This is from the MGMAT Equations Question Bank. I know the answer is there but it is not the way I went about solving the equation. I haven't provided the answer choices but the equation is here. I am stuck at the last step and I don't think I can get a ratio of X to Y from that point. Can someone please shed some light? Thanks


IF X^1/2+Y^1/2)/(X-Y) = 2(X^1/2+2Y^1/2)/(X+2(XY)^1/2+Y), what is the ratio of X to Y?

(X^1/2+Y^1/2)/(X-Y) = 2(X^1/2+2Y^1/2)/(X^1/2+Y^1/2)^2

(X^1/2+Y^1/2)/(X-Y)= 2/(X^1/2+Y^1/2)

(X^1/2+Y^1/2)^2 = 2(X-Y)

X+2(XY)^1/2+Y= 2X-2Y

2(XY)^1/2 = X-3Y

(X^1/2+Y^1/2)/(X-Y) = 2(X^1/2+2Y^1/2)/(X^1/2+Y^1/2)^2

(X^1/2+Y^1/2)/(X-Y)= 2/(X^1/2+Y^1/2)

(X^1/2+Y^1/2)^2 = 2(X-Y)


Continuing from your solution:

(X^1/2+Y^1/2)^2 = 2(X-Y)
(X^1/2+Y^1/2)^2 = 2((X^1/2+Y^1/2) (X^1/2 - Y^1/2)) ((a+b)(a-b))

(X^1/2+Y^1/2) = 2(X^1/2-Y^1/2)
X^1/2+Y^1/2 = 2X^1/2 - 2 Y^1/2
X^1/2 = 3 Y^1/2

So you can find X/Y by squaring....
Correct?

IF X^1/2+Y^1/2)/(X-Y) = 2(X^1/2+2Y^1/2)/(X+2(XY)^1/2+Y), what is the ratio of X to Y?

is the orginal equation written correctly? more specifically the second half?2(X^1/2+2Y^1/2) are you sure that it is 2(X^1/2+2Y^1/2) and not
(2X^1/2+2Y^1/2)

you have to convert all exponents into sqrt.

After you simplify you should get the ratio of x :y as 1:5.

Is that the correct answer?

The ratio should be 9:1

X^1/2+Y^1/2/(X-Y) = 2(X^1/2+2Y^1/2)/(X+2(XY)^1/2+Y)

(X-Y can be split as (X^1/2+Y^1/2) (X^1/2-Y^1/2 ) AND X+2(XY)^1/2+Y) = (X^1/2+Y^1/2) ^ 2 i.e (a+b)^2 = a^2+b^2+2ab)


X^1/2+Y^1/2 / (X^1/2+Y^1/2) (X^1/2-Y^1/2 ) = 2(X^1/2+Y^1/2) / (X^1/2+Y^1/2) ^ 2

= 1/(X^1/2-Y^1/2 ) = 2 / (X^1/2+Y^1/2)

Therfore
(X^1/2+Y^1/2) = 2((X^1/2-Y^1/2 ) (Cross Multiplying)

(X^1/2+Y^1/2) = 2x^1/2-2y^1/2
i.e -x^1/2 = -3y^1/2

Squaring both sides

x=9y
x/y = 9

ok yeah, i'm going to write this one with actual rad signs. i don't know whether the other readers of this forum have the patience to swim through all the fractional exponents in forum notation, but i sure don't.

(√x + √y) / (x - y) = (2√x + 2√y) / (x + 2√xy + y)

you need to recognize the difference of squares. here it's more subtle than normal: the things being squared are square roots in the first place, so the squares go into hiding. the difference of squares is (√x - √y)(√x + √y) = x - y. tricky stuff.

you should also recognize the perfect square in the right-hand denominator: x + 2√xy + y = (√x + √y)^2. this is also more difficult to recognize, because you have to realize that x is (√x)^2 and that y is (√y)^2.

so, this says

(√x + √y) / (√x + √y)(√x - √y) = (2√x + 2√y) / (√x + √y)^2

cancel to give
1 / (√x - √y) = 2 / (√x + √y)

cross multiply to give
√x + √y = 2√x - 2√y

combine terms to give
3√y = √x

square to give
9y = x

voila

Hi All,

I took me 4 mins to solve this problem the first time. On second look, I realized that
if you replace X^1/2 with a , Y^1/2 with b it becomes very easy.
(a+b)/(a^2-b^2) = 2(a+2b)/(a^2+2ab+b^2)

Solve this to find a^2=9b^2

Hope this helps. The square root was just to confuse the student.

absolutely.

this brings up a great topic: don't just use formulas for literal plug-ins; look at them as RELATIONSHIPS between the different quantities involved.
for instance, the difference of squares formula states, literally, that a^2 - b^2 = (a + b)(a - b).
most students approach this formula by literally looking to substitute for a and b. in other words, they look for actual quantities that they can plug directly into 'a' and 'b'. this works all well and good in some circumstances, but, in others, in renders recognition of the pattern difficult.

instead, you should just think about the difference of squares as a RELATIONSHIP. in other words, here's the formula:
a^2 - b^2 = (a - b)(a + b)
the thing to do here is not to think overly literally about what quantities should be plugged in for 'a' and 'b'. instead, think about the relationship: note that the orange things on the right are the SQUARE ROOT of the orange thing on the left, and the blue things on the right are the SQUARE ROOT of the blue thing on the left.

if you recognize this pattern, it is much simpler to recognize clever applications such as the one suggested by the above poster.

good stuff.