Eric was preparing for his grandpa's birthday party. He bought some food with 20% of his money. He bought some drinks with 30% of the remaining money and had $28 left. How much did Eric have at first?
Well, what we are looking for is theoriginal amount of money Eric had.....let's call that amount 'x'
He spent 20% of 'x' or 0.20x which left him with 80 percent of 'x' (or 0.8x) after his first purchase of drinks.
Then he spent 30% of what he had left (80 percent of 'x' ...or 0.8x )
SO for food he spent 0.3 (0.8x)
and THEN he had only $28 dolllars left.....in equation form this looks like this:
x - 02.x - .3(.8x) = 28 then just solve for 'x'
x - .2x -.24x = 28
x - .44x = 28
.56x = 28 (now divide both sides by .56)
(.56x)/.56 = 28/(.56)
x = 50 dollars
+118723 Well
70% of 80% of original is $28
You need to turn this into an equation and then solve it :)
Hint
Let original be X
Ooops....sorry, I had the words FOOD and DRINKS reversed in my original answer (but the math is the same)
Well, what we are looking for is theoriginal amount of money Eric had.....let's call that amount 'x'
He spent 20% of 'x' or 0.20x which left him with 80 percent of 'x' (or 0.8x) after his first purchase of FOOD.
Then he spent 30% of what he had left (80 percent of 'x' ...or 0.8x )
SO for DRINKS he spent 0.3 (0.8x)
and THEN he had only $28 dolllars left.....in equation form this looks like this:
x - 02.x - .3(.8x) = 28 then just solve for 'x'
x - .2x -.24x = 28
x - .44x = 28
.56x = 28 (now divide both sides by .56)
(.56x)/.56 = 28/(.56)
x = 50 dollars
