1. (f) 33 : 44 = 33/44 = 3/4 75 : 100 = 75/100 = 3/4=3:4 Since 33 : 44 = 75 : 100 Therefore, 33, 44, 75, 100 are in ratio. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

    (f) 33 : 44 = 33/44 = 3/4
    75 : 100 = 75/100 = 3/4=3:4
    Since 33 : 44 = 75 : 100
    Therefore, 33, 44, 75, 100 are in ratio.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

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  2. (a) 15:45 = 15/45=1/3=1:3 40 : 120 = 40/120=1/3=1:3 Since 15 : 45 = 40 : 120 Therefore, 15, 45, 40, 120 are in proportion. (b) 33 : 121 = 33/121=3/11=3:11 9 : 96 = 9/96=3/32=3:11 Since 33 : 121 ≠ 9 : 96 Therefore, 33, 121, 9, 96 are not in proportion. (c) 24 : 28 = 24/28 = 6/7 =6:7 36 : 48 = 36/48 =Read more

    (a) 15:45 = 15/45=1/3=1:3
    40 : 120 = 40/120=1/3=1:3
    Since 15 : 45 = 40 : 120
    Therefore, 15, 45, 40, 120 are in proportion.

    (b) 33 : 121 = 33/121=3/11=3:11
    9 : 96 = 9/96=3/32=3:11
    Since 33 : 121 ≠ 9 : 96
    Therefore, 33, 121, 9, 96 are not in proportion.

    (c) 24 : 28 = 24/28 = 6/7 =6:7
    36 : 48 = 36/48 = 3/4=3:4
    Since 24 : 28 ≠ 36 : 48
    Therefore, 24, 28, 36, 48 are not in proportion.

    (d) 32 : 48 = 32/48 = 2/3 = 2 : 3
    70 : 210 = 70/120 = 1/3 = 1:3
    Since 32 : 48 ≠ 70 : 210
    Therefore, 32, 48, 70, 210 are not in proportion.

    (e) 4 : 6 = 4/6= 2/3- 2:3
    8 : 12 = 8/12 = 2/3 = 2:3
    Since 4 : 6 = 8 : 12
    Therefore, 4, 6, 8, 12 are in proportion.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

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  3. (a) Ratio of father’s present age to that of son = 42/14 = 3/1 = 3 : 1 (b) When son was 12 years, i.e., 2 years ago, then father was (42 – 2) = 40 years Therefore, the ratio of their ages = 40/12 = 10/3 = 10:3 (c) Age of father after 10 years = 42 + 10 = 52 years Age of son after 10 years = 14 + 10Read more

    (a) Ratio of father’s present age to that of son = 42/14 = 3/1 = 3 : 1
    (b) When son was 12 years, i.e., 2 years ago, then father was (42 – 2) = 40 years
    Therefore, the ratio of their ages = 40/12 = 10/3 = 10:3
    (c) Age of father after 10 years = 42 + 10 = 52 years
    Age of son after 10 years = 14 + 10 = 24 years
    Therefore, ratio of their ages = 52/24 = 13/6 = 16:6
    (d) When father was 30 years old,
    i.e., 12 years ago, then son was (14 – 12) = 2 years old
    Therefore, the ratio of their ages = 30/2 = 15/1 = 15:1

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

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  4. Ratio of the age of Shreya to that of Bhoomika = 15/12 = 5/4 =5 : 4 Thus, ₹36 divide between Shreya and Bhoomika in the ratio of 5 : 4. Shreya gets = 5/9 of ₹36 = 5/9x36=₹20 Bhoomika gets = 4/9 of ₹36 = 4/9x36=₹16 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

    Ratio of the age of Shreya to that of Bhoomika = 15/12 = 5/4 =5 : 4
    Thus, ₹36 divide between Shreya and Bhoomika in the ratio of 5 : 4.
    Shreya gets = 5/9 of ₹36 = 5/9×36=₹20
    Bhoomika gets = 4/9 of ₹36 = 4/9×36=₹16

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

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  5. Ratio between Sheela and Sangeeta = 3 : 2 Total these terms = 3 + 2 = 5 Therefore, the part of Sheela = 3/5 of the total pens and the part of Sangeeta = 2/5 of total pens Thus, Sheela gets = 3/5x20=12pens and Sangeeta gets = 2/5x20=8 pens https://www.tiwariacademy.com/ncert-solutions/class-6/maths/cRead more

    Ratio between Sheela and Sangeeta = 3 : 2
    Total these terms = 3 + 2 = 5
    Therefore, the part of Sheela = 3/5 of the total pens
    and the part of Sangeeta = 2/5 of total pens
    Thus, Sheela gets = 3/5×20=12pens
    and Sangeeta gets = 2/5×20=8 pens

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-12/

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