Is X^4 + Y^4 > Z^4?
(1) X^2 + Y^2 > Z^2
(2) X + Y > Z
The answer is E. What is a quick and efficient way to prove this?
Thanks,
Alfredo
Data Sufficiency stumper
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Hi miguel,
This question has been discussed before in this forum a couple of weeks ago.
I think, as others here, that the only way to solve it is by trial and error (the same as picking numbers until you get something). If you pick the right numbers fast you get the answer within the 2min, if not you can stay the all test testing numbers.
Try with x=1; y=1; z=1 and x=3; y=3; z=4
(1) 1^2+1^2>1^2 -> true and 1^4+1^4>1^4 -> true
3^2+3^2>4^2 -> true and 3^4+3^4>4^4 -> false, so INSUFF
(2) 1+1>1 -> yes and 1^4+1^4>1^4 yes
3+3>4 -> yes and 3^4+3^4>4^4 -> false, so INSUFF
Perhaps our experts, Ian or Stuart, have something to say about it.
BTW: your name is portuguese. Are you portuguese?
This question has been discussed before in this forum a couple of weeks ago.
I think, as others here, that the only way to solve it is by trial and error (the same as picking numbers until you get something). If you pick the right numbers fast you get the answer within the 2min, if not you can stay the all test testing numbers.
Try with x=1; y=1; z=1 and x=3; y=3; z=4
(1) 1^2+1^2>1^2 -> true and 1^4+1^4>1^4 -> true
3^2+3^2>4^2 -> true and 3^4+3^4>4^4 -> false, so INSUFF
(2) 1+1>1 -> yes and 1^4+1^4>1^4 yes
3+3>4 -> yes and 3^4+3^4>4^4 -> false, so INSUFF
Perhaps our experts, Ian or Stuart, have something to say about it.
BTW: your name is portuguese. Are you portuguese?
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Well, the first thing I notice is that we have even exponents (and therefore non-negative numbers) in the question and in Statement 1, but not in Statement 2. So it should be easy to show that 2) is not enough- just make z a small negative number. For example, if x = y = 0 and z = -1000, then x+y > z, but x^4 + y^4 is much, much smaller than z^4.
Statement 1 is a lot more interesting. I'll suggest three ways to look at the problem:
You could look at it algebraically: note that neither side of the inequality can be negative. That means it is legitimate to square both sides:
(x^2 + y^2)^2 > (z^2)^2
x^4 + y^4 + 2(x^2)(y^2) > z^4
We do not find that x^4 + y^4 > z^4, and in fact, what we get above is the best we can do. If 2(x^2)(y^2) is large enough, relative to the other terms, x^4 + y^4 may be smaller than z^4.
Or we can choose numbers here if we like. I'd prefer to avoid large numbers, so I would not let x, y and z be integers here (don't want to worry about calculating z^4); instead I'd let x^2, y^2 and z^2 be integers. If we let, for example,
x^2 = y^2 = 2
and
z^2 = 3
then
x^2 + y^2 = 4 > z^2 = 3
but
x^4 + y^4 = 4 + 4 = 8 < z^4 = 9.
But the last solution I'll offer is the one I find most interesting. I guarantee you've seen this problem dozens of times before, perhaps without realizing it. Let's rename the unknowns (you'll see why I'm doing this in a second):
a = x^2
b = y^2
c = z^2
then
a^2 = x^4
b^2 = y^4
c^2 = z^4
Now statement 1) says that a+b > c. This is true whenever a, b and c are the sides of a triangle- the sum of two sides of a triangle is always greater than the third side. The question asks whether it must be true that a^2 + b^2 > c^2. If this were true, then in every triangle you've ever seen, the sum of the squares of the sides would be larger than the square of the other side. But we know that in some triangles, a^2 + b^2 is not greater than c^2; in right angled triangles, for example, if c is the hypotenuse, we know a^2 + b^2 = c^2- that's the most famous theorem in geometry. So from Pythagoras, Statement 1 is not enough.
Statement 1 is a lot more interesting. I'll suggest three ways to look at the problem:
You could look at it algebraically: note that neither side of the inequality can be negative. That means it is legitimate to square both sides:
(x^2 + y^2)^2 > (z^2)^2
x^4 + y^4 + 2(x^2)(y^2) > z^4
We do not find that x^4 + y^4 > z^4, and in fact, what we get above is the best we can do. If 2(x^2)(y^2) is large enough, relative to the other terms, x^4 + y^4 may be smaller than z^4.
Or we can choose numbers here if we like. I'd prefer to avoid large numbers, so I would not let x, y and z be integers here (don't want to worry about calculating z^4); instead I'd let x^2, y^2 and z^2 be integers. If we let, for example,
x^2 = y^2 = 2
and
z^2 = 3
then
x^2 + y^2 = 4 > z^2 = 3
but
x^4 + y^4 = 4 + 4 = 8 < z^4 = 9.
But the last solution I'll offer is the one I find most interesting. I guarantee you've seen this problem dozens of times before, perhaps without realizing it. Let's rename the unknowns (you'll see why I'm doing this in a second):
a = x^2
b = y^2
c = z^2
then
a^2 = x^4
b^2 = y^4
c^2 = z^4
Now statement 1) says that a+b > c. This is true whenever a, b and c are the sides of a triangle- the sum of two sides of a triangle is always greater than the third side. The question asks whether it must be true that a^2 + b^2 > c^2. If this were true, then in every triangle you've ever seen, the sum of the squares of the sides would be larger than the square of the other side. But we know that in some triangles, a^2 + b^2 is not greater than c^2; in right angled triangles, for example, if c is the hypotenuse, we know a^2 + b^2 = c^2- that's the most famous theorem in geometry. So from Pythagoras, Statement 1 is not enough.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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Ian, as always a great explanation. I am wondering if there is any rule when to consider + integers, -ve intergers and maybe, fractions as well when picking numbers to solve arithmetic DS questions involving comparisions. Also, are there any pointers or some kind of rules you would suggest for certain kind of comparisions, so we don't need to pick numbers to solve them.
for eg;
y-1 is always < y; fo all values of y
x^2 < x whne x is between 0 and 1,
similarly, x+1 > x; for all values of x
also, if x >y then 1/x < 1/y , provided that x and y are of the same sign for all values of x and y.
I'm wondering if there are more of these hard core rules that can save us time when we come across such comparisions. I would appreciate your response.
for eg;
y-1 is always < y; fo all values of y
x^2 < x whne x is between 0 and 1,
similarly, x+1 > x; for all values of x
also, if x >y then 1/x < 1/y , provided that x and y are of the same sign for all values of x and y.
I'm wondering if there are more of these hard core rules that can save us time when we come across such comparisions. I would appreciate your response.
Ian Stewart wrote:Well, the first thing I notice is that we have even exponents (and therefore non-negative numbers) in the question and in Statement 1, but not in Statement 2. So it should be easy to show that 2) is not enough- just make z a small negative number. For example, if x = y = 0 and z = -1000, then x+y > z, but x^4 + y^4 is much, much smaller than z^4.
Statement 1 is a lot more interesting. I'll suggest three ways to look at the problem:
You could look at it algebraically: note that neither side of the inequality can be negative. That means it is legitimate to square both sides:
(x^2 + y^2)^2 > (z^2)^2
x^4 + y^4 + 2(x^2)(y^2) > z^4
We do not find that x^4 + y^4 > z^4, and in fact, what we get above is the best we can do. If 2(x^2)(y^2) is large enough, relative to the other terms, x^4 + y^4 may be smaller than z^4.
Or we can choose numbers here if we like. I'd prefer to avoid large numbers, so I would not let x, y and z be integers here (don't want to worry about calculating z^4); instead I'd let x^2, y^2 and z^2 be integers. If we let, for example,
x^2 = y^2 = 2
and
z^2 = 3
then
x^2 + y^2 = 4 > z^2 = 3
but
x^4 + y^4 = 4 + 4 = 8 < z^4 = 9.
But the last solution I'll offer is the one I find most interesting. I guarantee you've seen this problem dozens of times before, perhaps without realizing it. Let's rename the unknowns (you'll see why I'm doing this in a second):
a = x^2
b = y^2
c = z^2
then
a^2 = x^4
b^2 = y^4
c^2 = z^4
Now statement 1) says that a+b > c. This is true whenever a, b and c are the sides of a triangle- the sum of two sides of a triangle is always greater than the third side. The question asks whether it must be true that a^2 + b^2 > c^2. If this were true, then in every triangle you've ever seen, the sum of the squares of the sides would be larger than the square of the other side. But we know that in some triangles, a^2 + b^2 is not greater than c^2; in right angled triangles, for example, if c is the hypotenuse, we know a^2 + b^2 = c^2- that's the most famous theorem in geometry. So from Pythagoras, Statement 1 is not enough.
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What numbers we need to pick, if we do pick numbers, really depends on the question being asked. In some questions, we might know that x is not negative- perhaps it measures a distance, or the number of oranges in a box. But, if x is allowed to be negative, always consider that possibility- otherwise you'll get into a lot of trouble on the GMAT!ildude02 wrote:Ian, as always a great explanation. I am wondering if there is any rule when to consider + integers, -ve intergers and maybe, fractions as well when picking numbers to solve arithmetic DS questions involving comparisions. Also, are there any pointers or some kind of rules you would suggest for certain kind of comparisions, so we don't need to pick numbers to solve them.
for eg;
y-1 is always < y; fo all values of y
x^2 < x whne x is between 0 and 1,
similarly, x+1 > x; for all values of x
also, if x >y then 1/x < 1/y , provided that x and y are of the same sign for all values of x and y.
I'm wondering if there are more of these hard core rules that can save us time when we come across such comparisions. I would appreciate your response.
We do often see statements that give inequalities involving powers of x- an inequality like "x^3 > x". If you are going to test numbers here, then you should always consider four different possible values for x:
x > 1
0 < x < 1
-1 < x < 0
x < -1
So you could try x = 2, x = 1/2, x = -1/2 and x = -2, for example. You'll see, with the example I've given, that if x^3 > x is true, then x might be larger than 1, but it might also be true that x is between -1 and 0. As a useful exercise, you might try looking at the inequality "x^3 > 1/x" in the same way- what does this tell you about x?
That type of statement is common enough that it's worthwhile learning how to decode it. There are, however, many other kinds of statements we can see on the GMAT, so it's impossible to give simple universal rules for what values need to be checked. In problems involving ratios, I'll normally consider extreme values -- what if x is very large? What if x is very small? In number theory questions (involving divisors or evens/odds), the guidelines would be completely different, and I'm not sure what to suggest there since I'd never choose numbers for a question involving evens and odds, or divisibility.
I'm sure you were hoping for a shorter answer, but it's not possible to give one, I'm afraid. I would strongly suggest, as you prepare, keeping notes of those DS questions you answer incorrectly because you didn't consider enough possible values for x. You may notice a pattern- perhaps you aren't considering negative values. for example- and you may then be able to learn from that.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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