Class 9: Simultaneous Linear Equations

Problems Based on Articles and their cost

$\displaystyle \textbf{Question 4: }5\text{ pens and }6\text{ pencils together cost Rs. }9\text{ and }3\text{ pens and } \\ 2\text{ pencils cost Rs. }5.\text{ Find the cost of one pen and one pencil.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the cost of one pen be Rs. }x\text{ and the cost of one pencil be Rs. }y.$
$\displaystyle \text{Then,}$
$\displaystyle 5x+6y=9 \qquad \ldots(1)$
$\displaystyle 3x+2y=5 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }3\text{ and equation (2) by }5,\text{ we get}$
$\displaystyle 15x+18y=27 \qquad \ldots(3)$
$\displaystyle 15x+10y=25 \qquad \ldots(4)$
$\displaystyle \text{Subtracting equation (4) from equation (3),}$
$\displaystyle 15x+18y=27$
$\displaystyle \underline{-\,(15x+10y=25)}$
$\displaystyle 8y=2$
$\displaystyle \Rightarrow y=\frac{2}{8}=\frac{1}{4}=0.25$
$\displaystyle \text{Therefore, the cost of one pencil is Rs. }0.25.$
$\displaystyle \text{Substituting }y=0.25\text{ in equation (1),}$
$\displaystyle 5x+6(0.25)=9$
$\displaystyle \Rightarrow 5x+1.50=9$
$\displaystyle \Rightarrow 5x=7.50$
$\displaystyle \Rightarrow x=1.50$
$\displaystyle \text{Therefore, the cost of one pen is Rs. }1.50.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{A person has pens and pencils which are together }40\text{ in number. If she had}$
$\displaystyle 5\text{ more pencils and }5\text{ fewer pens, then the number of pencils would be four}$
$\displaystyle \text{times the number of pens. Find the number of pens and pencils that the person had.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of pens be }x\text{ and the number of pencils be }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=40 \qquad \ldots(1)$
$\displaystyle y+5=4(x-5)$
$\displaystyle y+5=4x-20$
$\displaystyle 4x-y=25 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }4,\text{ we get}$
$\displaystyle 4x+4y=160 \qquad \ldots(3)$
$\displaystyle \text{Subtracting equation (2) from equation (3),}$
$\displaystyle (4x+4y)-(4x-y)=160-25$
$\displaystyle 5y=135$
$\displaystyle y=27$
$\displaystyle \text{Substituting }y=27\text{ in equation (1),}$
$\displaystyle x+27=40$
$\displaystyle x=13$
$\displaystyle \text{Verification: }27+5=32\text{ and }4(13-5)=4\times8=32.$
$\displaystyle \therefore \text{The person had }13\text{ pens and }27\text{ pencils.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Five books and seven pens together cost Rs. }79,\text{ whereas seven books}$
$\displaystyle \text{and five pens together cost Rs. }77.\text{ Find the cost of one book and two pens.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the cost of one book be Rs. }x\text{ and the cost of one pen be Rs. }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle 5x+7y=79 \qquad \ldots(1)$
$\displaystyle 7x+5y=77 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }7\text{ and equation (2) by }5,\text{ we get}$
$\displaystyle 35x+49y=553 \qquad \ldots(3)$
$\displaystyle 35x+25y=385 \qquad \ldots(4)$
$\displaystyle \text{Subtracting equation (4) from equation (3),}$
$\displaystyle (35x+49y)-(35x+25y)=553-385$
$\displaystyle 24y=168$
$\displaystyle y=7$
$\displaystyle \text{Substituting }y=7\text{ in equation (1),}$
$\displaystyle 5x+7(7)=79$
$\displaystyle 5x+49=79$
$\displaystyle 5x=30$
$\displaystyle x=6$
$\displaystyle \therefore \text{Cost of one book and two pens}=x+2y$
$\displaystyle =6+2(7)=6+14=\text{Rs. }20.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{On selling a TV at }5\%\text{ gain and a fridge at }10\%\text{ gain, a shopkeeper}$
$\displaystyle \text{gains Rs. }2000.\text{ If he sells the TV at }10\%\text{ gain and the fridge at }5\%\text{ loss,}$
$\displaystyle \text{he gains Rs. }1500\text{ on the transaction. Find the cost prices of the TV and fridge.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the cost price of the TV be Rs. }x\text{ and that of the fridge be Rs. }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{5}{100}x+\frac{10}{100}y=2000$
$\displaystyle 0.05x+0.10y=2000 \qquad \ldots(1)$
$\displaystyle \frac{10}{100}x-\frac{5}{100}y=1500$
$\displaystyle 0.10x-0.05y=1500 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (2) by }2,\text{ we get}$
$\displaystyle 0.20x-0.10y=3000 \qquad \ldots(3)$
$\displaystyle \text{Adding equations (1) and (3),}$
$\displaystyle (0.05x+0.10y)+(0.20x-0.10y)=2000+3000$
$\displaystyle 0.25x=5000$
$\displaystyle x=\frac{5000}{0.25}=20000$
$\displaystyle \text{Substituting }x=20000\text{ in equation (1),}$
$\displaystyle 0.05(20000)+0.10y=2000$
$\displaystyle 1000+0.10y=2000$
$\displaystyle 0.10y=1000$
$\displaystyle y=10000$
$\displaystyle \therefore \text{The cost price of the TV is Rs. }20000\text{ and that of the fridge is Rs. }10000.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{A lending library has a fixed charge for the first three days and an additional}$
$\displaystyle \text{charge for each extra day. Person A paid Rs. }27\text{ for a book kept for }7\text{ days, while}$
$\displaystyle \text{Person B paid Rs. }21\text{ for the same book kept for }5\text{ days. Find the fixed charge}$
$\displaystyle \text{and the charge for each additional day.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the fixed charge be Rs. }x\text{ and the charge for each additional day be Rs. }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+(7-3)y=27$
$\displaystyle x+4y=27 \qquad \ldots(1)$
$\displaystyle x+(5-3)y=21$
$\displaystyle x+2y=21 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (x+4y)-(x+2y)=27-21$
$\displaystyle 2y=6$
$\displaystyle y=3$
$\displaystyle \text{Substituting }y=3\text{ in equation (1),}$
$\displaystyle x+4(3)=27$
$\displaystyle x+12=27$
$\displaystyle x=15$
$\displaystyle \therefore \text{The fixed charge is Rs. }15\text{ and the charge for each additional day is Rs. }3.$
$\displaystyle \\$

Problems Based on Numbers

$\displaystyle \textbf{Question 9: }\text{The sum of the digits of a two-digit number is }8,\text{ and the difference}$
$\displaystyle \text{between the number and the number formed by reversing its digits is }18.\text{ Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the number}=10y+x$
$\displaystyle \text{and the number formed by reversing its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=8 \qquad \ldots(1)$
$\displaystyle (10y+x)-(10x+y)=18$
$\displaystyle 9y-9x=18$
$\displaystyle y-x=2 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle x+y+y-x=8+2$
$\displaystyle 2y=10$
$\displaystyle y=5$
$\displaystyle \text{Substituting }y=5\text{ in equation (1),}$
$\displaystyle x+5=8$
$\displaystyle x=3$
$\displaystyle \therefore \text{The required number}=10y+x=10(5)+3=53.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The sum of a two-digit number and the number obtained by reversing its}$
$\displaystyle \text{digits is }121,\text{ and the two digits differ by }3.\text{ Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the number}=10y+x$
$\displaystyle \text{and the number formed by reversing its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle (10y+x)+(10x+y)=121$
$\displaystyle 11x+11y=121$
$\displaystyle x+y=11 \qquad \ldots(1)$
$\displaystyle \text{Since the digits differ by }3,$
$\displaystyle x-y=\pm3$
$\displaystyle \text{Case I: }x-y=3$
$\displaystyle x+y=11 \qquad \ldots(1)$
$\displaystyle x-y=3 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle 2x=14$
$\displaystyle x=7$
$\displaystyle y=11-7=4$
$\displaystyle \therefore \text{The number}=10y+x=10(4)+7=47.$
$\displaystyle \text{Case II: }x-y=-3$
$\displaystyle x+y=11 \qquad \ldots(3)$
$\displaystyle x-y=-3 \qquad \ldots(4)$
$\displaystyle \text{Adding equations (3) and (4),}$
$\displaystyle 2x=8$
$\displaystyle x=4$
$\displaystyle y=11-4=7$
$\displaystyle \therefore \text{The number}=10y+x=10(7)+4=74.$
$\displaystyle \therefore \text{The required numbers are }47\text{ and }74.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{The sum of a two-digit number and the number formed by interchanging}$
$\displaystyle \text{its digits is }110.\text{ If }10\text{ is subtracted from the first number, the new number is}$
$\displaystyle 4\text{ more than five times the sum of the digits of the first number. Find the first number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the first number}=10y+x$
$\displaystyle \text{and the number formed by interchanging its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle (10y+x)+(10x+y)=110$
$\displaystyle 11x+11y=110$
$\displaystyle x+y=10 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle 10y+x-10=5(x+y)+4$
$\displaystyle 10y+x-10=5x+5y+4$
$\displaystyle 5y-4x=14 \qquad \ldots(2)$
$\displaystyle \text{From equation (1),}$
$\displaystyle x=10-y$
$\displaystyle \text{Substituting }x=10-y\text{ in equation (2),}$
$\displaystyle 5y-4(10-y)=14$
$\displaystyle 5y-40+4y=14$
$\displaystyle 9y=54$
$\displaystyle y=6$
$\displaystyle \text{Substituting }y=6\text{ in equation (1),}$
$\displaystyle x+6=10$
$\displaystyle x=4$
$\displaystyle \therefore \text{The first number}=10y+x=10(6)+4=64.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{The sum of two numbers is }8.\text{ If their sum is four times their}$
$\displaystyle \text{difference, find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the larger number be }x\text{ and the smaller number be }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=8 \qquad \ldots(1)$
$\displaystyle x+y=4(x-y)$
$\displaystyle x+y=4x-4y$
$\displaystyle 3x-5y=0 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }5,\text{ we get}$
$\displaystyle 5x+5y=40 \qquad \ldots(3)$
$\displaystyle \text{Adding equations (2) and (3),}$
$\displaystyle (3x-5y)+(5x+5y)=0+40$
$\displaystyle 8x=40$
$\displaystyle x=5$
$\displaystyle \text{Substituting }x=5\text{ in equation (1),}$
$\displaystyle 5+y=8$
$\displaystyle y=3$
$\displaystyle \therefore \text{The required numbers are }5\text{ and }3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{The sum of the digits of a two-digit number is }15.\text{ The number obtained}$
$\displaystyle \text{by reversing its digits exceeds the given number by }9.\text{ Find the given number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the given number}=10y+x$
$\displaystyle \text{and the number formed by reversing its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=15 \qquad \ldots(1)$
$\displaystyle 10x+y=(10y+x)+9$
$\displaystyle 10x+y=10y+x+9$
$\displaystyle 9x-9y=9$
$\displaystyle x-y=1 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle (x+y)+(x-y)=15+1$
$\displaystyle 2x=16$
$\displaystyle x=8$
$\displaystyle \text{Substituting }x=8\text{ in equation (1),}$
$\displaystyle 8+y=15$
$\displaystyle y=7$
$\displaystyle \therefore \text{The given number}=10y+x=10(7)+8=78.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{The sum of two numbers is }1000,\text{ and the difference between their}$
$\displaystyle \text{squares is }256000.\text{ Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the larger number be }x\text{ and the smaller number be }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=1000 \qquad \ldots(1)$
$\displaystyle x^2-y^2=256000$
$\displaystyle (x-y)(x+y)=256000$
$\displaystyle (x-y)(1000)=256000$
$\displaystyle x-y=256 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle (x+y)+(x-y)=1000+256$
$\displaystyle 2x=1256$
$\displaystyle x=628$
$\displaystyle \text{Substituting }x=628\text{ in equation (1),}$
$\displaystyle 628+y=1000$
$\displaystyle y=372$
$\displaystyle \therefore \text{The required numbers are }628\text{ and }372.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{A two-digit number is four times the sum of its digits. If }18\text{ is added}$
$\displaystyle \text{to the number, its digits are reversed. Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the given number}=10y+x$
$\displaystyle \text{and the number formed by reversing its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle 10y+x+18=10x+y$
$\displaystyle 9x-9y=18$
$\displaystyle x-y=2 \qquad \ldots(1)$
$\displaystyle \text{Also, the number is four times the sum of its digits.}$
$\displaystyle 10y+x=4(x+y)$
$\displaystyle 10y+x=4x+4y$
$\displaystyle 3x-6y=0$
$\displaystyle x-2y=0 \qquad \ldots(2)$
$\displaystyle \text{From equation (2),}$
$\displaystyle x=2y$
$\displaystyle \text{Substituting }x=2y\text{ in equation (1),}$
$\displaystyle 2y-y=2$
$\displaystyle y=2$
$\displaystyle x=2(2)=4$
$\displaystyle \therefore \text{The required number}=10y+x=10(2)+4=24.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{A two-digit number is such that the product of its digits is }20.\text{ If }9\text{ is}$
$\displaystyle \text{added to the number, its digits interchange their places. Find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the given number}=10y+x$
$\displaystyle \text{and the number formed by reversing its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle 10y+x+9=10x+y$
$\displaystyle 9x-9y=9$
$\displaystyle x-y=1 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle xy=20$
$\displaystyle y=\frac{20}{x} \qquad \ldots(2)$
$\displaystyle \text{Substituting equation (2) in equation (1),}$
$\displaystyle x-\frac{20}{x}=1$
$\displaystyle x^2-20=x$
$\displaystyle x^2-x-20=0$
$\displaystyle (x-5)(x+4)=0$
$\displaystyle x=5\text{ or }x=-4$
$\displaystyle \text{Since a digit cannot be negative, }x=-4\text{ is not possible.}$
$\displaystyle \therefore x=5$
$\displaystyle \text{Substituting }x=5\text{ in }xy=20,$
$\displaystyle 5y=20$
$\displaystyle y=4$
$\displaystyle \therefore \text{The required number}=10y+x=10(4)+5=45.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Seven times a two-digit number is equal to four times the number obtained}$
$\displaystyle \text{by reversing its digits. If the difference between the digits is }3,\text{ find the number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the units digit be }x\text{ and the tens digit be }y.$
$\displaystyle \text{Then, the given number}=10y+x$
$\displaystyle \text{and the number formed by reversing its digits}=10x+y.$
$\displaystyle \text{According to the question,}$
$\displaystyle 7(10y+x)=4(10x+y)$
$\displaystyle 70y+7x=40x+4y$
$\displaystyle 66y=33x$
$\displaystyle x=2y \qquad \ldots(1)$
$\displaystyle \text{Since }x=2y,\text{ the units digit is greater than the tens digit. Therefore,}$
$\displaystyle x-y=3 \qquad \ldots(2)$
$\displaystyle \text{Substituting }x=2y\text{ in equation (2),}$
$\displaystyle 2y-y=3$
$\displaystyle y=3$
$\displaystyle x=2(3)=6$
$\displaystyle \therefore \text{The required number}=10y+x=10(3)+6=36.$
$\displaystyle \\$

Problems Based on Fractions

$\displaystyle \textbf{Question 18: }\text{A fraction becomes }\frac{4}{5}\text{ when }1\text{ is added to both its numerator}$
$\displaystyle \text{and denominator. If }5\text{ is subtracted from both its numerator and denominator, the fraction}$
$\displaystyle \text{becomes }\frac{1}{2}.\text{ Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \text{Then, the fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{x+1}{y+1}=\frac{4}{5}$
$\displaystyle 5(x+1)=4(y+1)$
$\displaystyle 5x+5=4y+4$
$\displaystyle 5x-4y=-1 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle \frac{x-5}{y-5}=\frac{1}{2}$
$\displaystyle 2(x-5)=y-5$
$\displaystyle 2x-10=y-5$
$\displaystyle 2x-y=5 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (2) by }4,\text{ we get}$
$\displaystyle 8x-4y=20 \qquad \ldots(3)$
$\displaystyle \text{Subtracting equation (3) from equation (1),}$
$\displaystyle (5x-4y)-(8x-4y)=-1-20$
$\displaystyle -3x=-21$
$\displaystyle x=7$
$\displaystyle \text{Substituting }x=7\text{ in equation (1),}$
$\displaystyle 5(7)-4y=-1$
$\displaystyle 35-4y=-1$
$\displaystyle 4y=36$
$\displaystyle y=9$
$\displaystyle \therefore \text{The required fraction is }\frac{x}{y}=\frac{7}{9}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{The denominator of a fraction is }4\text{ more than twice its numerator.}$
$\displaystyle \text{When both the numerator and denominator are decreased by }6,\text{ the denominator becomes}$
$\displaystyle 12\text{ times the numerator. Determine the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \text{Then, the fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the question,}$
$\displaystyle y=2x+4$
$\displaystyle 2x-y=-4 \qquad \ldots(1)$
$\displaystyle \text{Also, after decreasing both by }6,$
$\displaystyle y-6=12(x-6)$
$\displaystyle y-6=12x-72$
$\displaystyle 12x-y=66 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (2x-y)-(12x-y)=-4-66$
$\displaystyle -10x=-70$
$\displaystyle x=7$
$\displaystyle \text{Substituting }x=7\text{ in }y=2x+4,$
$\displaystyle y=2(7)+4$
$\displaystyle y=18$
$\displaystyle \therefore \text{The required fraction is }\frac{x}{y}=\frac{7}{18}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{A fraction becomes }\frac{1}{3}\text{ if }1\text{ is subtracted from both its numerator}$
$\displaystyle \text{and denominator. If }1\text{ is added to both its numerator and denominator, it becomes }\frac{1}{2}.$
$\displaystyle \text{Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \text{Then, the fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{x-1}{y-1}=\frac{1}{3}$
$\displaystyle 3(x-1)=y-1$
$\displaystyle 3x-3=y-1$
$\displaystyle 3x-y=2 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle \frac{x+1}{y+1}=\frac{1}{2}$
$\displaystyle 2(x+1)=y+1$
$\displaystyle 2x+2=y+1$
$\displaystyle 2x-y=-1 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (3x-y)-(2x-y)=2-(-1)$
$\displaystyle x=3$
$\displaystyle \text{Substituting }x=3\text{ in equation (1),}$
$\displaystyle 3(3)-y=2$
$\displaystyle 9-y=2$
$\displaystyle y=7$
$\displaystyle \therefore \text{The required fraction is }\frac{x}{y}=\frac{3}{7}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{If the numerator of a fraction is multiplied by }2\text{ and its denominator is}$
$\displaystyle \text{reduced by }5,\text{ the fraction becomes }\frac{6}{5}.\text{ If the denominator is doubled and the}$
$\displaystyle \text{numerator is increased by }8,\text{ the fraction becomes }\frac{2}{5}.\text{ Determine the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \text{Then, the fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{2x}{y-5}=\frac{6}{5}$
$\displaystyle 10x=6(y-5)$
$\displaystyle 10x=6y-30$
$\displaystyle 5x-3y=-15 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle \frac{x+8}{2y}=\frac{2}{5}$
$\displaystyle 5(x+8)=4y$
$\displaystyle 5x+40=4y$
$\displaystyle 5x-4y=-40 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (5x-3y)-(5x-4y)=-15-(-40)$
$\displaystyle y=25$
$\displaystyle \text{Substituting }y=25\text{ in equation (1),}$
$\displaystyle 5x-3(25)=-15$
$\displaystyle 5x-75=-15$
$\displaystyle 5x=60$
$\displaystyle x=12$
$\displaystyle \therefore \text{The required fraction is }\frac{x}{y}=\frac{12}{25}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{The sum of the numerator and denominator of a fraction is }18.\text{ If the}$
$\displaystyle \text{denominator is increased by }2,\text{ the fraction reduces to }\frac{1}{3}.\text{ Find the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \text{Then, the fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=18 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle \frac{x}{y+2}=\frac{1}{3}$
$\displaystyle 3x=y+2$
$\displaystyle 3x-y=2 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle (x+y)+(3x-y)=18+2$
$\displaystyle 4x=20$
$\displaystyle x=5$
$\displaystyle \text{Substituting }x=5\text{ in equation (1),}$
$\displaystyle 5+y=18$
$\displaystyle y=13$
$\displaystyle \therefore \text{The required fraction is }\frac{x}{y}=\frac{5}{13}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{The sum of the numerator and denominator of a fraction is }3\text{ less than}$
$\displaystyle \text{twice the denominator. If the numerator and denominator are each decreased by }1,\text{ the}$
$\displaystyle \text{numerator becomes half the denominator. Determine the fraction.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator be }x\text{ and the denominator be }y.$
$\displaystyle \text{Then, the fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+y=2y-3$
$\displaystyle x-y=-3 \qquad \ldots(1)$
$\displaystyle \text{Also, after decreasing both by }1,$
$\displaystyle x-1=\frac{1}{2}(y-1)$
$\displaystyle 2x-2=y-1$
$\displaystyle 2x-y=1 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (x-y)-(2x-y)=-3-1$
$\displaystyle -x=-4$
$\displaystyle x=4$
$\displaystyle \text{Substituting }x=4\text{ in equation (1),}$
$\displaystyle 4-y=-3$
$\displaystyle y=7$
$\displaystyle \therefore \text{The required fraction is }\frac{x}{y}=\frac{4}{7}.$
$\displaystyle \\$

Problems Based on Ages

$\displaystyle \textbf{Question 24: }\text{If twice the son's age is added to the father's age, the sum is }70.$
$\displaystyle \text{If twice the father's age is added to the son's age, the sum is }95.\text{ Find their ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the son's age be }x\text{ years and the father's age be }y\text{ years.}$
$\displaystyle \text{According to the question,}$
$\displaystyle 2x+y=70 \qquad \ldots(1)$
$\displaystyle x+2y=95 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }2,\text{ we get}$
$\displaystyle 4x+2y=140 \qquad \ldots(3)$
$\displaystyle \text{Subtracting equation (2) from equation (3),}$
$\displaystyle (4x+2y)-(x+2y)=140-95$
$\displaystyle 3x=45$
$\displaystyle x=15$
$\displaystyle \text{Substituting }x=15\text{ in equation (1),}$
$\displaystyle 2(15)+y=70$
$\displaystyle 30+y=70$
$\displaystyle y=40$
$\displaystyle \therefore \text{The son's age is }15\text{ years and the father's age is }40\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Ten years ago, a father was twelve times as old as his son. Ten years}$
$\displaystyle \text{hence, he will be twice as old as his son. Find their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present ages of the son and the father be }x\text{ years and }y\text{ years, respectively.}$
$\displaystyle \begin{array}{c|c|c|c} &\text{Present age}&\text{Age 10 years ago}&\text{Age 10 years hence}\\ \hline \text{Son}&x&x-10&x+10\\ \text{Father}&y&y-10&y+10 \end{array}$
$\displaystyle \text{According to the question,}$
$\displaystyle y-10=12(x-10)$
$\displaystyle y-10=12x-120$
$\displaystyle 12x-y=110 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle y+10=2(x+10)$
$\displaystyle y+10=2x+20$
$\displaystyle 2x-y=-10 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (12x-y)-(2x-y)=110-(-10)$
$\displaystyle 10x=120$
$\displaystyle x=12$
$\displaystyle \text{Substituting }x=12\text{ in equation (2),}$
$\displaystyle 2(12)-y=-10$
$\displaystyle 24-y=-10$
$\displaystyle y=34$
$\displaystyle \therefore \text{The son's present age is }12\text{ years and the father's present age is }34\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Ten years hence, }A\text{ will be twice as old as }B.\text{ Five years ago,}$
$\displaystyle A\text{ was three times as old as }B.\text{ Find their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present ages of }A\text{ and }B\text{ be }x\text{ years and }y\text{ years, respectively.}$
$\displaystyle \begin{array}{c|c|c|c} &\text{Present age}&\text{Age 5 years ago}&\text{Age 10 years hence}\\ \hline A&x&x-5&x+10\\ B&y&y-5&y+10 \end{array}$
$\displaystyle \text{According to the question,}$
$\displaystyle x+10=2(y+10)$
$\displaystyle x+10=2y+20$
$\displaystyle x-2y=10 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle x-5=3(y-5)$
$\displaystyle x-5=3y-15$
$\displaystyle x-3y=-10 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (x-2y)-(x-3y)=10-(-10)$
$\displaystyle y=20$
$\displaystyle \text{Substituting }y=20\text{ in equation (1),}$
$\displaystyle x-2(20)=10$
$\displaystyle x-40=10$
$\displaystyle x=50$
$\displaystyle \therefore \text{The present ages of }A\text{ and }B\text{ are }50\text{ years and }20\text{ years, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Ten years ago, a father was twelve times as old as his son. Ten years}$
$\displaystyle \text{hence, he will be twice as old as his son. Find their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present ages of the father and the son be }x\text{ years and }y\text{ years, respectively.}$
$\displaystyle \begin{array}{c|c|c|c} &\text{Present age}&\text{Age 10 years ago}&\text{Age 10 years hence}\\ \hline \text{Father}&x&x-10&x+10\\ \text{Son}&y&y-10&y+10 \end{array}$
$\displaystyle \text{According to the question,}$
$\displaystyle x-10=12(y-10)$
$\displaystyle x-10=12y-120$
$\displaystyle x-12y=-110 \qquad \ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle x+10=2(y+10)$
$\displaystyle x+10=2y+20$
$\displaystyle x-2y=10 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (x-12y)-(x-2y)=-110-10$
$\displaystyle -10y=-120$
$\displaystyle y=12$
$\displaystyle \text{Substituting }y=12\text{ in equation (2),}$
$\displaystyle x-2(12)=10$
$\displaystyle x-24=10$
$\displaystyle x=34$
$\displaystyle \therefore \text{The father's present age is }34\text{ years and the son's present age is }12\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{The present age of a father is three years more than three times the}$
$\displaystyle \text{age of his son. Three years hence, the father's age will be }10\text{ years more than twice}$
$\displaystyle \text{the son's age. Determine their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present ages of the father and the son be }x\text{ years and }y\text{ years, respectively.}$
$\displaystyle \text{According to the question,}$
$\displaystyle x=3y+3$
$\displaystyle x-3y=3 \qquad \ldots(1)$
$\displaystyle \text{Also, three years hence,}$
$\displaystyle x+3=2(y+3)+10$
$\displaystyle x+3=2y+6+10$
$\displaystyle x-2y=13 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (2) from equation (1),}$
$\displaystyle (x-3y)-(x-2y)=3-13$
$\displaystyle -y=-10$
$\displaystyle y=10$
$\displaystyle \text{Substituting }y=10\text{ in equation (2),}$
$\displaystyle x-2(10)=13$
$\displaystyle x-20=13$
$\displaystyle x=33$
$\displaystyle \therefore \text{The son's present age is }10\text{ years and the father's present age is }33\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{A father's age is three times the sum of the ages of his two children.}$
$\displaystyle \text{After }5\text{ years, his age will be twice the sum of the ages of the two children. Find the}$
$\displaystyle \text{father's present age.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present ages of the father and his two children be }x,\ y\text{ and }z\text{ years, respectively.}$
$\displaystyle \text{According to the question,}$
$\displaystyle x=3(y+z)$
$\displaystyle y+z=\frac{x}{3} \qquad \ldots(1)$
$\displaystyle \text{After }5\text{ years, the sum of the ages of the two children will be}$
$\displaystyle (y+5)+(z+5)=y+z+10.$
$\displaystyle \text{Therefore,}$
$\displaystyle x+5=2(y+z+10)$
$\displaystyle x+5=2(y+z)+20$
$\displaystyle \text{Substituting }y+z=\frac{x}{3}\text{ from equation (1),}$
$\displaystyle x+5=\frac{2x}{3}+20$
$\displaystyle x-\frac{2x}{3}=15$
$\displaystyle \frac{x}{3}=15$
$\displaystyle x=45$
$\displaystyle \therefore \text{The father's present age is }45\text{ years.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{The ages of two friends }A\text{ and }B\text{ differ by }3\text{ years. }A\text{'s father }D$
$\displaystyle \text{is twice as old as }A,\text{ and }B\text{ is twice as old as her sister }C.\text{ The ages of }C\text{ and }D$
$\displaystyle \text{differ by }30\text{ years. Find the ages of }A\text{ and }B.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the ages of }A\text{ and }B\text{ be }x\text{ years and }y\text{ years, respectively.}$
$\displaystyle \text{Since their ages differ by }3\text{ years and }A\text{ is older than }B,$
$\displaystyle x-y=3 \qquad \ldots(1)$
$\displaystyle \text{Since }D\text{ is twice as old as }A,\text{ the age of }D=2x.$
$\displaystyle \text{Since }B\text{ is twice as old as }C,\text{ the age of }C=\frac{y}{2}.$
$\displaystyle \text{Since }D\text{ is }30\text{ years older than }C,$
$\displaystyle 2x-\frac{y}{2}=30$
$\displaystyle 4x-y=60 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (4x-y)-(x-y)=60-3$
$\displaystyle 3x=57$
$\displaystyle x=19$
$\displaystyle \text{Substituting }x=19\text{ in equation (1),}$
$\displaystyle 19-y=3$
$\displaystyle y=16$
$\displaystyle \therefore \text{The age of }A\text{ is }19\text{ years and the age of }B\text{ is }16\text{ years.}$
$\displaystyle \\$

Problems based on Time, Distance and Speed

$\displaystyle \textbf{Question 31: }\text{Points }A\text{ and }B\text{ are }90\text{ km apart on a road. A car starts from point }A$
$\displaystyle \text{and another car starts from point }B\text{ at the same time. If they travel in the same direction,}$
$\displaystyle \text{they meet after }9\text{ hours. If they travel in opposite directions, they meet after }\frac{9}{7}\text{ hours.}$
$\displaystyle \text{Find their speeds.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speeds of the cars starting from }A\text{ and }B\text{ be }x\text{ km/h and }y\text{ km/h, respectively.}$
$\displaystyle \text{Case I: When the cars travel in the same direction}$
$\displaystyle \text{Since the car starting from }A\text{ catches the car starting from }B,\text{ we take }x>y.$
$\displaystyle \text{Distance covered by the car from }A\text{ in }9\text{ hours}=9x\text{ km.}$
$\displaystyle \text{Distance covered by the car from }B\text{ in }9\text{ hours}=9y\text{ km.}$
$\displaystyle \text{Therefore,}$
$\displaystyle 9x-9y=90$
$\displaystyle x-y=10 \qquad \ldots(1)$
$\displaystyle \text{Case II: When the cars travel in opposite directions}$
$\displaystyle \text{Distance covered by the car from }A\text{ in }\frac{9}{7}\text{ hours}=\frac{9x}{7}\text{ km.}$
$\displaystyle \text{Distance covered by the car from }B\text{ in }\frac{9}{7}\text{ hours}=\frac{9y}{7}\text{ km.}$
$\displaystyle \text{Therefore,}$
$\displaystyle \frac{9x}{7}+\frac{9y}{7}=90$
$\displaystyle x+y=70 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle (x-y)+(x+y)=10+70$
$\displaystyle 2x=80$
$\displaystyle x=40$
$\displaystyle \text{Substituting }x=40\text{ in equation (2),}$
$\displaystyle 40+y=70$
$\displaystyle y=30$
$\displaystyle \therefore \text{The speeds of the cars starting from }A\text{ and }B\text{ are }40\text{ km/h and }30\text{ km/h, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Points }A\text{ and }B\text{ are }70\text{ km apart on a road. A car starts from point }A$
$\displaystyle \text{and another car starts from point }B\text{ at the same time. If they travel in the same direction,}$
$\displaystyle \text{they meet after }7\text{ hours. If they travel in opposite directions, they meet after }1\text{ hour.}$
$\displaystyle \text{Find their speeds.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speeds of the cars starting from }A\text{ and }B\text{ be }x\text{ km/h and }y\text{ km/h, respectively.}$
$\displaystyle \text{Case I: When the cars travel in the same direction}$
$\displaystyle \text{Since the car starting from }A\text{ catches the car starting from }B,\text{ we take }x>y.$
$\displaystyle \text{Distance covered by the car from }A\text{ in }7\text{ hours}=7x\text{ km.}$
$\displaystyle \text{Distance covered by the car from }B\text{ in }7\text{ hours}=7y\text{ km.}$
$\displaystyle \text{Therefore,}$
$\displaystyle 7x-7y=70$
$\displaystyle x-y=10 \qquad \ldots(1)$
$\displaystyle \text{Case II: When the cars travel in opposite directions}$
$\displaystyle \text{Distance covered by the car from }A\text{ in }1\text{ hour}=x\text{ km.}$
$\displaystyle \text{Distance covered by the car from }B\text{ in }1\text{ hour}=y\text{ km.}$
$\displaystyle \text{Therefore,}$
$\displaystyle x+y=70 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle (x-y)+(x+y)=10+70$
$\displaystyle 2x=80$
$\displaystyle x=40$
$\displaystyle \text{Substituting }x=40\text{ in equation (2),}$
$\displaystyle 40+y=70$
$\displaystyle y=30$
$\displaystyle \therefore \text{The speeds of the cars starting from }A\text{ and }B\text{ are }40\text{ km/h and }30\text{ km/h, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{A train covers a certain distance at a uniform speed. If the train had travelled}$
$\displaystyle 6\text{ km/h faster, it would have taken }4\text{ hours less than the scheduled time. If it had travelled}$
$\displaystyle 6\text{ km/h slower, it would have taken }6\text{ hours more than the scheduled time. Find the length}$
$\displaystyle \text{of the journey.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the actual speed of the train be }x\text{ km/h and the scheduled time be }y\text{ hours.}$
$\displaystyle \therefore \text{Length of the journey}=xy\text{ km.}$
$\displaystyle \text{If the speed were }6\text{ km/h more, the time taken would be }4\text{ hours less.}$
$\displaystyle xy=(x+6)(y-4)$
$\displaystyle xy=xy-4x+6y-24$
$\displaystyle 4x-6y=-24$
$\displaystyle -2x+3y=12 \qquad \ldots(1)$
$\displaystyle \text{If the speed were }6\text{ km/h less, the time taken would be }6\text{ hours more.}$
$\displaystyle xy=(x-6)(y+6)$
$\displaystyle xy=xy+6x-6y-36$
$\displaystyle 6x-6y=36$
$\displaystyle x-y=6 \qquad \ldots(2)$
$\displaystyle \text{From equation (2),}$
$\displaystyle x=y+6$
$\displaystyle \text{Substituting }x=y+6\text{ in equation (1),}$
$\displaystyle -2(y+6)+3y=12$
$\displaystyle -2y-12+3y=12$
$\displaystyle y=24$
$\displaystyle x=24+6=30$
$\displaystyle \therefore \text{Length of the journey}=xy=30(24)=720\text{ km.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{A man travels }370\text{ km partly by train and partly by car. If he covers }250\text{ km}$
$\displaystyle \text{by train and the remaining distance by car, the journey takes }4\text{ hours. If he covers }130\text{ km}$
$\displaystyle \text{by train and the remaining distance by car, it takes }18\text{ minutes longer. Find the speeds}$
$\displaystyle \text{of the train and the car.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speeds of the train and the car be }x\text{ km/h and }y\text{ km/h, respectively.}$
$\displaystyle \text{When he travels }250\text{ km by train and }120\text{ km by car,}$
$\displaystyle \frac{250}{x}+\frac{120}{y}=4$
$\displaystyle \frac{125}{x}+\frac{60}{y}=2 \qquad \ldots(1)$
$\displaystyle \text{When he travels }130\text{ km by train and }240\text{ km by car,}$
$\displaystyle \text{the total time taken}=4+\frac{18}{60}=\frac{43}{10}\text{ hours.}$
$\displaystyle \frac{130}{x}+\frac{240}{y}=\frac{43}{10} \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 125u+60v=2 \qquad \ldots(3)$
$\displaystyle 130u+240v=\frac{43}{10} \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }4,\text{ we get}$
$\displaystyle 500u+240v=8 \qquad \ldots(5)$
$\displaystyle \text{Subtracting equation (4) from equation (5),}$
$\displaystyle (500u+240v)-(130u+240v)=8-\frac{43}{10}$
$\displaystyle 370u=\frac{37}{10}$
$\displaystyle u=\frac{1}{100}$
$\displaystyle \text{Substituting }u=\frac{1}{100}\text{ in equation (3),}$
$\displaystyle 125\left(\frac{1}{100}\right)+60v=2$
$\displaystyle \frac{5}{4}+60v=2$
$\displaystyle 60v=\frac{3}{4}$
$\displaystyle v=\frac{1}{80}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{1}{100}\text{ and }\frac{1}{y}=v=\frac{1}{80},$
$\displaystyle x=100\text{ and }y=80.$
$\displaystyle \therefore \text{The speeds of the train and the car are }100\text{ km/h and }80\text{ km/h, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{A man travels }760\text{ km partly by train and partly by car. If he covers }160\text{ km}$
$\displaystyle \text{by train and the remaining distance by car, the journey takes }8\text{ hours. If he covers }240\text{ km}$
$\displaystyle \text{by train and the remaining distance by car, it takes }12\text{ minutes longer. Find the speeds}$
$\displaystyle \text{of the train and the car.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speeds of the train and the car be }x\text{ km/h and }y\text{ km/h, respectively.}$
$\displaystyle \text{When he travels }160\text{ km by train and }600\text{ km by car,}$
$\displaystyle \frac{160}{x}+\frac{600}{y}=8$
$\displaystyle \frac{20}{x}+\frac{75}{y}=1 \qquad \ldots(1)$
$\displaystyle \text{When he travels }240\text{ km by train and }520\text{ km by car,}$
$\displaystyle \text{the total time taken}=8+\frac{12}{60}=\frac{41}{5}\text{ hours.}$
$\displaystyle \frac{240}{x}+\frac{520}{y}=\frac{41}{5} \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 20u+75v=1 \qquad \ldots(3)$
$\displaystyle 240u+520v=\frac{41}{5} \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }12,\text{ we get}$
$\displaystyle 240u+900v=12 \qquad \ldots(5)$
$\displaystyle \text{Subtracting equation (4) from equation (5),}$
$\displaystyle (240u+900v)-(240u+520v)=12-\frac{41}{5}$
$\displaystyle 380v=\frac{19}{5}$
$\displaystyle v=\frac{1}{100}$
$\displaystyle \text{Substituting }v=\frac{1}{100}\text{ in equation (3),}$
$\displaystyle 20u+75\left(\frac{1}{100}\right)=1$
$\displaystyle 20u+\frac{3}{4}=1$
$\displaystyle 20u=\frac{1}{4}$
$\displaystyle u=\frac{1}{80}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{1}{80}\text{ and }\frac{1}{y}=v=\frac{1}{100},$
$\displaystyle x=80\text{ and }y=100.$
$\displaystyle \therefore \text{The speeds of the train and the car are }80\text{ km/h and }100\text{ km/h, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{A girl travels }300\text{ km partly by train and partly by car. If she covers }60\text{ km}$
$\displaystyle \text{by train and the remaining distance by car, the journey takes }4\text{ hours. If she covers }100\text{ km}$
$\displaystyle \text{by train and the remaining distance by car, it takes }10\text{ minutes longer. Find the speeds}$
$\displaystyle \text{of the train and the car.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speeds of the train and the car be }x\text{ km/h and }y\text{ km/h, respectively.}$
$\displaystyle \text{When she travels }60\text{ km by train and }240\text{ km by car,}$
$\displaystyle \frac{60}{x}+\frac{240}{y}=4$
$\displaystyle \frac{15}{x}+\frac{60}{y}=1 \qquad \ldots(1)$
$\displaystyle \text{When she travels }100\text{ km by train and }200\text{ km by car,}$
$\displaystyle \text{the total time taken}=4+\frac{10}{60}=\frac{25}{6}\text{ hours.}$
$\displaystyle \frac{100}{x}+\frac{200}{y}=\frac{25}{6} \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 15u+60v=1 \qquad \ldots(3)$
$\displaystyle 100u+200v=\frac{25}{6} \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }10,\text{ we get}$
$\displaystyle 150u+600v=10 \qquad \ldots(5)$
$\displaystyle \text{Multiplying equation (4) by }3,\text{ we get}$
$\displaystyle 300u+600v=\frac{25}{2} \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (5) from equation (6),}$
$\displaystyle (300u+600v)-(150u+600v)=\frac{25}{2}-10$
$\displaystyle 150u=\frac{5}{2}$
$\displaystyle u=\frac{1}{60}$
$\displaystyle \text{Substituting }u=\frac{1}{60}\text{ in equation (3),}$
$\displaystyle 15\left(\frac{1}{60}\right)+60v=1$
$\displaystyle \frac{1}{4}+60v=1$
$\displaystyle 60v=\frac{3}{4}$
$\displaystyle v=\frac{1}{80}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{1}{60}\text{ and }\frac{1}{y}=v=\frac{1}{80},$
$\displaystyle x=60\text{ and }y=80.$
$\displaystyle \therefore \text{The speeds of the train and the car are }60\text{ km/h and }80\text{ km/h, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{A boat covers }32\text{ km upstream and }36\text{ km downstream in }7\text{ hours.}$
$\displaystyle \text{It also covers }40\text{ km upstream and }48\text{ km downstream in }9\text{ hours. Find the speed}$
$\displaystyle \text{of the boat in still water and the speed of the stream.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/h and the speed of the stream be }y\text{ km/h.}$
$\displaystyle \therefore \text{Speed of the boat upstream}=(x-y)\text{ km/h.}$
$\displaystyle \text{Speed of the boat downstream}=(x+y)\text{ km/h.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{32}{x-y}+\frac{36}{x+y}=7 \qquad \ldots(1)$
$\displaystyle \frac{40}{x-y}+\frac{48}{x+y}=9 \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x-y}=u\text{ and }\frac{1}{x+y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 32u+36v=7 \qquad \ldots(3)$
$\displaystyle 40u+48v=9 \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }4,\text{ we get}$
$\displaystyle 128u+144v=28 \qquad \ldots(5)$
$\displaystyle \text{Multiplying equation (4) by }3,\text{ we get}$
$\displaystyle 120u+144v=27 \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (6) from equation (5),}$
$\displaystyle 8u=1$
$\displaystyle u=\frac{1}{8}$
$\displaystyle \text{Substituting }u=\frac{1}{8}\text{ in equation (3),}$
$\displaystyle 32\left(\frac{1}{8}\right)+36v=7$
$\displaystyle 4+36v=7$
$\displaystyle 36v=3$
$\displaystyle v=\frac{1}{12}$
$\displaystyle \text{Since }\frac{1}{x-y}=u=\frac{1}{8}\text{ and }\frac{1}{x+y}=v=\frac{1}{12},$
$\displaystyle x-y=8 \qquad \ldots(7)$
$\displaystyle x+y=12 \qquad \ldots(8)$
$\displaystyle \text{Adding equations (7) and (8),}$
$\displaystyle 2x=20$
$\displaystyle x=10$
$\displaystyle \text{Substituting }x=10\text{ in equation (8),}$
$\displaystyle 10+y=12$
$\displaystyle y=2$
$\displaystyle \therefore \text{The speed of the boat in still water is }10\text{ km/h and the speed of the stream is }2\text{ km/h.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{A sailor goes }8\text{ km downstream in }40\text{ minutes and returns in }1\text{ hour.}$
$\displaystyle \text{Determine the speed of the sailor in still water and the speed of the current.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the sailor in still water be }x\text{ km/h and the speed of the current be }y\text{ km/h.}$
$\displaystyle \therefore \text{Speed downstream}=(x+y)\text{ km/h.}$
$\displaystyle \text{Speed upstream}=(x-y)\text{ km/h.}$
$\displaystyle \text{Since }40\text{ minutes}=\frac{40}{60}=\frac{2}{3}\text{ hour,}$
$\displaystyle \frac{8}{x+y}=\frac{2}{3}$
$\displaystyle 2(x+y)=24$
$\displaystyle x+y=12 \qquad \ldots(1)$
$\displaystyle \text{For the upstream journey,}$
$\displaystyle \frac{8}{x-y}=1$
$\displaystyle x-y=8 \qquad \ldots(2)$
$\displaystyle \text{Adding equations (1) and (2),}$
$\displaystyle 2x=20$
$\displaystyle x=10$
$\displaystyle \text{Substituting }x=10\text{ in equation (1),}$
$\displaystyle 10+y=12$
$\displaystyle y=2$
$\displaystyle \therefore \text{The speed of the sailor in still water is }10\text{ km/h and the speed of the current is }2\text{ km/h.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\text{A boat covers }24\text{ km upstream and }28\text{ km downstream in }6\text{ hours.}$
$\displaystyle \text{It also covers }30\text{ km upstream and }21\text{ km downstream in }6.5\text{ hours. Find the speed}$
$\displaystyle \text{of the boat in still water and the speed of the stream.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/h and the speed of the stream be }y\text{ km/h.}$
$\displaystyle \therefore \text{Speed upstream}=(x-y)\text{ km/h.}$
$\displaystyle \text{Speed downstream}=(x+y)\text{ km/h.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{24}{x-y}+\frac{28}{x+y}=6 \qquad \ldots(1)$
$\displaystyle \frac{30}{x-y}+\frac{21}{x+y}=\frac{13}{2} \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x-y}=u\text{ and }\frac{1}{x+y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 24u+28v=6 \qquad \ldots(3)$
$\displaystyle 30u+21v=\frac{13}{2} \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }3,\text{ we get}$
$\displaystyle 72u+84v=18 \qquad \ldots(5)$
$\displaystyle \text{Multiplying equation (4) by }4,\text{ we get}$
$\displaystyle 120u+84v=26 \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (5) from equation (6),}$
$\displaystyle 48u=8$
$\displaystyle u=\frac{1}{6}$
$\displaystyle \text{Substituting }u=\frac{1}{6}\text{ in equation (3),}$
$\displaystyle 24\left(\frac{1}{6}\right)+28v=6$
$\displaystyle 4+28v=6$
$\displaystyle 28v=2$
$\displaystyle v=\frac{1}{14}$
$\displaystyle \text{Since }\frac{1}{x-y}=u=\frac{1}{6}\text{ and }\frac{1}{x+y}=v=\frac{1}{14},$
$\displaystyle x-y=6 \qquad \ldots(7)$
$\displaystyle x+y=14 \qquad \ldots(8)$
$\displaystyle \text{Adding equations (7) and (8),}$
$\displaystyle 2x=20$
$\displaystyle x=10$
$\displaystyle \text{Substituting }x=10\text{ in equation (8),}$
$\displaystyle 10+y=14$
$\displaystyle y=4$
$\displaystyle \therefore \text{The speed of the boat in still water is }10\text{ km/h and the speed of the stream is }4\text{ km/h.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: }\text{A boat covers }12\text{ km upstream and }40\text{ km downstream in }8\text{ hours.}$
$\displaystyle \text{It also covers }16\text{ km upstream and }32\text{ km downstream in }8\text{ hours. Find the speed}$
$\displaystyle \text{of the boat in still water and the speed of the stream.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/h and the speed of the stream be }y\text{ km/h.}$
$\displaystyle \therefore \text{Speed upstream}=(x-y)\text{ km/h.}$
$\displaystyle \text{Speed downstream}=(x+y)\text{ km/h.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{12}{x-y}+\frac{40}{x+y}=8 \qquad \ldots(1)$
$\displaystyle \frac{16}{x-y}+\frac{32}{x+y}=8 \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x-y}=u\text{ and }\frac{1}{x+y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 12u+40v=8$
$\displaystyle 3u+10v=2 \qquad \ldots(3)$
$\displaystyle 16u+32v=8$
$\displaystyle 2u+4v=1 \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }2,\text{ we get}$
$\displaystyle 6u+20v=4 \qquad \ldots(5)$
$\displaystyle \text{Multiplying equation (4) by }3,\text{ we get}$
$\displaystyle 6u+12v=3 \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (6) from equation (5),}$
$\displaystyle 8v=1$
$\displaystyle v=\frac{1}{8}$
$\displaystyle \text{Substituting }v=\frac{1}{8}\text{ in equation (4),}$
$\displaystyle 2u+4\left(\frac{1}{8}\right)=1$
$\displaystyle 2u+\frac{1}{2}=1$
$\displaystyle 2u=\frac{1}{2}$
$\displaystyle u=\frac{1}{4}$
$\displaystyle \text{Since }\frac{1}{x-y}=u=\frac{1}{4}\text{ and }\frac{1}{x+y}=v=\frac{1}{8},$
$\displaystyle x-y=4 \qquad \ldots(7)$
$\displaystyle x+y=8 \qquad \ldots(8)$
$\displaystyle \text{Adding equations (7) and (8),}$
$\displaystyle 2x=12$
$\displaystyle x=6$
$\displaystyle \text{Substituting }x=6\text{ in equation (8),}$
$\displaystyle 6+y=8$
$\displaystyle y=2$
$\displaystyle \therefore \text{The speed of the boat in still water is }6\text{ km/h and the speed of the stream is }2\text{ km/h.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: }X\text{ takes }3\text{ hours more than }Y\text{ to walk }30\text{ km. But if }X\text{ doubles}$
$\displaystyle \text{his speed, he takes }1\frac{1}{2}\text{ hours less than }Y.\text{ Find their walking speeds.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the walking speeds of }X\text{ and }Y\text{ be }x\text{ km/h and }y\text{ km/h, respectively.}$
$\displaystyle \text{Time taken by }X\text{ to walk }30\text{ km}=\frac{30}{x}\text{ hours.}$
$\displaystyle \text{Time taken by }Y\text{ to walk }30\text{ km}=\frac{30}{y}\text{ hours.}$
$\displaystyle \text{Since }X\text{ takes }3\text{ hours more than }Y,$
$\displaystyle \frac{30}{x}=\frac{30}{y}+3$
$\displaystyle \frac{10}{x}-\frac{10}{y}=1 \qquad \ldots(1)$
$\displaystyle \text{When }X\text{ doubles his speed, his speed becomes }2x\text{ km/h.}$
$\displaystyle \text{He then takes }1\frac{1}{2}\text{ hours less than }Y.$
$\displaystyle \frac{30}{2x}+\frac{3}{2}=\frac{30}{y}$
$\displaystyle \frac{15}{x}-\frac{30}{y}=-\frac{3}{2} \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 10u-10v=1 \qquad \ldots(3)$
$\displaystyle 15u-30v=-\frac{3}{2} \qquad \ldots(4)$
$\displaystyle \text{Dividing equation (3) by }10,\text{ we get}$
$\displaystyle u-v=\frac{1}{10} \qquad \ldots(5)$
$\displaystyle \text{Dividing equation (4) by }15,\text{ we get}$
$\displaystyle u-2v=-\frac{1}{10} \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (6) from equation (5),}$
$\displaystyle v=\frac{1}{5}$
$\displaystyle \text{Substituting }v=\frac{1}{5}\text{ in equation (5),}$
$\displaystyle u-\frac{1}{5}=\frac{1}{10}$
$\displaystyle u=\frac{3}{10}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{3}{10}\text{ and }\frac{1}{y}=v=\frac{1}{5},$
$\displaystyle x=\frac{10}{3}\text{ and }y=5.$
$\displaystyle \therefore \text{The walking speeds of }X\text{ and }Y\text{ are }\frac{10}{3}\text{ km/h and }5\text{ km/h, respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: }\text{A man walks a certain distance at a certain speed. If he walks }0.5\text{ km/h faster,}$
$\displaystyle \text{he takes }1\text{ hour less. If he walks }1\text{ km/h slower, he takes }3\text{ hours more. Find the}$
$\displaystyle \text{distance covered by the man and his walking speed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the actual speed of the man be }x\text{ km/h and the actual time taken be }y\text{ hours.}$
$\displaystyle \therefore \text{Distance covered}=xy\text{ km.}$
$\displaystyle \text{If his speed were }0.5\text{ km/h more, he would take }1\text{ hour less.}$
$\displaystyle xy=(x+0.5)(y-1)$
$\displaystyle xy=xy-x+0.5y-0.5$
$\displaystyle x-0.5y=-0.5$
$\displaystyle 2x-y=-1 \qquad \ldots(1)$
$\displaystyle \text{If his speed were }1\text{ km/h less, he would take }3\text{ hours more.}$
$\displaystyle xy=(x-1)(y+3)$
$\displaystyle xy=xy+3x-y-3$
$\displaystyle 3x-y=3 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (3x-y)-(2x-y)=3-(-1)$
$\displaystyle x=4$
$\displaystyle \text{Substituting }x=4\text{ in equation (1),}$
$\displaystyle 2(4)-y=-1$
$\displaystyle 8-y=-1$
$\displaystyle y=9$
$\displaystyle \therefore \text{Distance covered}=xy=4(9)=36\text{ km.}$
$\displaystyle \therefore \text{The man's walking speed is }4\text{ km/h and the distance covered is }36\text{ km.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: }\text{A train covers a certain distance at a uniform speed. If the train had travelled}$
$\displaystyle 10\text{ km/h faster, it would have taken }2\text{ hours less than the scheduled time. If it had travelled}$
$\displaystyle 10\text{ km/h slower, it would have taken }3\text{ hours more than the scheduled time. Find the distance}$
$\displaystyle \text{covered by the train.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the actual speed of the train be }x\text{ km/h and the scheduled time be }y\text{ hours.}$
$\displaystyle \therefore \text{Distance covered by the train}=xy\text{ km.}$
$\displaystyle \text{If the speed were }10\text{ km/h more, the time taken would be }2\text{ hours less.}$
$\displaystyle xy=(x+10)(y-2)$
$\displaystyle xy=xy-2x+10y-20$
$\displaystyle 2x-10y=-20 \qquad \ldots(1)$
$\displaystyle \text{If the speed were }10\text{ km/h less, the time taken would be }3\text{ hours more.}$
$\displaystyle xy=(x-10)(y+3)$
$\displaystyle xy=xy+3x-10y-30$
$\displaystyle 3x-10y=30 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (3x-10y)-(2x-10y)=30-(-20)$
$\displaystyle x=50$
$\displaystyle \text{Substituting }x=50\text{ in equation (1),}$
$\displaystyle 2(50)-10y=-20$
$\displaystyle 100-10y=-20$
$\displaystyle -10y=-120$
$\displaystyle y=12$
$\displaystyle \therefore \text{Distance covered}=xy=50(12)=600\text{ km.}$
$\displaystyle \therefore \text{The distance covered by the train is }600\text{ km.}$
$\displaystyle \\$

Miscellaneous Problems

$\displaystyle \textbf{Question 44: }\text{A taxi charge comprises a fixed charge and a variable charge per km.}$
$\displaystyle \text{For a journey of }10\text{ km, the charge is Rs. }75\text{ and for a journey of }15\text{ km,}$
$\displaystyle \text{the charge is Rs. }110.\text{ Find the charge for a journey of }25\text{ km.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the fixed charge be Rs. }x\text{ and the variable charge be Rs. }y\text{ per km.}$
$\displaystyle \text{According to the question,}$
$\displaystyle x+10y=75 \qquad \ldots(1)$
$\displaystyle x+15y=110 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (x+15y)-(x+10y)=110-75$
$\displaystyle 5y=35$
$\displaystyle y=7$
$\displaystyle \text{Substituting }y=7\text{ in equation (1),}$
$\displaystyle x+10(7)=75$
$\displaystyle x=5$
$\displaystyle \text{Therefore, the charge for a journey of }25\text{ km is}$
$\displaystyle x+25y=5+25(7)=180.$
$\displaystyle \therefore \text{The person will pay Rs. }180.$
$\displaystyle \\$

$\displaystyle \textbf{Question 45: }\text{A part of the monthly hostel charge is fixed and the other part depends on}$
$\displaystyle \text{the number of days stayed. A student staying for }20\text{ days pays Rs. }1000,\text{ while another}$
$\displaystyle \text{student staying for }26\text{ days pays Rs. }1180.\text{ Find the fixed charge and the daily charge.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the fixed hostel charge be Rs. }x\text{ and the daily charge be Rs. }y\text{ per day.}$
$\displaystyle \text{According to the question,}$
$\displaystyle x+20y=1000 \qquad \ldots(1)$
$\displaystyle x+26y=1180 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (x+26y)-(x+20y)=1180-1000$
$\displaystyle 6y=180$
$\displaystyle y=30$
$\displaystyle \text{Substituting }y=30\text{ in equation (1),}$
$\displaystyle x+20(30)=1000$
$\displaystyle x=400$
$\displaystyle \therefore \text{The fixed hostel charge is Rs. }400\text{ and the daily charge is Rs. }30\text{ per day.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 46: }\text{A man starts a job with a certain monthly salary and receives a fixed increment}$
$\displaystyle \text{every year. If his salary is Rs. }1500\text{ after }4\text{ years and Rs. }1800\text{ after }10\text{ years,}$
$\displaystyle \text{find his starting salary and annual increment.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the starting monthly salary be Rs. }x\text{ and the annual increment be Rs. }y.$
$\displaystyle \text{According to the question,}$
$\displaystyle x+4y=1500 \qquad \ldots(1)$
$\displaystyle x+10y=1800 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (x+10y)-(x+4y)=1800-1500$
$\displaystyle 6y=300$
$\displaystyle y=50$
$\displaystyle \text{Substituting }y=50\text{ in equation (1),}$
$\displaystyle x+4(50)=1500$
$\displaystyle x=1300$
$\displaystyle \therefore \text{His starting monthly salary was Rs. }1300\text{ and his annual increment was Rs. }50.$
$\displaystyle \\$

$\displaystyle \textbf{Question 47: }\text{Students of a class are arranged in rows. If there is one extra student in each row,}$
$\displaystyle \text{there would be }2\text{ fewer rows. If there is one student less in each row, there would be }3\text{ more}$
$\displaystyle \text{rows. Find the number of students in the class.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of students in each row be }x\text{ and the number of rows be }y.$
$\displaystyle \therefore \text{Total number of students}=xy.$
$\displaystyle \text{If there is one extra student in each row, there are }2\text{ fewer rows.}$
$\displaystyle xy=(x+1)(y-2)$
$\displaystyle xy=xy-2x+y-2$
$\displaystyle 2x-y=-2 \qquad \ldots(1)$
$\displaystyle \text{If there is one student less in each row, there are }3\text{ more rows.}$
$\displaystyle xy=(x-1)(y+3)$
$\displaystyle xy=xy+3x-y-3$
$\displaystyle 3x-y=3 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle (3x-y)-(2x-y)=3-(-2)$
$\displaystyle x=5$
$\displaystyle \text{Substituting }x=5\text{ in equation (1),}$
$\displaystyle 2(5)-y=-2$
$\displaystyle 10-y=-2$
$\displaystyle y=12$
$\displaystyle \therefore \text{Total number of students}=xy=5(12)=60.$
$\displaystyle \therefore \text{There are }60\text{ students in the class.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: }8\text{ men and }12\text{ boys can finish a work in }10\text{ days, while }6\text{ men}$
$\displaystyle \text{and }8\text{ boys can finish the same work in }14\text{ days. Find the time taken by one man alone}$
$\displaystyle \text{and by one boy alone to finish the work.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one man alone complete the work in }x\text{ days and one boy alone complete it in }y\text{ days.}$
$\displaystyle \therefore \text{One man's work in one day}=\frac{1}{x}.$
$\displaystyle \text{One boy's work in one day}=\frac{1}{y}.$
$\displaystyle \text{Since }8\text{ men and }12\text{ boys complete the work in }10\text{ days,}$
$\displaystyle 10\left(\frac{8}{x}+\frac{12}{y}\right)=1$
$\displaystyle \frac{80}{x}+\frac{120}{y}=1 \qquad \ldots(1)$
$\displaystyle \text{Since }6\text{ men and }8\text{ boys complete the work in }14\text{ days,}$
$\displaystyle 14\left(\frac{6}{x}+\frac{8}{y}\right)=1$
$\displaystyle \frac{84}{x}+\frac{112}{y}=1 \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 80u+120v=1 \qquad \ldots(3)$
$\displaystyle 84u+112v=1 \qquad \ldots(4)$
$\displaystyle \text{Dividing equation (3) by }40,\text{ we get}$
$\displaystyle 2u+3v=\frac{1}{40} \qquad \ldots(5)$
$\displaystyle \text{Dividing equation (4) by }28,\text{ we get}$
$\displaystyle 3u+4v=\frac{1}{28} \qquad \ldots(6)$
$\displaystyle \text{Multiplying equation (5) by }3,\text{ we get}$
$\displaystyle 6u+9v=\frac{3}{40} \qquad \ldots(7)$
$\displaystyle \text{Multiplying equation (6) by }2,\text{ we get}$
$\displaystyle 6u+8v=\frac{1}{14} \qquad \ldots(8)$
$\displaystyle \text{Subtracting equation (8) from equation (7),}$
$\displaystyle v=\frac{3}{40}-\frac{1}{14}$
$\displaystyle v=\frac{42-40}{560}=\frac{1}{280}$
$\displaystyle \text{Substituting }v=\frac{1}{280}\text{ in equation (5),}$
$\displaystyle 2u+3\left(\frac{1}{280}\right)=\frac{1}{40}$
$\displaystyle 2u=\frac{1}{40}-\frac{3}{280}=\frac{1}{70}$
$\displaystyle u=\frac{1}{140}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{1}{140}\text{ and }\frac{1}{y}=v=\frac{1}{280},$
$\displaystyle x=140\text{ and }y=280.$
$\displaystyle \therefore \text{One man alone takes }140\text{ days and one boy alone takes }280\text{ days to finish the work.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 49: }2\text{ men and }7\text{ boys can finish a work in }4\text{ days, while }4\text{ men}$
$\displaystyle \text{and }4\text{ boys can finish the same work in }3\text{ days. Find the time taken by one man alone}$
$\displaystyle \text{and by one boy alone to finish the work.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one man alone complete the work in }x\text{ days and one boy alone complete it in }y\text{ days.}$
$\displaystyle \therefore \text{One man's work in one day}=\frac{1}{x}.$
$\displaystyle \text{One boy's work in one day}=\frac{1}{y}.$
$\displaystyle \text{Since }2\text{ men and }7\text{ boys complete the work in }4\text{ days,}$
$\displaystyle 4\left(\frac{2}{x}+\frac{7}{y}\right)=1$
$\displaystyle \frac{8}{x}+\frac{28}{y}=1 \qquad \ldots(1)$
$\displaystyle \text{Since }4\text{ men and }4\text{ boys complete the work in }3\text{ days,}$
$\displaystyle 3\left(\frac{4}{x}+\frac{4}{y}\right)=1$
$\displaystyle \frac{12}{x}+\frac{12}{y}=1 \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 8u+28v=1 \qquad \ldots(3)$
$\displaystyle 12u+12v=1 \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }3,\text{ we get}$
$\displaystyle 24u+84v=3 \qquad \ldots(5)$
$\displaystyle \text{Multiplying equation (4) by }2,\text{ we get}$
$\displaystyle 24u+24v=2 \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (6) from equation (5),}$
$\displaystyle 60v=1$
$\displaystyle v=\frac{1}{60}$
$\displaystyle \text{Substituting }v=\frac{1}{60}\text{ in equation (4),}$
$\displaystyle 12u+12\left(\frac{1}{60}\right)=1$
$\displaystyle 12u+\frac{1}{5}=1$
$\displaystyle 12u=\frac{4}{5}$
$\displaystyle u=\frac{1}{15}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{1}{15}\text{ and }\frac{1}{y}=v=\frac{1}{60},$
$\displaystyle x=15\text{ and }y=60.$
$\displaystyle \therefore \text{One man alone takes }15\text{ days and one boy alone takes }60\text{ days to finish the work.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 50: }2\text{ women and }5\text{ men can finish a work in }4\text{ days, while }3\text{ women}$
$\displaystyle \text{and }6\text{ men can finish the same work in }3\text{ days. Find the time taken by one woman alone}$
$\displaystyle \text{and by one man alone to finish the work.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one woman alone complete the work in }x\text{ days and one man alone complete it in }y\text{ days.}$
$\displaystyle \therefore \text{One woman's work in one day}=\frac{1}{x}.$
$\displaystyle \text{One man's work in one day}=\frac{1}{y}.$
$\displaystyle \text{Since }2\text{ women and }5\text{ men complete the work in }4\text{ days,}$
$\displaystyle 4\left(\frac{2}{x}+\frac{5}{y}\right)=1$
$\displaystyle \frac{8}{x}+\frac{20}{y}=1 \qquad \ldots(1)$
$\displaystyle \text{Since }3\text{ women and }6\text{ men complete the work in }3\text{ days,}$
$\displaystyle 3\left(\frac{3}{x}+\frac{6}{y}\right)=1$
$\displaystyle \frac{9}{x}+\frac{18}{y}=1 \qquad \ldots(2)$
$\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.$
$\displaystyle \text{Then, equations (1) and (2) become}$
$\displaystyle 8u+20v=1 \qquad \ldots(3)$
$\displaystyle 9u+18v=1 \qquad \ldots(4)$
$\displaystyle \text{Multiplying equation (3) by }9,\text{ we get}$
$\displaystyle 72u+180v=9 \qquad \ldots(5)$
$\displaystyle \text{Multiplying equation (4) by }8,\text{ we get}$
$\displaystyle 72u+144v=8 \qquad \ldots(6)$
$\displaystyle \text{Subtracting equation (6) from equation (5),}$
$\displaystyle 36v=1$
$\displaystyle v=\frac{1}{36}$
$\displaystyle \text{Substituting }v=\frac{1}{36}\text{ in equation (4),}$
$\displaystyle 9u+18\left(\frac{1}{36}\right)=1$
$\displaystyle 9u+\frac{1}{2}=1$
$\displaystyle 9u=\frac{1}{2}$
$\displaystyle u=\frac{1}{18}$
$\displaystyle \text{Since }\frac{1}{x}=u=\frac{1}{18}\text{ and }\frac{1}{y}=v=\frac{1}{36},$
$\displaystyle x=18\text{ and }y=36.$
$\displaystyle \therefore \text{One woman alone takes }18\text{ days and one man alone takes }36\text{ days to finish the work.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 51: }\text{The ratio of the incomes of two people is }9:7\text{ and the ratio of their}$
$\displaystyle \text{expenditures is }4:3.\text{ If each saves Rs. }200\text{ per month, find their monthly incomes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the monthly incomes be Rs. }9x\text{ and Rs. }7x\text{ respectively.}$
$\displaystyle \text{Let the monthly expenditures be Rs. }4y\text{ and Rs. }3y\text{ respectively.}$
$\displaystyle \text{Since each saves Rs. }200\text{ per month,}$
$\displaystyle 9x-4y=200 \qquad \ldots(1)$
$\displaystyle 7x-3y=200 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }3\text{ and equation (2) by }4,\text{ we get}$
$\displaystyle 27x-12y=600 \qquad \ldots(3)$
$\displaystyle 28x-12y=800 \qquad \ldots(4)$
$\displaystyle \text{Subtracting equation (3) from equation (4),}$
$\displaystyle x=200$
$\displaystyle \text{Substituting }x=200\text{ in equation (2),}$
$\displaystyle 7(200)-3y=200$
$\displaystyle 1400-3y=200$
$\displaystyle 3y=1200$
$\displaystyle y=400$
$\displaystyle \therefore \text{The monthly incomes are Rs. }9(200)=\text{Rs. }1800\text{ and Rs. }7(200)=\text{Rs. }1400.$
$\displaystyle \\$

$\displaystyle \textbf{Question 52: }\text{The incomes of }X\text{ and }Y\text{ are in the ratio }8:7\text{ and their}$
$\displaystyle \text{expenditures are in the ratio }19:16.\text{ If each saves Rs. }1250,\text{ find their incomes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the monthly incomes of }X\text{ and }Y\text{ be Rs. }8x\text{ and Rs. }7x\text{ respectively.}$
$\displaystyle \text{Let their monthly expenditures be Rs. }19y\text{ and Rs. }16y\text{ respectively.}$
$\displaystyle \text{Since each saves Rs. }1250,$
$\displaystyle 8x-19y=1250 \qquad \ldots(1)$
$\displaystyle 7x-16y=1250 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }16\text{ and equation (2) by }19,\text{ we get}$
$\displaystyle 128x-304y=20000 \qquad \ldots(3)$
$\displaystyle 133x-304y=23750 \qquad \ldots(4)$
$\displaystyle \text{Subtracting equation (3) from equation (4),}$
$\displaystyle 5x=3750$
$\displaystyle x=750$
$\displaystyle \text{Substituting }x=750\text{ in equation (2),}$
$\displaystyle 7(750)-16y=1250$
$\displaystyle 5250-16y=1250$
$\displaystyle 16y=4000$
$\displaystyle y=250$
$\displaystyle \therefore \text{The monthly incomes are Rs. }8(750)=\text{Rs. }6000\text{ and Rs. }7(750)=\text{Rs. }5250.$
$\displaystyle \\$

$\displaystyle \textbf{Question 53: }\text{Find the four angles of a cyclic quadrilateral }ABCD\text{ in which}$
$\displaystyle \angle A=(2x-1)^\circ,\ \angle B=(y+5)^\circ,\ \angle C=(2y+15)^\circ\text{ and }\angle D=(4x-7)^\circ.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In a cyclic quadrilateral, opposite angles are supplementary.}$
$\displaystyle \therefore \angle A+\angle C=180^\circ\text{ and }\angle B+\angle D=180^\circ.$
$\displaystyle (2x-1)+(2y+15)=180$
$\displaystyle 2x+2y=166$
$\displaystyle x+y=83 \qquad \ldots(1)$
$\displaystyle (y+5)+(4x-7)=180$
$\displaystyle 4x+y=182 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle 3x=99$
$\displaystyle x=33$
$\displaystyle \text{Substituting }x=33\text{ in equation (1),}$
$\displaystyle 33+y=83$
$\displaystyle y=50$
$\displaystyle \therefore \angle A=(2\times33-1)^\circ=65^\circ,$
$\displaystyle \angle B=(50+5)^\circ=55^\circ,$
$\displaystyle \angle C=(2\times50+15)^\circ=115^\circ,$
$\displaystyle \angle D=(4\times33-7)^\circ=125^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 54: }\text{In }\triangle ABC,\ \angle A=x^\circ,\ \angle B=3x^\circ\text{ and }\angle C=y^\circ.$
$\displaystyle \text{If }3y-5x=30,\text{ prove that the triangle is right-angled.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In a triangle, the sum of the angles is }180^\circ.$
$\displaystyle x+3x+y=180$
$\displaystyle 4x+y=180 \qquad \ldots(1)$
$\displaystyle -5x+3y=30 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }3,\text{ we get}$
$\displaystyle 12x+3y=540 \qquad \ldots(3)$
$\displaystyle \text{Subtracting equation (2) from equation (3),}$
$\displaystyle 17x=510$
$\displaystyle x=30$
$\displaystyle \text{Substituting }x=30\text{ in equation (1),}$
$\displaystyle 4(30)+y=180$
$\displaystyle y=60$
$\displaystyle \therefore \angle A=30^\circ,\ \angle B=3\times30^\circ=90^\circ,\ \angle C=60^\circ.$
$\displaystyle \therefore \triangle ABC\text{ is a right-angled triangle.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 55: }\text{In a rectangle, if the length is increased by }2\text{ units and the breadth is}$
$\displaystyle \text{decreased by }2\text{ units, the area is reduced by }28\text{ square units. If the length is reduced}$
$\displaystyle \text{by }1\text{ unit and the breadth is increased by }2\text{ units, the area increases by }33\text{ square units.}$
$\displaystyle \text{Find the area of the rectangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the rectangle be }x\text{ units and its breadth be }y\text{ units.}$
$\displaystyle \therefore \text{Area of the rectangle}=xy\text{ square units.}$
$\displaystyle \text{When the length is increased by }2\text{ units and the breadth is decreased by }2\text{ units,}$
$\displaystyle xy-28=(x+2)(y-2)$
$\displaystyle xy-28=xy-2x+2y-4$
$\displaystyle 2x-2y=24$
$\displaystyle x-y=12 \qquad \ldots(1)$
$\displaystyle \text{When the length is reduced by }1\text{ unit and the breadth is increased by }2\text{ units,}$
$\displaystyle xy+33=(x-1)(y+2)$
$\displaystyle xy+33=xy+2x-y-2$
$\displaystyle 2x-y=35 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle x=23$
$\displaystyle \text{Substituting }x=23\text{ in equation (1),}$
$\displaystyle 23-y=12$
$\displaystyle y=11$
$\displaystyle \therefore \text{Area of the rectangle}=xy=23\times11=253\text{ square units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 56: }\text{Half the perimeter of a garden is }36\text{ m. Its length is }4\text{ m more}$
$\displaystyle \text{than its width. Find the dimensions of the garden.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the width of the garden be }x\text{ m.}$
$\displaystyle \therefore \text{Its length}=(x+4)\text{ m.}$
$\displaystyle \text{Since half the perimeter of the garden is }36\text{ m,}$
$\displaystyle \frac{1}{2}\left[2x+2(x+4)\right]=36$
$\displaystyle x+x+4=36$
$\displaystyle 2x=32$
$\displaystyle x=16$
$\displaystyle \therefore \text{Width}=16\text{ m}$
$\displaystyle \text{and length}=16+4=20\text{ m.}$
$\displaystyle \therefore \text{The dimensions of the garden are }20\text{ m}\times16\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 57: }\text{If }A\text{ gives Rs. }30\text{ to }B,\text{ then }B\text{ will have twice the amount}$
$\displaystyle \text{left with }A.\text{ If }B\text{ gives Rs. }10\text{ to }A,\text{ then }A\text{ will have thrice the amount}$
$\displaystyle \text{left with }B.\text{ How much money does each have?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A\text{ have Rs. }x\text{ and }B\text{ have Rs. }y.$
$\displaystyle \text{If }A\text{ gives Rs. }30\text{ to }B,\text{ then}$
$\displaystyle y+30=2(x-30)$
$\displaystyle 2x-y=90 \qquad \ldots(1)$
$\displaystyle \text{If }B\text{ gives Rs. }10\text{ to }A,\text{ then}$
$\displaystyle x+10=3(y-10)$
$\displaystyle x-3y=-40 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }3,\text{ we get}$
$\displaystyle 6x-3y=270 \qquad \ldots(3)$
$\displaystyle \text{Subtracting equation (2) from equation (3),}$
$\displaystyle 5x=310$
$\displaystyle x=62$
$\displaystyle \text{Substituting }x=62\text{ in equation (1),}$
$\displaystyle 2(62)-y=90$
$\displaystyle y=34$
$\displaystyle \therefore A\text{ has Rs. }62\text{ and }B\text{ has Rs. }34.$
$\displaystyle \\$

$\displaystyle \textbf{Question 58: }A\text{ scored }40\text{ marks in a test, getting }3\text{ marks for each correct answer}$
$\displaystyle \text{and losing }1\text{ mark for each wrong answer. Had }4\text{ marks been awarded for each correct answer}$
$\displaystyle \text{and }2\text{ marks deducted for each wrong answer, }A\text{ would have scored }50\text{ marks.}$
$\displaystyle \text{How many questions were there in the test?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of questions answered correctly be }x\text{ and the number answered incorrectly be }y.$
$\displaystyle \text{According to the first marking scheme,}$
$\displaystyle 3x-y=40 \qquad \ldots(1)$
$\displaystyle \text{According to the second marking scheme,}$
$\displaystyle 4x-2y=50 \qquad \ldots(2)$
$\displaystyle \text{Multiplying equation (1) by }2,\text{ we get}$
$\displaystyle 6x-2y=80 \qquad \ldots(3)$
$\displaystyle \text{Subtracting equation (2) from equation (3),}$
$\displaystyle 2x=30$
$\displaystyle x=15$
$\displaystyle \text{Substituting }x=15\text{ in equation (1),}$
$\displaystyle 3(15)-y=40$
$\displaystyle 45-y=40$
$\displaystyle y=5$
$\displaystyle \therefore \text{Total number of questions}=x+y=15+5=20.$
$\displaystyle \therefore \text{There were }20\text{ questions in the test.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 59: }\text{Students of a class are arranged in rows. If there are }3\text{ extra students in}$
$\displaystyle \text{each row, there would be }1\text{ row less. If there are }3\text{ fewer students in each row, there would}$
$\displaystyle \text{be }2\text{ more rows. Find the number of students in the class.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of students in each row be }x\text{ and the number of rows be }y.$
$\displaystyle \therefore \text{Total number of students}=xy.$
$\displaystyle \text{If there are }3\text{ extra students in each row, there is }1\text{ row less.}$
$\displaystyle xy=(x+3)(y-1)$
$\displaystyle xy=xy-x+3y-3$
$\displaystyle x-3y=-3 \qquad \ldots(1)$
$\displaystyle \text{If there are }3\text{ fewer students in each row, there are }2\text{ more rows.}$
$\displaystyle xy=(x-3)(y+2)$
$\displaystyle xy=xy+2x-3y-6$
$\displaystyle 2x-3y=6 \qquad \ldots(2)$
$\displaystyle \text{Subtracting equation (1) from equation (2),}$
$\displaystyle x=9$
$\displaystyle \text{Substituting }x=9\text{ in equation (1),}$
$\displaystyle 9-3y=-3$
$\displaystyle 3y=12$
$\displaystyle y=4$
$\displaystyle \therefore \text{Total number of students}=xy=9\times4=36.$
$\displaystyle \therefore \text{There are }36\text{ students in the class.}$
$\displaystyle \\$