Start learning 50% faster. Sign in now
Work done by ( A + C ) in 2 days = 2 (1/10 + 1/20 ) = 2 ((2 + 1)/20 ) = 6/20 = 3/10 Remaining work = 1 - 3/10 = 7/10 ( B + C )’s 1 day’s work = 1/15 + 1/20 = (4 + 3)/60 = 7/60 Time taken by ( B + C ) to finish 7/10 part of the work = 60/7 × 7/10 = 6 days Total time = 2 + 6 = 8 days Alternate method: Let total work = LCM (10,15,20) = 60 units So A’s 1 day work = 60/10 = 6 units B’s 1 day work = 60/15 = 4 units C’s 1 day work = 60/20 = 3 units A & C’s work for 2 days = (6+3)×2 = 18 units So remaining work = 60 – 18 = 42 units Now A is replaced by B So B & C together will do work in 1 day = 3 + 4 = 7 units So they can do the remaining job in = 42/ 7 = 6 days So total days = 6 + 2 = 8 days
In a typical instruction pipeline, which stage is responsible for fetching the next instruction from memory?
Index of first element of linked list default value?
What is the role of a device driver in an operating system?
What is the term for the minimum voltage required to turn on a transistor?
Which one of the following statements is FALSE?
Which data visualization technique is best suited for displaying hierarchical data with a tree-like structure?
What is the maximum number of unique IP addresses that can be assigned in IPv4?
What is a default route in routing?
Write through technique is used in which memory for data updating
___ mapping is used in cache organization which is the quickest and most supple organization.