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In ∆PQR, PQ = x, QR = x + 3, and PR = y. If x = y + 3, then
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28 Mar 2019, 13:14
28 Mar 2019, 13:14
Expert Reply
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Question Stats:
88% (00:53) correct
11% (00:00) wrong based on 9 sessions
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In \(∆PQR\), \(PQ = x\), \(QR = x + 3\), and \(PR = y\). If \(x = y + 3\), then which one of the following is true?
(A) \(∠P < ∠Q < ∠R\)
(B) \(∠Q < ∠R < ∠P\)
(C) \(∠R < ∠P < ∠Q\)
(D) \(∠P < ∠R < ∠Q\)
(E) \(∠Q < ∠P < ∠R\)
(A) \(∠P < ∠Q < ∠R\)
(B) \(∠Q < ∠R < ∠P\)
(C) \(∠R < ∠P < ∠Q\)
(D) \(∠P < ∠R < ∠Q\)
(E) \(∠Q < ∠P < ∠R\)
Re: In ∆PQR, PQ = x, QR = x + 3, and PR = y. If x = y + 3, then
[#permalink]
30 Mar 2019, 11:30
30 Mar 2019, 11:30
Expert Reply
If x = y + 3, then x is greater than y. So x + 3 is the longest side, followed by x and then y.
Opposite side QR is angle P. Since QR is the longest side, P is the largest angle.
PQ is the next largest side, and it is opposite angle R. So angle R is the second largest angle.
This leaves angle Q as the smallest. The proper order, then, should be ∠Q < ∠R < ∠P, as in choice B.
Opposite side QR is angle P. Since QR is the longest side, P is the largest angle.
PQ is the next largest side, and it is opposite angle R. So angle R is the second largest angle.
This leaves angle Q as the smallest. The proper order, then, should be ∠Q < ∠R < ∠P, as in choice B.
gmatclubot
Re: In ∆PQR, PQ = x, QR = x + 3, and PR = y. If x = y + 3, then [#permalink]
30 Mar 2019, 11:30
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