A's one day's work =`1/10` B's one day's work =`1/12` C's one day's work =`1/15` A, B and C's one day's work =`1/10+1/12+1/15` `=(6+5+4)/60=15/60=1/4` Let the work completed in x days ∴ A's 2 days work + B's (x − 3) days work + C's x days work = 1 `⇒2xx1/10+("x"-3)xx1/12+"x"xx1/15=1` `⇒1/5+("x"-3)/12+"x"/15=1` `(12+5"x"-15+4"x"=60)/60` ...(L.C.M of 5, 12, 15 = 60) ⇒ 9x = 60 − 12 + 15 ⇒ 9x = 63 `⇒"x"=63/9=7` ∴ Work will last for 7 days Open in App Other Quantitative Aptitude Suggest Corrections 0 Description for Correct answer: 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 Part of solved Time and work questions and answers : >> Aptitude >> Time and work Comments Similar Questions The completion of the work would have been scheduled assuming that A & B will work together for completing the work. Let us suppose it to be x days. but, B worked for x - 2 days. Let the total work = L. C. M (10, 15) = 30 units. A completes → \(\dfrac{30}{10}=3\) units/day B completes → \(\dfrac{30}{15}=2\) units /day \(x=\left(\frac{10×15}{10+15}\right)=\frac{150}{25}=6 \) days For 4 days, A & B worked together Work done = 4 × 5 = 20 units. Remaining work = 10 units. We need to find how much time A will take to complete 10 units of work. A completes 3 units in 1 day A completes 10 units in 10/3 = 3.33 days. Total days → \(4+3\frac{1}{3}=7\frac{1}{3}\) days Alternate Method Formula: W = RT Where W is work and T is time Given, A complete a work in 10 days A complete a work in 15 days Let, 1 work = 30 unit (for simplification: LCM of 10, 15)
If 'B' leaves 2 days before the scheduled completion of the work then A alone will do the work for the last 2 days.
Time = \(4+\frac{10}{3}=7\frac{1}{3}\) days |