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Profit And Loss Test - 2

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Profit And Loss Test - 2
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  • Question 1
    1 / -0

    By selling 12 articles, a shopkeeper gains a profit equal to the cost price of 2 articles. What is his profit percentage?

    Solution

    Assume CP of one article as Re 1.
    So, by selling 12 articles (worth Rs. 12), he can make a profit of Rs. 2.
    So, profit% = 2/12 ×100 = 16.66%

     

  • Question 2
    1 / -0

    Raghu bought 4 dozens of oranges at Rs. 12 per dozen and 2 dozens of oranges at Rs. 16 per dozen. He sold them all and earned a profit of 20%. At what price per dozen did he sell the oranges?

    Solution

    Total CP = Rs. (12 × 4 + 16 × 2) = Rs. (48 + 32) = Rs. 80
    SP after profit = Rs. (1.2 × 80) = Rs. 96
    SP per dozen = Rs. (96/6) = Rs. 16

     

  • Question 3
    1 / -0

    A cloth merchant used to give 10% extra cloth on every purchase. Now he increases the selling price by an amount equal to x% of the cost price and does not give any extra length. If the profit percentage before the increase was 10% and now the profit is 20%, find the value of x.

  • Question 4
    1 / -0

    A chair dealer incurs an expense of Rs. 225 for manufacturing every chair. He also incurs an additional expenditure of Rs. 40,000, which is independent of the number of chairs manufactured. If he is able to sell a chair during the season, he sells it for Rs. 350. If he fails to do so, he has to sell each chair for Rs. 125. If he manufactures 1850 chairs, what is the number of chairs that he must sell during the season in order to break even, given he is able to sell all the chairs manufactured?

    Solution

    Let the number of chairs required to be sold in the season be x.
    He sells (1850 - x) chairs out of season.

    The expenditure for manufacturing 1850 chairs is (1850 × 225) = Rs. 4,16,250.
    Add this to the fixed expenditure of Rs. 40,000.

    His total income is from x chairs sold at Rs. 350 and (1850 - x) chairs sold at Rs. 125.
    Thus, 456250 = (350)(x) + (125)(1850 - x)

    On solving, we get
    x = 1000

     

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