Question

fruit-seller bought x apples for 1200.
(1) Write the cost price of each apple in terms of x.
(ii) If 10 of the apples were rotten and he sold each of the rest at 3 more than the cost
price of each, write the selling price of (x - 10) apples.
(i) If he made a profit of 60 in this transaction, form an equation in x and solve it to
evaluate x.​

Answer 1

$\large{\frak{ \blue{answer}}}$

x apples are bought for 1200 Rs

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(i) cost of per apple = (1200 / x ) Rs

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(ii) 10 apples were rotten

hence, apple left for sale = x - 10

the cost at which apples will be sold is 3 more than the cost price of each apple

that will be = (1200/x) + 3 Rs

= ( 1200+3x) / x Rs

so, new selling price will be

$\tt \: = \frac{1200 + 3x}{x} \times (x - 10) \\ \\ \tt = \frac{1200x - 12000 + 3 {x}^{2} - 30x }{x} \\ \\ \tt = \frac{3 {x}^{2} + 1170x - 12000}{x} \: rs$

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(iii) it is given that the profit made during this transaction is 60 Rs

it means

$\tt \frac{3 {x}^{2} + 1170x - 12000 }{x} - 1200 = 60 \\ \\ \tt \frac{3 {x}^{2} + 1170x - 12000 - 1200x }{x} = 60 \\ \\ \tt 3 {x}^{2} - 30x - 12000 = 60x \\ \\ \tt 3 {x}^{2} - 90x - 12000 = 0 \\ \\ \tt \: (dividing \: by \: 3 \: both \: sides) \\ \\ \tt \: {x}^{2} - 30x - 4000 = 0 \\ \\ \tt \: {x}^{2} - 80x + 50x - 4000 = 0 \\ \\ \tt \: x(x -80) + 50(x - 80) = 0 \\ \\ \tt \: (x + 50)(x - 80) = 0 \\ \\ \tt \: x = - 50 \: and \: x = 80$

Cost should not be negative hence we will take value of x as

$\boxed{ \tt \: x = 80}$

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Answer 2

Step-by-step explanation:

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