Step 1: Define individual working times and combined rate


Let the times taken to complete the job alone be $A, B, C$ for Andy, Ben, and Carl, respectively:


Carl's time: C=T
Ben's time: B=T+z (since Ben takes z hours more than Carl) PNG
Andy's time: A=B+y=T+y+z (since Andy takes y hours more than Ben) PNG


Since y, z>0 , the individual times are all distinct and strictly ordered: A>B>C .
The combined working time x satisfies the work equation:



\(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{x}\)



Step 2: Express Quantities A and B



Quantity A: 9 x

Quantity B: A+B+C


\(\left(\frac{A+B+C}{3}\right) \geq \frac{3}{\frac{1}{A}+\frac{1}{B}+\frac{1}{C}}\)


Substitute \(\frac{1}{x}=\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\) :

\( \frac{A+B+C}{3} \geq \frac{3}{\frac{1}{x}}=3 x\)

Multiply both sides by 3:


\(A+B+C \geq 9 x\)



Step 3: Account for equality conditions
Equality in the AM-HM inequality holds if and only if A=B=C .
However, the problem specifies that y>0 and z>0 , meaning Andy, Ben, and Carl take strictly different amounts of time to complete the work (A>B>C) .



Because A, B, C cannot be equal, the inequality is strict:


\(A+B+C>9 x\)


Therefore:
Quantity B > Quantity A
Correct Answer: Quantity B is greater.

understood. How did the inequality thing in step 2 click to you tho as a starting point?

Let:
Carl's time alone $=c$
Ben's time alone $=c+z$
Andy's time alone $=c+z+y$


Their combined rate:


\(\frac{1}{c}+\frac{1}{c+z}+\frac{1}{c+z+y}=\frac{1}{x}\)


Quantity B is the sum of their individual times:


\(c+(c+z)+(c+z+y)=3 c+2 z+y\)


We need to compare 9 x with 3 c+2 z+y .

From the rate equation, x is less than the smallest individual time c . So:


\(x<c\)


The relationship for three positive numbers:


\(\frac{3}{\frac{1}{c}+\frac{1}{c+z}+\frac{1}{c+z+y}} < \frac{c+(c+z)+(c+z+y)}{3} \)

\(3 x < \frac{3 c+2 z+y}{3} \)

\(9 x < 3 c+2 z+y\)


Thus Quantity B is greater.

B) Quantity B is greater.