SOLUTION: Marilyn had a bag of gold coins. She gave 1/8 of them to her mother and then gave 1/2 of what was left to her brother. She then gave 2/7 of what left to her dad. If she then had 25

Question 100716: Marilyn had a bag of gold coins. She gave 1/8 of them to her mother and then gave 1/2 of what was left to her brother. She then gave 2/7 of what left to her dad. If she then had 25 coins left, how many did she have originally?
Found 2 solutions by edjones, checkley71:
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
let x= what she started with.
x*7/8=7x/8
7x/8*1/2=7x/16
7x/16*2/7=14x/112
14x/112=x/8
so 1/8 of what she started with=25 coins
x/8=25
x=25*8 multiply both sides times 8.
25*8=200 coins she started with.
Ed

Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
((X/8)+.5(X-X/8)+2/7(X-X/8)+25=X
X/8+.5(7X/8)+2/7(7X/8)+25=X
X/8+3.5X/8+2X/8+25=X
(X+3.5X+2X)/8+25=X
6.5X/8-X=-25
(6.5X-8X)/8=-25
-2.5X/8=-25
-2.5X=-25*8
-2.5X=200
X=-200/-2.5
X=80 THE TOTAL AMOUNT OF COINS SHE HAD.
PROOF
80*1/8=10 SHE GAVE T MOTHER. LEAVING HER 80-10=70
70*.5=35 SHE GAVE TO HER BROTHER. LEAVING HER 70-35=35
35*2/7=10 SHE GAVE TO HER DAD. LEAVIN HER WITH 25 COINS.
10+35+10+25=80 COINS.
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