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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Let x and y be two-digit positive integers with mean 60. What is the m

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Math Expert V
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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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

_________________
Senior Manager  P
Joined: 13 Jan 2018
Posts: 341
Location: India
Concentration: Operations, General Management
GMAT 1: 580 Q47 V23 GMAT 2: 640 Q49 V27 GPA: 4
WE: Consulting (Consulting)
Re: Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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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
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5447
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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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
Senior Manager  G
Joined: 12 Sep 2017
Posts: 308
Let x and y be two-digit positive integers with mean 60. What is the m  [#permalink]

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