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sets of numbers used here (10,6,4) which satisfies the equation.

a 20+5 = 25 = w but 20-5 = 15. not sufficient.

b 11-2 = 9 satisfies. but 13-2 = 11 does not satisfy.

a+b (10,6) satisfies. But (20,5) does not.
Thus E.
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fluke, can u give us some tips about how to choose the right numbers to plug in inequalities?

thanks!
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fluke, can u give us some tips about how to choose the right numbers to plug in inequalities?

thanks!

I would rather solve inequality using algebra than plugging in numbers. However, if push comes to shove reduce the inequality in its simplest form by bringing all its terms to the LHS and then try six ranges for the variables.

0 to 1
1 to 2
more than 2
0 to -1
-1 to -2
less than -2

PIN is really a bad option for inequalities with more than one variables and no limiting conditions, such as integers or positive numbers. Try learning gurpreetsingh/karishma's and Bunuel's methods to deal with inequality problems. If you spent approximately 3 minutes without any conclusive results on a problem, proceed with a guess. Let us know if you face problem on specific inequality problems and we'll see how to approach those.

Good luck!!
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I picked below numbers
3,4,5
and
5,12,13

Cause of Pythagoras therefrom.
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Bunuel

Tough and Tricky questions: properties of numbers.



If w, y, and z are positive integers, and w = y – z, is w a perfect square?

(1) y + z is a perfect square.
(2) z is even.

We need to find whether w is a perfect square

St 1: y+z is perfect square
Consider y=3,z=6 , y+z=9=w=perfect square, but y-z is not a perfect square
Considery=18,z=18,y+z=36=w=perfect square,but y-z is a perfect sqaure

St 1 is insufficient. Option A and D ruled out

st 2: z is even, We can consider the above 2 cases and get 2 answers..Option B ruled out

Combining again we get same information...C ruled out

E is the answer
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Bunuel, What is your solution?
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honchos
Bunuel, What is your solution?

I'd test values for this one.

First of all, forget about w, it only confuses us. So, the question asks whether the difference of two integers, y and z, is a perfect square. When combining we know that the sum of those integers IS a prefect square and lesser of the two integers is even.

5 + 4 = 3^2 IS a perfect square and 5 - 4 = 1^2 is also a prefect square. So, we have an YES answer.
7 + 2 = 3^2 IS a perfect square but 7 - 2 = 5 is NOT. So, we have a NO answer.

Answer: E.
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take case of 16

Case 1 :
y=10, z=6
10+6 =16 perfect square
also 10-6 =4 perfect square

Case 2 :
y=12, z=4,
12+4 = 16 perfect square
but 12-4 =12 not a perfect square

Hence statement 1 is not enough to answer
Also Statement 2 z is even does not help as in case 2 z is even but it does not matter.

Hence E
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ravi67741
If w, y, and z are positive integers, and w = y – z, is w a perfect square?
(1) y + z is a perfect square.
(2) z is even.


Please find solution as attached
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For this one I would use straight theory....

For w being a perfect square we need to know somehow that:

1) W has odd number of TOTAL factors OR 2) Even powers for its prime factors...

Both statements they dont give any clue for 1 or 2 so E
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E
1) Insufficient - diff cases work
2) Insufficient - no info
1 + 2 = doesn't work so E

Ndkms the theoretical approach you have taken is a good alternative approach
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In case where w = y - z, to confirm if w is a perfect square: you may use the following method

1. y + z is a perfect square:

These are the examples I chose:
(a) y = 10 and z = 6 : Therefore, y+z = 16 which is a perfect square and y-z = 4 which is also a perfect square (correct)
(b) y = 12 and z = 24 : Therefore, y+z = 36 which is a perfect square but y-z = 12 which is not (incorrect)
Hence Insufficient

2. z is even: Does not provide information about y and hence cannot determine w

Combine:
Use the same examples from above:
for (a) y = 10 and z = 6 (even) - correct
for (b) y = 12 and z = 24 (even) - incorrect

Hence E




ravi67741
If w, y, and z are positive integers, and w = y – z, is w a perfect square?

(1) y + z is a perfect square.
(2) z is even.
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