Aptitude - Time and Work - Discussion

Guys this is very very simple.

Time taken by A + B : 30 days.
They didn't mention the efficiency of neither A nor B.
So we can take 50/50 efficiency.
For example
Total: 30 candles.
A+B making 30 candles in 30 days.

So, A alone makes 30 candles in 60 Days.

1 represents whole work..i.e
16X + 44Y =1
Means A having work for 16 days and B remaining in 44 so whole work is represented by 1 always.

Supplementary argument.
Like wise if A work 4/15 ( 16/60) and B remaining work 11 /15 ( 1- 4/15) which is done in 44 days.

4 / 15 + 11/15= 1 makes whole work.
Here we write X for A because we don't know As whole work alone and similarly Y for B too. The only thing we know is the remaining work.

Hope understand better now.

Let work done by A in 1 day be a and the work done by B in 1 day be;

A and B together can do a piece of work in 30 days.
=> Work done by A and B in 1 day = 1/30.
a + b = 1/30...........(1),

Work done by A in 16 days + work done by B in 44 days = 1.
16a + 44b = 1 ...........(2).

Solve (1) and (2)

Multiply equation (1) with 16 => 16a + 16b = 16/30=8/15 .......(3)
Subtract equation (3) from (2) => 28b = 1 - 8/15=7/15,
b = 7/15 * 1/28=1/60.

i.e., B alone needs 60 days to finish the work.

Consider like this,

A and b completes a work in 30days. So this means both individually work for 30 days. Let us assume they do some bottles. If 60 bottles they have to do. It can be 30 bottles each or it can be 45 and 15 bottles or 40 & 20. So it depends upon each one capability.

So we can frame an equation like this,

30x+30y=1 (where x and why are the number of bottles per day. It is equated to 1 because for a day we are calculating).

Similarly, if A works for 16 days and B work for the 44 days,

So it will be 16x+44y = 1.

Solving and we will arrive at the answer.

My method:- Mixture

A -------------- B
16 -------------- 44
30 --------------
14 : 14
1 : 1

So, B completed the whole work( means A's work +B's work alone)=(16 x1 +44 x 1) = 60 days.

A + b=30.
Let the efficiency of a and b is 1.
Then total work will be 30*2=60.
A did 16 work then remaining work is,

60-16=44 which will be completed by b in 44 days so B's actual efficiency is 1 so B will finish 60 work in 60 days.

A+B completes work in 30 days, so,

=>unit of work done by both A and B in 1 day = 1/30,
Consider that total work done by both A and B = 30 units,
So,in 1 day = 1 unit work is done,
Since 1 day = 1 unit of work then A do 16 units of work,
Remaining work = 30-16 = 14 units,
This 14 units of work is done by B in remaining 28 days(44 days-16 days).

If we see read the question we get that "the total number of days required by B to complete the total 30 units" is asked.
So we do like this,
for 28 days = 14 units,
x days = 30 units,
we get 14*x = 28 * 30,
=>x = 28*30/14,
=>x=60 days.

Please, anyone help me out from this question--.

A and B can do a work together in 10 days, B and C can do the same work in 18 days, if A starts the work and after working 5 days B do the work for 15 days and remaining work in done by C than find in how many days will C complete the work alone?

Solving the equations is tiresome brothers.
Instead take like

16a +44 b = 30 a + 30b.
a/b = 1/1
So if a one day work is 1 and b one day work is one and when they work together they do 2 units of work every day.

this means there is 30x2 = 60 units of work.

B can do it in = (total unit of work)/(B's one day work) = 60 days.

LCM Method

(A + B) 30 = 16A + 44B

14A = 14B
A/B = 1/1

So one day work

A = 1 unit
B = 1 unit

Total work = 30 x (1 + 1) = 60 units
Time taken by B = 60/1 = 60 days.