Fractions


  1. A has twice as much money as B. They play together and at the end of the first game B wins one-third of A's money from A, what fraction of the sum that B now has, must A win black in the second game, so that they may have exactly equal money ?
    1. 1/3
    2. 1/5
    3. 1/4
    4. 1/10

  1. View Hint View Answer Discuss in Forum

    Let A has ₹ 200 at the starting of game, then B has ₹ 100 at the starting of game.
    After the game, money left with A = 200 - 200/3 = ₹ 400/3
    ∴ Total money with B = 100 + 200/3 = ₹ 500/3
    Let the fraction of money lost by B = N
    Then, 500/3 - N x 500/3 = 400/3 + N x 500/3

    Correct Option: D

    Let A has ₹ 200 at the starting of game, then B has ₹ 100 at the starting of game.
    After the game, money left with A = 200 - 200/3 = ₹ 400/3
    ∴ Total money with B = 100 + 200/3 = ₹ 500/3
    Let the fraction of money lost by B = N
    Then, 500/3 - N x 500/3 = 400/3 + N x 500/3
    ⇒ 500/3 - 400/3 = 500N/3 + 500N/3
    ⇒ 100/3 = 1000N/3
    ∴ N = (100/3) x (3/1000) = 1/10
    so, the fraction of money lost is 1/10.


  1. Sum of three fractions is 211/24. If the greatest fraction is divided by the smallest fraction, the result is 7/6, which is greater than the middle fraction by 1/3. find all the three fractions ?
    1. 3/5, 4/7, 2/3
    2. 7/8, 5/6, 3/4
    3. 7/9, 2/3, 3/5
    4. 7/8, 7/9, 7/10

  1. View Hint View Answer Discuss in Forum

    Let the fraction be x, y and z respectively in decreasing order.
    Then, according to the question,
    x/z = 7/6
    ⇒ x = 7z/6

    And y = 7/6 - 1/3 = 7-2/6 = 5/6

    Now, x + y + z = 211/24

    Correct Option: B

    Let the fraction be x, y and z respectively in decreasing order.
    Then, according to the question,
    x/z = 7/6
    ⇒ x = 7z/6

    And y = 7/6 - 1/3 = 7-2/6 = 5/6

    Now, x + y + z = 211/24
    ⇒ 7z/6 + 5/6 + z = 59/24
    ⇒ (13z + 5)/ 6 = 59/24
    ⇒ 13z = 59/4 - 5 =39/4
    ⇒ z = 3/4
    ∴ x =7z/6 = 7/8
    Hence, the fraction are 7/8, 5/6, 3/4.



  1. Sum of three fraction is 211/24 . If the greatest fraction is divided by the smallest fraction the result is 7/6, which is greater than the middle fraction by 1/3. Find all the three fractions.
    1. 3/5, 4/7, 2/3
    2. 7/8, 5/6, 3/4
    3. 7/9, 2/3, 3/5
    4. 7/8, 7/9, 7/10
    5. None of the above

  1. View Hint View Answer Discuss in Forum

    Let the greatest, middle and smallest fraction be P, Q and R respectively in decreasing order.

    According to the question.
    (Greatest fraction) / (Smallest fraction) = 7/6
    ⇒ P/R = 7/6
    ⇒ P = 7R/6 ....(i)
    and Q = 7/6 - 1/3 = 7 - 2/6 = 5/6 (ii)

    Now, P + Q + R = 211/24 (iii )

    Substitute the value of P and Q from Eqs. (i), (ii) and put in Eqs (iii),

    Correct Option: B

    Let the greatest, middle and smallest fraction be P, Q and R respectively in decreasing order.

    According to the question.
    (Greatest fraction) / (Smallest fraction) = 7/6
    ⇒ P/R = 7/6
    ⇒ P = 7R/6 ....(i)
    and Q = 7/6 - 1/3 = 7 - 2/6 = 5/6 (ii)

    Now, P + Q + R = 211/24 (iii )

    Substitute the value of P and Q from Eqs. (i), (ii) and put in Eqs (iii), we get
    ⇒ 7R/6 + 5/6 + R = 59/24
    ⇒ (7R + 5 + 6R)/6 = 59/24
    ⇒ (13R + 5) / 6 = 59/24
    ⇒ 13R = (59/4) - 5 = 39/4
    ⇒ R = 39 / (4 x 13)
    ∴ R = 3/4

    On putting the value of R in Eq. (i), we get
    P = 7R/6 = (7/6) x (3/4) = 7/8


  1. In the year 2011, Shantanu gets ₹ 3832.5 as his pocket allowance. Find his pocket allowance per day.
    1. ₹ 9.5
    2. ₹ 10.5
    3. ₹ 12.5
    4. ₹ 11.5
    5. None of the above

  1. View Hint View Answer Discuss in Forum

    Shantanu's pocket allowance = ₹ 3832.50
    Total days in 2011 (general year) = 365 days

    Correct Option: B

    Shantanu's pocket allowance = ₹ 3832.50
    Total days in 2011 (general year) = 365 days
    Allowance per day = 3832.5/365 = ₹ 10.5



  1. 1/8 part of a pencil is black and 1/2 part of the remaining is white. If the remaining part is blue and length of this blue part is 31/2 cm, then find the length of the pencil.
    1. 6 cm
    2. 7 cm
    3. 8 cm
    4. 9 cm
    5. None of the above

  1. View Hint View Answer Discuss in Forum

    Let total length of pencil be x cm.
    Then, black part = x/8
    Remaining part pencil after black part = x - x/8 = 7x/8
    then White part = 1/2 (7x/8)
    = 7x/16
    Now Remaining part of pencil is blue = total length of pencil - (length of black part + length of white part)

    put the value and solve the equation.

    Correct Option: C

    Let total length of pencil be x cm.
    Then, black part = x/8
    Remaining part pencil after black part = x - x/8 = 7x/8
    then White part = 1/2 (7x/8)
    = 7x/16
    Now Remaining part of pencil is blue = total length of pencil - (length of black part + length of white part)
    length of blue part of pencil = x - (x/8 + 7x/16)
    length of blue part of pencil = x - (2x + 7x)/16
    length of blue part of pencil = x - 9x/16
    length of blue part of pencil = (16 x - 9x)/16
    ∴ Length of blue part = 7x/16;

    According to question
    Length of blue part = 31/2
    7x/16 = 31/2
    ⇒ 7x/16 = 7/2
    ⇒ x/16 = 1/2
    ⇒ x = 16/2
    ∴ x = 8 cm