If x is a positive number less than 10, is z greater than the mean of x, and 10?
1. On the number line z is closer to 10 than to x
2. z = 5x
Number line Question
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Statement (1)
Sufficient
if z is closer to 10, 10-z > z-x
Keeping that in mind let's plug in some values.
x=2, z=7
Mean of x and 10 = 6 < z
If x=0.5, z = 5.26 (It has to be closer to 10. So it has to be > 0.5+5.25)
Mean = 0.5+10/2 = 5.25 < z
Therefore z is always going to be more than the mean of x and 10
If z is closer to 10 than it is to x it must be greater than the mean of 10 and x . So (1) is sufficient
Statement (2)
z=5x
x=1, z=5
Mean = 1+10/2 = 5.5 > z
x=1.5, z=7.5
Mean = 1+10/2 = 5.5 < z
So statement 2 is insufficient.
Ans A
Sufficient
if z is closer to 10, 10-z > z-x
Keeping that in mind let's plug in some values.
x=2, z=7
Mean of x and 10 = 6 < z
If x=0.5, z = 5.26 (It has to be closer to 10. So it has to be > 0.5+5.25)
Mean = 0.5+10/2 = 5.25 < z
Therefore z is always going to be more than the mean of x and 10
If z is closer to 10 than it is to x it must be greater than the mean of 10 and x . So (1) is sufficient
Statement (2)
z=5x
x=1, z=5
Mean = 1+10/2 = 5.5 > z
x=1.5, z=7.5
Mean = 1+10/2 = 5.5 < z
So statement 2 is insufficient.
Ans A
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you can solve it algebraically.
is z>x+10/2 -> 2z>x+10
1.10-z<z-x -> 10+x<2z -> suff
2.10x>x+10 -> 9x>10 -> x>9/10 -> x can be many different nums hence insuff
hope that helps.
is z>x+10/2 -> 2z>x+10
1.10-z<z-x -> 10+x<2z -> suff
2.10x>x+10 -> 9x>10 -> x>9/10 -> x can be many different nums hence insuff
hope that helps.
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okigbo wrote:What about when z=9 and x=8. Doesn't A become insufficient then??
I agree with you. I don't understand ?
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okigbo wrote:
What about when z=9 and x=8. Doesn't A become insufficient then??
in this case, z is as close to x as is to 10. so doesnot satisfy the statement 1.
Statement 1 is sufficient to ans this.[/spoiler]
What about when z=9 and x=8. Doesn't A become insufficient then??
in this case, z is as close to x as is to 10. so doesnot satisfy the statement 1.
Statement 1 is sufficient to ans this.[/spoiler]
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In "A", it states z is closer to 10 than x ..
If we take this example:
x=8, z=9. This cannot happen because z is equidistant from x and 10. Therefore, the maximum values we can take is x=7,z=9. Here Z is closer to 10 than x..
Hence, 2Z>x+10.. 2*9>7+10..18>17..
Answer is A.
I assumed answer to be "E". Let me know if my reasoning is wrong..
If we take this example:
x=8, z=9. This cannot happen because z is equidistant from x and 10. Therefore, the maximum values we can take is x=7,z=9. Here Z is closer to 10 than x..
Hence, 2Z>x+10.. 2*9>7+10..18>17..
Answer is A.
I assumed answer to be "E". Let me know if my reasoning is wrong..
"Native of" is used for a individual while "Native to" is used for a large group
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I do not know how to solve this algebraically. If the maximum value possible for x is 7, then trying any combination for x and z that meet statement 1 restriction will make it sufficient.viju9162 wrote:In "A", it states z is closer to 10 than x ..
If we take this example:
x=8, z=9. This cannot happen because z is equidistant from x and 10. Therefore, the maximum values we can take is x=7,z=9. Here Z is closer to 10 than x..
Hence, 2Z>x+10.. 2*9>7+10..18>17..
Answer is A.
I assumed answer to be "E". Let me know if my reasoning is wrong..
Statement two is not sufficient.
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We can also think imagistically.sakurle wrote:If x is a positive number less than 10, is z greater than the mean of x, and 10?
1. On the number line z is closer to 10 than to x
2. z = 5x
Let's draw a number line that includes x and 10 and the average of the two.
x------mean------10
We know that the mean of x and 10 is exactly halfway between the two; in other words, the average of two numbers will always lie equidistant from the numbers.
(1) z is closer to 10 than to x.
Looking at our number line, we can see that the only way z can be closer to 10 than to x is if z lies to the right of the mean. Therefore, z must be greater than the mean: sufficient.
(2) z = 5x
As others have shown, we can pick numbers to show that this may or may not give a value of z greater than the mean. For example, if we let x = .0001, z will still be far less than the mean of x and 10; if we let x = 9.999, z will be far greater than the mean of x and 10: insufficient.
(1) is sufficient, (2) isn't: choose A.
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