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Sat 30 Aug, 2014 06:20 pm
During it's first week in business, a store sold 108 (total) apples and oranges.
Week 2- Five times the number of apples and three times the number of oranges were sold.
A total of 452 apples and oranges were sold for week 2.
Calculate number of (each) apples and oranges sold the first week.
What is known:
A store sold a total of 108 apples and oranges the first week of business.
5 times the number of apples and 3 times the number of oranges were sold for week 2.
A total of 452 apples and oranges were sold for week 2.
Want to know:
Number of (each) apples and oranges sold the first week.
Let A= apples
B= oranges
A + B = 108 (first week)
5A + 3B = 452 (week 2)
5A + 3B = 452
- A + B = 108
---------------------
4A +2B = 344
4A = 344
4A/4 = 344/4
A = 86
2B = 344
2B/2 = 344/2
B = 172
That is not correct. Where did I make the error(s)?
Simultaneous equations
You know:
Equation (1) A + B = 108
Equation (2) 5A + 3B = 452
If you can get equation (1) to show 3B instead of just B then you can eliminate the Bs by subtraction.
You can do this by multiplying everything in equation (1) by 3:
3A + 3B = 3 x 108
3A + 3B = 324
So now you know:
5A + 3B = 452
3A + 3B = 324
Subtracting the smaller equation from the bigger one will make the Bs disappear:
5A - 3A = 2A
3B - 3B = 0 (B vanishes)
452 - 324 = 128
So:
2A = 128
A = 64
Now put this value of A into equation (1):
64 + B = 108
Re arrange this to get just B on one side and just numbers on the other:
B = 108 - 64
B = 44
Check your answer
A + B = 108 (64 + 44 = 108, correct)
5A + 3B = 452 (320 + 132 = 452, correct)
Of course you could have chosen to eliminate the As by multiplying the first equation by 5 (because of the 5A in equation (2)
Given:
(1) A + B = 108
(2) 5A + 3B = 452
5A + 5B = 108 x 5
5A + 5B = 540
Now you know:
5A + 5B = 540
5A + 3B = 452
5A - 5A = 0
5B - 3B = 2B
2B = 540 - 452
2B = 88
B = 88/2
B = 44
A + 44 = 108
A = 108 - 44
A = 64
@contrex,
Thanks. I did not do the solution quite right, but I was on the right track.
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