Practice MCQ Questions and Answer on Problems on Numbers

81.

A fraction is such that if the double of the numerator and the triple pf the denominator is changed by +10 percent and -30 percent respectively, then we get 11 percent of $$\frac{16}{21}$$. Find the fraction :

• (A) $$\frac{2}{25}$$
• (B) $$\frac{3}{25}$$
• (C) $$\frac{4}{25}$$
• (D) None of these

Show Answer

 Let the fraction be $$\frac{x}{y}$$ Then, $$\eqalign{ & \Leftrightarrow \frac{{110\% {\text{ of 2}}x}}{{70\% {\text{ of 3}}y}} = 11\% {\text{ of }}\frac{{16}}{{21}} \cr & \Leftrightarrow \frac{{22x}}{{21y}} = \frac{{11}}{{100}} \times \frac{{16}}{{21}} \cr & \Leftrightarrow \frac{x}{y} = \left( {\frac{{11}}{{100}} \times \frac{{16}}{{21}} \times \frac{{21}}{{22}}} \right) \cr & \Leftrightarrow \frac{x}{y} = \frac{2}{{25}} \cr} $$

82.

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

Show Answer

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

83.

The product of three consecutive even numbers when divided by 8 is 720. The product of their square roots is :

• (A) $$12\sqrt{10}$$
• (B) $$24\sqrt{10}$$
• (C) 120
• (D) None of these

Show Answer

 Let the numbers be x, x + 2 and x + 4 Then, $$\eqalign{ & \Leftrightarrow \frac{{x\left( {x + 2} \right)\left( {x + 4} \right)}}{8} = 720 \cr & \Leftrightarrow x\left( {x + 2} \right)\left( {x + 4} \right) = 5760 \cr & \therefore \sqrt x \times \sqrt {\left( {x + 2} \right)} \times \sqrt {\left( {x + 4} \right)} \cr & = \sqrt {x\left( {x + 2} \right)\left( {x + 4} \right)} \cr & = \sqrt {5760} \cr & = 24\sqrt {10} \cr} $$

84.

If the numerator of a fraction is increased by $$\frac{1}{4}$$ and the denominator is decreased byy $$\frac{1}{3}$$, the new fraction obtained is $$\frac{33}{64}$$. What was the original fraction ?

• (A) $$\frac{3}{7}$$
• (B) $$\frac{5}{7}$$
• (C) $$\frac{7}{9}$$
• (D) Cannot be determined

Show Answer

 Let the fraction be $$\frac{x}{y}$$ Then, $$\eqalign{ & \Leftrightarrow \frac{{x + \frac{1}{4}}}{{y - \frac{1}{3}}} = \frac{{33}}{{64}} \cr & \Leftrightarrow \frac{{3\left( {4x + 1} \right)}}{{4\left( {3y - 1} \right)}} = \frac{{33}}{{64}} \cr & \Leftrightarrow \frac{{4x + 1}}{{3y - 1}} = \frac{{33}}{{64}} \times \frac{4}{3} \cr & \Leftrightarrow \frac{{4x + 1}}{{3y - 1}} = \frac{{11}}{{16}} \cr & \Leftrightarrow 16\left( {4x + 1} \right) = 11\left( {3y - 1} \right) \cr & \Leftrightarrow 64x + 16 = 33y - 11 \cr & \Leftrightarrow 64x - 33y = - 27 \cr} $$ Which cannot be solved to find $$\frac{x}{y}$$ Hence, the original fraction cannot be determined from the given data.

85.

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

Show Answer

 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 $$\eqalign{ & \therefore \frac{{11x + 3}}{{2x + 3}} = \frac{4}{1} \cr & \Leftrightarrow 11x + 3 = 8x + 12 \cr & \Leftrightarrow 3x = 9 \cr & \Leftrightarrow x = 3 \cr} $$ Hence, Required number = 11x + 3 = 11 × 3 + 3 = 36

86.

The difference between a number and its three-fifths is 50, What is the number ?

• (A) 75
• (B) 100
• (C) 125
• (D) None of these

Show Answer

 Let the number be x Then, $$\eqalign{ & \Leftrightarrow x - \frac{3}{5}x = 50 \cr & \Leftrightarrow \frac{2}{5}x = 50 \cr & \Leftrightarrow x = \left( {\frac{{50 \times 5}}{2}} \right) \cr & \Leftrightarrow x = 125 \cr} $$

87.

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) $$289$$

Show Answer

 Let the numbers be a and b Then, ab = 17 ⇒ a = 1 and b = 17 So,$$\eqalign{ & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr & = \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} \cr & = \frac{{{1^2} + {{\left( {17} \right)}^2}}}{{{{\left( {1 \times 17} \right)}^2}}} \cr & = \frac{{290}}{{289}} \cr} $$

88.

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

Show Answer

 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

89.

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

Show Answer

 Let the hundred's digit = x Then, ten's digit = (x + 1) Unit's digit : $$\eqalign{ & = 75\% {\text{ of }}\left( {x + 1} \right) \cr & = \frac{3}{4}\left( {x + 1} \right) \cr} $$ $$\eqalign{ & \therefore \left( {x + 1} \right) + x = 15 \cr & \Leftrightarrow 2x = 14 \cr & \Leftrightarrow x = 7 \cr} $$ So, hundred's digit = 7 Ten's digit = 8 Unit's digit : $$\eqalign{ & = \frac{3}{4}\left( {x + 1} \right) \cr & = \frac{3}{4}\left( {7 + 1} \right) \cr & = \frac{3}{4}\left( 8 \right) \cr & = 6 \cr} $$ Hence, required number = 786

90.

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

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