Question

A shopkeeper sold 2 articles each for rs 1560 then he secured 20% profit on first and 30% profit on other, then what was the total CP of both articles

Answer 1

$<i><b>Hi !$

Here is your answer !!

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S.P. of one article = ₹ 1560

Profit % = 20 %

C.P. = ?

$\bf \: c.p. \: = \: \frac{100}{( \: 100 \: + \: profit \: \% \: )} \: \times \: s.p. \\ \\ = > \: \bf \frac{100}{100 \: + \: 20} \: \times \: 1560 \\ \\ = > \: \bf \frac{100}{120} \: \times \: 1560 \\ \\ = > \: \bf 1300$

C.P. of one article = ₹ 1300

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S.P. of other article = ₹ 1560

Profit % = 30 %

C.P. = ?

$\bf \: c.p. \: = \: \frac{100}{( \: 100 \: + \: 30 \: )} \: \times \: 1560 \\ \\ = > \: \bf \frac{100}{130} \: \times \: 1560 \\ \\ = > \bf\: 1200$

C.P. of other article = ₹ 1200

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Total C.P. of both the articles =

=> 1300 + 1200

$<u>=> ₹ 2500 ANS$

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Thanks !!!

Answer 2

✍✍HERE IS YOUR ANSWER...⬇⬇
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$\underline{SOLUTION}$ ➡

Let,

C.P. of the first article = Rs. x

C.P. of the sencond article = Rs. y

•°• Total C.P. of the both articles = Rs. ( x + y )

Given,

S.P. of the first article = Rs. 1560

Also given,

Profit on the first article = 20%

A.T.Q.,

$x + 20\% \: \: of \: \: x \: = 1560 \\ \\ = >x + \frac{20}{100} \times x = 1560 \\ \\ = > \frac{120x}{100} =156 0 \\ \\ = > 120x = 156000 \\ \\ = > x = \frac{156000}{120} \\ \\ = > x =1300$

•°• C.P. of the first article , x = Rs. 1300

Again,

Given,

S.P. of the second article = Rs. 1560

Also given,

Profit on the second article = 30%

A.T.Q.,

$y + 30\% \: \: of \: \: y \: = 1560 \\ \\ = >y + \frac{30}{100} \times y = 1560 \\ \\ = > \frac{130y}{100} = 1560 \\ \\ = >1 30y = 156000 \\ \\ = > y = \frac{156000}{130} \\ \\ = > y = 1200$

•°• C.P. of the second article , y = Rs. 1200

•°• Total C.P. of the both articles = Rs. ( x + y )

= Rs. ( 1300 + 1200 )

= Rs. 2500

$\underline{ANSWER}$ ➡ $\boxed{RS. 2500}$

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