If x, y and z are positive numbers, is z between x and y?

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Answer: C.

Hope it's clear.
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question- is "x < z < y" or "y < z < x" ????

1) x < 2z < y
since 2z<y ( and all are positive) we can infer that z<y . But we do not know weather x<z
Insufficient
(2) 2x < z < 2y
since 2x<z , we can infer x<z , but we cant say if z<y
Insufficient

Combine both 1 and 2 - x<z<y ... sufficient

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 4, 2z = 5, y = 6 --> 4 < 5 < 6 BUT z < x < y -- NO
If x = 4, 2z = 10, y = 12 --> 4 < 10 < 12 and x < z < y -- YES

Not sufficient.

(2) 2x < z < 2y.

If 2x = 4, z = 5, 2y = 6 --> 4 < 5 < 6 BUT x < y < z --> NO
If 2x = 4, z = 5, 2y = 18 --> 4 < 5 < 9 and x < z < y --> YES

Not sufficient.

(1)+(2)

x < 2x (always, for positive numbers)
2x < z (from 2)
z < 2z (always, for positive numbers)
2z < y (from 1)

Which gives x < 2x < z < 2z < y

Therefore x < z < y

Sufficient.

Answer: C.

I want to know what is wrong with the following method:

1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes
x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No

Insufficient.

Statement 2 :
Since all are positive I can reduce by 2 , we get x<(z/2)<y

x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes

x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no

1+2
Adding both of these

x < 2z < y
x<z/2<y

we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1

I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.

(1) x < 2z < y
(2) 2x < z < 2y

When you add the inequalities you get:

(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

Hope this helps.
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sorry I didn't get it. My question was since all are positive I can definitely reduce statement 2 by 2, right?
now statement 2 becomes x<(z/2)<y
why am I not getting the answer if I take statement 1 as x < 2z < y , and statement 2 as x<(z/2)<y

Adding 1 and 2 we get 2x<2.5z<2y

x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No

So how come this is not working out if after reducing stat.2 I add it to 1?

In the same way if we were to solve :

(Q2)
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

adding 1 and 2--> 2x<2.5z<2y so here the answer is E
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes
x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No

So what is the answer to Q2, C or E

Thank you for your help.

1. You can reduce 2x < z < 2y by 2 no matter whether x, y, and z are positive or negative. You can always reduce/multiply an inequality by a positive value (2 in our case).

2. Reducing by 2 and adding, though legit is not a correct way to solve. You can add the two inequalities without reducing and directly get (x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.

3. When you reduce and then add you'll get 2x < 2.5z < 2y but you cannot plug arbitrary value of x, y, and z there. The values must satisfy all the inequalities, while x=1, z=2, y=3 and x=2, z=2, y=3 does NOT satisfy neither x < 2z < y nor 2x < z < 2y.

Hope it's clear.
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If the inequality signs were not all neatly facing the same directions, would we still be able to add?

Thanks

No.

1. You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

2. You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Check for more here: inequalities-tips-and-hints-175001.html

Hope it helps.
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\(x,y,z\,\,\, > 0\,\,\,\left( * \right)\)

\(?\,\,\,\,:\,\,\,\,z\,\,{\text{between}}\,\,x\,\,{\text{and}}\,\,y\,\,\,\)

\(\left( 1 \right)\,\,\,x < 2z < y\,\,\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,2,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,2x < z < 2y\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\\\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,x\,\,\mathop < \limits^{\left( * \right)} \,\,\,2x\,\,\mathop < \limits^{\left( 2 \right)} \,\,z\,\,\mathop < \limits^{\left( * \right)} \,\,\,2z\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x < z < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Bunuel VeritasKarishma I have a doubt regarding an alternate method mentioned below. I would really appreciate your help

Let's say that we divided inequalities in stmt 2 by 2.
2x < z < 2y => x < z/2 < y

Now if we add this modified stmt 2 inequalities with the one in stmt 1, we get

2x < 5z/2 < 2y => 4x < 5z < 4y

From 5z < 4y, we can infer that z < y.
From 4x < 5z, we can infer anything whether x < z or x > z.

How can we infer x < z in this case? Am I doing something wrong here?

Thanks!

We know that x < z from (2): 2x < z. This means that x < z.
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VeritasKarishma is there some systematic concept to answer such type of DS questions without picking numbers ? method of picking numbers is kinda time consuming and prone to be erroneous

I often like to think of the number line in such questions. You usually need to write down nothing if you imagine the number line in your head.
The numbers are all positive - great. I am looking at the number line to the right of 0.

0 ----1--- 2-------4 --------- 6- ------8 ---------10 ---------

(1) x < 2z < y

So I know that 2z is between x and y. Now 2z could be any number and x and y could be very close to 2z such that z doesn't lie between them. Or x and y could be far far apart such that z as well as 2z lie between them. Say 2z = 8.

0 ------2----- 4---- 6---x---8 ---y------ 10- --
or

0 --x---2------ 4-----6------8 ------ 10- --------y--


(2) 2x < z < 2y
This means x < z/2 < y
Now we know that z/2 lies between x and y. Whether z lies or not, we cannot say because of the same logic as above.

Using both statements together, we know that z/2 and 2z both lie between x and y. Then obviously x and y are far apart to include both. z will lie between z/2 and 2z so it will automatically be between x and y.

Answer (C)
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Question data tells us that x, y and z are positive numbers. Quite often, a lot of test takers assume this to be positive integers and end up falling for trap answers. x, y and z could be fractions too.

In this question, plugging in numbers is a good way to prove/disprove the individual statements, while using a property is a good way of combining them.

From statement I alone, x<2z<y.

If x = 1, z = 2 and y = 5, 1<4<5. Is z between x and y? YES.
If x = 1, z = 0.75 and y = 2, 1<1.5<2. Is z between x and y? NO.

Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, 2x<z<2y.

If x = 0.5, z = 2 and y = 1.5, 1<2<3. Is z between x and y? NO.
If x = 1, z = 3 and y = 5, 2<3<10. Is z between x and y? YES.

Statement II alone is insufficient. Answer option B can be eliminated. Possible answer options are C or E.

Combining statements I and II, we have the following:

From statement I, x<2z<y; from statement II, 2x<z<2y.

Since the inequality signs are the same, the two inequalities can be added. Adding the inequalities, we have,
3x<3z<3y.
Dividing all terms of the inequality by 3, we have x<z<y.

Is z between x and y? YES.
The combination of statements is sufficient. Answer option E can be eliminated.

The correct answer option is C.

Hope that helps!
Aravind B T

Stem analysis:
From statement one we can conclude that X,Y,Z are all >0 and we want to know if Z is between X and Y, which can be represented as X<Z<Y.

Statement 1:
From statement one we can conclude that Z<Y but we can't conclude if X<Z. We can conclude that Z<Y because if TWO POSITIVE Z's are less than Y then ONE POSITIVE Z MUST be less than Y. But depending on the value of Z and X the Z could swing before or after the X.

Statement 2:
Statement two is similar in that we can now conclude that X<Z but not that Z<Y. We can conclude that X<Z because if TWO POSITIVE X is less than Z then ONE POSITIVE X must also be less than Z.

Statement 1 and 2 combined:
With both statement together we can combine the inequalities based on the transitive property of inequalities.

Take the equation in statement divide each term by 2 so that we now have X<Z/2<Y. From this we can conclude the following:

X<Z/2<Z<Y, which means that X<Z<Y, both statement together are sufficient to answer the question.

Solution

Given
• Three positive numbers x, y, and z
To find
• whether z is between x and y

Approach and Working out
• Question Stem Analysis:

o The question stem gives us no information about the comparison between x, y, and z. We just know that these three are positive numbers.

• Statement 1: x < 2z < y.

o 2z is between x and y.
o z itself need not be between x and y. Here are two examples:

 If x = 2, y = 8, and z = 3 --> answer YES.
 If x = 1, y = 5, and z = 2 --> answer NO.
o This statement is not sufficient alone.

• Statement (2): 2x < z < 2y.

o z is between 2x and 2y.
o z may or not be between x and y as well. Here are two examples:

 If x = 1, y = 10, and z = 3 --> answer YES.
 If x = 1, y = 2, and z = 3 --> answer NO.
o Statement 2 is not sufficient alone.

• Combining statements (1) and (2):

o Adding the two given inequalities, we get 3x < 3z < 3y.
o Now, dividing throughout by 3 gives: x < z < y.
o The statements are sufficient together.

Correct Answer: Option C
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There are several ways to solve it, but I would like to go with logic (normally I'd go with hard algebra here).

Statement 1 says that 2z is greater than x and smaller than y. In other words, is 2z is STILL smaller than y, than z is definitely smaller than y --- good!
However, we don't know about the r/s between x and z. If 2z is greater than x, who knows whether one z is greater than x?
so AD is out.

Statement 2 says that z is in between 2x and 2y. In other words, is z is greater than 2x, then z is definitely greater than 1x. --- good!
However, we don't know about the r/s between x and y. if 2y is greater than z, who knows whether one y is greater than z?
so B is out.

C, together, we know that z is in between 1x and 1y = sufficient.

What stipulates me from picking numbers for the final inequality to deem in unsatisfactory? SO confused by these questions.