A and B together can finish a work at their usual efficiencies in 36 days. If A had worked at 2/3 of

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25 Mar 2022 Time and Work

A and B together can finish a work at their usual efficiencies in 36 days. If A had worked at 2/3 of his usual efficiently and B had worked at twice his usual efficiently, then the work would have been completed in 30 days. How many days would A take to finish the task if he works alone at twice the usual efficiency?

OPtion

1) 90
2) 45
3) 24
4) 20
5) 72
6) 30
7) 15
8) 60
9) 36
10) None of these

Solution

Let A and B can do A unit and B unit of work per day respectively, then 36A + 36B = 1 ----(i)
Now, with A working with 2/3 efficiently and B with twice efficiency, it takes 30 days to finish => 30*(2/3)*A + 30*2B = 1
20A + 60B = 1 ----(ii)
Solving (i) & (ii), we get, A=1/60, B=1/90
When A works alone with twice effeciency, his 1 day work = 2*A = 2*1/60 = 1/30
.'. A would finish the work in 30 days.

Correct Option: 6 Best Solution (2)

G.S.Kantharao   2 years AGO

Let x,y are efficiencies
1/x+1/y=1/36
2/3x+2/y=1/30
4/3x=1/18 -1/30
1/x=3(5-3)/90*4=
1/x=1/60=>x=60
If the efficiency is doubled then work completes in 30days
My option is 6👈

Like? Yes (6) | No   

Pargat Pal Singh   2 years AGO

If per day efficiency of A and B are x and y respectively, then
36x+36y= 30*(2/3)*x +30*2y = 20x+60y
16x=24y
y=(2/3)x
Putting y in place of x ,
36x+36*(2/3)*x=1
36x+24x=1
60x=1
It means , with normal efficiency, A will take 60 days to complete the work.
At twice the usual efficiency, A will take 30 days.

Like? Yes (4) | No (1)