Asked by Sofiya

Two boys decided to buy a computer. The second boy had 5/6 of the money the first had. The first boy had 7/8 of the price of the computer. Together they had $696.00 more than they need to pay. What was the price for the computer?
Answered by Damon
b1 = 7/8 C
b2 = 5/6 b1 = 35/ 48 C

7/8 C + 35/48 C = C + 696
42/48 C + 35/48 C = 48/48 C + 696 * 48/48
(42 + 35 - 48 ) C = 696 * 48
29 C = 33408
C =1152.
Answered by henry2,
Computer: $X.
5x/6 + 7x/8 = x + 690.
20x/24 + 21x/24 = x + 690
41x/24 = x + 690
41x = 24x + 24*690
17x = 24*690
X = $974.12.
Answered by henry2,
Correction: 17x = 24*696
X = $982.60.

Answered by Michael
Let the price of the computer = x
This means:
1st boy = (7/8)x {because the 1st boy has 7/8 of the price of the computer}

2nd boy = (5/6*(7/8)x) {because the 2nd boy has 5/6 of the first boy who has 7/8 of the price of the computer}

2nd boy = (35/48)x {after multiplying (5/6)*(7/8)}

Now let's set up the equation:

(7/8)x + (35/48)x = x +696 {The total of the money between both boys is the price of the computer and an extra 696 dollars}

Next, we subtract (7/8)x from both sides to get rid of the first x on the left side of the equation

(35/48)x = (1/8)x + 696

Now, we need to subtract (1/8)x from both sides. This fraction is also (6/48)x

(29/48)x = 696

Now we need to multiply the reciprocal of 29/48 which is 48/29.

x = 1152

The computer is $1152
Answered by rsm sucks
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