The average age of x and y is 20 years. If z | vedclass

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(A) Given that the average age of $x$ and $y$ is $20$ years,the sum of their ages is $x + y = 20 	imes 2 = 40$ years.If $z$ replaces $x$,the average

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$22, 18, 20$

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(A) Given that the average age of $x$ and $y$ is $20$ years,the sum of their ages is $x + y = 20 	imes 2 = 40$ years.If $z$ replaces $x$,the average age of $z$ and $y$ is $19$ years,so $z + y = 19 	imes 2 = 38$ years.If $z$ replaces $y$,the average age of $x$ and $z$ is $21$ years,so $x + z = 21 	imes 2 = 42$ years.Adding these three equations: $(x + y) + (z + y) + (x + z) = 40 + 38 + 42$,which simplifies to $2(x + y + z) = 120$,so $x + y + z = 60$.Now,find the individual ages:$x = (x + y + z) - (z + y) = 60 - 38 = 22$ years.$y = (x + y + z) - (x + z) = 60 - 42 = 18$ years.$z = (x + y + z) - (x + y) = 60 - 40 = 20$ years.Thus,the ages are $x = 22, y = 18, z = 20$.

The average age of $x$ and $y$ is $20$ years. If $z$ were to replace $x$,the average would be $19$ and if $z$ were to replace $y$,the average would be $21$. Find the ages of $x, y$ and $z$.

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