A man bought two computers for rs 45000. he sold one at a loss of 5% and other at a profit of 5%. if the selling price of each set is same, find the cost price of each set
Answers
The cost price of first computer is Rs 23,625.
The cost price of second computer is Rs 21,375.
Given :
A man bought two computers for rs 45000. he sold one at a loss of 5% and other at a profit of 5%. The selling price of each set is the same.
To find :
find the cost price of each set.
Solution :
Let the cost price of one computer be x
The second computer be ( 45000-x)
Selling price :
he sold one at a loss of 5%:
100 - 5 = 95.
Loss on CP = 95x/100
He sold others at a profit of 5%:
100 + 5 = 105
Profit on CP = (45000-x)*105/100
Cost Price :
Given : Selling price is same for each computer:
=> 95x/100 = (45000-x)*105/100
=> 95x/100 = (45000 * 105)/100 - 105x/100
=> 95x/100 = 47250 - 105x/100
=> 95x/100 + 105x/100 = 47250
=> 200x/100 = 47250
=> 2x = 47250
=> x = 47250/2
=> x = 23625
So, the cost of first computer is Rs 23,625
The cost of second computer is 45000 - 23625 = Rs 21,375.
The cost price of each set of computers is Rs. 23625 and Rs. 21375 respectively.
Step-by-step explanation:
Let the C.P. of the first computer be x
As per the question,
The cost of both computers = Rs. 45000
So, the cost of the second computer becomes = Rs. 45000 - x
Loss on the first computer = 5%
∵ S.P. of the first computer will be = x * 95/100
= 95x/100
Gain on the second computer = 5%
∵ S.P. of the second computer = 45000 - x * 105/100
A.T.Q,
95x/100 = 45000 - x * 105/100 (being S.P. of each set the same)
95x = 4725000 - 105x
95x + 105x = 4725000
200x = 4725000
x = 4725000/200
x = Rs. 23625
Since x = C.P. of the first computer = Rs. 23625
The cost of the second computer = Rs. 45000 - Rs. 23625
= Rs. 21375
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