A does half as much work as B in three fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it?
\[\begin{align}
& \text{A}.\text{3}0\text{ days} \\
& \text{B}.\text{35 days} \\
& \text{C}.\text{4}0\text{ days} \\
& \text{D}.\text{None of these} \\
\end{align}\]
Answer
614.7k+ views
Hint: For this question, we will first suppose days taken by B as x using the information we will calculate the number of days taken by A. Then we calculate both A and B's one day work in terms of x. Then we will equate both their one day’s work equal to $ \dfrac{1}{18} $ (As per question). Solving the equation, we will find the value of x and hence our required answer.
Complete step by step answer:
Let us suppose that B completes its work in x days. Now, we are given that A takes $ {{\dfrac{3}{4}}^{th}} $ time of B's work to complete half of the work.
So we can say that A takes $ \dfrac{3}{4}\times x $ days to complete half of the work.
So for the full work, it will take twice the number of days to complete half the work.
Therefore, number of days that A take to complete work $ \dfrac{3}{4}\times x\times 2=\dfrac{3}{2}x\text{ days} $ .
Now let us calculate their one-day work.
Days taken by A to complete total work $ \dfrac{3}{2}x\text{ days} $ .
So we can say, A's one day work $ \dfrac{2}{3x}\text{ days} $ .
Similarly, B's one day work $ \dfrac{1}{x} $ .
Now we are given that A and B together take 18 days to complete the whole work. Therefore, their combined one day work becomes $ \dfrac{1}{18} $ .
According to all the information, thus we can say that A's one day work + B's one day work = $ \dfrac{1}{18} $.
Therefore, $ \dfrac{2}{3x}+\dfrac{1}{x}=\dfrac{1}{18} $ .
Taking LCM as 3x we get, $ \dfrac{2+3}{3x}=\dfrac{1}{18}\Rightarrow \dfrac{5}{3x}=\dfrac{1}{18} $ .
Cross multiplying we get, $ 18\times 5=3x $ .
Dividing both sides by 3 we get, $ \dfrac{18\times 5}{3}=x\Rightarrow x=6\times 5\Rightarrow x=30 $ .
Therefore, the time taken by B to complete the work is 30 days.
Hence option A is the correct answer.
Note:
Students can make the mistake of taking $ \dfrac{3}{2}x+x $ as 18. But this is wrong as this sum requires inverse proportion because as the amount of work per day increases, the lesser the number of days is required. Take care in the calculation.
Complete step by step answer:
Let us suppose that B completes its work in x days. Now, we are given that A takes $ {{\dfrac{3}{4}}^{th}} $ time of B's work to complete half of the work.
So we can say that A takes $ \dfrac{3}{4}\times x $ days to complete half of the work.
So for the full work, it will take twice the number of days to complete half the work.
Therefore, number of days that A take to complete work $ \dfrac{3}{4}\times x\times 2=\dfrac{3}{2}x\text{ days} $ .
Now let us calculate their one-day work.
Days taken by A to complete total work $ \dfrac{3}{2}x\text{ days} $ .
So we can say, A's one day work $ \dfrac{2}{3x}\text{ days} $ .
Similarly, B's one day work $ \dfrac{1}{x} $ .
Now we are given that A and B together take 18 days to complete the whole work. Therefore, their combined one day work becomes $ \dfrac{1}{18} $ .
According to all the information, thus we can say that A's one day work + B's one day work = $ \dfrac{1}{18} $.
Therefore, $ \dfrac{2}{3x}+\dfrac{1}{x}=\dfrac{1}{18} $ .
Taking LCM as 3x we get, $ \dfrac{2+3}{3x}=\dfrac{1}{18}\Rightarrow \dfrac{5}{3x}=\dfrac{1}{18} $ .
Cross multiplying we get, $ 18\times 5=3x $ .
Dividing both sides by 3 we get, $ \dfrac{18\times 5}{3}=x\Rightarrow x=6\times 5\Rightarrow x=30 $ .
Therefore, the time taken by B to complete the work is 30 days.
Hence option A is the correct answer.
Note:
Students can make the mistake of taking $ \dfrac{3}{2}x+x $ as 18. But this is wrong as this sum requires inverse proportion because as the amount of work per day increases, the lesser the number of days is required. Take care in the calculation.
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