Question

A man purchased a car and a bike for ₹2,60,000. He sells the bike at 20% loss and the car at 30% profit. On the whole transaction he earns a profit of ₹18,000. Find the SP of the bike.

₹120,000
₹180,000
₹140,000
₹96,0​

Answer 1

Given:

A man purchased a car and a bike for ₹2,60,000.

He sells the bike at 20% loss and the car at 30% profit.

On the whole transaction, he earns a profit of ₹18,000.

To find:

The SP of the bike.

Solution:

The total C.P. of a car and a bike = Rs. 260000

Let's assume,

"x" → as the C.P. of the bike

Then,

"(260000 - x)" → will be the C.P. of the car

The total S.P. of a car and bike is,

= [S.P. of bike] + [S.P. of car]

= $[\frac{100 - 20}{100} \times x] + [\frac{100 + 30}{100} \times (260000 - x)]$

= $[\frac{ 80}{100} \times x] + [\frac{130 }{100} \times (260000 - x)]$

= $[0.8x] + [1.3 \times (260000 - x)]$

= $0.8x + 338000 - 1.3x$

= $Rs. (- 0.5x + 338000)$

It is given that the man earns a profit of Rs. 18000 on the whole transaction, so we can form an equation as:

[Total S.P.] - [Total C.P.] = Profit on the whole transaction = 18000

$\implies (-0.5x + 338000) - \:260000 = 18000$

$\implies-0.5x + 338000 = 18000 + 260000$

$\implies-0.5x + 338000 = 278000$

$\implies-0.5x = 278000 - 338000$

$\implies-0.5x = -60000$

$\implies \bold{x = 120000}$ ← C.P. of the bike

Now,

The S.P. of the bike is,

= $\frac{100 - 20}{100}\times 120000$

= $\frac{80}{100}\times 120000$

= $80 \times 1200$

= $\bold{96000}$ ← option (D)

Thus, the S.P. of the bike is → Rs. 96000.

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Answer 2

$Given \: cost \:price \: of \: a \:car \: and$

$bike = Rs \: 2,60,000$

$Let \: cost \:price \:of \:the \:car = x$

$cost \:price \:of \:the \:bike = Rs \:(260000 - x)$

/* He sells the bike 20% loss */

$i) selling \:price \: of \:the \: bike$

$= c.p \Big( \frac{100-loss\% }{100}\Big)$

$= (260000-x) \times \frac{ 100-20}{100} \\= (260000-x) \times \frac{ 80}{100} \\= (260000-x) \times \frac{ 8}{10} \\= \frac{2080000-8x}{10} \:--(1)$

/* He sells the car at 30% profit */

$ii)Selling \:price \:of \:the \:car$

$= x\Big( \frac{100+30}{100}\Big)$

$= x \times \frac{130}{100}$

$= \frac{13x}{10} \: --(2)$

$\blue { Total \:profit = Rs 18000 }$

$Total \: s.p - Total \:c.p = 18000$

$\implies \frac{2080000-8x}{10} + \frac{13x}{10} - 260000= 18000$

$\implies 2080000 - 8x + 13x = 180000 + 260000$

$\implies 5x = 2780000 - 2080000$

$\implies 5x = 700000$

$\implies x = 140000$

/* Put x = 140000 in equation (1) , we get*/

$s.p \: of \: the \:bike \\= \frac{2080000-8\times 140000}{10}\\= \frac{2080000 - 1120000}{10}\\= \frac{960000}{10}\\= Rs \: 96000$

Therefore.,

$\red{s.p \: of \: the \:bike}\green { = Rs\:96000}$

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