# Aptitude Tips and Tricks for Time and Work problems for Competitive Exams

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Time and work are one of the most important topics in any competitive examinations. The questions asked on this topic in any competitive examination are easier. So, attempting this questions faster would save a plethora of time to other questions.

Solving Time and work problems using traditional approach is time taking and needs lots of practice to achieve a good grip on the topic.

Interview stuffs here provides you an easier way to solve time and Work problems with easy short cuts and fractions to remember for solving the problems in the blink of an eye.

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Let’s explore easy solving methodology with an example problem

Before solving these problems, remember the fraction values provided below for faster calculations:

** see Problem 1: If A completes work in 3 days, B completes the same work in 6 days, In how many days they both together complete the work?**

** Solution:**

If A takes 3 days to complete the work, the amount of work done by him in one day is 1/3

If A takes 6 days to complete the work, the amount of work done by him in one day is 1/6

To easily calculate these fraction values we will solve problems in terms of efficiency

Efficiency is the term in %, it’s the value that signifies the % of work he does in a single day.

Work efficiency of A = 1/3 = 33.33%, Work efficiency of B = 1/6 = 16.66%

Combined efficiency = 33.33 + 16.66 = 49.99 ~ 50% efficient, in other terms both A & B completes 50% of work in single day and the total work (100%) is completed by them in 2 days.

Hence 2 days is the answer to above problem!

To easily calculate Percentages for different days, remember below table, Interview stuffs provided you the tips to easily remember mostly used values 🙂

Remember Percentages easily from below table

Days taken to complete work Work done in one day Efficiency in % Tips to Remember fraction
n 1/n 100/n To get any % multiply fraction with 100
1 1/1 100
2 1/2 50 half of 100
3 1/3 33.33 denom 3: all 3's in %
4 1/4 25 Quarter of 100
5 1/5 20 100/5 = 20
6 1/6 16.66 denom 6: 3 6's in the %
7 1/7 14.28 denom 7:

(7*2).(7*4) = 14.28
8 1/8 12.5 denom 8: 1 + 2 + 5 -- 12.5
9 1/9 11.11 denom 9:

(20-9).(20-9) = 11.11
10 1/10 10 100/10
11 1/11 9.09 denom 11:

(20-11):(20-11) = 09.09
12 1/12 8.33 100/12
20 1/20 5 below all denom are multiplied by 10, so remember corresponding single digit % and divide them by 10
30 1/30 3.33 33.33/10 = 3.33
40 1/40 2.5 25/10
50 1/50 2 20/10
60 1/60 1.66 16.66/10
70 1/70 1.428 14.28/10
80 1/80 1.25 12.5/10
90 1/90 1.111 11.11/10
100 1/100 1 10/10

**Problem 2: What is the time to fill a tank when pipe fills the tank in 10 minutes and leakage empties the tank in 20 minutes?**

**Solution:**

Efficiency of the pipe = 1/10 = 10%

Leakage efficiency = 1/20 = 5%

To get total efficiency needed to fill the tank we need to deduct leakage efficiency from the efficiency of the pipe. i.e.., 10% – 5% = 5%

Work done in 1 minute by the pipe to fill the tank with leakage is 5% or 1/20, then 100% work (to fill the tank) is done in 20 minutes.

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**Problem 3: 4 men and 6 women working together can complete the work within 10 days. 3 men and 7 women working together will complete the same work within 8 days. In how many days 10 women will complete this work?**

**Solution:**

Assume efficiency of 1 man = x,

the efficiency of 1 women = y

Efficiency of 4 men and women = 4x+6y = 10

Efficiency of 3 men and 7 women = 3x+7y = 12.5

Solving equation we get x=-0.5, y=2

here y is the efficiency of one woman = 2%

effiecincy of 10 women = 20%

Work done by 10 women in single day is 20%

100% work is done in 5 days

hence answer is 5 days

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