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51. The sum of two numbers is 40 and their product is 375. What will be the sum of their reciprocals ?

A. $\frac{1}{40}$

B. $\frac{8}{75}$

C. $\frac{75}{4}$

D. $\frac{75}{8}$

Discuss

Solution:

Let the numbers be x and y
Then,
x + y = 40 and xy = 375

\begin{aligned} ∴ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{x y} \\ = \frac{40}{275} \\ = \frac{8}{75} \end{aligned}

52. A two-digit number is 7 times the sum of its two digits. The number that is formed by reversing its digits is 18 less than the original number. What is the number ?

A. 42

B. 52

C. 62

D. 72

Discuss

Solution:

Let the ten's digit be x and the unit's digit be y
Then, number = 10x + y

\begin{aligned} ∴ 10 x + y = 7 (x + y) \\ \Leftrightarrow 3 x = 6 y \\ \Leftrightarrow x = 2 y \end{aligned}

Number formed by reversing the digits = 10y + x

\begin{aligned} ∴ (10 x + y) - (10 y + x) = 18 \\ \Leftrightarrow 9 x - 9 y = 18 \\ \Leftrightarrow x - y = 2 \\ \Leftrightarrow 2 y - y = 2 \\ \Leftrightarrow y = 2 \\ So, x = 2 y = 4 \end{aligned}

Hence,
∴ Required number
= 10x + y
= 40 + 2
= 42

53. If the difference between the reciprocal of a positive proper fraction and the fraction itself be $\frac{9}{20}$ , then the fraction is :

A. $\frac{3}{5}$

B. $\frac{4}{5}$

C. $\frac{5}{4}$

D. $\frac{3}{10}$

Discuss

Solution:

Let the fraction be

\frac{a}{1}

Then,

\begin{aligned} \Leftrightarrow \frac{1}{a} - a = \frac{9}{20} \\ \Leftrightarrow \frac{1 - a^{2}}{a} = \frac{9}{20} \\ \Leftrightarrow 20 - 20 a^{2} = 9 a \\ \Leftrightarrow 20 a^{2} + 9 a - 20 = 0 \\ \Leftrightarrow 20 a^{2} + 25 a - 16 a - 20 = 0 \\ \Leftrightarrow 5 a (4 a + 5) - 4 (4 a + 5) = 0 \\ \Leftrightarrow (4 a + 5) (5 a - 4) = 0 \\ \Leftrightarrow a = \frac{4}{5} [∵ a \neq - \frac{5}{4}] \end{aligned}

54. The sum of two numbers is 75 and their difference is 25. The product of the two numbers is :

A. 1350

B. 1250

C. 125

D. 1000

Discuss

Solution:

Let the numbers be a and b
According to the question,

\begin{aligned} a + b = 75 \\ a - b = 25 \\ ∵ {(a + b)}^{2} - {(a - b)}^{2} = 4 a b \\ \Rightarrow 75^{2} - 25^{2} = 4 a b \\ \Rightarrow 4 a b = (75 + 25) (75 - 25) \\ [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ \Rightarrow 4 a b = 100 \times 50 \\ \Rightarrow a b = \frac{100 \times 50}{4} \\ \Rightarrow a b = 1250 \end{aligned}

55. Three-forth of a number is 60 more than its one-third. The number is :

A. 84

B. 108

C. 144

D. None of these

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Leftrightarrow \frac{3}{4} x - \frac{1}{3} x = 60 \\ \Leftrightarrow \frac{5 x}{12} = 60 \\ \Leftrightarrow x = (\frac{60 \times 12}{5}) \\ \Leftrightarrow x = 144 \end{aligned}

56. The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is :

A. $\frac{1}{289}$

B. $\frac{289}{290}$

C. $\frac{290}{289}$

D. $289$

Discuss

Solution:

Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,

\begin{aligned} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \\ = \frac{a^{2} + b^{2}}{a^{2} b^{2}} \\ = \frac{1^{2} + {(17)}^{2}}{{(1 \times 17)}^{2}} \\ = \frac{290}{289} \end{aligned}

57. A number whose fifth part increase 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{x}{5} + 4 = \frac{x}{4} - 10 \\ \Leftrightarrow \frac{x}{4} - \frac{x}{5} = 14 \\ \Leftrightarrow \frac{x}{20} = 14 \\ \Leftrightarrow x = 14 \times 20 \\ \Leftrightarrow x = 280 \end{aligned}

58. If $2 \frac{1}{2}$ is added tp a number and the sum multiplied by $4 \frac{1}{2}$ and 3 is added to the product and the sum is divided by $1 \frac{1}{5}$ , the quotient becomes 25. What is the number ?

A. $2 \frac{1}{2}$

B. $3 \frac{1}{2}$

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

D. $5 \frac{1}{2}$

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Leftrightarrow \frac{4 \frac{1}{2} (x + 2 \frac{1}{2}) + 3}{1 \frac{1}{5}} = 25 \\ \Leftrightarrow \frac{\frac{9}{2} (x + \frac{5}{2}) + 3}{\frac{6}{5}} = 25 \\ \Leftrightarrow \frac{9 x}{2} + \frac{45}{4} + 3 = 25 \times \frac{6}{5} \\ \Leftrightarrow \frac{9 x}{2} + \frac{45}{4} + 3 = 30 \\ \Leftrightarrow \frac{9 x}{2} = 30 - \frac{57}{4} \\ \Leftrightarrow \frac{9 x}{2} = \frac{63}{4} \\ \Leftrightarrow x = (\frac{63}{4} \times \frac{2}{9}) \\ \Leftrightarrow x = \frac{7}{2} \\ \Leftrightarrow x = 3 \frac{1}{2} \end{aligned}

59. The product of two numbers is 120 and the sum of their square is 289. The sum of the numbers is :

A. 20

B. 23

C. 169

D. None of these

Discuss

Solution:

Let the numbers be x and y
Then,
xy = 120 and x² + y² = 289

\begin{aligned} ∴ {(x + y)}^{2} \\ = x^{2} + y^{2} + 2 x y \\ = 289 + 240 \\ = 529 \\ ∴ x + y \\ = \sqrt{529} \\ = 23 \end{aligned}

60. If the digit in the unit's place of a two-digit number is halved and the digit in the ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true ?

A. Sum of the digits is a two-digit number

B. Digit in the unit's place is half of the digit in the ten's place

C. Digit in the unit's place and the ten's place are equal

D. Digit in the unit's place is twice the digit in the ten's place

Discuss

Solution:

Let the ten's digit be x and the unit's digit be y
Then, number = 10x + y
New number :

\begin{aligned} = 10 \times 2 x + \frac{y}{2} \\ = 20 x + \frac{y}{2} \end{aligned}

\begin{aligned} ∴ 20 x + \frac{y}{2} = 10 y + x \\ \Leftrightarrow 40 x + y = 20 y + 2 x \\ \Leftrightarrow 38 x = 19 y \\ \Leftrightarrow y = 2 x \end{aligned}

So, the unit's digit is twice the ten's digit.

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