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Arithmetical Reasoning

Important Instructions
11.

The total of the ages of Amar, Akbar and Anthony is 80 years. What was the total of their ages three years ago ?

Answer: A

Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.

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12.

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

Answer: C

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

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13.

In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio 3 : 2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports. If the number of boys participating in the sports is 15, then how many girls are there in the class ?

Answer: C

Let the number of boys and girls participating in sports be 3x and 2x respectively.

Then, 3x = 15 or x = 5.

So, number of girls participating in sports = 2x = 10.

Number of students not participating in sports = 60 - (15 + 10) = 35.

Let number of boys not participating in sports be y.

Then, number of girls not participating in sports = (35 -y).

Therefore (35 - y) = y + 5 

\(\Rightarrow  2y = 30\) 

\(\Rightarrow  y = 15.\)

So, number of girls not participating in sports = (35 - 15) = 20.

Hence, total number of girls in the class = (10 + 20) = 30

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14.

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 ?

Answer: D

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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15.

An institute organised a fete and 1/5 of the girls and 1/8 of the boys participated in the same. What fraction of the total number of students took part in the fete ?

Answer: C

No answer description available for this question

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16.

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 ?

Answer: B

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

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17.

A woman says, "If you reverse my own age, the figures represent my husband's age. He is, of course, senior to me and the difference between our ages is one-eleventh of their sum." The woman's age is

Answer: C

Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.

Then, woman's age = (10X + y) years; husband's age = (10y + x) years.

Therefore \((10y + x)- (10X + y) = (1/11) (10y + x + 10x + y)\)

\(\Rightarrow\) (9y-9x) = (1/11)(11y + 11x) = (x + y)

 \(\Rightarrow\) \(10x = 8y\)

 \(\Rightarrow\) \( x = (4/5)y\)

Clearly, y should be a single-digit multiple of 5, which is 5.

So, x = 4, y = 5.

Hence, woman's age = 10x + y = 45 years

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18.

What is the product of all the numbers in the dial of a telephone ?

Answer: D

Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0

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19.

In a class, there are 18 boys who are over 160 cm tall. If these constitute three-fourths of the boys and the total number of boys is two-thirds of the total number of students in the class, what is the number of girls in the class ?

Answer: B

Let the number of boys be x. Then, (3/4)x = 18 or x = 18 x(4/3) = 24.

If total number of students is y, then (2/3) y = 24 or y = 24 x (3/2) = 36.

Therefore Number of girls in the class = (36 - 24) = 12

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20.

A is 3 years older to B and 3 years younger to C, while B and D are twins. How many years older is C to D?

Answer: C

Since B and D are twins, so B = D.

Now, A = B + 3 and A = C - 3.

Thus, B + 3 = C - 3 

\(\Rightarrow  D + 3 = C-3\) 

\(\Rightarrow  C - D = 6.\)

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