Question

A mother is twice sa old as her son . if 20 years ago of the mother was 10 times the age of the son, what is the present age of the mother​

Answer 1

Answer:

45

Step-by-step explanation:

Let the present age of the Son be - x

then,

present age of Mother will be - 2x

Now, 20 years ago, their ages wil be 20 years less., i.e

Mother - (2x-20) & Son - (x - 20)

Then, according to given condition, Mother was 10 times the age of son, so the equation becomes :-

(2x - 20) = 10(x - 20)

(2x - 20) = (10x - 200)

8x = 180

x = 180/8

x = 22.5 (This is the age if Son)And Mother is twice as old as her son,

So, Mothers age will be - 2X

i.e., 2 * 22.5 = 45

Hence, Mother is 45 years old !!

Answer 2

Answer:

The present age of mother is 45 years .

Step-by-step explanation:

Present ages

Let the present age of son be : a

and let the present age of mother be :2a

now,

20 years ago

20 years ago the ages will be less therefore

age of mother 20 years ago was : 2a-20

age of son 20 years ago was :a-20

given:

mothers age 20 years ago was 10 times the age of son.

hence,

2a-20=10(a-20)

2a-10a=200-20

180=8a

a= 22.5

hence, the age of son is 22 years and 6 month.

whereas age of mother is (2a) = 22.5×2 =45 years

(#SPJ2)