Aptitude - Time and Work - Discussion

Given: A is 7/4 times as efficient as B.

A:B = 7/4:1 => A:B = 7:4(the ratio is in terms of efficiency).

As we know that efficiency is indirectly porporational time taken to finish the work done.

A:B = 4:7(the ratio is in terms of time taken to finish the work).

A+B = 7 (this is the time taken to finish the work together).
=> B = 7-A substituting in above equation.
=> A:(7-A) = 4:7.
=> A/7-A = 4/7.
=> 7A = 4(7-A).
=> 7A = 28-4A
=> 7A+4A = 28.
=> 11A = 28.
=> A = 28/11.
=> A = 2 6/11 is the answer.

Efficiency of A:B = 7/4k : k.
No of days taken by A: B = 4k/7k = 4/7 (no of days taken = ratio of the reciprocals of the efficiency).

Let the work done in one day = 1/s.
then , work done by A in one day = 4/s.
work done by B in one day = 7/s.
work done by A+B = 1/7.
4/s + 7/s = 1/7.
11/s = 1/7.
s= 77.

Work done by A in one day = 4/s.
= 4/77.
the efficiency of A = 7/4.
Hence, work done by A in one day = 4/77 * 7/4.
= 1/11
Thus, A can do the job in 11 days.

I think this would be an easy approach:

A=7/4B (A is as efficient as 1*3/4 times as B).

=> On cross multiplying we get, 4A=7B-----------> 1.
=> From question, A+B=1/7 (A,B is one days work so equating to 1/7).

Therefore again cross multiply,
We get, 7A+7B = 1.

=>from 1, we have 4A=7B.
So substitute 4A in 7B.

We get,
7A+4A=1 ==>11A=1===> A=1/11(which is A's one day's work).

Hence,
Total days A need==> 11 days.

Hope you'd have undersTood :).

Here is my simple approach
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A=7 (A's 1 day work)
B=4 (B's 1 day work)
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A+B = 11 (A+B's 1 day work).


A+B can finish work in 7 days,
So, Total work = 11*7=77 unit.

A can do 7 units of work in 1 day.

for 77 unit he will take 77/7= 11 days.
(7 ----> 1,
77 ---> ?) = 77/7.

Step 1.
One day work of A and B, (A + B) = 1/7 ----> (1)

Step 2.
A:B =4:7, according to time not work.

Step 3.
Let assume A and B one day work be 1/4x and 1/7x.
Now, put the value of A and B in equation (1).

A + B = 1/4x + 1/7x.
=1/7 and getting x value as 11/4.

Step 4.
Put the value of x in A's 1-day work, 4x = 4X(11/4) =11.

A's working capacity = X.
B's working capacity = Y.

Given, X + Y = 1/30 ................equation(1).

Also, X = 7/4 Y

i.e.. Y= 4/7 X .................equation (2).

PUTTING Y=4/7X from eqtn (2) in Eqtn(1), we get,

X= 1/11.

i.e.. A working capacity is 1/11( interpretation: "A" can do 1 work in 11 days).

Let,

A = x.
B = 7/4x.
Total work = 7x (i.e. Lcm of A &B).
A's one day work = 7.
B's one day work = 4.
(A + B)'s one day work = 11.
As we know that total work we have 7x.
Using formula of work --->(total work/(A+B) one day work) = (A+B) total work.
7x/11 = 7--->x=7*11/7--> x = 11 i.e total days of x.

A and B collectively can do the work in 7 days.
so their one day work is 1/7.

A+B = 1/7-------(1).
A is 7/4 times efficient than B which means.

A = (7/4)B ----------(2).
or
B = (4/7)A ------------(2).

Put 2 int 1 eq,
A+(4/7)A = 1/7.
A(11/7) = 1/7.

SO,
A CAN COMPLETE THE PIECE OF WORK IN 11 DAYS.

A and B collectively can do the work in 7 days
so their one day work is 1/7
A+B=1/7-------(1)
A is 7/4 times efficient then B which means

A=(7/4)B ----------
or
B=(4/7)A ------------(2)
put the value of (2)in (1)
(4/7)*B+B=1/7

B=1/11 THIS IS HIS ONE DAY WORK
ALONE B CAN FINISH IN 11 DAY

A is 7/4 efficient than B means A=7/4B =>B=4/7A.

A and B together can complete the work in 7 days.
[i.e.Time is directly proportional to work => 7days=work => 1 day=1/7th work].

So,

A + B = 1/7.
A + 4/7A= 1/7.
11/7A = 1/7.
A = 1/11.

Hence, A alone can work for 11 days.