Number line Question

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

What about when z=9 and x=8. Doesn't A become insufficient then??

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.

okigbo wrote:What about when z=9 and x=8. Doesn't A become insufficient then??

I agree with you. I don't understand ?

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]

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..

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..

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.

Statement two is not sufficient.

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

We can also think imagistically.

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.