Two machines are sawing wood but due to space limitations only one at

Machine A:

Time = 2 shifts of 8 hours each* 12 days= 16*12 hrs to finish work alone.

Rate A= 1/(16*12)

Similarly, Rate B= 1/(16*15)

Rate A+B= 27/(16*15*12)
Time to finish, working together = x days*12 hours a day= 12x

As T=1/r, we have
12x=(16*15*12)/27= 80/9=~9 days

Ans D

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A 1/24 of task per shift; also does half of one shift-> 1/48 during the half shift
B: 1/30 of task per shift; also does half of one shift -> 1/60 during the half shift

total of A work in the 1.5 shift - 1/24 +1/48 = 1/16
total of B work in the 1.5 shift - 1/30 +1/60 = 1/20

Work formula: T=W/R
1/ (1/16 +1/20) = 80/9
Approx 9 days.
D.

Bunuel

Let's assume the total units of work to be done is 60. Therefore W = 60

A = 60/12
= 5 Units / Day for two shifts

Therefore Machine A does 2.5 units / day - shift

B = 60/15
= 4 Units / Day for two shifts

Therefore Machine B does 2 units / day - shift

Now if A works first shift, B works second shift and both work half of third shift, then the total units of work on a day
= 2.5 + 2 + 2.5 + 2
= 9 Units / Day

Therefore, total number of days required is
= 60/9
= 6.66
~ 7 Days (Answer E)

Can you help me understand where I am going wrong?

Hi pikolo2510.

Look at highlighted portions above.

Great method. Just have to be careful in your execution.

We can solve this using the efficiency method.

Let the efficiency of machine A be \(a\) and that of machine B be \(b\).

Now, \(a * 12 * 2 = b * 15 * 4\)
Therefore,
\(\frac{a }{ b} = \frac{5}{4}\)

Thus, efficiency of \(a\) is 5 and that of \(b\) is 4.

Now substituting either of the above values in the equations above => 12*2*5 = 120 is the total work to be done.

Following the current schedule: \(1 * a + 1 * b + 0.5 * a + 0.5 *b\)
\(= 1.5 a + 1.5 b\) of the work will get done in 1 day.
Again substituting the values of \(a\) and \(b\) = 7.5 + 6 = 13.5 of the work is done is 1 day.

Now dividing \(\frac{120}{13.5}\) = approx. 9 days will be needed to complete the work.