Quantitative Aptitude - Arithmetic Ability Questions

Q:

When 721+722+723+724 is divided by 25 the remainder is:

A) 5 B) 17
C) 7 D) 0
 
Answer & Explanation Answer: D) 0

Explanation:

Given,721+722+723+724

721(1 + 7 + 49 + 343)
721(400)

Now, 400 is divisible by 25. The remainder will be 0

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Filed Under: Simplification
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2 4720
Q:

Three strategies P, Q and R have been initiated for cost cutting in a company producing respectively 20%, 30% and 10% savings. Assuming that they operate independently, what is the net saving achieved ?

A) 49.6% B) 50.4%
C) 33.67% D) 66.66%
 
Answer & Explanation Answer: A) 49.6%

Explanation:

As these three strategies P, Q and R are independent so these will cut cost one after the other.
If initial cost is Rs 100, then
20% cost is cut after initializing strategy P, then cost will remain 80% = 80
further 30% cost is cut after strategy Q, then cost will remain 70% of 80 = 56
further 10% cost is cut after strategy R, then cost will remain 90% of 56 = 50.4

Thus final cost remains 50.4 % of the original cost. Hence net saving is 100 - 50.4 = 49.6 %.

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Filed Under: Percentage
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3 4719
Q:

Jaclyn buys $50 000 worth of debentures in a company. She earns 9.5% p.a. simple interest, paid to her quarterly (that is, every 3 months). If the agreed period of the debenture was 18 months: calculate the amount of interest Jaclyn will earn for each quarter

A) 1187.50 B) 1234
C) 1289 D) 1345
 
Answer & Explanation Answer: A) 1187.50

Explanation:

I=(p x r x t)/100

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

0 4716
Q:

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

A) 4/3 B) 2/3
C) 3/2 D) 3/4
 
Answer & Explanation Answer: D) 3/4

Explanation:

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

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Filed Under: Probability
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6 4715
Q:

On the occasion of New Year, each student of a class sends greeting cards to the others. If there are 21 students in the class, what is the total number of greeting cards exchanged by the students?

A) 380 B) 420
C) 441 D) 400
 
Answer & Explanation Answer: B) 420

Explanation:

Given total number of students in the class = 21

 

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

 

Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

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11 4714
Q:

The length of a rectangle is 10 cm more than the breadth. Find the diagonal length of rectangle if perimeter is 44 cm ?

A) 12.50 cm B) 14.21 cm
C) 15.98 cm D) 17.08 cm
 
Answer & Explanation Answer: D) 17.08 cm

Explanation:

Let the length of the rectangle = L cm

 

Then the breadth of the rectangle = (L - 10) cm

 

Perimeter of a rectangle = 2(L + B) cm

 

=> 44 = 2(L + L - 10)

 

=> 44 = 4L - 20

 

=> 4L = 64

 

=> L = 16 cm

 

=> B = L - 10 = 6 cm

 

Diagonal = L2+B2 = 162+62 = 17.08 cm

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Filed Under: Area
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11 4704
Q:

Chennai express left Hyderabad for Chennai at 14 : 30 hours, travelling at a speed of 60 kmph and Charminar Express left Hyderabad for Chennai on the same day at 16 : 30 hours, travelling at a speed of 80 kmph. How far away from Hyderabad will the two trains meet? 

A) 360 kms B) 480 kms
C) 520 kms D) 240 kms
 
Answer & Explanation Answer: B) 480 kms

Explanation:

Now, the distance covered by Chennai express in 2 hrs = 60 x 2 = 120 kms

Let the Charminar Express takes 't' hrs to catch Chennai express

=> 80 x t = 60 x (2 + t)

=> 80 t = 120 + 60t

=> t = 6 hrs

 

Therefore, the distance away from Hyderabad the two trains meet = 80 x 6 = 480 kms.

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10 4699
Q:

In how many different ways can the letters of the word 'ABYSMAL' be arranged ?

A) 5040 B) 3650
C) 4150 D) 2520
 
Answer & Explanation Answer: D) 2520

Explanation:

Total number of letters in the word ABYSMAL are 7

 

Number of ways these 7 letters can be arranged are 7! ways

 

But the letter is repeated and this can be arranged in 2! ways

 

Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

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Filed Under: Permutations and Combinations
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