Class 10: Linear Equations

$\displaystyle \textbf{Question 1: }\text{The numerator of a fraction is 4 less than the}$
$\displaystyle \text{denominator. If the numerator is decreased by 2 and denominator is}$
$\displaystyle \text{increased by 1, then the denominator is eight times the numerator.}$
$\displaystyle \text{Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{The numerator is 4 less than the denominator.}$
$\displaystyle \therefore x=y-4$
$\displaystyle \Rightarrow x-y=-4\qquad\ldots\text{(i)}$
$\displaystyle \text{The numerator is decreased by 2 and denominator is increased by 1.}$
$\displaystyle \therefore y+1=8(x-2)$
$\displaystyle \Rightarrow 8x-y=17\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting equation (i) from equation (ii),}$
$\displaystyle 7x=21$
$\displaystyle \Rightarrow x=3$
$\displaystyle \text{Substituting }x=3\text{ in equation (i),}$
$\displaystyle 3-y=-4$
$\displaystyle \Rightarrow y=7$
$\displaystyle \therefore \text{The required fraction}=\frac{3}{7}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A fraction becomes }\frac{9}{11}\text{ if 2 is added to both}$
$\displaystyle \text{numerator and the denominator. If 3 is added to both the numerator and}$
$\displaystyle \text{the denominator it becomes }\frac{5}{6}.\text{ Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{If 2 is added to both numerator and denominator,}$
$\displaystyle \frac{x+2}{y+2}=\frac{9}{11}$
$\displaystyle \Rightarrow 11(x+2)=9(y+2)$
$\displaystyle \Rightarrow 11x-9y=-4\qquad\ldots\text{(i)}$
$\displaystyle \text{If 3 is added to both numerator and denominator,}$
$\displaystyle \frac{x+3}{y+3}=\frac{5}{6}$
$\displaystyle \Rightarrow 6(x+3)=5(y+3)$
$\displaystyle \Rightarrow 6x-5y=-3\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (i) by }5,$
$\displaystyle 55x-45y=-20\qquad\ldots\text{(iii)}$
$\displaystyle \text{Multiplying equation (ii) by }9,$
$\displaystyle 54x-45y=-27\qquad\ldots\text{(iv)}$
$\displaystyle \text{Subtracting equation (iv) from equation (iii),}$
$\displaystyle x=7$
$\displaystyle \text{Substituting }x=7\text{ in equation (ii),}$
$\displaystyle 42-5y=-3$
$\displaystyle \Rightarrow 5y=45$
$\displaystyle \Rightarrow y=9$
$\displaystyle \therefore \text{The required fraction}=\frac{7}{9}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If we add 1 to the numerator and subtract 1 from}$
$\displaystyle \text{the denominator, a fraction becomes }1.\text{ It also becomes }\frac{1}{2}\text{ if we only}$
$\displaystyle \text{add 1 to the denominator. What is the fraction?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{On adding 1 to the numerator and subtracting 1 from the denominator,}$
$\displaystyle \frac{x+1}{y-1}=1$
$\displaystyle \Rightarrow x+1=y-1$
$\displaystyle \Rightarrow x-y=-2\qquad\ldots\text{(i)}$
$\displaystyle \text{On adding 1 only to the denominator,}$
$\displaystyle \frac{x}{y+1}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x=y+1$
$\displaystyle \Rightarrow 2x-y=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting equation (i) from equation (ii),}$
$\displaystyle x=3$
$\displaystyle \text{Substituting }x=3\text{ in equation (i),}$
$\displaystyle 3-y=-2$
$\displaystyle \Rightarrow y=5$
$\displaystyle \therefore \text{The required fraction}=\frac{3}{5}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The sum of the numerator and denominator of a}$
$\displaystyle \text{fraction is }12.\text{ If the denominator is increased by }3\text{, the fraction becomes}$
$\displaystyle \frac{1}{2}.\text{ Find the fraction.}\hfill\text{[CBSE 2006C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{The sum of the numerator and denominator is }12.$
$\displaystyle \therefore x+y=12\qquad\ldots\text{(i)}$
$\displaystyle \text{On increasing the denominator by }3\text{, the fraction becomes }\frac{1}{2}.$
$\displaystyle \therefore \frac{x}{y+3}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x=y+3$
$\displaystyle \Rightarrow 2x-y=3\qquad\ldots\text{(ii)}$
$\displaystyle \text{Adding equations (i) and (ii),}$
$\displaystyle 3x=15$
$\displaystyle \Rightarrow x=5$
$\displaystyle \text{Substituting }x=5\text{ in equation (i),}$
$\displaystyle 5+y=12$
$\displaystyle \Rightarrow y=7$
$\displaystyle \therefore \text{The required fraction}=\frac{5}{7}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{When 3 is added to the denominator and 2 is}$
$\displaystyle \text{subtracted from the numerator, a fraction becomes }\frac{1}{4}.\text{ And, when 6 is}$
$\displaystyle \text{added to numerator and the denominator is multiplied by 3, it becomes}$
$\displaystyle \frac{2}{3}.\text{ Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{On subtracting 2 from numerator and adding 3 to denominator,}$
$\displaystyle \frac{x-2}{y+3}=\frac{1}{4}$
$\displaystyle \Rightarrow 4(x-2)=y+3$
$\displaystyle \Rightarrow 4x-y=11\qquad\ldots\text{(i)}$
$\displaystyle \text{On adding 6 to numerator and multiplying denominator by 3,}$
$\displaystyle \frac{x+6}{3y}=\frac{2}{3}$
$\displaystyle \Rightarrow 3(x+6)=6y$
$\displaystyle \Rightarrow x-2y=-6\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (ii) by }4,$
$\displaystyle 4x-8y=-24\qquad\ldots\text{(iii)}$
$\displaystyle \text{Subtracting equation (iii) from equation (i),}$
$\displaystyle 7y=35$
$\displaystyle \Rightarrow y=5$
$\displaystyle \text{Substituting }y=5\text{ in equation (ii),}$
$\displaystyle x-10=-6$
$\displaystyle \Rightarrow x=4$
$\displaystyle \therefore \text{The required fraction}=\frac{4}{5}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A fraction becomes }\frac{1}{3}\text{ when 2 is subtracted from}$
$\displaystyle \text{the numerator, and it becomes }\frac{1}{2}\text{ when 1 is subtracted from the}$
$\displaystyle \text{denominator. Find the fraction.}\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{On subtracting 2 from the numerator, the fraction becomes }\frac{1}{3}.$
$\displaystyle \therefore \frac{x-2}{y}=\frac{1}{3}$
$\displaystyle \Rightarrow 3(x-2)=y$
$\displaystyle \Rightarrow 3x-y=6\qquad\ldots\text{(i)}$
$\displaystyle \text{On subtracting 1 from the denominator, the fraction becomes }\frac{1}{2}.$
$\displaystyle \therefore \frac{x}{y-1}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x=y-1$
$\displaystyle \Rightarrow 2x-y=-1\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting equation (ii) from equation (i),}$
$\displaystyle x=7$
$\displaystyle \text{Substituting }x=7\text{ in equation (ii),}$
$\displaystyle 14-y=-1$
$\displaystyle \Rightarrow y=15$
$\displaystyle \therefore \text{The required fraction}=\frac{7}{15}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The sum of the numerator and denominator of a}$
$\displaystyle \text{fraction is 4 more than twice the numerator. If the numerator and}$
$\displaystyle \text{denominator are increased by 3, they are in the ratio }2:3.\text{ Determine}$
$\displaystyle \text{the fraction.}\hfill\text{[CBSE 2001C, 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{The sum of numerator and denominator is 4 more than twice the numerator.}$
$\displaystyle \therefore x+y=2x+4$
$\displaystyle \Rightarrow -x+y=4\qquad\ldots\text{(i)}$
$\displaystyle \text{On increasing the numerator and denominator by }3,$
$\displaystyle \frac{x+3}{y+3}=\frac{2}{3}$
$\displaystyle \Rightarrow 3(x+3)=2(y+3)$
$\displaystyle \Rightarrow 3x-2y=-3\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (i) by }2,$
$\displaystyle -2x+2y=8\qquad\ldots\text{(iii)}$
$\displaystyle \text{Adding equations (ii) and (iii),}$
$\displaystyle x=5$
$\displaystyle \text{Substituting }x=5\text{ in equation (i),}$
$\displaystyle -5+y=4$
$\displaystyle \Rightarrow y=9$
$\displaystyle \therefore \text{The required fraction}=\frac{5}{9}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If the numerator of a fraction is multiplied by 2}$
$\displaystyle \text{and the denominator is reduced by 5, the fraction becomes }\frac{6}{5}.\text{ And, if}$
$\displaystyle \text{the denominator is doubled and the numerator is increased by 8, the}$
$\displaystyle \text{fraction becomes }\frac{2}{5}.\text{ Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{On multiplying the numerator by 2 and reducing denominator by 5,}$
$\displaystyle \frac{2x}{y-5}=\frac{6}{5}$
$\displaystyle \Rightarrow 10x=6y-30$
$\displaystyle \Rightarrow 5x-3y=-15\qquad\ldots\text{(i)}$
$\displaystyle \text{On doubling the denominator and increasing numerator by 8,}$
$\displaystyle \frac{x+8}{2y}=\frac{2}{5}$
$\displaystyle \Rightarrow 5(x+8)=4y$
$\displaystyle \Rightarrow 5x-4y=-40\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting equation (ii) from equation (i),}$
$\displaystyle y=25$
$\displaystyle \text{Substituting }y=25\text{ in equation (i),}$
$\displaystyle 5x-3(25)=-15$
$\displaystyle \Rightarrow 5x=60$
$\displaystyle \Rightarrow x=12$
$\displaystyle \therefore \text{The required fraction}=\frac{12}{25}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The sum of the numerator and denominator of a}$
$\displaystyle \text{fraction is 3 less than twice the denominator. If the numerator and}$
$\displaystyle \text{denominator are decreased by 1, the numerator becomes half the}$
$\displaystyle \text{denominator. Determine the fraction.}\hfill\text{[CBSE 2001C, 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{y}$
$\displaystyle \text{The sum of numerator and denominator is 3 less than twice denominator.}$
$\displaystyle \therefore x+y=2y-3$
$\displaystyle \Rightarrow x-y=-3\qquad\ldots\text{(i)}$
$\displaystyle \text{On decreasing the numerator and denominator by }1,$
$\displaystyle x-1=\frac{1}{2}(y-1)$
$\displaystyle \Rightarrow 2x-2=y-1$
$\displaystyle \Rightarrow 2x-y=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting equation (i) from equation (ii),}$
$\displaystyle x=4$
$\displaystyle \text{Substituting }x=4\text{ in equation (i),}$
$\displaystyle 4-y=-3$
$\displaystyle \Rightarrow y=7$
$\displaystyle \therefore \text{The required fraction}=\frac{4}{7}.$
$\displaystyle \\$