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The number of n satisfying

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by Brent@GMATPrepNow » Wed Oct 18, 2017 7:49 am
LUANDATO wrote:The number of n satisfying -n+2≥0 and 2n≤5 is
A. 0
B. 1
C. 2
D. 3
E. None of the above.
Given: -n+2≥0
Subtract 2 from both sides to get: -n ≥ -2
Divide both sides by -1 to get: n ≤2 [notice that, since we divided by a NEGATIVE value, we REVERSED the direction of the inequality.]

Given: 2n≤5
Divide both sides by 2 to get: n ≤2.5

So, we know that n ≤2 AND n ≤2.5
There are infinitely many values of n that satisfy BOTH inequalities.
Some possible values of n are: 2, 1, 0, -1, -2, -3, 1.9, -7.32123, etc
Answer: E

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Brent
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by EconomistGMATTutor » Wed Oct 18, 2017 9:14 am
The number of n satisfying -n+2≥0 and 2n≤5 is
A. 0
B. 1
C. 2
D. 3
E. None of the above.

The OA is E.

Can any expert illustrate this PS question with an example please?
Hi LUANDATO ,
Let's take a look at your question.

$$-n+2\ge0$$
$$-n\ge-2$$
the inequality is reversed when it is multiplied by a negative number. Therefore,
$$n\le2$$

The next inequality given in the question is,
$$2n\le5$$
$$n\le\frac{5}{2}$$
$$n\le2.5$$

Therefore, n should be restricted to:
$$n\le2$$ and $$n\le2.5$$

By combining both of these inequalities, the values of n should be less than or equal to 2.
It means that n has all the values less than or equal to 2 i.e. n has unlimited number of values.
Therefore, Option E is correct.

Hope this helps.
I am available if you'd like any follow up.
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