Ratio, Proportion


  1. Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, then they are in the ratio 12 : 23. Find the smaller number.









  1. View Hint View Answer Discuss in Forum

    Let the numbers be 3x and 5x.

    ∴ 
    3x − 9
    =
    12
    5x − 923

    ⇒  69x – 60x = 207 – 108
    ⇒  x =
    99
    = 11
    9

    ∴  The smaller number
    = 3x = 33
    Second Method :
    Here, a = 3, b = 5, x= 9, c = 12, d = 23
    1st Number =
    xa(d − c)
    ad − bc

    =
    9 × 3(23 − 12)
    3 × 23 − 5 × 12

    =
    27 × 11
    69 − 60

    =
    27 × 11
    = 33
    9

    2nd Number=
    xb(d − c)
    ad − bc

    =
    9 × 5(23 − 12)
    3 × 23 − 5 × 12

    =
    45 × 11
    69 − 60

    =
    45 × 11
    = 55
    9

    ∴  Smallest number = 33

    Correct Option: B

    Let the numbers be 3x and 5x.

    ∴ 
    3x − 9
    =
    12
    5x − 923

    ⇒  69x – 60x = 207 – 108
    ⇒  x =
    99
    = 11
    9

    ∴  The smaller number
    = 3x = 33
    Second Method :
    Here, a = 3, b = 5, x= 9, c = 12, d = 23
    1st Number =
    xa(d − c)
    ad − bc

    =
    9 × 3(23 − 12)
    3 × 23 − 5 × 12

    =
    27 × 11
    69 − 60

    =
    27 × 11
    = 33
    9

    2nd Number=
    xb(d − c)
    ad − bc

    =
    9 × 5(23 − 12)
    3 × 23 − 5 × 12

    =
    45 × 11
    69 − 60

    =
    45 × 11
    = 55
    9

    ∴  Smallest number = 33


  1. Two numbers are in the ratio 5 : 7. If 9 is subtracted from each of them, their ratio becomes 7 : 11. The difference of the numbers is









  1. View Hint View Answer Discuss in Forum

    Let the numbers be 5x and 7x.

    Now, 
    5x − 9
    =
    7
    7x − 911

    ⇒  11 (5x – 9) = 7 (7x – 9)
    ⇒  55x – 99 = 49x – 63
    ⇒  55x – 49x = 99 – 63
    ⇒  6x = 36
    ⇒  x = 6
    ∴  Required difference
    = 7x – 5x = 2x = 2 × 6 = 12
    Second Method :
    Here, a = 5, b = 7, x = 9, c = 7, d = 11
    1st Number =
    xa(d − c)
    ad − bc

    =
    9 × 5(11 − 7)
    5 × 11 − 7 × 7

    =
    45 × 4
    55 − 49

    =
    45 × 4
    = 30
    6

    2nd Number
    =
    xb(d − c)
    ad − bc

    =
    9 × 7(11 − 7)
    5 × 11 − 7 × 7

    =
    63 × 4
    55 − 49

    =
    63 × 4
    = 42
    6

    Correct Option: B

    Let the numbers be 5x and 7x.

    Now, 
    5x − 9
    =
    7
    7x − 911

    ⇒  11 (5x – 9) = 7 (7x – 9)
    ⇒  55x – 99 = 49x – 63
    ⇒  55x – 49x = 99 – 63
    ⇒  6x = 36
    ⇒  x = 6
    ∴  Required difference
    = 7x – 5x = 2x = 2 × 6 = 12
    Second Method :
    Here, a = 5, b = 7, x = 9, c = 7, d = 11
    1st Number =
    xa(d − c)
    ad − bc

    =
    9 × 5(11 − 7)
    5 × 11 − 7 × 7

    =
    45 × 4
    55 − 49

    =
    45 × 4
    = 30
    6

    2nd Number
    =
    xb(d − c)
    ad − bc

    =
    9 × 7(11 − 7)
    5 × 11 − 7 × 7

    =
    63 × 4
    55 − 49

    =
    63 × 4
    = 42
    6



  1. The students in three classes are in the ratio 2 : 3 : 5. If 40 students are increased in each class, the ratio changes to 4 : 5 : 7. Originally, the total number of students was :









  1. View Hint View Answer Discuss in Forum

    Let the number of students in three classes be 2x, 3x and 5x respectively.
    Due to increase of 40 students in each class, we have

    =
    2x + 40
    =
    4
    3x + 405

    ⇒  10x + 200 = 12x + 160
    ⇒  2x = 200 – 160 ⇒ 2x = 40
    ⇒  x = 20
    ∴  Original strength
    = 10x = 10 × 20 = 200

    Correct Option: C

    Let the number of students in three classes be 2x, 3x and 5x respectively.
    Due to increase of 40 students in each class, we have

    =
    2x + 40
    =
    4
    3x + 405

    ⇒  10x + 200 = 12x + 160
    ⇒  2x = 200 – 160 ⇒ 2x = 40
    ⇒  x = 20
    ∴  Original strength
    = 10x = 10 × 20 = 200


  1. Two numbers are in the ratio 1
    1
    : 2
    2
    .
    23
    Three numbers are in the ratio
    1
    :
    2
    :
    3
    .
    234
    The greater of the numbers is :









  1. View Hint View Answer Discuss in Forum

    Let the numbers be

    3
    x and
    8
    x
    23

    According to question,

    =
    3
    x + 15 (5/3)/(5/2)
    2
    8x
    + 5
    3

    ⇒  
    3x + 30
    = 2/3
    2
    8x + 45
    3

    ⇒ 
    3(3x + 30)
    =
    2
    2(8x + 45)3

    ⇒ 
    9x + 90
    =
    2
    16x + 903

    ⇒  27x + 270 = 32x + 180
    ⇒  32x – 27x = 270 – 180 = 90
    ⇒  5x = 90 ⇒ x = 18
    ∴  The greater number
    =
    8
    x =
    8
    × 18 = 48
    33

    Correct Option: C

    Let the numbers be

    3
    x and
    8
    x
    23

    According to question,

    =
    3
    x + 15 (5/3)/(5/2)
    2
    8x
    + 5
    3

    ⇒  
    3x + 30
    = 2/3
    2
    8x + 45
    3

    ⇒ 
    3(3x + 30)
    =
    2
    2(8x + 45)3

    ⇒ 
    9x + 90
    =
    2
    16x + 903

    ⇒  27x + 270 = 32x + 180
    ⇒  32x – 27x = 270 – 180 = 90
    ⇒  5x = 90 ⇒ x = 18
    ∴  The greater number
    =
    8
    x =
    8
    × 18 = 48
    33



  1. The ratio of the number of boys and girls of a school with 504 students is 13 : 11. What will be the new ratio if 12 more girls are admitted?









  1. View Hint View Answer Discuss in Forum

    Total numbers of girls in the school

    = 504 ×
    11
    13 + 11

    = 504 ×
    11
    = 231
    24

    Total numbers of boys in the school
    = 504 ×
    13
    13 + 11

    = 504 ×
    13
    = 273
    29

    Now, total no. of girls when 12 more girls are admitted
    = 231 + 12 = 243
    ∴  Required ratio = 273 : 243 = 91 : 81

    Correct Option: A

    Total numbers of girls in the school

    = 504 ×
    11
    13 + 11

    = 504 ×
    11
    = 231
    24

    Total numbers of boys in the school
    = 504 ×
    13
    13 + 11

    = 504 ×
    13
    = 273
    29

    Now, total no. of girls when 12 more girls are admitted
    = 231 + 12 = 243
    ∴  Required ratio = 273 : 243 = 91 : 81