Work time efficiency based problem tips
One of the important topics in aptitude solving is work-time-efficiency based problems but we should be able to manage time which is possible by solving more and more problems.
Important points to remember:
If A can finish work in m minutes and B can finish the same work in n minutes then both
A and B together can finish the work in t minutes then 1/t=(1/m+1/n)
If there are n persons A1, A2...An each take times t1,t2.........tn minutes respectively to finish their work indivudually then All together can work in T minutes then 1/T=1/t1+1/t2+...........+1/tn
If M1 persons work together for H1 hours a day to make a work of quantity W1 within D1 days,then
M2 persons work together for H2 hours a day to make a work of quantity W2 within D days then
(M1*D1*H1)/W1=(M2*D2*H2)/W2
*********************************************************************************
Example 1:
A can do a work in 27 days, B's efficiency is 3 times that of A and C's efficiency is 5 times that of A how much time all A, B, C take to finish the work.
Solution:
Person No.of days to finish the work
A 27
B 27/3
C 27/5
Let T be the time taken by all A, B, C to finish the work then
1/T=1/27+1/(27/3)+1/(27/5)
1/T=1/27+3/27+5/27
1/T=9/27
T=27/9=3 days
Example 2:
If A, B, C together finishes some work in 20 minutes, A and B together can finish the same work in 40 minutes, find how much time C needs to finish the same amount of work.
solution:
1/T=1/A+1/B+1/C
1/c=1/T-(1/A+1/B)
1/C=1/20-1/40
1/c=1/40
C=40
Hence C can alone finish the work in 40 minutes
Example 3:
3 Men worked 5 hours a day for 10 days to make 20 articles, How many articles can be produced 2 Men who worked 6 hours a day for 15 days.
solution:
M1=3 H1=5 D1=20 W1=20
M2=2 H2=6 D2=15 W2=(Unknown)
From the equation
(M1*H1*D1)/W1=(M2*H2*D2)/W2
(3*5*20)/20=(2*6*15)/W2
15=(12*15)/W2
W2=12 Articles.
Important points to remember:
If A can finish work in m minutes and B can finish the same work in n minutes then both
A and B together can finish the work in t minutes then 1/t=(1/m+1/n)
If there are n persons A1, A2...An each take times t1,t2.........tn minutes respectively to finish their work indivudually then All together can work in T minutes then 1/T=1/t1+1/t2+...........+1/tn
If M1 persons work together for H1 hours a day to make a work of quantity W1 within D1 days,then
M2 persons work together for H2 hours a day to make a work of quantity W2 within D days then
(M1*D1*H1)/W1=(M2*D2*H2)/W2
*********************************************************************************
Example 1:
A can do a work in 27 days, B's efficiency is 3 times that of A and C's efficiency is 5 times that of A how much time all A, B, C take to finish the work.
Solution:
Person No.of days to finish the work
A 27
B 27/3
C 27/5
Let T be the time taken by all A, B, C to finish the work then
1/T=1/27+1/(27/3)+1/(27/5)
1/T=1/27+3/27+5/27
1/T=9/27
T=27/9=3 days
Example 2:
If A, B, C together finishes some work in 20 minutes, A and B together can finish the same work in 40 minutes, find how much time C needs to finish the same amount of work.
solution:
1/T=1/A+1/B+1/C
1/c=1/T-(1/A+1/B)
1/C=1/20-1/40
1/c=1/40
C=40
Hence C can alone finish the work in 40 minutes
Example 3:
3 Men worked 5 hours a day for 10 days to make 20 articles, How many articles can be produced 2 Men who worked 6 hours a day for 15 days.
solution:
M1=3 H1=5 D1=20 W1=20
M2=2 H2=6 D2=15 W2=(Unknown)
From the equation
(M1*H1*D1)/W1=(M2*H2*D2)/W2
(3*5*20)/20=(2*6*15)/W2
15=(12*15)/W2
W2=12 Articles.
Comments
Post a Comment