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Post subject: GMAT Geometry Posted: Mon Feb 21, 2011 6:55 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

Let q represent the integer length of a side of a triangle. If r represents the number of distinct values for q such that we can create a triangle with lengths q, 9, and 13, what is the value of r? A. 5 B. 17 C. 18 D. 22 E. 29
(B) From the triangle inequality rule, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. So, the three inequalities are q + 9 > 13, 9 + 13 > q, and q + 13 > 9.
The third equation holds for any positive value of q. So solve the first two inequalities for q, so q > 4 and 22 > q. So the value of q is 4 < q < 22.
Since the lengths of the sides of the triangle are of integer value, the lowest possible integer/side can be 5 and the highest possible integer/side can be 21 (q = 5, 6, 7, 8, 9 ..., 19, 20, or 21). So, r, the number of numbers in this set, must equal 21 – (5 – 1) = 17. The correct answer is B.  Why don't you solve the third equation so that 4 < q < 22? (thus becoming 0 < q < 22)


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Gennadiy

Post subject: Re: math (test 1, question 22): geometry, inequalities Posted: Mon Feb 21, 2011 7:19 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

Solving the third inequality doesn't give us anything new. We know that q > 0 from the original statement.
The range for q must comply with all three inequalities, NOT just some of them. The first inequality, q + 9 > 13, yields q > 4. Therefore in the solution range q must be greater than 4. You can check that q = 1, 2, 3 or 4 do not fit in the first inequality, q + 9 > 13.
The set of all the possible values of q is {5, 6, 7, ... ,20, 21}.


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lizonnat

Post subject: Re: math (test 1, question 22): geometry, inequalities Posted: Wed Mar 02, 2011 5:44 pm 

Joined: Wed Mar 02, 2011 5:30 pm Posts: 2

Ok great thanks, I see it now.
P.s. why is the formula to determine the number of numbers in a set equal to (last number  first number + 1 ) ?


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Gennadiy

Post subject: Re: math (test 1, question 22): geometry, inequalities Posted: Sat Mar 12, 2011 6:15 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

When we count the number of consecutive integers from m to k inclusive, we use the following formula:
k – (m – 1) I advise you to remember it in this way and here is the explanation:
Imagine we count consecutive integers from 1 to n. It's clear that there are n such integers. (Note, by the way, that n = n – (1 – 1), NOT n = n – 1)
If we number the elements of the set {m, m+1, ..., k – 1, k} by 1, 2, ... the last integer will be k – (m – 1), because any integer that we number an element with is (m – 1) units less than that element.
m, m + 1, m + 2, ..., k – 1, k 1, 2, 3, ... , (k – 1) – (m – 1), k – (m – 1)
Similarly, as with consecutive integers from 1 to n, the're k – (m – 1) such integers.


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