Let the total work be \( L \) (in hours).
\[ \frac{L}{72} + \frac{L}{64} \]
Finding the LCM of 72 and 64, which is 4608:
\[ \frac{64L + 72L}{4608} = \frac{136L}{4608} \]
Work done per hour:
\[ \frac{L}{33.88} \]
\[ \frac{L}{64} \div \frac{L}{33.88} \times 100 \]
Substituting values:
\[ \frac{136L}{4608} \times 100 \]
\[ = 53\% \]
Thus, the correct answer is 53% (Option B).
List-I | List-II |
---|---|
(A) Confidence level | (I) Percentage of all possible samples that can be expected to include the true population parameter |
(B) Significance level | (III) The probability of making a wrong decision when the null hypothesis is true |
(C) Confidence interval | (II) Range that could be expected to contain the population parameter of interest |
(D) Standard error | (IV) The standard deviation of the sampling distribution of a statistic |