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

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Math Expert
Joined: 02 Sep 2009
Posts: 59590
Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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31 Mar 2019, 22:44
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Difficulty:

55% (hard)

Question Stats:

60% (01:53) correct 40% (02:08) wrong based on 86 sessions

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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

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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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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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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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Joined: 12 Sep 2017
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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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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
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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