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A person bought two bicycles for ₹1600 and sold the first at 10% profit and the second at 20% profit. If he sold the first at 20% profit and the second at 10% profit, he would get ₹5 more. The difference in the cost price of the two bicycles was

#### Options

₹25

₹75

₹50

₹40

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#### Solution

**₹50**

**Explanation :**

Let the cost price of the first bicycle be ₹ x.

Then, the cost price of second bicycle = ₹(1600 - x)

According to the given condition,

20% of x + 10% of (1600 - x) - [10% of x + 20% of (1600 - x)] = 5

`⇒ [(20xxx)/100+(10xx(1600-x))/100] -[(10xxx)/100+(20xx(1600-x))/100]=5`

⇒`(x/5+(1600-x)/10)- (x/10+(1600-x)/5)=5`

⇒`x/5-x/10+((1600-x))/10-((1600-x))/5=5`

⇒`(2x-x)/10+((1600-x)-2(1600-x))/10=5`

⇒`(x+1600-x-3200+2x)/10=5`

⇒`(-1600+2x)/10=5 `

⇒`2x=1600+50`

⇒`x=1650/2=825`

∴ Cost of second bicycle = (1600 - 825) = ₹775

Required difference = 825 - 775

= ₹50

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