QUANTITY A
Take \(x=12y+25\)
Replace \(y\) with \(7z+5\) to get: \(x=12(7z+5)+25\)
Expand and simplify: \(x=84z+85\)
Rewrite as: \(x=84z+84 + 1\)
Factor the first part to get: \(x=42(2z+2)+1\)
This tells us that x is 1 greater than some multiple of 42
So, when we divide x by 42, the remainder will be 1
QUANTITY A: 1


QUANTITY B
Case i: If \(z = 1\), then \(y=7(1)+5 = 12\). When \(12\) is divided by \(2\), the remainder is 0
So, in this case, Quantity A is greater than Quantity B

Case ii: If \(z = 2\), then \(y=7(2)+5 = 19\). When \(19\) is divided by \(2\), the remainder is 1
In this case, Quantity A is equal to Quantity B

Answer: D

Cheers,
Brent

\(x = 84z + 85\)

Dividing this by 42 will always leave behind a remainder of 1 because of the 85 at the end.

However, Quantity B is
\(= 7z + 5
= (odd)(odd/even) + (odd)\)

When z is odd:
\(= (odd) + (odd) = (even)\)
Results in remainder 0 when divided by 2

When z is even
\(= (odd)(even) + (odd)
= (even) + (odd) = (odd)\)
Results in remainder 1 when divided by 2

Hence, D

Carcass, in general what do you suggest the amount of time I should ideally be taking for questions like these?

1 or 1.30 at the very most .