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p is 6 times as large as q. The % that q is less than p , is:

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p is 6 times as large as q. The % that q is less than p , is:  [#permalink]

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New post 06 Feb 2019, 15:25
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Difficulty:

  25% (medium)

Question Stats:

78% (01:36) correct 22% (00:44) wrong based on 18 sessions

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p is 6 times as large as q. The % that q is less than p , is:

A)83\(\frac{1}{3}\)
B)16\(\frac{2}{3}\)
C)90
D)60
E)0.02
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Re: p is 6 times as large as q. The % that q is less than p , is:  [#permalink]

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New post 06 Feb 2019, 22:40
selim wrote:
p is 6 times as large as q. The % that q is less than p , is:

A)83\(\frac{1}{3}\)
B)16\(\frac{2}{3}\)
C)90
D)60
E)0.02


Key word : p is 6 times as large as q

q = 10 p =60

% that q is less than p = \(\frac{{60 - 10}}{60}\) * 100

= 83\(\frac{1}{3}\)
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Re: p is 6 times as large as q. The % that q is less than p , is:  [#permalink]

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New post 14 Feb 2019, 09:24
Can you please elaborate it. I am not getting it.
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Re: p is 6 times as large as q. The % that q is less than p , is:  [#permalink]

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New post 14 Feb 2019, 09:48
selim wrote:
p is 6 times as large as q. The % that q is less than p , is:

A)83\(\frac{1}{3}\)
B)16\(\frac{2}{3}\)
C)90
D)60
E)0.02


Take q=1,p=6
q is 83.33% less
So A
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Re: p is 6 times as large as q. The % that q is less than p , is:  [#permalink]

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New post 14 Feb 2019, 09:50
selim wrote:
p is 6 times as large as q. The % that q is less than p , is:

A)83\(\frac{1}{3}\)
B)16\(\frac{2}{3}\)
C)90
D)60
E)0.02
vinodsai wrote:
Can you please elaborate it. I am not getting it.


\(p = 6\) & \(q = 1\)

So, \(q\) is \(5\) less than \(p\)

The % that q is less than p , is: \(\frac{5}{6}*100\) = \(83\frac{1}{3}\), Answer must be (A)
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Re: p is 6 times as large as q. The % that q is less than p , is:  [#permalink]

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New post 14 Feb 2019, 09:56
selim wrote:
p is 6 times as large as q. The % that q is less than p , is:

A)83\(\frac{1}{3}\)
B)16\(\frac{2}{3}\)
C)90
D)60
E)0.02


p = 6q => q = p/6

->(p-q)/p *100
->5p/6p *100
-> 83.333
so option A


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Re: p is 6 times as large as q. The % that q is less than p , is:   [#permalink] 14 Feb 2019, 09:56
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