Question

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

Answer 1

Answer: The price for the computer was $1,152.

Step-by-step Explanation: Let the price of the computer be y. The first boy had 7/8 of the price of the computer, which can be expressed as 7/8 of y or simply 7y/8.

The second boy had 5/6 of the money the first boy had, that means he had 5/6 of 7y/8 and this too can be expressed as,

= 5/6 x 7y/8

= 35y/48

So we have determined that both boys have 7y/8 and 35y/48. Also we know that the total money in their possession is 696 more than they need to pay for the computer, that is, 696 more than y, or simply put,

696 + y

Therefore the addition of their money can be expressed as;

7y/8 + 35y/48 = 696 + y

(35y + 42y)/48 = 696 + y

By cross multiplication we now have

35y + 42y = 48(696 + y)

77y = 33408 + 48y

Subtract 48y from both sides of the equation

29y = 33408

Divide both sides of the equation by 29

y = 1152

Therefore the price of the computer was $1,152

Answer 2

Answer: The price for the computer = $1152

Explanation:

Let the first boy = F

Let the second boy = S

Let the price of computer = C

The second boy had 5/ 6 of the money the first had. I.e

S = 5/6 F

Then F = 6/5S ......(1)

The first boy had 7 /8 of the price of the computer. That is

F = 7/8C ....... (2)

Substitute F in (1) into (2)

6/5S = 7/8C

S = 5/6 × 7/8C

S = 35/48C ......(3)

Together they had $696.00 more than they need to pay. That is

F + S = C + 696 ........ (4)

Substitute equation 2 and 3 into 4

7/8C + 35/48C = C + 696

0.875C + 0.729C = C + 696

0.604C = 696

C = 696/0.604

C = 1152 dollars

Therefore, the price for the computer = $1152