Aptitude - Time and Work - Discussion

Avoid fractions! they mess up the calculations.

Solution:

Total amount of work = LCM (10, 50) = 50 units.

Given: A's time = (B+C)'s time.

Hence, A's rate = B's rate + C's rate.

(Rate is the rate of doing work, not the monetary rate).

(A+B)'s time = 10 days.

Hence A's rate + B's rate = 50/10 = 5 units/day.

C's time = 50 days.

Hence C's rate = 50/50 = 1 unit/day.

Hence, A's rate + B's rate + C's rate = 5 + 1 = 6 units/day.

Now, using A's rate = B's rate + C's rate,

We get A's rate = 6/2 = 3 units/day.

Hence B's rate = 5-3 = 2 units/day.

Hence B's time = Total work/B's rate = 50/2 = 25 days.

Given : AB=10; C=50; find B=?;

LCM of 50, 10 is "50"(total work),
Efficiency of 'AB' =total work : 50/10 = 5,
Efficiency of 'C' =total work : 50/50 = 1.

From efficiency of 'AB & C' is ---> A+B+C : 5 +1 = 6.
Question itself given that : A=B+C;
We know A+B+C = 6--->(1) , in question they said A=B+C -----> (2) ;
Substitute (2) in (1):
We get (1).
B+C+B+C =6;
2B+2C=6; (w.k.t efficiency of C = 1)
2B+2(1)=6;
2B+2=6;
2B=6-2 ; ---> 2B=4; ---> B=2;
Hence, total B alone worked : (50)/2 = '25' days.

Suppose A's one day work=1/x=(B+C) one day work.
A+B one day work=1/10,
C's one day work=1/50,
B's work=?
B's one day work= (A+B)'s one-day work-A's one day work.
=1/10-1/x.

Again, for B's one day work=(B+c)'s one-day work- c's one day work.
= 1/x-1/50.
by comparing these, we get;
1/10-1/x = 1/x-1/50,
by solving this we get x=50/3,
Hence A's one day work=3/50,
B's one day work=1/10-3/50,
B's one day work= 1/25,
So, B alone can do it in 25 days.

A+B can finish work in 10 days so its efficiency is 5 unit per day.
c can finish work in 50 days so its efficiency will be 1 unit per day.
So from here, we can calculate the total work, which will be 50 unit.
now according to the question,

Efficiency of A =efficiency of B+C
Adding B both sides,
A+B=2B+C
5=2B+1.

Therefore the efficiency of B will be,2 unit per day.

Now since the total work is 50 unit,we can calculate the time,
Time = total work/efficiency
Time=25 days.

A = B+C( A work is equal to B and C work)---> 1
A+ B = 10 days (A and B can do a work in 10 days)--> 2
C=50 days( C alone do work in 50 days)----> 3.

Let's assume work as a 50 parts.(take LCM of 10,50).

A+B=5parts(A anb B can do 5 parts a day work)--> 4.
C=1 part(C can do work 1 part a day I,e 50 parts 50 days)---> 5.

From equation 1: A=B+C
A = B + 1(from eq 5).
3 + 2 = 5(from eq 4).
B = 2 parts.
50 parts of work can do in 25 days. B can do parts a day.

Hey, I can explain in a simple way.

1st it says 'A' can do work which is done by 'B' and 'C'
there fore,A=B+C------->eq 1
then,
A + B = 10
c = 50.

Find the LCM for 10 and 50 LCM = 50units.

A + B=5units[50/10].
C = 1units[50/50].
Find B alone?
A + B = 5units
B + C + B = 5 [according to equation 1 A = B + C]
2B + C = 5
2B + 1 = 5
2B = 4
B = 2.

Therefore, 50units/2units = 25days.
I hope u understood this problem solving method.

A = B+C.
A+B->10.
C->50.

We can write C as A-B.
Therefore, A-B->50.
LCM of 10 and 50 is 50.

As we know that efficiency x time = work ------> (1)
Here work = lcm of 10 and 50 = 50

Therefore the efficiency of A+B is 50/10 = 5(use formula 1)
The efficiency of C i.e A-B is 50/50 = 1,
Solving both equations we get A=3 B = 2.

Now time is taken by B is=50/2 =25 (by using formula 1).
So, the time taken by B alone is 25days.

Given A = B+C ------ Eq.1.

& A+B = 1/10.

& C = 1/50.

NOW, A+B+C = (1/10+1/50).

= 3/25.

But in the question he is asking only 'B',

To get B we have to know the individual of A.

So,

SINCE B+C = A (given).

A+B+C = 3/25;.

A+A = 3/25;.

2A = 3/25.

A = 3/50.

HERE WE KNOW A = 3/50; C = 1/10 then.

A = B+C; 3/50 = B+1/50;.

B = 3/50 - 1/50;.

B = 2/50;.

B = 1/50;.

Therefore B can only do in 25 days.

Hey there is also a easy way..

A'1 day work = (B+C)'1 day work
(A+B) TOGETHER DO IT IN= 1/10.... equ(1)
C= 1/50..... equ(2)
put the value of A in equ(1)
that is (B+C+B)= 1/10
THEN (2B+C)=1/10
put the value of C in this equation
2B+1/50=1/10
then it becomes 2B= 1/10-1/50
2B= 4/50
2B= 2/25
B= 1/25( the value 2 is cancel by divided by 2)

Hi Guys, sorry for the late and here is the explanation about you doubt.

In my previous explanation I mentioned

(a+b+c)'s one day work = 3/25
and a+b+c = 2a here b+c=a means a's one day work = b+c's one day work
then a's one day work+(b+c)'s one day work = 3/25
so a+a = 3/25
2a = 3/25
so here 2* a's one days work = 3/25

So I think u can under stand now

keep practicing.....