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x+y=4
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08 Sep 2018, 22:35
08 Sep 2018, 22:35
Expert Reply
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Question Stats:
94% (00:20) correct
5% (01:06) wrong based on 19 sessions
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\(x+y=4\)
\(xy=0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
\(xy=0\)
Quantity A |
Quantity B |
\(x\) |
\(y\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x+y=4
[#permalink]
09 Sep 2018, 07:01
09 Sep 2018, 07:01
2
Carcass wrote:
\(x+y=4\)
\(xy=0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
\(xy=0\)
Quantity A |
Quantity B |
\(x\) |
\(y\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Kudos for R.A.E
GIVEN: x + y = 4 and xy = 0
If xy = 0, then EITHER x = 0 OR y = 0
Let's examine each possible case:
Case i) if x = 0, then y = 4 (since we're also told that x + y = 4)
Case ii) if y = 0, then x = 4 (since we're also told that x + y = 4)
In case i, Quantity A is greater than Quantity B
In case ii, Quantity B is greater than Quantity A
Answer: D
Cheers,
Brent
Re: x+y=4
[#permalink]
09 Sep 2018, 09:11
09 Sep 2018, 09:11
1
if X+Y = 4 & XY = 0
We can write as
= (X-Y)^2
= X^2+Y^2-2XY
= (X+Y)^2-4XY ( Bifurcating the equation)
= 4^2-4*0
= 16
THEREFORE
X-Y = 16^0.5
(Taking square root of 16)
X-Y = (+/-) 4
Now we have
X+Y = 4 .....1
X-Y = 4 .....2
On solving the equation 1 & 2 we get as follows:
X = 4
Y = 0
AND
X-Y = 4 .....2
X-Y = -4 ...4
On solving the equation 3 & 4 we get as follows:
X = 0
Y = 4
Since both X & Y can take both values 0 and 4 which is contradictory statements regarding the large value among X & Y.
Hence the answer is "D"
HENCE PROVED
We can write as
= (X-Y)^2
= X^2+Y^2-2XY
= (X+Y)^2-4XY ( Bifurcating the equation)
= 4^2-4*0
= 16
THEREFORE
X-Y = 16^0.5
(Taking square root of 16)
X-Y = (+/-) 4
Now we have
X+Y = 4 .....1
X-Y = 4 .....2
On solving the equation 1 & 2 we get as follows:
X = 4
Y = 0
AND
X-Y = 4 .....2
X-Y = -4 ...4
On solving the equation 3 & 4 we get as follows:
X = 0
Y = 4
Since both X & Y can take both values 0 and 4 which is contradictory statements regarding the large value among X & Y.
Hence the answer is "D"
HENCE PROVED
