急!急!急! 超難數學題(5)

Question

<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 % .

Answer 1

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%

Answer 2

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%

Answer 3

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%.