A hawker bought 120 oranges at $1 each.However 5% of them were damaged and had to be thrown away. Another 20% of them were in poor quality and had to be sold at half of the marked price. He then sold the rest at a profit of 40%.
* Find the total cost
AND
* Find the overall profit percentage
PLZ show the step thz
F3 MATHS PERCENTAGE
2007-08-10 12:02 am
回答 (3)
2007-08-10 12:09 am
✔ 最佳答案
Total cost=$120Overall profit: 120(1-5%-20%)(1+40%)+120(20%)*0.5-120
=18
profit %=18/120*100%
=15%
2007-08-10 12:16 am
Total cost = 120 x$1 = $120
Loss to damaged oranges = 120x5% = $6
Number of poor quality oranges = 120x20%=24
Number of good quality oranges = 120-24-6=90
Selling revenue from these 90 oranges = $90 x (1+40%)
=$126
Selling revenue from 24 poor quality oranges = 24x$1x(+40%)x0.5 = $16.8
Overall profit percentage = (126+16.8)/120 x 100% -1
= 19%
Loss to damaged oranges = 120x5% = $6
Number of poor quality oranges = 120x20%=24
Number of good quality oranges = 120-24-6=90
Selling revenue from these 90 oranges = $90 x (1+40%)
=$126
Selling revenue from 24 poor quality oranges = 24x$1x(+40%)x0.5 = $16.8
Overall profit percentage = (126+16.8)/120 x 100% -1
= 19%
2007-08-10 12:09 am
Total cost
= Money spent to buy the orange
= 120 x $1
= $120
Let $y be the marked price.
20% was sold at half of the marked price:
Money earned
= 120 x 20% x 0.5y
= 12y
The rest was sold at a profit of 40%
Money earned
= 120 x (1 - 20% - 5%) x (1 + 40%)y
= 126y
Overall profit percentage
= (12y + 126y - 120y) / 120y x 100%
= 15%
= Money spent to buy the orange
= 120 x $1
= $120
Let $y be the marked price.
20% was sold at half of the marked price:
Money earned
= 120 x 20% x 0.5y
= 12y
The rest was sold at a profit of 40%
Money earned
= 120 x (1 - 20% - 5%) x (1 + 40%)y
= 126y
Overall profit percentage
= (12y + 126y - 120y) / 120y x 100%
= 15%
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