2. A can do a piece of work in 40 days; B can do the same in 30 days. A started alone but left the work after 10 days, then B
worked at it for 10 days. C finished the remaining work in 10 days. Calone can do the whole work in?
A. 24 days B. 30 days C. 44 days D. 17 1/2 days
3. A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can
do it in:
A.20 days B.22 days C.25 days D.30 days E. None of these
4. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time
taken by 15 men and 20 boys in doing the same type of work will be:
1.7 days
left the job. In how many
Us
Answers
2a.24days
3b.22whole1\2
4.c. 4days
Step-by-step explanation:
2.Let the total work be x units.
A can do the work in 40 days. 1 day work of A = x/40
B can do the work in 30 days. 1 day work of B = x/30 =
For the first 10 days A does the work alone.
10 ays work of A = 10 x x/40 = x/4
For the next 10 days B does the work alone.
10 days work of B = x/3C does the remaining work left in the last 10 days .
x = 10 days work of C + x/3 + x/4
=> 10 days work of C = 5 x/12
C can do 5x/12 units of work in 10 days. C can do x units of work in = (x x 10)/(5x/ 12)
= 24 days
3.Ratio of times taken by A and B =1:3
The time difference is (3−1)2 days while B takes 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes (
2
3
×60)=90 days.
So, A takes 30 days to do the work.
A's 1 day's work =
30
1
B's 1 day's work =
90
1
(A+B)'s 1 day's work =(
30
1
+
90
1
)=
90
4
=
45
2
∴ A and B can do the work in
2
45
=22
2
1
days.
4.Given that
6 men and 8 boys can do a piece of work in 10 days
26 men and 48 boys can do the same in 2 days
As the work done is equal,
10(6M + 8B) = 2(26M + 48B)
60M + 80B = 52M + 96B
=> M = 2B
=> B = M/2 ……(1)
Now Put (1) in 15M + 20B
=> 15M + 10M = 25M
Now, 6M + 8B in 10 days
=> (6M + 4M) 10 = 100M
Then D(25M) = 100M
=> D = 4 days.