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  1. Asked: In:

    Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

    Ranamita

    Let Baichung’s age be x years, then Baichung’s father’s age = (x+29) years and Baichung’s granddaughter’s age = (x+29+26) = (x+55) years. According to condition, x+x+29+x+55=135 ⇒ 3x+84=135 ⇒ 3x+84-84=135-84 [Subtracting 84 from both sides] ⇒ 3x=51 ⇒ 3x/3=51/3 [Dividing both sides by 3] ⇒ x=17 yearsRead more

    Let Baichung’s age be x years, then Baichung’s father’s age = (x+29) years and
    Baichung’s granddaughter’s age = (x+29+26) = (x+55) years.
    According to condition, x+x+29+x+55=135
    ⇒ 3x+84=135
    ⇒ 3x+84-84=135-84 [Subtracting 84 from both sides]
    ⇒ 3x=51
    ⇒ 3x/3=51/3 [Dividing both sides by 3]
    ⇒ x=17 years

    Hence, Baichung’s age = 17 years, Baichung’s father’s age = 17 + 29 = 46 years and
    Baichung’s granddaughter’s age = 17 + 29 + 26 = 72 years.

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/

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  2. Asked: In:

    The number of boys and girls in a class are in the ratio 7: 5. The number of boys is 8 more than the number of girls. What is the total class strength?

    Ranamita

    Let the number of girls be x. Then, the number of boys = x+8. According to the question, x+8/x=7/5 ⇒ 5(x+8)=7x ⇒ 5x+40=7x ⇒ 5x-7x = -40 [Transposing 7x to L.H.S. and 40 to R.H.S.] ⇒ -2x=-40 ⇒ -2x/-2=-40/-2 [Dividing both sides by 2] ⇒ x=20 Hence the number of girls = 20 and number of boys = 20 + 8Read more

    Let the number of girls be x. Then, the number of boys = x+8.
    According to the question, x+8/x=7/5
    ⇒ 5(x+8)=7x ⇒ 5x+40=7x
    ⇒ 5x-7x = -40 [Transposing 7x to L.H.S. and 40 to R.H.S.]
    ⇒ -2x=-40
    ⇒ -2x/-2=-40/-2 [Dividing both sides by 2]
    ⇒ x=20

    Hence the number of girls = 20 and number of boys = 20 + 8 = 28.

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/

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  3. Asked: In:

    The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

    Ranamita

    Let the present ages of Rahul and Haroon be 5x years and 7x years respectively. According to condition, (5x+4)+(7x+4)=56 ⇒ 12x+8=56 ⇒ 12x+8-8=56-8 [Subtracting 8 from both sides] ⇒ 12x/8 ⇒ 12x/12=48/12 [Dividing both sides by 12] ⇒ x=4 Hence, present age of Rahul = 5 x 4 = 20 years and present age oRead more

    Let the present ages of Rahul and Haroon be 5x years and 7x years respectively.
    According to condition, (5x+4)+(7x+4)=56
    ⇒ 12x+8=56
    ⇒ 12x+8-8=56-8 [Subtracting 8 from both sides]
    ⇒ 12x/8
    ⇒ 12x/12=48/12 [Dividing both sides by 12]
    ⇒ x=4
    Hence, present age of Rahul = 5 x 4 = 20 years and present age of Haroon = 7 x 4 = 28 years.

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/

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  4. Asked: In:

    Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

    Ranamita

    Let the three consecutive integers be x,x+1 and x+2 According to the question, 2x+3(x+1)+4(x+2)=74 ⇒ 2x+3x+3+4x+8=74 ⇒ 9x+11=74 ⇒ 9x+11-11=74-11 [Subtracting 11 from both sides] ⇒ 9x+63 ⇒ 9x/9=63/9 [Dividing both sides by 9] ⇒ x=7 Hence first integer = 7, second integer = 7 + 1 = 8 and third integerRead more

    Let the three consecutive integers be x,x+1 and x+2
    According to the question, 2x+3(x+1)+4(x+2)=74
    ⇒ 2x+3x+3+4x+8=74
    ⇒ 9x+11=74
    ⇒ 9x+11-11=74-11 [Subtracting 11 from both sides]
    ⇒ 9x+63
    ⇒ 9x/9=63/9 [Dividing both sides by 9]
    ⇒ x=7

    Hence first integer = 7, second integer = 7 + 1 = 8 and third integer = 7 + 2 = 9.

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/

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  6. Asked: In:

    The sum of three consecutive multiples of 8 is 888. Find the multiples.

    Ranamita

    Let the three consecutive multiples of 8 be x,x+8 and x+16. According to question, x+x+8+x+16=888 ⇒ 3x+24=888 ⇒ 3x+24-24=888-24 [Subtracting 24 from both sides] ⇒ 3x=864 ⇒ 3x/3 = 864/3 ⇒ x=288 Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 =Read more

    Let the three consecutive multiples of 8 be x,x+8 and x+16.
    According to question, x+x+8+x+16=888
    ⇒ 3x+24=888
    ⇒ 3x+24-24=888-24 [Subtracting 24 from both sides]
    ⇒ 3x=864
    ⇒ 3x/3 = 864/3
    ⇒ x=288
    Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 = 304.

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-2/

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