Unitary Method Test

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Unitary Method Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

Renu can row her boat at the speed of 4 km/hrin still water. If it takes one hour more to row the boat 4.8 km upstreamthan to return downstream. Find the speed of the stream.

A
9 km/hr

B
11 km/hr

C
10 km/hr

D
5 km/hr

Solution

$\left(\frac{4.8}{4-x}\right)=\left(\frac{4.8}{4+x}\right)+1$
$\left(\frac{4.8(4+x)-4.8(4-x)}{16-x^{2}}\right)=1$
or $9.6 x=16-x^{2}$
or $x^{2}-9.6 x-16=0$
$\therefore x=11 \mathrm{km} / \mathrm{hr}$
Therefore, option b is correct.
Question 2
1 / -0

Two person’s P and Q start at the same time fromcity A for city B, 80 km away. P travels 6 km/h slower than Q. Q reaches city Band at once turns back meeting P, 14 km from city B. What is speed of P?

A
14 km/h

B
14.1 km/h

C
14.5 km/h

D
15 km/h

Solution

AB = 80 km, BC = 14 km,
AC = (80 – 14) km = 66km
Let the speed of P be x kmph so that speed ofQ is (x+6) kmph. Time for P to cover a distance AC and time for Q tocover (AB+BC) are equal.
$∴ \frac{94}{x + 6} = \frac{66}{x}$
94 x = 66 x + 396
28 x = 396
x = 14.1 km/h

Therefore, option b is correct.
Question 3
1 / -0

Two stations A and B are 170 km apart on astraight line. One train starts from A at 6 a.m. and travels towards B at 30km/h. Another train starts from B at 7 a.m. and travels towards A at a speed of40 km/h. At what time will they meet?

A
6 am

B
7 am

C
9 am

D
10 am

Solution

Suppose they meet x hours after 6 a.m.
Distance covered by A in x hours = 30 x km.
Distance covered by B in (x - 1) hours = 40(x- 1) km.
= 30x + 40(x-1) = 170
= 70x = 210
= x = 3
So, they meet at 9 am
Therefore,option c is correct.
Question 4
1 / -0

Out of a total distance covered of 300 km,first half is covered at the speed of 50 km / hr and the remaininghalf is covered at 25 km/hr. Find the total time taken to cover thewhole journey.

A
3 hours

B
4 hours

C
5 hours

D
6 hours

Solution

Time taken to cover first 100 Km at50km/h = 100/50 = 2
Time taken to cover last 100 Km at 25km/h = 100/25 = 4
Total time taken to cover 300 Km = 2 + 4 = 6 hrs
Therefore, option d is correct.
Question 5
1 / -0

Two trains start at the same time from A andB and proceed toward each other at the speed of 60 km hr and 40 km/hrrespectively. When both meet at a point in between, one train was found to havetravelled 125 km more than the other. Find the distance between A andB.

A
625 km

B
650 km

C
700 km

D
800 km

Solution

Let the distance travelled by slower train bex km. Then distance travelled by faster train is (x+125) km.
$\frac{s 1}{s 2} = \frac{d 1}{d 2}$ (time is constant)
$\frac{60}{40} = \frac{x + 125}{x} \Rightarrow \frac{3}{2} = \frac{x + 125}{x}$
3X = 2X × 250
= X = 250
Total distance = x + x + 125
= 2x + 125 = 2 × 250 + 125 = 500 + 125 km
= 625 km
Therefore, option a is correct.
Question 6
1 / -0

A train travels 102.5 km/hr. How many meterswill it travel in 20 minutes?

A
34160

B
3416

C
416

D
45721

Solution

Speed of train = 102.5 km/h
Train travel in 20 minutes 102.5 × 20/60
= 34.16 km
⇒34.16 km = 34.16 × 1000m = 34160 m
Therefore, option a is correct.
Question 7
1 / -0

The speed of a boat in still water in 20 km/hrand the rate of current is 5 km/hr. The distance travelled downstream in 15minutes is:

A
4.56 km

B
3.56 km

C
5.25 km

D
6.25 km

Solution

The speed of a boat in still water = 20 km/hr
Rate of current = 5 km/hr = 5 km/hr
Speed in downstream = 20 + 5 = 25 km/hr
Time given = 15min = 15/60hr = 1/4hr
Distance travelled = Speed ×Time = =6.25 km
Therefore, option d is correct.
Question 8
1 / -0

What is the distance travelled at the speed of50 km/hr for $\frac{1}{5}$ hours?

A
10km

B
15km

C
25km

D
20km

Solution

Distance = Speed × Time
$50 \times \frac{1}{5} = 50 \times \frac{1}{5} = 10 k m$
Therefore, option a is correct.
Question 9
1 / -0

Man makes a trip by automobile at an average speed of 80 km/hr. He returns over the same route at an average speed of 40 km/hr. His average speed for the entire trip is:

A
45

B
43.33

C
46.33

D
53.33

Solution

Let distance ne x
Speed in 1st trip = 80km/hr
Time taken x/80hrs.
speed in return trip = 40 km/hr
Timetaken = x/40hrs.
Avg.speed =Total Distance / Total time
$= \frac{x + x}{\frac{x}{80} + \frac{x}{40}} = \frac{2 x \times 80}{x + 2 x}$
= $\frac{2 \times 80}{3} = 53.33$
Therefore, option d is correct.
Question 10
1 / -0

Simran drove 60 miles to see hercousin and at 40 mph. The trip took Simran 1 hour. Then,Simran drove from her cousin’s house and drove another 50 milesto the store at a speed of 20 mph. It took Simran 4 hoursto arrive at the store. What was Simran’s average speed for the trip?

A
20 mph

B
22 mph

C
24 mph

D
28 mph

Solution

$a v e r a g e s p e e d o f s i m r a n = \frac{t o t a l d i s \tan c e t r a v e l l e d}{t o t a l t i m e t a k e n}$
= $\frac{(60 + 50) m i l e s}{(1 + 4) h r s} = \frac{110}{5} m p h = 22 m p h$
Therefore, option b is correct.
Question 11
1 / -0

A train 300 m long is running at a speed of 54km/hr. In what time will it pass a bridge 150 m long?

A
20 sec

B
30 sec

C
40 sec

D
50 sec

Solution

Formula for converting from km/hr tom/s= X km/hr = $(x \times \frac{5}{18}) m / s$
Therefore,
Speed= $(54 \times \frac{5}{18}) m / s e c$ =16m/sec
Total distance to be covered =(300 + 150)m =450 m
Formula for finding time =(Distance/Speed)
∴ RequiredTime = $\frac{450}{15} s e c$ = 30 sec
Therefore, option b is correct.
Question 12
1 / -0

A car travelling at a speed of 60 km/ hour cancomplete a journey in 10 hours. How long will it take to travel the samedistance at 80 km/hour?

A
7 hours

B
7.5 hours

C
8 hours

D
8.5 hours

Solution

Distance=Speed × Time
When speed is 60 km/h
Distance travelled by car = 60 × 10
= 600 km
When speed is 80 km/h
Time taken to complete the journey =600/80
=7.5 hours
Therefore, option b is correct.
Question 13
1 / -0

A man covers a certain distance between hishouse and office on a scooter. Having an average speed of $20 k m / h r$ , he reachesoffice late by $15 m i n u t e s$ . However, with a speed of $50 k m / h r$ , he reaches hisoffice $10 m i n u t e s$ earlier. The distance between his house and office is-

A
$2.7 k m$

B
$2.5 k m$

C
$2 k m$

D
$13.88 k m$

Solution

When speed is 20 km/hr he reaches 15 minutes late and when speed is 50 km/hr he reaches 10 minutes earlier.
Let the distance travelled be d,
Time in first case $- 15 m i n u t e s$ = Time in second case $+ 10 m i n u t e s$
$\frac{d}{20} - \frac{15}{60} = \frac{d}{50} + \frac{10}{60}$
$\frac{d}{20} - \frac{d}{50} = \frac{10}{60} + \frac{15}{60}$
$\frac{3 d}{100} = \frac{5}{12}$
$d = 13.88 k m$
Hence, the correct option is (D).
Question 14
1 / -0

It takes 10 seconds for a train travelling at 50km/hr to cross entirely another train half its length and travelling inopposite direction at 42 km/hr. It also passes a bridge in 41 seconds. Thelength of the bridge is:

A
500 m

B
502 m

C
502.48 m

D
510.45 m

Solution

Let length of the train is X and length of thebridge is Y.
Speed of the First train = 50 km = (50 × 5)/18= 13.88 m/sec
Trains are moving in opposite direction, so theirrelative speed would be,
S = 50 + 42 = 92 km/h = (92 × 5)/18 = 25.55m/sec
train crosses another train having half of itslength in 10 sec.
So, Train covered (X + X/2) = 3X/2 m in 10sec.
Relative Speed = 3X/ (2 × 10)
3X/20 = 20/3 m/sec
X = 6.66 m/sec = 66.6 m
Length of the train = 66.6 m
Train crosses bridge in 41 sec. That is
Train needs to cover (66.6 + Y) m in 41 sec.
Speed of the train = (66.6 + Y)/41 m/sec
13.88 = (66.6 + Y)/41
569.08 = 66.6 + Y
Y =569.08 – 66.6 = 502.48 m
Length of the bridge = 502.48 m.
Therefore, option c is correct.
Question 15
1 / -0

When a person leaves his home for picnic by hiscar, the meter reads 14250 km. When he returns home after two hoursthe reading is 14315 km, during journey he stays for 20 minutes atmidway. What is the average speed of the car during this period?

A
39 km/hr

B
40 km/hr

C
49 km/hr

D
50 km/hr

Solution

$T o t a l t r a v e l = = 2 \times 60 - 20 = 100 m i n$
$= \frac{100}{60} h r s$
$A v e r a g e s p e e d = \frac{14315 - 14250}{\frac{100}{60}}$ = $\frac{65 \times 60}{105}$ =39 km/hr
Therefore, option a is correct.

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Unitary Method Test - 4

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Unitary Method Test - 4

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Question 1

9 km/hr

11 km/hr

10 km/hr

5 km/hr

Solution

Question 2

14 km/h

14.1 km/h

14.5 km/h

15 km/h

Solution

Question 3

6 am

7 am

9 am

10 am

Solution

Question 4

3 hours

4 hours

5 hours

6 hours

Solution

Question 5

625 km

650 km

700 km

800 km

Solution

Question 6

34160

3416

416

45721

Solution

Question 7

4.56 km

3.56 km

5.25 km

6.25 km

Solution

Question 8

10km

15km

25km

20km

Solution

Question 9

45

43.33

46.33

53.33

Solution

Question 10

20 mph

22 mph

24 mph

28 mph

Solution

Question 11

20 sec

30 sec

40 sec

50 sec

Solution

Question 12

7 hours

7.5 hours

8 hours

8.5 hours

Solution

$2.7 k m$

$2.5 k m$

$2 k m$

$13.88 k m$