Math, asked by sachin9715, 5 hours ago


Three years ago, the ratio of ages of A and B was 7:1. After three years from noww, the ratio of their ages will be 7:5 What is the difference between their ages (in years) after seven years from now?​.

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Answers

Answered by yadavaadi6904
1

Answer:

A-3÷b-3= 7

A+3÷b+3= 4

A- 7b= -18

A-4b= 9

B=9

A=45

seven years from now

A's age = 52

B's= 16

different is 36 answer.

Answered by itzcutechandan
4

GIVEN :

Three years ago, the ratio of ages of A and B was 7:1. After three years from noww, the ratio of their ages will be 7:5.

TO FIND :

What is the difference between their ages (in years) after seven years from now?

SOLUTION:

Let us assume that ,three years ago ,The age of A was 7x years and B was 1x years.

Than,

Age of A is = (7x +3 ) years

Present age of B is = ( 1x + 3) years

So,

Age of A ,3 years from now = (7x + 3) + 3 years = ( 7x + 6) years

Age of B ,3 years from now = (1x + 3) + 3 = (1x + 6) years

Given that ratio will be 4 : 1.

Hence,

(7x + 6 ) / (1x + 6) = 4/1

( 7x + 6 ) = 4(1x + 6)

7x + 6 = 4x + 24

7x - 4x = 24 -6

3x = 18

x = 18 /3

x = 6

Therefore,

Present age of A is =( 7x + 3 )= (7*6 +3)

= 42 + 3 = 45 years.

present age of B is = (1x + 3) = (1*6 + 3)

=6 + 3 = 9 years.

Hence,

Age of A after 7 years from now = 45 + 7 = 52 years.

Age of B after 7 years from now = 9 + 7 = 16 years.

SO,

Difference between their ages = 52 - 16 = 36 years.

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