Question 1041022
Let {{{ a }}} = father's age now
Let {{{ b }}} = son's age now
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The difference in their ages never changes
It is {{{ a - b }}}
{{{ a - ( a  - b ) = b }}}
So, {{{ a-b }}} years ago, the father was the
son's age
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{{{ a - b }}} years ago, the son was
{{{ b - ( a-b ) = 2b - a }}} yrs old
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You are given that:
{{{ a = 8*( 2b - a ) }}}
{{{ a = 16b - 8a }}}
{{{ 9a = 16b }}}
(1) {{{ b = ( 9/16 )*a }}}
Also given is:
(2) {{{ a + b = 75 }}}
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Substitute (1) into (2)
(2) {{{ a + (9/16)*a = 75 }}}
(2) {{{ ( 25/16 )*a = 75 }}}
(2) {{{ a = ( 16/25 )*75 }}}
(2) {{{ a = 48 }}}
and
(2) {{{ a + b = 75 }}}
(2) {{{ 48 + b = 75 }}}
(2) {{{ b = 27 }}}
The father is 48 and the son is 27
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check:
The difference in their ages is {{{ 48 - 27 = 21 }}}
When the father was {{{ 27 }}}, the son was {{{ 27 - 21 = 6 }}}
The father's age is 8 times that:
{{{ 6*8 = 48 }}}
OK
(1) {{{ b = (9/16)*a }}}
(1) {{{ 27 = (9/16)*48 }}}
(1) {{{ 27/48 = 9/16 }}}
(1) {{{ .5625 = .5625 }}}
OK