Time and Work shortcuts, tips and tricks

Hello and Welcome everyone!!

I always thought that after completing my masters (MCA) I'll distribute my so valued notes to other needy people, especially students. So why not create a blog and make them available to all YOU dear students. 

So here I am to blog many Quantitative-Reasoning topics with SHORTCUTS, Computer related academic topics in simple language and some well crafted ANSWERS collected from different sources (books and internet) and may be any random topic which I feel to share with you all.

Right now I am appearing for Bank examination and I know why shortcuts are so important for these kinds of exams. So my first topic will be regarding shortcuts to solve Time and Work questions.

LETS START!!!

TIME AND WORK

Time and work is one of those topics that can surely fetch marks. But when done in traditional way it is time consuming, so we will look at these problems in different way. There are different types of questions asked in Time and work, let’s take them one by one.

TYPE I:

EG 1 - A completes work in 6 days and B completes the same work in 4 days. So in how many days A and B together complete the same work??

SHORTCUT 1 [Use for small numbers of days]

Given,

A=6 days,

B=4 days,

A+B=?

So,

A+B = (6*4) / (6+4)

         = 24/10

         = 12/5 days

SHORTCUT 2 [Use for large numbers of days]

Take LCM of 6 and 4 which is 12.

Now let us assume, 12 = total number of units of work done.

Therefore,

A’s 1 day work = 12/6 = 2units and

B’s 1 day work = 12/4 = 3units

So,

A+B’s 1 day work = 2+3 = 5units

Hence,

A+B together will complete 12 units of work in = 12/5 days.

TYPE II:

EG 1 - A completes work in 20 days and A and B together competes the same work in 5 days. So in how many days B alone can complete the same work??

[This question is variation of type I.]

SHORTCUT

Given,

A=20 days,

A+B=5 days,

B=?

So,

B = (20*5) / (20-5)

    = 100/15

    = 20/3 days

TYPE III:

EG 1 - A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will B alone finish the work??

SHORCUT

In this type of question where two or more workmen are compared (eg. A is twice as good as workman as B) to each other, we have to consider efficiency of these workmen. Efficiency(E) is always inversely proportional to Days(D).

i.e,

If E increases, then D decreases.

If D increases, then E decreases.

NOTE: When D decreases, Least Efficiency / High efficiency.

              When D increases, High Efficiency / Least efficiency.

(Don’t worry!! You will know what that means ones you look at the solution below.)

Given,

A = 2B,  A+B = 18 days,  B=?

Now,

            A = 2B = 2 [2 is LCM of 1 and 2]

Therefore,

            Efficiency of A = 2/1 = 2

            Efficiency of B = 2/2 = 1 and

            Efficiency of A+B = 2+1 = 3

This is how you can think while solving :

	EFFICIENCY	DAYS
A+B	3	18(given)
B	1	?

Here we can see for B :

Efficiency decreases (3 to 1)

Therefore, Days will INCREASE. (Remember the note??!! Read it one more time now)

 So calculate,

            18*(3/1) = 54 days

EG 2 - A is twice as good as B and thrice as good a C. B and C together finish a piece of work in 44 days. In how many days will A, B and C together finishes the work. 

SHORCUT

Given,

A = 2B = 3C,  B+C = 44 days,  A+B+C=?

Now,

            A = 2B = 3C = 6 [6 is LCM of 1, 2 and 3]

Therefore,

            Efficiency of A = 6/1 = 6

            Efficiency of B =6/2 = 3

            Efficiency of C = 6/3 = 2  

            Efficiency of B+C = 3+2 = 5 and

            Efficiency of A+B+C = 6+3+2 = 11

This is how you can think while solving :

	EFFICIENCY	DAYS
B+C	5	44(given)
A+B+C	11	?

Here we can see for A+B+C :

Efficiency increases (5 to 11)

Therefore, Days will DECREASE. (Remember the note??!! Read it one more time again)

 So calculate,

            44*(5/11) = 20 days

TYPE IV:

EG 1- If 3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to finish the work??

SHORCUT

Given,

3 men completes work in 12 days è 3*12 = 36 men completes work in 1 day

5 women complete work in 12 days è 5*12 = 60 women completes work in 1 day

Therefore,

                        36M = 60W

            i.e       3M =5W ---(i)

We want,

            6men and 5 women completes work in ?? days

            6M + 5W

= 6M + 3M --- (from (i))

            = 9M

So,

            3M takes 12 days

            9M takes 12 / 3 = 4 days

 

EG 2- If 2 men and 3 boys can do a work in 10 days. 3 men and 2 boys can do a work in 8 days. How long will 2 men and 1 boy take to finish the work??

SHORCUT

Given,

2M + 3B = 10 days

3M + 2B = 8 days,

2M + 1B = ??

2M + 3B = 10 days è (2M*10) + (3B*10)è 20M + 30B

3M + 2B = 8 days è (3M*10) + (2B*10) è 24M + 16B

Therefore,

            20M + 30B = 24M + 16B

            14B = 4M

            7B = 2M   ---(i)

Substitute eq(i) in 20M + 30B we get,

            70B + 30B = 100B  ---(ii)

We want,

            2men and 1 boy completes work in ?? days

            2M + 1B è 7B + 1B--- (from (i))

            = 8B

So,

            100B / 8B   ---(from (ii))

= 25/2 days

= 12.5 days

That’s all for the today..... Next blog will be continuation of Time and Work...... Till then keep smiling!!