Bunuel wrote:
abhishekkhosla wrote:
Three men A, B and C working together can do a work in 30 days. They start the work together and A works for 3 days and takes rest on the fourth day, B works for 5 days and takes rest on the next two days and C works for 7 days and takes rest on the next three days. In how many days will the work be completed ? (A, B, C work at the same rate)
A. 39
B. 40
C. 42
D. 41
E. 45
Total work is 3*30 = 90 units.
A's schedule WWW
R 3 workdays in 4.
B's schedule WWWWW
RR 5 workdays in 7.
C's schedule WWWWWWW
RRR 7 workdays in 10.
Test the options. In 40 days:
A will do 10*3=30 units (10 complete cycles).
B will do 5*5+5=30 units (5 complete cycles and 5 days of work).
C will do 4*7=28 units (4 complete cycles).
Total = 88 units.
On 41st day, A and C will work and do 2 units of work, so in 41 days 88+2=90 units of work will be done.
Answer: D.
P.S. Not a GMAT question.
I took a slightly different approach and got C as the answer. Here it goes:
A's schedule lasts 4 days, B's schedule lasts 7 days, C's schedule lasts 10 days. Now, LCM of 4,10 and 7 is 140 days.
For A: In 4 days, A works for 3 days. Therefore in 140 days, A will work for \(105 days\).
For B: In 7 days, B works for 5 days. Therefore in 140 days, B will work for \(100 days\).
For C: In 10 days, C works for 7 days. Therefore in 140 days, C will work for \(98 days\).
Now, the total days worked = \((105 + 100 + 98)= 303 workdays\)
To achieve 303 workdays we need 140 days. Therefore to achieve 90 workdays we need \(\frac{90*140}{303}\)= 41.6(approx) = 42 days.