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Arithmetic : Level 2 Test -7
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Question 1 |
A train travels a distance of 600 km at a constant speed. If the speed of the train is increased by 5 km/h, the journey would take 4 h less. Find the speed of the train.
100 km/h | |
25 km/h | |
50 km/h | |
None of these |
Question 1 Explanation:
Distance = 600 km/h
Let the speed of the train = s km/h
Now increased speed = (s + 5) km/h
So 600/s = 600/(s+5) + 4
600/s - 600/(s+5) = 4
600[5/s(s+5)]= 4
s = 25 km/h
Let the speed of the train = s km/h
Now increased speed = (s + 5) km/h
So 600/s = 600/(s+5) + 4
600/s - 600/(s+5) = 4
600[5/s(s+5)]= 4
s = 25 km/h
Question 2 |
Three pipes A, B and C can fill a tank in 20 min, 10 min and 30 min respectively. When the tank is empty, all the three pipes are opened. A, Band C discharge chemical solutions x, y and z respectively. The proportion of solution y in the liquid in the tank after 3 min is
6/11 | |
7/11 | |
8/11 | |
5/11 |
Question 2 Explanation:
Work done by the pipes in one minutes = 1/20 + 1/10 +1/30
So the work done by the pipes in 3 minutes = 3/20 + 3/10 +3/30 = 11/20
The work done by the 2nd pipe in one minute = 1/10
The work done by the 2nd pipe in three minutes = 3/10
So the required ratio will be = (3/10) / (11/20) = 6/11
So the work done by the pipes in 3 minutes = 3/20 + 3/10 +3/30 = 11/20
The work done by the 2nd pipe in one minute = 1/10
The work done by the 2nd pipe in three minutes = 3/10
So the required ratio will be = (3/10) / (11/20) = 6/11
Question 3 |
Three taps A, Band C can fill a tank in 12, 15 and 20 h respectively. If A is open all the time and B and C is open for one hour each alternatively; the tank will be filled in
6 h | |
7 h | |
5 h | |
None of these |
Question 3 Explanation:
Tap A and B can fill the tank in =1/12 + 1/15 = 3/20
Tap A and C can fill the tank in =1/12 + 1/20 = 2/15
Work done by all the taps in 2 hours = 3/20 + 2/15 = 17/60
Filling done in 6 hours = 17/60 x 3 = 51/60
So remaining work is = 1 – 51/60 = 3/20
Now the taps which are opened are A and B and the work done by the taps in one hour is 3/20 as discussed earlier,
so the total time for the whole filling would be = 7 hours
Tap A and C can fill the tank in =1/12 + 1/20 = 2/15
Work done by all the taps in 2 hours = 3/20 + 2/15 = 17/60
Filling done in 6 hours = 17/60 x 3 = 51/60
So remaining work is = 1 – 51/60 = 3/20
Now the taps which are opened are A and B and the work done by the taps in one hour is 3/20 as discussed earlier,
so the total time for the whole filling would be = 7 hours
Question 4 |
A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time it increases its speed by 250 km/h. What is its original speed?
1000 km/h | |
750 km/h | |
600 km/h | |
800 km/h |
Question 4 Explanation:
Let the actual speed of the plane = S km/h
Let the actual time taken = T h
Distance = 1500 km
Therefore T = 1500/S…………………………………a
Now acc. to question
1500/(S + 250) = T – 30/60 ………………………..b
1500/ (s+ 250) = 1500/s – 30/60
Solving the above equation S = 750 km/h
Let the actual time taken = T h
Distance = 1500 km
Therefore T = 1500/S…………………………………a
Now acc. to question
1500/(S + 250) = T – 30/60 ………………………..b
1500/ (s+ 250) = 1500/s – 30/60
Solving the above equation S = 750 km/h
Question 5 |
A computer can perform 30 identical tasks in 6 h. At that rate, what is the approximately minimum number of computers that should be assigned to complete 80 of the tasks within 3 h?
12 | |
6.66 | |
5.33 | |
16 |
Question 5 Explanation:
6 hour = 30 tasks
1 hour = 5 tasks
3 hour = 15 tasks
So 15 article are done by 1 computer in 3 hours
15N = 80
N = 80 / 15 = approx. 5.33 times
1 hour = 5 tasks
3 hour = 15 tasks
So 15 article are done by 1 computer in 3 hours
15N = 80
N = 80 / 15 = approx. 5.33 times
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