# CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Four runners A, B, C and D are running around in a circle, 2430 meter in circumference, at 30 mpm, 24 mpm, 20 mpm and 12 mpm respectively. If they start at the same time from the same point in the same direction, when will they be together again?

- A.
12 hours and 35 minutes

- B.
10 hours and 30 minutes

- C.
16 hours and 45 minutes

- D.
20 hours and 15 minutes

- E.
24 hours and 15 minutes

Answer: Option D

**Explanation** :

A, meets D in every

$\left[\frac{2430}{\left(30-12\right)}\right]$ = 135 minutes

A meets B in every

$\left[\frac{2430}{\left(30-24\right)}\right]$ = 405 minutes

A meets C in every

$\left[\frac{2430}{\left(30-20\right)}\right]$ = 243 minutes

∴ A will meet B, C and D in every LCM (135, 405 and 243) i.e. 1215 minutes or 20 hours and 15 minutes.

Hence, option (d).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Two runners are running on a circular track in the same direction. The speed of slower runner is 10 m/s and that of faster runner is 12 m/s. If they meet after 6 minutes for the first time, then what is the length of the track?

- A.
600 m

- B.
720 m

- C.
800 m

- D.
900 m

- E.
200 m

Answer: Option B

**Explanation** :

Let they meet for the first time at point P on the track.

It is to be noted that the faster runner would have covered the same distance as the slower runner plus 1 full round of the track from P to point when they first meet.

Now, for every second, the faster runner covers 2 m more.

So, in 6 min, he can cover 720 m more.

This extra distance must be the length of the track.

Hence, option (b).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Four runners A, B, C and D are running around in a circle, 1680 meter in circumference, at 8 mpm, 12 mpm, 16 mpm and 20 mpm respectively. If they start at the same time from the same point in the same direction, when will they be together again at the starting point for the first time?

- A.
20 hours

- B.
16 hours

- C.
12 hours

- D.
8 hours

- E.
7 hours

Answer: Option E

**Explanation** :

A completes one round of the circle in (1680/8) = 210 minutes

B completes one round of the circle in (1680/12) = 140 minutes

C completes one round of the circle in (1680/16) = 105 minutes

D completes one round of the circle in (1680/20) = 84 minutes

They will meet at the starting point for the first time at LCM (210, 140, 105, 84) i.e. 420 minutes or 7 hours.

Hence, option (e).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Two cars running on a circular track, 96 meter in circumference meet after every 24 minutes when they are in the same direction. However, when they are in opposite directions, the faster car crosses the slower one every 8 minutes. Find the speeds of the cars.

- A.
12 mpm and 8 mpm

- B.
8 mpm and 4 mpm

- C.
4 mpm and 2 mpm

- D.
16 mpm and 10 mpm

- E.
20 mpm and 10 mpm

Answer: Option B

**Explanation** :

Let the speed of the first car or the faster car be S_{1} while that of the second car be S_{2}.

When in the same direction, they meet after every 24 minutes,

∴ S_{1} – S_{2} = 96/24 = 4 mpm

When in opposite directions, the faster car crosses the slower one every 8 minutes,

∴ S_{1} + S_{2} = 96/8 = 12 mpm

Solving both the equations simultaneously,

S_{1} = 8 mpm and S_{2}= 4 mpm

Hence, option (b).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Aakash and Bikash are running on a 1000 m circular track. Their speeds are 16 km/hr and 10 km/hr, respectively. After what time will they meet for the first time at the starting point if they start simultaneously?

- A.
60 minutes

- B.
45 minutes

- C.
30 minutes

- D.
20 minutes

- E.
15 minutes

Answer: Option C

**Explanation** :

Let us first calculate the time Aakash and Bikash take to cover one full circle.

Time taken by Aakash = $\frac{1000}{16\times \frac{5}{18}}$ = 225 seconds

Time taken by Bikash = $\frac{1000}{10\times \frac{5}{18}}$ = 360 seconds

Hence, after every 225 seconds, Aakash would be at the starting point and after every 360 seconds, Bikash would be at the starting point. The time when they will be together again at the starting point simultaneously for the first time would be the smallest multiple of both 225 and 360, which is the LCM of 225 and 360.

Hence, they would both be together at the starting point for the first time after (LCM of 225, 360) = 1800 seconds = 0.5 hour.

Thus, every half an hour, they would meet at the starting point.

Hence, option (c).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Two men are running in the same direction along a circle. They meet for the first time at a point diametrically opposite the starting point, when the faster one is in his fourth round. Find the ratio of their speeds.

- A.
4 : 5

- B.
7 : 6

- C.
6 : 5

- D.
7 : 5

- E.
None of these

Answer: Option D

**Explanation** :

It is to be noted that when they meet for the first time, the faster man has covered the same distance as that covered by the slower man plus 1 full round.

When they meet, the faster man has completed 3½ rounds, whereas the slower man has completed 2½ rounds.

Since their distances have been covered in the same time, the ratio of distances is the same as the ratio of speeds = 3.5 : 2.5 = 7 : 5.

Hence, option (d).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Three men A, B and C go around a circle, 3520 metres in circumference, at the speeds of 320, 240 and 210 metres per minute, respectively. If they all start together, after what time will they meet again?

- A.
4 hours 35 minutes

- B.
5 hours 52 minutes

- C.
6 hours 24 minutes

- D.
1 hour 16 minutes

- E.
3 hours 25 minutes

Answer: Option B

**Explanation** :

A and B move at a relative speed of (320 - 240) = 80 m/min.

A and B come together in 3520/80 = 44 min.

A and C move together at a relative speed of (320 - 210) = 110 m/min.

A and C come together in 3520/110 = 32 min.

A, B and C come together in 352 min (LCM of 44 and 32), i.e.,

A, B and C come together in 5 hrs 52 mins.

Hence, option (b).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

Three runners A, B and C are running around a circular track. The ratio of their speeds is 3 : 4 : 5. They meet exactly 12th time at the end of an hour at the starting point. A covers 9840 m less distance than C in 1 hour.

What is the distance travelled by B in 1 hour?

- A.
19840 m

- B.
19680 m

- C.
19920 m

- D.
18920 m

- E.
None of these

Answer: Option B

**Explanation** :

Let speeds of A, B and C be 3x, 4x and 5x.

A covers 9840 meters less than C in 1 hour. Let distance covered by C in 1 hour = C meters.

∴ (C - 9840)/3x = C/5x = 1

Solving the above for C and x

⇒ x = 4920

Thus, distance travelled by B in 1 hr = 1 × 4x = 1 × 4 × 4920 = 19,680 m

**Alternately:**

Let the total distance, travelled by A, B and C = 3d, 4d and 5d.

⇒ 5d – 3d = 9840

⇒ 2d = 9840

∴ Distance travelled by B = 4d = 2 × 2d = 2 × 9840 = 19680 meters.

Hence, option (b).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

A and B are running along a circular track in opposite directions. They meet at a point 900 m from the starting point and continue running. They now meet again at a point 600 m from the starting point, but in the opposite direction to before. What is the length of the track?

- A.
2400 m

- B.
1300 m

- C.
1400 m

- D.
1600 m

Answer: Option A

**Explanation** :

Let S be the starting point of the race, and A and B run at speeds of S_{A} and S_{B}, respectively.

a is the point where A and B meet for the first time and b is the point where A and B meet for the second time.

When they meet, A and B have covered different distances in the same time.

So, the ratio of the distances covered by them is the same as the ratio of their speeds.

When they meet for the first time, distance travelled by A = 900 and let X be the distance travelled by B.

So, $\frac{{S}_{A}}{{S}_{B}}=\frac{900}{X}$ …(i)

Now when they meet for the second time, distance travelled by A is X - 600 and distance travelled by A = 900 + 600.

So, $\frac{{S}_{A}}{{S}_{B}}=\frac{X-600}{900+600}=\frac{X-600}{1500}$ …(ii)

∴ from the 2 equations

900/X = (X - 600)/1500

⇒ X^{2} – 600X – 1500 × 900 = 0

⇒ X = 1500

∴ Length of the track = 900 + 1500 = 2400 m

Hence, option (a).

Workspace:

**CRE 6 - Circular Race | Arithmetic - Time, Speed & Distance**

A and B are running around a 40 km circle at respective speeds of 16 kmph and 20 kmph. The number of distinct meeting points for A and B when they run in same direction is x and when they run in opposite directions is y. Find x + y

Answer: 10

**Explanation** :

When the simplified ratio of speed of two people running around a circle is a : b

No. of distinct meeting points when they run in:

(i) same direction = |a - b|

(ii) opposite direction = a + b

Here, simplified ratio of speeds in the given question = 16 : 20 = 4 : 5.

∴ x = 1 and y = 9

Hence, 10.

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**