Math, asked by abc22222, 1 year ago

a man bought a number of bicycles for rs 10000. he kept one for his own use and then sold the rest at a price of rs 50 more than the cost price. besides getting his own bicycle for nothing he made a profit of rs 450. how many bicycles did he buy?

Answers

Answered by siddhartharao77
58

Answer:

20

Step-by-step explanation:

Let the number of bicycles he bought for 10000 be x.

∴ Cost price of each bicycle = (10000/x)

Given that the man kept one for his own use.

∴ Number of bicycles sold = x - 1.

Selling price of each bicycle = (10000/x + 50)

According to the given condition,

⇒ (x - 1) + (10000/x + 50) - 10000 = 450

⇒ (x - 1) + (10000 + 50x/x) = 450 + 10000

⇒ 10000x + 50x² - 10000 - 50x = 10450x

⇒ 50x² - 500x - 10000 = 0

⇒ x² - 10x - 200 = 0

⇒ x² - 20x + 10x - 200 = 0

⇒ x(x - 20) + 10(x - 20) = 0

⇒ (x - 20)(x + 10) = 0

⇒ x = 20,-10{∴ number of bicycles cannot be negative}

⇒ x = 20.

Therefore, the number of bicycles he bought = 20.


Hope it helps!

Answered by Anonymous
37

Answer:

Number of cycles is 20.


Step-by-step explanation:


Let number of bicycles bought be x .

So here C.P of x cycles = 10,000  .

C.P of 1 cycle = 10,000 / x


One vehicle is not sold and rest are sold.

So x - 1 cycles are sold .


Now S.P of 1 cycles = 50 more than 10,000 / x

= > 50 + 10,000 / x


S.P = C.P + Profit .

So , S.P = 10,000 + 450

= > S.P = 10,450


So : S.P of x - 1 cycles = 10,450


( x - 1 )( 50 + 10,000/x ) = 10450

= > 50 x + 10,000 - 50 - 10,000 / x = 10450

= > 50 x² + 10,000 x - 10,000 = 10500 x

= > 50 x² - 500 x - 10,000 = 0

= > x² - 10 x - 200 = 0

= > x² - 20 x + 10 x - 200 = 0

= > x ( x - 20 ) + 10 ( x - 20 ) = 0

= > ( x + 10 )( x - 20 ) = 0

x = 20

x = - 10

But no of cycle can't be -ve.

So x = 20.

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