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Arithmetic : Level 2 Test -9

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Question 1
Two men undertake to do a piece of work for Rs. 1,400. First man alone can do this work in 7 days while the second man alone can do this work in 8 days. If they working together complete this work in 3 days with the help of a boy, what would be the share of boy ?
A
 Rs 250
B
Rs 275
C
Rs 300
D
Rs 375
Question 1 Explanation:
Work done by first man in 3 days = 3/7 of the total work
Work done by the second man in 3 days = 3/8 of the work
Work done by the boy in 3 days = 1 – ( 3/7+ 3/8) = 11/56 of the work
Therefore the ratio of profit = ratio of there work done = 3/7 : 3/8 : 11/56 = 24 : 21 : 11
Therefore the share of boy would be  = 11 / 56 x 1400 = Rs. 275
Question 2
Two taps can fill a tank in 20 minutes and 30 minutes respectively. There is an outlet tap at exactly half level of that rectangular tank which can pump out 100 litres of water per minute. If the outlet tap is open, then it takes 24 minutes to fill an empty tank. What is the volume of the tank?
A
1800 litres
B
1500 litres
C
1200 litres
D
2400 litres
Question 2 Explanation:
Work done by the taps in one minute = 1/20 + 1/30 = 1/12
So both will fill the tank in 12 minutes
So half the tank will be filled in 6 minutes
The statement that it took 24 minutes to fill when we open the outlet pipe gives us 24 – 6 = 18 minutes .
But if the tap is closed then it will take usual 6 mins . Hence the tap can empty half tank in 1/18 -1/6 = 12/108 = 1/9 = 9 minutes
Therefore the volume of the tank is = 2 x water displaced in the time taken by empty tap = 100 x 2 x 9 = 1800 litres
Question 3
15 men can complete a work in 210 days. They started the work but at the end of 10 days 15 additional men, with double efficiency, were inducted. How many days, in whole, did they take to finish the work?
A
60 days
B
63.33days
C
66.66 days
D
70 days
Question 3 Explanation:
No. of unit work completed by 15 men in 210 days = 1
No. of unit work completed by 15 men in 10 days = 1/21
Work left = 1 – 1/21 = 20/21
15 additional men, with double efficiency, were inducted
Therefore we have
M1 = 15
M2 = 15 + 30 = 45
W1 = 1
W2 =20/21
D1 =210
D2 = d
Therefore
M1 X D1 / W1 = M2 X D2 / W2
By putting values we get the value of D2 = (15 * 210 *20) /( 1* 21 * 45)  = 200/3 = 66.66 days
Question 4
There are two taps to fill a tank while a third to empty it. When the third tap is closed, they can fill the tank in 10 minutes and 12 minutes, respectively. If all the three taps be opened, the tank is filled in 15 minutes. If the first two taps are closed, in what time can the third tap empty the tank when it is full?
A
8 min and 34 sec
B
9 min and 32 sec
C
7 min
D
6 min
Question 4 Explanation:
Let us find the speed of the third tap.
Let capacity of tank be C.
Speed of Tap 1 = C/10 Per minute
Speed of Tap 2 = C/12 Per minute
Let speed of Tap 3 = x (this would be negative value)
The three taps fill the tank in 15 mins together, that is in one minute, they fill C/15 of the tank.
Thus, (C/10) + (C/12) + x = C/15
x = 1/15 – 1/12 – 1/10 = -7/60
Remember, tap 3 takes a negative value as it is emptying the tank.
Therefore tap 3 will take 60/7 minutes to empty the tank.
Therefore option A is the right answer
Question 5
A cistern has two taps, which fill it in 12 minutes and 15 minutes respectively. There is also a waste pipe in the cistern. When all the pipes are opened, the empty cistern is full in 20 minutes. How long will the waste pipe take to empty a full cistern?
A
12 minutes
B
10 minutes
C
8 minutes
D
16 minutes
Question 5 Explanation:
Work done by the two taps in one minute = 1/12 + 1/15 = 9/60
So the work done by waste pipe in one minute = 1/20 – 9/60 = (3 – 9) / 60
= -6/60 = -1/10
Negative sign shows that the pipe empties the tank
If the waste pipe empties 1/10 part of the tank in 1 minute then to empties the whole tank it will take 10 minute so the answer is (b)
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