Bunuel wrote:

Together Ari and Izy have less than P colored markers. Ari has P - 2 colored markers and Izy has P - 20 colored markers. If each girl has at least one colored marker, what is the value of P?

(A) 21

(B) 22

(C) 22

(D) 24

(E) 25

I looked at this problem three times. I could not understand the question. As inline text, I did not see P - 2 and P - 20 as (P - 2) and (P - 20) until

akadiyan posted. +1

Ari, A = (P - 2) markers

Izy, I = (P - 20) markers

Together, they have fewer than P markers.

Each girl has at least one marker.

\(A + I < P\)

\((P - 2) + (P - 20) < P\)

\(P - 2 + P - 20 < P\)

\(2P - 22 < P\)

\(P < 22\)Only one answer, A, is < 22

P = 21

You could stop here.

If not sure, check answers A and B.

A) 21. If P = 21, Ari = 19 and Izy = 1. Together = 20

20 < 21. That works.

B) 22. P cannot be 22. Ari = 20 and Izy = 2. Together they would have 22. Not allowed*

P must = 21

Answer A

*Greater than 22? No. The girls' total will exceed P by greater and greater numbers.

The absolute value of the girls' combined integer difference is 22: (-2 + (-20)) = (-2 - 20) = -22, |-22| = 22

After P = 22, the girls' total will exceed that absolute value of combined difference (22), by +2, +4, +6.

If P = 23: A = 21, I = 3, both = 24 (= 22 + 2). 24 > 23. Not allowed. If P = 24, together = 26 ( = 22 + 4).
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