# Time and Work Short Tricks Solved Problems for Competitive Exams

Time and Work is an important topic in the Quantitative Aptitude section for Bank and SSC Exam. In this topic, 2 to 3 questions are asked every year in competition exam. You should learn the shortcut tricks for saving time in the competitive exams.

Here is a detailed introduction of Time and Work along with short tricks which will help the candidates to prepare the topic well. For more practice, you can visit on **Time and work problems with solutions.**

Here you can get all the topics of** ****Time and Work Aptitude**** **by click here.

**What is Time and work?**

Performing or doing any work involves efforts of person over a period of time. Therefore, the number of person (P), the quantity of work (W) and the period of time taken (T) are important variables in problem-related to time and work.

**Time and Work Short Tricks Solved Problems**

**Example 1:** **Chandni and Divakar can do a piece of work in 9 days and 12 days, respectively. If they work for a day alternatively, Chandni beginning in how many days, the work will be completed?**

(A)

(B)

(C) 11.11

(D) 10

**Solution **

. Chandni’s 1 days’s work = 1/9.

Divakar’s 1 day’s work = 1/12.

Chandni and Divakar’s (1+1) = 2.

Day’s work = (1/9)+(1/12) = 7/36

So, in 10 days they do = (7×5)/36 = 35/36.

So, the remaining 1/36 {= 1-35/36}

Work will be done by Chandni

= (1/36)/(1/9) = 1/4.

Thus, total number of required days

= 10+(1/4) = days

**Short Tricks **

Efficiency of Chandni = 11.11%

Efficiency of Divakar = 8.33%

They do 19.44% work in 2 days.

⸫ They need 10 days to do 97.22% work.

Now, the rest work (2.78) was done by Chandni in 2.78/11.11 = 1/4 day

Therefore, total number of days

Required = 10+1/4 =

**Example 2:** **Pipe A can fill an empty tank in 30 h while B can fill it in 45 h. Pipe A and B are opened and closed alternatively i.e., first pipe A is opened, then B, again A and then B and so on for 1 h each time without any time lapse. In how many hours the tank will be filled when it was empty, initially?**

(A) 36

(B) 54

(C) 48

(D) 60

**Solution **

In 1 h pipe A can fill = 1/30 part of the tank

In 1 h pipe B can fill = 1/45 part of the tank

In 2 h pipes A and B can fill = 1/18 part of the tank

Therefore, in 36 h the tank will be completely filled.

**Short Tricks **

Efficiency of pipe A = 3.33%

Efficiency of pipe B = 2.22%

And combined efficiency = 5.55%

Therefore, in 2 h pipes A and B fill 5.55%. Thus, to fill 100% tank, these pipes will take 36 h.

**Example 3:** **If 20 persons can do a piece of work in 7 days, then calculate the number of persons required to complete the work in 28 days.**

(A) 5

(B) 10

(C) 15

(D) 25

**Solution **

Number of persons × days = work.

20 × 7 = 140 man-days.

Now, X × 28 = 140 man-days.

→ X = 5

Therefore, in second case the required number of person is 5.

**Short Tricks**

Since, work in constant, therefore

M1×D1 = M2×D2 = Work done

20 × 7 = M2 × 28 → M2 = 5.

**Example 4: **A can do a piece of work in 14 days, while B can do it in 21 days. In how many days, working together they will complete the whole work?

(A) 10.5

(B) 8

(C) 8.4

(D) 9

**Solution **

Efficiency of A = 7.14%

Efficiency of B = 4.76%

Efficiency of (A+B) = 11.9%

⸫ Number of days required by A and B, working together = 100/11.9 = 8.4 days

Hint you can see, that there is only one option between 8 and 9, which is 8.4.

For 8 days denominator should be 12.5 as for 9 days denominator should be almost 11.

**Short Trick **

1 days work of A and B = 1/14 + 1/21 = 5/42

⸫ Required number of days = 42/5 = 8.4 days.

**Example 5:** **16 men finished one-third work in 6 days. The number of additional men are required to complete the job in next 6 days.**

(A) 10

(B) 8

(C) 16

(D) 32

**Solution **

M × D = W,

16 × 6 = 1/3 W

Rest work = 2/3 W

For double work in same time 32 men are required to do the work or 16 more men required to do the work.

**Shor trick **

2×(16×6) = 6×M → M = 32.

⸫ 16 more men are required.

**Example 6:** A group of men decided to do a job in 4 days. But since 20 men dropped out every day, the job completed the at the end of the 7^{th} day. How many men were there at the beginning?

(A) 240

(B) 140

(C) 280

(D) 150

Let n be the initial number of men, then n × 4 = n +(n-20)+(n-40)+…….+(n-120)

→ 4n = 7n – 240 → 3n = 420

⸫ n = 140 men.

**Short Tricks**

Go through option

140 × 4 = ( 140 + 120 +100+….+20)

560 = 560

**Example 7:** A is twice as good a workman as B and therefore A takes 6 days less than B to finish the work individually. If A and B working together complete the work in 4 days, then how many days are required by B to complete the work alone?

(A) 12

(B) 18

(C) 8

(D) 6

**Solution **

A B

Efficiency → 2 : 1

Days → 1 : 2

{Days α 1/Efficiency}

Now, let A requires X days, then B requires 2X days

⸫ Difference in number of days

{=(2X – X) = X = 6 → X = 6

⸫ B requires 2X = 2×6 = 12 days.

### Second Method:

If A takes X days, then B takes ( X+6) days.

Now, A’s 1 day’s work = 1/X.

B’s 1 day’s work = 1/(X+6) ⸫ (1/X)/1/(X+6) = 2/1.

(Since, A does twice the work as B does)

→ X = 6

⸫ B takes 2x = 12 days.

### Third Method:

(1/X) + {1/(X+6)} = 1/4.

→ X = 6 and 2X = 12 days (required by B).

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