a2zinterviews

6.	A tyre has punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat?
	A.  1x1/2 minutes
	B.  3x1/2 minutes
	C.  3x3/5 minutes
	D.  4x1/4 minutes

Solution

1 minute's work of both the punctures = (1/9+1/6) = 5/18 So, both the punctures will make the tyre flat in = 18/5 = 3x3/7 days.

7.	10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
	A.  1
	B.  3
	C.  5
	D.  7

Solution

1 woman's 1 day's work = 1/70 1 child's 1 day's work = 1/140. (5 women +10 children)'s 1 day's work = (5/70+10/140) = (1/14+1/14) = 1/7. 5 women + 10 children will complete the work in 7 days.

8.	A does a work in 10 days and B does the same work in 15 days. In how many days they together will do the same work?
	A.  5 days
	B.  6 days
	C.  8 days
	D.  9 days

Solution

A's 1 day's work = 1/10 and B's 1 day's work = 1/15. (A+B)'s 1 day's work = (1/10+1/15) = 1/6.

9.	10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
	A.  90
	B.  145
	C.  150
	D.  225

Solution

1 man's 1 day's work = 1/100. 10 men + 15 women)'s 1 day's work = 1/6 15 women's 1 day's work, = (1/6-10/100) =(1/6-1/10) = 1/15. 1 woman's 1 day's work = 1/225. 1 woman alone can complete the work in 225 days.

10.	A takes twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days, can do the work alone in
	A.  4 days
	B.  6 days
	C.  12 days
	D.  15 days

Solution

Suppose A,B and C take x,x/2 and x/3 hours respectively to finish the work. Then, (1/x+2/x+3/x) = 1/2 <=> 6/x=1/2 <=> 12. So, B takes 6 hours to finish the work.