Practice | Hashtable

content Computer Science Arithmetic Aptitude Data Interpretation Logical Reasoning Verbal Reasoning Non-Verbal Reasoning

About About Privacy Policy

Hashtable

Home
Practice
PYQ
Mock Test
Register Free

Home Practice PYQ Mock Test Register Free

Home > Practice > Arithmetic Aptitude > Problem on Numbers > Miscellaneous

31. If a number is multiplied by two-third of itself the value so obtained is 864. What is the number ?

A. 34

B. 36

C. 38

D. 44

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Leftrightarrow x \times \frac{2}{3} x = 864 \\ \Leftrightarrow \frac{2}{3} x^{2} = 864 \\ \Leftrightarrow x^{2} = (\frac{864 \times 3}{2}) \\ \Leftrightarrow x^{2} = 1296 \\ \Leftrightarrow x = \sqrt{1296} \\ \Leftrightarrow x = 36 \end{aligned}

32. The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is :

A. 380

B. 395

C. 400

D. 425

Discuss

Solution:

Let the numbers be x and y
Then,

x y = 9375 and \frac{x}{y} = 15

\begin{aligned} \Leftrightarrow \frac{x y}{(\frac{x}{y})} = \frac{9375}{15} \\ \Leftrightarrow y^{2} = 625 \\ \Leftrightarrow y = 25 \\ ∴ x = 15 y \\ \Rightarrow x = 15 \times 25 \\ \Rightarrow x = 375 \end{aligned}

∴ Sum of the numbers :
= 375 + 25
= 400

33. If the product of three consecutive integers is 120, then the sum of the integers is :

A. 9

B. 12

C. 14

D. 15

Discuss

Solution:

\begin{aligned} 120 : \\ = 2 \times 2 \times 2 \times 3 \times 5 \\ = (2 \times 2) \times 5 \times (2 \times 3) \\ = 4 \times 5 \times 6 \end{aligned}

Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6
Required sum :
= 4 + 5 + 6
= 15

34. A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is :

A. 18

B. 24

C. 42

D. 81

Discuss

Solution:

Let the ten's and unit's digits be

x

and

\frac{8}{x}

respectively
Then,

\begin{aligned} \Leftrightarrow (10 x + \frac{8}{x}) + 18 = 10 \times \frac{8}{x} + x \\ \Leftrightarrow 10 x^{2} + 8 + 18 x = 80 + x^{2} \\ \Leftrightarrow 9 x^{2} + 18 x - 72 = 0 \\ \Leftrightarrow x^{2} + 2 x - 8 = 0 \\ \Leftrightarrow (x + 4) (x - 2) = 0 \\ \Leftrightarrow x = 2 \end{aligned}

So, ten's digit = 2 and unit's digit = 4
Hence, required number = 24

35. The sum of the squares of three numbers is 138, while the sum of their product taken two at a time is 131. Their sum is :

A. 20

B. 30

C. 40

D. None of these

Discuss

Solution:

Let the numbers be a, b and c.
Then, a² + b² + c² = 138
And (ab + bc + ca) = 131
∴ (a + b + c)²
= a² + b² + c² + 2(ab + bc + ca)
= 138 + 2 × 131
= 400
⇒ (a + b + c) =

\sqrt{400}

= 20

36. The sum of the squares of two positive integers is 100 and the difference of their squares is 28. The sum of the numbers is ?

A. 12

B. 13

C. 14

D. 15

Discuss

Solution:

Let the positive integers be a and b where a > b
According to the question,

\begin{aligned} a^{2} + b^{2} = 100..... (i) \\ a^{2} - b^{2} = 28..... (ii) \end{aligned}

By adding (i) and (ii), we get :

\begin{aligned} ∴ a^{2} + b^{2} + a^{2} - b^{2} = 100 + 28 \\ \Rightarrow 2 a^{2} = 128 \\ \Rightarrow a^{2} = \frac{128}{2} \\ \Rightarrow a = \sqrt{64} \\ \Rightarrow a = 8 \end{aligned}

From equation (i)

\begin{aligned} \Rightarrow 8^{2} + b^{2} = 100 \\ \Rightarrow b^{2} = 100 - 64 \\ \Rightarrow b = \sqrt{36} \\ \Rightarrow b = 6 \\ ∴ a + b = 8 + 6 \\ = 14 \end{aligned}

37. A number when multiplied by 13 is increased by 180. The number is :

A. 5

B. 12

C. 15

D. 45

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Rightarrow 13 x = x + 180 \\ \Rightarrow 12 x = 180 \\ \Rightarrow x = \frac{180}{12} \\ \Rightarrow x = 15 \end{aligned}

38. A positive number when decreased by 4 is equal to 21 times the reciprocal of the number. The number is ?

A. 3

B. 5

C. 7

D. 9

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Leftrightarrow x - 4 = \frac{21}{x} \\ \Leftrightarrow x^{2} - 4 x - 21 = 0 \\ \Leftrightarrow (x - 7) (x + 3) = 0 \\ \Leftrightarrow x = 7 \end{aligned}

39. The difference between two numbers is 1365. When the large number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is :

A. 240

B. 270

C. 295

D. 360

Discuss

Solution:

Let the numbers be x and (x + 1365)
Then,
⇒ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270

40. A number of two digits has 3 for its unit's digit, and the sum of digits is $\frac{1}{7}$ of the itself. The number is :

A. 43

B. 63

C. 53

D. 73

Discuss

Solution:

Let the ten's digit be x
Then, number = 10x + 3 and
Sum of digits = (x + 3)
So, (x + 3) =

\frac{1}{7}

(10x + 3)
⇔ 7x + 21 = 10x + 3
⇔ 3x = 18
⇔ x = 6
Hence, the number is :
= (10x + 3)
= (10 × 6 + 3)
= 63

1 2 3 4 5 6 7 8 9 10

Sign up for our newsletter

Are you passionate about opportunities in the realm of Computer Science within the government sector? Don't miss out on crucial updates, job openings, and career prospects tailored specifically straight to your inbox.

Email address

By subscribing, you agree with our Terms of Service and Privacy Policy.

Resources

Practice PYQ Mock Test

Company

About us Contact us Privacy policy Terms of service

Hashtable