# CRE 1 - Basics | Arithmetic - Time, Speed & Distance

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A person crosses a 1200 m long street in 20 minutes. What is his speed in km per hour?

- A.
3.6

- B.
7.2

- C.
8.4

- D.
10

Answer: Option A

**Explanation** :

Speed = $\frac{1200m}{20\times 60sec}$ = 1 m/sec.

Converting m/sec to km/hr = (1 × 18/5) km/hr = 3.6 kmph

Hence, option (a).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Aakash walks at 5 kmph and reaches his school, 2 km away. How long will he take (in minutes) to reach the school?

Answer: 24

**Explanation** :

T = D/S = 2/5 hours = 0.4 × 60 mins = 24 mins.

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A train travels 184.8km/h. How many meters will it travel in 5 minutes?

- A.
15400

- B.
1540

- C.
154

- D.
15.40

Answer: Option A

**Explanation** :

Speed of train = $\left(184.8\times \frac{5}{18}\right)=\frac{154}{3}$ m/sec.

Distance covered in (5 × 60) sec = $\frac{154}{3}$× 5 × 60 = 15400 meters

Hence, option (a).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Aakash takes 5 sec to cover certain distance at 54 km/hr. What is the total distance covered by him in those 5 seconds (in meters)?

Answer: 75

**Explanation** :

Speed = 54 × (5/18) = 15 m/sec

D = S × T = 15 × 5 = 75 m

Hence, 75.

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Aakash walks at 10 kmph from his office and reaches his home in 1 hour 12 mins. How far is his home from his office (in km)?

Answer: 12

**Explanation** :

1 hour 12 minutes = 1 + 12/60 hours = 1.2 hours

D = S × T= 10 × 1.20 = 12 km

Hence, 12.

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A man covers a certain distance in 2 hours 45 min, when he walks at the rate of 8 km/h. How much time will he take to cover the same distance if he runs at a speed of 33 km/h?

- A.
40 min

- B.
41 min 15 sec

- C.
45 min

- D.
100 min

Answer: Option A

**Explanation** :

2 hours 45 mins = 2 + 45/60 = 11/4 hours.

∴ Distance = Speed × time = 8 × $\frac{11}{4}$= 22 kms.

⇒ Time taken while running at 33 kmph = $\frac{22}{33}$ hrs = $\frac{22}{33}$ × 60 mins = 40 mins

Hence, option (a).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Ankush can cover a certain distance in 21 mins. He covers two third of it at 8 km an hour and the remaining at 10 km an hour. The total distance is:

- A.
2.5 km

- B.
4.6 km

- C.
4 km

- D.
3 km

Answer: Option D

**Explanation** :

Let the distance be x km.

Then, total time taken = $\frac{{\displaystyle \frac{2x}{3}}}{8}+\frac{{\displaystyle \frac{x}{3}}}{10}$

$\Rightarrow \left(\frac{2x}{3}\times \frac{1}{8}\right)+\left(\frac{x}{3}\times \frac{1}{10}\right)=\frac{21}{60}$

⇒ x = 3 kms.

Hence, option (d).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A man goes to a place at the rate of 8 km/h. He comes back on a bicycle at 32 km/h. His average speed for the entire journey is:

- A.
10 km/h

- B.
12.8 km/h

- C.
17 km/h

- D.
20 km/h

Answer: Option B

**Explanation** :

Distance is same while going and coming back.

Average speed = $\frac{2D}{{\displaystyle \frac{D}{8}}+{\displaystyle \frac{D}{32}}}=\frac{2\times 8\times 32}{8+32}=\frac{64}{5}$ km/h = 12.8 km/h

Hence, option (b).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A boy goes to school with the speed of 6 km/h. and returns with a speed of 4 km/h. If he takes 10 hours in all. The distance in kilometers between the village and the school is:

- A.
24

- B.
28

- C.
32

- D.
36

Answer: Option A

**Explanation** :

Let distance be d km.

Then, average speed = $\frac{2\times 6\times 4}{6+4}=\frac{2d}{10}$

∴ d = 24 km

Hence, option (a).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A car covers four successive 60 km. stretches at speeds of 10 km/h, 20 km/h, 30 km/h and 60 km/h respectively. Its average speed over this distance is:

- A.
10 km/h

- B.
20 km/h

- C.
30 km/h

- D.
25 km/h

Answer: Option B

**Explanation** :

Total time taken = $\left(\frac{60}{10}+\frac{60}{20}+\frac{60}{30}+\frac{60}{60}\right)$ = 12 hrs.

Avg. speed = $\frac{240}{12}$ = 20 km/h.

Hence, option (b).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A car completes a certain journey in 16 hours. It covers half the distance at 20 km/h and the rest at 30 km/h. The length of the journey in km is :

- A. 350
- B.
384

- C.
400

- D.
420

Answer: Option B

**Explanation** :

Let the distance travelled during each half be x km.

Then $\frac{x}{20}+\frac{x}{30}=16$

⇒x=192

∴ Length of journey = 2 × 192 = 384 km

Hence, option (b).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A Train covers a certain distance at a speed of 120 kmph in 10 hours . To cover twice the distance in $3\frac{1}{3}$ hours, it must travel at a speed of:

- A.
1440 kmph

- B.
360 kmph

- C.
600 kmph

- D.
720 kmph

Answer: Option B

**Explanation** :

Distance = (120 ×10) = 1200 km.

Speed = Distance/Time

Speed = 2400/(10/3) km/hr

⇒ Required speed =2400 × 3/10km/hr= 720 km/hr.

Hence, option (b).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

If a person walks at 28 km/hr instead of 20 km/hr, he would have walked 80 km more. The actual distance travelled by him is:

- A.
200 km

- B.
224 km

- C.
280 km

- D.
320 km

Answer: Option A

**Explanation** :

Let the actual distance travelled be x km.

Then, $\frac{x}{20}=\frac{x+80}{28}$

⇒ 28x = 20x + 1600

⇒8x = 1600

⇒ x = 200 km.

Hence, option (a)

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Aakash travels the first 4 hours of his journey at 120 mph speed and remaining 6 hours at 50 mph speed. What is the average speed of Aakash’s travel (in mph)?

Answer: 78

**Explanation** :

Average speed = (Total Distance)/(Total Time)

Total Distance = Distance covered in first 4 hours + Distance covered in next 6 hours.

Distance covered in first 4 hours = 4 × 120 = 480 miles

Distance covered in next 5 hours = 6 × 50 = 300 miles

∴ Total distance = 480 + 300 = 780 miles

Total time = 4 + 6 = 10 hours

Average speed = 780/10 = 78 mph

Hence, 78.

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A man completes a journey in 10 hours. He travels first half of the journey at the rate of 42 km/hr and second half at the rate of 48 km/hr. Find the total journey in km.

- A.
440 km

- B.
448 km

- C.
460 km

- D.
468 km

Answer: Option B

**Explanation** :

Let x be the total distance travelled.

$\frac{\left({\displaystyle \frac{1}{2}}\right)x}{42}+\frac{\left({\displaystyle \frac{1}{2}}\right)x}{48}=10\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{42}+\frac{x}{48}=20$

⇒ x = 448 kms

Hence, option (b).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Aakash travels with a speed of 40 kmph by bus for 2 hours and then takes an auto which travels at 60 kmph for another 2 hours. Find his average speed (in kmph) for entire journey.

Answer: 50

**Explanation** :

Average speed =$\frac{totaldis\mathrm{tan}ce}{totaltime}=\frac{40\times 2+60\times 2}{4}=\frac{200}{4}=50$kmph

Alternately,

When time is constant average speed = Arithmetic mean of speeds = (40 + 60)/2 = 50kmph

Hence, 50.

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

The ratio between the speeds of two trains is 7 : 8. If the second train runs 200 km in 1 hours, then the speed of the first train is:

- A.
140 km/hr

- B.
15 km/hr

- C.
168 km/hr

- D.
175 km/hr

Answer: Option D

**Explanation** :

Let the speed of two trains be 7x and 8x km/hr.

Then, 8x =200/1= 200

∴ x =200/8 = 25

⇒ Speed of first train = (7 × 25) km/hr = 175 km/hr.

Hence, option (d).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Siraj travelled 600 kms by air which formed (2/5) of his trip. The part, which was one third of the whole trip, he travelled by car. The rest of the journey was performed by train. The distance travelled by train was :

- A.
800 km

- B.
400 km

- C.
240 km

- D.
900 km

Answer: Option B

**Explanation** :

Let the total distance be x km.

Then,$\frac{2}{5}x$ = 600 ⇒ x = 1500 km

Distance covered by car $1500\times \frac{1}{3}$= 500 km.

∴ Distance travelled by train = [1500 – (500 + 600)] = 400 km.

Hence, option (b).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

Two men start together to walk to a certain destination, one at 7.5 km an hour and another at 6 km an hour. The former arrives 15 minutes before the latter. The distance is :

- A.
9.5 km

- B.
8 km

- C.
7.5 km

- D.
6 km

Answer: Option C

**Explanation** :

Let the distance be x km.

Then, $\frac{x}{6}-\frac{x}{7.5}=\frac{1}{4}$

⇒x=7.5 km

Hence, option (c).

Workspace:

**CRE 1 - Basics | Arithmetic - Time, Speed & Distance**

A train goes from a station A to another station B at a speed of 64 km/h but returns to A at a slower speed. If its average speed for the trip is 56 km/h., the return speed of the train is nearly :

- A.
48 km/h

- B.
50 km/h

- C.
52 km/h

- D.
47.4 km/h

Answer: Option B

**Explanation** :

Let the return speed of train be y km/h. Then,

$\frac{2\times 64\times y}{64+y}=56$

⇒ y = 49.87 ≅ 50 km/h.

Hence, option (b).

Workspace:

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