(x + 2) ^2 = 9 and (y + 3) ^2 = 25

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(x + 2) ^2 = 9 and (y + 3) ^2 = 25

by sanju09 » Fri Mar 19, 2010 5:58 am
If (x + 2) ^2 = 9 and (y + 3) ^2 = 25, then what is the maximum value of x/y?
(A) 1/8
(B) 1/2
(C) 5/8
(D) 5/2
(E) 10/3
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by harshavardhanc » Fri Mar 19, 2010 6:06 am
sanju09 wrote:If (x + 2) ^2 = 9 and (y + 3) ^2 = 25, then what is the maximum value of x/y?
(A) 1/8
(B) 1/2
(C) 5/8
(D) 5/2
(E) 10/3
x = 1 or -5 and y = 2 or -8

Max of x/y will be when x = -5 and y = -8. Hence, max value will be 5/8. IMO C.
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by kstv » Fri Mar 19, 2010 6:11 am
for max value x and y have to be the same sign
(x+2) = +- 3 and y+3 = +- 5
if they are +ve x = 1 and y = 2 so x/y = 1/2
if they are -ve x = -5 and y = -8 so x/y = 5/8
IMO C

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by Scott@TargetTestPrep » Wed Dec 13, 2017 3:43 pm
sanju09 wrote:If (x + 2) ^2 = 9 and (y + 3) ^2 = 25, then what is the maximum value of x/y?
(A) 1/8
(B) 1/2
(C) 5/8
(D) 5/2
(E) 10/3
Let's determine values for x and y.

(x + 2)^2 = 9

√(x + 2)^2 = √9

|x + 2| = 3

We need to determine the value of x when (x + 2) is positive and when (x + 2) is negative.

x + 2 = 3

x = 1

OR

-(x + 2) = 3

-x - 2 = 3

-x = 5

x = -5

So x = 1 or x = -5.

In a similar fashion, we determine the value of y.

(y + 3)^2 = 25

√(y + 3)^2 =√25

|y + 3| = 5

We need to determine the value of y when (y + 3) is positive and when (y + 3) is negative.

y + 3 = 5

y = 2

OR

-(y + 3) = 5

-y - 3 = 5

-y = 8

y = -8

y = 2 or y = -8

We can maximize the value of x/y if x and y are both positive or if x and y are both negative. If x and y are both positive, then x/y = 1/2. If x and y are both negative, then x/y = -5/-8 = 5/8. Since 5/8 > 1/2, the maximum value of x/y is 5/8.

Answer: C

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