\displaystyle \textbf{Question 1: }\text{Two people are 16 km apart on a straight road. They}
\displaystyle \text{start walking at the same time. If they walk towards each other with}
\displaystyle \text{different speeds, they will meet in 2 hours. Had they walked in the same}
\displaystyle \text{direction with same speeds as before, they would have met in 8 hours.}
\displaystyle \text{Find their walking speeds.}\hfill\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let their walking speeds be }x\text{ km/h and }y\text{ km/h, where }x>y.
\displaystyle \text{When they walk towards each other, their relative speed}=x+y.
\displaystyle \text{Distance covered in 2 hours}=16\text{ km.}
\displaystyle \therefore 2(x+y)=16
\displaystyle \Rightarrow x+y=8\qquad\ldots\text{(i)}
\displaystyle \text{When they walk in the same direction, their relative speed}=x-y.
\displaystyle \text{They meet after 8 hours, covering the initial gap of }16\text{ km.}
\displaystyle \therefore 8(x-y)=16
\displaystyle \Rightarrow x-y=2\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=10
\displaystyle \Rightarrow x=5
\displaystyle \text{Substituting }x=5\text{ in equation (i),}
\displaystyle 5+y=8
\displaystyle \Rightarrow y=3
\displaystyle \therefore \text{Their walking speeds are }5\text{ km/h and }3\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The boat goes 30 km upstream and 44 km}
\displaystyle \text{downstream in 10 hours. In 13 hours, it can go 40 km upstream and}
\displaystyle \text{55 km downstream. Determine the speed of stream and that of the boat}
\displaystyle \text{in still water.}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/h.}
\displaystyle \text{Let the speed of the stream be }y\text{ km/h.}
\displaystyle \therefore \text{Speed upstream}=(x-y)\text{ km/h}
\displaystyle \text{and speed downstream}=(x+y)\text{ km/h}.
\displaystyle \frac{30}{x-y}+\frac{44}{x+y}=10\qquad\ldots\text{(i)}
\displaystyle \frac{40}{x-y}+\frac{55}{x+y}=13\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x-y}=u\text{ and }\frac{1}{x+y}=v.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle 30u+44v=10\qquad\ldots\text{(iii)}
\displaystyle 40u+55v=13\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iii) by }4,
\displaystyle 120u+176v=40\qquad\ldots\text{(v)}
\displaystyle \text{Multiplying equation (iv) by }3,
\displaystyle 120u+165v=39\qquad\ldots\text{(vi)}
\displaystyle \text{Subtracting equation (vi) from equation (v),}
\displaystyle 11v=1
\displaystyle \Rightarrow v=\frac{1}{11}
\displaystyle \text{Substituting }v=\frac{1}{11}\text{ in equation (iii),}
\displaystyle 30u+44\left(\frac{1}{11}\right)=10
\displaystyle \Rightarrow 30u=6
\displaystyle \Rightarrow u=\frac{1}{5}
\displaystyle \therefore \frac{1}{x-y}=\frac{1}{5}\Rightarrow x-y=5\qquad\ldots\text{(vii)}
\displaystyle \frac{1}{x+y}=\frac{1}{11}\Rightarrow x+y=11\qquad\ldots\text{(viii)}
\displaystyle \text{Adding equations (vii) and (viii),}
\displaystyle 2x=16
\displaystyle \Rightarrow x=8
\displaystyle \text{Substituting }x=8\text{ in equation (viii),}
\displaystyle 8+y=11
\displaystyle \Rightarrow y=3
\displaystyle \therefore \text{Speed of the stream}=3\text{ km/h and speed of the boat in still water}=8\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A person rowing at the rate of 5 km/h in still}
\displaystyle \text{water, takes thrice as much time in going 40 km upstream as in going}
\displaystyle \text{40 km downstream. Find the speed of the stream.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the stream be }x\text{ km/h.}
\displaystyle \therefore \text{Speed upstream}=(5-x)\text{ km/h}
\displaystyle \text{and speed downstream}=(5+x)\text{ km/h}.
\displaystyle \text{Time taken to go 40 km upstream}=\frac{40}{5-x}\text{ hours.}
\displaystyle \text{Time taken to go 40 km downstream}=\frac{40}{5+x}\text{ hours.}
\displaystyle \text{Upstream time is three times the downstream time.}
\displaystyle \therefore \frac{40}{5-x}=3\left(\frac{40}{5+x}\right)
\displaystyle \Rightarrow \frac{1}{5-x}=\frac{3}{5+x}
\displaystyle \Rightarrow 5+x=3(5-x)
\displaystyle \Rightarrow 5+x=15-3x
\displaystyle \Rightarrow 4x=10
\displaystyle \Rightarrow x=\frac{5}{2}=2.5
\displaystyle \therefore \text{The speed of the stream is }2.5\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A man travels 600 km partly by train and}
\displaystyle \text{partly by car. If he covers 400 km by train and the rest by car, it takes}
\displaystyle \text{him 6 hours and 30 minutes. But, if he travels 200 km by train and the}
\displaystyle \text{rest by car, he takes half an hour longer. Find the speed of the train}
\displaystyle \text{and that of the car.}\hfill\text{[CBSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the train be }x\text{ km/h and that of the car be }y\text{ km/h.}
\displaystyle \text{When 400 km is covered by train, 200 km is covered by car.}
\displaystyle \text{Total time}=6\text{ h }30\text{ min}=\frac{13}{2}\text{ h}.
\displaystyle \therefore \frac{400}{x}+\frac{200}{y}=\frac{13}{2}\qquad\ldots\text{(i)}
\displaystyle \text{When 200 km is covered by train, 400 km is covered by car.}
\displaystyle \text{The time taken is half an hour longer, i.e. }7\text{ hours.}
\displaystyle \therefore \frac{200}{x}+\frac{400}{y}=7\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Equation (i) becomes}
\displaystyle 800u+400v=13
\displaystyle \Rightarrow 2u+v=\frac{13}{400}\qquad\ldots\text{(iii)}
\displaystyle \text{Equation (ii) becomes}
\displaystyle 200u+400v=7
\displaystyle \Rightarrow u+2v=\frac{7}{200}\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iii) by }2,
\displaystyle 4u+2v=\frac{13}{200}\qquad\ldots\text{(v)}
\displaystyle \text{Subtracting equation (iv) from equation (v),}
\displaystyle 3u=\frac{6}{200}
\displaystyle \Rightarrow u=\frac{1}{100}
\displaystyle \text{Substituting }u=\frac{1}{100}\text{ in equation (iv),}
\displaystyle \frac{1}{100}+2v=\frac{7}{200}
\displaystyle \Rightarrow 2v=\frac{5}{200}
\displaystyle \Rightarrow v=\frac{1}{80}
\displaystyle \therefore \frac{1}{x}=\frac{1}{100}\Rightarrow x=100
\displaystyle \text{and }\frac{1}{y}=\frac{1}{80}\Rightarrow y=80.
\displaystyle \therefore \text{The speed of the train is }100\text{ km/h and that of the car is }80\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Ritu can row downstream 20 km in 2 hours, and}
\displaystyle \text{upstream 4 km in 2 hours. Find her speed of rowing in still water and}
\displaystyle \text{the speed of the current.} 
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of Ritu in still water be }x\text{ km/h.}
\displaystyle \text{Let the speed of the current be }y\text{ km/h.}
\displaystyle \text{Downstream speed}=x+y
\displaystyle =\frac{20}{2}=10\text{ km/h}.
\displaystyle \therefore x+y=10\qquad\ldots\text{(i)}
\displaystyle \text{Upstream speed}=x-y
\displaystyle =\frac{4}{2}=2\text{ km/h}.
\displaystyle \therefore x-y=2\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=12
\displaystyle \Rightarrow x=6
\displaystyle \text{Substituting }x=6\text{ in equation (i),}
\displaystyle 6+y=10
\displaystyle \Rightarrow y=4
\displaystyle \therefore \text{Ritu's speed in still water is }6\text{ km/h and the speed}
\displaystyle \text{of the current is }4\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A motor boat can travel 30 km upstream and}
\displaystyle \text{28 km downstream in 7 hours. It can travel 21 km upstream and return}
\displaystyle \text{in 5 hours. Find the speed of the boat in still water and the speed of}
\displaystyle \text{the stream.} 
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/h.}
\displaystyle \text{Let the speed of the stream be }y\text{ km/h.}
\displaystyle \therefore \text{Speed upstream}=(x-y)\text{ km/h}
\displaystyle \text{and speed downstream}=(x+y)\text{ km/h}.
\displaystyle \frac{30}{x-y}+\frac{28}{x+y}=7\qquad\ldots\text{(i)}
\displaystyle \frac{21}{x-y}+\frac{21}{x+y}=5\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x-y}=u\text{ and }\frac{1}{x+y}=v.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle 30u+28v=7\qquad\ldots\text{(iii)}
\displaystyle 21u+21v=5
\displaystyle \Rightarrow u+v=\frac{5}{21}\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iv) by }28,
\displaystyle 28u+28v=\frac{20}{3}\qquad\ldots\text{(v)}
\displaystyle \text{Subtracting equation (v) from equation (iii),}
\displaystyle 2u=\frac{1}{3}
\displaystyle \Rightarrow u=\frac{1}{6}
\displaystyle \text{Substituting }u=\frac{1}{6}\text{ in equation (iv),}
\displaystyle \frac{1}{6}+v=\frac{5}{21}
\displaystyle \Rightarrow v=\frac{1}{14}
\displaystyle \therefore \frac{1}{x-y}=\frac{1}{6}\Rightarrow x-y=6\qquad\ldots\text{(vi)}
\displaystyle \frac{1}{x+y}=\frac{1}{14}\Rightarrow x+y=14\qquad\ldots\text{(vii)}
\displaystyle \text{Adding equations (vi) and (vii),}
\displaystyle 2x=20
\displaystyle \Rightarrow x=10
\displaystyle \text{Substituting }x=10\text{ in equation (vii),}
\displaystyle 10+y=14
\displaystyle \Rightarrow y=4
\displaystyle \therefore \text{Speed of the boat in still water is }10\text{ km/h and speed}
\displaystyle \text{of the stream is }4\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Abdul travelled 300 km by train and 200 km by taxi,}
\displaystyle \text{it took him 5 hours 30 minutes. But if he travels 260 km by train and}
\displaystyle \text{240 km by taxi, he takes 6 minutes longer. Find the speed of the train}
\displaystyle \text{and that of the taxi.}\hfill\text{[CBSE 2006C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the train be }x\text{ km/h and that of the taxi be }y\text{ km/h.}
\displaystyle 5\text{ h }30\text{ min}=5+\frac{30}{60}=\frac{11}{2}\text{ h}.
\displaystyle \therefore \frac{300}{x}+\frac{200}{y}=\frac{11}{2}\qquad\ldots\text{(i)}
\displaystyle 5\text{ h }36\text{ min}=5+\frac{36}{60}=\frac{28}{5}\text{ h}.
\displaystyle \therefore \frac{260}{x}+\frac{240}{y}=\frac{28}{5}\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Equation (i) becomes}
\displaystyle 600u+400v=11
\displaystyle \Rightarrow 3u+2v=\frac{11}{200}\qquad\ldots\text{(iii)}
\displaystyle \text{Equation (ii) becomes}
\displaystyle 1300u+1200v=28
\displaystyle \Rightarrow 13u+12v=\frac{7}{25}\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iii) by }6,
\displaystyle 18u+12v=\frac{33}{100}\qquad\ldots\text{(v)}
\displaystyle \text{Subtracting equation (iv) from equation (v),}
\displaystyle 5u=\frac{33}{100}-\frac{28}{100}=\frac{1}{20}
\displaystyle \Rightarrow u=\frac{1}{100}
\displaystyle \text{Substituting }u=\frac{1}{100}\text{ in equation (iii),}
\displaystyle \frac{3}{100}+2v=\frac{11}{200}
\displaystyle \Rightarrow 2v=\frac{1}{40}
\displaystyle \Rightarrow v=\frac{1}{80}
\displaystyle \therefore \frac{1}{x}=\frac{1}{100}\Rightarrow x=100
\displaystyle \text{and }\frac{1}{y}=\frac{1}{80}\Rightarrow y=80.
\displaystyle \therefore \text{Speed of the train is }100\text{ km/h and speed of the taxi is }80\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Places }A\text{ and }B\text{ are 100 km apart on a highway.}
\displaystyle \text{One car starts from }A\text{ and another from }B\text{ at the same time. If the}
\displaystyle \text{cars travel in the same direction at different speeds, they meet in 5 hours.}
\displaystyle \text{If they travel towards each other, they meet in 1 hour. What are the}
\displaystyle \text{speeds of the two cars?}\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speeds of the two cars be }x\text{ km/h and }y\text{ km/h, where }x>y.
\displaystyle \text{When they travel in the same direction, their relative speed}=x-y.
\displaystyle \text{They meet in 5 hours after covering the initial gap of }100\text{ km.}
\displaystyle \therefore 5(x-y)=100
\displaystyle \Rightarrow x-y=20\qquad\ldots\text{(i)}
\displaystyle \text{When they travel towards each other, their relative speed}=x+y.
\displaystyle \text{They meet in 1 hour after covering a total distance of }100\text{ km.}
\displaystyle \therefore x+y=100\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2x=120
\displaystyle \Rightarrow x=60
\displaystyle \text{Substituting }x=60\text{ in equation (ii),}
\displaystyle 60+y=100
\displaystyle \Rightarrow y=40
\displaystyle \therefore \text{The speeds of the two cars are }60\text{ km/h and }40\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A train covered a certain distance at a uniform}
\displaystyle \text{speed. If the train could have been 10 km/h faster, it would have taken}
\displaystyle \text{2 hours less than the scheduled time. And, if the train were slower by}
\displaystyle \text{10 km/h, it would have taken 3 hours more than the scheduled time. Find}
\displaystyle \text{the distance covered by the train.}\hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original speed of the train be }x\text{ km/h.}
\displaystyle \text{Let the scheduled time of the journey be }y\text{ hours.}
\displaystyle \therefore \text{Distance covered}=xy\text{ km.}
\displaystyle \text{If the speed is increased by }10\text{ km/h, time is reduced by }2\text{ hours.}
\displaystyle \therefore xy=(x+10)(y-2)
\displaystyle \Rightarrow xy=xy-2x+10y-20
\displaystyle \Rightarrow x-5y=-10\qquad\ldots\text{(i)}
\displaystyle \text{If the speed is reduced by }10\text{ km/h, time increases by }3\text{ hours.}
\displaystyle \therefore xy=(x-10)(y+3)
\displaystyle \Rightarrow xy=xy+3x-10y-30
\displaystyle \Rightarrow 3x-10y=30\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 2x-10y=-20\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (ii),}
\displaystyle x=50
\displaystyle \text{Substituting }x=50\text{ in equation (i),}
\displaystyle 50-5y=-10
\displaystyle \Rightarrow 5y=60
\displaystyle \Rightarrow y=12
\displaystyle \therefore \text{Distance covered}=xy=50\times12=600\text{ km.}
\displaystyle \therefore \text{The distance covered by the train is }600\text{ km.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{While covering a distance of 30 km, Ajeet}
\displaystyle \text{takes 2 hours more than Amit. If Ajeet doubles his speed, he would take}
\displaystyle \text{1 hour less than Amit. Find their speeds of walking.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speeds of Ajeet and Amit be }x\text{ km/h and }y\text{ km/h, respectively.}
\displaystyle \text{Time taken by Ajeet to cover 30 km}=\frac{30}{x}\text{ hours.}
\displaystyle \text{Time taken by Amit to cover 30 km}=\frac{30}{y}\text{ hours.}
\displaystyle \text{Ajeet takes 2 hours more than Amit.}
\displaystyle \therefore \frac{30}{x}-\frac{30}{y}=2\qquad\ldots\text{(i)}
\displaystyle \text{If Ajeet doubles his speed, his speed becomes }2x\text{ km/h.}
\displaystyle \text{He then takes 1 hour less than Amit.}
\displaystyle \therefore \frac{30}{2x}=\frac{30}{y}-1
\displaystyle \Rightarrow \frac{15}{x}-\frac{30}{y}=-1\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle 30u-30v=2\qquad\ldots\text{(iii)}
\displaystyle 15u-30v=-1\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle 15u=3
\displaystyle \Rightarrow u=\frac{1}{5}
\displaystyle \text{Substituting }u=\frac{1}{5}\text{ in equation (iii),}
\displaystyle 30\left(\frac{1}{5}\right)-30v=2
\displaystyle \Rightarrow 6-30v=2
\displaystyle \Rightarrow v=\frac{2}{15}
\displaystyle \therefore \frac{1}{x}=\frac{1}{5}\Rightarrow x=5
\displaystyle \text{and }\frac{1}{y}=\frac{2}{15}\Rightarrow y=\frac{15}{2}=7.5.
\displaystyle \therefore \text{Ajeet's speed is }5\text{ km/h and Amit's speed is }7.5\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A man walks a certain distance with certain speed.}
\displaystyle \text{If he walks }\frac{1}{2}\text{ km an hour faster, he takes 1 hour less. But, if he}
\displaystyle \text{walks 1 km an hour slower, he takes 3 more hours. Find the distance}
\displaystyle \text{covered by the man and his original rate of walking.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original speed of the man be }x\text{ km/h.}
\displaystyle \text{Let the original time taken be }y\text{ hours.}
\displaystyle \therefore \text{Distance covered}=xy\text{ km.}
\displaystyle \text{If his speed is increased by }\frac{1}{2}\text{ km/h, time is reduced by 1 hour.}
\displaystyle \therefore xy=\left(x+\frac{1}{2}\right)(y-1)
\displaystyle \Rightarrow xy=xy-x+\frac{y}{2}-\frac{1}{2}
\displaystyle \Rightarrow 2x-y=-1\qquad\ldots\text{(i)}
\displaystyle \text{If his speed is reduced by 1 km/h, time increases by 3 hours.}
\displaystyle \therefore xy=(x-1)(y+3)
\displaystyle \Rightarrow xy=xy+3x-y-3
\displaystyle \Rightarrow 3x-y=3\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle x=4
\displaystyle \text{Substituting }x=4\text{ in equation (i),}
\displaystyle 8-y=-1
\displaystyle \Rightarrow y=9
\displaystyle \therefore \text{Distance covered}=xy=4\times9=36\text{ km.}
\displaystyle \therefore \text{The distance is }36\text{ km and the original speed is }4\text{ km/h.}
\displaystyle \\

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