Unitary Method Test

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Unitary Method Test - 8

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Question 1
1 / -0

The speeds of two trains are in the ratio of 3:5. If the speed of the 1st train is 15 km/hr, then what is the speed of the second train?

A
25 kmph

B
30 kmph

C
2.5 kmph

D
100 kmph

Solution

Let the speed of the first train be $S_{1}=15 \mathrm{kmph}$
Let the speed of the second train be $S_{2}$
Ratio is given as $\frac{S_{1}}{S_{2}}=\frac{3}{5}$
$\frac{15}{S_{2}}=\frac{3}{5}$
$S_{2}=\frac{15 \times 5}{3}$
$S_{2}=25 \mathrm{kmph}$
Therefore speed of the second train is 25 kmph.
Question 2
1 / -0

Two stations A and B are 100 km apart on astraight line. One train starts from A at 7 a.m. and travels towards B at 10km/h. Another train starts from B at 8 a.m. and travels towards A at a speed of20 km/h. At what time will they meet?

A
9am

B
10am

C
11am

D
11:30am

Solution

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 10 x km.
Distance covered by B in (x - 1) hours = 20(x - 1) km.
$∴ 10 x + 20 (x - 1) = 100$
$\Rightarrow 30 x = 120$
$\Rightarrow x = 4$
So, they meet at 11 am
Therefore, option c is correct.
Question 3
1 / -0

Acar travels the first half of a certain distance with a speed of 20 km/hr, thenext half with a speed of 20 km/hr and the last half distance with a speed of 30km/hr The average speed of the car for the whole journey is

A
15km/hr

B
16km/hr

C
17km/hr

D
18 km/hr

Solution

Let the total distance travelled be $x$ km.
Then, total time taken = $\frac{\frac{x}{2}}{20}+\frac{\frac{x}{2}}{20}+\frac{\frac{x}{2}}{30}=\frac{x}{40}+\frac{x}{40}+\frac{x}{60}$
$=\frac{3 x+3 x+2 x}{120}=8 \frac{x}{120}=\frac{x}{15}$ hrs
Then average speed $=\frac{\text { Total distance travelled }}{\text { Total time taken }}$
$=\frac{x}{\frac{x}{15}} km/hr=15 km/hr$
Therefore, option a is correct.
Question 4
1 / -0

A car starts from Bengaluru, goes 100 km in astraight line towards south, immediately turns around and returns to Bengaluru.The time taken for this round trip is 4 hours. The magnitude of the averagevelocity of the car for this round trip

A
25 km/hr

B
30 km/hr

C
35 km/hr

D
40 km/hr

Solution

Velocity = displacement / time
Given time = 4 hr
Displacement = 100 km
Average velocity = total displacement / totaltime
Average velocity = 100/4 = 25km/hr
Therefore, option a is correct.
Question 5
1 / -0

A man car covers 14 km in 8 hours.His speed in m/min is:

A
29.16

B
30

C
30.5

D
23.5

Solution

14 km = 14 × 1000 = 14000 m
8 hours = 8 × 60 = 480 minutes
Speed = $\frac{14000}{480}$ = 29.16 m/min
Therefore, option a is correct.
Question 6
1 / -0

Two persons A and B are 400 m apart.If A can run at the speed of 5 m/sec and B can run at the speedof 3 m/sec; Find in how much time will they meet each other, if theyrun in the opposite directions?

A
35 sec

B
45 sec

C
55 sec

D
50 sec

Solution

Time = distance/speed
When two objects are moving in oppositedirections at x m/s and y m/s, then theirrelative speed = (x+y) m/s
Therefore, relative speed = 5+3 = 8 m/s
Distance = 400 m
⇒ time = 400/8 = 50 sec
Therefore, option d is correct.
Question 7
1 / -0

Speed of boat in still water is 4 km/h.The river is flowing with a speed of 3 km/h and time taken tocover a certain distance upstream is 2 h more than time taken tocover the same distance downstream. Find the distance.

A
2 km

B
2.1 km

C
2.3 km

D
2.5 km

Solution

Let distance be x km
Speed downstream = 4 + 3 = 7 km /hr.
Speed upstream = 4 - 3 = 1 km/hr.
So,
$\frac{x}{1}-\frac{x}{7}=2$
$7 x-x=2 \times 7$
$6 x=14$
$x=\frac{14}{6}=2.3 \mathrm{km} / \mathrm{h}$
Therefore, option c is correct.
Question 8
1 / -0

A car travels 72 km in 2 hrs.The distance that can be travelled by the car in 4 hr is

A
140 km

B
144 km

C
148 km

D
150 km

Solution

Speed of car = 72/2 km/hr
= 36 km/hr
Now, Distance travelled by Car = 36×4
= 144 km
Therefore, option b is correct.
Question 9
1 / -0

Ifa student walks from his house to school at 4 km/h, he is late by 35 minutes.However, if he walks at 5 km/h, he is late by 5 minutes only. The distance ofhis school from his house is:

A
10km

B
5km

C
15km

D
20km

Solution

Let therequired distance be x km.
Then,
$\frac{x}{4} - \frac{x}{5} = \frac{30}{60}$
$5 x - 4 x = \frac{30 \times 20}{60}$
x = 10 km
Therefore, option a is correct
Question 10
1 / -0

Excluding stoppages, the speed of a bus is 60km/h and including stoppages, it is 54 km/h. For how many minutes does the busstop per hour?

A
5 min

B
6 min

C
10 min

D
20min

Solution

Due to stoppages, it covers 6 km less [perhour.
Timetaken to cover 6 km = $(\frac{6}{60} \times 60)$ = 6 min
Therefore, option b is correct.
Question 11
1 / -0

A can run 1 km in 5 min 54 sec and B in 6 min.How many metres start can A can give B in a km race so that the race may end ina dead heat?

A
16.66 m

B
18.66 m

C
20.66 m

D
22 m

Solution

In the race, A will win by (6 min - 5min 54sec) = (360 – 354) = 6sec
B covers 1000 m in 6 min
B covers 1000 m in 360 sec
Distance covered by B in 6 sec = $(\frac{1000}{360} \times 6)$ = $\frac{50}{3} m$
Therefore, A beats B by 16.66 m.
For a dead-heat race, A must give B a start of16.66 m.
Therefore, option a is correct.
Question 12
1 / -0

A car travels a distance of 600 km at a uniformspeed. If the speed of the car is 10 kmph more, it takes three hours less tocover the same distance. The original speed of the car is

A
40 km/h

B
50 km/h

C
60 km/h

D
70 km/h

Solution

Let the original speed of the car be x kmph $⟹ \frac{600}{x} - \frac{600}{x + 10} = 3$
$\frac{600 x + 6000 - 600 x}{x (x + 10)} = 3$
6000 = 3x2 + 30x
= 3x² + 30x - 6000
= x²+ 10x – 2000
= x=40 km/h
Therefore, option a is correct.
Question 13
1 / -0

A farmertravelled a distance of 51 km in 10 hours. He travelled partly on foot at 5km/hr and partly on bicycle at 10 km/hr. The distance travelled on foot is

A
49km

B
47km

C
46km

D
45km

Solution

Let thedistance travelled on foot be x km.
Then, distancetravelled on bicycle = (51-x) km.
$So. \frac{x}{5} + \frac{(51 - x)}{10} = 10$
$\Rightarrow 2 x + (51 - x) = 10 \times 10$
$\Rightarrow x + 51 = 100$
→ x = 49 km
Therefore, option a is correct.
Question 14
1 / -0

In a 350 m race A beats Bby 17.5 m or 5 seconds. B's time over the course is:

A
100 sec

B
180 sec

C
120 sec

D
150 sec

Solution

B runs in 35/2 m in 5 sec.
Therefore,B covers 350 m in $(5 \times \frac{2}{35} \times 350) sec = 100 sec.$
Therefore, option a is correct.
Question 15
1 / -0

A train 150 m long, passes a telegraphpost in 15 sec. Find time taken by it to pass a platform 180 mlong.

A
33 sec

B
35 sec

C
36 sec

D
37 sec

Solution

We know,
Speed = distance / time
Distance = 150 m
Time = 15sec
Speed = 150/15 = 10 m/s
Distance of platform = 180 m
Total distance = 150 + 180 = 330 m
$time = \frac{distance}{speed}$
time $=\frac{330}{10}=33 \mathrm{sec}$
Therefore, option a is correct.

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Re-Attempt Weekly Quiz Competition

Unitary Method Test - 8

Result

Unitary Method Test - 8

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SHARING IS CARING

Question 1

25 kmph

30 kmph

2.5 kmph

100 kmph

Solution

Question 2

9am

10am

11am

11:30am

Solution

Question 3

15km/hr

16km/hr

17km/hr

18 km/hr

Solution

Question 4

25 km/hr

30 km/hr

35 km/hr

40 km/hr

Solution

Question 5

29.16

30

30.5

23.5

Solution

Question 6

35 sec

45 sec

55 sec

50 sec

Solution

Question 7

2 km

2.1 km

2.3 km

2.5 km

Solution

Question 8

140 km

144 km

148 km

150 km

Solution

Question 9

10km

5km

15km

20km

Solution

Question 10

5 min

6 min

10 min

20min

Solution

Question 11

16.66 m

18.66 m

20.66 m

22 m

Solution

Question 12

40 km/h

50 km/h

60 km/h

70 km/h

Solution