A and B, working together, can complete a work in 16 days, C and A together can complete it in 32 days, B and C together can completeit in 24 days. They worked togetherfor 12 days. In how many days will C alone complete the remaining work ?

Let A, B and C can finish the work in x, y and z days respectively.

So, A's one day's work $$=\dfrac{1}{x}$$

B's one day's work $$=\dfrac{1}{y}$$

C's one day's work $$=\dfrac{1}{z}$$

A and B working together can complete the work $$=16$$ days.

So, A and B together will finish the work in one day, $$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}$$ ------(i)

C and A together can finish the work in $$=32$$ days

So, C and A together will finish the work in one day $$\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{32}$$ ------(ii)

B and C together can finish the work in $$=24$$ days

So, B and C together can finish the work in one day $$\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{24}$$-----(iii)

Now, From equation (i), (ii) and (iii)

$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{24}$$

$$\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}=\dfrac{13}{96}$$

$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{13}{192}$$

Hence, if they are working together, they can finish the work in one day $$=\dfrac{13}{192}$$

Substituting the value from the equation (i)

$$\dfrac{1}{z}=\dfrac{13}{192}-(\dfrac{1}{16})$$

$$\dfrac{1}{z}=\dfrac{13-12}{192}$$

$$\dfrac{1}{z}=\dfrac{1}{192}$$

So, C can finish one work in $$192$$ days.

As per the question, they worked for$$12$$days,

So, remaining work after $$12$$days $$=1-\dfrac{12\times13}{192}$$

 $$=\dfrac{192-156}{192}$$

 $$=\dfrac{36}{192}$$

 C can finish one work in 192 days

Hence, C can finish $$\dfrac{36}{192}$$ work in $$=\dfrac{192\times36}{192} =36$$days