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A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of $24$. If three times the number of red balls exceeds the number of blue balls by $20$, find the number of red and blue balls.
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Let number of red balls be $x$
& number of blue balls be $y$
A.T.Q.
$\frac{10x}{100} + \frac{20y}{100} = 24$
or $x + 2y = 240$ .....(i)
Also, $3x - y = 20$ ......(ii)
Solving (i) and (ii), we get
$x = 40, y = 100$
$\therefore$ Number of red balls = $40$ and Number of blue balls = $100$
& number of blue balls be $y$
A.T.Q.
$\frac{10x}{100} + \frac{20y}{100} = 24$
or $x + 2y = 240$ .....(i)
Also, $3x - y = 20$ ......(ii)
Solving (i) and (ii), we get
$x = 40, y = 100$
$\therefore$ Number of red balls = $40$ and Number of blue balls = $100$