Answer
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Hint: In this question, we are given the number of days required by A to do some work and we need to find the number of days required by B to do the same work if B is 50% more efficient than A. For this, we will consider the amount of work as x and then find A's one-day work. Then B's one day workshop will be 50% more than A's one day work which means 150% of A's one day work. Using B's one day workshop, we will then calculate the total number of days.
$\text{Total days}=\dfrac{\text{Total work}}{\text{One day work}}$.
Complete step-by-step solution
Let us suppose that, total work that has to be done is equal to x.
Now to make use of efficiency, we must find A's one day work using his total work and total days taken.
We know $\text{Total days}=\dfrac{\text{Total work}}{\text{One day work}}$.
Therefore, for A who took 9 days to complete the work we have $9=\dfrac{\text{x}}{\text{One day work}}$.
One day work of A $\Rightarrow \dfrac{x}{9}$.
Now, we know that B is 50% more efficient than A.
We know that the efficiency of any person is considered as 100%, so let us consider the efficiency of A as 100%. Since the efficiency of B is 50% more than A's efficiency. Hence, the efficiency of B is 150% of A's efficiency.
If A's one day work is $\dfrac{x}{9}$ then B's one day workshop will be 150% of $\dfrac{x}{9}$
So we get B's one day work:
$\Rightarrow \dfrac{150}{100}\times \dfrac{x}{9}=\dfrac{3\times x}{2\times 9}=\dfrac{x}{6}$.
Hence B's one day work is $\dfrac{x}{6}$.
Now, let us calculate the number of days taken by B to complete x work.
We know, $\text{Total days}=\dfrac{\text{Total work}}{\text{One day work}}$.
Therefore, the total number of days taken by B $\Rightarrow \dfrac{x}{\dfrac{x}{6}}=x\times \dfrac{6}{x}=6$.
Hence, the total number of days taken by B is equal to 6 days.
Hence option C is the correct answer.
Note: Students often make the mistake of choosing option B by taking 50% of the number of days but this is wrong. Students can remember the following formula for these questions: $\text{Time taken by B}=\text{Time taken B}\times \dfrac{100}{100+R}$ where R is the increased efficiency of B.
Here, Time taken by B $\Rightarrow 9\times \dfrac{100}{100+5}=\dfrac{900}{105}=6$.
Hence 6 days are required for B to do the same work.
$\text{Total days}=\dfrac{\text{Total work}}{\text{One day work}}$.
Complete step-by-step solution
Let us suppose that, total work that has to be done is equal to x.
Now to make use of efficiency, we must find A's one day work using his total work and total days taken.
We know $\text{Total days}=\dfrac{\text{Total work}}{\text{One day work}}$.
Therefore, for A who took 9 days to complete the work we have $9=\dfrac{\text{x}}{\text{One day work}}$.
One day work of A $\Rightarrow \dfrac{x}{9}$.
Now, we know that B is 50% more efficient than A.
We know that the efficiency of any person is considered as 100%, so let us consider the efficiency of A as 100%. Since the efficiency of B is 50% more than A's efficiency. Hence, the efficiency of B is 150% of A's efficiency.
If A's one day work is $\dfrac{x}{9}$ then B's one day workshop will be 150% of $\dfrac{x}{9}$
So we get B's one day work:
$\Rightarrow \dfrac{150}{100}\times \dfrac{x}{9}=\dfrac{3\times x}{2\times 9}=\dfrac{x}{6}$.
Hence B's one day work is $\dfrac{x}{6}$.
Now, let us calculate the number of days taken by B to complete x work.
We know, $\text{Total days}=\dfrac{\text{Total work}}{\text{One day work}}$.
Therefore, the total number of days taken by B $\Rightarrow \dfrac{x}{\dfrac{x}{6}}=x\times \dfrac{6}{x}=6$.
Hence, the total number of days taken by B is equal to 6 days.
Hence option C is the correct answer.
Note: Students often make the mistake of choosing option B by taking 50% of the number of days but this is wrong. Students can remember the following formula for these questions: $\text{Time taken by B}=\text{Time taken B}\times \dfrac{100}{100+R}$ where R is the increased efficiency of B.
Here, Time taken by B $\Rightarrow 9\times \dfrac{100}{100+5}=\dfrac{900}{105}=6$.
Hence 6 days are required for B to do the same work.
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