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

Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Tue Oct 09, 2018 6:04 am

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BTGmoderatorDC wrote:If x, y, and z are positive numbers, Is z between x and y?

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

OA C

Source: GMAT Prep

Given: x, y, and z are positive numbers
Question: Is z between x and y?

Let's take each statement one by one.

(1) x < 2z < y

Case 1: Say x = z = 1 and y = 3, then x < 2z < y => 1 < 2*1 < 3 => 1 < 2 < 3. We see that z is NOT between x and y. The answer is No.
Case 2: Say x = 1, z = 2 and y = 5, then x < 2z < y => 1 < 2*2 < 5 => 1 < 4 < 5. We see that z is between x and y. The answer is Yes.

No unique answer. Insufficient.

(2) 2x < z < 2y

Case 1: Say x = 1/4, and z = y = 1, then x < 2z < y => 2*(1/4) < 1 < 2*1 => 1/2 < 1 < 2. We see that z is NOT between x and y. The answer is No.
Case 2: Say x = 1/2, z = 2, y = 3, then x < 2z < y => 2*(1/2) < 2 < 2*3 => 1 < 2 < 6. We see that z is between x and y. The answer is Yes.

No unique answer. Insufficient.

(1) and (2) together

Adding the two inequalities given in the statements, we have,

(x < 2z < y) + (2x < z < 2y) => 3x < 3y < 3y => x < z < y. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Tue Oct 09, 2018 6:08 am

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BTGmoderatorDC wrote:If x, y, and z are positive numbers, Is z between x and y?

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

OA C

Source: GMAT Prep

Target question: Is z between x and y?

Statement 1: x < 2z < y
There are several sets of values of x, y and z that satisfy this condition. Here are two:
Case a: x = 3, y = 10, and z = 2, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 2x < z < 2y
There are several sets of values of x, y and z that satisfy this condition. Here are two:
Case a: x = 1, y = 2, and z = 3, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x < 2z < y
Statement 2: 2x < z < 2y
Since the two inequalities are facing the same direction, we can add them to get:
3x < 3z < 3y
Divide all three parts by 3 to get: x < z < y
As we can see, z IS definitely between x and y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Tue Oct 09, 2018 6:15 am

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BTGmoderatorDC wrote:If x, y, and z are positive numbers, Is z between x and y?

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

Statement 1: x < 2z < y
Let x=2 and y=10.
In this case:
2 < 2z < 10
1 < z < 5.
If z=4, then it is between x=2 and y=10.
If z=1.5, then it is not between x=2 and y=10.
INSUFFICIENT.

Statement 2: 2x < z < 2y
Let x=2 and y=10.
In this case:
2*2 < z < 2*10
4 < z < 20
If z=5, then it is between x=2 and y=10.
If z=19, then it is not between x=2 and y=10.
INSUFFICIENT.

Statements combined:
Inequalities can be ADDED TOGETHER.
Adding together x < 2z < y and 2x < z < 2y, we get:
(x+2x) < (2z+z) < (y+2y)
3x < 3z < 3y
x < z < y.
Thus, z is between x and y.
SUFFICIENT.

The correct answer is C.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pm; Thanked: 59 times; Followed by:33 members

by fskilnik@GMATH » Tue Oct 09, 2018 10:18 am

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BTGmoderatorDC wrote:If x, y, and z are positive numbers, Is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y
Source: GMAT Prep

(At least to my students) There is no need to add both statements to deal with (1+2), although this is a very good idea used by all other experts above.
Could you find another approach? (One possibility is presented in my solution.)
\[x,y,z\,\,\, > 0\,\,\,\left( * \right)\]
\[?\,\,\,\,:\,\,\,\,z\,\,{\text{between}}\,\,x\,\,{\text{and}}\,\,y\,\,\,\]
\[\left( 1 \right)\,\,\,x < 2z < y\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,2,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,2x < z < 2y\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,\,x\,\,\mathop < \limits^{\left( * \right)} \,\,\,2x\,\,\mathop < \limits^{\left( 2 \right)} \,\,z\,\,\mathop < \limits^{\left( * \right)} \,\,\,2z\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x < z < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

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

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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