Question

A and B can do a job in 8 days, B and C in 12 days and A, B and C together in 6 days. How long can A and C work together? ​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Given That :-

• (A + B) can do a job in = 8 Days
• (B + C) can do same job in = 12 Days..
• (A + B + C) can do same job in = 6 Days.

LCM of 8,12,6 = 24 units = Let Total work.

So,

→ Efficiency of (A + B) = Total work / Total Days = 24/8 = 3 units/Day .

→ Efficiency of (A + B) = Total work / Total Days = 24/12 = 2 units/Day .

→ Efficiency of (A + B + C) = Total work / Total Days = 24/6 = 4 units/Day .

Now,

→ 2(A + B + C) - (A + B) - (B + C) = (A + C)

→ 2 * 4 - 3 - 2 = (A + C)

→ 8 - 5 = (A + C)

→ (A + C) = 3 units / Days.

Hence,

→ Time Taken by (A + C) to complete same work = (Total work) / (Efficiency) = (24/3) = 8 Days. (Ans.)

Answer 2

Answer:

When we will Add all, it will be 2(A + B + C). That's why we will already Increase all Work time by Twice.

⠀

⠀⠀⠀✩ (A + B) = 8 Days

⠀⠀⠀✩ (B + C) = 12 Days

⠀⠀⠀✩ (C + A) = x Days

⠀⠀⠀✩ (A + B + C) = 6 Days

⠀

But for us it will be 16 Days, 24 Days & 2x respectively.

$\rule{120}{1.2}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question :}}$

$:\implies\sf \dfrac{Time}{(A+B)}+\dfrac{Time}{(B+C)}+\dfrac{Time}{(C+A)}=Total\:Work\\\\\\:\implies\sf \dfrac{6}{16} + \dfrac{6}{24} + \dfrac{6}{2x} = 1\\\\\\:\implies\sf \dfrac{3}{8} + \dfrac{3}{12} + \dfrac{3}{x} = 1\\\\\\:\implies\sf \dfrac{(3 \times 3) + (3 \times 2)}{24} + \dfrac{3}{x} = 1\\\\\\:\implies\sf \dfrac{9 + 6}{24} + \dfrac{3}{x} = 1\\\\\\:\implies\sf \dfrac{15}{24} + \dfrac{3}{x} = 1\\\\\\:\implies\sf \dfrac{3}{x} = 1 -\dfrac{15}{24}\\\\\\:\implies\sf \dfrac{3}{x} = \dfrac{9}{24}\\\\\\:\implies\sf 3 \times \dfrac{24}{9} = x \\\\\\:\implies\underline{\boxed{\sf x = 8\:Days}}$

⠀

$\therefore\:\underline{\textsf{Hence, A \& C will work \textbf{8 Days} together}}.$