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P can complete $$\frac{1}{4}$$ of a work in 10 days, Q can complete 40% of the same work in 145 days. R, complete $$\frac{1}{3}$$ of the work in 13 days and S, $$\frac{1}{6}$$ of the work in 7 days. Who will be able complete the work first ?
Answer & Solution
Correct Answer:
Option
C
P completes $$\frac{1}{4}$$ of work in 10 days
P completes full of work in
$$\eqalign{ & = \frac{{10}}{1} \times 4 \cr & = 40{\text{ days}} \cr} $$
Q completes 40% of work in 145 days
Q completes full 100% of work in
$$\eqalign{ & = \frac{{145}}{{40}} \times 100 \cr & = 362.5{\text{ days}} \cr} $$
R completes $$\frac{1}{3}$$ of work in 13 days
R completes full of work in
$$\eqalign{ & = \frac{{13}}{1} \times 3 \cr & = 39{\text{ days}} \cr} $$
S completes $$\frac{1}{6}$$ of work in 7 days
S completes full of work in
$$\eqalign{ & = \frac{7}{1} \times 6 \cr & = 42{\text{ days}} \cr} $$
Clearly, we can see R completes the work first
P completes full of work in
$$\eqalign{ & = \frac{{10}}{1} \times 4 \cr & = 40{\text{ days}} \cr} $$
Q completes 40% of work in 145 days
Q completes full 100% of work in
$$\eqalign{ & = \frac{{145}}{{40}} \times 100 \cr & = 362.5{\text{ days}} \cr} $$
R completes $$\frac{1}{3}$$ of work in 13 days
R completes full of work in
$$\eqalign{ & = \frac{{13}}{1} \times 3 \cr & = 39{\text{ days}} \cr} $$
S completes $$\frac{1}{6}$$ of work in 7 days
S completes full of work in
$$\eqalign{ & = \frac{7}{1} \times 6 \cr & = 42{\text{ days}} \cr} $$
Clearly, we can see R completes the work first
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