## Quantitative Aptitude - Problems on Time and Work |

Important Things to know for solving Problems on Time and Work
Example: If A can do a piece of work in 4 days,then A’s 1 day’s work = 1/4. If A’s 1 day’s work = 1/5, then A can finish the work in 5 days
- Number of workers and their wages. If the number of workers increases, their total wages increase. If the number of days reduced, there will be less work. If the number of days is increased, there will be more work. Therefore, here we have direct proportion or direct variation.
- Number workers and days required to do a certain work is an example of inverse variation. If more men are employed, they will require fewer days and if there are less number of workers, more days are required.
- There is an inverse proportion between the daily hours of a work and the days required. If the number of hours is increased, less number of days are required and if the number of hours is reduced, more days are required.
More Men - Less Days and Conversely More Day - Less Men. Since the total work is assumed to be one(unit), the number of days required to complete the given work would be the reciprocal of one day’s work. Sometimes, the problems on time and work can be solved using the proportional rule ((man*days*hours)/work) in another situation.
Problems on Time and Work
Solution: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
Solution: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
Solution: Given that 5 women is equal to 8 girls to complete a work
Solution: A’s one hour work = 1/8.
Solution: Given that B alone can complete the same work in days = half the time
Solution: if A takes x days to do a work then
Solution: Ratio of time taken by A & B = 160:100 = 8:5
Solution: A can complete the work in (7*9) = 63 days
Solution: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
Solution: Whole work will be done by X in 10*4 = 40 days.
Solution: C’s one day’s work = (1/3)-(1/6+1/8) = 1/24
Solution: Work done by A in 10 days = 10/80 = 1/8
Solution: Let the number of pages typed in one hour by P, Q and R be x,y and z respectively. Then
Solution: Number of pages typed by Ronald in one hour = 32/6 = 16/3
Solution: (1/(x+8))+(1/(x+(9/2))) = 1/x
Solution: (A+B)’s one day’s work = 1/12;
Solution: (A+B+C)’s one day’s work = 1/6;
Solution:(A+B)’s one day’s work = 1/10;
Solution:Ratio of times taken by A and B = 1:3. Solution: Work Done by A n 10 days = 10/80 = 1/8
Solution: M1*D1/W1 = M2*D2/W2
Solution: A’s one day’s work:B’s one days work = 150:100 = 3:2
Solution: B’s 10 day’s work = 10/15 = 2/3
Solution: (B+C)’s one day’s work = 1/9+1/12 = 7/36
Solution: work done by X in 4 days = 4/20 = 1/5
Solution: Whole work is done by A in 20*5/4 = 25 days
Solution: (A+B)’s one day’s work = 1/45+1/40 = 17/360
Solution: Part of work done by A = 5/10 = 1/2
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days Solution: (A+B)’s 2 days work = 1/16 + 1/12 = 7/48
Solution: A’s two day’s work = 2/20 = 1/10
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none Solution: 7*12 men complete the work in 1 day.
(a) 2 days (b) 3 days (c) 4 days (d) 5days Solution: 1 man’s 1 day work = 1/108
(a). 1 (b). 4 (c). 19 (d). 41 Solution:(1 man + 1 woman)’s 1 days work = 1/3+1/4 = 7/12
(A) 15 (B) 18 (C) 22 (D) data inadequate Solution: one man’s one day’s work = 1/48
(A) 3 (B) 4 ( C) 6 (D) 8 Solution: one child’s one day work = 1/192;
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days Solution: one man’s one day’s work = 1/192
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men Solution: one man’s one day’s work = 1/384
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days Solution: Let 1 man’s 1 day’s work = x days and
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs Solution: Let 1 man’s 1 hour’s work = x |

Posted by: Administrator - Mon, Mar 22, 2010 at 11:55 AM. This article has been viewed 9983 times. |

Online URL: http://www.articlediary.com/article/quantitative-aptitude-problems-on-time-and-work-400.html |

Powered by PHPKB (Knowledge Base Software)