Question

A and B together can do a piece of work in 12 days,while b alone can finish it in 30 days.how many days can A alone finish the work?​

Answer 1

Step-by-step explanation:

A and B together 1day work = 1/12

B alone 1day work = 1/30

A alone 1day work= 1/12-1/30

= (5-2)/60

= 3/60

= 1/20

A alone finish the work in 20days.

Answer 2

Answer:

• (A + B) can complete = 12 days
• B can complete alone = 30 days
• A can complete alone = ?

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf A=\dfrac{B \times (A+B)}{B-(A+B)}\\\\\\:\implies\sf A=\dfrac{30 \times 12}{30-12}\\\\\\:\implies\sf A=\dfrac{30 \times 12}{18}\\\\\\:\implies\sf A=\dfrac{5 \times 12}{3}\\\\\\:\implies\sf A = 5 \times 4\\\\\\:\implies\sf A = 20 \:Days$

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$\therefore\:\underline{\textsf{A can alone complete the work in \textbf{20 Days}}}.$

$\rule{200}{1}$

Important Formulae :

1. Let A can do a work in x days and B can do the same work in y days. They'll do the same work together in :

$\begin{aligned}\dashrightarrow\sf (A+B)=\dfrac{1}{x}+\dfrac{1}{y}=1\\\\\\\dashrightarrow\sf (A+B)=\dfrac{x+y}{xy}=1\\\\\\\dashrightarrow\sf (A+B)=\dfrac{xy}{x+y}\end{aligned}$

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2. Let A, B and C can do a work in x, y and z days respectively. They'll do the same work together in :

$\begin{aligned}\dashrightarrow\sf (A+B+C)=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\\\\\\\dashrightarrow\sf (A+B+C)=\dfrac{xy+yz+zx}{xyz}=1\\\\\\\dashrightarrow\sf (A+B+C)=\dfrac{xyz}{xy+yz+zx}\end{aligned}$

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3. Let (A + B) can do a work in x days and A can do the same work in y days. B will do the same work in :

$\begin{aligned}\dashrightarrow\sf B=\dfrac{1}{x}-\dfrac{1}{y}=1\\\\\\\dashrightarrow\sf B=\dfrac{y-x}{xy}=1\\\\\\\dashrightarrow\sf B=\dfrac{xy}{y-x}\end{aligned}$