Profit and Loss


  1. A man bought a horse and a carriage for $ 40,000. He sold the horse at a gain of 10 % and the carriage at a loss of 5%. He gained 1% on his whole transaction. The cost price of the horse was :









  1. View Hint View Answer Discuss in Forum

    Given in question , Total CP of a horse and a carriage = $ 40,000.
    Suppose the C.P. of horse be $ y, then
    C.P. of carriage = $ (40000 – y)
    Then,

    110 × y
    +
    (40000 - y) × 95
    =
    40000 × 101
    100100100

    ⇒ 110y + 3800000 – 95y = 4040000

    Correct Option: B

    Given in question , Total CP of a horse and a carriage = $ 40,000.
    Suppose the C.P. of horse be $ y, then
    C.P. of carriage = $ (40000 – y)
    Then,

    110 × y
    +
    (40000 - y) × 95
    =
    40000 × 101
    100100100

    ⇒ 110y + 3800000 – 95y = 4040000
    ⇒ 15y = 4040000 – 3800000
    ⇒ 15y = 240000
    ⇒ y =
    240000
    = $ 16000
    15


  1. A man sold two articles at $ 375 each. On one, he gains 25% and on the other, he loses 25%. The gain or loss% on the whole transaction is









  1. View Hint View Answer Discuss in Forum

    Here, both the articles are sold at the same price. Hence, there is always loss.

    ∴ Required loss% =
    y2
    100

    Here , y = 25%

    Correct Option: D

    Here, both the articles are sold at the same price. Hence, there is always loss.

    ∴ Required loss% =
    y2
    100

    Here , y = 25%
    Loss percent =
    25 × 25
    =
    25
    = 6
    1
    %
    10044



  1. A person bought two articles A and B for $ 5,000. He sold A at 20% profit and B at 10% loss. He thus gained 2% on his outlay. The cost price of A was









  1. View Hint View Answer Discuss in Forum

    Given that , CP of two articles A and B = $ 5,000
    Let the CP of article A be $ y
    ∴ CP of article B = $ (5000 – y)
    According to the question,
    120% of y + 90% of (5000 – y) = 102% of 5000
    ⇒ 120y + 450000 – 90y = 510000
    ⇒ 30y = 510000 – 450000 = 60000

    Correct Option: C

    Given that , CP of two articles A and B = $ 5,000
    Let the CP of article A be $ y
    ∴ CP of article B = $ (5000 – y)
    According to the question,
    120% of y + 90% of (5000 – y) = 102% of 5000
    ⇒ 120y + 450000 – 90y = 510000
    ⇒ 30y = 510000 – 450000 = 60000

    ⇒ y =
    60000
    = $ 2000
    30

    Hence , The cost price of A is $ 2000 .


  1. A trader bought two horses for $ 19,500. He sold one at a loss of 20% and the other at a profit of 15%. If the selling price of each horse is the same, then their cost price are respectively.









  1. View Hint View Answer Discuss in Forum

    Given that , Total CP of two horses ( y ) = $ 19,500.
    One of them is sold at a loss of a% and other is sold at a gain of b% and their S.P. is same.

    ∴ C.P. of article sold at a loss of a% =
    100 + b
    × y
    200 - a + b

    Here , a = 20 , b = 15%
    C.P. of first horse =
    100 + 15
    × 19500
    200 - 20 + 15

    Correct Option: B

    Given that , Total CP of two horses ( y ) = $ 19,500.
    One of them is sold at a loss of a% and other is sold at a gain of b% and their S.P. is same.

    ∴ C.P. of article sold at a loss of a% =
    100 + b
    × y
    200 - a + b

    Here , a = 20 , b = 15%
    C.P. of first horse =
    100 + 15
    × 19500
    200 - 20 + 15

    C.P. of first horse =
    115
    × 19500 = $ 11500
    195

    ⇒ C.P. of second horse = 19500 - 11500 = $ 8000



  1. A man buys a field of agricultural land for $ 3,60,000. He sells one- third at a loss of 20% and two- fifths at a gain of 25%. At what price must he sell the remaining field so as to make an overall profit of 10 %?









  1. View Hint View Answer Discuss in Forum

    Here , CP of total agricultural field = $ 3,60,000

    SP of total agricultural field = $360000 ×110= $ 396000
    100

    SP of one–third of the field =
    1
    × 360000 ×
    80
    = $ 96000
    3100

    SP of
    2
    th of the field
    5

    Correct Option: C

    Here , CP of total agricultural field = $ 3,60,000

    SP of total agricultural field = $360000 ×110= $ 396000
    100

    SP of one–third of the field =
    1
    × 360000 ×
    80
    = $ 96000
    3100

    SP of
    2
    th of the field
    5

    Or SP of two –fifth of the field =
    2
    × 360000 ×
    125
    = $ 180000
    5100

    ∴ SP of the remaining field = $ (396000 – 96000 – 180000) = $ 120000