Question

if b can do a piece of work in 6 hrs ,d snd c together can do it in 4 hrs and a,b and c together in 2 2/3 hrs. In how many hours can a and b together do the same piece of work​

Answer 1

$\sf\large\underline{Given:}$

$\tt{\implies A\: can\:do\:a\: work=6\: hours}$

$\tt{\implies B+C\: can\:do\:a\: work=4\: hours}$

$\tt{\implies A+B+C\: can\:do\:a\: work=2\dfrac{2}{3}\: hours}$

$\sf\large\underline{To\: Find:}$

$\tt{\implies A+B\: can\:do\:a\: work=?}$

$\sf\large\underline{Let:}$

$\tt{\implies A's\:one\:hour\: work=\frac{1}{6}}$

$\tt{\implies (B+C)'s\:one\:hour\: work=\frac{1}{4}}$

$\tt{\implies (A+B+C)'s\:one\:hour\: work=\frac{3}{8}}$

$\sf\large\underline{Solution:}$

• Here we have to calculate 1st work done by A]

$\tt{\implies (A+B+C)-(B+C)}$

$\tt{\implies \dfrac{3}{8}-\dfrac{1}{4}}$

$\tt{\implies \dfrac{3-2}{8}}$

$\tt{\implies \dfrac{1}{8}}$

$\tt{\implies \therefore\:A's\:one\: hour\:work=\dfrac{1}{8}}$

• Now, calculate work done by A+B here]

$\tt{\implies A+B\:can\:do\:a\: work}$

$\tt{\implies \dfrac{1}{8}+\dfrac{1}{6}}$

$\tt{\implies \dfrac{3+4}{24}}$

$\tt{\implies \dfrac{7}{24}}$

$\tt{\implies \dfrac{24}{7}}$

$\tt{\implies 3\dfrac{3}{7}\: hours}$

$\sf\large{Hence,}$

$\tt{\implies A+B\: can\:do\:a\: work=3\dfrac{3}{7}\:hours}$

Answer 2

Answer:

Given:

\tt{\implies A\: can\:do\:a\: work=6\: hours}⟹Acandoawork=6hours

\tt{\implies B+C\: can\:do\:a\: work=4\: hours}⟹B+Ccandoawork=4hours

\tt{\implies A+B+C\: can\:do\:a\: work=2\dfrac{2}{3}\: hours}⟹A+B+Ccandoawork=2

3

2

hours

\sf\large\underline{To\: Find:}

ToFind:

\tt{\implies A+B\: can\:do\:a\: work=?}⟹A+Bcandoawork=?

\sf\large\underline{Let:}

Let:

\tt{\implies A's\:one\:hour\: work=\frac{1}{6}}⟹A

′

sonehourwork=

6

1

\tt{\implies (B+C)'s\:one\:hour\: work=\frac{1}{4}}⟹(B+C)

′

sonehourwork=

4

1

\tt{\implies (A+B+C)'s\:one\:hour\: work=\frac{3}{8}}⟹(A+B+C)

′

sonehourwork=

8

3

\sf\large\underline{Solution:}

Solution:

Here we have to calculate 1st work done by A]

\tt{\implies (A+B+C)-(B+C)}⟹(A+B+C)−(B+C)

\tt{\implies \dfrac{3}{8}-\dfrac{1}{4}}⟹

8

3

−

4

1

\tt{\implies \dfrac{3-2}{8}}⟹

8

3−2

\tt{\implies \dfrac{1}{8}}⟹

8

1

\tt{\implies \therefore\:A's\:one\: hour\:work=\dfrac{1}{8}}⟹∴A

′

sonehourwork=

8

1

Now, calculate work done by A+B here]

\tt{\implies A+B\:can\:do\:a\: work}⟹A+Bcandoawork

\tt{\implies \dfrac{1}{8}+\dfrac{1}{6}}⟹

8

1

+

6

1

\tt{\implies \dfrac{3+4}{24}}⟹

24

3+4

\tt{\implies \dfrac{7}{24}}⟹

24

7

\tt{\implies \dfrac{24}{7}}⟹

7

24

\tt{\implies 3\dfrac{3}{7}\: hours}⟹3

7

3

hours

\sf\large{Hence,}Hence,

\tt{\implies A+B\: can\:do\:a\: work=3\dfrac{3}{7}\:hours}⟹A+Bcandoawork=3

7

3

hours

Explanation:

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