Below are three different problems based on the concept of time and work.
Question 1
There are 4 machines namely P, Q, R and S in a factory. P and Q running together can finish an order in 10 days. If R works twice as P and S works 1/3 as much as Q then the same order of work can be finished in 6 days. Find the time taken by P alone to complete the same order.
a) 11.5 days
b) 12.5 days
c) 13.5 days
d) 14.5 days
Answer : b) 12.5days
Solution :
Let P's 1 day work be X and Q's 1 day work be Y ...(1)
Given that, the time taken to complete the order by (P+Q) = 10 days.
Then (P+Q)'s 1 day work = 1/10 …(2)
Therefore, from (1) and (2), X+Y = 1/10 …(3)
Suppose, R works twice as P then R's 1 day work 2X.
And, S works 1/3 as much as Q then S's 1 day work be Y/3.
Now, the time taken to complete the order by (R+S) = 6 days
2X+Y/3 = 1/6
12X+2Y = 1 ….(4)
Solving the above two equations(3) and (4), we get X = 4/50.
Thus, P's 1 day work = 4/50.
Hence P alone can complete the entire order of work in 50/4 days = 12.5 days.
Question 2
X and Y individually can complete a task in 30 days and 40 days respectively.If X and Y worked together for 12 days and B alone worked for the remaining part of task. Then how many part of the task is completed by B alone ?
a) 2/7
b) 3/10
c) 8/15
d) 1/3
Answer : b) 3/10.
Solution :
X can complete the task in 30 days.
Then X's 1 day work = 1/30
Y can complete the task in 40 days then Y's 1 day work = 1/40.
Therefore, (X+Y)'s 1 day's work = 1/30 + 1/40 = 7/120.
And (X+Y)'s 12 days work = 12 x 7/120 = 7/10.
Remaining work = 1 - 7/10 = 3/10.
Hence, B alone completes 3/10 of the given task.
Question 3
Two pipes P and Q together can fill a tank in 18 hours.If P alone takes 1/2 of the time of thrice Q's then the time taken by Q alone to fill the tank is:
a)29 hours
b)32 hours
c)28 hours
d)30 hours.
Answer : d)30 hours.
Solution :
Suppose Q alone takes X hours to fill the tank.
Then P takes = 1/2 of 3X = 3X/2 hours.
Now, Q's 1 hours work = 1/X and P's 1 hours work = 2/3X.
Given that, (P+Q) takes = 18 hours.
Then (P+Q)'s 1 hour's work = 1/18.
Therefore, 1/18 = 1/X + 2/3X.
5/3X = 1/18
X = 30.
Hence the answer is 30 hours.
Question 1
There are 4 machines namely P, Q, R and S in a factory. P and Q running together can finish an order in 10 days. If R works twice as P and S works 1/3 as much as Q then the same order of work can be finished in 6 days. Find the time taken by P alone to complete the same order.
a) 11.5 days
b) 12.5 days
c) 13.5 days
d) 14.5 days
Answer : b) 12.5days
Solution :
Let P's 1 day work be X and Q's 1 day work be Y ...(1)
Given that, the time taken to complete the order by (P+Q) = 10 days.
Then (P+Q)'s 1 day work = 1/10 …(2)
Therefore, from (1) and (2), X+Y = 1/10 …(3)
Suppose, R works twice as P then R's 1 day work 2X.
And, S works 1/3 as much as Q then S's 1 day work be Y/3.
Now, the time taken to complete the order by (R+S) = 6 days
2X+Y/3 = 1/6
12X+2Y = 1 ….(4)
Solving the above two equations(3) and (4), we get X = 4/50.
Thus, P's 1 day work = 4/50.
Hence P alone can complete the entire order of work in 50/4 days = 12.5 days.
Question 2
X and Y individually can complete a task in 30 days and 40 days respectively.If X and Y worked together for 12 days and B alone worked for the remaining part of task. Then how many part of the task is completed by B alone ?
a) 2/7
b) 3/10
c) 8/15
d) 1/3
Answer : b) 3/10.
Solution :
X can complete the task in 30 days.
Then X's 1 day work = 1/30
Y can complete the task in 40 days then Y's 1 day work = 1/40.
Therefore, (X+Y)'s 1 day's work = 1/30 + 1/40 = 7/120.
And (X+Y)'s 12 days work = 12 x 7/120 = 7/10.
Remaining work = 1 - 7/10 = 3/10.
Hence, B alone completes 3/10 of the given task.
Question 3
Two pipes P and Q together can fill a tank in 18 hours.If P alone takes 1/2 of the time of thrice Q's then the time taken by Q alone to fill the tank is:
a)29 hours
b)32 hours
c)28 hours
d)30 hours.
Answer : d)30 hours.
Solution :
Suppose Q alone takes X hours to fill the tank.
Then P takes = 1/2 of 3X = 3X/2 hours.
Now, Q's 1 hours work = 1/X and P's 1 hours work = 2/3X.
Given that, (P+Q) takes = 18 hours.
Then (P+Q)'s 1 hour's work = 1/18.
Therefore, 1/18 = 1/X + 2/3X.
5/3X = 1/18
X = 30.
Hence the answer is 30 hours.
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