Machine A takes 10 hours to complete a certain job and starts that job

A does the job in 10 hrs, so in 1 hr it will complete 1/10 of the total work
When A is joined by B at 10 AM, 9/10 (1 - 1/10)of the total work is left. They both complete this work in 7 hrs.
Therefore, they would have completed the whole work in 70/9 hrs.
From this we can say that:
1/10 + 1/B = 9/70 (where B is the number of hours B takes to complete the job alone)
=> 1/B = 9/70 - 1/10 = 2/70 = 1/35
=> B = 35 hours

Hence the answer is E.

VERITAS PREP OFFICIAL SOLUTION:

Using only algebra and the memorized formula for Work problems (\(Work = Rate * Time\) or \(Rate = \frac{Work}{Time}\)), we can break this problem into 3 parts: Machine A alone from 9 AM to 10 AM , Machines A and B from 10 AM to 5 PM, and then the hypothetical Machine B alone. Let’s set up these three steps to see how we can solve this on the scenic path.

Machine A takes 10 hours to do the job, so each hour it works finishes 1/10th of the total job. From 9 AM to 10 AM, Machine A works alone, so at 10 AM, machine B kicks in and 1/10th of the job is done. Ergo, 9/10th of the job is left to complete for both machines.

From 10 AM to 5 PM, 7 hours pass, during which 9/10th of the job gets completed. Thus we can calculate the rate of the machines working together: \(\frac{9}{10} = Rate_{A+B} * 7\) hours. \(Rate_{A+B} = \frac{9}{70}\).

Since we know that \(Rate_A + Rate_B = Rate_{A+B}\) (i.e. rates are additive), we can leverage the fact that we know 2 of these 3 rates to find the third using basic fraction addition. \(\frac{1}{10} + Rate_B = \frac{9}{70}\). Putting them all on a common denominator: \(\frac{7}{70} + Rate_B = \frac{9}{70}\), so \(Rate_B = \frac{2}{70}\), or 1/35.

Now that we have B’s rate of 1/35, we can easily tell that it would take 35 hours to complete the entire job.

The correct answer is (E).

The algebraic solution works fine and gets you the right answer, but there are many moving parts to keep track of and many opportunities for mistakes. Can we get to the same answer but faster using conceptual understanding and avoid the scenic route entirely?

If machine A does 1/10th of the work in an hour, and it works from 9 to 5, then it works for 8 hours and accomplishes 80% of the job on its own. This means that machine B only accomplishes 20% of the job, and it does so in 7 hours (10 AM to 5 PM). If the machine does 1/5 of the job in 7 hours, it will take (7*5) 35 hours to complete 5/5 of the job. Answer choice E, using almost no math whatsoever, but rather by exploiting the logic of the question.

In general, if you see how to solve a problem via algebra and are confident you can solve it in 3 minutes or less, then by all means go for it. However, you can save some time if you really understand how questions are set up and what they are testing. It may not be possible to come up with a handy shortcut on test day because of nerves and stress, but during your preparation take a look at how problems are solved and see if you can find a more elegant solution. All roads lead to Rome, and the more routes you know, the less likely you are to get stuck in unfamiliar territory.

Rate of Machine A = 1/10 job/h

Rate of Machine B = 1/B

In one hour Machine A completed = 1/10 job

Remaining job = 1 - 1/10 = 9/10

Combined rate of Machine A and B = 1/10 + 1/B

They worked combinedly for 7 hours.

Therefore, (1/10 + 1/B) * 7 = 9/10

1/B = 9/70 - 1/10

1/B = 1/35

Time taken by B to complete the job = 35

(E)

A alone can finish work in 10 hours..
Therefore a can do 1/10th of work in 1 hour..

A starts at 9 a.m and works alone for 1 hour. A finishes 1/10th of the work.
Remaining work = 1 - (1/10) = 9/10

A and B together then finish the remaining work (9/10) from 10a.m to 5 p.m = 7 hours
therefore in 1 hour together they finish 9/70 work

Adding Rate of A and B

1/A + 1/B = 9/70
1/10 + 1/B = 9/70
B = 35

Answer E

Let x be the time taken by machine B to complete the work alone. Then it's 1 hour work will be 1/x.
Total work done by machine A and B in 1 hour will be
\(\frac{1}{10}+\frac{1}{x}\)
=\(\frac{x+10}{10x}\)

So time taken to complete whole work will be \(\frac{10x}{x+10}\)

Machine A worked for 1 hour before Machine B joined.
Work done by A in 1 hour is 1/10
work left to do is 9/10
Machine A and B working together take 7 hours to complete 9/10 work.
\(\frac{9}{10}*\frac{10x}{x+10}=7\)

or \(9x=7x+70\)
or \(2x=70\)
or \(x=35\)

Answer:- E

A completes 1/10 part of the work in 1 hour
The remaining work is 9/10; therefore, 7(1/10 + 1/y) = 9/10
Solving y = 35 hours.

Alternatively,
Machine A completes 4/5 of the total work. Remaining 1/5 of the work is completed by Machine B in 7 hours. Therefore, B will take 35 hours to complete the whole work.

rate of A=1/10=7/70
rate of A+B=9/70
rate of B=2/70=1/35
B will complete job alone in 35 days

let the total work be x units
rate of work for A = x/10
let the rate for B be y
A worked for 8 hrs and b worked for 7 hrs

therefore
8x/10 +7y=x
7y = 2x/10
y = x/35

thefore time taken by B = x/y = x/x/35 = 35

Option E

We let n = the number of hours that it takes machine B to complete the job by itself. We know that machine A worked for 8 hours and machine B worked for 7 hours to complete the entire job. Thus, we can create the equation:

(1/10)(8) + (1/n)(7) = 1

8/10 + 7/n = 1

4/5 + 7/n = 1

7/n = 1/5

35 = n

Answer: E

1=[1/10](1) + [1/10 + (Rate of B)] (7)

Simplifying the above gives Rate of B as 1/35. Implies that answer is 35