If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4.

Question

If  3x - 2y = 32  and  \sqrt{3x} - \sqrt{2y} = 4 , what is the value of  x - y ?

(A) 4
(B) 8
(C) 10
(D) 12
(E) 14

gmatophobia · Accepted Answer

Given:

3x - 2y = 32   
  \sqrt{3x} - \sqrt{2y} = 4  ------ (i)

3x - 2y = 32  ⇒ Assume that the equation is the form of  a^2 - b^2

(\sqrt{3x} - \sqrt{2y})(\sqrt{3x} + \sqrt{2y}) = 32

4*(\sqrt{3x} + \sqrt{2y}) = 32

\sqrt{3x} + \sqrt{2y} = 8  ------ (ii)

Adding (i) and (ii)

2\sqrt{3x} = 12

\sqrt{3x} = 6

Squaring both sides

3x = 36

x = 12

We can replace the value of  x  in equation  3x - 2y = 32

36 - 32 = 2y

y = 2

x - y = 10

Option C

gmatophobia · Answer

AnkurGMAT20  The value is taken from the question stem. Check if this alternate representation of the solution helps -

MrWhite · Answer

My approach 3x - 2y = 32 1.5x - y = 16 Dividing by 2 (to get something similar / close to desired x-y) => (x-y) + \frac{x}{2} = 16 Condition 1 \sqrt{3x} - \sqrt{2y} = 4 Condition 2 Looking for values of (x-y) from answer choices which yield x & y such that they satisfy Conditions 1 & 2 Cycling through options for (x-y) . Only when (x-y) = 10 solutions of x & y meet both conditions, since \sqrt{3x} becomes perfect square! (C) I am not too sure whether this is the intended / optimal way to approach the problem, so am open to other solution methods

adewale223 · Answer

The problem can be solved pretty quickly by looking at square root 3x - square root 2y = 4

It becomes obvious that the two have to be perfect squares. If x is 3 and y is 2 answer will be 5 not acceptable.

But 36 and 4 are possible so x is 12 and y is 2. Plug in the first equation to see that it is also true. So 12-2 = 10. Option C answer.

AnkurGMAT20 · Answer

Bunuel   chetan2u   gmatphobia  I'm not able to understand how you got  4  in  4∗(3x−−√+2y−−√)=32

Can you please explain?

KarishmaB · Answer

Intuitive Method:

Since difference of x and y in an integer (from options) and
 \sqrt{3x} - \sqrt{2y} = 4  which is also an integer, most likely x is a multiple of 3 with another perfect square and y is a multiple of 2 with another perfect square. 
So y must be at least 2*1 = 2 and since

\sqrt{3x} = 4 + \sqrt{2y} ,

x cannot be just 3. Let's try 3*4 = 12 (the next perfect square after 1 is 4)

It works! All numbers over here and simple and small hence it wouldn't require many iterations in any case.

So x = 12 and y = 2. 
x - y = 10

Answer (C)

Regor60 · Answer

Set 3X = v^2, 2Y=w^2

Substituting:

V^2-W^2=(V+W)(V-W)=32

V-W=4

Since V-W=4, V+W = 8

Adding the two:

2V=12, so V=6 and W=2

So X=V^2/3=36/3 =

12

and Y=W^2/2=4/2 =

2

So X-Y = 10

Posted from my mobile device

bumpbot · Answer

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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4.

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If 3x - 2y = 32 and sqrt (3x) - sqrt (2y) = 4.

Prep Toolkit

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