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41. In a three-digit number, the digit in the unit's place is 75% of the digit in the ten's place. The digit in the ten's place is greater than the digit in the hundred's place by 1. If the sum of the digits in the ten's place and the hundred's place is 15. What is the number ?

A. 687

B. 786

C. 795

D. Cannot be determined

Discuss

Solution:

Let the hundred's digit = x
Then, ten's digit = (x + 1)
Unit's digit :

\begin{aligned} = 75 % of (x + 1) \\ = \frac{3}{4} (x + 1) \end{aligned}

\begin{aligned} ∴ (x + 1) + x = 15 \\ \Leftrightarrow 2 x = 14 \\ \Leftrightarrow x = 7 \end{aligned}

So, hundred's digit = 7
Ten's digit = 8
Unit's digit :

\begin{aligned} = \frac{3}{4} (x + 1) \\ = \frac{3}{4} (7 + 1) \\ = \frac{3}{4} (8) \\ = 6 \end{aligned}

Hence, required number = 786

42. The sum of two numbers is 37 and the difference of their squares is 185, then the difference between the two numbers is :

A. 10

B. 4

C. 5

D. 3

Discuss

Solution:

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

\begin{aligned} a + b = 37 & a^{2} - b^{2} = 185 \\ \Rightarrow (a + b) (a - b) = 185 \\ \Rightarrow 37 (a - b) = 185 \\ \Rightarrow a - b = \frac{185}{37} \\ \Rightarrow a - b = 5 \end{aligned}

43. The difference between $\frac{3}{5}$ th of $\frac{2}{3}$ rd of a number and $\frac{2}{5}$ th of $\frac{1}{4}$ th of the same number is 288. What is the number ?

A. 850

B. 895

C. 955

D. 960

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \frac{3}{5} of \frac{2}{3} of x - \frac{2}{5} of \frac{1}{4} of x = 288 \\ \Leftrightarrow (x \times \frac{3}{5} \times \frac{2}{3}) - (x \times \frac{2}{5} \times \frac{1}{4}) = 288 \\ \Leftrightarrow \frac{2}{5} x - \frac{1}{10} x = 288 \\ \Leftrightarrow \frac{3 x}{10} = 288 \\ \Leftrightarrow x = (\frac{288 \times 10}{3}) \\ \Leftrightarrow x = 960 \end{aligned}

44. The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals will be :

A. $\frac{12}{35}$

B. $\frac{1}{35}$

C. $\frac{35}{8}$

D. $\frac{7}{32}$

Discuss

Solution:

Let the numbers be x and y
Then,

\begin{aligned} x + y = 12 & x y = 35 \\ ∴ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{x y} = \frac{12}{35} \end{aligned}

45. The difference between two numbers is 16. If one-third of the smaller number is greater than one-seventh of the larger number by 4, then the two numbers are :

A. 9 and 25

B. 12 and 28

C. 33 and 49

D. 56 and 72

Discuss

Solution:

Let the numbers be x and (x + 16)
Then,

\begin{aligned} \Leftrightarrow \frac{x}{3} - \frac{(x + 16)}{7} = 4 \\ \Leftrightarrow 7 x - 3 (x + 16) = 84 \\ \Leftrightarrow 4 x = 84 + 48 \\ \Leftrightarrow 4 x = 132 \\ \Leftrightarrow x = 33 \end{aligned}

Hence, the numbers are 33 and 49

46. If a number of two digits is k times the sum of its digits, then the number formed by interchanging the digits is the sum of the digits multiplied by :

A. k - 1

B. 11 - k

C. 9 + k

D. 10 - k

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 = k (x + y) \\ \Rightarrow k = \frac{10 x + y}{x + y} \end{aligned}

Number formed by interchanging the digits = 10y + x
Let, 10y + x = h(x + y)
Then,

\begin{aligned} h = \frac{10 y + x}{x + y} \\ = \frac{11 (x + y) - (10 x + y)}{x + y} \\ = 11 - \frac{10 x + y}{x + y} \\ = 11 - k \end{aligned}

47. The product of two fractions is $\frac{14}{15}$ and their quotient is $\frac{35}{24}$ . The greater fraction is :

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

B. $\frac{7}{6}$

C. $\frac{7}{4}$

D. $\frac{7}{3}$

Discuss

Solution:

Let the two fractions be a and b
Then,

\begin{aligned} a b = \frac{14}{15} & \frac{a}{b} = \frac{35}{24} \\ \Leftrightarrow \frac{a b}{(\frac{a}{b})} = (\frac{14}{15} \times \frac{24}{35}) \\ \Leftrightarrow b^{2} = \frac{16}{25} \\ \Leftrightarrow b = \frac{4}{5} \\ So, \\ \Leftrightarrow a b = \frac{14}{15} \\ \Leftrightarrow a = (\frac{14}{15} \times \frac{5}{4}) \\ \Leftrightarrow a = \frac{7}{6} \end{aligned}

Since a > b,
So, greater fraction is

\frac{7}{6}

48. A man bought some eggs of which 10% are rotten. He gives 80% of the remainder to his neighbours. Now he is left out with 36 eggs. How many eggs he bought ?

A. 40

B. 100

C. 200

D. 72

Discuss

Solution:

Let the total number of eggs bought be a
10% of eggs are rotten
∴ Remaining eggs :

\begin{aligned} = a - 10 % of a \\ = a - \frac{10 a}{100} \\ = \frac{100 a - 10 a}{100} \\ = \frac{90 a}{100} \\ = \frac{9 a}{10} \end{aligned}

Man gives 80% of

\frac{9 a}{100}

eggs to his neighbour

\begin{aligned} = \frac{80}{100} \times \frac{9 a}{10} \\ = \frac{72 a}{100} \end{aligned}

Remaining eggs :

\begin{aligned} = \frac{9 a}{10} - \frac{72 a}{100} \\ = \frac{90 a - 72 a}{100} \\ = \frac{18 a}{100} \\ = \frac{9 a}{50} \end{aligned}

According the question,

\begin{aligned} \Rightarrow \frac{9 a}{50} = 36 \\ \Rightarrow 9 a = 36 \times 50 \\ \Rightarrow a = \frac{36 \times 50}{9} \\ \Rightarrow a = 200 \end{aligned}

Hence, the total number of eggs bought be = 200

49. A number is double and 9 is added. If the resultant is trebled, it becomes 75. What is that number ?

A. 3.5

B. 6

C. 8

D. None of these

Discuss

Solution:

Let the number be x
Then,
⇔ 3(2x + 9) = 75
⇔ 2x + 9 = 25
⇔ 2x = 16
⇔ x = 8

50. The sum of a positive number and its reciprocal is thrice the difference of the number and its reciprocal. The number is :

A. $\sqrt{2}$

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

C. $\sqrt{3}$

D. $\frac{1}{\sqrt{3}}$

Discuss

Solution:

Let the number be x
Then,

\begin{aligned} \Leftrightarrow x + \frac{1}{x} = 3 (x - \frac{1}{x}) \\ \Leftrightarrow \frac{x^{2} + 1}{x} = 3 (\frac{x^{2} - 1}{x}) \\ \Leftrightarrow x^{2} + 1 = 3 x^{2} - 3 \\ \Leftrightarrow 2 x^{2} = 4 \\ \Leftrightarrow x^{2} = 2 \\ \Leftrightarrow x = \sqrt{2} \end{aligned}

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