The GMAT Quant section only tests concepts that you have seen until high school mostly. There are no integrals, derivatives, trigonometry questions or anything that requires a cientific calculator. So why doesn't everyone score 51, the highest score?
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The first thing to understand is that not all of the scores are distributed equally. According to GMAT statistics, the median score is at about 44, and the biggest jumps happen from score 48 to 51 as you can see in the following chart:
Score | Percentile |
---|---|
51 | 96% |
50 | 85% |
49 | 74% |
48 | 67% |
47 | 60% |
46 | 57% |
45 | 54% |
There are a couple of factors that separate students that are at level 45 from students at level 48 or 51.
In the following example I will show how a student from each level solves the question differently.
The triangles in the figure above are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2 to 1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?
a) \(\frac { 3 }{ 4 } K\)
b) \(\frac { 2 }{ 3 } K\)
c) \(\frac { 1 }{ 2 } K\)
d) \(\frac { 1 }{ 3 } K\)
e) \(\frac { 1 }{ 4 } K\)
First identify the formula \(Area=\frac { b\times h }{ 2 }\)
In order to find the height, use the Pythagorean theorem. If we call the side \(L\), than dividing the triangle in two gives me a base of \(\frac { L }{ 2 }\). Using the formula the height is \(\frac { L }{ 2 } \sqrt { 3 }\).
Let's say the large triangle has side \(2x\) and the small triangle has side \(x\).
The area \(K\) of the large triangle would be \(K=\frac { (2x)(x\sqrt { 3 } ) }{ 2 }\). Simplifying: \(K={ x }^{ 2 }\sqrt { 3 }\)
The shaded area is the large area minus the small area, therefore:
\(\frac { (2x)(x\sqrt { 3 } ) }{ 2 } -\frac { (x)(\frac { x }{ 2 } \sqrt { 3 } ) }{ 2 }\)
\({ x }^{ 2 }\sqrt { 3 } -\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 } =\frac { 3({ x }^{ 2 }\sqrt { 3 } ) }{ 4 }\)
â€‹Replacing \(K\):
\(\frac { 3 }{ 4 } K\)
First identify the direct formula for area of equilateral triangles of side \(L\): \(Area=\frac { { L }^{ 2 }\sqrt { 3 } }{ 4 }\)
Let's say the large triangle has side \(2x\) and the small triangle has side \(x\).
The area \(K\) of the large triangle would be \(K=\frac { { (2x) }^{ 2 }\sqrt { 3 } }{ 4 } ={ x }^{ 2 }\sqrt { 3 }\)
The shaded area is the large area minus the small area, and we can replace \(K\) wherever it appears:
\(\frac { { (2x) }^{ 2 }\sqrt { 3 } }{ 4 } -\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 } =\frac { 3 }{ 4 } { x }^{ 2 }\sqrt { 3 } =\frac { 3 }{ 4 } K\)
(Mostly in his or her mind) First recognize the most efficient property: proportional figures have areas that are in the same relation as the sides, but squared. If the sides have a proportion of 2 to 1, the areas have a proportion 4 to 1. Therefore the shaded region is \(\frac { 3 }{ 4 }\) the area of the large triangle \(K\).
Answer: \(\frac { 3 }{ 4 } K\)
What really separates the students that score well versus those that don't is not so much the formulas, but other factors such as using efficient properties, operating less and having the agility that is gained from a lot of practice and reflection.
Whatever your current level, now you have a clear idea of what is expected at a higher level, so you can work towards achieving that new level. Also, if you want more in depth training, you can check out my complete GMAT Quant course here.