SOLUTION: There were 218 more coins Box X than in Box Y. After 76 coins were removed from Box Y and 34 coins were removed from Box X, the number of coins in Box Y was 1/6 of the total numb

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: There were 218 more coins Box X than in Box Y. After 76 coins were removed from Box Y and 34 coins were removed from Box X, the number of coins in Box Y was 1/6 of the total numb      Log On


   

Question 1186592: There were 218 more coins Box X than in Box Y.
After 76 coins were removed from Box Y and 34 coins were removed
from Box X, the number of coins in Box Y was 1/6 of the total number of
coins in the two boxes. How many coins were there in Box Y at first?
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
let the number of coins in Box X be x and the number of coins in Box Y be y
total number x%2By
if there were 218 more coins Box X than in Box Y, we have
x=y%2B218.....eq.1


After 76 coins were removed from Box Y, we have y-76
and 34 coins were removed from Box X, we have x-34

the number of coins in Box Y was 1/6 of the total number of coins in the two boxes.
y-76=%281%2F6%29%28%28x-34%29%2B%28y-76%29%29 ..........substitute x from eq.1
y-76=%281%2F6%29%28y%2B218-34%2By-76%29
y-76=%281%2F6%29%282y%2B108%29.........both sides multiply by 6
6y-456=2y%2B108
6y-2y=456%2B108
4y=564
y=141


answer: highlight%28141%29 coins were there in Box Y at first
check:
x=141%2B218=359 coins were there in Box X
total: 359%2B141=500
After 76 coins were removed from Box Y, we have 141-76=65
and 34 coins were removed from Box X, we have 359-34=325
total: 325%2B65=390
now in Box Y is 65 and it should be 1%2F6 of 390: 390%2F6=65 ->true

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

There were 218 more coins Box X than in Box Y.
After 76 coins were removed from Box Y and 34 coins were removed
from Box X, the number of coins in Box Y was 1/6 of the total number of
coins in the two boxes. How many coins were there in Box Y at first?
Let number in Box Y, be Y
Then number in Box X = Y + 218

Removing 76 from Y leaves: Y  -  76
Removing 34 from X leaves Y + 218 - 34 = Y + 184

We then get: 

         6Y - 6(76) = 2Y + 108 ------- Multiplying by LCD, 6
            6Y - 2Y = 108 + 456
                 4Y = 564
Original number in Box Y, or highlight_green%28matrix%281%2C5%2C+Y%2C+%22=%22%2C+564%2F4%2C+%22=%22%2C+141%29%29