Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{The sum of the squares of two consecutive odd positive integers is }394.$
$\displaystyle \text{Find them.}\hfill\text{[CBSE 2009, 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller odd positive integer be }x.$
$\displaystyle \therefore \text{The next consecutive odd positive integer is }x+2.$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2+(x+2)^2=394$
$\displaystyle \Rightarrow x^2+x^2+4x+4=394$
$\displaystyle \Rightarrow 2x^2+4x-390=0$
$\displaystyle \Rightarrow x^2+2x-195=0$
$\displaystyle \Rightarrow x^2+15x-13x-195=0$
$\displaystyle \Rightarrow x(x+15)-13(x+15)=0$
$\displaystyle \Rightarrow (x+15)(x-13)=0$
$\displaystyle \Rightarrow x=-15\text{ or }x=13$
$\displaystyle \text{Since }x\text{ is a positive integer, }x=-15\text{ is rejected.}$
$\displaystyle \therefore x=13,\quad x+2=15.$
$\displaystyle \therefore \text{The required integers are }13\text{ and }15.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The sum of two numbers is }8\text{ and }15\text{ times the sum of their reciprocals}$
$\displaystyle \text{is also }8.\text{ Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one number be }x.$
$\displaystyle \therefore \text{The other number is }8-x.$
$\displaystyle \text{According to the question,}$
$\displaystyle 15\left(\frac{1}{x}+\frac{1}{8-x}\right)=8$
$\displaystyle \Rightarrow 15\left(\frac{8-x+x}{x(8-x)}\right)=8$
$\displaystyle \Rightarrow \frac{120}{x(8-x)}=8$
$\displaystyle \Rightarrow 120=8x(8-x)$
$\displaystyle \Rightarrow 15=x(8-x)$
$\displaystyle \Rightarrow x^2-8x+15=0$
$\displaystyle \Rightarrow (x-3)(x-5)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=5$
$\displaystyle \therefore \text{The required numbers are }3\text{ and }5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Two numbers differ by }3\text{ and their product is }504.\text{ Find the numbers.}$
$\displaystyle \hfill\text{[CBSE 2002 C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller number be }x.$
$\displaystyle \therefore \text{The other number is }x+3.$
$\displaystyle \text{According to the question,}$
$\displaystyle x(x+3)=504$
$\displaystyle \Rightarrow x^2+3x-504=0$
$\displaystyle \Rightarrow x^2+24x-21x-504=0$
$\displaystyle \Rightarrow x(x+24)-21(x+24)=0$
$\displaystyle \Rightarrow (x+24)(x-21)=0$
$\displaystyle \Rightarrow x=-24\text{ or }x=21$
$\displaystyle \text{If }x=21,\text{ the other number is }24.$
$\displaystyle \text{If }x=-24,\text{ the other number is }-21.$
$\displaystyle \therefore \text{The numbers are }21,24\text{ or }-24,-21.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The sum of two numbers }a\text{ and }b\text{ is }15,\text{ and the sum of their}$
$\displaystyle \text{reciprocals }\frac{1}{a}\text{ and }\frac{1}{b}\text{ is }\frac{3}{10}.\text{ Find the numbers }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle a+b=15$
$\displaystyle \text{Also, }\frac{1}{a}+\frac{1}{b}=\frac{3}{10}$
$\displaystyle \Rightarrow \frac{a+b}{ab}=\frac{3}{10}$
$\displaystyle \Rightarrow \frac{15}{ab}=\frac{3}{10}$
$\displaystyle \Rightarrow ab=50$
$\displaystyle \text{Let }a=x.\text{ Then }b=15-x.$
$\displaystyle x(15-x)=50$
$\displaystyle \Rightarrow x^2-15x+50=0$
$\displaystyle \Rightarrow (x-5)(x-10)=0$
$\displaystyle \Rightarrow x=5\text{ or }x=10$
$\displaystyle \therefore \text{The required numbers are }5\text{ and }10.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The sum of two numbers is }9.\text{ The sum of their reciprocals is }\frac{1}{2}.$
$\displaystyle \text{Find the numbers.}\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one number be }x.$
$\displaystyle \therefore \text{The other number is }9-x.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{9}{x(9-x)}=\frac{1}{2}$
$\displaystyle \Rightarrow x(9-x)=18$
$\displaystyle \Rightarrow x^2-9x+18=0$
$\displaystyle \Rightarrow (x-3)(x-6)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=6$
$\displaystyle \therefore \text{The required numbers are }3\text{ and }6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Three consecutive positive integers are such that the sum of the square}$
$\displaystyle \text{of the first and the product of the other two is }46.\text{ Find the integers.}$
$\displaystyle \hfill\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three consecutive positive integers be }x,\ x+1,\ x+2.$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2+(x+1)(x+2)=46$
$\displaystyle \Rightarrow x^2+x^2+3x+2=46$
$\displaystyle \Rightarrow 2x^2+3x-44=0$
$\displaystyle \Rightarrow 2x^2+11x-8x-44=0$
$\displaystyle \Rightarrow x(2x+11)-4(2x+11)=0$
$\displaystyle \Rightarrow (2x+11)(x-4)=0$
$\displaystyle \Rightarrow x=-\frac{11}{2}\text{ or }x=4$
$\displaystyle \text{Since }x\text{ is a positive integer, }x=-\frac{11}{2}\text{ is rejected.}$
$\displaystyle \therefore x=4,\quad x+1=5,\quad x+2=6.$
$\displaystyle \therefore \text{The required integers are }4,\ 5\text{ and }6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The difference of squares of two numbers is }88.\text{ If the larger number is}$
$\displaystyle 5\text{ less than twice the smaller number, then find the two numbers.}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller number be }x.$
$\displaystyle \therefore \text{The larger number is }2x-5.$
$\displaystyle \text{According to the question,}$
$\displaystyle (2x-5)^2-x^2=88$
$\displaystyle \Rightarrow 4x^2-20x+25-x^2=88$
$\displaystyle \Rightarrow 3x^2-20x-63=0$
$\displaystyle \Rightarrow 3x^2-27x+7x-63=0$
$\displaystyle \Rightarrow 3x(x-9)+7(x-9)=0$
$\displaystyle \Rightarrow (x-9)(3x+7)=0$
$\displaystyle \Rightarrow x=9\text{ or }x=-\frac{7}{3}$
$\displaystyle \text{For }x=9,\text{ the larger number}=2(9)-5=13.$
$\displaystyle \text{For }x=-\frac{7}{3},\quad 2x-5=-\frac{29}{3}<-\frac{7}{3},$
$\displaystyle \text{so }2x-5\text{ is not the larger number. Hence, this value is rejected.}$
$\displaystyle \therefore \text{The required numbers are }9\text{ and }13.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{The sum of the squares of two consecutive odd numbers is }394.$
$\displaystyle \text{Find the numbers.}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller odd number be }x.$
$\displaystyle \therefore \text{The next consecutive odd number is }x+2.$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2+(x+2)^2=394$
$\displaystyle \Rightarrow 2x^2+4x+4=394$
$\displaystyle \Rightarrow x^2+2x-195=0$
$\displaystyle \Rightarrow x^2+15x-13x-195=0$
$\displaystyle \Rightarrow x(x+15)-13(x+15)=0$
$\displaystyle \Rightarrow (x+15)(x-13)=0$
$\displaystyle \Rightarrow x=-15\text{ or }x=13$
$\displaystyle \text{If }x=13,\text{ the numbers are }13\text{ and }15.$
$\displaystyle \text{If }x=-15,\text{ the numbers are }-15\text{ and }-13.$
$\displaystyle \therefore \text{The required numbers are }13,15\text{ or }-15,-13.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The sum of the squares of two consecutive multiples of }7\text{ is }637.$
$\displaystyle \text{Find the multiples.}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the consecutive multiples of }7\text{ be }7x\text{ and }7(x+1).$
$\displaystyle \text{According to the question,}$
$\displaystyle (7x)^2+\{7(x+1)\}^2=637$
$\displaystyle \Rightarrow 49x^2+49(x+1)^2=637$
$\displaystyle \Rightarrow x^2+(x+1)^2=13$
$\displaystyle \Rightarrow 2x^2+2x+1=13$
$\displaystyle \Rightarrow x^2+x-6=0$
$\displaystyle \Rightarrow (x+3)(x-2)=0$
$\displaystyle \Rightarrow x=-3\text{ or }x=2$
$\displaystyle \text{For }x=2,\text{ the multiples are }14\text{ and }21.$
$\displaystyle \text{For }x=-3,\text{ the multiples are }-21\text{ and }-14.$
$\displaystyle \therefore \text{The required multiples are }14,21\text{ or }-21,-14.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The sum of the squares of two consecutive even numbers is }340.$
$\displaystyle \text{Find the numbers.}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller even number be }x.$
$\displaystyle \therefore \text{The next consecutive even number is }x+2.$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2+(x+2)^2=340$
$\displaystyle \Rightarrow 2x^2+4x+4=340$
$\displaystyle \Rightarrow x^2+2x-168=0$
$\displaystyle \Rightarrow x^2+14x-12x-168=0$
$\displaystyle \Rightarrow x(x+14)-12(x+14)=0$
$\displaystyle \Rightarrow (x+14)(x-12)=0$
$\displaystyle \Rightarrow x=-14\text{ or }x=12$
$\displaystyle \text{If }x=12,\text{ the numbers are }12\text{ and }14.$
$\displaystyle \text{If }x=-14,\text{ the numbers are }-14\text{ and }-12.$
$\displaystyle \therefore \text{The required numbers are }12,14\text{ or }-14,-12.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{The difference of two numbers is }4.\text{ If the difference of their}$
$\displaystyle \text{reciprocals is }\frac{4}{21},\text{ find the numbers.}\hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller number be }x.$
$\displaystyle \therefore \text{The larger number is }x+4.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1}{x}-\frac{1}{x+4}=\frac{4}{21}$
$\displaystyle \Rightarrow \frac{x+4-x}{x(x+4)}=\frac{4}{21}$
$\displaystyle \Rightarrow \frac{4}{x(x+4)}=\frac{4}{21}$
$\displaystyle \Rightarrow x(x+4)=21$
$\displaystyle \Rightarrow x^2+4x-21=0$
$\displaystyle \Rightarrow (x+7)(x-3)=0$
$\displaystyle \Rightarrow x=-7\text{ or }x=3$
$\displaystyle \text{If }x=3,\text{ the numbers are }3\text{ and }7.$
$\displaystyle \text{If }x=-7,\text{ the numbers are }-7\text{ and }-3.$
$\displaystyle \therefore \text{The required numbers are }3,7\text{ or }-7,-3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{A two digit number is }4\text{ times the sum of its digits and twice the}$
$\displaystyle \text{product of its digits. Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.$
$\displaystyle \therefore \text{The number}=10x+y.$
$\displaystyle \text{According to the first condition,}$
$\displaystyle 10x+y=4(x+y)$
$\displaystyle \Rightarrow 10x+y=4x+4y$
$\displaystyle \Rightarrow 6x=3y$
$\displaystyle \Rightarrow y=2x$
$\displaystyle \text{According to the second condition,}$
$\displaystyle 10x+y=2xy$
$\displaystyle \text{Substituting }y=2x,$
$\displaystyle 10x+2x=2x(2x)$
$\displaystyle \Rightarrow 12x=4x^2$
$\displaystyle \Rightarrow 4x(x-3)=0$
$\displaystyle \Rightarrow x=0\text{ or }x=3$
$\displaystyle \text{Since }x\text{ is the tens digit, }x\ne0.$
$\displaystyle \therefore x=3,\quad y=2x=6.$
$\displaystyle \therefore \text{The required number}=10(3)+6=36.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{The difference of the squares of two positive integers is }180.\text{ The}$
$\displaystyle \text{square of the smaller number is }8\text{ times the larger. Find the numbers.}$
$\displaystyle \hfill\text{[CBSE 2014, 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the larger positive integer be }x.$
$\displaystyle \therefore \text{Square of the smaller integer}=8x.$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2-8x=180$
$\displaystyle \Rightarrow x^2-8x-180=0$
$\displaystyle \Rightarrow x^2-18x+10x-180=0$
$\displaystyle \Rightarrow x(x-18)+10(x-18)=0$
$\displaystyle \Rightarrow (x-18)(x+10)=0$
$\displaystyle \Rightarrow x=18\text{ or }x=-10$
$\displaystyle \text{Since }x\text{ is a positive integer, }x=-10\text{ is rejected.}$
$\displaystyle \therefore \text{Larger integer}=18.$
$\displaystyle \text{Square of the smaller integer}=8\times18=144.$
$\displaystyle \therefore \text{Smaller integer}=\sqrt{144}=12.$
$\displaystyle \therefore \text{The required integers are }12\text{ and }18.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{The difference of two natural numbers is }3\text{ and the difference of their}$
$\displaystyle \text{reciprocals is }\frac{3}{28}.\text{ Find the numbers.}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller natural number be }x.$
$\displaystyle \therefore \text{The larger natural number is }x+3.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1}{x}-\frac{1}{x+3}=\frac{3}{28}$
$\displaystyle \Rightarrow \frac{x+3-x}{x(x+3)}=\frac{3}{28}$
$\displaystyle \Rightarrow \frac{3}{x(x+3)}=\frac{3}{28}$
$\displaystyle \Rightarrow x(x+3)=28$
$\displaystyle \Rightarrow x^2+3x-28=0$
$\displaystyle \Rightarrow (x+7)(x-4)=0$
$\displaystyle \Rightarrow x=-7\text{ or }x=4$
$\displaystyle \text{Since }x\text{ is a natural number, }x=-7\text{ is rejected.}$
$\displaystyle \therefore x=4,\quad x+3=7.$
$\displaystyle \therefore \text{The required numbers are }4\text{ and }7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The numerator of a fraction is }3\text{ less than the denominator. If }2\text{ is}$
$\displaystyle \text{added to both the numerator and the denominator, then the sum of the new fraction and}$
$\displaystyle \text{the original fraction is }\frac{29}{20}.\text{ Find the original fraction.}\hfill\text{[CBSE 2015, 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the denominator of the original fraction be }x.$
$\displaystyle \therefore \text{Its numerator}=x-3.$
$\displaystyle \therefore \text{Original fraction}=\frac{x-3}{x}.$
$\displaystyle \text{After adding }2\text{ to the numerator and denominator, the new fraction is }\frac{x-1}{x+2}.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{x-3}{x}+\frac{x-1}{x+2}=\frac{29}{20}$
$\displaystyle \Rightarrow 20\{(x-3)(x+2)+x(x-1)\}=29x(x+2)$
$\displaystyle \Rightarrow 20(2x^2-2x-6)=29x^2+58x$
$\displaystyle \Rightarrow 11x^2-98x-120=0$
$\displaystyle \Rightarrow 11x^2-110x+12x-120=0$
$\displaystyle \Rightarrow 11x(x-10)+12(x-10)=0$
$\displaystyle \Rightarrow (x-10)(11x+12)=0$
$\displaystyle \Rightarrow x=10\text{ or }x=-\frac{12}{11}$
$\displaystyle \text{Since the numerator and denominator are positive integers, }x=-\frac{12}{11}\text{ is rejected.}$
$\displaystyle \therefore \text{Denominator}=10,\quad \text{numerator}=10-3=7.$
$\displaystyle \therefore \text{The original fraction is }\frac{7}{10}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Find a natural number whose square diminished by }84\text{ is equal to thrice}$
$\displaystyle \text{of }8\text{ more than the given number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the natural number be }x.$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2-84=3(x+8)$
$\displaystyle \Rightarrow x^2-84=3x+24$
$\displaystyle \Rightarrow x^2-3x-108=0$
$\displaystyle \Rightarrow x^2-12x+9x-108=0$
$\displaystyle \Rightarrow x(x-12)+9(x-12)=0$
$\displaystyle \Rightarrow (x-12)(x+9)=0$
$\displaystyle \Rightarrow x=12\text{ or }x=-9$
$\displaystyle \text{Since }x\text{ is a natural number, }x=-9\text{ is rejected.}$
$\displaystyle \therefore \text{The required natural number is }12.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{A natural number when increased by }12\text{ equals }160\text{ times its}$
$\displaystyle \text{reciprocal. Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the natural number be }x.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+12=\frac{160}{x}$
$\displaystyle \Rightarrow x(x+12)=160$
$\displaystyle \Rightarrow x^2+12x-160=0$
$\displaystyle \Rightarrow x^2+20x-8x-160=0$
$\displaystyle \Rightarrow x(x+20)-8(x+20)=0$
$\displaystyle \Rightarrow (x+20)(x-8)=0$
$\displaystyle \Rightarrow x=-20\text{ or }x=8$
$\displaystyle \text{Since }x\text{ is a natural number, }x=-20\text{ is rejected.}$
$\displaystyle \therefore \text{The required natural number is }8.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{A }2\text{-digit number is such that the product of its digits is }24.\text{ If }18$
$\displaystyle \text{is subtracted from the number, the digits interchange their places. Find the number.}$
$\displaystyle \hfill\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.$
$\displaystyle \therefore \text{The number}=10x+y.$
$\displaystyle \text{The number obtained by interchanging the digits}=10y+x.$
$\displaystyle \text{According to the question,}$
$\displaystyle xy=24\qquad\ldots\text{(i)}$
$\displaystyle 10x+y-18=10y+x$
$\displaystyle \Rightarrow 9x-9y=18$
$\displaystyle \Rightarrow x-y=2$
$\displaystyle \Rightarrow x=y+2$
$\displaystyle \text{Substituting in (i),}$
$\displaystyle y(y+2)=24$
$\displaystyle \Rightarrow y^2+2y-24=0$
$\displaystyle \Rightarrow (y+6)(y-4)=0$
$\displaystyle \Rightarrow y=-6\text{ or }y=4$
$\displaystyle \text{Since }y\text{ is a digit, }y=-6\text{ is rejected.}$
$\displaystyle \therefore y=4,\quad x=y+2=6.$
$\displaystyle \therefore \text{The required number}=10(6)+4=64.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{In a two-digit number, the digit at unit's place is }5\text{ less than the}$
$\displaystyle \text{digit at ten's place. The product of digits is }36.\text{ Find the number.}\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the digit at ten's place be }x.$
$\displaystyle \therefore \text{The digit at unit's place}=x-5.$
$\displaystyle \text{According to the question,}$
$\displaystyle x(x-5)=36$
$\displaystyle \Rightarrow x^2-5x-36=0$
$\displaystyle \Rightarrow x^2-9x+4x-36=0$
$\displaystyle \Rightarrow x(x-9)+4(x-9)=0$
$\displaystyle \Rightarrow (x-9)(x+4)=0$
$\displaystyle \Rightarrow x=9\text{ or }x=-4$
$\displaystyle \text{Since }x\text{ is a digit, }x=-4\text{ is rejected.}$
$\displaystyle \therefore \text{Digit at ten's place}=9,\quad \text{digit at unit's place}=9-5=4.$
$\displaystyle \therefore \text{The required number is }94.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{The difference of squares of two positive integers is }400.\text{ Find the}$
$\displaystyle \text{integers if twice the smaller integer is }5\text{ more than the greater integer.}$
$\displaystyle \hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller positive integer be }x.$
$\displaystyle \therefore \text{The greater integer}=2x-5.$
$\displaystyle \text{According to the question,}$
$\displaystyle (2x-5)^2-x^2=400$
$\displaystyle \Rightarrow 4x^2-20x+25-x^2=400$
$\displaystyle \Rightarrow 3x^2-20x-375=0$
$\displaystyle \Rightarrow 3x^2-45x+25x-375=0$
$\displaystyle \Rightarrow 3x(x-15)+25(x-15)=0$
$\displaystyle \Rightarrow (x-15)(3x+25)=0$
$\displaystyle \Rightarrow x=15\text{ or }x=-\frac{25}{3}$
$\displaystyle \text{Since }x\text{ is a positive integer, }x=-\frac{25}{3}\text{ is rejected.}$
$\displaystyle \therefore \text{Smaller integer}=15.$
$\displaystyle \therefore \text{Greater integer}=2(15)-5=25.$
$\displaystyle \therefore \text{The required integers are }15\text{ and }25.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{A two-digit number is such that the product of its digits is }12.$
$\displaystyle \text{When }36\text{ is added to this number, the digits interchange their places. Find the number.}$
$\displaystyle \hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.$
$\displaystyle \therefore \text{The number}=10x+y.$
$\displaystyle \text{The number obtained by interchanging the digits}=10y+x.$
$\displaystyle \text{According to the question,}$
$\displaystyle xy=12\qquad\ldots\text{(i)}$
$\displaystyle 10x+y+36=10y+x$
$\displaystyle \Rightarrow 9y-9x=36$
$\displaystyle \Rightarrow y-x=4$
$\displaystyle \Rightarrow y=x+4$
$\displaystyle \text{Substituting in (i),}$
$\displaystyle x(x+4)=12$
$\displaystyle \Rightarrow x^2+4x-12=0$
$\displaystyle \Rightarrow (x+6)(x-2)=0$
$\displaystyle \Rightarrow x=-6\text{ or }x=2$
$\displaystyle \text{Since }x\text{ is a digit, }x=-6\text{ is rejected.}$
$\displaystyle \therefore x=2,\quad y=x+4=6.$
$\displaystyle \therefore \text{The required number}=10(2)+6=26.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{A two-digit number is seven times the sum of its digits and two more}$
$\displaystyle \text{than }5\text{ times the product of its digits. Find the number.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the tens digit be }x\text{ and the units digit be }y.$
$\displaystyle \therefore \text{The number}=10x+y.$
$\displaystyle \text{According to the first condition,}$
$\displaystyle 10x+y=7(x+y)$
$\displaystyle \Rightarrow 10x+y=7x+7y$
$\displaystyle \Rightarrow 3x=6y$
$\displaystyle \Rightarrow x=2y$
$\displaystyle \text{According to the second condition,}$
$\displaystyle 10x+y=5xy+2$
$\displaystyle \text{Substituting }x=2y,$
$\displaystyle 20y+y=10y^2+2$
$\displaystyle \Rightarrow 10y^2-21y+2=0$
$\displaystyle \Rightarrow 10y^2-20y-y+2=0$
$\displaystyle \Rightarrow 10y(y-2)-1(y-2)=0$
$\displaystyle \Rightarrow (y-2)(10y-1)=0$
$\displaystyle \Rightarrow y=2\text{ or }y=\frac{1}{10}$
$\displaystyle \text{Since }y\text{ is a digit, }y=\frac{1}{10}\text{ is rejected.}$
$\displaystyle \therefore y=2,\quad x=2y=4.$
$\displaystyle \therefore \text{The required number}=10(4)+2=42.$
$\displaystyle \\$