Problems On Numbers — Part 1 MCQs with Explanations

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then}},\,\frac{1}{3}\,{\text{of}}\,\frac{1}{{4\,}}\,{\text{of}}\,x = 15\cr & \Rightarrow \, x = 15 \times 12 = 180 \cr & {\text{So,}}\,{\text{required}}\,{\text{number}} \cr & = {\frac{3}{{10}} \times 180} \cr & = 54 \cr} $$

Let the three integers be x , x + 2 and x + 4 Then, 3 x = 2( x + 4) + 3 ⇔ x = 11 ∴ Third integer = x + 4 = 15

Let the ten's digit be x and unit's digit be y . Then, (10 x + y ) - (10 y + x ) = 36 ⇒ 9( x - y ) = 36 ⇒ x - y = 4

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit. Let ten's and unit's digits be 2 x and x respectively. Then, (10 × 2 x + x ) - (10 x + 2 x ) = 36 ⇒ 9 x = 36 ⇒ x = 4 ∴ Required difference = (2 x + x ) - (2 x - x ) = 2 x = 8

Let the ten's and unit digit be x and $$\frac{8}{x}$$ respectively $$\eqalign{ & {\text{Then,}} \cr & \left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Rightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Rightarrow 9{x^2} + 18x - 72 = 0 \cr & \Rightarrow {x^2} + 2x - 8 = 0 \cr & \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr} $$ ∴ first digit will be 2 and second digit will be 4. i.e digit is 24.

Let the ten's digit be x and unit's digit be y Then, x + y = 15 and x - y = 3 or y - x = 3 Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6 Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9 So, the number is either 96 or 69 Hence, the number cannot be determined.

Let the numbers be a , b and c Then, a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 (a + b + c) 2 = a 2 + b 2 + c 2 + 2( ab + bc + ca ) = 138 + 2 x 131 = 400 ⇒ ( a + b + c ) = $$\sqrt {400} $$ = 20

Let the ten's digit be x and unit's digit be y . Then, number = 10 x + y Number obtained by interchanging the digits = 10 y + x ∴ (10 x + y ) + (10 y + x ) = 11( x + y ), which is divisible by 11

Let the ten's digit be x Then, unit's digit = x + 2 Number = 10 x + ( x + 2) = 11 x + 2 Sum of digits = x + ( x + 2) = 2 x + 2 ∴ (11 x + 2)(2 x + 2) = 144 ⇒ 22 x 2 + 26 x - 140 = 0 ⇒ 11 x 2 + 13 x - 70 = 0 ⇒ 11 x 2 + (35 - 22)x - 70 = 0 ⇒ 11 x 2 + 35x - 22x - 70 = 0 ⇒ ( x - 2)(11 x + 35) = 0 ⇒ x = 2 Hence, required number = 11 x + 2 = 24

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then}},\,x + 17 = \frac{{60}}{x} \cr & \Rightarrow {x^2} + 17x - 60 = 0 \cr & \Rightarrow (x + 20)(x - 3) = 0 \cr & \Rightarrow x = 3 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,x\,{\text{and}}\,y \cr & {\text{Then}},\,xy = 9375\,{\text{and}}\,\frac{x}{y} = 15 \cr & \frac{{xy}}{{\left( {x/y} \right)}} = \frac{{9375}}{{15}} \cr & \Rightarrow {y^2} = 625 \cr & \Rightarrow y = 25 \cr & \Rightarrow x = 15y = \left( {15 \times 25} \right) = 375 \cr & \therefore {\text{Sum}}\,{\text{of}}\,{\text{the}}\,{\text{number}} \cr & = x + y = 375 + 25 = 400 \cr} $$

Let the numbers be x and y Then, x y = 120 and x 2 + y 2 = 289 ∴ ( x + y ) 2 = x 2 + y 2 + 2 x y = 289 + (2 x 120) = 529 ∴ x + y = $$\sqrt {529} $$ = 23

Let the middle digit be x Then, 2 x = 10 or x = 5. So, the number is either 253 or 352 Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit. Hence, required number = 253

Let the numbers be x and y Then, x + y = 25 and x - y = 13 4 x y = ( x + y ) 2 - ( x - y ) 2 = (25) 2 - (13) 2 = (625 - 169) = 456 ∴ x y = 114

Let the numbers be x and x + 2. Then, ( x + 2) 2 - x 2 = 84 ⇒ 4 x + 4 = 84 ⇒ 4 x = 80 ⇒ x = 20. ∴ The required sum = x + ( x + 2) = 2 x + 2 = 42

Let the number be x Then, $$\eqalign{ & \Leftrightarrow 2x + 3 \times 42 = 238 \cr & \Leftrightarrow 2x + 126 = 238 \cr & \Leftrightarrow 2x = 112 \cr & \Leftrightarrow x = 56 \cr} $$ ∴ Required sum : $$\eqalign{ & = 3x + 2 \times 42 \cr & = 3 \times 56 + 2 \times 42 \cr & = 168 + 84 \cr & = 252 \cr} $$

$$\eqalign{ & = \frac{7}{8}{\text{ of }}1008 - \frac{3}{4}{\text{ of }}568 \cr & = \left( {1008 \times \frac{7}{8}} \right) - \left( {568 \times \frac{3}{4}} \right) \cr & = 882 - 426 \cr & = 456 \cr} $$

Let the number be x Then, $$\eqalign{ & \Leftrightarrow 3{x^2} - 4x = x + 50 \cr & \Leftrightarrow 3{x^2} - 5x - 50 = 0 \cr & \Leftrightarrow \left( {3x + 10} \right)\left( {x - 5} \right) = 0 \cr & \Leftrightarrow x = 5 \cr} $$

Let the numbers be x and y Then, $$xy = 1092.....(i)$$ And, $$\eqalign{ & \Leftrightarrow \left( {x + y} \right) - \left( {x - y} \right) = 42 \cr & \Leftrightarrow 2y = 42 \cr & \Leftrightarrow y = 21 \cr} $$ Putting y = 21 in (i), we get : $$x = \frac{{1092}}{{21}} = 52$$ Hence, greater number = 52

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

Let the ten's digit be x and unit's digit be y Then, $$\eqalign{ & \Leftrightarrow \frac{{10x + y}}{2} = 10y + \left( {x + 1} \right) \cr & \Leftrightarrow 10x + y = 20y + 2x + 2 \cr & \Leftrightarrow 8x - 19y = 2.....(i) \cr & {\text{And}} \Leftrightarrow x + y = 7.....(ii) \cr} $$ Solving (i) and (ii), we get : x = 5, y = 2 Hence, required number = 52

Let the numbers be x, y and z Then, $$\eqalign{ & {\bf{x + y = 73}} \cr & \Leftrightarrow y = 73 - x \cr & {\bf{y + z = 77 }} \cr & \Leftrightarrow z = 77 - y \cr & \Leftrightarrow z = 77 - \left( {73 - x} \right) \cr & \Leftrightarrow z = 4 + x \cr & {\bf{3x + z = 104}} \cr & \Leftrightarrow 3x + 4 + x = 104 \cr & \Leftrightarrow 4x = 100 \cr & \Leftrightarrow x = 25 \cr & \because y = \left( {73 - 25} \right) \cr & \Leftrightarrow y = 48 \cr & \because z = \left( {4 + 25} \right) \cr & \Leftrightarrow z = 29 \cr} $$ ∴ Third number = 29

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

Let the number be x Then, $$\eqalign{ & \frac{2}{3}x = \frac{25}{216} \times \frac{1}{x} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{216}} \times \frac{3}{2} \cr & \Leftrightarrow {x^2} = \frac{{25}}{{144}} \cr & \Leftrightarrow x = \sqrt {\frac{{25}}{{144}}} \cr & \Leftrightarrow x = \frac{5}{{12}} \cr} $$

Let the integers be x and (x + 5) Then, $$\eqalign{ & \Leftrightarrow x\left( {x + 5} \right) = 500 \cr & \Leftrightarrow {x^2} + 5x - 500 = 0 \cr & \Leftrightarrow \left( {x + 25} \right)\left( {x - 20} \right) = 0 \cr & \Leftrightarrow x = 20 \cr} $$ So, the numbers are 20 and 25

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

Let two numbers be a and b Given ab = 5 and a = $$\frac{3}{2}$$ $$\eqalign{ & \Rightarrow b = \frac{5}{a} \cr & \Rightarrow b = \frac{5}{{\frac{3}{2}}} \cr & \Rightarrow b = \frac{{5 \times 2}}{3} \cr & \Rightarrow b = \frac{{10}}{3} \cr} $$ Required sum of : ⇒ a + b = $$\frac{{3}}{2}$$ + $$\frac{{10}}{3}$$ L.C.M. of 2 and 3 = 6 ⇒ a + b = $$\frac{{9 + 20}}{6}$$ ⇒ a + b = $$\frac{{29}}{6}$$ ⇒ a + b = $$4\frac{{5}}{6}$$

Let the number be x Then, $$\eqalign{ & \Leftrightarrow x + \frac{2}{5}x = 455 \cr & \Leftrightarrow \frac{7}{5}x = 455 \cr & \Leftrightarrow x = \left( {\frac{{455 \times 5}}{7}} \right) \cr & \Leftrightarrow x = 325 \cr} $$

Let the required number be x Then, $$\eqalign{ & \Leftrightarrow x + 20 = \frac{{69}}{x} \cr & \Leftrightarrow {x^2} + 20x - 69 = 0 \cr & \Leftrightarrow {x^2} + 23x - 3x - 69 = 0 \cr & \Leftrightarrow x\left( {x + 23} \right) - 3\left( {x + 23} \right) = 0 \cr & \Leftrightarrow \left( {x + 23} \right)\left( {x - 3} \right) = 0 \cr & \Leftrightarrow x = 3\,\,\,\,\,\,\,\,\,\left[ {\because x \ne - 23} \right] \cr} $$

Let the number be x Then, $$\eqalign{ & \Leftrightarrow x \times \frac{2}{3}x = 864 \cr & \Leftrightarrow \frac{2}{3}{x^2} = 864 \cr & \Leftrightarrow {x^2} = \left( {\frac{{864 \times 3}}{2}} \right) \cr & \Leftrightarrow {x^2} = 1296 \cr & \Leftrightarrow x = \sqrt {1296} \cr & \Leftrightarrow x = 36 \cr} $$

$$\eqalign{ & 120: \cr & = 2 \times 2 \times 2 \times 3 \times 5 \cr & = \left( {2 \times 2} \right) \times 5 \times \left( {2 \times 3} \right) \cr & = 4 \times 5 \times 6 \cr} $$ Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6 Required sum : = 4 + 5 + 6 = 15

Let the numbers be a, b and c. Then, a 2 + b 2 + c 2 = 138 And (ab + bc + ca) = 131 ∴ (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) = 138 + 2 × 131 = 400 ⇒ (a + b + c) = $$\sqrt {400} $$ = 20

Let the positive integers be a and b where a > b According to the question, $$\eqalign{ & {a^2} + {b^2} = 100.....(\text{i}) \cr & {a^2} - {b^2} = 28.....(\text{ii}) \cr} $$ By adding (i) and (ii), we get : $$\eqalign{ & \therefore {a^2} + {b^2} + {a^2} - {b^2} = 100 + 28 \cr & \Rightarrow 2{a^2} = 128 \cr & \Rightarrow {a^2} = \frac{{128}}{2} \cr & \Rightarrow a = \sqrt {64} \cr & \Rightarrow a = 8 \cr} $$ From equation (i) $$\eqalign{ & \Rightarrow {8^2} + {b^2} = 100 \cr & \Rightarrow {b^2} = 100 - 64 \cr & \Rightarrow b = \sqrt {36} \cr & \Rightarrow b = 6 \cr & \therefore a + b = 8 + 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$

Let the number be x Then, $$\eqalign{ & \Rightarrow 13x = x + 180 \cr & \Rightarrow 12x = 180 \cr & \Rightarrow x = \frac{{180}}{{12}} \cr & \Rightarrow x = 15 \cr} $$

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

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

Let the ten's digit be x Then, number = 10x + 3 and Sum of digits = (x + 3) So, (x + 3) = $$\frac{1}{7}$$(10x + 3) ⇔ 7x + 21 = 10x + 3 ⇔ 3x = 18 ⇔ x = 6 Hence, the number is : = (10x + 3) = (10 × 6 + 3) = 63

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

Let the numbers be a and b, where a > b According to the question, $$\eqalign{ & a + b = 37\& {a^2} - {b^2} = 185 \cr & \Rightarrow \left( {a + b} \right)\left( {a - b} \right) = 185 \cr & \Rightarrow 37\left( {a - b} \right) = 185 \cr & \Rightarrow a - b = \frac{{185}}{{37}} \cr & \Rightarrow a - b = 5 \cr} $$

Let the number be x Then, $$\eqalign{ & \frac{3}{5}{\text{of }}\frac{2}{3}{\text{of }}x - \frac{2}{5}{\text{of }}\frac{1}{4}{\text{of }}x = 288 \cr & \Leftrightarrow \left( {x \times \frac{3}{5} \times \frac{2}{3}} \right) - \left( {x \times \frac{2}{5} \times \frac{1}{4}} \right) = 288 \cr & \Leftrightarrow \frac{2}{5}x - \frac{1}{{10}}x = 288 \cr & \Leftrightarrow \frac{{3x}}{{10}} = 288 \cr & \Leftrightarrow x = \left( {\frac{{288 \times 10}}{3}} \right) \cr & \Leftrightarrow x = 960 \cr} $$