SOLUTION: Please help me solve, Iam unsure how to set up the problem... it shows in the book also but I cannot understand the explanation #10. A student finishes the first half of an exa

Question 203761: Please help me solve, Iam unsure how to set up the problem... it shows in the book also but I cannot understand the explanation
#10. A student finishes the first half of an exam in 2/3 the time it takes to finish the second half . If the entire exam takes an hour, how many minutes does she spend on the first half of the exam?

The back of the book sets up the problem like this... 5/3S = 60, then
3/5 * 5/3S = 3/5*60, then S=36....

I do not understand how they came up with the 5/3, then I figured 2/3 + 2/3 + 1hr = 5/3, but why did they multiply the recipricols and then place 3/5 on the other side *60?
Found 2 solutions by Earlsdon, solver91311:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let F = the time taken to finish the first half of the exam.
Let S = the time taken to finish the second half of the exam.
F+S = 1 hour or...
1) minutes. and...
2) "...student finishes the first half of the exam (F) in the time it takes to finish the second half of the exam .
From this we can write the equation:
Substitute from equation 2)
Add the S-terms on the left side.
Multiply both sides by the multiplicative inverse of which is
Evaluate this.
so...

minutes.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Let represent the time, in minutes, it took to finish the second half of the exam. Then we are given that is the time, in minutes, it took to finish the first half of the exam. Add these two quantities for the total time, again in minutes, to finish the exam which is 60 because there are 60 minutes in 1 hour. So:

But , so:

So far, so good.

There are two ways to go about solving from here. You can either multiply both sides by 3:

And then multiply both sides by :

Or, you can combine those two steps into one, as your book shows, but multiplying by the reciprocal of the fractional coefficient on the variable. (That is because we know that )

Achieving the same result, as one would expect.

John

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