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Question 1210397: Andrew bought some apples and pears. The ratio of the number of apples bought to the number of pears bought was 7:4. He spent $61.20. He paid $22.80 more for the apples than the pears. Each apple was $0.30 more than each pears.
(a) How much did he spend on the pears?
(b) How many pears did he buy?
Found 4 solutions by josgarithmetic, ikleyn, greenestamps, MathTherapy: Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! This is not very neat and much likely not the best way, but this should work.
PRICE QUANTY COST
Apple p+0.3 7x 7x(p+0.3)
Pear p 4x 4x*p
Totals 61.20
and too, the 22.8 dollars difference
Answer by ikleyn(52781) (Show Source): Answer by greenestamps(13200) (Show Source): Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
Andrew bought some apples and pears. The ratio of the number of apples bought to the number of pears bought was 7:4. He spent $61.20. He paid $22.80 more for the apples than the pears. Each apple was $0.30 more than each pears.
(a) How much did he spend on the pears?
(b) How many pears did he buy?
Let multiplicative factor be x
As ratio of apples to pears is 7:4, number of apples and pears purchased = 7x and 4x, respectively
Let amount spent on pears be p
Since $22.80 more was spent on apples than pears, then amount spent on apples = p + $22.80
A total of $61.20 was spent, so we then have: p + p + 22.8 = 61.2
2p = 61.2 - 22.8
2p = 38.4
p, or amount spent on pears =
Since $19.20 was spent on “4x” pears, then cost of each pear =
As $19.20 was spent on pears, $61.20 - $19.20, or $42 was spent on “7x” apples, and so, each apple cost
Now, since each apple was $0.30 more than each pear, we get:
6 = .3x + 4.8 ---- Multiplying by LCD, x
6 - 4.8 = .3x
1.2 = .3x
x, or multiplicative factor =
Number of pears purchased: 4x = 4(4) = 16
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