Problems On Numbers — Part 2 MCQs with Explanations

Let the numbers be x and y Then, $$\eqalign{ & x + y = 12\,\, \& \,\, xy = 35 \cr & \therefore \frac{1}{x} + \frac{1}{y} = \frac{{x + y}}{{xy}} = \frac{{12}}{{35}} \cr} $$

Let the numbers be x and (x + 16) Then, $$\eqalign{ & \Leftrightarrow \frac{x}{3} - \frac{{\left( {x + 16} \right)}}{7} = 4 \cr & \Leftrightarrow 7x - 3\left( {x + 16} \right) = 84 \cr & \Leftrightarrow 4x = 84 + 48 \cr & \Leftrightarrow 4x = 132 \cr & \Leftrightarrow x = 33 \cr} $$ Hence, the numbers are 33 and 49

Let the ten's digit be x and the unit's digit be y Then, number = 10x + y $$\eqalign{ & \therefore 10x + y = k\left( {x + y} \right) \cr & \Rightarrow k = \frac{{10x + y}}{{x + y}} \cr} $$ Number formed by interchanging the digits = 10y + x Let, 10y + x = h(x + y) Then, $$\eqalign{ & h = \frac{{10y + x}}{{x + y}} \cr & \,\,\,\,\, = \frac{{11\left( {x + y} \right) - \left( {10x + y} \right)}}{{x + y}} \cr & \,\,\,\,\, = 11 - \frac{{10x + y}}{{x + y}} \cr & \,\,\,\,\, = 11 - k \cr} $$

Let the two fractions be a and b Then, $$\eqalign{ & ab = \frac{{14}}{{15}}\, \& \, \frac{a}{b} = \frac{{35}}{{24}} \cr & \Leftrightarrow \frac{{ab}}{{\left( {\frac{a}{b}} \right)}} = \left( {\frac{{14}}{{15}} \times \frac{{24}}{{35}}} \right) \cr & \Leftrightarrow {b^2} = \frac{{16}}{{25}} \cr & \Leftrightarrow b = \frac{4}{5} \cr & {\text{So, }} \cr & \Leftrightarrow ab = \frac{{14}}{{15}} \cr & \Leftrightarrow a = \left( {\frac{{14}}{{15}} \times \frac{5}{4}} \right) \cr & \Leftrightarrow a = \frac{7}{6} \cr} $$ Since a > b, So, greater fraction is $$\frac{7}{6}$$

Let the total number of eggs bought be a 10% of eggs are rotten ∴ Remaining eggs : $$\eqalign{ & = a - 10\% {\text{ of }}a \cr & = a - \frac{{10a}}{{100}} \cr & = \frac{{100a - 10a}}{{100}} \cr & = \frac{{90a}}{{100}} \cr & = \frac{{9a}}{{10}} \cr} $$ Man gives 80% of $$\frac{{9a}}{{100}}$$ eggs to his neighbour $$\eqalign{ & = \frac{{80}}{{100}} \times \frac{{9a}}{{10}} \cr & = \frac{{72a}}{{100}} \cr} $$ Remaining eggs : $$\eqalign{ & = \frac{{9a}}{{10}} - \frac{{72a}}{{100}} \cr & = \frac{{90a - 72a}}{{100}} \cr & = \frac{{18a}}{{100}} \cr & = \frac{{9a}}{{50}} \cr} $$ According the question, $$\eqalign{ & \Rightarrow \frac{{9a}}{{50}} = 36 \cr & \Rightarrow 9a = 36 \times 50 \cr & \Rightarrow a = \frac{{36 \times 50}}{9} \cr & \Rightarrow a = 200 \cr} $$ Hence, the total number of eggs bought be = 200

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

Let the number be x Then, $$\eqalign{ & \Leftrightarrow x + \frac{1}{x} = 3\left( {x - \frac{1}{x}} \right) \cr & \Leftrightarrow \frac{{{x^2} + 1}}{x} = 3\left( {\frac{{{x^2} - 1}}{x}} \right) \cr & \Leftrightarrow {x^2} + 1 = 3{x^2} - 3 \cr & \Leftrightarrow 2{x^2} = 4 \cr & \Leftrightarrow {x^2} = 2 \cr & \Leftrightarrow x = \sqrt 2 \cr} $$

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

Let the ten's digit be x and the unit's digit be y Then, number = 10x + y $$\eqalign{ & \therefore 10x + y = 7\left( {x + y} \right) \cr & \Leftrightarrow 3x = 6y \cr & \Leftrightarrow x = 2y \cr} $$ Number formed by reversing the digits = 10y + x $$\eqalign{ & \therefore \left( {10x + y} \right) - \left( {10y + x} \right) = 18 \cr & \Leftrightarrow 9x - 9y = 18 \cr & \Leftrightarrow x - y = 2 \cr & \Leftrightarrow 2y - y = 2 \cr & \Leftrightarrow y = 2 \cr & {\text{So, }}x = 2y = 4 \cr} $$ Hence, ∴ Required number = 10x + y = 40 + 2 = 42

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

Let the number be x Then, $$\eqalign{ & \Leftrightarrow \frac{3}{4}x - \frac{1}{3}x = 60 \cr & \Leftrightarrow \frac{{5x}}{{12}} = 60 \cr & \Leftrightarrow x = \left( {\frac{{60 \times 12}}{5}} \right) \cr & \Leftrightarrow x = 144 \cr} $$

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

Let the number be x Then, $$\eqalign{ & \Leftrightarrow \frac{x}{5} + 4 = \frac{x}{4} - 10 \cr & \Leftrightarrow \frac{x}{4} - \frac{x}{5} = 14 \cr & \Leftrightarrow \frac{x}{{20}} = 14 \cr & \Leftrightarrow x = 14 \times 20 \cr & \Leftrightarrow x = 280 \cr} $$

Let the number be x Then, $$\eqalign{ & \Leftrightarrow \frac{{4\frac{1}{2}\left( {x + 2\frac{1}{2}} \right) + 3}}{{1\frac{1}{5}}} = 25 \cr & \Leftrightarrow \frac{{\frac{9}{2}\left( {x + \frac{5}{2}} \right) + 3}}{{\frac{6}{5}}} = 25 \cr & \Leftrightarrow \frac{{9x}}{2} + \frac{{45}}{4} + 3 = 25 \times \frac{6}{5} \cr & \Leftrightarrow \frac{{9x}}{2} + \frac{{45}}{4} + 3 = 30 \cr & \Leftrightarrow \frac{{9x}}{2} = 30 - \frac{{57}}{4} \cr & \Leftrightarrow \frac{{9x}}{2} = \frac{{63}}{4} \cr & \Leftrightarrow x = \left( {\frac{{63}}{4} \times \frac{2}{9}} \right) \cr & \Leftrightarrow x = \frac{7}{2} \cr & \Leftrightarrow x = 3\frac{1}{2} \cr} $$

Let the numbers be x and y Then, xy = 120 and x 2 + y 2 = 289 $$\eqalign{ & \therefore {\left( {x + y} \right)^2} \cr & = {x^2} + {y^2} + 2xy \cr & = 289 + 240 \cr & = 529 \cr & \therefore x + y \cr & = \sqrt {529} \cr & = 23 \cr} $$

Let the ten's digit be x and the unit's digit be y Then, number = 10x + y New number : $$\eqalign{ & = 10 \times 2x + \frac{y}{2} \cr & = 20x + \frac{y}{2} \cr} $$ $$\eqalign{ & \therefore 20x + \frac{y}{2} = 10y + x \cr & \Leftrightarrow 40x + y = 20y + 2x \cr & \Leftrightarrow 38x = 19y \cr & \Leftrightarrow y = 2x \cr} $$ So, the unit's digit is twice the ten's digit.

Let the fraction be $$\frac{x}{y}$$ Then, $$\eqalign{ & \Leftrightarrow x + y = 11.....(\text{i}) \cr & \Leftrightarrow \frac{{x + 1}}{{y - 2}} = \frac{2}{3} \cr & \Leftrightarrow 3\left( {x + 1} \right) = 2\left( {y - 2} \right) \cr & \Leftrightarrow 3x - 2y = - 7.....(\text{ii}) \cr} $$ Solving (i) and (ii), we get: x = 3 and y = 8 So, the fraction is $$\frac{3}{8}$$

Let the five successive odd number be, x, x + 2, x + 4, x + 6, x + 8 Then, according to given information, 185 = x + x + 2 + x + 4 + x + 6 + x + 8 $$\eqalign{ & \Leftrightarrow 185 = 5x + 20 \cr & \Leftrightarrow 5x = 185 - 20 \cr & \Leftrightarrow 5x = 165 \cr & \Leftrightarrow x = 33 \cr} $$ Highest number = 33 + 8 = 41

Let the numbers be 3x, 4x and 6x Then, $$\eqalign{ & \Leftrightarrow 3x \times 4x \times 6x = 1944 \cr & \Leftrightarrow 72{x^3} = 1944 \cr & \Leftrightarrow {x^3} = 27 \cr & \Leftrightarrow x = 3 \cr} $$ ∴ Largest number = 6x = 18

Let the numbers be x and y. Then, x 2 + y 2 = 3341..... (i) And, x 2 - y 2 = 891..... (ii) Adding (i) and (ii), we get : 2x 2 = 4232 or x 2 = 2116 or x = 46 Subtracting (ii) from (i), we get : 2y 2 = 2450 or y 2 = 1225 or y = 35 So, the numbers are 35 and 46

Let the ten's digit be x Then, unit's digit = x + 2 ∴ Number : = 10x + (x + 2) = 11x + 2 Sum of digits : = x + (x + 2) = 2x + 2 $$\eqalign{ & \therefore \left( {11x + 2} \right)\left( {2x + 2} \right) = 144 \cr & \Leftrightarrow 22{x^2} + 26x - 140 = 0 \cr & \Leftrightarrow 11{x^2} + 13x - 70 = 0 \cr & \Leftrightarrow \left( {x - 2} \right)\left( {11x + 35} \right) = 0 \cr & \Leftrightarrow x = 2 \cr} $$ Hence, required number: = 11x + 2 = 11× 2 + 2 = 24

Let the denominator be x Then, numerator = x + 5 Now, $$\eqalign{ & \Leftrightarrow \frac{{x + 5}}{x} - \frac{{x + 5}}{{x + 5}} = \frac{5}{4} \cr & \Leftrightarrow \frac{{x + 5}}{x} = \frac{5}{4} + 1 \cr & \Leftrightarrow \frac{{x + 5}}{x} = \frac{9}{4} \cr & \Leftrightarrow \frac{{x + 5}}{x} = 2\frac{1}{4} \cr} $$ So, the fraction is $$2\frac{1}{4}$$

Let the number be x Then, $$\eqalign{ & \Leftrightarrow \left( {\frac{1}{5}x + 4} \right) = \left( {\frac{1}{4}x - 10} \right) \cr & \Leftrightarrow \frac{x}{{20}} = 14 \cr & \Leftrightarrow x = 14 \times 20 \cr & \Leftrightarrow x = 280 \cr} $$

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

Let the numbers be x and (x + 3) Then, $$\eqalign{ & \Leftrightarrow {x^2} + {\left( {x + 3} \right)^2} = 369 \cr & \Leftrightarrow {x^2} + {x^2} + 9 + 6x = 369 \cr & \Leftrightarrow 2{x^2} + 6x - 360 = 0 \cr & \Leftrightarrow {x^2} + 3x - 180 = 0 \cr & \Leftrightarrow \left( {x + 15} \right)\left( {x - 12} \right) = 0 \cr & \Leftrightarrow x = 12 \cr} $$ So, the numbers are 12 and 15 ∴ Required sum = (12 + 15) = 27

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

Let the fraction be $$\frac{x}{y}$$ Then, $$\eqalign{ & \Leftrightarrow \frac{{x + 2}}{{y + 3}} = \frac{7}{9} \cr & \Leftrightarrow 9x - 7y = 3.....(i) \cr & {\text{And,}} \cr & \Leftrightarrow \frac{{x - 1}}{{y - 1}} = \frac{4}{5} \cr & \Leftrightarrow 5x - 4y = 1.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 5 and y = 6 Hence, the original fraction is $$\frac{5}{6}$$

Let the number be x Then, $$\eqalign{ & \Leftrightarrow x - 12 = 20\% {\text{ of }}x \cr & \Leftrightarrow x - \frac{x}{5} = 12 \cr & \Leftrightarrow \frac{{4x}}{5} = 12 \cr & \Leftrightarrow x = \frac{{12 \times 5}}{4} \cr & \Leftrightarrow x = 15 \cr} $$

Let the numbers be 4x and 7x Then, $$\eqalign{ & \Leftrightarrow \frac{{4x + 4}}{{7x + 4}} = \frac{3}{5} \cr & \Leftrightarrow 5\left( {4x + 4} \right) = 3\left( {7x + 4} \right) \cr & \Leftrightarrow x = 8 \cr} $$ ∴ Larger number : = 7x = 7 × 8 = 56

Let the number be x and y Then, $${x^2} - {y^2} = 63\,\,\,\,\& \,\,\,\,x - y = 3$$ On dividing, we get: x + y = 21 Solving x + y = 21 and x - y = 3, We get: x = 12 and y = 9 ∴ Larger number = 12

Let the integer be x Them, $$\eqalign{ & \Leftrightarrow {x^2} - 20x = 96 \cr & \Leftrightarrow {x^2} - 20x - 96 = 0 \cr & \Leftrightarrow \left( {x + 4} \right)\left( {x - 24} \right) = 0 \cr & \Leftrightarrow x = 24 \cr} $$

Let the numbers be x and y Then, 2x + 3y = 100 ..... (i) And, 3x + 2y = 120 ..... (ii) Adding (i) and (ii), we get : 5x + 5y = 220 or x + y = 44 ..... (iii) Subtracting (i) from (ii): we get : x - y = 20 ..... (iv) Adding (iii) and (iv), we get : 2x = 64 or x = 32 Putting x = 32 in (iii), we get : y = 12 Hence, larger number = 32

Let the number be x Then, $$\eqalign{ & \Rightarrow \frac{1}{7}x - \frac{1}{{11}}x = 100 \cr & \Rightarrow \frac{{4x}}{{77}} = 100 \cr & \Rightarrow x = \frac{{7700}}{4} \cr & \Rightarrow x = 1925 \cr} $$

Let the second number be x Then, first number = 2x, And third number = $$\frac{2x}{3}$$ $$\eqalign{ & \therefore 2x + x + \frac{{2x}}{3} = 264 \cr & \Leftrightarrow \frac{{11x}}{3} = 264 \cr & \Leftrightarrow x = \left( {\frac{{264 \times 3}}{{11}}} \right) \cr & \Leftrightarrow x = 72 \cr} $$

Let A = x B = x + 2 C = x + 4 D = x + 6 and E = x + 8 Then, $$\eqalign{ & \Rightarrow A + C = 146 \cr & \Rightarrow x + \left( {x + 4} \right) = 146 \cr & \Rightarrow 2x = 142 \cr & \Rightarrow x = 71 \cr & \therefore E = x + 8 \cr & \,\,\,\,\,\,\,\,\,\,\, = 71 + 8 \cr & \,\,\,\,\,\,\,\,\,\,\, = 79 \cr} $$

Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \left( {10x + y} \right) - \left( {10y + x} \right) = 36 \cr & \Leftrightarrow 9\left( {x - y} \right) = 36 \cr & \Leftrightarrow x - y = 4 \cr} $$

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.

Let the number be x Then, ⇔ 2x + 20 = 8x - 4 ⇔ 6x = 24 ⇔ x = 4

Let the six numbers be, x , (x + 1), (x + 2), (x + 3), (x + 4) and (x + 5) Then, ⇔ x + (x + 1) + (x + 2) = 27 ⇔ 3x + 3 = 27 Required sum : = (x + 3) + (x + 4) + (x + 5) = 3x + 12 = (3x + 3) + 9 = 27 + 9 = 36

Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \left( {10x + y} \right) - \left( {10y + x} \right) = 63 \cr & \Leftrightarrow 9\left( {x - y} \right) = 63 \cr & \Leftrightarrow x - y = 7 \cr} $$