>> Aptitude >> Time and work

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1). A and B together can do a piece of work in 12 days, while B alone can do finish it in 30 days. A alone can finish the work in
(A + B)'s 1 day's work = 1/12 B's 1 day's work = 1 /30 A's 1 day's work = \( \Large\frac {1}{12}- \frac{1}{30} = \frac{5-2}{60} = \frac{1}{20}\) | ||||

2). A and B can do a piece of work in 6 and 12 days, respectively. They (both) will complete the work in how many days?
A's 1 day's work = 1 / 6 B's 1 day's work = 1 / 12 (A + B)'s 1 day's work =\( \Large\frac{1}{6} + \frac{1}{12} \) \( \Large\frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4}\) Hence, both will complete the work in 4 days. | ||||

3). A alone can complete a work in 12 days and B alone can complete the same work in 24 days. In how many days can A and B together complete the same work?
A's 1 day's work = 1 / 12 B's 1 day's work = 1/ 24 (A + B)'s 1 day's work = \( \Large\frac{1}{12} + \frac{1}{24} \) =\( \Large\frac{2 + 1}{24} = \frac{3}{24} = \frac{1}{8}\) (A + B) will finish the work in 8 days. | ||||

4). A can do a piece of work in x days and B can do the same work 3x days. To finish the work together they take 12 days. What is the value of x?
A's 1 day's work A = 1 / x B's 1 day's work B = 1 / 3x (A + B)'s 1 day's work = \( \Large\frac{1}{x} + \frac{1}{3x} = \frac{4}{3x} \) and given one day work of both A and B = 1/12 \( \Large\frac{4}{3x} = \frac{1}{12} = > 3x = 48 = > x = 16 \) | ||||

5). A can do a piece of work in 4 days and B can complete the same work in 12 days. What is the number of days required to do the same work together?
A's 1 day's work = 1/ 4 B's 1 day's work= 1/12 = 1 / (A + B)'s 1 day's work = 3 days | ||||

6). A can do a piece of work in 8 days, B can do it in 10 days and C can do it in 20 days. In how many days can A, B and C together complete the work?
A's 1 day's work = 1 / 8 B's 1 day's work = 1 / 10 C's 1 day's work = 1 / 20 | ||||

7). A and B can do a piece of work in 72 days. B and C can do it in 120 days. A and C can do it in 90 days. In what time can A alone do it?
(A + B)'s 1 day's work = 1 / 72 (B + C)'s 1 day's work = 1/120 (A + C)'s 1 day's work = 1 /90 2(A + B + C)'s 1 day's work =\( \Large\frac{1}{72} + \frac{1}{120} + \frac{1}{90} \) (A + B + C)'s 1 day's work = \( \Large\frac{5 + 3 + 4 }{360 X 2} = \frac{12}{360 X 2} = \frac{1}{60} \) A's 1 day's work = (A + B + C)'s 1 day's work - (B + C)'s 1 day's work =\( \Large\frac{1}{60} - \frac{1}{120} = \frac{2-1}{120} = \frac{1}{120} \) A alone can finish the work in 120 days. | ||||

8). A and B can do a piece of work in 10 h, B and C can do it in 15 h, while A and C take 12 h to complete the work. B independently can complete the work in
A's and B's 1 h work = 1/10 B's and C's 1 h work = 1/15 and A's and C's 1 h work = 1/12 A's, B's and C's 1 h work = \( \Large\frac{1}{2}\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{12}\right) \) =\( \Large\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \) Hence, B's work in 1 h = \( \Large\frac{1}{8} - \frac{1}{12} = \frac{1}{24}\) B independently can complete the work in 24 h. | ||||

9). A can do a piece of work in 10 days and B in 20 days. They begin together but A leaves 2 days before the completion of the work. The whole work will be done in
Let the required days be x. A works for (x - 2) days, while B works for x days. According to the question, \( \Large\frac{x - 2}{10} + \frac{x}{20} = 1 \) 2x- 4 + x = 20 3x = 24 x = 8 days | ||||

10). A and B together can complete a work in 3 days. They started together but after 2 days, B left the work. If the work is completed after 2 more days, B alone could do the work in how many days?
(A + B)'s 2 day's work = \( \Large 2 \times \frac{1}{3} = \frac{2}{3} \) Remaining work = \( \Large 1 - \frac{2}{3} \)= 1/3 A will complete 1/ 3 work in 2 A will complete 1 work in 6 A's 1 day's work = 1/6 B's 1 day's work = 1/ 3 - 1/6 = 1/6 B will take 6 days to complete the work alone. |