Four times $B^{'} s$ age exceeds $A^{'} s$ age by twenty years, and one-third of $A^{'} s$ age is less than $B^{'} s$ age by two years; find their ages.

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Solution

Given
$4 B = A + 20$ ……… $(1)$
$B - \frac{1}{3} A = 2$ ……….. $(2)$
$4 B = A + 20$ ………. $(1)$
from $(2)$ $3 B - A = 6$
from $(1)$ $4 B - A = 20$
$(-)$
_________________________
$- B + 0 = - 14$
$B = 14$
$A = 4 B - 20 = 56 - 20$
$A = 36$ .

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Four times B′s age exceeds A′s age by twenty years, and one-third of A′s age is less than B′s age by two years; find their ages.

Given4B=A+20 ………(1)B−13A=2 ………..(2)4B=A+20 ……….(1)from (2) 3B−A=6from (1) 4B−A=20 (−)_________________________ −B+0=−14B=14A=4B−20=56−20A=36.

Four times $B^{'} s$ age exceeds $A^{'} s$ age by twenty years, and one-third of $A^{'} s$ age is less than $B^{'} s$ age by two years; find their ages.

Given
$4 B = A + 20$ ……… $(1)$
$B - \frac{1}{3} A = 2$ ……….. $(2)$
$4 B = A + 20$ ………. $(1)$
from $(2)$ $3 B - A = 6$
from $(1)$ $4 B - A = 20$
$(-)$
_________________________
$- B + 0 = - 14$
$B = 14$
$A = 4 B - 20 = 56 - 20$
$A = 36$ .