Aptitude - Time and Work - Discussion

A = 1.75 times efficient means
B = 1.
Together completed = 7days.
If B works alone;
1 * 7 = 7days.

Remaining A work also done by B (efficiency of A) 1.75 * 7 =12.25 days
Total days taken by B = 7 + 12.25 = 19.25days;
If A alone 19.25/1.75(efficiency of A).
= 11 days.

Let A+B = 7 days.

Then (A+B)'s one day work = 1/7,
Let X represent B,
=> 7/4X + X = 1/7,
=>7X + 4X = 4/7,
=>11X = 4/7,
=>X = 4/77,
But A = 7/4X.
=> A = (7/4*4/77),
=> A = 1/11,
Giving us 11 days.

(A+B) 1day work is 1/7 --------------> (1)
Let 1-day work of A is 1/A, such as 1-day work of B is 1/B
Then 1/A + 1/B = 1A + 1/A * 7/4.

1/A + 4/7A = 11/7A--------------(2)
11/7A = 1/7,
7A = 77,
A = 11.

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.

Thank you @Jimmy John.

(A+B)'s 1 day = 1/7
And given that,
B=(4/7)A.

(A+ (4/7)A) 1 day = 1/7
(11/7) A 1 day = 1/7
11 A 1 day= 1
A's 1 day= 1/11.

Therefore, A needs 11 days alone to complete the required work.

Can you please explain me? Please.

Nice explanation @Jimmy John.

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.

Why have they taken efficiency as such without converting it to number of days?