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61. The sum of the numerator and denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from the denominator, it becomes $\frac{2}{3}$ . The fraction is :

A. $\frac{5}{6}$

B. $\frac{6}{5}$

C. $\frac{3}{8}$

D. $\frac{8}{3}$

Discuss

Solution:

Let the fraction be

\frac{x}{y}

Then,

\begin{aligned} \Leftrightarrow x + y = 11..... (i) \\ \Leftrightarrow \frac{x + 1}{y - 2} = \frac{2}{3} \\ \Leftrightarrow 3 (x + 1) = 2 (y - 2) \\ \Leftrightarrow 3 x - 2 y = - 7..... (ii) \end{aligned}

Solving (i) and (ii), we get:
x = 3 and y = 8
So, the fraction is

\frac{3}{8}

62. In a Mathematics examination the number scored by 5 candidates are 5 successive odd integers. If their total marks are 185, the highest score is :

A. 39

B. 43

C. 41

D. 47

Discuss

Solution:

Let the five successive odd number be,
x, x + 2, x + 4, x + 6, x + 8
Then, according to given information,
185 = x + x + 2 + x + 4 + x + 6 + x + 8

\begin{aligned} \Leftrightarrow 185 = 5 x + 20 \\ \Leftrightarrow 5 x = 185 - 20 \\ \Leftrightarrow 5 x = 165 \\ \Leftrightarrow x = 33 \end{aligned}

Highest number = 33 + 8 = 41

63. Three numbers are in in the ratio of 3 : 4 : 6 and their product is 1944. The largest of these numbers is :

A. 6

B. 12

C. 18

D. None of these

Discuss

Solution:

Let the numbers be 3x, 4x and 6x
Then,

\begin{aligned} \Leftrightarrow 3 x \times 4 x \times 6 x = 1944 \\ \Leftrightarrow 72 x^{3} = 1944 \\ \Leftrightarrow x^{3} = 27 \\ \Leftrightarrow x = 3 \end{aligned}

∴ Largest number = 6x = 18

64. The sum of the squares of two numbers is 3341 and the difference of their squares is 891. The numbers are :

A. 25, 36

B. 25, 46

C. 35, 46

D. None of these

Discuss

Solution:

Let the numbers be x and y.
Then,
x² + y² = 3341..... (i)
And,
x² - y² = 891..... (ii)
Adding (i) and (ii), we get :
2x² = 4232
or x² = 2116
or x = 46
Subtracting (ii) from (i), we get :
2y² = 2450
or y² = 1225
or y = 35
So, the numbers are 35 and 46

65. In a two-digit number, if it is known that its unit's digits exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is :

A. 24

B. 26

C. 42

D. 46

Discuss

Solution:

Let the ten's digit be x
Then, unit's digit = x + 2
∴ Number :
= 10x + (x + 2)
= 11x + 2
Sum of digits :
= x + (x + 2)
= 2x + 2

\begin{aligned} ∴ (11 x + 2) (2 x + 2) = 144 \\ \Leftrightarrow 22 x^{2} + 26 x - 140 = 0 \\ \Leftrightarrow 11 x^{2} + 13 x - 70 = 0 \\ \Leftrightarrow (x - 2) (11 x + 35) = 0 \\ \Leftrightarrow x = 2 \end{aligned}

Hence, required number:
= 11x + 2
= 11× 2 + 2
= 24

66. The difference between the numerator and the denominator of a fraction is 5. If 5 is added to its denominator, the fraction is decreased by $1 \frac{1}{4}$ . Find the value of the fraction.

A. $\frac{1}{6}$

B. $2 \frac{1}{4}$

C. $3 \frac{1}{4}$

D. $6$

Discuss

Solution:

Let the denominator be x
Then, numerator = x + 5
Now,

\begin{aligned} \Leftrightarrow \frac{x + 5}{x} - \frac{x + 5}{x + 5} = \frac{5}{4} \\ \Leftrightarrow \frac{x + 5}{x} = \frac{5}{4} + 1 \\ \Leftrightarrow \frac{x + 5}{x} = \frac{9}{4} \\ \Leftrightarrow \frac{x + 5}{x} = 2 \frac{1}{4} \end{aligned}

So, the fraction is

2 \frac{1}{4}

67. A number whose fifth part increased by 4 is equal to its fourth part diminished by 10, is :

A. 240

B. 260

C. 270

D. 280

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Leftrightarrow (\frac{1}{5} x + 4) = (\frac{1}{4} x - 10) \\ \Leftrightarrow \frac{x}{20} = 14 \\ \Leftrightarrow x = 14 \times 20 \\ \Leftrightarrow x = 280 \end{aligned}

68. The ratio between a two-digit number and the sum of the digits of that number is 4 : 1. If the digit in the unit's place is 3 more than the digit in the ten's place, then the number is ?

A. 24

B. 36

C. 63

D. 96

Discuss

Solution:

Let the ten's digit be x
Then, units digit = x + 3
Number = 10x + (x + 3)
= 11x + 3
Sum of digits = x + (x + 3)
= 2x + 3

\begin{aligned} ∴ \frac{11 x + 3}{2 x + 3} = \frac{4}{1} \\ \Leftrightarrow 11 x + 3 = 8 x + 12 \\ \Leftrightarrow 3 x = 9 \\ \Leftrightarrow x = 3 \end{aligned}

Hence, Required number
= 11x + 3
= 11 × 3 + 3
= 36

69. The difference between two positive integers is 3. If the sum of their squares is 369, then the sum of the numbers is :

A. 25

B. 27

C. 33

D. 81

Discuss

Solution:

Let the numbers be x and (x + 3)
Then,

\begin{aligned} \Leftrightarrow x^{2} + {(x + 3)}^{2} = 369 \\ \Leftrightarrow x^{2} + x^{2} + 9 + 6 x = 369 \\ \Leftrightarrow 2 x^{2} + 6 x - 360 = 0 \\ \Leftrightarrow x^{2} + 3 x - 180 = 0 \\ \Leftrightarrow (x + 15) (x - 12) = 0 \\ \Leftrightarrow x = 12 \end{aligned}

So, the numbers are 12 and 15
∴ Required sum = (12 + 15) = 27

70. A number consists of two digits. If the digits interchange place and the new number is added to the original number, then the resulting number will be divisible by :

A. 3

B. 5

C. 9

D. 11

Discuss

Solution:

Let the ten's digit be x and unit's digit be y
Then, number = 10x + y
Number obtained by interchanging the digits = 10y + x
∴ (10x + y) + (10y + x)
= 11(x + y), which is divisible by 11

1 2 3 4 5 6 7 8 9 10

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