Aptitude - Time and Work - Discussion

When dealing with addition or subtraction of fractions you consider the denominators i.e
15&20 and then you find the L.C.M which is 60 or multiply them together.

15*20 = 300 so,

(1/15 + 1/20)/300 = (20 + 15)/300.

=25/300 reduce to the lowest term and you get 7/60.

This 7/60 is the amount of work A & B will complete in one day.
Hence for 4 days we have;

7/60 * 4 = 7/15 of the total work.

Since we don't know the real value of the total work we then assume total work to be done to be 1.

Therefore,

Remaining work left will be,

1- (work done)

1-7/15 ; the denominator here is 15 & 1 (since 1 =1/1).

Do the math and you get our final answer to be 8/15.

NOTE !

If you are subtracting a fraction from a whole number, just multiply the denominator by the whole number and then subtract from the numerator. The same rule applies for addition.

@L Raj Scope

Please understand this concept first.

LET CONSIDER A SIMPLE STATEMENT:

A can do a work in 2 days.

That means A's 1 day work = 1/2.

This is the methodology we wanna apply on the data which is given in question.

From the question.

A (CAN D0 A JOB) ----15 days.
B (CAN DO A JOB) ----20 days.

Therefore by the above concept A's 1 day work = 1/15.

B's 1 day work = 1/20.

Therefore both A&B 1 work = (1/15)+1/20) = 7/60.

NOW READ THE QUESTION.

If A&B work together for 4 days then what the fraction of work left.

Consider:

Total work = 1.

A&B (4 DAYS WORK) = (7/60)*4 = 7/15.

REMAIN B WORK = (TOTAL WORK)-(COMPLETED WORK BY A&B FOR 4 DAYS).

= 1-7/15.

= 8/15.

Hope this is helpful.

LCM of A's and B's:
A-15
B-20.

LCM is 60 (It is also total Work) and Efficiency is TOTAL WORK/ TIME i.e.
60/15 = 4 <<<A's 1 day Efficiency
60/20 = 3 <<<B's 1 day Efficiency.

Combining both 1-day Efficiency then we get 4+3=7 (It means A and B can do 7 units in 1 day)

According to the question:
if they have 7 unit Efficiency they work for 4 days then we get:

7 x 4 and divide by total work (60).
7 x 4/60= 7/15 << it is the part of work they will finish in 4 days.

But the question was asked about how much fraction is left then simply we know that total work is 1 then subtract total work 1-7/15.

We get 8/15 that is the answer.

@All.

According to me, it can be solve by 2 methods.

Method- 1:
A's 1 day work = 1/15.
B's 1 day work= 1/40.

(A+B)'s 1 day work= (1/15)+(1/40).
= (8+6) /120.
= 14/120.
= 7/60,

(A+B)'s 4 days work= (7*4)/600,
= 7/15,

Left work is= 1- (7/15 ),
= 8/15 (Ans).

Method-2:
(A+B)'s 4 days work= (4/15)+(4/20),
= (4/15)+(1/5),
= 7/15.
Left work= 1- (7/15).
= 8/15 (Ans).

A can complete work in 15 days.
B can complete work in 20 days.

total work (in units) = L.C.M of 15 and 20 => 60 units.

Unit work of A = Total work (in units)/Number of days A take to complete work => 60/15 = 4.

Unit work of B = Total work (in units)/Number of days B take to complete work => 60/20 = 3.

Work done in 1 day by A and B together = Unit work done by A + Unit work done by B => 4+3 = 7.

Work done by A and B together in 4 days = Work done by A and B together in 1 day * 4 => 7 * 4=28.

Work left = Total unit of work - work done by A and B together in 4 days => 60 - 28 = 32.

@Sassy.

First, we have to calculate the work done for 1 day. For that we have to divide the number of days by 1.

So, A's 1 day work = 1/15 and B's 1 day work = 1/20.

Work done by A and B for 1 day is (1/15)+(1/20) = 7/60.

Here, L.C.M is taken.

As per question given, For 4 days, multiple with 4 then we can get.

Work done by A and B for 4 days = 4(7/60) = 7/15.

Here too L.C.M is taken.

Total work is 1(Assumption).

So,Work left = Total work - Work done by A and B.

Therefore, work left = 1-(7/15) = 8/15.

This is the answer.

I hope this helps you.

For 7/60.

It is necessary to add both 1/15 and 1/20.
But the denominators are different.
So first we equals the denominators by multiplying with.
Two different alternatives.

i.e.
1/15*4/4.

Because whenever we want to gave an alternative for multiplying.
It should applicable to both numerator and denominator.
That's why we use 4/4 for 1/15.

As the same way for 1/20.
We multiplying with 3.
i.e.

1/20*3/3.

Why I'm particularly take 4/4 and 3/3 particularly is.
This is the minimum stage to equal both denominators, understand.
Now it is easy to add.

We can solve this problem by another way also..let us see..

A can do a work in 15 days
B can do a work in 20 days
Take LCM for 15 & 20 i.e Total work = 60

Then,
A's capacity = 60/15 = 4
B's capacity = 60/20 = 3
They work together for 4 days,

Then, A's capacity + B's capacity = 4 + 3 =7
AB's one day capacity = 7
since they work for 4 days, they have done 4x7 =28 work
Work left = Total work - work done by AB
= 60 - 28 = 32
Remaining work / total work = 32 / 60 = 8 / 15

This method will take less time to compute guys...please try it.

A's 1 day's work = 1/15;
B's 1 day's work = 1/20;

(A + B)'s 1 day's work = (1/15 + 1/20)

LCM = 60

5 | 15 , 20
__________
3 , 4
= 5*3*4 = 60

(A + B)'s 1 day's work = (4/60 + 3/60) ( 4/60 = 1/15 AND 3/60 =1/20 )
= 7/60

(A + B)'s 4 day's work = 7/60*4
= 28/60
= 7/15 .

Therefore, Remaining work = (1 - 7/15)
= (15-7)/15
= 8/15.

Since our target is to crack for the competitive exam which means we must do problems as fast as we can to save time, in other words in a short cut way.

Now, for this question, lets suppose they are making chairs (the work).
A takes 15 days. B takes 20 days. Then LCM of 15, 20 is 60.
Capacity of A = 4 chair/day and capacity of B = 3 chairs/day.
So A + B capacity in one day = 4 + 3 = 7 chairs/day,
The question says for 4 days they worked together. ie (7 * 4 = 28).
Remaining = 60 - 28 = 32.
Fraction of work remaining = 32/60 = 8/15.