Time and Work Problem with Explanations:
Time and Work is an important topic for all competitive exams. At least 2-3 problems are come from Time and Work chapter in all competitive exams like SSC and bank exams (SBI and IBPS-clerk and PO). This section will teach you to understand Time and Work common formulas to solve problems quickly.
Time and Work- Study material for all competitive exams
Time and work is a part of quantitative or numerical aptitude section for all competitive exams. This chapter will give brief explanation to Time and Work Problems. This will be helpful to crack all competitive exams with more marks. “Practice makes a man perfect” so please practice more problems with different models. All the best for your preparations and success.
Important Formulas for Time & Work:
If A can do a piece of work in n days, work done by A in 1 day = 1/n
If A does 1/n work in a day, A can finish the work in n days
If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then M1 D1 H1 / W1 = M2 D2 H2 / W2
If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days
If A is thrice as good as B in work, then Ratio of work done by A and B = 3:1 Ratio of time taken to finish a work by A and B = 1:3
Time & Work Problems with Explanations
The below Time and Work questions are model for preparing all competitive exams. Time and Work is an important chapter for IBPS and SBI bank exams, SSC exams, Insurance sector exams and other state government competitive exams. This is an easy chapter to learn and get full marks.
Before go through the problems of Time and Work. Please understand and learn basic formulas of Time and Work.Please note question complexity is differ for each exams. While learning solve high difficulty questions to boost your knowledge and success your exams.
Problems with Explanation:
1) 2 groups A and B contain some people each. Efficiencies of all people in group A is same while that in group B is same. 3 workers from group A and 6 from group B can complete the work in 20 days. 8 workers from group A and 4 from group B can complete the work in 10 days. Find the number of days in which 1 person from each group can complete the work working together.
A) 96 days
B) 72 days
C) 90 days
D) 54 days
E) None of these
The question is same as 3 m and 6 w complete work in 20 days. And 8 m and 4 w complete work in 10 days.
So 3A + 6B = 20 => 60 A + 120B = 1
And 8A + 4B = 10 => 80 A + 40B = 1
So 60 A + 120B = 80 A + 40B
80B = 20A
B = A/4
So – 8A + 4 * A/4 = 10
or 9A = 10
So 9 workers of group A can complete work in 10 days
Required to find A + B = A + A/4 = 5A/4
9 * 10 = 5/4 * x
Solve, x = 72 days
2) A, B and C have to complete a work. They decide to divide work in the ratio 2 : 3 : 5 respectively. Their rates of work is in the ratio 1 : 2 : 3. If it takes 12 days by A to complete his part of work, then how much of work can they complete in 8 days?
Let total work = 2 +3 + 5 = 10
So A completes 2 units of work in 12 days, so whole 10 units he can do in 10/2 *12 = 60 days
Now ratio of their efficiencies = 1 : 2 : 3
So days ratio = 1/1 : 1/2 : 1/3 = 6 : 3 : 2
So 6x = 60, x = 10
So A can complete work in 60 days, B in 3*10 = 30 days, C in 2*10 = 20 days
So work together in 8 days = (1/60 + 1/30 + 1/20) * 8 = 4/5
3) A and B can complete a work in 12 and 20 days respectively. After 4 days, they are joined by C who can complete the same work in 24 days, how much work will remain uncompleted after 2 more days?
They work for 4 days. So complete
(1/12 + 1/20) × 4 = 8/15 of work
Now: In next 2 days all A, B, C completed
(1/12 + 1/20 + 1/24) × 2 = 7/20 of work
So total work completed = 8/15 + 7/20 = 53/60
So remaining work = 1 – 53/60 = 7/60.
4) 16 men can complete a job in 7 days. The same work can be completed by 20 women in 9 days. In how many days 14 men and 15 women can complete the job all working together?
A) 2 1/3 days
B) 7 days
C) 4 4/5 days
D) 8 days
E) 6 1/3 days
16 m in 7 days so 14 m in 16×7/14 = 8 days.
20w in 9 days so 15 w in 20×9/15 = 12 days.
So together 1/8 + 1/12 = 5/24
So 24/5 days.
5) A, B and C can complete a work in 12, 15 and 25 days respectively. A started the work, C and B joined him after 3 and 6 days respectively. How much work did A complete?
E) None of these
Let total work completed in x days. So A worked for all x days. B for (x-6) days and C for (x-3) days. So
x/12 + (x-6)/15 + (x-3)/25 = 1.
Solving for x, we get x = 8
So A completed x/12 = 8/12 = 2/3rd of work.
6) A and B produced similar chairs. When A worked for 2 hrs and B for 5 hrs, they completed half of the work. Next they worked together for 3 hrs. If 1/20th of the work is yet to be completed, then find in how many hours can B complete the work working alone?
Let A can complete work in a days and B in b days.
So 2/a + 5/b = 1/2 …..(1)
Now 1/20 work is left after further 3 days. So in 3 days work completed is 1 – (1/2 + 1/20) = 9/20
So 3/a + 3/b = 9/20 ……(2)
Solve equations 1 and 2. b = 15
7) 4 men and 5 women can complete a work in 8 days. The same work can be completed by 2 men and 3 women in 14 days. In how many days 1 women will complete the work?
A) 32 days
B) 45 days
C) 38 days
D) 56 days
E) 28 days
4m + 5w = 8dSo 8 × (4m + 5w) = 8/8 d
32m + 40w = 1 …… (1)
2m + 3w = 14d
So 28m + 42w = 1 ……(2)Now equate (1) and (2)
32m + 40w = 28m + 42w
Solve, 2m = 1w
Put in any of equations above.2w + 5w = 8d
7 w in 8 days. So 1 w in 7 × 8 = 56 days
8) A and B can complete a job in 16 hrs and 24 hrs respectively. In how many hours they together can produce 10 such jobs?
E) None of these
1/16 + 1/24 = 5/48
So they complete 5 jobs in 48 hrs.
Therefore, 10 jobs in 48×2 = 96 hrs.
9) A and B can complete a work in 20 days and 30 days respectively. A starts the job and B joins him on alternate days starting from 2nd day. In how many days they can together complete the work?
A) 10 days
B) 12 2/3 days
C) 14 days
D) 34 days
E) 15 1/5 days
On 1st day A completes 1/20 of work.On 2nd day,
A and B together completes 1/20 + 1/30 = 1/12 of work.
Using LCM method:
Total work = LCM of 20 and 12 = 60
So efficiency of A = 60/20 = 3.
And efficiency of A and B on 2nd day = 60/12 = 5
So in 2 days they complete 3+5 = 8 works
Multiply by 7 both sides.
In 14 days they complete 56 works.
Work left = 60 – 56 = 4
Now on 15th days A, turn. He completes 3 of the work out if 4 left.
Now remaining work = 4 – 3 = 1.
On 16th day, A and B’s turn. So they complete remaining 1 work in 1/5 day.
So total 15 1/5 days.
10) A and B can complete a work in 12 and ‘n’ days respectively. They worked for 3 days and found 3/5th of the work as pending. If C who can complete the same work in 24 days also works with them, how much work will remain uncompleted after 4 days?
3/5th is pending means 2/5th is completed. So
(1/12 + 1/n) × 3 = 2/5
Solve, B can complete work in n = 20 days.
Now: in 4 days all A, B, C completed
(1/12 + 1/20 + 1/24) × 4 = 7/10 of work
So pending work = 1 – 7/10 = 3/10 of work.