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Maximum value

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by amitdgr » Mon Aug 04, 2008 5:51 pm
Please help me solve this q

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Q1) If (x + 2)^2 = 9 and (y + 3)^2 = 25, then the maximum value of x / y is

1) 1 / 2
2) 5 / 2
3) 5 / 8
4) 1 / 8
5) None

Guys, I am sorry but I do not know what the correct answer is :?

i tried this way , i divided the x expression with y expression and ended up with (x+2)/(y+3) = 3/5 .

Stuck. Is the approach right ? If yes then how do i proceed :( ?

Thanks
Amit
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Source: — Problem Solving |

Re: Maximum value

by Kirby » Mon Aug 04, 2008 7:07 pm
amitdgr wrote:Please help me solve this q

----------------------------------------------------------------------------
Q1) If (x + 2)^2 = 9 and (y + 3)^2 = 25, then the maximum value of x / y is

1) 1 / 2
2) 5 / 2
3) 5 / 8
4) 1 / 8
5) None

Guys, I am sorry but I do not know what the correct answer is :?

i tried this way , i divided the x expression with y expression and ended up with (x+2)/(y+3) = 3/5 .

Stuck. Is the approach right ? If yes then how do i proceed :( ?

Thanks
Amit

I think this is a simple "Plug-and-Chug"
First you have to solve for X.

(x+2)^2 = 9

We also know that (plus or minus 3)^2 = 9
So X must equal (1) or (-5)
Because (1+2)=3 and (-5+2)=-3

We know the values for X, now lets move on to Y

(y + 3)^2 = 25

Just like before, we know that (plus or minus 5)^2 = 25
So Y must equal (2) or (-8)
Because (2+3)=5 and (-8+3)=-5


As for the maximum value, you can try combinations of the numbers (x/y) to find the highest value.
In this case it is (-5)/(-8)
and the negatives cancel out to (5)/(8)
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5/8

by hakyology » Mon Aug 04, 2008 8:21 pm
yup 5/8. i got the same result.
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by amitdgr » Mon Aug 04, 2008 8:33 pm
Thanks guys .. :D :D
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