Practice MCQ Questions and Answer on Problems on Numbers

51.

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) $$\frac{3}{10}$$

Show Answer

 Let the fraction be $$\frac{a}{1}$$ Then, $$\eqalign{ & \Leftrightarrow \frac{1}{a} - a = \frac{9}{{20}} \cr & \Leftrightarrow \frac{{1 - {a^2}}}{a} = \frac{9}{{20}} \cr & \Leftrightarrow 20 - 20{a^2} = 9a \cr & \Leftrightarrow 20{a^2} + 9a - 20 = 0 \cr & \Leftrightarrow 20{a^2} + 25a - 16a - 20 = 0 \cr & \Leftrightarrow 5a\left( {4a + 5} \right) - 4\left( {4a + 5} \right) = 0 \cr & \Leftrightarrow \left( {4a + 5} \right)\left( {5a - 4} \right) = 0 \cr & \Leftrightarrow a = \frac{4}{5}\,\,\,\,\,\,\,\,\left[ {\because a \ne - \frac{5}{4}} \right] \cr} $$

52.

A number consists of two digits. If the digits interchange places 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

Show Answer

 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

53.

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) $$\frac{75}{8}$$

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 Let the numbers be x and y Then, x + y = 40 and xy = 375 $$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40}}{{275}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{75}} \cr} $$

54.

A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is ?

• (A) 23
• (B) 27
• (C) 32
• (D) 72

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 Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Rightarrow 10x + y = 3\left( {x + y} \right) \cr & \Rightarrow 7x - 2y = 0.....(i) \cr & 10x + y + 45 = 10y + x \cr & \Rightarrow y - x = 5.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 2 and y = 7 ∴ Required number = 27

55.

A student was asked to divide the half of a certain number by 6 and the other half by 4 and then to add the two quantities so obtained. Instead of doing so the student divided the number by 5 and the result fell short by 4. The given number was ?

• (A) 240
• (B) 288
• (C) 384
• (D) 480

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 Let the number be x Then, $$\eqalign{ & \Leftrightarrow \left[ {\frac{{\left( {\frac{x}{2}} \right)}}{6} + \frac{{\left( {\frac{x}{2}} \right)}}{4}} \right] - \frac{x}{5} = 4 \cr & \Leftrightarrow \frac{x}{{12}} + \frac{x}{8} - \frac{x}{5} = 4 \cr & \Leftrightarrow \frac{{10x + 15x - 24x}}{{120}} = 4 \cr & \Leftrightarrow x = 4 \times 120 \cr & \Leftrightarrow x = 480 \cr} $$

56.

In a two-digit positive number, the digit in the unit's place is equal to the square of the digit in ten's place, and the difference between the number and the number obtained by interchanging the digits is 54. What is 40% of the original number ?

• (A) 15.6
• (B) 24
• (C) 37.2
• (D) 39

Show Answer

 Let ten's digit = x Then, unit's digit = x2 Then, number = 10x + x2 Clearly, since x2 > x, So, the number formed by interchanging the digits is greater than the original number. $$\eqalign{ & \therefore \left( {10{x^2} + x} \right) - \left( {10x + {x^2}} \right) = 54 \cr & \Leftrightarrow 9{x^2} - 9x = 54 \cr & \Leftrightarrow {x^2} - x = 6 \cr & \Leftrightarrow {x^2} - x - 6 = 0 \cr & \Leftrightarrow {x^2} - 3x + 2x - 6 = 0 \cr & \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0 \cr & \Leftrightarrow x = 3 \cr} $$ So. ten's digit = 3, unit's digit = 32 = 9 ∴ Original number = 39 Required result : = 40% of 39 = 15.6

57.

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

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

58.

The sum of five consecutive odd numbers is 575. What is the sum of the next set of five consecutive odd numbers ?

• (A) 595
• (B) 615
• (C) 635
• (D) Cannot be determined

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Answer & Solution Answer: Option E Solution: Let the five numbers be x, (x + 2), (x + 4), (x + 6) and (x + 8) Then, ⇔ x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 575 ⇔ 5x + 20 = 575 ⇔ 5x = 555 ⇔ x = 111 ∴ Required sum : = (x + 10) + (x + 12) + (x + 14) + (x + 16) + (x + 18) = 5x + 70 = 5 × 111 + 70 = 555 + 70 = 625

59.

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

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 Let the number be x Then, $$\eqalign{ & \Leftrightarrow x - 4 = \frac{{21}}{x} \cr & \Leftrightarrow {x^2} - 4x - 21 = 0 \cr & \Leftrightarrow \left( {x - 7} \right)\left( {x + 3} \right) = 0 \cr & \Leftrightarrow x = 7 \cr} $$

60.

The number obtained by interchanging the two digits of a two-digit number is lesser than the original number by 54. If the sum of the two digit of the number is 12, then what is the original number ?

• (A) 28
• (B) 39
• (C) 82
• (D) Cannot be determined

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Answer & Solution Answer: Option E Solution: Let ten's digit = x Then, unit's digit = (12 - x) $$\therefore \left[ {10x + \left( {12 - x} \right)} \right] - $$    $$\left[ {10\left( {12 - x} \right) + x} \right]$$   $$ = 54$$ $$\eqalign{ & \Leftrightarrow 18x - 108 = 54 \cr & \Leftrightarrow 18x = 162 \cr & \Leftrightarrow x = 9 \cr} $$ So, ten's digit = 9 and unit's digit = 3 Hence, original number = 93