6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.

A) 19/34	B) 5/4
C) 20/34	D) 25/34

Answer & Explanation Answer: D) 25/34

Explanation:

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.

So the required probability:
= $(\frac{8}{17} * \frac{9}{16}) + (\frac{9}{17} * \frac{8}{16}) + (\frac{8}{17} * \frac{7}{16})$

= $\frac{9}{34} + \frac{9}{34} + \frac{7}{34}$
= $\frac{25}{34}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

21 24020

Q:

Rahul went to purchase a Nokia mobile handset, the shopkeeper told him to pay 20% tax if he asked the bill. Rahul manages to get the discount of 5% on the actual saleprice of the mobile and he paid the shopkeeper Rs. 3325 without tax. Besides he manages to avoid to pay 20% tax on the already discounted price, what is the amount of discount that he has gotten?

A) 750	B) 375
C) 875	D) 525

Answer & Explanation Answer: C) 875

Explanation:

CP = 100, SP (with tax) =120

New SP = 100 - 5 = 95

Effective discount = 120 - 95 = 25

So, at SP of 95 ----> discount = 25

and at SP of 3325 -----> discount = $\frac{25}{95} \times 3325 = 875$

Report Error

View Answer Report Error Discuss

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

54 24002

Q:

15% of 60 is 45% of __

A) 20	B) 30
C) 180	D) 40

Answer & Explanation Answer: A) 20

Explanation:

Report Error

View Answer Report Error Discuss

Filed Under: Percentage
Exam Prep: Bank Exams

4 23976

Q:

If the arithmetic mean of seventy-five numbers is calculated, it is 35. if each number is increased by 5, then mean of new numbers is :

A) 30	B) 40
C) 50	D) 60

Answer & Explanation Answer: B) 40

Explanation:

Let the numbers are $x_{1,} x_{2,} . . . . . . . x_{75}$

By Hypothesis , $\frac{x_{1} + x_{2} + . . . . x_{75}}{75} = 35$ ......(i)

Mean of new numbers = $\frac{(x_{1} + 5) + (x_{2} + 5) + . . . . . . . (x_{75} + 5)}{75}$

$\frac{x_{1} + x_{2} + . . . . . . x_{75} + 5 * 75}{75}$

$\frac{x_{1} + x_{2} + . . . . . . x_{75}}{75} + \frac{375}{75} = 35 + 5 = 40$ By (i)

Report Error

View Answer Report Error Discuss

Filed Under: Average

43 23974

Q:

Rs. 5887 is divided between Shyam and Ram, such that Shyam's share at the end of 9 years is equal to Ram's share at the end of 11 years, compounded annually at the rate of 5%. Find the share of Shyam.

A) 3567	B) 3452
C) 3087	D) 3544

Answer & Explanation Answer: C) 3087

Explanation:

Shyam's share * (1+0.05)9 = Ram's share * (1 + 0.05)11

Shyam's share / Ram's share = (1 + 0.05)11 / (1+ 0.05)9 = (1+ 0.05)2 = 441/400

Therefore Shyam's share = (441/841) * 5887 = 3087

Report Error

View Answer Report Error Discuss

Filed Under: Compound Interest
Exam Prep: Bank Exams
Job Role: Bank PO

17 23962

Q:

Consider the following statements:

1) The perimeter of a triangle is greater than the sum of its three medinas.

2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.

Which of the above statements is/are correct?

A) 1 only	B) 2 only
C) Both 1 and 2	D) Neither 1 nor 2

Answer & Explanation Answer: C) Both 1 and 2

Explanation:

Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.

Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)

Hence in ΔABD, AD is a median

⇒

AB + AC > 2(AD)

Similarly, we get

BC + AC > 2CF

BC + AB > 2BE

On adding the above inequations, we get

(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE

2(AB + BC + AC) > 2(AD + BE + CF)

∴

AB + BC + AC > AD + BE +CF

2.

To prove: AB + BC + CA > 2AD

Construction: AD is joined

Proof: In triangle ABD,

AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the

third side]

----

1

In triangle ADC,

AC + DC > AD [because, the sum of any two

sides of a tri

angle is always greater than the

third side]

----

2

Adding 1 and 2 we get,

AB + BD + AC + DC > AD + AD

=> AB + (BD + DC) + AC > 2AD

=> AB + BC + AC > 2AD

Hence proved

Report Error

View Answer Report Error Discuss

Filed Under: Volume and Surface Area
Exam Prep: Bank Exams

1 23949

Q:

A man walks 6 km at a speed of $1 \frac{1}{2}$ kmph, runs 8 km at a speed of 2 kmph and goes by bus another 32 km. Speed of the bus is 8 kmph. Find the average speed of the man in kmph.

$A) 4 \frac{3}{14}$

B) $4 \frac{2}{13}$

C) $2 \frac{4}{13}$

D) $3 \frac{4}{14}$

A) Option A	B) Option B
C) Option C	D) Option D

Answer & Explanation Answer: B) Option B

Explanation:

Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 40 km at a speed of 8kmph
time taken indivisually:
=> 6/1.5 = 4 m
=> 8/2 = 4 m
=> 40/8 = 5 m

Average speed of man= total distance/ total time
=> 54/13 = $4 \frac{2}{13}$ kmph

Report Error

View Answer Report Error Discuss

Filed Under: Time and Distance
Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk , Analyst

17 23923

Q:

Select the pair that does NOT belong in the following group.

$(\sqrt{64}, 9), (\sqrt{81}, 10), (\sqrt{36}, 8), (\sqrt{121}, 12)$

A) 121, 12	B) 36, 8
C) 81, 10	D) 64, 9

Quantitative Aptitude - Arithmetic Ability Questions

Quick Links