Question 1201576
<pre>
Suppose A can do 1 job in x days at 100% efficiency, so his work rate is 1/x
jobs per day.

B can do 1 job in y days at 100% efficiency, so his work rate is 1/y jobs per
day

>>>Both A and B working together [at 100% efficiency] can complete the work in
6 days.<<<

{{{6/x + 6/y = 1}}}

At 10% [{{{1/10}}}] 1efficiency, B's work rate is {{{1/(10y)}}} jobs per day.

At 75% [{{{3/4}}}] more than his efficiency, A's work rate is 175% [{{{1&3/4}}}
or {{{7/4}}}]
 or {{{7/(4x)}}} jobs per day.



>>>116.6% of the work done by B at 10% of his efficiency<<<

That means in any given amount of time, say 1 unit of time, so it says

116.6% of the work done by B at 10% of his efficiency in 1 unit of time:

[I'm going to consider that as 116.6666666...% or {{{"1.166666666..." = 1&1/6}}}
or {{{7/6}}}, to avoid god-awful decimals or fractions.]

That's {{{(7/6)(1/(10y)*1)}}}, or {{{7/(60y)}}}

>>>is equal to<<< 

>>>25% [1/4] of the work done by A when he works at 75% more than his 
efficiency.<<<

That means in the same amount of time, say 1 unit of time, so it says

So {{{1/4}}} of A's work rate at {{{1&3/4}}}{{{""*""}}}{{{1}}}

That's {{{(1/4)(7/(4x)) = 7/(16x)}}} jobs per day.

So we have the equation

{{{7/(60y) = 7/(16x)}}}

{{{7(16x)=7(60y)}}}

{{{16x=60y}}}

{{{4x=15y}}}

{{{x=15y/4}}}

Substitute in

{{{6/x + 6/y = 1}}}

{{{6^""/(15y/4) + 6/y = 1}}}

Multiply the first term by {{{4/4}}}

{{{24/(15y) + 6/y = 1}}}
{{{8/(5y) + 6/y = 1}}}
{{{expr(1/y)(8/5+6)=1}}}
{{{expr(1/y)(38/5)=1}}}
Multiply through by 5/38
{{{1/y=5/38}}}
Take reciprocals
{{{y=38/5}}}
{{{y=7.6}}}  <--- Answer: B can do 1 job in 7.6 days.

Edwin</pre>