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Joined: 20 Feb 2017
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If 2/5 p + q = 3q - 10, and if q = 10 - p, then p/q =
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20 Mar 2020, 23:42
20 Mar 2020, 23:42
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If \(\frac{2}{5} p + q = 3q - 10,\) and if \(q = 10 - p,\) then \(\frac{p}{q} =\)
(A) 5/7
(B) 3/2
(C) 5/3
(D) 25/6
(E) 6
(A) 5/7
(B) 3/2
(C) 5/3
(D) 25/6
(E) 6
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Re: If 2/5 p + q = 3q - 10, and if q = 10 - p, then p/q =
[#permalink]
12 May 2020, 23:56
12 May 2020, 23:56
1
GeminiHeat wrote:
If \(\frac{2}{5} p + q = 3q - 10,\) and if \(q = 10 - p,\) then \(\frac{p}{q} =\)
(A) 5/7
(B) 3/2
(C) 5/3
(D) 25/6
(E) 6
(A) 5/7
(B) 3/2
(C) 5/3
(D) 25/6
(E) 6
We have: 2p/5 + q = 3q - 10 => 2p/5 = 2q - 10 => 2p = 10q - 50 => 10q - 2p = 50
Also: q = 10 - p => p + q = 10
Multiplying 5: 5p + 5q = 50
Thus: 10q - 2p = 5p + 5q (since both are equal to 50)
=> 5q = 7p
=> p/q = 5/7
Alternative: We can always solve for p and q and find the ratio
Answer A
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If 2/5 p + q = 3q - 10, and if q = 10 - p, then p/q =
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03 Sep 2022, 01:47
03 Sep 2022, 01:47
Given that \(\frac{2}{5} p + q = 3q - 10\) and q = 10 - p and we need to find the value of \(\frac{p}{q} =\)
\(\frac{2}{5} p + q = 3q - 10\)
=> \(\frac{2}{5}\) p = 3q - 10 - q = 2q - 10 ...(1)
and q = 10 - p (given)
Putting value of q in (1) we get
\(\frac{2}{5}\) p = 2*(10 - p) - 10 = 20 - 2p - 10 = 10 - 2p
=> 2p = 5*(10-2p)
=> 2p = 50 - 10p
=> 12p = 50
=> p = \(\frac{50}{12}\) = \(\frac{25}{6}\)
q = 10 - p = 10 - \(\frac{25}{6}\) = \(\frac{60 - 25}{6}\) = \(\frac{35}{6}\)
\(\frac{p}{q}\) = (\(\frac{25}{6}\)) / (\(\frac{35}{6}\)) = \(\frac{25}{35}\) = \(\frac{5}{7}\)
So, Answer will be A
Hope it helps!
\(\frac{2}{5} p + q = 3q - 10\)
=> \(\frac{2}{5}\) p = 3q - 10 - q = 2q - 10 ...(1)
and q = 10 - p (given)
Putting value of q in (1) we get
\(\frac{2}{5}\) p = 2*(10 - p) - 10 = 20 - 2p - 10 = 10 - 2p
=> 2p = 5*(10-2p)
=> 2p = 50 - 10p
=> 12p = 50
=> p = \(\frac{50}{12}\) = \(\frac{25}{6}\)
q = 10 - p = 10 - \(\frac{25}{6}\) = \(\frac{60 - 25}{6}\) = \(\frac{35}{6}\)
\(\frac{p}{q}\) = (\(\frac{25}{6}\)) / (\(\frac{35}{6}\)) = \(\frac{25}{35}\) = \(\frac{5}{7}\)
So, Answer will be A
Hope it helps!
