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Given four rods of length 1 meter, 3 meters, 5 meters, and 7

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Given four rods of length 1 meter, 3 meters, 5 meters, and 7 [#permalink] New post 03 Jan 2019, 10:29
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Given four rods of length 1 meter, 3 meters, 5 meters, and 7 meters, how many different triangles can be made using one rod for each side?

A. 6
B. 4
C. 3
D. 2
E. 1
[Reveal] Spoiler: OA

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Re: Given four rods of length 1 meter, 3 meters, 5 meters, and 7 [#permalink] New post 03 Jan 2019, 11:39
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ANS=E

this question can best be solved using the triangle strategy that states that: given 2 sides of a triangle, the 3rd side will be greater than the difference of the 2 known sides but less than their sum. To begin solving this question, take the two extreme values provided in the question, 1 and 7. With them, find the range for the 3rd side.
For 1 and 7,
7-1 < 3rd side < 7+1
6 < 3rd side < 8

Since neither 3 nor 5 fall in this range, the rod combinations of 1-5-7 and 1-3-7 cannot form a triangle. Definitely, 1 cannot be a side of the triangle.
Now try 3 and 7

7-3 < 3rd side < 7+3
4 < 3rd side < 10

Here, we observe that 5 falls within this range. Therefore the rod combination of 3-5-7 does indeed form a triangle. That counts as 1 triangle so far.

At this point you can infer and notice that when you try sides 5 and 7, you will obviously end up with a side being 3. However, let's just do it for the love of it.

Trying 5 and 7,

7-5 < 3rd side < 7+5
2 < 3rd side < 12

As we can see, 3 falls within this range. Note that the triangle formed is still the one with sides 3-5-7.
Therefore only 1 triangle can be formed from the given rods. Thus the answer is(E)
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Re: Given four rods of length 1 meter, 3 meters, 5 meters, and 7 [#permalink] New post 04 Jan 2019, 07:08
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GreenlightTestPrep wrote:
Given four rods of length 1 meter, 3 meters, 5 meters, and 7 meters, how many different triangles can be made using one rod for each side?

A. 6
B. 4
C. 3
D. 2
E. 1


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

Let's focus on this part: length of third side < SUM of A and B
We can also say that the length of LONGEST side must be less than the SUM of the other two sides

Let's systematically go through all possible combinations of 3 sides

case a) the LONGEST side has a length of 7 meters
So, 7 must be less than the SUM of the other two sides
This means the remaining 2 sides must have lengths 3 and 5 meters
So, a triangle with lengths 3-5-7 is POSSIBLE
This is the ONLY possible configuration in which the LONGEST side has a length of 7 meters

case b) the LONGEST side has a length of 5 meters
So, 5 must be less than the SUM of the other two (shorter) sides
If 5 is the longest side, then the other 2 sides must have lengths of 1 and 3 meters
HOWEVER, this breaks our rule that says the length of LONGEST side must be less than the SUM of the other two sides
So, we CANNOT have a triangle in which the LONGEST side has a length of 5 meters

case c) the LONGEST side has a length of 3 meters
This cannot work, since there's only one rod that has a length that's less than 1

case d) the LONGEST side has a length of 1 meters
This cannot work

So, there's only 1 possible triangle that can be created.

Answer: E

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Re: Given four rods of length 1 meter, 3 meters, 5 meters, and 7 [#permalink] New post 07 Feb 2019, 18:40
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GreenlightTestPrep wrote:
Given four rods of length 1 meter, 3 meters, 5 meters, and 7 meters, how many different triangles can be made using one rod for each side?

A. 6
B. 4
C. 3
D. 2
E. 1


Since the sum of 2 sides of a triangle must be greater than the 3rd, the only option for the three sides is {3, 5, 7}. We cannot use the rod of length 1 meter in forming any triangles; we can verify that in any choice of three sides including the rod of length 1, there are two sides where the sum of the lengths is less than the length of the third side.

Answer: E
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Re: Given four rods of length 1 meter, 3 meters, 5 meters, and 7   [#permalink] 07 Feb 2019, 18:40
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