$A$ and $B$ working together complete a job in $10$ days. If $B$ worked twice as fast as he actually does,…
$A$ and $B$ working together complete a job in $10$ days. If $B$ worked twice as fast as he actually does, they would finish in $6$ days. $A$ alone takes:
$24$ days
$28$ days
$30$ days
$36$ days
Solution
Let $A$'s rate $= a$, $B$'s rate $= b$. Then $a + b = \dfrac{1}{10}$ and $a + 2b = \dfrac{1}{6}$. Subtract: $b = \dfrac{1}{6} - \dfrac{1}{10} = \dfrac{5 - 3}{30} = \dfrac{1}{15}$. So $a = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$. $A$ alone $= 30$ days.