Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: } \text{The speed of a slow train is }x\text{ km/hr and that of}$
$\displaystyle \text{a fast train is }(x+25)\text{ km/hr. Find the time taken by each train}$
$\displaystyle \text{to cover }300\text{ km. If the slow train takes }2\text{ hours more than the fast train,}$
$\displaystyle \text{calculate the speed of the fast train.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Time taken by slow train}=\frac{300}{x}\text{ hours}$
$\displaystyle \text{Time taken by fast train}=\frac{300}{x+25}\text{ hours}$
$\displaystyle \text{Given, }\frac{300}{x}=\frac{300}{x+25}+2$
$\displaystyle \frac{300}{x}-\frac{300}{x+25}=2$
$\displaystyle \frac{300(x+25)-300x}{x(x+25)}=2$
$\displaystyle \frac{7500}{x(x+25)}=2$
$\displaystyle 2x(x+25)=7500$
$\displaystyle 2x^2+50x-7500=0$
$\displaystyle x^2+25x-3750=0$
$\displaystyle (x-50)(x+75)=0$
$\displaystyle \Rightarrow x=50\text{ or }x=-75$
$\displaystyle \text{Since speed cannot be negative, }x\neq -75.$
$\displaystyle \therefore x=50$
$\displaystyle \text{Speed of fast train}=x+25=75\text{ km/hr}$
$\displaystyle \therefore \text{The speed of the fast train is }75\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 2: } \text{If the speed of a car is increased by }10\text{ km/hr,}$
$\displaystyle \text{it takes }18\text{ minutes less to cover a distance of }36\text{ km.}$
$\displaystyle \text{Find the speed of the car.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the car be }x\text{ km/hr.}$
$\displaystyle \text{Increased speed}=(x+10)\text{ km/hr}$
$\displaystyle 18\text{ minutes}=\frac{18}{60}\text{ hour}=\frac{3}{10}\text{ hour}$
$\displaystyle \therefore \frac{36}{x}=\frac{36}{x+10}+\frac{3}{10}$
$\displaystyle \frac{36}{x}-\frac{36}{x+10}=\frac{3}{10}$
$\displaystyle \frac{36(x+10)-36x}{x(x+10)}=\frac{3}{10}$
$\displaystyle \frac{360}{x(x+10)}=\frac{3}{10}$
$\displaystyle 3600=3x(x+10)$
$\displaystyle 3x^2+30x-3600=0$
$\displaystyle x^2+10x-1200=0$
$\displaystyle (x-30)(x+40)=0$
$\displaystyle \Rightarrow x=30\text{ or }x=-40$
$\displaystyle \text{Since speed cannot be negative, }x\neq -40.$
$\displaystyle \therefore x=30$
$\displaystyle \therefore \text{The speed of the car is }30\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 3: } \text{If the speed of an airplane is reduced by }40\text{ km/hr,}$
$\displaystyle \text{it takes }20\text{ minutes more to cover }1200\text{ km. Find the speed of the airplane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the airplane be }x\text{ km/hr.}$
$\displaystyle \text{Reduced speed}=(x-40)\text{ km/hr}$
$\displaystyle 20\text{ minutes}=\frac{20}{60}\text{ hour}=\frac{1}{3}\text{ hour}$
$\displaystyle \therefore \frac{1200}{x}+\frac{1}{3}=\frac{1200}{x-40}$
$\displaystyle \frac{1200}{x-40}-\frac{1200}{x}=\frac{1}{3}$
$\displaystyle \frac{1200x-1200(x-40)}{x(x-40)}=\frac{1}{3}$
$\displaystyle \frac{48000}{x(x-40)}=\frac{1}{3}$
$\displaystyle x(x-40)=144000$
$\displaystyle x^2-40x-144000=0$
$\displaystyle (x-400)(x+360)=0$
$\displaystyle \Rightarrow x=400\text{ or }x=-360$
$\displaystyle \text{Since speed cannot be negative, }x\neq -360.$
$\displaystyle \therefore x=400$
$\displaystyle \therefore \text{The speed of the airplane is }400\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 4: } \text{A car covers a distance of }400\text{ km at a certain speed.}$
$\displaystyle \text{Had the speed been }12\text{ km/hr more, the time taken for the journey}$
$\displaystyle \text{would have been }1\text{ hour }40\text{ minutes less. Find the original speed}$
$\displaystyle \text{of the car.}\hfill \text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the car be }x\text{ km/hr.}$
$\displaystyle \text{Increased speed}=(x+12)\text{ km/hr}$
$\displaystyle 1\text{ hour }40\text{ minutes}=1+\frac{40}{60}=\frac{5}{3}\text{ hours}$
$\displaystyle \therefore \frac{400}{x}=\frac{400}{x+12}+\frac{5}{3}$
$\displaystyle \frac{400}{x}-\frac{400}{x+12}=\frac{5}{3}$
$\displaystyle \frac{400(x+12)-400x}{x(x+12)}=\frac{5}{3}$
$\displaystyle \frac{4800}{x(x+12)}=\frac{5}{3}$
$\displaystyle 5x(x+12)=14400$
$\displaystyle x^2+12x-2880=0$
$\displaystyle (x-48)(x+60)=0$
$\displaystyle \Rightarrow x=48\text{ or }x=-60$
$\displaystyle \text{Since speed cannot be negative, }x\neq -60.$
$\displaystyle \therefore x=48$
$\displaystyle \therefore \text{The original speed of the car is }48\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 5: } \text{A girl goes to her friend's house, which is at a distance}$
$\displaystyle \text{of }12\text{ km. She covers half the distance at a speed of }x\text{ km/hr}$
$\displaystyle \text{and the remaining distance at a speed of }(x+2)\text{ km/hr. If she takes}$
$\displaystyle 2\text{ hours }30\text{ minutes to cover the whole distance, find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle 2\text{ hours }30\text{ minutes}=\frac{5}{2}\text{ hours}$
$\displaystyle \therefore \frac{6}{x}+\frac{6}{x+2}=\frac{5}{2}$
$\displaystyle \frac{6(x+2)+6x}{x(x+2)}=\frac{5}{2}$
$\displaystyle \frac{12x+12}{x(x+2)}=\frac{5}{2}$
$\displaystyle 2(12x+12)=5x(x+2)$
$\displaystyle 24x+24=5x^2+10x$
$\displaystyle 5x^2-14x-24=0$
$\displaystyle 5x^2-20x+6x-24=0$
$\displaystyle 5x(x-4)+6(x-4)=0$
$\displaystyle (x-4)(5x+6)=0$
$\displaystyle \Rightarrow x=4\text{ or }x=-\frac{6}{5}$
$\displaystyle \text{Since speed cannot be negative, }x\neq -\frac{6}{5}.$
$\displaystyle \therefore x=4$
$\displaystyle \therefore \text{The value of }x\text{ is }4.$
$\\$

$\displaystyle \textbf{Question 6: } \text{A car made a run of }390\text{ km in }x\text{ hours.}$
$\displaystyle \text{If the speed had been }4\text{ km/hr more, it would have taken }2\text{ hours less}$
$\displaystyle \text{for the journey. Find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Original time}=x\text{ hours}$
$\displaystyle \therefore \text{Original speed}=\frac{390}{x}\text{ km/hr}$
$\displaystyle \text{New time}=(x-2)\text{ hours}$
$\displaystyle \therefore \text{New speed}=\frac{390}{x-2}\text{ km/hr}$
$\displaystyle \text{Given, }\frac{390}{x-2}=\frac{390}{x}+4$
$\displaystyle \frac{390}{x-2}-\frac{390}{x}=4$
$\displaystyle \frac{390x-390(x-2)}{x(x-2)}=4$
$\displaystyle \frac{780}{x(x-2)}=4$
$\displaystyle 4x(x-2)=780$
$\displaystyle x^2-2x-195=0$
$\displaystyle (x-15)(x+13)=0$
$\displaystyle \Rightarrow x=15\text{ or }x=-13$
$\displaystyle \text{Since time cannot be negative, }x\neq -13.$
$\displaystyle \therefore x=15$
$\displaystyle \therefore \text{The value of }x\text{ is }15\text{ hours}.$
$\\$

$\displaystyle \textbf{Question 7: } \text{A goods train leaves a station at }6\text{ pm followed by}$
$\displaystyle \text{an express train which leaves at }8\text{ pm and travels }20\text{ km/hr faster}$
$\displaystyle \text{than the goods train. The express train arrives at a station, }1040\text{ km away,}$
$\displaystyle 36\text{ minutes before the goods train. Assuming that the speed of both trains}$
$\displaystyle \text{remains constant, calculate their speeds.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the goods train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Speed of the express train}=(x+20)\text{ km/hr}$
$\displaystyle \text{The express train leaves }2\text{ hours later and arrives }36\text{ minutes earlier.}$
$\displaystyle \therefore \text{Goods train takes }2\text{ hours }36\text{ minutes more than the express train.}$
$\displaystyle 2\text{ hours }36\text{ minutes}=2+\frac{36}{60}=\frac{13}{5}\text{ hours}$
$\displaystyle \therefore \frac{1040}{x}=\frac{1040}{x+20}+\frac{13}{5}$
$\displaystyle \frac{1040}{x}-\frac{1040}{x+20}=\frac{13}{5}$
$\displaystyle \frac{1040(x+20)-1040x}{x(x+20)}=\frac{13}{5}$
$\displaystyle \frac{20800}{x(x+20)}=\frac{13}{5}$
$\displaystyle 104000=13x(x+20)$
$\displaystyle x^2+20x-8000=0$
$\displaystyle (x-80)(x+100)=0$
$\displaystyle \Rightarrow x=80\text{ or }x=-100$
$\displaystyle \text{Since speed cannot be negative, }x\neq -100.$
$\displaystyle \therefore x=80$
$\displaystyle \text{Speed of express train}=x+20=100\text{ km/hr}$
$\displaystyle \therefore \text{The speeds are }80\text{ km/hr and }100\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 8: } \text{A man bought an article for Rs. }x\text{ and sold it}$
$\displaystyle \text{for Rs. }16.\text{ If his loss was }x\%\text{, find the cost price of the article.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Cost price}=\text{Rs. }x$
$\displaystyle \text{Selling price}=\text{Rs. }16$
$\displaystyle \text{Loss}=x\%\text{ of Rs. }x=\text{Rs. }\frac{x^2}{100}$
$\displaystyle \therefore x-\frac{x^2}{100}=16$
$\displaystyle 100x-x^2=1600$
$\displaystyle x^2-100x+1600=0$
$\displaystyle (x-80)(x-20)=0$
$\displaystyle \Rightarrow x=80\text{ or }x=20$
$\displaystyle \therefore \text{The cost price of the article is Rs. }80\text{ or Rs. }20.$
$\\$

$\displaystyle \textbf{Question 9: } \text{A trader bought an article for Rs. }x\text{ and sold it}$
$\displaystyle \text{for Rs. }52,\text{ thereby making a profit of }(x-10)\%\text{ on his outlay.}$
$\displaystyle \text{Calculate the cost price.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Cost price}=\text{Rs. }x$
$\displaystyle \text{Selling price}=\text{Rs. }52$
$\displaystyle \text{Profit}=(x-10)\%\text{ of Rs. }x$
$\displaystyle =\text{Rs. }\frac{x(x-10)}{100}$
$\displaystyle \therefore 52-x=\frac{x(x-10)}{100}$
$\displaystyle 5200-100x=x^2-10x$
$\displaystyle x^2+90x-5200=0$
$\displaystyle (x-40)(x+130)=0$
$\displaystyle \Rightarrow x=40\text{ or }x=-130$
$\displaystyle \text{Since cost price cannot be negative, }x\neq -130.$
$\displaystyle \therefore x=40$
$\displaystyle \therefore \text{The cost price is Rs. }40.$
$\\$

$\displaystyle \textbf{Question 10: } \text{By selling a chair for Rs. }75,\text{ a person gained}$
$\displaystyle \text{as much percent as its cost. Calculate the cost of the chair.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the cost price of the chair be Rs. }x.$
$\displaystyle \text{Selling price}=\text{Rs. }75$
$\displaystyle \text{Gain}=x\%\text{ of Rs. }x$
$\displaystyle =\text{Rs. }\frac{x^2}{100}$
$\displaystyle \therefore 75-x=\frac{x^2}{100}$
$\displaystyle 7500-100x=x^2$
$\displaystyle x^2+100x-7500=0$
$\displaystyle (x-50)(x+150)=0$
$\displaystyle \Rightarrow x=50\text{ or }x=-150$
$\displaystyle \text{Since cost price cannot be negative, }x\neq -150.$
$\displaystyle \therefore x=50$
$\displaystyle \therefore \text{The cost price of the chair is Rs. }50.$
$\\$