Explanation / Important formulas:
Important formulas:
- km/hr to m/s conversion:
a km/hr = a x 5/18 m/s.
- m/s to km/hr conversion:
a m/s = a x 18/5 km/hr.
Formulas for finding Speed, Time and Distance:
- Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.
- Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.
- Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u – v) m/s.
- Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
- If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:
The time taken by the trains to cross each other = (a + b) / (u + v) sec.
- If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:
The time taken by the faster train to cross the slower train = (a + b) / (u -v) sec.
- If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:
(A’s speed) : (B’s speed) = (b : a)
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Time, Work and Distance - Test
Time, Work and Distance - Question and Answers
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Question 1
|
A
|
120 meters
|
B
|
180 meters
|
C
|
324 meters
|
D
|
150 meters
|
Length of the train = (Speed * Time)
Length of the train = ((50/3) * 9) m = 150 m
Question 2
|
A
|
45 km/hr
|
B
|
54 km/hr
|
C
|
50 km/hr
|
D
|
55 km/hr
|
=> 25/2 m/sec
=> 25/2 * 18/5 km/hr
=> 45 km/hr
Let the speed of the train be x km/hr
Then, relative speed = (x - 5) km/hr
=> x - 5 = 45
=> x = 50 km/hr
Question 3
|
A
|
8/15
|
B
|
7/15
|
C
|
11/15
|
D
|
2/11
|
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 * (7/60) = 7/15
Fraction of work left = 1 – 7/15 = 8/15
Question 4
|
A
|
9 (3/5) days
|
B
|
9 (1/5) days
|
C
|
9 (2/5) days
|
D
|
10 days
|
Amount of work Q can do in 1 day = 1/12
Amount of work P, Q and R can together do in 1 day = 1/4
Amount of work R can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48
=> Hence R can do the job on 48/5 days = 9 (3/5) days
Question 5
|
A
|
10 days
|
B
|
14 days
|
C
|
15 days
|
D
|
9 days
|
Amount of work Q can do in 1 day = 1/30
Amount of work R can do in 1 day = 1/60
P is working alone and every third day Q and R is helping him
Work completed in every three days = 2 * (1/20) + (1/20 + 1/30 + 1/60) = 1/5
So work completed in 15 days = 5 * 1/5 = 1
I.e, the work will be done in 15 days
Question 6
|
A
|
18 days
|
B
|
22 1/2 days
|
C
|
24 days
|
D
|
26 days
|
Hence, if the difference is 2 days, B can complete the work in 3 days
=> if the difference is 60 days, B can complete the work in 90 days
=> Amount of work B can do in 1 day = 1/90
Amount of work A can do in 1 day = 3 * (1/90) = 1/30
Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45
=> A and B together can do the work in 45/2 days = 22 ½ days
Question 7
|
A
|
380
|
B
|
600
|
C
|
420
|
D
|
400
|
Amount of work B can do in 1 day = 1/8
Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24
Amount of work A + B + C can do = 1/3
Amount of work C can do in 1 day = 1/3 - 7/24 = 1/24
Work A can do in 1 day : work B can do in 1 day : work C can do in 1 day
= 1/6 : 1/8 : 1/24 = 4 : 3 : 1
Amount to be paid to C = 3200 * (1/8) = 400
Question 8
|
A
|
4 days
|
B
|
6 days
|
C
|
2 days
|
D
|
8 days
|
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1 --- (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1--- (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
=> 15/100 + 20/200 = 1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days
Question 9
|
A
|
12 hours
|
B
|
6 hours
|
C
|
8 hours
|
D
|
10 hours
|
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = 1/4 + 1/3 = 7/12
Work done by B in 1 hour = 7/12 – 1/2 = 1/12
=> B alone can complete the work in 12 hours
Question 10
|
A
|
30 days
|
B
|
25 days
|
C
|
20 days
|
D
|
15 days
|
Work done by R in 1 day = 1/50
Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day
Hence the above equation can be written as
Work done by P in 1 day * 2 = 6/50
=> Work done by P in 1 day = 3/50
=> Work done by Q and R in 1 day = 3/50
Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25
So Q alone can do the work in 25 days
Question 11
|
A
|
37 1/2 days
|
B
|
22 days
|
C
|
31 days
|
D
|
27 days
|
Work done by A in 1 day = (4/5)/20 = 4/100 = 1/25 --- (1)
Work done by A and B in 3 days = 20/100 = 1/5
(Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 ---(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 ½ days
Question 12
|
A
|
3 PM
|
B
|
1PM
|
C
|
11 AM
|
D
|
2 PM
|
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating
i.e, they all operated for 2 hours and work completed = 2 * (37/120) = 37/60
Pending work = 1 - 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; i.e. around 1 pm
Question 13
|
A
|
8
|
B
|
5
|
C
|
4
|
D
|
6
|
Work done by Q in 1 day = 1/15
Work done by Q in 10 days = 10/15 = 2/3
Remaining work = 1 – 2/3 = 1/3
Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6
Question 14
|
A
|
50
|
B
|
40
|
C
|
30
|
D
|
20
|
Work done by 3 men and 7 women in 1 day = 1/10
Let 1 man does m work in 1 day and 1 woman does w work in 1 day
The above equations can be written as
4m + 6w = 1/8 ---(1)
3m + 7w = 1/10 ---(2)
Solving equation (1) and (2) , we get m=11/400 and w=1/400
Amount of work 10 women can do in a day = 10 * (1/400) = 1/40
i.e, 10 women can complete the work in 40 days
Question 15
|
A
|
60
|
B
|
50
|
C
|
40
|
D
|
30
|
Amount of work done by A and B in 20 days = 20 *(1/30) = 20/30 = 2/3
Remaining work – 1– 2/3 = 1/3
A completes 1/3 work in 20 days
Amount of work A can do in 1 day = (1/3)/20 = 1/60
=> A can complete the work in 60 days
Question 16
|
A
|
5 5/11
|
B
|
4 5/11
|
C
|
6 4/11
|
D
|
6 5/11
|
=> Number of hours A can complete the work = 12*8 = 96 hours
=> Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day
=> Number of hours B can complete the work = 8*10 = 80 hours
=> Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480
=> A and B can complete the work in 480/11 hours
A and B works 8 hours a day
Hence total days to complete the work with A and B working together
= (480/11)/ (8) = 60/11 days = 5 5 ⁄ 11 days
Question 17
|
A
|
9
|
B
|
11
|
C
|
13
|
D
|
15
|
Let work done by Q in 1 day = q
q * (130/100) = 1/23
=> q = 100 / (23*130) = 10 / (23*13)
Work done by P and Q in 1 day = 1/23 + 10 / (23*13) = 23 / (23*13) = 1/13
=> P and Q together can do the work in 13 days
Question 18
|
A
|
2
|
B
|
3 3/7
|
C
|
4 1/4
|
D
|
5
|
Work done by Q in 1 day = 1/6
Work done by R in 1 day = 1/12
Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24
=> Working together, they will complete the work in 24/7 days = 3 3⁄7 days
Question 19
|
A
|
5
|
B
|
6
|
C
|
7
|
D
|
8
|
Work done by 1 man in 1 day = (1/7)/10 = 1/70
Work done by 10 women in 1 day = 1/14
Work done by 1 woman in 1 day = 1/140
Work done by 5 men and 10 women in 1 day = 5 × (1/70) + 10 × (1/140)
= 5/70 + 10/140 = 1/7
=> 5 men and 10 women can complete the work in 7 days
Question 20
|
A
|
14
|
B
|
16
|
C
|
18
|
D
|
20
|
Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16
=> Suresh can complete the work in 16 days
Question 21
|
A
|
7 hour 15 minutes
|
B
|
7 hour 30 minutes
|
C
|
8 hour 15 minutes
|
D
|
8 hour 30 minutes
|
Pages typed by Suresh in 1 hour = 40/5 = 8
Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3
Time taken to type 110 pages when Anil and Suresh work together = 110 * 3 /40 = 33/4
= 8 ¼ hours = 8 hour 15 minutes
Question 22
|
A
|
5 days
|
B
|
10 days
|
C
|
14 days
|
D
|
22 days
|
Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5
Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15
Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6
Total days = 4 + 6 = 10
Question 23
|
A
|
4
|
B
|
5
|
C
|
6
|
D
|
7
|
Then Q takes x/2 days and R takes x/3 days to finish the work
Amount of work P does in 1 day = 1/x
Amount of work Q does in 1 day = 2/x
Amount of work R does in 1 day = 3/x
Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x
6/x = 2
=> x = 12
=> Q takes 12/2 days = 6 days to complete the work
Question 24
|
A
|
12
|
B
|
16
|
C
|
20
|
D
|
24
|
Work done by Q in 1 day = 1/10
Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6
Work done by P and Q in 2 days = 2 * (1/6) = 1/3
Remaining work = 1 – 1/3 = 2/3
Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days
Total time taken = 2 + 10 = 12 days
Question 25
|
A
|
10
|
B
|
15
|
C
|
18
|
D
|
22
|
Work done by Q in 1 day = q
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 10 days = 10 × (1/16) = 10/16 = 5/8
Remaining work = 1 = 5/8 = 3/8
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (3/8) / (1/48) = 18
Question 26
|
P and Q can finish the work in how many days?
A
|
1
|
B
|
2
|
C
|
3
|
D
|
4
|
Work done by P in 1 day = 2 * (1/12) = 1/6
Work done by P and Q in 1 day = 1/12 + 1/6 = ¼
=> P and Q can finish the work in 4 days
Question 27
|
A
|
1:3
|
B
|
4:3
|
C
|
2:3
|
D
|
2:1
|
Work done by 1 woman in 1 day = 1/(16*20)
Work done by 16 men in 1 day = 1/15
Work done by 1 man in 1 day = 1/(15*16)
Ratio of the capacity of a man and woman =1/(15*16) : 1/(16*20) = 1/15 : 1/20
= 1/3 :1/4 = 4:3
Question 28
|
A
|
3
|
B
|
8
|
C
|
12
|
D
|
4
|
work done by Q in 1 day = q
Work done by R in 1 day = r
p + q = 1/8 ---(1)
q + r = 1/12 ---(2)
p + q + r = 1/6 ---(3)
(3) – (2) => p = 1/6 - 1/12 = 1/12
(3) – (1) => r = 1/6 – 1/8 = 1/24
p + r = 1/12 + 1/24 = 3/24 = 1/8
=> P and R will finish the work in 8 days
Question 29
|
A
|
7
|
B
|
8
|
C
|
9
|
D
|
10
|
Work done by Q in 1 day = 1/9
Work done by R in 1 day = 1/12
Work done by Q and R in 1 day = 1/9 + 1/12 = 7/36
Work done by Q and R in 3 days = 3*7/36 = 7/12
Remaining work = 1 – 7/12 = 5/12
Number of days in which P can finish the remaining work = (5/12) / (1/24) = 10
Question 30
|
A
|
12
|
B
|
14
|
C
|
16
|
D
|
18
|
Wages of 1 man for 1 day = 21600×240*30
Wages of 1 man for 25 days = 21600*2*2540*30
Number of men = 14400(21600*2*2540*30) = 144(216*5040*30) = 1449 = 16
Question 31
|
A
|
Rs. 40
|
B
|
Rs. 70
|
C
|
Rs. 90
|
D
|
Rs. 100
|
Amount Earned by P and R in 1 day = 600/5 = 120 ---(2)
Amount Earned by Q and R in 1 day = 910/7 = 130 ---(3)
(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day
Amount Earned by P,Q and R in 1 day = 120+130-180 = 70
=> Amount Earned by R in 1 day = 70
Question 32
|
A
|
Rs. 1104
|
B
|
Rs. 1000
|
C
|
Rs. 934
|
D
|
Rs. 1210
|
cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 ---(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c = 23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56*23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12*92 = 1104
Question 33
|
A
|
3 1⁄4 days
|
B
|
4 1⁄3 days
|
C
|
5 1⁄6 days
|
D
|
6 1⁄5 days
|
Work completed in 2nd day = (1/16) + (1/16) = 2/16
Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16
An easy way to attack such problems is from the choices. You can see the choices are
very close to each other. So just see one by one
For instance, The first choice given in 3 1⁄4
The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16
The work done in 4 days = (1+2+3+4)/16 = 10/16
The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn't it?
The work done in 6 days = (1+2+3+4+5+6)/16 > 1
Hence the answer is less than 6, but greater than 5
Hence the answer is 5 1⁄6 days
(Just for your reference, work done in 5 days = 15/16
Pending work in 6th day = 1 – 15/16 = 1/16
In 6th day, 6 people are working and work done = 6/16
To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days
Hence total time required = 5 + 1/6 = 5 1⁄6 days
Question 34
|
A
|
25 km
|
B
|
10 km
|
C
|
50 km
|
D
|
30 km
|
= (20+15) * 1/2
=25
Question 35
|
A
|
11 hrs
|
B
|
8 hrs 45 min
|
C
|
7 hrs 45 min
|
D
|
9 hrs 20 min
|
From this, we can understand that time needed for
riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
Question 36
|
A
|
12
|
B
|
11
|
C
|
10
|
D
|
9
|
Time taken to cover 9 km = (9/54 * 60) min = 10 min
Question 37
|
A
|
121
|
B
|
242
|
C
|
224
|
D
|
112
|
x/21 + x/24 = 20
15x = 168 * 20
x = (168 * 20 / 15) = 224 km
Question 38
|
A
|
30 km/hr
|
B
|
35 km/hr
|
C
|
25 km/hr
|
D
|
40 km/hr
|
Let the actual speed be x km/hr
Then, 5/7x * 126/75 = 42
x = (42 * 7 * 75/5 * 126) = 35 km/hr
Question 39
|
A
|
36
|
B
|
38
|
C
|
40
|
D
|
42
|
Then, x/y – x /(y+3) = 40/60 => 2y(y+3) = 9x ….(1)
And, x/(y-2) – x/y = 40/60 => y(y-2) = 3x ………(2)
On dividing (1) by (2), we get x = 40
Question 40
|
A
|
5
|
B
|
6
|
C
|
7
|
D
|
8
|
= 2 + 3 = 5 rounds per hour
=> They cross each other 5 times in 1 hour and 2 times in 1/2 hour
Time duration from 8 am to 9.30 am = 1.5 hour
Hence they cross each other 7 times before 9.30 am
Question 41
|
A
|
17 hr
|
B
|
14 hr
|
C
|
12 hr
|
D
|
19 hr
|
Distance = 8.5 km
Time = Distance * Speed = 8.5 * 0.5 = 17 hr
Question 42
|
A
|
8 kmph
|
B
|
5 kmph
|
C
|
4 kmph
|
D
|
7 kmph
|
Then, 30/x – 30/2x = 3
6x = 30
x = 5 km/hr
Question 43
|
A
|
70.24 km/hr
|
B
|
74. 24 km/hr
|
C
|
71.11 km/hr
|
D
|
72.21 km/hr
|
So, total time taken = (160/64 + 160/80) = 9/2 hours
Time taken for 320 km = 320 * 2/9 = 71.11
= 71.11 km/hr
Question 44
|
A
|
12 km
|
B
|
14 km
|
C
|
16 km
|
D
|
18 km
|
Then the distance he travelled on bicycle = (61-x) km
So, x/4 + (61−x)/9 = 9
9x + 4(61−x) = 9 * 36
5x= 80
x = 16 km
Question 45
|
A
|
1 hr 42 min
|
B
|
1 hr
|
C
|
2 hr
|
D
|
1 hr 12 min
|
Speed and time are inversely proportional
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time - usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 * 6 = 72 minutes
=> 1 hour 12 minutes
Question 46
|
A
|
3 km
|
B
|
4 km
|
C
|
5 km
|
D
|
6 km
|
An Equal distance at y kmph, the average speed of the whole journey = 2xy/x+y kmph
Hence, average speed = (2*3*2) /2+3 = 12/5 km/hr
Total time taken = 5 hours
⇒ Distance travelled = 12/5*5 = 12 km
⇒ Distance between his house and office = 12/2 = 6 km
Question 47
|
A
|
11.2 kmph
|
B
|
10 kmph
|
C
|
10.2 kmph
|
D
|
10.8 kmph
|
Time taken to travel 10 km at an average speed of 12 km/hr = distance * speed = 10/12 hr
Time taken to travel 12 km at an average speed of 10 km/hr = distance * speed = 12/10 hr
Total time taken = [10/12 + 12/10] hr = 61/30 hrs
Average speed = [22*30/61] km/hr = 10.8 km/hr
Question 48
|
A
|
660 km/hr
|
B
|
680 km/hr
|
C
|
700 km/hr
|
D
|
720 km/hr
|
Speed = Distance/Time
Speed = 1200 / (5/3) km/hr
Therefore required speed = (1200 * 3/5) km/hr
= 720 km/hr
Question 49
|
A
|
80 kmph
|
B
|
102 kmph
|
C
|
120 kmph
|
D
|
140 kmph
|
Then, speed of the train = 150/100x = (3/2x) kmph
75/x - 75/(3/2)x = 125/10 * 60
75/x – 50/x = 5/24
x = (25 * 24 / 5) = 120 kmph
Question 50
|
A
|
2 hour
|
B
|
112 hour
|
C
|
12 hour
|
D
|
1 hour
|
Then, 600/x – 600/x + (1/2) = 200
600/x – 1200/2x + 1 = 200
x(2x + 1) = 3
2*2 + x - 3 = 0
(2x + 3)(x - 1) = 0
x = 1 hr [neglecting the -ve value of x]
Question 51
|
A
|
80 km
|
B
|
70 km
|
C
|
60 km
|
D
|
50 km
|
Then, x / 10 = x+20 / 14
14x = 10x + 200
4x = 200
x = 50 km
Question 52
|
A
|
85 km/hr
|
B
|
87.5 km/hr
|
C
|
90 km/hr
|
D
|
92.5 km/hr
|
Then, 8x = 400/4) = 100
x = (100/8) = 12.5
Speed of first train = (7 * 12.5) km/hr = 87.5 km/hr
Question 53
|
A
|
210 miles
|
B
|
230 miles
|
C
|
250 miles
|
D
|
260 miles
|
Time1 = 212 hour = 52 Hour
Distance1 = Speed1 × Time1
=> 50 × 52 = 25 × 5 = 125 Miles
Speed2 = 70 miles/hour
Time2 = 112 hour = 32 hour
Distance2 = Speed2 × Time2 = 70*32 = 35*3 = 105 miles
Total Distance = Distance1 + Distance2
= 125 + 105 = 230 miles
Question 54
|
A
|
422 km
|
B
|
552 km
|
C
|
502 km
|
D
|
492 km
|
And speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km
Hence distance travelled in 3rd hour = 39 km
Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)]
This is an Arithmetic Progression(AP) with
first term, a = 35, number of terms ,n = 12 and common difference, d = 2
The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . .
. is called an Arithmetic Progression(AP)where a is the first term and d is the common difference of the AP
Sum of the first n terms of an Arithmetic Progression(AP),
Sn=n2[2a+(n−1)d]where n = number of terms
Hence, [35+37+39+... (12 terms)] = S12 = 122[2*35+(12−1)2] = 6[70+22] = 6 * 92 = 552
Hence the total distance travelled = 552 km
Question 55
|
A
|
1800 ft
|
B
|
2810 ft
|
C
|
3020 ft
|
D
|
2420 ft
|
Time = 11/5 second
Distance = Speed * Time = 1100 *115 = 220*11 = 2420 ft
Question 56
|
A
|
10.30 a.m
|
B
|
10 a.m.
|
C
|
12 noon
|
D
|
11 a.m.
|
Then, train1, starting from A , travelling towards B, travels x hours till the trains meet
⇒ Distance travelled by train1 in x hours = Speed *Time = 60x
Then, train2, starting from B , travelling towards A, travels (x-1) hours till the trains meet
⇒ Distance travelled by train2 in (x-1) hours = Speed *Time = 75(x-1)
Total distance travelled = Distance travelled by train1 + Distance travelled by train2
=> 330 = 60x + 75(x-1)
=> 12x + 15(x-1) = 66
=> 12x + 15x - 15 = 66
=> 27x = 66 + 15 = 81
=> 3x = 9
=> x = 3
Hence the trains meet 3 hours after 8 a.m., i.e. at 11 a.m
Question 57
|
A
|
1250
|
B
|
1280
|
C
|
1320
|
D
|
1340
|
So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60
Speed = 5*5/18 = 25/18 m/sec
Time = 15 * 60 seconds = 900 seconds
Distance = Time * Speed
Distance = (25 / 18) * 900 = 1250 meter
Question 58
|
A
|
6 hrs 21 min
|
B
|
7 hrs 14 min
|
C
|
7 hrs 22 min
|
D
|
6 hrs
|
With this speed, train can cover 600 km in 6 hours but train used to stop for 3 minutes after every 75 km
So, Train will stop = 600/75 = 8 times in the whole journey
But, you need to understand that after 7th stopping train will be at the destiny in next stop. Therefore, train will stop 7 times in journey
So, extra time = 7 *3 = 21 minutes
Total time taken in the journey = 6 hour 21 minutes
Question 59
|
A
|
60 km/hr
|
B
|
56 km/hr
|
C
|
52 km/hr
|
D
|
48 km/hr
|
Speed while returning = 150% 0f 40 = 60 km/hr
Average speed = 2xy/x+y
= (2*40*60/40+60)
= 4800/100 = 48 km/hr
Question 60
|
A
|
3.6
|
B
|
7.2
|
C
|
8.4
|
D
|
10
|
(Converting m/sec to km/hr (see important formulas section)
= [2* (18 / 5)] km/hr
= 7.2 km/hr
Question 61
|
A
|
50 km
|
B
|
80 km
|
C
|
70 km
|
D
|
90 km
|
x / 10 = x + 20/14
14x = 10x + 200
4x = 200
x = 50 km
Question 62
|
A
|
3 : 4
|
B
|
2 : 3
|
C
|
1 : 2
|
D
|
1 : 3
|
Then, 120/x + 480/y = 8 => 1/x + 4/y = 1/15 …..(1)
And, 200/x + 400/y = 25/3 => 1/x +2/y =1/24 …..(2)
Solving (1) and (2), we get x = 60 and y = 80
Ratio of speeds = 60 : 80 = 3 : 4
Question 63
|
A
|
8 kmph
|
B
|
10 kmph
|
C
|
12 kmph
|
D
|
14 kmph
|
Then, x/10 – x/15 = 2
= 3x - 2x = 60 => x = 60 km
Time taken to travel 60 km at 10 km/hr = (60/10)hrs = 6 hrs
So, Arun started 6 hours before 2 P.M i.e.., at 8 A.M
Therefore Required speed = (60/5)kmph = 12 kmph
Question 64
|
A
|
20 km/hr
|
B
|
25 km/hr
|
C
|
27.5 km/hr
|
D
|
30 km/hr
|
=> 200/24*185 km/hr
=> 40*34 km/hr = 10*3 km/hr = 30 km/hr
Question 65
|
A
|
2229 m/s
|
B
|
22 m/s
|
C
|
2119 m/sec
|
D
|
21 m/s
|
=> 40×59 m/s = 2009 m/s
=> 2229 m/s
Question 66
|
A
|
61 km/hr
|
B
|
56 km/hr
|
C
|
63 km/hr
|
D
|
60 km/hr
|
Distance between them = 50m
Total distance = 20*50m = 1000m = 1 km
Time taken is = 1min = 1/60 hr
Speed = Distance / time
Speed = 1/(1/60) = 1*60 = 60 kmph
Question 67
|
A
|
2 : 1
|
B
|
1 : 2
|
C
|
4 : 3
|
D
|
3 : 4
|
= 33 : 44 = 3 : 4
Question 68
|
A
|
14 km/hr
|
B
|
12 km/hr
|
C
|
10 km/hr
|
D
|
8 km/hr
|
Given that he covers one-half of the distance in two-thirds of the total time
He covers half of 6 km in two-thirds of 45 minutes
He covers 3 km in 30 minutes
Hence, now he need to cover the remaining 3 km in the remaining 15 minutes
Distance = 3 km
Time = 15 minutes = 1/4 hour
Required Speed = Distance/Time = 3/(1/4) = 12 km/hr
Question 69
|
A
|
300 kmph
|
B
|
360 kmph
|
C
|
600 kmph
|
D
|
720 kmph
|
Speed = Distance/Time
Speed = 1200 / (5/3) km/hr [We can write 1 2/3 hours as 5/3 hours]
Therefore required speed = (1200 * 3/5) km/hr = 720 km/hr
Question 70
|
A
|
8 kmph
|
B
|
11 kmph
|
C
|
12 kmph
|
D
|
14 kmph
|
Then, x/10 – x/15 = 2
3x – 2x = 60
x = 60 km
Time taken to travel 60 km at 10 km/hr = (60/10)hrs = 6 hrs
So, Robert started 6 hours before 2 P.M
i.e., at 8 A.M
Therefore, Required speed = (60/5) kmph = 12 kmph
Question 71
|
A
|
14 km
|
B
|
15 km
|
C
|
16 km
|
D
|
17 km
|
Then, distance travelled on bicycle = (61 -x) km
So, x / 4 + (61-x) / 9 = 9
9x + 4(61-x) = 9 * 36
5x = 80
x = 16 km
Question 72
|
A
|
35.55 km/hr
|
B
|
36 km/hr
|
C
|
71.11 km/hr
|
D
|
71 km/hr
|
Average speed = (320 * 2/9) km/hr = 71.11 km/hr
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