Question: Comparing Side Lengths

Comment on Comparing Side Lengths

What if y=5, x=4.11, and w=.9? Then 2w+x would equal 4.92, which is less than 5. Shouldn't the answer be D?
greenlight-admin's picture

Not quite. Plugging in your values, we get:
2w + x = 2(0.9) + 4.11
= 1.8 + 4.11
= 5.91 (which is greater than 5)

What I fail to understand is how can you just assume Y is the bigger side? Let's say w : x : y are 2 : 5 : 6 respectively in this case yes A is the answer but what if w : x : y were the opposite 1: 6 : 6?
greenlight-admin's picture

We aren't assuming that Y is the longest side. We're just using the rule that says "The sum of ANY two sides of a triangle will always be greater than the 3rd side."

Also, in both of your examples, Quantity A is greater than Quantity B.

In your first example, w = 2, x = 5 and y = 6. In this case, 2w + x = 2(2) + 5 = 9. This means Quantity A is 9 and Quantity B is 6 (so, Quantity A is greater)

In your second example, w = 1, x = 6 and y = 6. In this case, 2w + x = 2(1) + 6 = 8. This means Quantity A is 8 and Quantity B is 6 (so, Quantity A is greater)

Can i subtract X from both sides and use the fact that y - x < w
and from there reason that 2w must be greater?
greenlight-admin's picture

Yes! Awesome approach!

In case of equilateral triangle we can say that the difference between two sides is equal to third side....then this above mentioned rule will fail.
greenlight-admin's picture

The rule, |A - B| < 3rd side < |A + B|, applies to all triangles, including equilateral triangles.

Let's say the sides of an equilateral triangle have lengths 7, 7, and x (x is the 3rd side)

So, we get: |7 - 7| < x < |7 + 7|

Simplify: 0 < x < 14

Since x = 7, we get: 0 < 7 < 14.

So, the rule still holds true.

for this, can we use the special triangles we know like 3,4,5 and test it where we make Y different values .. ex: sc1 would be x=3, w=4, y=5
sc2: x=4 w=5 y=3 . Using this approach you can see that in both cases, A is greater.
greenlight-admin's picture

That's a valid approach. However, the only problem with the plugging in numbers approach is that testing two sets of values is not enough to answer the question conclusively, UNLESS they yield two different results.

For more on this, start watching at 2:50 of the following video:


Before watching your solution, here is what I tried.

2w + x versus x

Then using the 3rd side rule I said that w must be greater than the difference of the other two sides, so the minimum w could be is |(x-y)|. So, I have:

2|(x-y)| + x versus x

subtract x from both sides and I get

2|(x-y)| versus 0

which would mean A is greater.

Is this a valid approach? It feels like there might be something off about it.
greenlight-admin's picture

That's a perfect solution. Very inventive!
Nice work!!

Oh, I see what I did. I mis-copied the problem. I should have been comparing 2w + x and Y, not X.

Oh, well. I'll just practice your approach. It seems straight forward, and accurate.
greenlight-admin's picture

Oops. I just assumed your were comparing 2w + x vs x
If that WERE the question, your approach would have been perfect :-)


The Sum of Length of any two sides of a triangle must be greater than the third side.

Based on above law and the question asked-

We need a valid triangle - W,X,Y are side given

Now, if we assume W =1, X=2 then 1+2=3 and 3>Y => possible values are Y= 2,1,0.

Y=0 and Y=1 is not valid as 0 formed no side and if Y=1 then W+x=3 greater than 1, if X+Y=3>W but when W+Y=2 is not greater than X where X=2.

Finally possible triangle values are W=1, X=2, Y=2.

Now Quant A -> 2W+X and Qunat B -> Y

2W+X = 5 and Y=1

Clearly A>B

Brent, is this a valid one?
Also, all your replies are super useful so far. I'm glad to learn MATH being you as my instructor.
greenlight-admin's picture

Hi Vineet,

Although you arrived at the correct answer, there are a couple of issues with your solution that could get you into trouble with future questions:

1) When it comes to Quantitative comparison questions, testing values will never yield definitive results, unless we achieve conflicting results, which means the correct answer is D. I cover this at 2:50 in the following video:

2) Your solution assumes that w, x and y are integers, but the question does not state that this is so.


the rule of the sum and difference of 2 sides, can be applied for all sides regardless whether which one is longest, and still makes sense.

Thanks Brent, my exam is on 9th August wish me luck. :)
greenlight-admin's picture

Good luck! Please let me know how you do.

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