A certain number of men can complete a work in six hours less than the time taken by some women. Work completed by one man in one hour is the same as the work completed by one woman in one hour.

Which one of the following ratios of number of men to number of women can satisfy the above given condition?

(i) 5:6

(ii) 10:3

(iii) 8:5

(iv) 10:7

Question

A certain number of men can complete a work in six hours less than the time taken by some women. Work completed by one man in one hour is the same as the work completed by one woman in one hour.

Which one of the following ratios of number of men to number of women can satisfy the above given condition?

(i) 5:6

(ii) 10:3

(iii) 8:5

(iv) 10:7

Accepted Answer

**Solution:**

1. Let the number of men be *x* and the number of women be *y*.
2. Let the time taken by men to complete the work be *(a - 6)* hours and by women be *a* hours.
3. Since one man and one woman have the same work rate, total work = number of workers × time.

Therefore, total work by men = *x (a - 6)*

Total work by women = *y a*

4. Equate the total work:

$$x (a - 6) = y a$$

5. For each ratio, substitute *x* and *y* and solve for *a*:

- (i) 5:6

$$5 (a - 6) = 6 a \ 5a - 30 = 6a \ 6a - 5a = 30 \ a = 30$$

But substituting back, this gives a negative time difference (not possible), so discard.

- (ii) 10:3

$$10 (a - 6) = 3 a \ 10a - 60 = 3a \ 7a = 60 \ a = \frac{60}{7} \approx 8.57$$

Valid positive time, so possible.

- (iii) 8:5

$$8 (a - 6) = 5 a \ 8a - 48 = 5a \ 3a = 48 \ a = 16$$

Valid positive time, so possible.

- (iv) 10:7

$$10 (a - 6) = 7 a \ 10a - 60 = 7a \ 3a = 60 \ a = 20$$

Valid positive time, so possible.

**Answer:** (e) only (ii), (iii) and (iv)

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