collection of tough problems from G PREP - 28

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by shashank.ism » Mon Feb 08, 2010 3:58 am
abhasjha wrote:If x is a positive number less than 10, is z greater than the average (arithmetic mean) of x and 10 ?

1) On the number line, z is closer to 10 than it is to x.

2) z = 5x
x<10
avg of x and 10 = (x+10)/2
st 1: if z is closer to 10 than x , surely z is greater than average of x and 10 ....sufficient.
st2: Z=5x if x>1 then z > 5 then it is greater tahn average of x,10
if x<1 then z is more towards x and hence is less than average of x,10

so ans is A
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by ajith » Tue Feb 09, 2010 3:31 am
abhasjha wrote:If x is a positive number less than 10, is z greater than the average (arithmetic mean) of x and 10 ?

1) On the number line, z is closer to 10 than it is to x.

2) z = 5x
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by sanju09 » Tue Feb 09, 2010 5:29 am
abhasjha wrote:If x is a positive number less than 10, is z greater than the average (arithmetic mean) of x and 10 ?

1) On the number line, z is closer to 10 than it is to x.

2) z = 5x
We are given with 0 < x < 10 and are asked, "Is 2 z > x + 10?"

(1) With x already 0 < x < 10, the statement, "on the number line, z is closer to 10 than it is to x", proves that 0 < x < z < 10. Something as underneath


__0_______x________z____10_____

If we talk positive integers only, though the stem with (1) never confirms that, following are our observations:

When x = 1, minimum z is 6, and 2 z > x + 10 and it doesn't hold till x = 5 on same z. Insufficient

[spoiler]What is true for numbers is not necessarily true for integers, but what is true for integers is true for all numbers, on the real number line. I'm really clueless about what I wrote in italics. Isn't it so?[/spoiler]

(2) With x already 0 < x < 10, the statement, "z = 5 x" makes us write 0 < z < 50. Now, is 2 z > x + 10? In more simple words, did 2 z cross the range 10 to 20 or did z cross the range 5 to 10? With z already known to be in the range 0 to 50, we can never answer to that any firmly. Insufficient

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