A and B together can complete one-ninth of a job in a day.

A pure algebraic way can be this:

\(rate*time=work\)

\((A+B)*1=\frac{1}{9}\) and \(A*t=\)1 and \(B*k=1\) where t and k are the days A, B would take to finish the job and are integers.

From those we get: \(\frac{1}{t}+\frac{1}{k}=\frac{1}{9}\) or \(t+k=\frac{tk}{9}\). Lets study t (the time taken by A alone) \(t(1-\frac{k}{9})=-k\) or \(t=\frac{-9k}{9-k}\). How can we read this? Since k is a number the represents a number of days, must be positive=> -9k will be negative so because t must be positive, the denominator must be negative.
\(+ve=\frac{-ve}{-ve}\).
\(9-k<0\) so \(k>9\), but since k is an integer => THE LEAST number of days B can take to finish the job is 10.

So if this is THE LEAST for B, it must be the MOST for A => \(t=\frac{-9*10}{9-10}=90\).

You can repeat the whole analysis just by switching k and t and you find \(90-10=80\).

Hope I've explained myself well (and hope you enjoy algebra )

Thanks Zarrolou

Regards
Vinni

solution:

1/A + 1/B = 1/9
when B max then A min. when B min A max. For minimum positive value(real), put B = 10 and consequently A=90. and vice versa. so difference= 90 - 10 = 80 (Answer)

1/A + 1/B = 1/9

so A = 9B/(B-9)

A will be max when denominator is minimum (1) so B = 10 and A = 90
For A to be minimum these values will just be reverse so A = 10
Difference 80

Given, A & B together can complete 1/9th of the job in one day.

Let the job consist of 90 units. Therefore in one day A & B together can produce 10 units.

A & B together work at a rate of 10 units/day

So the combinations for the no. of units per day produced by A & B can be (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)

So Maximum units produced per day by A = 9 units/ day at its highest efficiency.

Hence to complete 90 units, A will take (90/9) = 10 days, this is the fastest for A at its highest efficiency.

So Minimum Days to finish the job = 10 days

Now, Similarly, minimum units produced per day by A = 1 unit/ day at its lowest efficiency.

Hence to complete 90 units, A will take (90/1) = 90 days, this is the slowest for a its lowest efficiency.

So Maximum Days to finish the job = 90 days

Required is Difference of Max Days & Min Days = 90 - 10 = 80

Answer C

We a = the number of days A needs to take to complete the job alone and b = the number of days B needs to take to complete the job alone. So A and B’s rates are 1/a and 1/b, respectively and we have:

1/a + 1/b = 1/9

1/a = 1/9 - 1/b

1/a = (b - 9)/(9b)

a = 9b/(b - 9)

Since a and b are positive integers, we see that the smallest positive integer for b is 10. In that case, a = 9(10)/(10 - 9) = 90 and this must be the maximum value of a since b is minimum.

Notice that in the above we’ve solved a in terms of b. If we solve b in terms of a, we should have:
b = 9a/(a - 9)

In this case we see that the minimum value of a is 10. Therefore, the difference between the maximum and minimum number of days A takes to complete the job alone is 90 - 10 = 80.

Answer: C

I did in a different way.
R(A+B) = 1/(9 )=10/90=(1+9)/90=1/90+9/90=1/90+1/10
so, their rates can be divided as 1/90 & 1/10
i.e. Time TA & TB can be divided as 90 & 10
Thus, Time(Max – MIN) = 90-10 = 80 days.
Bunuel, is this approach OK?

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