Question

9. Two cycles were bought at the same cost. One was sold at a profit of 15% and another was sold
at a loss of 10%. If the difference in selling price is rs. 200, find the cost of the cycles.

the answer is rs. 800 on my book

i know that correct answer is 800

but give me explanation plz that how to do it ....but the answer should be 800

those who know only that answer...otherwise not answer ​

Answer 1

${\underline {\underline {\huge {\sf { Required \: Answer: } }}}}$

Procedure to solve this:

• Here, we are given that two cycles were bought at the same cost, this means their cost price is same, let's assume it as x.
• Then, we are given that the first one was sold at a profit of 15%,

⇢ Profit = 15% of x. [First cycle]

⇢ S.P = x + Profit [ First cycle]

• Also, we are given that the 2nd one was sold at a loss of 10%,

⇢ Loss = 10% of x. [Second cycle]

⇢ S.P = x - Loss [ Second cycle]

• After that, we are given that the difference in selling price is 200.
• So, we will form a suitable equation to find the cost of the cycles (x).

Solution:

Given:

• Two cycles were bought at the same cost.

• One was sold at a profit of 15% and another was sold at a loss of 10%.

• Difference in selling price is Rs.200.

To calculate:

• Cost of the cycles (Cost price)

Calculation:

Let's assume the cost price of the both cycles as x.

In the case of first cycle:

• Profit = 15% of x

$\sf { Profit = \dfrac{15}{100}x }$

So,

• S.P = x + Profit

$\sf { S.P =x + \dfrac{15}{100}x }$...........(i)

In the case of second cycle:

• Loss = 10% of x

$\sf { Loss = \dfrac{10 }{100}x }$

So,

• S.P = x - loss

$\sf { S.P =x - \dfrac{10}{100}x }$ ............... (ii)

According to the question:–

• Difference in selling price of both cycles is Rs.200. ( Put the values from eq. (i) and (ii)

$\sf { ⇢ \Bigg( x + \dfrac{15}{100}x \Bigg) - \Bigg( x - \dfrac{10}{100}x \Bigg) = 200 }$

⠀⠀⠀

$\sf { ⇢ \Bigg( \dfrac{100x + 15x}{100} \Bigg) - \Bigg( \dfrac{100x-10x}{100} \Bigg) = 200 }$

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$\sf { ⇢ \Bigg( \dfrac{115x}{100} \Bigg) - \Bigg( \dfrac{90x}{100} \Bigg) = 200 }$

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$\sf { ⇢ \dfrac{115x}{100} - \dfrac{90x}{100} = 200 }$

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$\sf { ⇢ \dfrac{115x-90x}{100} = 200 }$

⠀⠀⠀

$\sf { ⇢ \dfrac{25x}{100} = 200 }$

⠀⠀⠀

$\sf { ⇢ 25x = 200 \times 100}$

⠀⠀⠀

$\sf { ⇢ 25x = 20000}$

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$\sf { ⇢ x = \dfrac{ 20000}{25} }$

⠀⠀⠀

$\sf { ⇢ x = Rs. 800}$

⠀⠀⠀

$\boxed { \sf \pink { ⇢ C.P= Rs. 800}}$

Therefore, C.P of both cycles were Rs. 800.

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Answer 2

Answer:

200

Step-by-step explanation:

Let cp of both be Rs x each

So

SP1 = x + 15% x

SP2 = 10% of x - x

SP1 - SP2 = 200

x + 1.15x - (x - 0.9x)

= 200 .