# Arithmetical Reasoning Questions

Q:

Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A ?

 A) Rs. 4, Rs. 23 B) Rs. 13, Rs. 17 C) Rs. 15, Rs. 14 D) Rs. 17, Rs. 13

Explanation:

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

716 79008
Q:

If you write down all the numbers from 1 to 100, then how many times do you write 3 ?

 A) 11 B) 18 C) 20 D) 21

Explanation:

Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.

749 64397
Q:

A tailor had a number of shirt pieces to cut from a roll of fabric. He cut each roll of equal length into 10 pieces. He cut at the rate of 45 cuts a minute. How many rolls would be cut in 24 minutes ?

 A) 32 rolls B) 54 rolls C) 108 rolls D) 120 rolls

Explanation:

Number of cuts made to cut a roll into 10 pieces = 9.

Therefore, Required number of rolls = (45 x 24)/9 = 120.

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Q:

Today is Varun's birthday. One year, from today he will be twice as old as he was 12 years ago. How old is Varun today ?

 A) 20 years B) 22 years C) 25 years D) 27 years

Explanation:

Let Varun's age today = x years.

Then, Varun's age after 1 year = (x + 1) years.

Therefore x + 1 = 2 (x - 12) =>  x + 1 = 2x - 24 =>  x = 25.

333 40225
Q:

7 is to 34 as 5 is to

 A) 33 B) 24 C) 54 D) 41

Explanation:

We can get this by any of two explanations.

7 = 3 + 4  = 34

Similarly, 5 = 4 + 1  = 41

7 - 3 = 4  => 34

Similarly, 5 - 4 = 1  => 41

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Q:

In three coloured boxes - Red, Green and Blue, 108 balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box ?

 A) 18 B) 36 C) 45 D) None of these

Explanation:

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then,

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108 =>  6R = 108 => R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

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Q:

Two old friends, Jack and Bill, meet after a long time.

Jack: Hey, how are you man?
Bill: Not bad, got married and I have three kids now

Jack: That’s awesome. How old are they?
Bill: The product of their ages is 72 and the sum of their ages is the same as your birth date.

Jack: Cool … But I still don’t know.
Bill : My eldest kid just started taking piano lessons.

Jack: Oh now I get it.
How old are Bill’s kids ?

 A) 3, 3, 8 B) 2, 6, 6 C) 1, 6, 12 D) 1, 8, 9

Explanation:

Let’s break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 – sum(2, 2, 18) = 22
2, 4, 9 – sum(2, 4, 9) = 15
2, 6, 6 – sum(2, 6, 6) = 14
2, 3, 12 – sum(2, 3, 12) = 17
3, 4, 6 – sum(3, 4, 6) = 13
3, 3, 8 – sum(3, 3, 8 ) = 14
1, 8, 9 – sum(1,8,9) = 18
1, 3, 24 – sum(1, 3, 24) = 28
1, 4, 18 – sum(1, 4, 18) = 23
1, 2, 36 – sum(1, 2, 36) = 39
1, 6, 12 – sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 – sum(2, 6, 6) = 14
3, 3, 8 – sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

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Q:

The following question is based on the given data for an examination.

A. Candidates appeared                 10500

B. Passed in all the five subjects     5685

C. Passed in three subjects only     1498

D. Passed in two subjects only       1250

E. Passed in one subject only           835

F. Failed in English only                       78

G. Failed in Maths only                      275

H. Failed in Physics only                    149

I. Failed in Chemistry only                147

J. Failed in Biology only                     221

How many candidates passed in at least four subjects ?

 A) 6555 B) 5685 C) 1705 D) 870

Explanation:

Candidates passed in atleast four subjects

= (Candidates passed in 4 subjects) + (Candidates Passed in all 5 subjects)

= (Candidates failed in only 1 subject ) + ( Candidates passed in all subjects)

= (78 + 275 + 149 + 147 + 221) + 5685 = 870 + 5685 = 6555