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If p + q = 12 and pq = 35, then 1 p + 1 q =
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02 Apr 2019, 04:26
02 Apr 2019, 04:26
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Question Stats:
90% (00:32) correct
9% (01:37) wrong based on 11 sessions
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If \(p + q = 12\) and \(pq = 35\), then \(\frac{1}{p} + \frac{1}{q} =\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{35}\)
(D) \(\frac{12}{35}\)
(E) \(\frac{23}{35}\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{35}\)
(D) \(\frac{12}{35}\)
(E) \(\frac{23}{35}\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If p + q = 12 and pq = 35, then 1 p + 1 q =
[#permalink]
02 Apr 2019, 05:04
02 Apr 2019, 05:04
Carcass wrote:
If \(p + q = 12\) and \(pq = 35\), then \(\frac{1}{p} + \frac{1}{q} =\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{35}\)
(D) \(\frac{12}{35}\)
(E) \(\frac{23}{35}\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{35}\)
(D) \(\frac{12}{35}\)
(E) \(\frac{23}{35}\)
One approach is to look for two values that have a SUM of 12 and a PRODUCT of 35
We get 5 and 7
So, EITHER p = 5 and q = 7 OR p = 7 and q = 5
Each case will yield the same value for \(\frac{1}{p} + \frac{1}{q}\)
\(\frac{1}{p} + \frac{1}{q} = \frac{1}{5} + \frac{1}{7}\)
\(= \frac{7}{35} + \frac{5}{35}\)
\(= \frac{12}{35}\)
Answer: D
Cheers,
Brent
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If p + q = 12 and pq = 35, then 1 p + 1 q =
[#permalink]
02 Apr 2019, 05:06
02 Apr 2019, 05:06
Carcass wrote:
If \(p + q = 12\) and \(pq = 35\), then \(\frac{1}{p} + \frac{1}{q} =\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{35}\)
(D) \(\frac{12}{35}\)
(E) \(\frac{23}{35}\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{35}\)
(D) \(\frac{12}{35}\)
(E) \(\frac{23}{35}\)
Approach #2
\(\frac{1}{p} + \frac{1}{q} = \frac{q}{pq} + \frac{p}{pq}\)
\(= \frac{q+p}{pq}\)
Now replace q+p and pq with the given values to get:
\(= \frac{12}{35}\)
Answer: D
Cheers,
Brent
