<1> A man sells two machineat $19 800 each. He makes a profit of 10% on one machine and suffers a loss of 10% on the other. On the hole, find
(a) his profit or loss,
(b) his profit % or loss % .
急!急!急! 超難數學題(5)
2006-12-02 6:48 pm
回答 (3)
2006-12-02 7:02 pm
✔ 最佳答案
Total cost = $19800/(100+10)% + $19800/(100-10)%
= $40000
Total selling price
= $19800 x 2
= $39600
Total cost > Total selling price
Hence, Loss
= $(40000 - 39600)
= $400
Loss %
= ($400/40000) x 100%
= 1%
2006-12-02 7:14 pm
The cost of Machine 1 (10% profit) : 19,800/1.1 = 18,000
The cost of Machine 2 (10% loss) : 19,800/0.9 = 22,000
Total cost: 18,000+22,000 = 40,000
Total selling : 19,800+19,800 = 39.600
a, It is in loss
b, it is in 40,000 + 39,600 = 400.
The loss % is in 1%
The cost of Machine 2 (10% loss) : 19,800/0.9 = 22,000
Total cost: 18,000+22,000 = 40,000
Total selling : 19,800+19,800 = 39.600
a, It is in loss
b, it is in 40,000 + 39,600 = 400.
The loss % is in 1%
2006-12-02 7:01 pm
Let the original price of Machine A be X, and that of Machine B be Y.
X * 1.1 = 19800, and
Y * 0.9 = 19800.
X = 19800/1.1 = 18000, and Y = 19800/0.9 = 22000.
Therefore the original total price paid by this man = X + Y = 18000 + 22000 = 40000.
The money he obtained from selling the 2 machines = 19800 * 2 = 39600.
Therefore, his loss is 40000 - 39600 = 400,
and his loss % is 400/40000 * 100% = 1%.
X * 1.1 = 19800, and
Y * 0.9 = 19800.
X = 19800/1.1 = 18000, and Y = 19800/0.9 = 22000.
Therefore the original total price paid by this man = X + Y = 18000 + 22000 = 40000.
The money he obtained from selling the 2 machines = 19800 * 2 = 39600.
Therefore, his loss is 40000 - 39600 = 400,
and his loss % is 400/40000 * 100% = 1%.
參考: myself
收錄日期: 2021-04-25 19:45:08
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