Question

HIIIII !!!!
I WILL MAKE YOU BRAINLIST
ANSWER THIS QUESTION....❤️❤️❤️

Answer 1

Let the cost price of one computer be 'x' and the cost price of another be 45000-x.

(i)

He sold one at a loss of 5%.

Selling price = 95% of x

                     = (95/100)x.

(ii)

He sold other at a profit of 5%.

Selling price = 105% of (45000 - x)

                     = (105/100) * (45000 - x)

Given that selling price of each set is same.

⇒ (95/100)x = (105/100) * (45000 - x)

⇒ 95x = 105(45000 - x)

⇒ 95x = 4725000 - 105x

⇒ 200x = 4725000

⇒ x = 23625.

Then:

⇒ 45000 - x

⇒ 45000 - 23625

⇒ 21375.

Therefore:

⇒ Cost price of 1st computer = 23625.

⇒ Cost price of 2nd computer = 21375.

Hope it helps!

Answer 2

cost price of two computers = ₹45000


let the cost price of 1st computer be '₹x'


cost price of 2nd computer


c.p2 = ₹(45000 - x)


On First computer:
----------------------------



loss = 5% of x = ₹ 0.05x




Find the s.p1 :
--------------------



selling price of 1st computer

s.p1 = ₹x - ₹0.05x = ₹0.95x


On 2nd computer:
---------------------------



profit = 5% of (45000 - x )



= 0.05 ( 45000 - x )



= 2250 - 0.05x




Find the s.p2:
---------------------


Selling price of 2nd computer



s.p2 = (45000 - x) + 2250 - 0.05x



= 47250 - ( x + 0.05x)


= 47250 - 1.05x



since ,the selling price of 1st computer
= selling price of 2nd computer ,so then


s.p1 = s.p2


0.95x = 47250 - 1.05x


2x = 47250 => x = ₹23625


Answer:
--------------

cost price of 1st computer = ₹23625


cost price of 2nd computer = ₹45000 - x


=45000 - ₹23625 = ₹21375
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