(Paper) Quantitative Aptitude (Problems on Time and Work)

Quantitative Aptitude - Problems on Time and Work

1. If A can do a piece of work in n days, then A’s 1 day work = 1/n

2. If A’s 1 day’s work = 1/n, then A can finish the work in n days.
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

3. If A is thrice as good workman as B,then: Ratio of work done by A and B = 3:1. Ratio of time taken by A and B to finish a work = 1:3

4. Definition of Variation: The change in two different variables follow some definite rule. It said that the two variables vary directly or inversely. Its notation is X/Y = k, where k is called constant. This variation is called direct variation. XY = k. This variation is called inverse variation.

5. Some Pairs of Variables:
1. 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.

2. 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.

3. 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.

6. Some Important Tips:
More Men - Less Days and Conversely More Day - Less Men.
More Men - More Work and Conversely More Work - More Men.
More Days - More Work and Conversely More Work - More Days.
Number of days required to complete the given work = Total work/One day’s work.

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.

7. If men is fixed,work is proportional to time. If work is fixed, then time is inversely proportional to men therefore, (M1*T1/W1) = (M2*T2/W2)

Problems on Time and Work

1) If 9 men working 6 hours a day can do a work in 88 days. Then 6 men working 8 hours a day can do it in how many days?

Solution: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
so (9*6*88/1) = (6*8*d/1)
on solving, d = 99 days.

2) If 34 men completed 2/5th of a work in 8 days working 9 hours a day. How many more man should be engaged to finish the rest of the work in 6 days working 9 hours a day?

Solution: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
so, (34*8*9/(2/5)) = (x*6*9/(3/5))
so x = 136 men
number of men to be added to finish the work = 136-34 = 102 men

3) If 5 women or 8 girls can do a work in 84 days. In how many days can 10 women and 5 girls can do the same work?

Solution: Given that 5 women is equal to 8 girls to complete a work
so, 10 women = 16 girls.
Therefore 10women +5girls = 16girls+5girls = 21 girls.
8 girls can do a work in 84 days