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OG 13 Problem

by oquiella » Tue Dec 22, 2015 3:08 pm
If K is an integer and 2<k<7, for how many different values of k is there a triangle with sides of lengths 2,7, and k?

A. One
B. Two
C. Three
D. Four
E. Five

Answer: A


Why please explain your reasoning
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by MartyMurray » Tue Dec 22, 2015 10:47 pm
If K is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2,7, and k?

A. One
B. Two
C. Three
D. Four
E. Five
For three line segments to work as a triangle, the lengths of any two of the segments have to add up to a sum greater than the length of the third. Otherwise two sides will not be long enough to connect with each other and with the ends of the third side.

The integers k such that 2 < k < 7 are 3, 4, 5 and 6.

See how many of those could form a triangle with sides 2, 7, and k.

If the lengths of the three segments were 2, 7 and 3, the segments 2 and 3 long would not be long enough to connect with each other and with the ends of the segment 7 long to form a triangle.

2 + 4 < 7 Doesn't work.

2 + 5 = 7 The only way to connect all the ends is to put them together flat. That's not a triangle.

2 + 6 > 7 -- 7 + 2 > 6 -- 7 + 6 > 2 That works.

So only K = 6 works and the correct answer is A.
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by [email protected] » Thu Feb 15, 2018 7:05 pm
Hi All,

We're told that K is an integer and 2 < K < 7. We're asked for the number of different values of K that would create a triangle with sides of lengths 2, 7 and K.

This question is based on a math rule called the Triangle Inequality Theorem. In simple terms, it means that if you're dealing with a triangle, then the sum of ANY 2 sides will be greater than the 3rd side. For example, with a 3/4/5 right triangle....

3 + 4 > 5
3 + 5 > 4
4 + 5 > 3
So these 3 lengths (and these 3 inequalities) PROVE that we're dealing with an actual triangle.

In that same way, we can use the rule to determine when we're NOT dealing with an actual triangle... For example, you CANNOT have a triangle with sides of 1, 1 and 100 (because 1+1 is NOT greater than 100).

With this question, we're given two sides: 2 and 7 and we're asked to determine what the third side (K) could be. We're given a range of values for K and we're told that K must be an INTEGER. Since 2 + K must be GREATER than 7, the only possible value that fits all of the given information is K=6. Thus, there's only one value for K.

Final Answer: A

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by Scott@TargetTestPrep » Mon Apr 16, 2018 3:48 pm
oquiella wrote:If K is an integer and 2<k<7, for how many different values of k is there a triangle with sides of lengths 2,7, and k?

A. One
B. Two
C. Three
D. Four
E. Five
Recall the triangle inequality, which states that in order for three values to be the lengths of the sides of a triangle, the sum of the two smaller values must be greater than the third (the largest value).

Since k is an integer between 2 and 7, k can be 3, 4, 5 or 6.

If k = 3, {2, 3, 7} can't be the lengths of the sides of a triangle since 2 + 3 = 5 is not greater than 7.

If k = 4, {2, 4, 7} can't be the lengths of the sides of a triangle since 2 + 4 = 6 is not greater than 7.

If k = 5, {2, 5, 7} can't be the lengths of the sides of a triangle since 2 + 3 = 7 is not greater than 7.

If k = 6, {2, 6, 7} CAN be the lengths of the sides of a triangle since 2 + 6 = 8 is greater than 7.

Answer: A

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Re: OG 13 Problem

by Brent@GMATPrepNow » Wed Jan 29, 2020 12:16 pm
oquiella wrote: ↑
Tue Dec 22, 2015 3:08 pm
 If K is an integer and 2<k<7, for how many different values of k is there a triangle with sides of lengths 2,7, and k?

A. One
B. Two
C. Three
D. Four
E. Five

Answer: A


Why please explain your reasoning
IMPORTANT RULE: If two sides of a triangle have lengths A and B, then . . .
DIFFERENCE between A and B < length of third side < SUM of A and B

We're told that the two KNOWN sides have lengths 2 and 7
So, we can write: (7 - 2) < k < (7 + 2)
Simplify: 5 < k < 9
Since k is an INTEGER, k can equal 6, 7 or 8

However, we're also told that 2 < k < 7, so 6 is the only value that k can have.

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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