Time Speed and Distance Questions for CAT

Question 26: A and B start on a journey at the same time. B travels at 4/7th of A's rate, and arrives 3 hours 15 minutes after him. How long did each take to complete the whole journey?
 $\frac{{91}}{{12}}$hours, $\frac{{13}}{3}$ hours
 $\frac{{89}}{{12}}$ hours,$\frac{{13}}{5}$ hours
 $\frac{{91}}{{10}}$ hours,$\frac{{13}}{3}$ hours
 $\frac{{91}}{{12}}$ hours,$\frac{{13}}{7}$ hours

Option: 1

Explanation:

A's speed : B's speed = 7: 4.

A’s time : B's time = 4 : 7 = 4X, 7X.

$\Rightarrow 7X - 4X = 3\frac{{15}}{{60}} \Rightarrow X = \frac{{13}}{{12}}$

$\Rightarrow 7X = \frac{{91}}{{12}}{\rm{ hrs}},{\rm{ }}4X = \frac{{13}}{3}$ hrs.

Question 27: A ship 156 km from the shore springs a leak which admits 2.5 metric tons of water In $6_2^1$ minutes. A quantity of 68 metric tons would suffice to sink it, but the pumps can throw out 15 metric tons in an hour. The average rate of sailing so that it just reaches the shore as it begins to sink should be
 18 kmph
 60 kmph
 15 kmph
 10 kmph

Option: 1

Explanation:

In one minute, amount flowing in = $\frac{{15}}{{39}}$ MT.

In one minute, amount thrown out = $\frac{{15}}{{60}} = \frac{1}{4}$MT.

Effective rate of filling in one hour= $\left( {\frac{{15}}{{39}} - \frac{1}{4}} \right)MT = \frac{{21}}{{56}}$ MT/Min.

Time till it just begins to sink = $\frac{{68}}{{21/156}} = 505$ minutes.

Speed required = (156/505) = 0.3 km/min = 0.3 x 60 km/hr = 18 km/hr.

Question 28: A train 90 m long is travelling at the rate of 40 kmph. A man is running at a speed of 5 kmph in a direction opposite to the motion of the train. The time taken by the train to cross the man is
 7.2 sec
 92 sec
 2 sec
 66 sec

Option: 1

Explanation:

Distance to be covered = 90 m. Relative speed = 40 + 5 = 45 km/hr.

Time = 0.09/(45) = 7.2 sec.

Question 29: A train travelling at 25 kmph leaves Delhi at 9 am and another train travelling at 35 kmph starts at 2 pm in the same direction. How many km from Delhi are they together?
 $437_2^1$ km
 43.7 km
 35 km
 12.5 km

Option: 1

Explanation:

The first train has a start of 25 x 5 km and

the second train gains (35 - 25) or 10 km per hour.

=> The second train will gain 25 x 5 km in 25 x 5 /10 or $12_2^1$ hrs.

=> The required distance from Delhi = $12_2^1 \times 35{\rm{ km }} = 437_2^1km$.

Question 30: Walking at 3/4 of his usual rate, a man is $1_2^1$ hours too late. Find his usual time.
 $4_2^1$ hours
 ( $1_2^1$ x 3) hours
 (3 + ) $1_2^1$) hours
 All of these

Option: 4

Explanation:

Since the man walks at 3/4 of his usual rate, the time that he takes is 4/3 of his usual time.

=> 4/3 of usual time= usual time + $1_2^1$hours.

=> 1/3 of usual time = $1_2^1$ hours.

Usual time = ($1_2^1$ x 3) hours = ( 3 + $1_2^1$ ) hours = $4_2^1$ hours.

Time Speed and Distance

Time Speed and Distance Questions for CAT
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