Question

Divya and ranu together can do piece of work in 15 days Divya alone can finish the work in 20 days in how many days can ranu alone finish the work​

Answer 1

Answer:

(Divya + ranu) = 15

Divya = 2O

Ranu = B

( 1\2O + 1\B ) = 1\15

1\B = 1\15 - 1\2O

1\B = 4 - 3\6O

1\B = 1\6O

B = 6O Days ( Ans )

Answer 2

Answer:

• Divya & Ranu together = 15 days
• Divya alone = 20 days
• Ranu alone = ?

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

$:\implies\sf Ranu=\dfrac{(Divya\:\& \:Ranu) \times Divya}{Divya-(Divya\:\& \:Ranu)}\\\\\\:\implies\sf Ranu = \dfrac{15 \times 20}{20 - 15} \\\\\\:\implies\sf Ranu = \dfrac{15 \times 20}{5}\\\\\\:\implies\sf Ranu = 3 \times 20\\\\\\:\implies\sf Ranu = 60 \: Days$

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$\therefore\:\underline{\textsf{Hence, Ranu can alone do work in \textbf{60 Days}}}.$

$\rule{200}{1}$

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 :

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