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

Question

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

eswarchethu135 · Accepted Answer

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

Archit3110 · Answer

x+y= 120
x= 99 max and y = 21
so x/y ; 33/7
IMO B

jfranciscocuencag · Answer

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

DonBosco7 · Answer

A friendly tip to all those who picked D (i am one of those 😁), remember both x & y are  2 digit positive integers

So when we select option D = 9 for x/y

this gives y = 12 & x = 108 which is a 3 digit integer, hence wrong.

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