Suppose $A$ and $B$ can complete a work together in 10 days. If $B$ alone can complete the work in 15 days, then in how many days can $A$ alone finish the work?
Let $R_A$ be the work rate of A (amount of work A can do in one day).
Let $R_B$ be the work rate of B (amount of work B can do in one day).
Work done = Rate × Time.
Rate = Work / Time.
A and B together complete the work in 10 days.
Their combined rate is $R_A + R_B$.
$(R_A + R_B) \times 10 = W$
$R_A + R_B = \frac{W}{10}$
B alone completes the work in 15 days.
B’s rate is $R_B$.
$R_B \times 15 = W$
$R_B = \frac{W}{15}$
We want to find the time it takes for A alone to finish the work. Let this time be $T_A$.
$R_A \times T_A = W$
$T_A = \frac{W}{R_A}$
Substitute the value of $R_B$ into the combined rate equation:
$R_A + \frac{W}{15} = \frac{W}{10}$
Solve for $R_A$:
$R_A = \frac{W}{10} – \frac{W}{15}$
Find a common denominator for the fractions (LCM of 10 and 15 is 30):
$R_A = \frac{3W}{30} – \frac{2W}{30}$
$R_A = \frac{3W – 2W}{30} = \frac{W}{30}$
Now, calculate the time taken for A alone:
$T_A = \frac{W}{R_A} = \frac{W}{\frac{W}{30}}$
$T_A = W \times \frac{30}{W} = 30$
A alone can finish the work in 30 days.
– Combined work rate is the sum of individual work rates.
– Solving for the unknown individual work rate and then calculating the time.
Combined rate = 1/10 per day.
B’s rate = 1/15 per day.
A’s rate = Combined rate – B’s rate = 1/10 – 1/15 = (3 – 2)/30 = 1/30 per day.
Time taken by A alone = 1 / A’s rate = 1 / (1/30) = 30 days.
This approach simplifies calculations by normalizing the total work to 1.