7

# How long will a boy take to run round a square field of side 35 meters, If he runs at the rate of 9 km/hr?

 A) 50 sec B) 52 sec C) 54 sec D) 56 sec

Explanation:

Speed = 9 km/hr =     ${\color{Blue}&space;\left&space;(&space;9\times&space;\frac{5}{18}&space;\right&space;)&space;m/sec&space;=\frac{5}{2}m/sec}$

Distance = (35 x 4) m = 140 m.

Time taken =${\color{Blue}&space;\left&space;(&space;140\times&space;\frac{2}{5}&space;\right&space;)&space;sec&space;=56&space;sec}$

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• Related Questions

A and B runs around a circular track. A beats B by one round or 10 minutes. In this race, they had completed 4 rounds. If the race was only of one round, find the A's time over the course:

 A) 8 min B) 7.5 min C) 12.5 min D) 12 min

Explanation:

B runs around the track in 10 min.

i.e ,    Speed of B = 10 min per round

$\inline \fn_jvn \therefore$ A beats B by 1 round

Time taken  by A to complete 4 rounds

= Time taken by B to complete 3 rounds

= 30 min

$\inline \fn_jvn \therefore$ A's speed = 30/4 min per round

= 7.5 min per round

Hence, if the race is only of one round A's time over the course = 7 min 30 sec

Subject: Time and Distance - Quantitative Aptitude - Arithmetic Ability

12

A train with 120 wagons crosses Arun who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Arun ( he is riding on the horse) before it starts overtaking the Sriram( who is also riding on his horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Sriram do the Arun meets to Sriram?

 A) 3560 sec B) 3600 sec C) 3576 sec D) can't be determined

Explanation:

Let the length of the train be L metres and speeds of the train Arun and Sriram be R, A and S respectively, then

$\inline \fn_jvn \frac{L}{R-A}=36$   ---------- (i)

and $\inline \fn_jvn \frac{L}{(R+K)}=24$ ---------(ii)

From eq.(i) and (ii)

3(R - A ) = 2 (R + K)

$\inline \fn_jvn \Rightarrow$    R = 3A + 2K

In 30 minutes (i.e 1800 seconds), the train covers 1800R (distance) but the Arun also covers 1800 A (distance) in the same time. Therefore distance between Arun and Sriram, when the train has just crossed Sriram

= 1800 ( R - A) - 24 ( A + K)

$\inline \fn_jvn \therefore$ Time required =$\inline \fn_jvn \frac{1800(R - A)-24(A+K)}{(A+K)}$

= (3600 - 24) = 3576 s

Subject: Time and Distance - Quantitative Aptitude - Arithmetic Ability

10

Two trains A and B start simultaneously in the opposite direction from two points P and Q and arrive at their destinations 16 and 9 hours respectively after their meeting each other. At what speed does the second train B travel if the first train travels at 120 km/h

 A) 90 km/h B) 160 km/h C) 67.5 km/h D) None of these

Explanation:

$\inline \fn_jvn \frac{s1}{s2}=\sqrt{\frac{t2}{t1}}$

$\inline \fn_jvn \Rightarrow \frac{120}{s2}=\sqrt{\frac{9}{16}}=\frac{3}{4}$

$\inline \fn_jvn \Rightarrow s2= 160 km/h$

Subject: Time and Distance - Quantitative Aptitude - Arithmetic Ability

16

Akash leaves mumbai at 6 am and reaches Bangalore  at  10 am . Prakash leaves Bangalore at 8 am and reaches Mumbai at 11:30 am. At what time do they cross each other?

 A) 10 am B) 8:32 am C) 8:56 am D) 9:20 am

Explanation:

Time taken by Akash = 4 h

Time taken by Prakash = 3.5 h

For your convenience take the product of times taken by both as a distance.

Then  the distance = 14km

Since, Akash covers half of the distance  in 2 hours(i.e at 8 am)

Now, the rest half (i.e 7 km) will be coverd by both prakash and akash

Time taken by them = 7/7.5 = 56 min

Thus , they will cross each other at  8 : 56am.

Subject: Time and Distance - Quantitative Aptitude - Arithmetic Ability

8

In a race of 800m Kaushik gives Piyush a start of 200m and then loses the race by 20 seconds. What is the speed of Piyush and Kaushik respectively ? If the ratio of respective speeds be 3:2.

Let the speed of Kaushik be $\inline \fn_jvn S_{k}$ and speed of Piyush be $\inline \fn_jvn S_{p}$ and let the time taken by Piyush be t second then,

$\inline \fn_jvn \frac{S_{K}}{S_{P}}=\frac{\frac{800}{(t+20)}}{\frac{600}{t}}=\frac{2}{3}$

$\inline \fn_jvn \frac{800}{600}\times \frac{t}{(t+20)}=\frac{2}{3}$

t = 20 s

$\inline \fn_jvn \therefore$ Speed of Piyush = $\inline \fn_jvn \frac{600}{20} = 30 m/s$

and    Speed of Kaushik = $\inline \fn_jvn \frac{800}{40} = 20 m/s$