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Let x and y be two-digit positive integers with mean 60. What is the m

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Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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New post 31 Mar 2019, 22:44
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

60% (01:52) correct 40% (02:08) wrong based on 85 sessions

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Re: Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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New post 31 Mar 2019, 22:50
1
x + y = 120

In order for \(\frac{x}{y}\) to be maximum, x has to be the maximum possible two digit number and y has to be minimum possible two digit number.

If we take x as 99, which is the maximum possible two digit number, then y has to be 21

\(\frac{x}{y} = \frac{99}{21}\)

= \(\frac{33}{7}\)

OPTION: B
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Re: Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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New post 01 Apr 2019, 10:11
Bunuel wrote:
Let x and y be two-digit positive integers with average (arithmetic mean) 60. What is the maximum value of the ratio x/y?

A. 3
B. 33/7
C. 39/7
D. 9
E. 99/10


x+y= 120
x= 99 max and y = 21
so x/y ; 33/7
IMO B
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Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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New post 18 Apr 2019, 14:43
Bunuel wrote:
Let x and y be two-digit positive integers with average (arithmetic mean) 60. What is the maximum value of the ratio x/y?

A. 3
B. 33/7
C. 39/7
D. 9
E. 99/10


Another way to do it is just to prove with the given solutions:

a) \(90/30 =3\)

b) \(\frac{33}{7} =\frac{66}{14} = \frac{99}{21}\) ..- Hold (it is more than 3 so you can eliminate A)

c) \(\frac{39}{7} =\frac{78}{14} = 78 + 14 = 92\) ... Discard

d) \(\frac{120}{9}\) Not possible ... Discard

d) \(\frac{99}{10} = 99 + 10 = 109\) ... Discard

B
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Let x and y be two-digit positive integers with mean 60. What is the m   [#permalink] 18 Apr 2019, 14:43
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