Math, asked by Elsa1221, 5 months ago

10 years before, the ages of a mother and her daughter were in the ratio 3: 1. On another 10 years from now, the ratio of their ages will be 13: 7. What are their present ages?​

Answers

Answered by Anonymous
1

Answer:

) 27 yr

Description for Correct answer:

Let the ages of A and B 10 yr before were 13x yr and 17x yr, respectively.

Then, present age of A = 13x + 10 and present age of B = 17x + 10

According to the question,

\( \large\frac{13x + 10 + 17}{17x + 10 + 17} = \frac{10}{11} \)

\( \large\frac{13x + 27 }{17x + 27} = \frac{10}{11} \)

Hence, present age of B = 17 x 1 + 10 = 27yr

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Answered by 1184ashish
1

Answer:

let mother's age before 10 years be 3x and daughter's age be x.

at present

mother's age = 3x+10

daughter's age = x+10

after 10 years

mother's age = 3x+20

daughter's age = x+20

by question their age will be in ratio 13:7

therefore, 3x+20/x+20 = 13/7

= 7(3x+20) = 13(x+20)

= 21x+140 = 13x+260

= 21x-13x = 260-140

= 8x = 120

= x = 120/8 = 15

therefore mother's present age = 3x +10

= 3*15+10 = 55 years

daughter's present age = x+10 = 25 years

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