Percentage


  1. A number is increased by x %; to get back to the original number, it is to be reduced by









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    Initial value
    increasing value

    P × x
    → increased value → P +
    Px
    100100

    = P
    100 + x
    100

    ∴ Required answer
    =
    x
    × 100%
    100 + x

    Correct Option: B

    Initial value
    increasing value

    P × x
    → increased value → P +
    Px
    100100

    = P
    100 + x
    100

    ∴ Required answer
    =
    x
    × 100%
    100 + x


  1. A number is first decreased by 20%. The decreased number is then increased by 20%. The resulting number is less than the original number by 20. Then the original number is









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    Effective percentage

    =- 20 + 20 -
    20 × 20
    = – 4%
    100

    If the number be x, then
    4% of x = 20
    ⇒ x ×
    4
    = 20
    100

    ⇒ x =
    20 × 100
    = 500
    4

    Aliter : Using Rule 3,
    Let the number be x
    Decrease % =
    202
    = 20
    100

    x – 4% of x = x – 20
    4x
    = + 20
    100

    x = 500

    Correct Option: C

    Effective percentage

    =- 20 + 20 -
    20 × 20
    = – 4%
    100

    If the number be x, then
    4% of x = 20
    ⇒ x ×
    4
    = 20
    100

    ⇒ x =
    20 × 100
    = 500
    4

    Aliter : Using Rule 3,
    Let the number be x
    Decrease % =
    202
    = 20
    100

    x – 4% of x = x – 20
    4x
    = + 20
    100

    x = 500



  1. The difference between the value of the number increased by 20% and the value of the number decreased by 25% is 36. Find the number.









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    Let the number be x.
    ∴ (20 + 25)% of x = 36

    45x
    = 36
    100

    ⇒ x =
    36 × 100
    = 80
    45

    Correct Option: D

    Let the number be x.
    ∴ (20 + 25)% of x = 36

    45x
    = 36
    100

    ⇒ x =
    36 × 100
    = 80
    45


  1. The strength of a school increases and decreases in every alternate year by 10%. It started with increase in 2000. Then the strength of the school in 2003 as compared to that in 2000 was









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    Using Rule 4,
    Increase in first year = 10%
    Decrease in 2nd year = 10%
    Effective result

    =10 - 10 -
    10 × 10
    % = –1%
    100

    Increase in 3rd year = 10%
    ∴ Effective result
    =10 - 1 -
    10 × 1
    %
    100

    = (9 – 0.1)% = 8.9% (increase)

    Correct Option: A

    Using Rule 4,
    Increase in first year = 10%
    Decrease in 2nd year = 10%
    Effective result

    =10 - 10 -
    10 × 10
    % = –1%
    100

    Increase in 3rd year = 10%
    ∴ Effective result
    =10 - 1 -
    10 × 1
    %
    100

    = (9 – 0.1)% = 8.9% (increase)



  1. If each side of a cube is increased by 10% the volume of the cube will increase by









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    Single equivalent increase for 10% and 10%

    =10 + 10 +
    10 × 10
    % = 21%
    100

    Again, single equivalent increase for 21% and 10%
    =21 + 10 +
    21 × 10
    %
    100

    = 31 + 2.1 = 33.1%
    Aliter : Using Rule 14,
    Increase % in volume
    =3 × 10 +
    3 × 102
    +
    102
    %
    100(100)2

    =30 + 3 +
    1
    % = 33.1%
    100

    Note : Volume of cube = (Edge)3
    Hence, formulax + y +
    xy
    % should be used twice.
    100

    Correct Option: C

    Single equivalent increase for 10% and 10%

    =10 + 10 +
    10 × 10
    % = 21%
    100

    Again, single equivalent increase for 21% and 10%
    =21 + 10 +
    21 × 10
    %
    100

    = 31 + 2.1 = 33.1%
    Aliter : Using Rule 14,
    Increase % in volume
    =3 × 10 +
    3 × 102
    +
    102
    %
    100(100)2

    =30 + 3 +
    1
    % = 33.1%
    100

    Note : Volume of cube = (Edge)3
    Hence, formulax + y +
    xy
    % should be used twice.
    100