Chin went to a store where he spent one half of his money and$14 more.He then went to another store where he spent one third of his money and $14 more. If he then had no money left, how much did he have when he entered the first store?
money spent in first store = (1/2)x + 14
money spent in 2nd store = (1/3)x + 14
so (1/2)x + 14 + (1/3)x + 14 =x
times 6 , the LCD, to clear fractions
3x + 84 + 2x + 84 = 6x
x = 168
check:
amount spent in store 1 = 84 + 14 = 98
money spent in store 2 = 56+14 =70
he spent 98+70 = 168
Remaining Money (y) from the first store is x - 1/2x +14
The Money spent in the second store is 1/3y + 14
So solving for initial money value gives us $70
Nicely done
To solve this problem, we need to set up an equation based on the given information. Let's start by breaking down the problem step by step:
1. Chin spent one half of his money plus $14 at the first store.
2. He then spent one third of his remaining money plus $14 at the second store.
3. After these two purchases, he had no money left.
Let's represent Chin's initial amount of money as "x."
1. Chin spent one half of his money plus $14 at the first store.
This means he spent (1/2)x + $14 at the first store.
After this purchase, he had (1/2)x - [(1/2)x + $14] = (1/2)x - (1/2)x - $14 = $14 left.
2. Chin spent one third of his remaining money plus $14 at the second store.
This means he spent (1/3) * $14 plus $14 at the second store.
After this purchase, he had $14 - [(1/3) * $14 + $14] = $14 - (14/3) = $14 - $4.67 ≈ $9.33 left.
3. Since Chin had no money left after the second purchase, we can set up the equation:
$9.33 = 0
To solve for "x," we can equate the equation from step 3 to zero:
$14 - $4.67 ≈ $9.33 = 0
Therefore, Chin initially had $9.33 when he entered the first store.