Question

Rahul, Rakshita and Gurmeet, working together, would have taken more than 7 days to finish a job. On the other hand, Rahul and Gurmeet, working together would have taken less than 15 days to finish the job. However, they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. If Rakshita had worked alone on the job then the number of days she would have taken to finish the job, cannot be

• A. 16
• B. 21
• C. 17
• D. 20

Answer 1

The Correct Option is B

Let the work rates of Rahul, Rakshita, and Gurmeet be $a$, $b$, and $c$ units of work per day, respectively.

Let the total work be $W$. 

According to the problem statement:
 

• The three individuals worked collaboratively for 6 days.
• Rakshita independently completed the remaining work in an additional 3 days.

The total work done can be expressed as:
$W = 6(a + b + c) + 3b$    --- (1) 

Additional information provided:
 

• The work could not be completed by all three individuals together in 7 days:
  $7(a + b + c) < W$    --- (2)
• Rahul and Gurmeet working together would have completed the work in less than 15 days:
  $15(a + c) > W$    --- (3)

Combining inequalities (2) and (3) yields:
$15(a + c) < W < 7(a + b + c)$    --- (4) 

Substitute equation (1) into inequality (4): 
$15(a + c) < 6(a + b + c) + 3b < 7(a + b + c)$ 

Simplify the middle expression:
$6(a + b + c) + 3b = 6a + 6b + 6c + 3b = 6a + 9b + 6c$ 

Now, analyze the two resulting inequalities:
$15(a + c) < 6a + 9b + 6c$    --- (5) 
$6a + 9b + 6c < 7(a + b + c)$    --- (6) 

Expand the right side of inequality (6): 
$7(a + b + c) = 7a + 7b + 7c$ 

Rewrite inequality (6) with the expanded term:
$6a + 9b + 6c < 7a + 7b + 7c$ 
Subtracting terms from both sides results in:
$-a + 2b - c < 0 \implies a + c > 2b$    --- (7) 

Simplify inequality (5): 
$15(a + c) < 6a + 9b + 6c$
$ \Rightarrow 15a + 15c < 6a + 9b + 6c$
$ \Rightarrow 9a + 9c < 9b$
$ \Rightarrow a + c < b$    --- (8) 

The derived inequalities are:

• From (7): $a + c > 2b$
• From (8): $a + c < b$

Inequality (8) contradicts inequality (7), indicating a calculation error occurred previously. 
Re-evaluating inequality (5) accurately: 

$15(a + c) < 6a + 9b + 6c$
$ \Rightarrow 15a + 15c < 6a + 9b + 6c$
$ \Rightarrow 9a + 9c < 9b$
$ \Rightarrow a + c < b$    This result still contradicts (7), suggesting an earlier sign error. 

Let's reconcile the findings from the inequalities. Using the relationships derived from the problem constraints:
From $7(a + b + c)<W$ and $W<15(a + c)$, we infer:

• $7(a + b + c)<15(a + c)$

From $W = 6(a + b + c) + 3b$, and $W<7(a + b + c)$: $6(a + b + c) + 3b<7(a + b + c)$ $6a + 6b + 6c + 3b<7a + 7b + 7c$ $6a + 9b + 6c<7a + 7b + 7c$ $2b<a + c$    --- (7') (Corrected inequality (7))

From $W = 6(a + b + c) + 3b$, and $W>15(a + c)$: $6(a + b + c) + 3b>15(a + c)$ $6a + 6b + 6c + 3b>15a + 15c$ $6a + 9b + 6c>15a + 15c$ $9b>9a + 9c$ $b>a + c$    --- (8') (Corrected inequality (8))

Combining the corrected inequalities (7') and (8'):
We have $b>a + c$ and $a + c>2b$. This implies $b>2b$, which is only possible if $b$ is negative, which is not feasible for a work rate. 

Let's re-examine the inequalities from the prompt statement:
$7(a + b + c)<W$ and $15(a + c)>W$. This implies $7(a+b+c)<15(a+c)$. $7a + 7b + 7c<15a + 15c$ $7b<8a + 8c$. 

Using $W = 6(a + b + c) + 3b$: From $W<7(a+b+c)$: $6(a+b+c) + 3b<7(a+b+c)$ $6a + 9b + 6c<7a + 7b + 7c$ $2b<a+c$. From $W>15(a+c)$: $6(a+b+c) + 3b>15(a+c)$ $6a + 9b + 6c>15a + 15c$ $9b>9a + 9c$ $b>a+c$.

The derived inequalities are $2b<a+c$ and $b>a+c$. These inequalities are contradictory. Let's reconsider the initial problem statement implications.

If $7(a+b+c)<W$, then $W/7>a+b+c$. If $15(a+c)>W$, then $W/15<a+c$. If Rakshita worked alone for 3 days to finish, $3b = W - 6(a+b+c)$. So $b = (W - 6(a+b+c))/3 = W/3 - 2(a+b+c)$.

Substitute $a+c$ from $W/15<a+c$: $b>W/15$. Substitute $a+b+c$ from $W/7>a+b+c$: $b = W/3 - 2(a+b+c)>W/3 - 2(W/7) = W/3 - 2W/7 = (7W - 6W)/21 = W/21$.

So, $b>W/15$ and $b>W/21$. The tighter lower bound is $b>W/15$.

Now consider the upper bound for $b$. From $7(a+b+c)<W$, we have $a+b+c<W/7$. From $W = 6(a+b+c) + 3b$, we have $3b = W - 6(a+b+c)$. $3b>W - 6(W/7) = W - 6W/7 = W/7$. $b>W/21$. This confirms the lower bound.

From $15(a+c)>W$, we have $a+c>W/15$. $W = 6a + 6b + 6c + 3b = 6(a+c) + 6b + 3b = 6(a+c) + 9b$. Since $a+c>W/15$, $W = 6(a+c) + 9b>6(W/15) + 9b = 2W/5 + 9b$. $W - 2W/5>9b$ $3W/5>9b$ $b<3W/45 = W/15$.

So we have $b>W/21$ and $b<W/15$. Therefore, $b$ lies in the range $\dfrac{W}{21} < b < \dfrac{W}{15}$. This means the number of days for Rakshita is between 15 and 21. 

The correct answer is (B): 15 to 21 days.