Sarah receives 10z coins in addition to what she already...

AAPL wrote:Sarah receives 10z coins in addition to what she already had. She now has 5y+1 times as many coins as she had originally. In terms of y and z, how many did Sarah have originally?

A. 10z(5y + 1)
B. (5y+1)/10z
C. 2z/y
D. 10/(5y + 1)
E. 10/(5y + 10)

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

We could try some easy numbers. Say y = 1. That means that after she receives more coins, she'll have 5*1 + 1 = 6 times as many coins as she had originally.

Let's say she started with 1 coin. Then we know she'll end up with 1*6 = 6 coins, meaning she must have received 5 more coins. If she received 5 more coins, then z = 1/2, so that 10*(1/2).= 5.

To summarize:
- she starts with 1 coin
- z = 1/2
- y = 1

We want the number of coins Sara had initially, or 1. So the correct answer will give us a value of 1 when we plug '1/2' in for z and '1' in place of y. Only C works, as 2*(1/2)/1 = 1

AAPL wrote:Sarah receives 10z coins in addition to what she already had. She now has 5y+1 times as many coins as she had originally. In terms of y and z, how many did Sarah have originally?

A. 10z(5y + 1)
B. (5y+1)/10z
C. 2z/y
D. 10/(5y + 1)
E. 10/(5y + 10)

The OA is C.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

And there's always good old-fashioned algebra. Say she starts with T coins. After she receives 10z more, she'll have T + 10z. And we know that her new number of coins (T + 10z) is '5y +1' times the old value of coins, T. Algebraically, we'd get

(T + 10z) = (5y + 1)(T)
T + 10z = 5yT + T
10z = 5yT
10z/5y = T
2z/y = T
The answer is C