functions

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functions

by Gurpinder » Fri Aug 20, 2010 12:41 pm
If [x] denotes the greatest integer less than or equal to
x, is [x] = 0 ?
(1) 5x + 1 = 3 + 2x
(2) 0 < x < 1
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by Makushr1 » Fri Aug 20, 2010 3:11 pm
Is the answer both are suff?

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by Stuart@KaplanGMAT » Fri Aug 20, 2010 6:23 pm
Gurpinder wrote:If [x] denotes the greatest integer less than or equal to
x, is [x] = 0 ?
(1) 5x + 1 = 3 + 2x
(2) 0 < x < 1
Step 1 of the Kaplan method for DS: Analyze the Stem

We're told that [x] is the biggest integer less than or equal to x. So, for example, [1.4]=1 (sometimes plugging in numbers helps you understand these wacky functions).

Our question: Is [x]=0?

Yes/no question: if [x] always equals 0, sufficient; if it never equals 0, sufficient. If sometimes it does and sometimes it doesn't, insufficient.

When will [x]=0? If 0 <=x < 1

Step 2 of the Kaplan method for DS: Evaluate the Statements

(1) allows us to calculate an exact value for x. If we can calculate an exact value, we can certainly answer the question: sufficient.

(2) tells us that x is definitely inside the range we defined: sufficient.

Each statement is sufficient alone: choose (D).
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by Gurpinder » Fri Aug 20, 2010 7:21 pm
please check if my reasoning is right for picking (D)

Stmt (1)

since you have the equation, we can solve both sides to see if they are equal. if x=0, both sides are not equal. therefore [x] not qual to 0.

so this one is sufficient.

stmt (2)

tells us straight that x is greater than 0, so [x] not equal to zero. therefore this one is sufficient too.

?
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by Rahul@gurome » Fri Aug 20, 2010 8:01 pm
To Gurpinder,

Your reasoning with statement (1) is not correct.
Now [x] = 0 for any value of x such that 0 <= x < 1.
There is no single value of x for which [x] is zero. You are assuming [x] = 0 only if x = 0.
So you have to first solve the equation in statement (1), test whether it is more than or equal to 0 and less than 1 and then decide whether [x] = 0 or not.
Since (1) is giving x as 2/3, [x] = [2/3] = 0.

From statement (2), we have 0 < x < 1. So [x] = 0.
Your conclusion is incorrect for statement (2).

The correct answer is hence (D).
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by Stuart@KaplanGMAT » Sat Aug 21, 2010 8:17 am
Rahul@gurome wrote:To Gurpinder,


So you have to first solve the equation in statement (1), test whether it is more than or equal to 0 and less than 1 and then decide whether [x] = 0 or not.
I don't often disagree with other GMAT experts, but I'm going to do so here.

A key to saving time in DS is to do as few calculations as possible. Remember, we don't care what the answer is, we just care whether it's possible to get an answer.

So, you 100% do NOT need (or want) to solve the equation in statement (1). As soon as you see that (1) will generate a precise value for x, you know that it's also possible to get a precise value for [x]. If we can get a precise value for [x], then we can certainly answer the question "does [x]=0?"

How do we know that (1) will generate a precise value for x? We have 1 linear equation and 1 variable, that's how!
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by [email protected] » Fri Mar 16, 2012 11:01 pm
But there are times in DS when you have to do the specific calculations and see whether there is only one specific answer. Especially with the quadratic equations and parabolas.

There are times when in quadratic equations you have 2 roots both positive and both real. Hence the answer is such a case would be insufficient and if you answer it sufficient, you are committing a mistake.
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