Time Speed and Distance Questions for CAT

Question 31: 1 have to be at a certain place at a certain time and find that I shall be 40 minutes late, if I walk at the rate of 4 kmph and 40 min early if I walk at the rate of 5 kmph. What is the usual time required by me to reach my destination?
[1] 7 hrs
[2] 14 hrs
[3] 6 hrs
[4] 3.5 hrs

Option: 3

Explanation:

Let t be the usual time.

:. Distance covered D $= 4 \times \left( {t + \frac{2}{3}{\rm{ }}} \right) = 5 \times \left( {t - \frac{2}{3}{\rm{ }}} \right)$

$\Rightarrow t = 6$

Question 32: Two men A and B walk from P to Q, a distance of 21 km, at 3 and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. The distance from P to R is
[1] 12 km
[2] 18 km
[3] 10 km
[4] 24 km

Option: 2

Explanation:

When B meets A at R, B has walked the distance PQ + QR and A the distance PR.

That is, both of them have together walked twice the distance from P to Q, i.e. 42 km.

Now, the speed of A and B are in the ratio 3 : 4, and they have walked 42 km.

Hence, the distance PR travelled by A = 3/7 of 42 km = 18 km.

Question 33: A person can row at the rate of 3 kmph in still water. If the river is running at 1.2 kmph, then it takes him 2 hours to row to a place and back. How far is the place?
[1] 5 km
[2] 2.5 km
[3] 6 km
[4] 8 km

Option: 2

Explanation:

X = 3 kmph, Y = 1.2 kmph. Since time taken = Distance/Speed = 2 hrs.

=> D/4.2 + D/1.8 = 2 =>(1.8 D + 4.2 D)(4.2 x 1.8).

=> D = 2.52 km.

Question 34: The current of a stream runs at 5 kmph. A motor launch goes 25 km upstream and back again to the starting point in 275/2 minutes. The speed of the motor launch in the absence of any current is
[1] 22.8 kmph
[2] 10.1 kmph
[3] 31.6 kmph
[4] 6 kmph

Option: 1

Explanation:

Let X = Speed of motor launch.

Y = Speed of current = 5 kmph. Distance = 25 km.

Time required = 275/(2 x 60) hrs = 25/(X + 5) +25/(X — 5).

$\Rightarrow 2.3 = \frac{{25\left( {X - 5} \right) + 25\left( {X + 5} \right)}}{{\left( {X + 5} \right)\left( {X - 5} \right)}} =$>$2.3 = 50{\rm{X}}/\left( {{X^2} - 25} \right)$

$\Rightarrow 2.3{X^2} - 50X - 57.2 = 0$

Apply the quadratic formula to get.

X= speed of launch = (50 ± 55)/4.6.

Neglect the negative sign => X= 105/4.6 = 22.82 kmph.

Question 35: A man can row with the stream at the rate of 20 kmph and against the stream at 5 kmph. The man's rate in still water is
[1] 7.5 kmph
[2] 12.5 kmph
[3] 17.5 kmph
[4] 15.5 kmph