Math, asked by vishnusuthar437, 8 months ago

2. Amit brought two cars. He then sold the
1st car at 10% profit and 2nd one at 25%
profit. The selling price of the 2nd car is
25% more than the SP of the 1st car. What
is the approx. profit % in both the cars
together?
अमित ने दो कार खरीदी ।उसने बाद में पहली
कार 10% लाभ और दूसरी कार को 25% लाभ पर
बेच दिया। दूसरी कार का विक्रयमूल्य पहली कार
के विक्रयमूल्य से 25% अधिक है । तो उसका कुल
लाभ% ज्ञात करे।​

Answers

Answered by bhagyashreechowdhury
0

Given:

Amit bought 2 cars

He sold 1st at 10% profit and 2nd at 25% profit

The selling price of the 2nd car is  25% more than the SP of the 1st car

To find:

What  is the approx. profit % in both the cars  together?

Solution:

Let's assume,

"C.P.₁" = "x" → cost price of 1st car

"C.P.₂" = "y" → cost price of 2nd car

"S.P.₁" → selling price of 1st car

"S.P.₂" → selling price of 2nd car

So,

S.P.₁ = \frac{100 + G\%}{100}\times C.P._1 = \frac{100 + 10}{100}\times x = \frac{110}{100}\times x = 1.1x

and

S.P.₂ = \frac{100 + G\%}{100}\times C.P._1 = \frac{100 + 25}{100}\times y = \frac{125}{100}\times y = 1.25y

Also, the selling price of the 2nd car is 25% more than that of the 1st car

∴ S.P.₂ = S.P.₁ + 25% of S.P.₁

\implies 1.25y = 1.1x + [\frac{25}{100}\times 1.1x ]

\implies 1.25y = 1.1x + 0.275x

\implies 1.25y = 1.375x

\implies y = \frac{1.375x}{1.25}

\implies y = 1.1x

∴ Total C.P. of the two cars is,

= C.P.₁ + C.P.₂

= x + y

= x + 1.1x

= 2.1x

and

∴ Total S.P. of the two cars is,

= S.P.₁ + S.P.₂

= 1.1x + 1.25y

= 1.1x + (1.25 × 1.1x)

= 1.1x + 1.375x

= 2.475x

Now,

The profit % in both cars together is,

= \frac{S.P. - C.P.}{C.P.} \times 100

= \frac{2.475x - 2.1 x}{2.1x} \times 100

= \frac{0.375x }{2.1x} \times 100

= 0.1785 \times 100

= \bold{17.85\%}

Thus, the approx. profit % in both cars  together is \underline{\underline{17.85\%}}.

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