Quantitative Aptitude - Arithmetic Ability Questions

Q:

How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines?

A) 215 B) 315
C) 415 D) 115
 
Answer & Explanation Answer: B) 315

Explanation:
Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
 

Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
 

Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315
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Filed Under: Permutations and Combinations
Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
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30 18623
Q:

Find the appropriate relation for quantity 1 and quantity 2 in the following question:

Quantity I: the unit digit in (6817)754

Quantity II: the unit digit in (365×659×771)

A) Quantity I > Quantity II B) Quantity I < Quantity II
C) Quantity I ≥ Quantity II D) Quantity I ≤ Quantity II
 
Answer & Explanation Answer: A) Quantity I > Quantity II

Explanation:
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1 18612
Q:

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?

A) 69.5 km/hr B) 70 km/hr
C) 79 km/hr D) 79.2 km/hr
 
Answer & Explanation Answer: D) 79.2 km/hr

Explanation:

Let the length of the train be x metres and its speed by y m/sec.


Then, x/y = 8 => x = 8y


Now, (x+264)/20 = y
8y + 264 = 20y
y = 22.
Speed = 22 m/sec = (22*18/5) km/hr = 79.2 km/hr.

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Filed Under: Problems on Trains

60 18595
Q:

What value should come in place of the question mark (?) in the following question?

1637 + 1832 = 452 + (?)2

A) 38 B) 42
C) 46 D) 54
 
Answer & Explanation Answer: A) 38

Explanation:

3469 = 2025+?x?
? x ?= 3469-2025
? x ?= 1444
? = 38

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0 18589
Q:

If the C.I. on a sum for 2 years at 12 1/2 % per annum is Rs. 510, the S.I. on the same sum at the same rate for the same period of time is   ?

A) Rs. 580 B) Rs. 480
C) Rs. 550 D) Rs. 470
 
Answer & Explanation Answer: B) Rs. 480

Explanation:

Let the sum be Rs. P. Then,

 

P1+252x1002-P= 510

 

P[982- 1] = 510.

 

Sum = Rs. 1920

 

So, S.I. = (1920 x 25 x 2) / (2 x 100) = Rs. 480

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Filed Under: Simple Interest
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39 18573
Q:

The denominator of a fraction is 3 more than the numerator. If the numerator as well as the denominator is increased by 4, the fraction becomes 4/5. What was the original fraction ?

A) 7/11 B) 8/11
C) 9/11 D) 10/11
 
Answer & Explanation Answer: B) 8/11

Explanation:

Let the numerator be x Then, denominator =  x + 3.

Now. (x + 4)/(x + 3) +4 = 4/5  <=> 5 (x + 4) =  4(x + 7)

=>  x = 8.

The fraction is  8/11.

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Filed Under: Problems on Numbers

24 18539
Q:

What value should come in place of question mark (?) in the following question?

442 + 232 = (?)2 + 44

A) 47 B) 65
C) 52 D) 41
 
Answer & Explanation Answer: A) 47

Explanation:

1936 + 529 = ?2 + 256 2465 = ?2 + 256 ?2 = 2465 - 256 ?2 = 2209 ?=47

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0 18537
Q:

A contractor undertakes to complete a work in 130 days. He employs 150 men for 25 days and they complete 1/4 of the work . He then reduces the number of men to 100, who work for 60 days, after which there are 10 days holidays.How many men must be employed for the remaining period to finish the work?

Answer

150 men in 25 days do = 14 work  


Let 1 man in 1 day does = x  work  


Total work done by 150 men in 25 days = 150x *  25 = 14   work => x = 115000   


Therefore, 100 men in 60 days do = 100 * 60x = 6000x work = 6/15 = 2/5


Total work done =14 + 25 = 1320 


Therefore, Remaining work = 1 - 1320 = 720  


Remaining time = 130 - (25+60+10) = 35 days 


Therefore, work is done in 25 days by 150 men.


Therefore, Work is done in 35 days by 150 men.


Hence, he should employ 50 more men.

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Subject: Time and Work Exam Prep: Bank Exams , AIEEE
Job Role: Bank PO

47 18521