3 Views

Question : $\mathrm{N}_1$ alone can do 20% of a piece of work in 6 days, $\mathrm{N}_2$ alone can do 20% of the same work in 4 days, and $\mathrm{N}_3$ alone can do 50% of the same work in 12 days. In how many days can they complete the work, if all of them work together?

Option 1: 10 days

Option 2: $\frac{120}{13}$ days

Option 3: 8 days

Option 4: $\frac{120}{17}$ days


Team Careers360 16th Jan, 2024
Answer (1)
Team Careers360 19th Jan, 2024

Correct Answer: 8 days


Solution : $N_1$ alone can do 20% of a piece of work in 6 days.
= 100% = 30 days
$N_2$ alone can do 20% of the same work in 4 days.
= 100% = 20 days
$N_3$ alone can do 50% of the same work in 12 days.
= 100% = 24 days
Total work = Time × Efficiency
Total work = LCM(30, 20, 24) = 120 units
Efficiency of $N_1$= $\frac{120}{30}$ = 4
Efficiency of $N_2$ = $\frac{120}{20}$ = 6
Efficiency of $N_3$ = $\frac{120}{24}$ = 5
Total of efficiency = 4 + 6 + 5 = 15
All of them working together finish the work = $\frac{120}{15}$ = 8 days
Hence, the correct answer is 8 days.

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
GRE ® Registrations 2024
Apply
Apply for GRE® Test now & save 10% with ApplyShop Gift Card | World's most used Admission Test for Graduate & Professional Schools
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books