	D) 3:16 4/11

Answer & Explanation Answer: D) 3:16 4/11

Explanation:

Coincide means 00 angle.

This can be calculated using the formulafor time A to B means [11m/2 - 30 (A)]

Here m gives minutes after A the both hands coincides.

Here A = 3, B = 4

0 =11m/2 –30 × 3
11m = 90 × 2 = 180
m= 180/11 = 16 ⁴/₁₁

So time = 3 : 16 ⁴/₁₁

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Filed Under: Clocks
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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33 22731

Q:

If Mar 18th,1994 falls on Friday then Feb 25th,1995 falls on which day?

A) Wednesday	B) Monday
C) Saturday	D) Sunday

Answer & Explanation Answer: C) Saturday

Explanation:

First,we count the number of odd days for the left over days in the given period.Here,given period is 18.3.1994 to 25.2.1995

Month	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec	Jan	Feb
Days	13	30	31	30	31	31	30	31	30	31	31	25
Odd Days	6	2	3	2	3	3	2	3	2	3	3	4

Therefore, No. of Odd Days = 6 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 4 = 36 = 1 odd day

So, given day Friday + 1 = Saturday is the required result.

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Filed Under: Calendar

103 22730

Q:

A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

A) 35/96	B) 19/90
C) 19/96	D) None of these

Answer & Explanation Answer: B) 19/90

Explanation:

$A S S I S T A N T \to A A I N S S S T T$

$S T A T I S T I C S \to A C I I S S S T T T$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{1_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{9_{C_{1}}} \times \frac{2_{C_{1}}}{10_{C_{1}}}$ = 1/45

Probability of choosing S = $\frac{3_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/10

Probability of choosing T = $\frac{2_{C_{1}}}{9_{C_{1}}} \times \frac{3_{C_{1}}}{10_{C_{1}}}$ = 1/15

Hence, Required probability = $\frac{1}{45} + \frac{1}{45} + \frac{1}{10} + \frac{1}{15} = \frac{19}{90}$

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Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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97 22678

Q:

In a class of 78 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either course?

A) 6	B) 12
C) 24	D) 18

Answer & Explanation Answer: C) 24

Explanation:

You could solve this by drawing a Venn diagram. A simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking French to find the number taking only French. Likewise find those taking only German. Then we have:Total = only French + only German + both + neither

Quantitative Aptitude - Arithmetic Ability Questions

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