Practice Time and distance - quantitative aptitude Online Quiz (set-2) For All Competitive Exams

Q-1)   An athlete runs 200 metres race in 24 seconds. His speed (in km/ hr) is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = $\text"Distance"/\text"Time"$

= $200/24$ m/s

$200/24$ m/s = $200/24 × 18/5$

= 30 km/h [Since, x m/s = $18/5$ x km/h]


Q-2)   A man crosses a road 250 metres wide in 75 seconds. His speed in km/hr is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = $\text"Distance"/\text"Time" = 250/75$

= $10/3$ m/sec = $10/3 × 18/5$ km/hr

[Since, 1 m/s = $18/5$ km/hr]

= 2 × 6 km/hr. = 12 km/hr.


Q-3)   The speed 3$1/3$ m/sec when expressed in km/hour becomes

(a)

(b)

(c)

(d)

Explanation:

Since, 1 m/sec = $18/5$ kmph

$10/3$ m/sec

= $18/5 × 10/3$ = 12 kmph


Q-4)   A train is moving with the speed of 180 km/hr. Its speed (in metres per second) is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed = 180 kmph

= ${180 × 5}/18$ m/sec = 50 m/sec

[Since, 1 km/hr = $5/18$ m/s]


Q-5)   A car goes 10 metres in a second. Find its speed in km/hour.

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed of car = 10 m/sec.

Required speed in kmph

=${10 × 18}/5 = 36$ km/hr


Q-6)   Two trains started at the same time, one from A to B and the other from B to A. If they arrived at B and A respectively 4 hours and 9 hours after they passed each other, the ratio of the speed of the two trains was

(a)

(b)

(c)

(d)

Explanation:

Using Rule 11,
Time taken by 1st man to reach B after meeting 2nd man at C is '$t_1$' and time taken by 2nd man to reach A after meeting 1st man at C is '$t_2$' then:
${\text"Speed of 1st man"(s_1)}/{\text"Speed of 2nd man"(s_2)} = √{t_2/t_1}$
Distance from A to B = $s_1t_1 + S_2t_2$

Required ratio of the speed of two trains

= $√{9}/√{4} = 3/2$ or 3 : 2


Q-7)   A car travels 80 km. in 2 hours and a train travels 180 km. in 3 hours. The ratio of the speed of the car to that of the train is :

(a)

(b)

(c)

(d)

Explanation:

Speed = $\text"Distance"/ \text"Time"$

∴ Speed of car : Speed of train

= $80/2 : 180/3$ = 40 : 60 = 2 : 3


Q-8)   It takes 8 hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed of train = x kmph

Speed of car = y kmph

Case I,

$120/x + {600 - 120}/y$ = 8

$120/x + 480/y$ = 8

$15/x + 60/y$ = 1 ...(i)

Case II,

$200/x + 400/y$ = 8 hours 20 minutes

$200/x + 400/y = 8{1}/3$ hours = $25/3$

$8/x + 16/y = 1/3$

$24/x + 48/y$ = 1 ...(ii)

$15/x + 60/y = 24/x + 48/y$

$24/x - 15/x = 60/y - 48/y$

$9/x = 12/y$

$x/y = 9/12 = 3/4$ = 3 : 4


Q-9)   A truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minutes. The ratio of their speed is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Speed of truck = 550m/minute

Speed of bus = $33000/45$ m/minute

or $2200/3$ m/minute

Required ratio = 550 : $2200/3$

= $1 : 4/3 = 3 : 4$


Q-10)   In a one-kilometre race A, B and C are the three participants. A can give B a start of 50 m. and C a start of 69 m. The start, which B can allow C is

(a)

(b)

(c)

(d)

Explanation:

Let the time taken to complete the race by A,B, and C be x minutes.

Speed of A = $1000/x$,

B = ${1000 - 50}/x = 950/x$

C = ${1000 - 69}/x = 931/x$

Now, time taken to complete the race by

B = $1000/{950/x} = {1000 × x}/950$

and distance travelled by C in

${1000x}/950$ min = ${1000x}/950 × 931/x$ = 980 km.

B can allow C

= 1000 - 980 = 20 m


Q-11)   The speed of A and B are in the ratio 3 : 4. A takes 20 minutes more than B to reach a destination. In what time does A reach the destination ?

(a)

(b)

(c)

(d)

Explanation:

Ratio of speed = 3 : 4

Ratio of time taken = 4 : 3

Let the time taken by A and B be 4x hours and 3 x hours respectively.

Then, $4x - 3x = 20/60 ⇒ x = 1/3$

Time taken by A = 4x hours

= $(4 × 1/3)$ hours = 1$1/3$ hours

Using Rule 9,

Here, $S_1 = 3x, S_2 = 4x$

$t_2 = y, t_1 = y + 20/60 = y + 1/3$

$S_1t_1 = S_2t_2$

$3x(y + 1/3) = 4xy$

3y + 1 = 4y, y = 1

∴ Time taken by A

= 1 + $1/3 = 1{1}/3$ hours


Q-12)   A jeep is chasing a car which is 5km ahead. Their respective speed are 90 km/hr and 75 km/ hr. After how many minutes will the jeep catch the car ?

(a)

(b)

(c)

(d)

Explanation:

Relative speed

= 95 - 75 = 15 kmph

Required time

= $\text"Distance"/ \text"Relative speed"$

= $5/15$ hours = $5/15 × 60$ minutes

= 20 minutes


Q-13)   Walking at the rate of 4 kmph a man covers certain distance in 2 hrs 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in how many minutes ?

(a)

(b)

(c)

(d)

Explanation:

2 hours 45 minutes

= $(2 + 45/60)$ hours

= $(2 + 3/4)$ hours = $11/4$ hours

Distance = Speed × Time

= 4 × $11/4$ = 11 km.

Time taken in covering 11 km at 16.5 kmph

= $11/{16.5}$ hour

= $({11 × 10 × 60}/165)$ minutes

= 40 minutes


Q-14)   A car driver leaves Bangalore at 8.30 A.M. and expects to reach a place 300 km from Bangalore at 12.30 P.M. At 10.30 he finds that he has covered only 40% of the distance. By how much he has to increase the speed of the car in order to keep up his schedule?

(a)

(b)

(c)

(d)

Explanation:

Distance covered by car in2hours

=${300 × 40}/100$ = 120km

Remaining distance

= 300 - 120 = 180 km

Remaining time= 4 - 2= 2 hours

Required speed

=$180/2$= 90kmph

Original speed of car

=$120/2$= 60kmph

Required increase in speed

= 90 - 60 = 30 kmph


Q-15)   A train is travelling at the rate of 45km/hr. How many seconds it will take to cover a distance of $4/5$ km ?

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,
Distance = Speed × Time
Speed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$
1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s

Time taken = $\text"Distance"/ \text"time"$

= ${4/5}/45$ hour

= ${4 × 60 × 60}/{5 × 45}$ sec. = 64 seconds


Q-16)   Three cars travelled distance in the ratio 1 : 2 : 3. If the ratio of the time of travel is 3 : 2 : 1, then the ratio of their speed is

(a)

(b)

(c)

(d)

Explanation:

Required ratio = $1/3 : 2/2 : 3/1$

= $1/3$ : 1 : 3

$1/3$ × 3 : 1 × 3 : 3 × 3

[Since, Speed = $\text"Distance"/\text"Time"$]

= 1 : 3 : 9


Q-17)   In a race of 200 metres, B can give a start of 10 metres to A, and C can give a start of 20 metres to B. The start that C can give to A, in the same race, is

(a)

(b)

(c)

(d)

Explanation:

According to the question,

Since, When B runs 200 m metres, A runs 190 metres

When B runs 180 metres, A runs

= $190/200 × 180$ = 171 metres

When C runs 200m, B runs 180 metres.

Hence, C will give a start to A by

= 200 - 171 = 29 metres


Q-18)   In a 100m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 Kmph, then the speed of Bimal is

(a)

(b)

(c)

(d)

Explanation:

Time taken by Kamal

= $100/{18 × 5/18}$ = 20 seconds

Time taken by Bimal

= 20 + 5 = 25 seconds

Bimal's speed

= $100/25$ =4 m/sec = ${4 × 18}/5$ kmph

= 14.4 kmph.


Q-19)   A car covers a certain distance in 25 hours. If it reduces the speed by $1/5$th, the car covers 200 km. less in that time. The speed of car is

(a)

(b)

(c)

(d)

Explanation:

Speed of car = x kmph.

Distance = Speed × Time = 25x km.

Case II,

Speed of car = ${4x}/5$ kmph.

Distance covered

= ${4x}/5 × 25$ = 20x km.

According to the question,

25x - 20x = 200

5x = 200

x = $200/5$ = 40 kmph.


Q-20)   A car covers four successive 7 km distances at speeds of 10 km/hour, 20 km/hour, 30 km/ hour and 60 km/hour respectively. Its average speed over this distance is

(a)

(b)

(c)

(d)

Explanation:

Total distance

= 7 × 4 = 28 km.

Total time

= $(7/10 + 7/20 + 7/30 + 7/60)$ hours

= $({42 + 21 + 14 + 7}/60)$ hours

= $84/60$ hours = $7/5$ hours

Average speed

= $\text"Total distance"/ \text"Total time" = (28/{7/5})$ kmph

= ${28 × 5}/7$ = 20 kmph