Quiz Level-2

Hint

Let the speeds of two trains be \(7 x\) and \(8 x \mathrm{~km} / \mathrm{hr}\).
Then, \(8 x=\frac{400}{4}=100 \Rightarrow x=\left(\frac{100}{8}\right)=12.5\).
\(\therefore\) Speed of first train \(=(7 \times 12.5) \mathrm{km} / \mathrm{hr}=87.5 \mathrm{~km} / \mathrm{hr}\).

Hint

\(
\begin{aligned}
& \text { Total distance travelled }=\left[\left(50 \times 2 \frac{1}{2}\right)+\left(70 \times 1 \frac{1}{2}\right)\right] \\
& \text { miles }=(125+105) \text { miles }=230 \text { miles. }
\end{aligned}
\)

Hint

The number of gaps between 21 telephone posts \(=20\).
Distance travelled in 1 minute \(=(50 \times 20) \mathrm{m}=1000 \mathrm{~m}=1 \mathrm{~km}\) \(\therefore\) Speed \(=60 \mathrm{~km} / \mathrm{hr}\)

Hint

\(
\text { Distance }=\left(1100 \times \frac{11}{5}\right) \text { feet }=2420 \text { feet. }
\)

Hint

\(
\text { Time taken to cover } 600 \mathrm{~km}=\left(\frac{600}{100}\right) \mathrm{hrs}=6 \mathrm{hrs} \text {. }
\)
Number of stoppages \(=\frac{600}{75}-1=7\)
Total time of stoppage \(=(3 \times 7) \mathrm{min}=21 \mathrm{~min}\).
Hence, total time taken \(=6 \mathrm{hrs} 21 \mathrm{~min}\).

Hint

Let the distance covered by the cyclist be \(x\) and the time taken be \(y\). Then,
\(
\text { Required ratio }=\frac{\frac{1}{2} x}{2 y}: \frac{x}{y}=\frac{1}{4}: 1=1: 4 \text {. }
\)

Hint

Distance covered in first 2 hours \(=(70 \times 2) \mathrm{km}=140 \mathrm{~km}\). Distance covered in next 2 hours \(=(80 \times 2) \mathrm{km}=160 \mathrm{~km}\). Remaining distance \(=345-(140+160)=45 \mathrm{~km}\) Speed in the fifth hour \(=90 \mathrm{~km} / \mathrm{hr}\).
Time taken to cover \(45 \mathrm{~km}=\left(\frac{45}{90}\right) \mathrm{hr}=\frac{1}{2} \mathrm{hr}\).
\(\therefore\) Total time taken \(=\left(2+2+\frac{1}{2}\right)=4 \frac{1}{2} \mathrm{hrs}\).

Hint

This is a question on uniform retardation (as it is given that the car slows down at a fixed rate)
If \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the uniform acceleration (or retardation), \(t\) is the time and \(s\) is the distance covered, we have :
\(
v=u+a t \quad \text { and } \quad s=u t+\frac{1}{2} a t^2
\)
Here, \(v=0, u=24 \mathrm{~m} / \mathrm{s}, a=-3 \mathrm{~m} / \mathrm{s}^2\)
\(
\begin{aligned}
& \therefore \quad 0=24-3 t \Rightarrow 3 t=24 \Rightarrow t=8 . \\
& \text { And, } s=\left[24 \times 8+\frac{1}{2} \times(-3) \times 8^2\right] \mathrm{m}=(192-96) \mathrm{m}=96 \mathrm{~m} .
\end{aligned}
\)

Hint

Total time taken
\(
\begin{aligned}
& =\left[\frac{x}{y}+\frac{x}{\left(\frac{3 y}{5}\right)}+\frac{x}{\left(\frac{2 y}{5}\right)}\right] \text { hours } \\
& =\left(\frac{x}{y}+\frac{5 x}{3 y}+\frac{5 x}{2 y}\right) \text { hours }=\left(\frac{6 x+10 x+15 x}{6 y}\right) \text { hours } \\
& =\left(\frac{31 x}{6 y}\right) \text { hours. } \\
& \therefore \frac{31 x}{6 y}=1 \Leftrightarrow \frac{x}{y}=\frac{6}{31} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Average speed }=\frac{\text { Total distance covered }}{\text { Total time taken }} \\
& \\
& \quad=\left(\frac{4 \times 9 \times 400}{88+96+89+87}\right) \mathrm{m} / \mathrm{min}=\left(\frac{14400}{360}\right) \mathrm{m} / \mathrm{min} \\
& =40 \mathrm{~m} / \mathrm{min} .
\end{aligned}
\)

Hint

Time taken by the express train to cover \(600 \mathrm{~km}\)
\(
=\left(\frac{600}{100}\right) \text { hrs }=6 \text { hrs. }
\)
Number of stoppages \(=(600 \div 75)-1=7\).
Duration of stoppage \(=(3 \times 7) \mathrm{min}=21 \mathrm{~min}\).
Total time taken \(=6\) hrs \(21 \mathrm{~min}\).
Total time taken by local train to cover \(50 \mathrm{~km}\) (with stoppages) \(=1 \mathrm{hr} 2 \mathrm{~min}\).
So, the local train covers \((50 \times 6)=300 \mathrm{~km}\) in \(6 \mathrm{hr} 12 \mathrm{~min}\). In remaining \(9 \mathrm{~min}\), it covers \(\left(\frac{50}{60} \times 9\right) \mathrm{km}=7.5 \mathrm{~km}\). \(\therefore\) Required distance \(=(300+7.5) \mathrm{km}=307.5 \mathrm{~km}\)

Hint

Let the required number of hours be \(n\).
Clearly, the car covers \(40 \mathrm{~km}\) in first hour, \(45 \mathrm{~km}\) in the second hour, \(50 \mathrm{~km}\) in the third hour, and so on.
Thus, we have : \(40+45+50+\ldots \ldots .\). upto \(n\) terms \(=385\).
This is an AP. with first term \(a=40\), common difference \(d=5\).
\(
\therefore S_n=\frac{n}{2}[2 \times 40+(n-1) 5]
\)
So, \(\frac{n}{2}[80+5(n-1)]=385 \Leftrightarrow 80 n+5 n^2-5 n=770 \Leftrightarrow[latex] [latex]5 n^2+75 n-770=0\)
\(\Leftrightarrow n^2+15 n-154=0 \Leftrightarrow n^2+22 n-7 n-154=0\)
\(\Leftrightarrow n(n+22)-7(n+22)=0 \Leftrightarrow(n+22)(n-7)\) \(=0 \Leftrightarrow n=7\).
Hence, the required number of hours \(=7\).

Hint

Total distance travelled in 12 hours \(=(35+37+39+\ldots\). upto 12 terms).
This is an AP. with first term, \(a=35\), number of terms, \(n=12\), common difference, \(d=2\).
\(\therefore\) Required distance
\(
=\frac{12}{2}[2 \times 35+(12-1) \times 2]=6(70+22)=552 \mathrm{~km} \text {. }
\)

Hint

Time taken to cover a distance is inversely proportional to the speed.
Let the required time be \(x \mathrm{~min}\).
Then, \(50: 55:: x: 44 \Leftrightarrow 55 x=50 \times 44\)
\(
\Leftrightarrow x=\left(\frac{50 \times 44}{55}\right) \min =40 \mathrm{~min} .
\)

Hint

Ratio of speeds \(=2: 3: 4\).
\(
\therefore \text { Ratio of times taken }=\frac{1}{2}: \frac{1}{3}: \frac{1}{4}=6: 4: 3 \text {. }
\)

Hint

Ratio of speeds \(=3: 4\).
Ratio of times taken \(=\frac{1}{3}: \frac{1}{4}=4: 3\).
Let \(A\) and \(B\) take \(4 x\) and \(3 x\) minutes respectively to reach a destination.
Then, \(4 x-3 x=20 \Leftrightarrow x=20\).
\(\therefore\) Time taken by \(A=4 x=(4 \times 20) \mathrm{min}=80 \mathrm{~min}=1 \frac{1}{3} \mathrm{hr}\).

Hint

Let the speed of steam engine train be \(x\).
Then, speed of electric train \(=125 \%\) of \(x=\frac{5 x}{4}\).
Time taken bysteam engine \(=4 \mathrm{hr} 25 \mathrm{~min}=4 \frac{25}{60} \mathrm{hr}=4 \frac{5}{12} \mathrm{hr}\)
Let the time taken by electric train be \(t\) hours.
Then, \(x: \frac{5 x}{4}:: t: 4 \frac{5}{12} \Leftrightarrow 1: \frac{5}{4}:: t: \frac{53}{12}\)
\(
\Leftrightarrow \frac{5}{4} t=\frac{53}{12} \Leftrightarrow t=\left(\frac{53}{12} \times \frac{4}{5}\right)=\frac{53}{15}=3 \frac{8}{15} \mathrm{hr} .
\)

Hint

Suppose \(B\) takes \(x\) hours to walk \(d \mathrm{~km}\)
Then, \(A\) takes \((x+2)\) hours to walk \(d \mathrm{~km}\)
A’s speed \(=\left(\frac{d}{x+2}\right) \mathrm{km} / \mathrm{hr}\) and B’s speed \(=\left(\frac{d}{x}\right) \mathrm{km} / \mathrm{hr}\).
A’s new speed \(=\left(\frac{2 d}{x+2}\right) \mathrm{km} / \mathrm{hr}\).
\(\therefore \frac{d}{\left(\frac{d}{x}\right)}-\frac{d}{\left(\frac{2 d}{x+2}\right)}=1 \Leftrightarrow x-\left(\frac{x+2}{2}\right)=1\)
\(\Leftrightarrow x-2=2 \Leftrightarrow x=4\).

Hint

\(
\begin{aligned}
& \text { Speed }=\left(10 \times \frac{60}{12}\right) \mathrm{km} / \mathrm{hr}=50 \mathrm{~km} / \mathrm{hr} . \\
& \text { New speed }=(50-5) \mathrm{km} / \mathrm{hr}=45 \mathrm{~km} / \mathrm{hr} . \\
& \begin{aligned}
\therefore \text { Time taken } & =\left(\frac{10}{45}\right) \mathrm{hr}=\left(\frac{2}{9} \times 60\right) \mathrm{min}=13 \frac{1}{3} \mathrm{~min} \\
& =13 \mathrm{~min} 20 \mathrm{sec} .
\end{aligned}
\end{aligned}
\)

Hint

Distance covered in \(2 \mathrm{hrs} 15 \mathrm{~min}\), i.e.,
\(
2 \frac{1}{4} \text { hrs }=\left(80 \times \frac{9}{4}\right) \text { hrs }=180 \text { hrs }
\)
Time taken to cover remaining distance
\(
=\left(\frac{350-180}{60}\right) \mathrm{hrs}=\frac{17}{6} \mathrm{hrs}=2 \frac{5}{6} \mathrm{hrs}=2 \mathrm{hrs} 50 \mathrm{~min} \text {. }
\)
Total time taken \(=(2 \mathrm{hrs} 15 \mathrm{~min}+2 \mathrm{hrs} 50 \mathrm{~min})=5 \mathrm{hrs} 5 \mathrm{~min}\).
So, Anna reached city A at \(10.25 \mathrm{am}\)

Hint

\(
\begin{aligned}
& \text { Time required }=(2 \mathrm{hrs} 30 \mathrm{~min}-50 \mathrm{~min}) \\
& \qquad=1 \mathrm{hr} 40 \mathrm{~min}=1 \frac{2}{3} \mathrm{hrs} . \\
& \therefore \text { Required speed }=\left(50 \times \frac{3}{5}\right) \mathrm{km} / \mathrm{hr}=30 \mathrm{~km} / \mathrm{hr} . \\
& \qquad \text { Original speed }=\left(50 \times \frac{2}{5}\right) \mathrm{km} / \mathrm{hr}=20 \mathrm{~km} / \mathrm{hr} .
\end{aligned}
\)

Hint

Remaining distance \(=3 \mathrm{~km}\)
and Remaining time \(=\left(\frac{1}{3} \times 45\right) \mathrm{min}=15 \mathrm{~min}=\frac{1}{4}\) hour.
\(\therefore\) Required speed \(=(3 \times 4) \mathrm{km} / \mathrm{hr}=12 \mathrm{~km} / \mathrm{hr}\)

Hint

Let the total journey be \(x \mathrm{~km}\)
Then
\(
\begin{aligned}
\frac{3 x}{5}+\frac{7 x}{20}+6.5 & =x \Leftrightarrow 12 x+7 x+20 \times 6.5 \\
& =20 x \Leftrightarrow x=130 \mathrm{~km}
\end{aligned}
\)

Hint

\(
\text { Time taken to travel } \begin{aligned}
45 \mathrm{~km} & =\left(\frac{45}{20}\right) \mathrm{hr}=\frac{9}{4} \mathrm{hr}=2 \frac{1}{4} \mathrm{hr} \\
& =2 \mathrm{hr} 15 \mathrm{~min} .
\end{aligned}
\)

Remaining time \(=(3 \mathrm{hr}-2 \mathrm{hr} 15 \mathrm{~min})=45 \mathrm{~min}\).
Hence, required speed \(=\left(\frac{45}{45}\right) \mathrm{km} / \mathrm{min}=1 \mathrm{~km} / \mathrm{min}\).

Hint

Time taken to travel \(25 \mathrm{~km}=\left(\frac{25}{40}\right) \mathrm{hr}=\frac{5}{8} \mathrm{hr}\)
Remaining time \(=\left(1-\frac{5}{8}\right) h r=\frac{3}{8} h r\).
\(\therefore\) Required speed \(=\left(21 \times \frac{8}{3}\right) \mathrm{km} / \mathrm{hr}=56 \mathrm{~km} / \mathrm{hr}\).

Hint

Total distance \(=(70+50)[latex] miles [latex]=120\) miles.
Average speed \(=50\) miles \(/\) hour.
Required time of journey \(=\left(\frac{120}{50}\right) \mathrm{hr}=\frac{12}{5}=\mathrm{hr} 2 \frac{2}{5} \mathrm{hr}\) \(=2 \mathrm{hr} 24 \mathrm{~min}\).
Time taken to cover 50 miles
\(
=1 \frac{1}{2} \mathrm{hr}=1 \mathrm{hr} 30 \mathrm{~min} \text {. }
\)
\(\therefore\) Remaining time \(=(2 \mathrm{hr} 24 \mathrm{~min}-1 \mathrm{hr} 30 \mathrm{~min})=54 \mathrm{~min}\).

Hint

Let the total distance be \(x \mathrm{~km}\). Then,
\(
\begin{aligned}
& \frac{\frac{1}{2} x}{21}+\frac{\frac{1}{2} x}{24}=10 \Rightarrow \frac{x}{21}+\frac{x}{24}=20 \\
\Rightarrow \quad & 15 x=168 \times 20 \Rightarrow x=\left(\frac{168 \times 20}{15}\right)=224 \mathrm{~km}
\end{aligned}
\)

Hint

Let the distance travelled by the train be \(x \mathrm{~km}\) Then, distance travelled by bus \(=(285-x) \mathrm{km}\).
\(
\begin{aligned}
& \therefore \quad\left(\frac{285-x}{40}\right)+\frac{x}{55}=6 \Leftrightarrow \frac{(285-x)}{8}+\frac{x}{11}=30 \\
& \Leftrightarrow \quad \frac{11(285-x)+8 x}{88}=30 \\
& \Leftrightarrow \quad 3135-11 x+8 x=2640 \Leftrightarrow 3 x=495 \Leftrightarrow x=165 .
\end{aligned}
\)
Hence, distance travelled by train \(=165 \mathrm{~km}\)

Hint

Let A’s speed \(=x \mathrm{~km} / \mathrm{hr}\). Then, B’s speed \(=(7-x) \mathrm{km} / \mathrm{hr}\).
\(
\begin{aligned}
& \text { So, } \frac{24}{x}+\frac{24}{(7-x)}=14 \Leftrightarrow 24(7-x)+24 x=14 x(7-x) \\
& \Leftrightarrow 14 x^2-98 x+168=0 \\
& \Leftrightarrow x^2-7 x+12=0 \\
& \Leftrightarrow \quad(x-3)(x-4)=0 \\
& \Leftrightarrow \quad x=3 \text { or } x=4 .
\end{aligned}
\)
Since \(A\) is faster than \(B\), so A’s speed \(=4 \mathrm{~km} / \mathrm{hr}\) and B’s speed \(=3 \mathrm{~km} / \mathrm{hr}\).

Hint

Speed on return trip \(=150 \%\) of \(40=60 \mathrm{kmph}\).
\(\therefore\) Average speed
\(
=\left(\frac{2 \times 40 \times 60}{40+60}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{4800}{100}\right) \mathrm{km} / \mathrm{hr}=48 \mathrm{~km} / \mathrm{hr} .
\)

Hint

\(
\begin{aligned}
\text { Average speed } & =\left(\frac{2 \times 500 \times 700}{500+700}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{1750}{3}\right) \mathrm{km} / \mathrm{hr} \\
& =583 \frac{1}{3} \mathrm{~km} / \mathrm{hr} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
\text { Average speed } & =\left(\frac{2 \times 56 \times 53}{56+53}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{5936}{109}\right) \mathrm{km} / \mathrm{hr} \\
& =54.45 \mathrm{~km} / \mathrm{hr} \approx 54.5 \mathrm{~km} / \mathrm{hr} .
\end{aligned}
\)

Hint

Average speed
\(
=\frac{\left(2 \times \frac{5}{2} \times \frac{13}{4}\right)}{\left(\frac{5}{2}+\frac{13}{4}\right)} \mathrm{km} / \mathrm{hr}=\left(\frac{65}{4} \times \frac{4}{23}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{65}{23}\right) \mathrm{km} / \mathrm{hr} \text {. }
\)
Total time taken \(=4 \mathrm{hr} 36 \mathrm{~min}=4 \frac{36}{60} \mathrm{hr}=4 \frac{3}{5} \mathrm{hr}=\frac{23}{5} \mathrm{hr}\).
Total distance covered uphill and downhill
\(
=\left(\frac{65}{23} \times \frac{23}{5}\right) \mathrm{km}=13 \mathrm{~km} \text {. }
\)
\(\therefore\) Distance walked uphill \(=\left(\frac{13}{2}\right) \mathrm{km}=6 \frac{1}{2} \mathrm{~km}\).

Hint

Time taken to cover \(150 \mathrm{~km}\) in going trip \(=3 \mathrm{hr} 20 \mathrm{~min}=3 \frac{20}{60} \mathrm{hr}=3 \frac{1}{3} \mathrm{hr}=\frac{10}{3} \mathrm{hr}\).
Speed in going trip \(=\left(150 \times \frac{3}{10}\right) \mathrm{km} / \mathrm{hr}=45 \mathrm{~km} / \mathrm{hr}\).
Time taken to cover \(150 \mathrm{~km}\) in return trip \(=4 \mathrm{hr} 10 \mathrm{~min}\).
\(
=4 \frac{1}{6} \mathrm{hr}=\frac{25}{6} \mathrm{hr} \text {. }
\)
Speed in return trip \(=\left(150 \times \frac{6}{25}\right) \mathrm{km} / \mathrm{hr}=36 \mathrm{~km} / \mathrm{hr}\).
\(\therefore\) Average speed
\(=\left(\frac{2 \times 45 \times 36}{45+36}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{2 \times 45 \times 36}{81}\right) \mathrm{km} / \mathrm{hr}=40 \mathrm{~km} / \mathrm{hr}\).
Required difference \(=(45-40) \mathrm{km} / \mathrm{hr}=5 \mathrm{~km} / \mathrm{hr}\).

Hint

Let the speed in return journey be \(x \mathrm{~km} / \mathrm{hr}\). Then, speed in onward journey \(=\frac{125}{100} x=\left(\frac{5}{4} x\right) \mathrm{km} / \mathrm{hr}\). Average speed \(=\left(\frac{2 \times \frac{5}{4} x \times x}{\frac{5}{4} x+x}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{10 x}{9}\right) \mathrm{km} / \mathrm{hr}\). \(\therefore \quad\left(800 \times \frac{9}{10 x}\right)=16 \Leftrightarrow x=\left(\frac{800 \times 9}{16 \times 10}\right)=45\).
So, speed in onward journey
\(
=\left(\frac{5}{4} \times 45\right) \mathrm{km} / \mathrm{hr}=56.25 \mathrm{~km} / \mathrm{hr} .
\)

Hint

\(
\text { Time taken }=5 \mathrm{hrs} 25 \mathrm{~min}=\frac{65}{12} \text { hrs. }
\)
Let the required distance be \(x \mathrm{~km}\).
Then, \(\frac{x}{10}+\frac{x}{1}=\frac{65}{12} \Leftrightarrow 11 x=\frac{650}{12} \Leftrightarrow x=\frac{325}{66}=4 \frac{61}{66} \mathrm{~km}\).

Hint

\(
\begin{aligned}
& \text { Total distance travelled }=(50 \times 1+48 \times 2+52 \times 3) \mathrm{km} \\
& =302 \mathrm{~km} \\
& \text { Total time taken }=6 \mathrm{hrs} \text {. } \\
& \therefore \text { Mean speed }=\left(\frac{302}{6}\right) \mathrm{km} / \mathrm{hr}=50 \frac{1}{3} \mathrm{~km} / \mathrm{hr} . \\
&
\end{aligned}
\)

Hint

Total distance travelled \(=(39+25) \mathrm{km}=64 \mathrm{~km}\)
Total time taken \(=(45+35) \mathrm{min}=80 \mathrm{~min}=\frac{4}{3} \mathrm{hr}\).
\(\therefore\) Average speed \(=\left(64 \times \frac{3}{4}\right) \mathrm{km} / \mathrm{hr}=48 \mathrm{~km} / \mathrm{hr}\).

Hint

\(
\begin{aligned}
& \text { Total distance travelled }=\left(30 \times \frac{12}{60}+45 \times \frac{8}{60}\right) \mathrm{km}=12 \mathrm{~km} \text {. } \\
& \text { Total time taken }=(12+8) \mathrm{min}=20 \mathrm{~min}=\frac{1}{3} \mathrm{hr} \\
& \therefore \text { Average speed }=(12 \times 3) \mathrm{km} / \mathrm{hr}=36 \mathrm{~km} / \mathrm{hr}
\end{aligned}
\)

Hint

Total time taken \(=\left(\frac{160}{64}+\frac{160}{80}\right) \mathrm{hrs}=\frac{9}{2}\) hrs.
\(
\therefore \text { Average speed }=\left(320 \times \frac{2}{9}\right) \mathrm{km} / \mathrm{hr}=71.11 \mathrm{~km} / \mathrm{hr} .
\)

Hint

Total distance travelled \(=(10+12) \mathrm{km} / \mathrm{hr}=22 \mathrm{~km} / \mathrm{hr}\).
Total time taken \(=\left(\frac{10}{12}+\frac{12}{10}\right) \mathrm{hrs}=\frac{61}{30}\) hrs.
\(\therefore\) Average speed \(=\left(22 \times \frac{30}{61}\right) \mathrm{km} / \mathrm{hr}=10.8 \mathrm{~km} / \mathrm{hr}\).

Hint

\(
\begin{aligned}
& \text { Total distance travelled }=(600+800+500+100) \mathrm{km} \\
& =2000 \mathrm{~km} \text {. } \\
& \text { Total time taken }=\left(\frac{600}{80}+\frac{800}{40}+\frac{500}{400}+\frac{100}{50}\right) \mathrm{hrs}=\frac{123}{4} \mathrm{hrs} . \\
& \therefore \text { Average speed }=\left(2000 \times \frac{4}{123}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{8000}{123}\right) \mathrm{km} / \mathrm{hr} \\
& =65 \frac{5}{123} \mathrm{~km} / \mathrm{hr} \text {. } \\
&
\end{aligned}
\)

Hint

Total distance travelled \(=(24+36) \mathrm{km}=60 \mathrm{~km}\) Total time taken \(=\left(\frac{24}{16}+\frac{36}{15}\right) \mathrm{hr}=\left(\frac{3}{2}+\frac{12}{5}\right) \mathrm{hr}=\frac{39}{10} \mathrm{hrs}\).
\(
\begin{aligned}
\therefore \text { Average speed } & =\left(60 \times \frac{10}{39}\right) \mathrm{km} / \mathrm{hr}=\left(\frac{200}{13}\right) \mathrm{km} / \mathrm{hr} \\
& =15.38 \mathrm{~km} / \mathrm{hr} .
\end{aligned}
\)

Hint

Let each distance be equal to \(d\). Then, Total distance travelled \(=3 d\).
\(
\begin{aligned}
& \text { Total time taken }=\left(\frac{d}{x}+\frac{d}{y}+\frac{d}{z}\right) \mathrm{hr}=\frac{d(x y+y z+z x)}{x y z} \mathrm{hr} \text {. } \\
& \therefore \text { Average speed }=\left[3 d \times \frac{x y z}{d(x y+y z+z x)}\right] \mathrm{km} / \mathrm{hr} \\
& =\frac{3 x y z}{(x y+y z+z x)} \mathrm{km} / \mathrm{hr} \text {. } \\
&
\end{aligned}
\)

Hint

Let the whole distance travelled be \(x \mathrm{~km}\) and the average speed of the car for the whole journey be \(y \mathrm{~km} / \mathrm{hr}\).
Then, \(\frac{(x / 3)}{10}+\frac{(x / 3)}{20}+\frac{(x / 3)}{60}=\frac{x}{y}\)
\(\)
\Leftrightarrow \frac{x}{30}+\frac{x}{60}+\frac{x}{180}=\frac{x}{y} \Leftrightarrow \frac{1}{18} y=1 \Leftrightarrow y=18 \mathrm{~km} / \mathrm{hr} .
/latex]

Hint

\(
\begin{aligned}
& x \times \frac{15}{60}+2 x \times \frac{20}{60}+x \times \frac{10}{60}=39 \\
& \Rightarrow \frac{x}{4}+\frac{2 x}{3}+\frac{x}{6}=39 \Rightarrow 3 x+8 x+2 x=468 \Rightarrow x=36 .
\end{aligned}
\)

Hint

Let speed of jogging be \(x \mathrm{~km} / \mathrm{hr}\).
Total time taken \(=\left(\frac{9}{6}\right.\) hrs +1.5 hrs \()=3\) hrs.
Total distance covered \(=(9+1.5 x) \mathrm{km}\).
\(
\begin{aligned}
& \therefore \quad \frac{9+1.5 x}{3}=9 \Leftrightarrow 9+1.5 x=27 \Leftrightarrow \frac{3}{2} x=18 \\
& \Leftrightarrow \quad x=\left(18 \times \frac{2}{3}\right)=12 \mathrm{kmph} .
\end{aligned}
\)

Hint

Ratio of speeds while going and returning \(=40: 30=4: 3\). Ratio of times taken while going and returning \(=3: 4\). Time taken while going \(=\left(\frac{3}{7} \times 8\right) \mathrm{hr}=\frac{24}{7} \mathrm{hr}\). Time taken while returning \(=\left(\frac{4}{7} \times 8\right) \mathrm{hr}=\frac{32}{7} \mathrm{hr}\).
\(
\begin{aligned}
\therefore \text { Required distance } & =\left(40 \times \frac{24}{7}\right) \text { miles }=\frac{960}{7} \text { miles } \\
& =137.14 \text { miles } .
\end{aligned}
\)

Hint

Time taken \(=1 \mathrm{hr} 40 \mathrm{~min} 48 \mathrm{sec}\)
\(
=1 \mathrm{hr} 40 \frac{4}{5} \mathrm{~min}=1 \frac{51}{75} \mathrm{hrs}=\frac{126}{75} \mathrm{hrs} .
\)
Let the actual speed be \(x \mathrm{~km} / \mathrm{hr}\).
Then, \(\frac{5}{7} x \times \frac{126}{75}=42\) or \(x=\left(\frac{42 \times 7 \times 75}{5 \times 126}\right)=35 \mathrm{~km} / \mathrm{hr}\).

Hint

New speed \(=\frac{7}{11}\) of usual speed.
\(\therefore \quad\) New time \(=\frac{11}{7}\) of usual time.
So, \(\frac{11}{7}\) of usual time \(=22 \mathrm{hrs}\)
\(\Rightarrow \quad\) usual time \(=\left(\frac{22 \times 7}{11}\right)=14 \mathrm{hrs}\).
Hence, time saved \(=(22-14)=8 \mathrm{hrs}\).

Hint

Let the speed be \(x \mathrm{~km} / \mathrm{hr}\).
Then, \(30 x-30 \times \frac{14}{15} x=10 \Leftrightarrow 2 x=10 \Leftrightarrow x=5 \mathrm{~km} / \mathrm{hr}\).

Hint

New speed \(=\frac{6}{7}\) of usual speed.
New time \(=\frac{7}{6}\) of usual time
\(
\begin{aligned}
& \therefore \quad\left(\frac{7}{6} \text { of usual time }\right)-\text { (usual time) }=\frac{1}{5} \mathrm{hr} . \\
& \Rightarrow \quad \frac{1}{6} \text { of usual time }=\frac{1}{5} \mathrm{hr} \\
& \Rightarrow \quad \text { usual time }=\frac{6}{5} \mathrm{hr}=1 \mathrm{hr} 12 \mathrm{~min} .
\end{aligned}
\)

Hint

Let the average speed on the onward journey be \(x \mathrm{~km} / \mathrm{hr}\). Then, average speed on return journey
\(
\begin{aligned}
& =(80 \% \text { of } x) \mathrm{km} / \mathrm{hr}=\left(\frac{4 x}{5}\right) \mathrm{km} / \mathrm{hr} . \\
& \therefore \quad \frac{500}{x}+\frac{500}{\left(\frac{4 x}{5}\right)}+\frac{1}{2}=23 \Rightarrow \frac{500}{x}+\frac{625}{x}=\frac{45}{2} \\
& \Rightarrow \quad \frac{1125}{x}=\frac{45}{2} \Rightarrow x=\frac{1125 \times 2}{45}=50 .
\end{aligned}
\)
Hence, speed on return journey
\(
=\left(\frac{4 x}{5}\right)=\left(\frac{4 \times 50}{5}\right) \mathrm{km} / \mathrm{hr}=40 \mathrm{~km} / \mathrm{hr} .
\)

Hint

Let the normal speed of the train be \(x \mathrm{~km} / \mathrm{hr}\).
\(
\text { Then, new speed }=\left(112 \frac{1}{2} \% \text { of } x\right) \mathrm{km} / \mathrm{hr}
\)
\(
=\left(\frac{225}{2} \times \frac{1}{100} \times x\right) \mathrm{km} / \mathrm{hr}=\left(\frac{9}{8} x\right) \mathrm{km} / \mathrm{hr} .
\)
Let the distance covered be \(d \mathrm{~km}\).
\(
\begin{aligned}
& \text { Then, } \frac{d}{x}-\frac{d}{\left(\frac{9 x}{8}\right)}=\frac{20}{60}=\frac{1}{3} \Rightarrow \frac{d}{x}-\frac{8 d}{9 x}=\frac{1}{3} \\
& \Rightarrow \frac{d}{9 x}=\frac{1}{3} \Rightarrow d=3 x .
\end{aligned}
\)
\(\therefore\) Actual time taken \(=\frac{d}{x}=\frac{3 x}{x}=3\) hours \(=180 \mathrm{~min}\).

Hint

Let the distance be \(x \mathrm{~km}\)
Difference in timings \(=12 \mathrm{~min}=\frac{12}{60} \mathrm{hr}=\frac{1}{5} \mathrm{hr}\).
\(
\begin{aligned}
\therefore \quad \frac{x}{\left(\frac{5}{2}\right)}-\frac{x}{\left(\frac{7}{2}\right)}=\frac{1}{5} \Leftrightarrow \frac{2 x}{5}- & \frac{2 x}{7}=\frac{1}{5} \\
& \Leftrightarrow 14 x-10 x=7 \Leftrightarrow x=1 \frac{3}{4} \mathrm{~km} .
\end{aligned}
\)

Hint

Difference between timings \(=24 \mathrm{~min}=\frac{24}{60} \mathrm{hr}=\frac{2}{5} \mathrm{hr}\).
Let the length of the journey be \(x \mathrm{~km}\).
Then, \(\frac{x}{40}-\frac{x}{50}=\frac{2}{5} \Leftrightarrow \frac{x}{200}=\frac{2}{5} \Leftrightarrow x=\left(\frac{2}{5} \times 200\right)=80 \mathrm{~km}\).

Hint

Let the correct time to complete the journey be \(x\) min.
Distance covered in \((x+11) \mathrm{min}[latex]. at [latex]40 \mathrm{kmph}\)
\(=\) Distance covered in \((x+5)[latex] min. at [latex]50 \mathrm{kmph}\)
\(
\therefore \quad \frac{(x+11)}{60} \times 40=\frac{(x+5)}{60} \times 50 \Leftrightarrow x=19 \mathrm{~min} .
\)

Hint

\(
\text { Let the distance travelled be } x \mathrm{~km} \text {. }
\)
Then, \(\frac{x}{10}-\frac{x}{15}=2 \Leftrightarrow 3 x-2 x=60 \Leftrightarrow x=60 \mathrm{~km}\).
Time taken to travel \(60 \mathrm{~km}\) at \(10 \mathrm{~km} / \mathrm{hr}=\left(\frac{60}{10}\right) \mathrm{hrs}=6 \mathrm{hrs}\).
So, Robert started 6 hours before \(2 \mathrm{PM}\). i.e., at 8 AM.
\(\therefore\) Required speed \(=\left(\frac{60}{5}\right) \mathrm{kmph}=12 \mathrm{kmph}\)

Hint

Let the to and fro distance to the mall be \(x \mathrm{~km}\) Then, \(\frac{x}{10}-\frac{x}{15}=\frac{30}{60}=\frac{1}{2} \Rightarrow \frac{x}{30}=\frac{1}{2} \Rightarrow x=15\).
Time taken to travel \(15 \mathrm{~km}\) at \(10 \mathrm{~km} / \mathrm{hr}\) \(=\left(\frac{15}{10}\right) \mathrm{hr}=\frac{3}{2} \mathrm{hrs}=1 \frac{1}{2} \mathrm{hrs}\).
Since 30 minutes were spent in shopping, so Ravi started for the mall 2 hours before \(19.00 \mathrm{hrs}\) i.e., at \(17.00 \mathrm{hrs}\).
Now, required time for to and fro journey \(=(18.15 \mathrm{hrs}-17.00 \mathrm{hrs})-30 \mathrm{~min}=45 \mathrm{~min}=\frac{3}{4}\) hrs.
Hence, required speed \(=\left(15 \times \frac{4}{3}\right) \mathrm{km} / \mathrm{hr}=20 \mathrm{~km} / \mathrm{hr}\).

Hint

Let the distance travelled by bus be \(x \mathrm{~km}\) Then, distance travelled by train \(=(285-x) \mathrm{km}\).
\(
\begin{aligned}
& \therefore \quad \frac{x}{40}+\frac{(285-x)}{55}=6 \Rightarrow \frac{11 x+8(285-x)}{440}=6 \\
& \Rightarrow \quad 11 x-8 x+2280=2640 \Rightarrow 3 x=360 \Rightarrow x=120 .
\end{aligned}
\)
Hence, distance travelled by train \(=(285-120) \mathrm{km}=165 \mathrm{~km}\)

Hint

Let distance \(=x \mathrm{~km}\) and usual rate \(=y \mathrm{kmph}\).
\(\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60}[latex] or [latex]2 y(y+3)=9 x \dots(i)[latex]
And, [latex]\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60}[latex] or [latex]y(y-2)=3 x \dots(ii)\)
On dividing (i) by (ii), we get :
\(
\begin{aligned}
2(y+3) & =3(y-2) \Rightarrow y=12 \\
\therefore \text { Distance }=x & =\frac{2 y(y+3)}{9}=\left(\frac{2 \times 12 \times 15}{9}\right) \mathrm{km}=40 \mathrm{~km}
\end{aligned}
\)

Hint

Let distance \(=x \mathrm{~km}\) and usual rate \(=y \mathrm{kmph}\).
Then, \(\frac{x}{y}-\frac{x}{y+10}=1[latex] or [latex]y(y+10)=10 x \dots(i)\)
And \(\frac{x}{y}-\frac{x}{y+20}=\frac{7}{4}[latex] or [latex]y(y+20)=\frac{80 x}{7} \dots(ii)\)
On dividing (i) by (ii), we get : \(y=60\).
\(
\text { Substituting } y=60 \text { in (i), we get : } x=420 \mathrm{~km}
\)

Hint

Let speed of the car be \(x \mathrm{kmph}\).
Then, speed of the train \(=\frac{150}{100} x=\left(\frac{3}{2} x\right) \mathrm{kmph}\).
\(
\begin{aligned}
\therefore \frac{75}{x}-\frac{75}{\frac{3}{2} x} & =\frac{125}{10 \times 60} \Leftrightarrow \frac{75}{x}-\frac{50}{x} \\
& =\frac{5}{24} \Leftrightarrow x=\left(\frac{25 \times 24}{5}\right)=120 \mathrm{kmph}
\end{aligned}
\)

Hint

Due to stoppages, it covers \(9 \mathrm{~km}\) less.
Time taken to cover \(9 \mathrm{~km}=\left(\frac{9}{54} \times 60\right) \mathrm{min}=10 \mathrm{~min}\).

Hint

Due to stoppage, Kingfisher flight covers (700 – 560) \(=140 \mathrm{~km}\) less per hour.
Time taken to cover \(140 \mathrm{~km}=\left(\frac{140}{700} \times 60\right) \mathrm{min}=12 \mathrm{~min}\).
Hence, stoppage time per hour \(=12 \mathrm{~min}\).

Hint

Let the distance between village \(A\) and \(B\) be \(x \mathrm{~km}\). Then, \(\frac{x}{40}-\frac{x}{60}=2 \Rightarrow x=240\).

Hint

After t hours, the two trains will have traveled distances D1 and D2 (in
miles) given by
\(\mathrm{D} 1=72 \mathrm{t}\) and \(\mathrm{D} 2=78 \mathrm{t}\)
After t hours total distance \(D\) traveled by the two trains is given by
\(D=D 1+D 2=72 t+78 t=150 t\)
When distance \(D\) is equal to 900 miles, the two trains pass each other.
\(150 t=900\)
Solve the above equation for \(t\)
\(t=6\) hours.

Hint

\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { driving } & d & 30 & t \\
\hline \text { flying } & 150-d & 60 & 3-t \\
\hline \text { total } & 150 & — & 3 \\
\hline
\end{array}
\)

The first row gives me the equation \(d=30 t\). The second row is messier, giving me the equation:
\(
150-d=60(3-t)
\)
There are various ways we can go from here; we think we’ll solve this second equation for the variable \(d\), and then set the results equal to each other.
\(
\begin{aligned}
& 150-d=60(3-t) \\
& 150-60(3-t)=d
\end{aligned}
\)
Setting equal these two expressions for \(d\), we get:
\(
30 t=150-60(3-t)
\)
\(
\begin{aligned}
& 30 t=150-60(3-t) \\
& 30 t=150-180+60 t \\
& 30 t=-30+60 t \\
& 30=30 t \\
& 1=t
\end{aligned}
\)
He drove for 1 hour.
He drove for one hour at \(30 \mathrm{mph}\), covering 30 miles. This left \(150-30=\) 120 miles to fly. The airport is 120 miles from the offices.

Hint

\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { car } & d+20 & 2 r-30 & 2 \\
\hline \text { bus } & d & r & 2 \\
\hline \text { total } & — & — & — \\
\hline
\end{array}
\)
(As it turns out, I won’t need the “total” row this time.) The car’s row gives me:
\(
d+20=2(2 r-30) \dots(i)
\)
This is not terribly helpful. The second row gives me:
\(
d=2 r \dots(ii)
\)
Solving (i) and (ii) we get
\(
\begin{aligned}
& 2 r+20=2(2 r-30) \\
& 2 r+20=4 r-60 \\
& 20+60=4 r-2 r \\
& 80=2 r \\
& 40=r
\end{aligned}
\)
The bus was going 40 miles per hour.
\(
\begin{aligned}
& 2 r-30=2(40)-30 \\
& 2 r-30=80-30 \\
& 2 r-30=50
\end{aligned}
\)
The car was going 50 miles an hour.

Hint

The freight train left two hours before the passenger train, and the passenger (or “PAX”) train took three hours to catch up. This means
Distance Word Problems that the freight train travelled for \(2+3=5\) hours.
For the purposes of this exercise, I care only about the distance between the depot (where each train started) and the point at which the passenger train caught up. So, for the purposes of this exercise, the distance travelled by the trains is the same. (I neither know nor care how far the trains might have gone once the second train passed the first.)
Whatever was the speed \(r\) of the passenger train, the freight train was \(20 \mathrm{mph}\) slower, or \(r-20\).
Using variables for the unknowns of rate and distance, I create this table:
\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { PAX } & d & r & 3 \\
\hline \text { freight } & d & r-20 & 5 \\
\hline \text { total } & & — & \\
\hline
\end{array}
\)
\(
\begin{aligned}
& 3 r=5(r-20) \\
& 3 r=5 r-100 \\
& 100=2 r \\
& 50=r
\end{aligned}
\)
The passenger train is going 50 miles an hour.
passenger train: \(50 \mathrm{mph}\), freight train: \(30 \mathrm{mph}\)

Hint

However long it takes the two cyclists to meet, they will have taken the same amount of time to get to their meeting point. I don’t know (yet) how much time this is, so I’ll use a variable.
However much distance \(d\) the slower guy covers, the faster guy covered how much was left of the course, or \(45-d\).
They have given me the two speeds. So my grid looks like this:
\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { slow } & d & 14 & t \\
\hline \text { fast } & 45-d & 16 & t \\
\hline \text { total } & 45 & — & \\
\hline
\end{array}
\)
Using \(d=r t\), I get \(d=14 t\) from the first row, and \(45-d=16 t\) from the second row. Since these distances add up to 45, I will add the distance expressions and set equal to the given total:
\(
45=14 t+16 t
\)
\(
\begin{aligned}
& 45=30 \mathrm{t} \\
& 3 / 2=t=1.5
\end{aligned}
\)
They will meet in an hour and a half.

Hint

Whatever the distance was, it was the same in each direction. I’ll use a variable to represent this. I don’t know what the boat’s speedometer was saying; l’ll use the variable \(b\) (for “boat”) to stand for this rate. They’ve given me the times in each direction, and they’ve given me the speed of the current, so I can do addition and subtraction to get expressions for the net speed upstream and downstream. My grid looks like this:
\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { downstream } & d & b+3 & 3 \\
\hline \text { upstream } & d & b-3 & 4 \\
\hline \text { total } & 2 d & — & 7 \\
\hline
\end{array}
\)
Using \(d=r t\), the downstream row gives me:
\(
d=3(b+3)
\)
The upstream row gives me:
\(
d=4(b-3)
\)
Since these distances are the same, I will set them equal:
\(
3(b+3)=4(b-3)
\)
\(
\begin{aligned}
& 3 b+9=4 b-12 \\
& 9+12=4 b-3 b \\
& 21=b
\end{aligned}
\)
The boat’s speedometer reads ‘ 21 ‘ mph.
\(
d=3(21+3)=3 \times 24=72
\)
The boat goes 72 miles each way.

Hint

Here I am really dealing with two rates together: the plane’s speedometer reading, and the wind’s speed. When the plane turns around, the wind is no longer pushing the plane to go faster but is instead pushing against the plane to slow it down. I don’t know either of the speeds; to keep them straight, I’ll use the variables \(p\) and \(w\) for the speed of the plane and the wind, respectively. They have given me the distances and the times. Adding this info, my table looks like this:
\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { tailwind } & 1120 & p+w & 7 \\
\hline \text { headwind } & 1120 & p-w & 8 \\
\hline \text { total } & 2240 & — & 15 \\
\hline
\end{array}
\)
The tailwind row gives me:
\(
1120=7(p+w)
\)
The headwind row gives me:
\(
1120=8(p-w)
\)
The temptation is to just set these equal, like I did with the last problem, but that just gives me:
\(
7(p+w)=8(p-w)
\)
I’ll take that tailwind equation:
\(
1120=7(p+w)
\)
and divide through by 7 and solve for the variable \(p\) :
\(
\begin{aligned}
& 160=p+w \\
& 160-w=p
\end{aligned}
\)
I’ll do the same thing with headwind equation, this time dividing through by 8 :
\(
\begin{aligned}
& 140=p-w \\
& 140+w=p
\end{aligned}
\)
Then I’ll set the two left-hand sides equal to each other:
\(
160-w=140+w
\)
\(
\begin{aligned}
& 160-w=140+w \\
& 20=2 w \\
& 10=w
\end{aligned}
\)
The wind is blowing at \(10 \mathrm{mph}\).
\(
\begin{aligned}
& 160=p+w=p+10 \\
& 150=p
\end{aligned}
\)
The plane’s speedometer reads ‘ 150 ‘ mph.

Hint

However far away that spike is, the distance d travelled by the sound, either through the air or through the rail, is the same. They’ve given me the rates. However long the sound took to reach me through the air, the sound reached me through the steel six seconds earlier. (The travel-speed is greater, so the travel-time must be lesser.) Putting this together, my grid looks like this:

\(
\begin{array}{|c|c|c|c|}
\hline & \boldsymbol{d} & \boldsymbol{r} & \boldsymbol{t} \\
\hline \text { air } & d & 1,100 & t \\
\hline \text { steel } & d & 16,500 & t-6 \\
\hline \text { total } & — & — & — \\
\hline
\end{array}
\)

Multiplying across, the air and steel rows give me these two equations:
\(
\begin{aligned}
& d=1100 t \\
& d=16500(t-6)
\end{aligned}
\)
Since the distances are the same, I set the distance expressions equal to get:
\(
1100 t=16,500(t-6)
\)
divide through by \(100:\)
\(
11 t=165(t-6)
\)
then:
\(
\begin{aligned}
& 11 t=165 t-990 \\
& 990=154 t \\
& 990 / 154=t=45 / 7
\end{aligned}
\)
\(
d=1100(45 / 7)=7071.428571 \ldots
\)
The spike is about \(7,071.4\) feet away.

Hint

Note that he drives at 50 miles per hour after the first hour and continues doing so until he arrives.
Let \(d\) be the distance still needed to travel after 1 hour. We have that \(\frac{d}{50}+1.5=\frac{d}{35}\), where the 1.5 comes from 1 hour late decreased to 0.5 hours early.
Simplifying gives \(7 d+525=10 d\), or \(d=175\).
Now, we must add an extra 35 miles traveled in the first hour, giving a total of 210 miles.