1. A man starts from L to M, another from M to L at the same time. After passing each other they complete their journey in $3\dfrac{1}{3}$ hours and $4\dfrac{4}{5}$ hours respectively. Find the speed of the second man if the speed of the first is 24 km/hr?
a. 15 kmph
b. 18 kmph
c. 20 kmph
d. 24 kmph

Answer: C

Explanation:
Speed of first man : Speed of second man = $\displaystyle\sqrt {\frac{b}{a}} $ where a and b are the time taken by 1st and 2nd man respectively.
Speed of first man : Speed of second man = $\sqrt {\dfrac{{4{\raise0.5ex\hbox{$\scriptstyle 4$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 5$}}}}{{3{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}}}} $ = $\sqrt {\dfrac{{24/5}}{{10/3}}}$ $= \sqrt {\dfrac{{24}}{5} \times \dfrac{3}{{10}}}$ $= \sqrt {\dfrac{{36}}{{25}}} = \dfrac{6}{5}$
Thus second man’s speed = $\displaystyle\frac{5}{6}$ × 24 = 20 km/hr.

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2. A hare makes 9 leaps in the same time as a dog makes 5. But the dog’s leap is 2m while hare’s is only 1 m. How many leaps will the dog have to make before catching up with the hare if the hare has a head start of 16 m?
a. 70 leaps
b. 75 leaps
c. 80 leaps
d. 85 leaps

Answer: C

Explanation:
Distance covered by dog in 5 leaps = 5 × 2 = 10 m
Distance covered by hare in 9 leaps = 9 × 1 = 9 m
Distance gained by the dog in 5 leaps = 1 m. Hence, for 1 m gain he has to make 5 leaps.
Number of leaps required by the dog to gain 16 m = 5 × 16 = 80 leaps.

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3. A man is walking at a speed of 10 km/h. After every km, he takes rest for 4 minutes. How much time will he take to cover a distance of 10 km?
a. 96
b. 84
c. 90
d. 80

Answer: A

Explanation:
He covers 10 km in 1 hour (i.e. in 60 minutes)
Therefore He will take 6 minutes in covering 1 km.
He rests for 4 minutes after every km.
Time taken = (6 + 4) minutes = 10 minutes for every km.
Therefore, Time taken (for first 9 km) = 9 x 10 = 90 minutes.
Time taken to cover 10th km = 6 minutes
Therefore, Total time taken = 90 + 6 = 96 minutes.
Hint: Rest time after 10th km is not added as he has reached his destination.

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4. A train left station A for station B at a certain speed. After travelling for 100 km, the train meets with an accident and could travel at $\displaystyle\frac{4}{5}$th of the original speed and reaches 45 minutes late at station B. Had the accident taken place 50 km further on, it would have reached 30 minutes late at station B. What is the distance between station A and B?
a. 200
b. 250
c. 300
d. 350

Answer: C

Explanation:
Let, initial speed of the train = 5 km/h
Then, speed after the accident = $\displaystyle\frac{4}{5} \times 5$ = $4$ km/h
Time taken to cover 50 km @ 5 km/h = 10 hours
Time taken to cover 50 km @ 4 km/h = $12\displaystyle\frac{1}{2}$ hours
Difference between times taken = $12\displaystyle\frac{1}{2} - 10$ = $2\displaystyle\frac{1}{2}$ hours = 150 minutes
But, actual difference = (45 - 30) minutes = 15 minutes = $\displaystyle\frac{1}{{10}}$ of 150 minutes
Therefore, Speed is 10 times of assumed speed.
Therefore, Speed before accident = 10 x 5 = 50 km/h
And, Speed after accident = 10 x 4 = 40 km/h
Distance after accident is covered @ 40 km/h instead of 50 km/h
And, time difference = 45 minutes = $\displaystyle\frac{3}{4}$ hour
Let the distance between place of accident and B = $d$
$ \Rightarrow \dfrac{d}{{40}} - \dfrac{d}{{50}} = \dfrac{3}{4}$
$ \Rightarrow \dfrac{{5d - 4d}}{{200}} = \dfrac{3}{4}$
$ \Rightarrow d = \dfrac{{3 \times 200}}{4} = 150$ km
Therefore, Distance between A and B = 100 + 150 = 250 km.

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5. A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speeds is :
a. 4 : 1
b. 1 : 4
c. 2 : 1
d. 1 : 2

Answer: A

Explanation:
Let x km be covered in y hrs. Then,
1st speed = $\left( {\displaystyle\frac{x}{y}} \right)$ km/hr.
2nd speed = $\left( {\displaystyle\frac{x}{2} \div 2y} \right)$km/hr = $\left( {\displaystyle\frac{x}{{4y}}} \right)$ km/hr
Ratio of speed = ${\displaystyle\frac{x}{y}:\displaystyle\frac{x}{{4y}} = 1:\displaystyle\frac{1}{4} = 4:1}$

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6. A train leaves Vijayawada at 6 a.m and reaches Rajahmundry at 10 a.m. Another train leaves Rajahmundry at 8 a.m and reaches Vijayawada at 11.30 a.m. At what time do the two trains cross one another ?
a. 9.26 a.m
b. 9 a.m
c. 8.36 a.m
d. 8.56 a.m

Answer: B

Explanation:
Let the distance between Vijayawada and Rajahmundry be $y$ km.
Time taken by the first train to cover the distance ＝ $4$ hrs.
Time taken by the second train to cover the distance ＝ $3\dfrac{1}{2}$ ＝ $\frac{7}{2}$ hrs.
Average speed of the train leaving Vijayawada ＝ $\left( {\displaystyle\frac{y}{4}} \right)$ km/hr
Average speed of the train leaving Rajahmundry ＝ $\dfrac{y}{{\left( {7/2} \right)}} = \left( {\dfrac{{2y}}{7}} \right)$ km/hr
Suppose they meet $x$ hours after 6 a.m
Then, $x\left( {\dfrac{y}{4}} \right) + \left( {x - 2} \right)\dfrac{{2y}}{7} = y$
or $\dfrac{x}{4} + \left( {x - 2} \right)\dfrac{2}{7} = 1$ ($\because$ cancelling $y$ on both sides)
$15x = 44$
$x = {\displaystyle\frac{{44}}{{15}}}$ = $2\dfrac{{14}}{{15}}$ ＝ 2 hrs 56 min
($\because$ $\dfrac{{14}}{{15}}$ hours = $\dfrac{{14}}{{15}} \times 60 = 56$ minutes)
So, the trains meet at 8.56 a.m.