Cameron can paint a room in c hours. Cameron and Mackenzie, working together, can paint the room in d hours. In terms of c and d, how long in hours would it take Mackenzie, working alone, to paint the room?
$$A.\ 2d-c$$
$$B.\ \frac{c+d}{cd}$$
$$C.\ \frac{c-d}{cd}$$
$$D.\ \frac{cd}{c+d}$$
$$E.\ \frac{cd}{c-d}$$
The OA is E.
Can I solve it as follow,
Cameron's rate:
$$Cam's\ rate=\frac{1}{c}$$
Cameron and Mackenzie's rate when working together:
$$Combined\ rate=\frac{1}{d}$$
Mackenzie's rate:
$$Mack's\ rate=\frac{1}{d}-\frac{1}{c}$$
Finally, Mackenzie's time:
$$T(Mackenzie)=\frac{1}{\frac{1}{d}-\frac{1}{c}}=\frac{1}{\frac{c-d}{cd}}=\frac{cd}{c-d}$$
Is there a strategic approach to this question? Can any experts help, please?
$$A.\ 2d-c$$
$$B.\ \frac{c+d}{cd}$$
$$C.\ \frac{c-d}{cd}$$
$$D.\ \frac{cd}{c+d}$$
$$E.\ \frac{cd}{c-d}$$
The OA is E.
Can I solve it as follow,
Cameron's rate:
$$Cam's\ rate=\frac{1}{c}$$
Cameron and Mackenzie's rate when working together:
$$Combined\ rate=\frac{1}{d}$$
Mackenzie's rate:
$$Mack's\ rate=\frac{1}{d}-\frac{1}{c}$$
Finally, Mackenzie's time:
$$T(Mackenzie)=\frac{1}{\frac{1}{d}-\frac{1}{c}}=\frac{1}{\frac{c-d}{cd}}=\frac{cd}{c-d}$$
Is there a strategic approach to this question? Can any experts help, please?
