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Profit And Loss Test - 3

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Profit And Loss Test - 3
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  • Question 1
    1 / -0

    A total of 500 people voted for a resolution; but after some discussion, the opponents were increased by 100%. The motion was then rejected by a majority, three times as great as that by which it was formerly passed. How many voted for and how many voted against the resolution initially?

    Solution

    Let the number of opponents be x.
    So, number of supporters = 500 - x

    After discussion, number of opponents = 2x
    And number of supporters = 500 - 2x

    According to the question,
    500 - x = 3(500 - 2x)

    1500 - 6x = 500 - x

    1000 = 5x

    x = 1000/5 = 200
    Number of supporters = 500 - 200 = 300

     

  • Question 2
    1 / -0

    A tradesman marks an article at a price which would give him a profit of 20% on the cost price. To the favoured customers, he makes a deduction of 5% from the marked price. What actual profit does he receive from the sale of an article to a favoured customer for which the latter pays him $28.50?

    Solution

    Let the CP of the article be $x.
    MP = x + 20% of x = $1.2x

    When a discount is given on this MP,
    SP of the article = (95/100) × 1.2x = $1.14x

    But 1.14x = 28.50
    So, x = 25

    CP of the article = $25
    SP = $28.50

    Profit = $3.50

     

  • Question 3
    1 / -0

    A man sells an article at a gain of 10%. If he had bought it at 10% less and sold it for Rs. 132 less, he would have still gained 10%. The cost price of the article is

    Solution

    SP = 1.1CP
    New cost price = 0.9CP
    New selling price = SP - 132

    According to the given condition,
    [(new SP - new CP)/(new CP)]x100 = profit %

    (1.1CP - 132 - 0.9CP)/(0.9CP) = 10/100
    0.2CP - 132 = 0.09CP

    0.11CP = 132
    CP = Rs. 1200

     

  • Question 4
    1 / -0

    According to a survey, the population of a city increases by 10% every year for two years and then decreases by 10% every year for two years. If the population 4 years ago was 100000, what will it be after four years?

    Solution

    Percentage increase (R1) = 10%; n = 2
    Percentage decrease (R2) = 10%; n = 2

    Population 4 years ago = 100000

    Population after 4 years = 100000 (1 + 10/100)2 (1 - 10/100)2
    = 100000 × 110/100 × 110/100 × 90/100 × 90/100 = 98010

     

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