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number property

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by goalevan » Fri Jul 22, 2011 5:29 pm
This is easiest to visualize by drawing a number line and splitting the interval between 5 and 15 into thirds. You can calculate that (15 - 10) / 3 = 3 1/3, so 5 + (3 1/3 * 2) = 11 2/3.

You can also represent this algebraically with absolute values:

"the distance between x and 5": |x - 5|

"is": =

"twice as far as": 2 *

"the distance between x and 15": |x - 15|

|x - 5| = 2|x - 15|

This absolute value equation has the following two solutions:

x - 5 = 2(x - 15)
x - 5 = 2x - 30
x = 25

-(x - 5) = 2(x - 15)
5 - x = 2x - 30
3x = 35
x = 35/3 = 11 2/3

The problem states that 5 < x < 15, so x = 11 2/3.
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by sumasajja » Fri Jul 22, 2011 5:55 pm
goalevan wrote:This is easiest to visualize by drawing a number line and splitting the interval between 5 and 15 into thirds. You can calculate that (15 - 10) / 3 = 3 1/3, so 5 + (3 1/3 * 2) = 11 2/3.

You can also represent this algebraically with absolute values:

"the distance between x and 5": |x - 5|

"is": =

"twice as far as": 2 *

"the distance between x and 15": |x - 15|

|x - 5| = 2|x - 15|

This absolute value equation has the following two solutions:

x - 5 = 2(x - 15)
x - 5 = 2x - 30
x = 25

-(x - 5) = 2(x - 15)
5 - x = 2x - 30
3x = 35
x = 35/3 = 11 2/3

The problem states that 5 < x < 15, so x = 11 2/3.

thanks for the explaination ....but my first thought on that was diverted by seeing the lines "the distance between x and 5" in the problem i interpreted it to be from 6 to 14..is it usually taken in that way while solving number line problems?
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by BlindVision » Fri Jul 22, 2011 6:47 pm
sumasajja wrote:if x is a number on the number line between 5 and 15 that is twice as far from 5 as from 15,then x is?
a] 5 2/3
b] 10
c] 11 2/3
d] 12 1/2
e] 13 1/3
the answer is c.
I took the route of using less algebra...

5 - 15 = 11 numbers

11/3 (twice as far + 1)

= 10/3 (1 part of 3 equal distances)

10/3(2) = 20/3 (twice as far)

20/3 + 5 (starting point)

= 35/3 = 11 2/3
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by Ozlemg » Mon Jul 25, 2011 3:57 am
I just take the middle number which is located in C, and able to find it quickly. But I may not be as lucky as this one on real GMAT!

thank you golaevan for your excellent explanation!
The more you suffer before the test, the less you will do so in the test! :)
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by navami » Mon Jul 25, 2011 7:05 am
ans is C.

consider a number scale as below
---------------------------------
5 15

now point X lies somewhere in this scale dividing this scale in 2:1 ratio

that means a scale of length 10, {15-5} is divided in 2:1 ratio
means point X is 10*2/3 away from point 5.
means point X is 5 + 10*2/3 away from point 0 in number scale.
This time no looking back!!!
Navami
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by ECMoyano » Mon Jul 25, 2011 12:59 pm
Ozlemg wrote:I just take the middle number which is located in C, and able to find it quickly. But I may not be as lucky as this one on real GMAT!

thank you golaevan for your excellent explanation!
I did the same. I too am afraid that I won't be that lucky on the GMAT.
To solve by plugging in:
I started at 10, and than moved down to 11 2/3 and plugged number in.
Its 6+2/3 from 5/ or 3+1/3 from 15. 6+2/3 needs to be twice (2 times)3+1/3
2/1 * 10/3 = 20/3 or 6+2/3

Answer is C :)
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by SticklorForDetails » Tue Jul 26, 2011 7:52 am
ECMoyano wrote:
Ozlemg wrote:I just take the middle number which is located in C, and able to find it quickly. But I may not be as lucky as this one on real GMAT!

thank you golaevan for your excellent explanation!
I did the same. I too am afraid that I won't be that lucky on the GMAT.
To solve by plugging in:
I started at 10, and than moved down to 11 2/3 and plugged number in.
Its 6+2/3 from 5/ or 3+1/3 from 15. 6+2/3 needs to be twice (2 times)3+1/3
2/1 * 10/3 = 20/3 or 6+2/3

Answer is C :)
Plugging in, while it may feel less satisfying, is actually a pretty good approach on this problem. The expert way to do it is always plug in (B) and then (D), because then even if they don't work, you'll know the answer for sure. Either it must be less than (B), it must be bigger than (D), or it must be between the two. That way "luck" doesn't have anything to do with it and you know you'll get the right answer in a maximum of 2 tries. On this one:

(B) 10 -- is half-way between the two so it's wrong (equally far instead of "twice as far").

(D) 12 1/2 -- is 7.5 away from 5 and 2.5 away from 15. That's wrong (3x as far).

Since (B) was once as far and (D) was thrice as far, it must be (C), between them. No luck involved, just smart strategy!

The nicest thing about this approach is that it's much easier to interpret weird language like "x is twice as far from 5 as from 15" when you have an actual number involved, so sumasajja's confusion above probably wouldn't have happened.
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