Mixing
How many cases on a GRE question
$begingroup$
This is more or less a question about how many cases I should be trying in general on the quantitative section of the GRE.
I got this question wrong and it is because I did not try enough cases, so I was wondering if anyone has any tips on how to see that a given condition may be true without trying 5 different cases.
Question:
If x, y, and z are positive numbers such that $3x < 2y < 4z$, which of the following statements could be true? Indicate all such statements.
a. $x=y$
b. $y=z$
c. $y>z$
d. $x>z$
My work:
For a)
$x=2, y=2, z=3$
$3x=6, 2y=4, 4z=12$
$6<4<12$ is false.
a) cannot be true.
For b)
$x=1, y=4, z=4$
$3x=3, 2y=8, 4z=16$
$3<8<16$ is true.
b) can be true.
For c)
$x=1, y=4, z=3$
$3x=3, 2y=8, 4z=12$
$3<8<12$ is true.
c) can be true.
Now d) was the reason for missing this question. I plugged in one case as I did with a) yet there is in fact a threshold such that this works.
For example:
$x=7, y=11, z=6$
$3x=21, 2y=22, 4z=24$
$21<22<24$ is true.
d) can be true.
However, during the practice test (this was the last question I was crunched for time and had to the second to last because this one looked easier) I only plugged in one case and it didn't work so I quit.
Tips?
gre-exam
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
$endgroup$
add a comment |
$begingroup$
This is more or less a question about how many cases I should be trying in general on the quantitative section of the GRE.
I got this question wrong and it is because I did not try enough cases, so I was wondering if anyone has any tips on how to see that a given condition may be true without trying 5 different cases.
Question:
If x, y, and z are positive numbers such that $3x < 2y < 4z$, which of the following statements could be true? Indicate all such statements.
a. $x=y$
b. $y=z$
c. $y>z$
d. $x>z$
My work:
For a)
$x=2, y=2, z=3$
$3x=6, 2y=4, 4z=12$
$6<4<12$ is false.
a) cannot be true.
For b)
$x=1, y=4, z=4$
$3x=3, 2y=8, 4z=16$
$3<8<16$ is true.
b) can be true.
For c)
$x=1, y=4, z=3$
$3x=3, 2y=8, 4z=12$
$3<8<12$ is true.
c) can be true.
Now d) was the reason for missing this question. I plugged in one case as I did with a) yet there is in fact a threshold such that this works.
For example:
$x=7, y=11, z=6$
$3x=21, 2y=22, 4z=24$
$21<22<24$ is true.
d) can be true.
However, during the practice test (this was the last question I was crunched for time and had to the second to last because this one looked easier) I only plugged in one case and it didn't work so I quit.
Tips?
gre-exam
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
$endgroup$
$begingroup$
It seems like you can save more time by excluding the third variable in each of your tests. You can always take one to go to a very large value or very small value, depending on what you want
$endgroup$
– Goldname
Jul 4 '18 at 20:25
$begingroup$
Don't plug in numbers: think conceptually. For example, to decide whether $y>z$ could be true, focus on $y$ and $z$. The given statement implies $y<2z$, and this is all we know about how $y$ and $z$ compare. This just means $y$ is less than twice $z$. So of course it's possible that $y$ could be more than $z$, because there are numbers between $z$ and $2z$.
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:25
$begingroup$
Similarly, the statement implies $x<(4/3)z$. So all we know is that $x$ is smaller than a number bigger than $z$. So of course $x$ could be bigger than $z$. (Keep in mind the assumption that all numbers are positive is important here.)
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:27
$begingroup$
Didn't see these before marking the best answer! Thank you both for the comments, the problem is much more clear with both suggestions! :D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:36
add a comment |
$begingroup$
This is more or less a question about how many cases I should be trying in general on the quantitative section of the GRE.
I got this question wrong and it is because I did not try enough cases, so I was wondering if anyone has any tips on how to see that a given condition may be true without trying 5 different cases.
Question:
If x, y, and z are positive numbers such that $3x < 2y < 4z$, which of the following statements could be true? Indicate all such statements.
a. $x=y$
b. $y=z$
c. $y>z$
d. $x>z$
My work:
For a)
$x=2, y=2, z=3$
$3x=6, 2y=4, 4z=12$
$6<4<12$ is false.
a) cannot be true.
For b)
$x=1, y=4, z=4$
$3x=3, 2y=8, 4z=16$
$3<8<16$ is true.
b) can be true.
For c)
$x=1, y=4, z=3$
$3x=3, 2y=8, 4z=12$
$3<8<12$ is true.
c) can be true.
Now d) was the reason for missing this question. I plugged in one case as I did with a) yet there is in fact a threshold such that this works.
For example:
$x=7, y=11, z=6$
$3x=21, 2y=22, 4z=24$
$21<22<24$ is true.
d) can be true.
However, during the practice test (this was the last question I was crunched for time and had to the second to last because this one looked easier) I only plugged in one case and it didn't work so I quit.
Tips?
gre-exam
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
$endgroup$
This is more or less a question about how many cases I should be trying in general on the quantitative section of the GRE.
I got this question wrong and it is because I did not try enough cases, so I was wondering if anyone has any tips on how to see that a given condition may be true without trying 5 different cases.
Question:
If x, y, and z are positive numbers such that $3x < 2y < 4z$, which of the following statements could be true? Indicate all such statements.
a. $x=y$
b. $y=z$
c. $y>z$
d. $x>z$
My work:
For a)
$x=2, y=2, z=3$
$3x=6, 2y=4, 4z=12$
$6<4<12$ is false.
a) cannot be true.
For b)
$x=1, y=4, z=4$
$3x=3, 2y=8, 4z=16$
$3<8<16$ is true.
b) can be true.
For c)
$x=1, y=4, z=3$
$3x=3, 2y=8, 4z=12$
$3<8<12$ is true.
c) can be true.
Now d) was the reason for missing this question. I plugged in one case as I did with a) yet there is in fact a threshold such that this works.
For example:
$x=7, y=11, z=6$
$3x=21, 2y=22, 4z=24$
$21<22<24$ is true.
d) can be true.
However, during the practice test (this was the last question I was crunched for time and had to the second to last because this one looked easier) I only plugged in one case and it didn't work so I quit.
Tips?
gre-exam
gre-exam
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
edited Jan 31 at 0:04
Lopey Tall
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
asked Jul 4 '18 at 20:20
Lopey TallLopey Tall
616
616
$begingroup$
It seems like you can save more time by excluding the third variable in each of your tests. You can always take one to go to a very large value or very small value, depending on what you want
$endgroup$
– Goldname
Jul 4 '18 at 20:25
$begingroup$
Don't plug in numbers: think conceptually. For example, to decide whether $y>z$ could be true, focus on $y$ and $z$. The given statement implies $y<2z$, and this is all we know about how $y$ and $z$ compare. This just means $y$ is less than twice $z$. So of course it's possible that $y$ could be more than $z$, because there are numbers between $z$ and $2z$.
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:25
$begingroup$
Similarly, the statement implies $x<(4/3)z$. So all we know is that $x$ is smaller than a number bigger than $z$. So of course $x$ could be bigger than $z$. (Keep in mind the assumption that all numbers are positive is important here.)
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:27
$begingroup$
Didn't see these before marking the best answer! Thank you both for the comments, the problem is much more clear with both suggestions! :D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:36
add a comment |
$begingroup$
It seems like you can save more time by excluding the third variable in each of your tests. You can always take one to go to a very large value or very small value, depending on what you want
$endgroup$
– Goldname
Jul 4 '18 at 20:25
$begingroup$
Don't plug in numbers: think conceptually. For example, to decide whether $y>z$ could be true, focus on $y$ and $z$. The given statement implies $y<2z$, and this is all we know about how $y$ and $z$ compare. This just means $y$ is less than twice $z$. So of course it's possible that $y$ could be more than $z$, because there are numbers between $z$ and $2z$.
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:25
$begingroup$
Similarly, the statement implies $x<(4/3)z$. So all we know is that $x$ is smaller than a number bigger than $z$. So of course $x$ could be bigger than $z$. (Keep in mind the assumption that all numbers are positive is important here.)
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:27
$begingroup$
Didn't see these before marking the best answer! Thank you both for the comments, the problem is much more clear with both suggestions! :D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:36
$begingroup$
It seems like you can save more time by excluding the third variable in each of your tests. You can always take one to go to a very large value or very small value, depending on what you want
$endgroup$
– Goldname
Jul 4 '18 at 20:25
$begingroup$
It seems like you can save more time by excluding the third variable in each of your tests. You can always take one to go to a very large value or very small value, depending on what you want
$endgroup$
– Goldname
Jul 4 '18 at 20:25
$begingroup$
Don't plug in numbers: think conceptually. For example, to decide whether $y>z$ could be true, focus on $y$ and $z$. The given statement implies $y<2z$, and this is all we know about how $y$ and $z$ compare. This just means $y$ is less than twice $z$. So of course it's possible that $y$ could be more than $z$, because there are numbers between $z$ and $2z$.
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:25
$begingroup$
Don't plug in numbers: think conceptually. For example, to decide whether $y>z$ could be true, focus on $y$ and $z$. The given statement implies $y<2z$, and this is all we know about how $y$ and $z$ compare. This just means $y$ is less than twice $z$. So of course it's possible that $y$ could be more than $z$, because there are numbers between $z$ and $2z$.
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:25
$begingroup$
Similarly, the statement implies $x<(4/3)z$. So all we know is that $x$ is smaller than a number bigger than $z$. So of course $x$ could be bigger than $z$. (Keep in mind the assumption that all numbers are positive is important here.)
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:27
$begingroup$
Similarly, the statement implies $x<(4/3)z$. So all we know is that $x$ is smaller than a number bigger than $z$. So of course $x$ could be bigger than $z$. (Keep in mind the assumption that all numbers are positive is important here.)
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:27
$begingroup$
Didn't see these before marking the best answer! Thank you both for the comments, the problem is much more clear with both suggestions! :D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:36
$begingroup$
Didn't see these before marking the best answer! Thank you both for the comments, the problem is much more clear with both suggestions! :D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:36
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Simplify the inequality to three separate inequalities:
$$(1)quad x<dfrac{2}{3}y$$
$$(2)quad y<2z$$
$$(3)quad x<dfrac{4}{3}z$$
Now we conclude easily
(a) false by (1)
(b) true by (2) since $z=y<2z$ still holds.
(c) true by (2) since $z<1.5z=y<2z$ is possible
(d) true by (3) since $z<1.1z=x<dfrac{4}{3}z$ is possible.
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
add a comment |
$begingroup$
Your work for part (a) is useless. It is given that $x, y, z$ always satisfy $3x < 2 y < 4 z$. If you choose $x,y,z$ that violate that constraint, they are not relevant to the problem.
Part (a): $x < frac{2}{3}y$, so $x$ is never as large as $y$, so $x neq y$ always.
For parts (b) and (c), you made valid choices of $x,y,z$, so have these answers.
For part (d), $3x<4z$ is the given relation (that is, $x < frac{4}{3}z$) and $x>z$ can be satisfied if $z < x < frac{4}{3} z$, which is easy to arrange for large enough $z$. You also require a multiple of $2$, $2y$, between $3x$ and $4z$, so you may have to go a little bigger.
In a bit more detail on (d): $z < x < frac{4}{3} z$, so look at multiples of $3$ for $z$ so I don't have to waste time on that fraction. $z=3$ is too small because there's no room for $x$ between $3$ and $4$. $z = 6$ forces $x = 7$. Is there an even number between $21$ and $24$? Yes, $22$, so $y=11$.
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
$endgroup$
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
add a comment |
$begingroup$
In a), you try one specific set of numbers only. So at best you show that a) is sometimes false. This does not exclude the possibility that it is sometimes true. Instead, if $x,y,z$ are positive and $3x<2y<4z$, then
$$ 3x<2y<2y+y=3y$$
and hence $x<y$ (and not $x=y$). We see that a) is always false and never true* under the given assumptions.
Of course ouy are right that for the three cases, where one has to show that the asked statement can be true, a single specific example suffices. However, it does not suffice to plug in just any valid triple - after all, it need not be (and is not) the case that the statement is always true. Thus it suffices to plug in one set of numbers, but it has to be a smartly chosen one. In other words, you have to contemplate the given conditions and the target property as to what "obstacles" there might be for the target property and if these obstables can be shown to always be there or if they disappear in certain achievable situations ...
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2841135%2fhow-many-cases-on-a-gre-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Simplify the inequality to three separate inequalities:
$$(1)quad x<dfrac{2}{3}y$$
$$(2)quad y<2z$$
$$(3)quad x<dfrac{4}{3}z$$
Now we conclude easily
(a) false by (1)
(b) true by (2) since $z=y<2z$ still holds.
(c) true by (2) since $z<1.5z=y<2z$ is possible
(d) true by (3) since $z<1.1z=x<dfrac{4}{3}z$ is possible.
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
add a comment |
$begingroup$
Simplify the inequality to three separate inequalities:
$$(1)quad x<dfrac{2}{3}y$$
$$(2)quad y<2z$$
$$(3)quad x<dfrac{4}{3}z$$
Now we conclude easily
(a) false by (1)
(b) true by (2) since $z=y<2z$ still holds.
(c) true by (2) since $z<1.5z=y<2z$ is possible
(d) true by (3) since $z<1.1z=x<dfrac{4}{3}z$ is possible.
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
add a comment |
$begingroup$
Simplify the inequality to three separate inequalities:
$$(1)quad x<dfrac{2}{3}y$$
$$(2)quad y<2z$$
$$(3)quad x<dfrac{4}{3}z$$
Now we conclude easily
(a) false by (1)
(b) true by (2) since $z=y<2z$ still holds.
(c) true by (2) since $z<1.5z=y<2z$ is possible
(d) true by (3) since $z<1.1z=x<dfrac{4}{3}z$ is possible.
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
Simplify the inequality to three separate inequalities:
$$(1)quad x<dfrac{2}{3}y$$
$$(2)quad y<2z$$
$$(3)quad x<dfrac{4}{3}z$$
Now we conclude easily
(a) false by (1)
(b) true by (2) since $z=y<2z$ still holds.
(c) true by (2) since $z<1.5z=y<2z$ is possible
(d) true by (3) since $z<1.1z=x<dfrac{4}{3}z$ is possible.
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
answered Jul 4 '18 at 20:40
InterstellarProbeInterstellarProbe
3,154728
3,154728
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
add a comment |
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
$begingroup$
Thank you so much! This is very clear, thorough, and concise!
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:35
add a comment |
$begingroup$
Your work for part (a) is useless. It is given that $x, y, z$ always satisfy $3x < 2 y < 4 z$. If you choose $x,y,z$ that violate that constraint, they are not relevant to the problem.
Part (a): $x < frac{2}{3}y$, so $x$ is never as large as $y$, so $x neq y$ always.
For parts (b) and (c), you made valid choices of $x,y,z$, so have these answers.
For part (d), $3x<4z$ is the given relation (that is, $x < frac{4}{3}z$) and $x>z$ can be satisfied if $z < x < frac{4}{3} z$, which is easy to arrange for large enough $z$. You also require a multiple of $2$, $2y$, between $3x$ and $4z$, so you may have to go a little bigger.
In a bit more detail on (d): $z < x < frac{4}{3} z$, so look at multiples of $3$ for $z$ so I don't have to waste time on that fraction. $z=3$ is too small because there's no room for $x$ between $3$ and $4$. $z = 6$ forces $x = 7$. Is there an even number between $21$ and $24$? Yes, $22$, so $y=11$.
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
$endgroup$
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
add a comment |
$begingroup$
Your work for part (a) is useless. It is given that $x, y, z$ always satisfy $3x < 2 y < 4 z$. If you choose $x,y,z$ that violate that constraint, they are not relevant to the problem.
Part (a): $x < frac{2}{3}y$, so $x$ is never as large as $y$, so $x neq y$ always.
For parts (b) and (c), you made valid choices of $x,y,z$, so have these answers.
For part (d), $3x<4z$ is the given relation (that is, $x < frac{4}{3}z$) and $x>z$ can be satisfied if $z < x < frac{4}{3} z$, which is easy to arrange for large enough $z$. You also require a multiple of $2$, $2y$, between $3x$ and $4z$, so you may have to go a little bigger.
In a bit more detail on (d): $z < x < frac{4}{3} z$, so look at multiples of $3$ for $z$ so I don't have to waste time on that fraction. $z=3$ is too small because there's no room for $x$ between $3$ and $4$. $z = 6$ forces $x = 7$. Is there an even number between $21$ and $24$? Yes, $22$, so $y=11$.
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
$endgroup$
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
add a comment |
$begingroup$
Your work for part (a) is useless. It is given that $x, y, z$ always satisfy $3x < 2 y < 4 z$. If you choose $x,y,z$ that violate that constraint, they are not relevant to the problem.
Part (a): $x < frac{2}{3}y$, so $x$ is never as large as $y$, so $x neq y$ always.
For parts (b) and (c), you made valid choices of $x,y,z$, so have these answers.
For part (d), $3x<4z$ is the given relation (that is, $x < frac{4}{3}z$) and $x>z$ can be satisfied if $z < x < frac{4}{3} z$, which is easy to arrange for large enough $z$. You also require a multiple of $2$, $2y$, between $3x$ and $4z$, so you may have to go a little bigger.
In a bit more detail on (d): $z < x < frac{4}{3} z$, so look at multiples of $3$ for $z$ so I don't have to waste time on that fraction. $z=3$ is too small because there's no room for $x$ between $3$ and $4$. $z = 6$ forces $x = 7$. Is there an even number between $21$ and $24$? Yes, $22$, so $y=11$.
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
$endgroup$
Your work for part (a) is useless. It is given that $x, y, z$ always satisfy $3x < 2 y < 4 z$. If you choose $x,y,z$ that violate that constraint, they are not relevant to the problem.
Part (a): $x < frac{2}{3}y$, so $x$ is never as large as $y$, so $x neq y$ always.
For parts (b) and (c), you made valid choices of $x,y,z$, so have these answers.
For part (d), $3x<4z$ is the given relation (that is, $x < frac{4}{3}z$) and $x>z$ can be satisfied if $z < x < frac{4}{3} z$, which is easy to arrange for large enough $z$. You also require a multiple of $2$, $2y$, between $3x$ and $4z$, so you may have to go a little bigger.
In a bit more detail on (d): $z < x < frac{4}{3} z$, so look at multiples of $3$ for $z$ so I don't have to waste time on that fraction. $z=3$ is too small because there's no room for $x$ between $3$ and $4$. $z = 6$ forces $x = 7$. Is there an even number between $21$ and $24$? Yes, $22$, so $y=11$.
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
answered Jul 4 '18 at 20:35
Eric TowersEric Towers
33.5k22370
33.5k22370
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
add a comment |
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
Thank you so much this is so much clearer! I gave the answer to @InterstellarProbe because i) their answer is shorter and more likely for me to remember the next time I encounter a similar problem, ii) they answered first, and iii) they have less points under their profile name so they could use the answer points:D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:33
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
@LopeyTall : You can accept any answer you like. As a detail, InterstellarProbe did not answer first.
$endgroup$
– Eric Towers
Jul 4 '18 at 21:36
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
$begingroup$
Oh gosh your right, I'm blind! haha :) Thanks again mate!:)
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:38
add a comment |
$begingroup$
In a), you try one specific set of numbers only. So at best you show that a) is sometimes false. This does not exclude the possibility that it is sometimes true. Instead, if $x,y,z$ are positive and $3x<2y<4z$, then
$$ 3x<2y<2y+y=3y$$
and hence $x<y$ (and not $x=y$). We see that a) is always false and never true* under the given assumptions.
Of course ouy are right that for the three cases, where one has to show that the asked statement can be true, a single specific example suffices. However, it does not suffice to plug in just any valid triple - after all, it need not be (and is not) the case that the statement is always true. Thus it suffices to plug in one set of numbers, but it has to be a smartly chosen one. In other words, you have to contemplate the given conditions and the target property as to what "obstacles" there might be for the target property and if these obstables can be shown to always be there or if they disappear in certain achievable situations ...
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
In a), you try one specific set of numbers only. So at best you show that a) is sometimes false. This does not exclude the possibility that it is sometimes true. Instead, if $x,y,z$ are positive and $3x<2y<4z$, then
$$ 3x<2y<2y+y=3y$$
and hence $x<y$ (and not $x=y$). We see that a) is always false and never true* under the given assumptions.
Of course ouy are right that for the three cases, where one has to show that the asked statement can be true, a single specific example suffices. However, it does not suffice to plug in just any valid triple - after all, it need not be (and is not) the case that the statement is always true. Thus it suffices to plug in one set of numbers, but it has to be a smartly chosen one. In other words, you have to contemplate the given conditions and the target property as to what "obstacles" there might be for the target property and if these obstables can be shown to always be there or if they disappear in certain achievable situations ...
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
In a), you try one specific set of numbers only. So at best you show that a) is sometimes false. This does not exclude the possibility that it is sometimes true. Instead, if $x,y,z$ are positive and $3x<2y<4z$, then
$$ 3x<2y<2y+y=3y$$
and hence $x<y$ (and not $x=y$). We see that a) is always false and never true* under the given assumptions.
Of course ouy are right that for the three cases, where one has to show that the asked statement can be true, a single specific example suffices. However, it does not suffice to plug in just any valid triple - after all, it need not be (and is not) the case that the statement is always true. Thus it suffices to plug in one set of numbers, but it has to be a smartly chosen one. In other words, you have to contemplate the given conditions and the target property as to what "obstacles" there might be for the target property and if these obstables can be shown to always be there or if they disappear in certain achievable situations ...
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
In a), you try one specific set of numbers only. So at best you show that a) is sometimes false. This does not exclude the possibility that it is sometimes true. Instead, if $x,y,z$ are positive and $3x<2y<4z$, then
$$ 3x<2y<2y+y=3y$$
and hence $x<y$ (and not $x=y$). We see that a) is always false and never true* under the given assumptions.
Of course ouy are right that for the three cases, where one has to show that the asked statement can be true, a single specific example suffices. However, it does not suffice to plug in just any valid triple - after all, it need not be (and is not) the case that the statement is always true. Thus it suffices to plug in one set of numbers, but it has to be a smartly chosen one. In other words, you have to contemplate the given conditions and the target property as to what "obstacles" there might be for the target property and if these obstables can be shown to always be there or if they disappear in certain achievable situations ...
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jul 4 '18 at 20:43
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2841135%2fhow-many-cases-on-a-gre-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
It seems like you can save more time by excluding the third variable in each of your tests. You can always take one to go to a very large value or very small value, depending on what you want
$endgroup$
– Goldname
Jul 4 '18 at 20:25
$begingroup$
Don't plug in numbers: think conceptually. For example, to decide whether $y>z$ could be true, focus on $y$ and $z$. The given statement implies $y<2z$, and this is all we know about how $y$ and $z$ compare. This just means $y$ is less than twice $z$. So of course it's possible that $y$ could be more than $z$, because there are numbers between $z$ and $2z$.
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:25
$begingroup$
Similarly, the statement implies $x<(4/3)z$. So all we know is that $x$ is smaller than a number bigger than $z$. So of course $x$ could be bigger than $z$. (Keep in mind the assumption that all numbers are positive is important here.)
$endgroup$
– symplectomorphic
Jul 4 '18 at 20:27
$begingroup$
Didn't see these before marking the best answer! Thank you both for the comments, the problem is much more clear with both suggestions! :D
$endgroup$
– Lopey Tall
Jul 4 '18 at 21:36