1 point
27. Ten years ago, the age of
mother was three times the
age of her son. After ten years,
mother's age will be twice that
of his son. Find the ratio of
their present ages
O :
11:7
09:5
O 7:4
O 7:3
Answers
Answer is given in this attachment.
Answer:
=> 7 : 3
Explaination :-
let the age of mother be = x
and let the age of son be = y
ten years before the age of mother will be = x-10
and the age of son will be= y-10
now the age mother will be= 3(x-10)
taking them both equal we'll get= 3(x-10) = y-1 = 3x-30 = y- 10.
= 3x-y = 20
now taking an another equation where age of mother will be = 2(x+10)
and the age of mother will be twice the age of son = y+10
now making an equation = 2(x+10) = y+10
= 2x+20 = y+10
= 2x-y = -20+10.
= 2x-y = -10
now putting these two equations in elimination method:- 3x-y= 20 -----(1)
2x-y= -10------ (2)
=> 3x- y= 20 -----(1) X 2
=> 2x+ y= -10 -----(2) X 3
so the modified equations will be like :-. 6x- 2y = 40---(1)
6x+ 3y = -30--(2)
- + +.
____________
y = 70
Now if we found the value of y putting this in equation (2) we will get = 2x -70 = -10
=> 2x = -10+70
=> x = 30
putting them in ratio = y : x
will give us = 7 : 3
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