Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{A train covers a distance of }90\text{ km at a uniform speed. Had the speed been}$
$\displaystyle 15\text{ km/hour more, it would have taken }30\text{ minutes less for the journey. Find the original}$
$\displaystyle \text{speed of the train.}\hfill\text{[CBSE 2006C, 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }90\text{ km}=\frac{90}{x}\text{ hours.}$
$\displaystyle \text{Increased speed}=(x+15)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the increased speed}=\frac{90}{x+15}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{90}{x}-\frac{90}{x+15}=\frac{30}{60}$
$\displaystyle \Rightarrow \frac{90(x+15)-90x}{x(x+15)}=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1350}{x(x+15)}=\frac{1}{2}$
$\displaystyle \Rightarrow x(x+15)=2700$
$\displaystyle \Rightarrow x^2+15x-2700=0$
$\displaystyle \Rightarrow x^2+60x-45x-2700=0$
$\displaystyle \Rightarrow x(x+60)-45(x+60)=0$
$\displaystyle \Rightarrow (x+60)(x-45)=0$
$\displaystyle \Rightarrow x=-60\text{ or }x=45$
$\displaystyle \text{Since speed cannot be negative, }x=-60\text{ is rejected.}$
$\displaystyle \therefore \text{The original speed of the train is }45\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A train travels }360\text{ km at a uniform speed. If the speed had been }5\text{ km/hr}$
$\displaystyle \text{more, it would have taken }1\text{ hour less for the same journey. Find the speed of the train.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }360\text{ km}=\frac{360}{x}\text{ hours.}$
$\displaystyle \text{Increased speed}=(x+5)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the increased speed}=\frac{360}{x+5}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{360}{x}-\frac{360}{x+5}=1$
$\displaystyle \Rightarrow \frac{360(x+5)-360x}{x(x+5)}=1$
$\displaystyle \Rightarrow x(x+5)=1800$
$\displaystyle \Rightarrow x^2+5x-1800=0$
$\displaystyle \Rightarrow x^2+45x-40x-1800=0$
$\displaystyle \Rightarrow x(x+45)-40(x+45)=0$
$\displaystyle \Rightarrow (x+45)(x-40)=0$
$\displaystyle \Rightarrow x=-45\text{ or }x=40$
$\displaystyle \text{Since speed cannot be negative, }x=-45\text{ is rejected.}$
$\displaystyle \therefore \text{The speed of the train is }40\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{A train travels a distance of }480\text{ km at a uniform speed. If the speed had}$
$\displaystyle \text{been }8\text{ km/h less, then it would have taken }3\text{ hours more to cover the same distance.}$
$\displaystyle \text{Find the speed of the train.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }480\text{ km}=\frac{480}{x}\text{ hours.}$
$\displaystyle \text{Reduced speed}=(x-8)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the reduced speed}=\frac{480}{x-8}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{480}{x-8}-\frac{480}{x}=3$
$\displaystyle \Rightarrow \frac{480x-480(x-8)}{x(x-8)}=3$
$\displaystyle \Rightarrow \frac{3840}{x(x-8)}=3$
$\displaystyle \Rightarrow x(x-8)=1280$
$\displaystyle \Rightarrow x^2-8x-1280=0$
$\displaystyle \Rightarrow x^2-40x+32x-1280=0$
$\displaystyle \Rightarrow x(x-40)+32(x-40)=0$
$\displaystyle \Rightarrow (x-40)(x+32)=0$
$\displaystyle \Rightarrow x=40\text{ or }x=-32$
$\displaystyle \text{Since speed cannot be negative, }x=-32\text{ is rejected.}$
$\displaystyle \therefore \text{The speed of the train is }40\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{An express train takes }1\text{ hour less than a passenger train to travel}$
$\displaystyle 132\text{ km between Mysore and Bangalore, without taking into consideration the time they stop}$
$\displaystyle \text{at intermediate stations. If the average speed of the express train is }11\text{ km/hr more}$
$\displaystyle \text{than that of the passenger train, find the average speeds of the two trains.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the average speed of the passenger train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Average speed of the express train}=(x+11)\text{ km/hr.}$
$\displaystyle \text{Time taken by the passenger train}=\frac{132}{x}\text{ hours.}$
$\displaystyle \text{Time taken by the express train}=\frac{132}{x+11}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{132}{x}-\frac{132}{x+11}=1$
$\displaystyle \Rightarrow \frac{132(x+11)-132x}{x(x+11)}=1$
$\displaystyle \Rightarrow \frac{1452}{x(x+11)}=1$
$\displaystyle \Rightarrow x(x+11)=1452$
$\displaystyle \Rightarrow x^2+11x-1452=0$
$\displaystyle \Rightarrow x^2+44x-33x-1452=0$
$\displaystyle \Rightarrow x(x+44)-33(x+44)=0$
$\displaystyle \Rightarrow (x+44)(x-33)=0$
$\displaystyle \Rightarrow x=-44\text{ or }x=33$
$\displaystyle \text{Since speed cannot be negative, }x=-44\text{ is rejected.}$
$\displaystyle \therefore \text{Speed of the passenger train}=33\text{ km/hr.}$
$\displaystyle \therefore \text{Speed of the express train}=33+11=44\text{ km/hr.}$
$\displaystyle \therefore \text{The average speeds are }33\text{ km/hr and }44\text{ km/hr respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{A motor boat whose speed in still water is }18\text{ km/hr takes }1\text{ hour more}$
$\displaystyle \text{to go }24\text{ km upstream than to return downstream to the same spot. Find the speed}$
$\displaystyle \text{of the stream.}\hfill\text{[CBSE 2014, 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the stream be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Upstream speed}=(18-x)\text{ km/hr.}$
$\displaystyle \text{Downstream speed}=(18+x)\text{ km/hr.}$
$\displaystyle \text{Time taken to travel }24\text{ km upstream}=\frac{24}{18-x}\text{ hours.}$
$\displaystyle \text{Time taken to travel }24\text{ km downstream}=\frac{24}{18+x}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{24}{18-x}-\frac{24}{18+x}=1$
$\displaystyle \Rightarrow \frac{24(18+x)-24(18-x)}{(18-x)(18+x)}=1$
$\displaystyle \Rightarrow \frac{48x}{324-x^2}=1$
$\displaystyle \Rightarrow x^2+48x-324=0$
$\displaystyle \Rightarrow x^2+54x-6x-324=0$
$\displaystyle \Rightarrow x(x+54)-6(x+54)=0$
$\displaystyle \Rightarrow (x+54)(x-6)=0$
$\displaystyle \Rightarrow x=-54\text{ or }x=6$
$\displaystyle \text{Since the speed of the stream must be positive, }x=-54\text{ is rejected.}$
$\displaystyle \therefore \text{The speed of the stream is }6\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A motor boat whose speed in still water is }9\text{ km/hr goes }15\text{ km}$
$\displaystyle \text{downstream and comes back to the same spot, in a total time of }3\text{ hours }45\text{ minutes.}$
$\displaystyle \text{Find the speed of the stream.}\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the stream be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Downstream speed}=(9+x)\text{ km/hr.}$
$\displaystyle \text{Upstream speed}=(9-x)\text{ km/hr.}$
$\displaystyle 3\text{ hours }45\text{ minutes}=3+\frac{45}{60}=\frac{15}{4}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{15}{9+x}+\frac{15}{9-x}=\frac{15}{4}$
$\displaystyle \Rightarrow \frac{15(9-x)+15(9+x)}{(9+x)(9-x)}=\frac{15}{4}$
$\displaystyle \Rightarrow \frac{270}{81-x^2}=\frac{15}{4}$
$\displaystyle \Rightarrow 1080=1215-15x^2$
$\displaystyle \Rightarrow 15x^2=135$
$\displaystyle \Rightarrow x^2=9$
$\displaystyle \Rightarrow x=3\text{ or }x=-3$
$\displaystyle \text{Since the speed of the stream cannot be negative, }x=-3\text{ is rejected.}$
$\displaystyle \therefore \text{The speed of the stream is }3\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A train travelling at a uniform speed for }360\text{ km, would have taken }48\text{ minutes}$
$\displaystyle \text{less to travel the same distance if its speed were }5\text{ km/hr more. Find the original speed}$
$\displaystyle \text{of the train.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }360\text{ km}=\frac{360}{x}\text{ hours.}$
$\displaystyle \text{Increased speed}=(x+5)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the increased speed}=\frac{360}{x+5}\text{ hours.}$
$\displaystyle 48\text{ minutes}=\frac{48}{60}=\frac{4}{5}\text{ hour.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{360}{x}-\frac{360}{x+5}=\frac{4}{5}$
$\displaystyle \Rightarrow \frac{1800}{x(x+5)}=\frac{4}{5}$
$\displaystyle \Rightarrow 9000=4x(x+5)$
$\displaystyle \Rightarrow x^2+5x-2250=0$
$\displaystyle \Rightarrow x^2+50x-45x-2250=0$
$\displaystyle \Rightarrow x(x+50)-45(x+50)=0$
$\displaystyle \Rightarrow (x+50)(x-45)=0$
$\displaystyle \Rightarrow x=-50\text{ or }x=45$
$\displaystyle \text{Since speed cannot be negative, }x=-50\text{ is rejected.}$
$\displaystyle \therefore \text{The original speed of the train is }45\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{A plane left }40\text{ minutes late due to bad weather and in order to reach its destination,}$
$\displaystyle 1600\text{ km away in time, it had to increase its speed by }400\text{ km/hr from its usual speed.}$
$\displaystyle \text{Find the usual speed of the plane.}\hfill\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the usual speed of the plane be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }1600\text{ km at the usual speed}=\frac{1600}{x}\text{ hours.}$
$\displaystyle \text{Increased speed}=(x+400)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the increased speed}=\frac{1600}{x+400}\text{ hours.}$
$\displaystyle 40\text{ minutes}=\frac{40}{60}=\frac{2}{3}\text{ hour.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1600}{x}-\frac{1600}{x+400}=\frac{2}{3}$
$\displaystyle \Rightarrow \frac{1600(x+400)-1600x}{x(x+400)}=\frac{2}{3}$
$\displaystyle \Rightarrow \frac{640000}{x(x+400)}=\frac{2}{3}$
$\displaystyle \Rightarrow 1920000=2x(x+400)$
$\displaystyle \Rightarrow x^2+400x-960000=0$
$\displaystyle \Rightarrow x^2+1200x-800x-960000=0$
$\displaystyle \Rightarrow x(x+1200)-800(x+1200)=0$
$\displaystyle \Rightarrow (x+1200)(x-800)=0$
$\displaystyle \Rightarrow x=-1200\text{ or }x=800$
$\displaystyle \text{Since speed cannot be negative, }x=-1200\text{ is rejected.}$
$\displaystyle \therefore \text{The usual speed of the plane is }800\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{A train travels at a certain average speed for a distance }63\text{ km and then travels}$
$\displaystyle \text{a distance of }72\text{ km at an average speed of }6\text{ km/hr more than the original speed.}$
$\displaystyle \text{If it takes }3\text{ hours to complete total journey, what is its original average speed?}$
$\displaystyle \hfill\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original average speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to travel }63\text{ km}=\frac{63}{x}\text{ hours.}$
$\displaystyle \text{Increased average speed}=(x+6)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to travel }72\text{ km}=\frac{72}{x+6}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{63}{x}+\frac{72}{x+6}=3$
$\displaystyle \Rightarrow \frac{63(x+6)+72x}{x(x+6)}=3$
$\displaystyle \Rightarrow 63x+378+72x=3x(x+6)$
$\displaystyle \Rightarrow 135x+378=3x^2+18x$
$\displaystyle \Rightarrow 3x^2-117x-378=0$
$\displaystyle \Rightarrow x^2-39x-126=0$
$\displaystyle \Rightarrow x^2-42x+3x-126=0$
$\displaystyle \Rightarrow x(x-42)+3(x-42)=0$
$\displaystyle \Rightarrow (x-42)(x+3)=0$
$\displaystyle \Rightarrow x=42\text{ or }x=-3$
$\displaystyle \text{Since speed cannot be negative, }x=-3\text{ is rejected.}$
$\displaystyle \therefore \text{The original average speed of the train is }42\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{An aeroplane left }50\text{ minutes later than its scheduled time, and in order to reach}$
$\displaystyle \text{the destination, }1250\text{ km away, in time, it had to increase its speed by }250\text{ km/hr from}$
$\displaystyle \text{its usual speed. Find its usual speed.}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the usual speed of the aeroplane be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }1250\text{ km at the usual speed}=\frac{1250}{x}\text{ hours.}$
$\displaystyle \text{Increased speed}=(x+250)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the increased speed}=\frac{1250}{x+250}\text{ hours.}$
$\displaystyle 50\text{ minutes}=\frac{50}{60}=\frac{5}{6}\text{ hour.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1250}{x}-\frac{1250}{x+250}=\frac{5}{6}$
$\displaystyle \Rightarrow \frac{312500}{x(x+250)}=\frac{5}{6}$
$\displaystyle \Rightarrow 1875000=5x(x+250)$
$\displaystyle \Rightarrow x^2+250x-375000=0$
$\displaystyle \Rightarrow x^2+750x-500x-375000=0$
$\displaystyle \Rightarrow x(x+750)-500(x+750)=0$
$\displaystyle \Rightarrow (x+750)(x-500)=0$
$\displaystyle \Rightarrow x=-750\text{ or }x=500$
$\displaystyle \text{Since speed cannot be negative, }x=-750\text{ is rejected.}$
$\displaystyle \therefore \text{The usual speed of the aeroplane is }500\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{While boarding an aeroplane, a passenger got hurt. The pilot, showing promptness}$
$\displaystyle \text{and concern, made arrangements to hospitalise the injured and so the plane started late by}$
$\displaystyle 30\text{ minutes. To reach the destination, }1500\text{ km away in time, the pilot increased the speed}$
$\displaystyle \text{by }100\text{ km/hr. Find the original speed/hour of the plane.}\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the plane be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to cover }1500\text{ km at the original speed}=\frac{1500}{x}\text{ hours.}$
$\displaystyle \text{Increased speed}=(x+100)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken at the increased speed}=\frac{1500}{x+100}\text{ hours.}$
$\displaystyle 30\text{ minutes}=\frac{30}{60}=\frac{1}{2}\text{ hour.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{1500}{x}-\frac{1500}{x+100}=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{150000}{x(x+100)}=\frac{1}{2}$
$\displaystyle \Rightarrow x(x+100)=300000$
$\displaystyle \Rightarrow x^2+100x-300000=0$
$\displaystyle \Rightarrow x^2+600x-500x-300000=0$
$\displaystyle \Rightarrow x(x+600)-500(x+600)=0$
$\displaystyle \Rightarrow (x+600)(x-500)=0$
$\displaystyle \Rightarrow x=-600\text{ or }x=500$
$\displaystyle \text{Since speed cannot be negative, }x=-600\text{ is rejected.}$
$\displaystyle \therefore \text{The original speed of the plane is }500\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{A car moves a distance of }2592\text{ km with uniform speed. The}$
$\displaystyle \text{number of hours taken for the journey is one-half the number representing the}$
$\displaystyle \text{speed in km/hour. Find the time taken to cover the distance.}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed of the car be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken for the journey}=\frac{x}{2}\text{ hours.}$
$\displaystyle \text{We know that, Distance}=\text{Speed}\times\text{Time}.$
$\displaystyle \therefore x\times\frac{x}{2}=2592$
$\displaystyle \Rightarrow x^2=5184$
$\displaystyle \Rightarrow x=\pm72$
$\displaystyle \text{Since speed cannot be negative, }x=-72\text{ is rejected.}$
$\displaystyle \therefore \text{Speed of the car}=72\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken}=\frac{72}{2}=36\text{ hours.}$
$\displaystyle \therefore \text{The time taken to cover the distance is }36\text{ hours.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{A train travels at a certain average speed for a distance of }54\text{ km and then}$
$\displaystyle \text{travels a distance of }63\text{ km at an average speed of }6\text{ km/hr more than the first speed.}$
$\displaystyle \text{If it takes 3 hours to complete the journey, what was its first average speed? } \hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the first average speed of the train be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to travel }54\text{ km}=\frac{54}{x}\text{ hours.}$
$\displaystyle \text{Second average speed}=(x+6)\text{ km/hr.}$
$\displaystyle \therefore \text{Time taken to travel }63\text{ km}=\frac{63}{x+6}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{54}{x}+\frac{63}{x+6}=3$
$\displaystyle \Rightarrow 54(x+6)+63x=3x(x+6)$
$\displaystyle \Rightarrow 54x+324+63x=3x^2+18x$
$\displaystyle \Rightarrow 3x^2-99x-324=0$
$\displaystyle \Rightarrow x^2-33x-108=0$
$\displaystyle \Rightarrow x^2-36x+3x-108=0$
$\displaystyle \Rightarrow x(x-36)+3(x-36)=0$
$\displaystyle \Rightarrow (x-36)(x+3)=0$
$\displaystyle \Rightarrow x=36\text{ or }x=-3$
$\displaystyle \text{Since speed cannot be negative, }x=-3\text{ is rejected.}$
$\displaystyle \therefore \text{The first average speed of the train is }36\text{ km/hr.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{The time taken by a person to travel an upward distance of }150\text{ km was}$
$\displaystyle 2\frac{1}{2}\text{ hours more than the time taken in the downward return journey. If he returned at a}$
$\displaystyle \text{speed of }10\text{ km/hr more than the speed while going up, find the speeds in each direction.}$
$\displaystyle \hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the speed while going upward be }x\text{ km/hr.}$
$\displaystyle \therefore \text{Speed during the downward journey}=(x+10)\text{ km/hr.}$
$\displaystyle \text{Time taken for the upward journey}=\frac{150}{x}\text{ hours.}$
$\displaystyle \text{Time taken for the downward journey}=\frac{150}{x+10}\text{ hours.}$
$\displaystyle 2\frac{1}{2}\text{ hours}=\frac{5}{2}\text{ hours.}$
$\displaystyle \text{According to the question,}$
$\displaystyle \frac{150}{x}-\frac{150}{x+10}=\frac{5}{2}$
$\displaystyle \Rightarrow \frac{150(x+10)-150x}{x(x+10)}=\frac{5}{2}$
$\displaystyle \Rightarrow \frac{1500}{x(x+10)}=\frac{5}{2}$
$\displaystyle \Rightarrow x(x+10)=600$
$\displaystyle \Rightarrow x^2+10x-600=0$
$\displaystyle \Rightarrow x^2+30x-20x-600=0$
$\displaystyle \Rightarrow x(x+30)-20(x+30)=0$
$\displaystyle \Rightarrow (x+30)(x-20)=0$
$\displaystyle \Rightarrow x=-30\text{ or }x=20$
$\displaystyle \text{Since speed cannot be negative, }x=-30\text{ is rejected.}$
$\displaystyle \therefore \text{Speed while going upward}=20\text{ km/hr.}$
$\displaystyle \therefore \text{Speed during the downward journey}=20+10=30\text{ km/hr.}$
$\displaystyle \\$