(Q 159) GRE Quantitative Comparison

(Col A) the area of a triangle with sides 2, 7, and 7
(Col B) the area of a triangle with sides 4, 4, and 4

These are dimensions of isosceles triangles. And since isosceles triangles can be broken down into two identical right triangles, it is easy to compute their areas when you know the side lengths. In the drawing below, Triangle A has a height of x, and Triangle B has a height of y.

We can find x by solving 7² = x² + 1², or x = sqrt(48). Similarly, we can find y by solving 4² = y² + 2², which gives us y = sqrt(12). Using the base-height formula, we compute the area of Triangle A as sqrt(48), and the area of Triangle B as 2sqrt(12).

Both sqrt(48) and 2sqrt(12) can be simplified to 4sqrt(3), so the areas of these triangles are equal. Thus, the answer is C.

(Q 158) GMAT Data Sufficiency

Source: http://gmatgremath.blogspot.com

If m and n are positive numbers, is sqrt(m²n² + 676) ≥ sqrt(4m²n² + 1) ?

(1) 24 ≥ mn
(2) m ≥ 16/n

As it is phrased, the question is not in the most simplified form. The statements will be much easier to evaluate if we do some algebra to simplify the original expression. One must be careful in simplifying the square roots correctly; this question is made to trip test takers who don't know the correct rules. To illustrate where the trap is, we will simplify it the right way and the wrong way.

Right Way:
First, square both sides of "sqrt(m²n² + 676) ≥ sqrt(4m²n² + 1) ?" to get
"m²n² + 676 ≥ 4m²n² + 1 ?"
Then, subtract m²n² and 1 from both sides to get
"675 ≥ 3m²n² ?"
Now, divide both sides by 3 to get
"225 ≥ m²n² ?"
Lastly, take the square root of both sides to get
"15 ≥ mn ?"
So, what the question is really asking is whether 15 ≥ mn. Pretty easy!

Statement (1) is insufficient because it only tells us that mn is less than or equal to 24, but that doesn't tell us whether mn is less than or equal to 15.
Statement (2) is equivalent to mn ≥ 16; this is sufficient to answer the question with a definite "no." Thus, the correct answer is B.

Here is the Wrong Way:
Break up "sqrt(m²n² + 676) ≥ sqrt(4m²n² + 1) ?" into
"sqrt(m²n²) + sqrt(676) ≥ sqrt(4m²n²) + sqrt(1) ?"
Then, evaluate all the square roots to get
"mn + 26 ≥ 2mn + 1 ?"
Now, subtract mn and 1 from both sides to get
"25 ≥ mn"

Notice that if you ended up with this inequality, Statement (1) would be sufficient, and Statement (2) would be insufficient. That's why it's important to simplify square roots the right way!

(Q 157) GMAT/GRE Angles

Source: http://gmatgremath.blogspot.com

In the figure above, ABCD is a semi circular arc, AD is a diameter, and R is the midpoint of AD. What is the value of ∠RBD + ∠RCA?

(A) 22.5°
(B) 30°
(C) 40°
(D) 45°
(E) not enough info

For this problem, it is helpful (in fact, necessary!) to know a key property of interior angles within circles. No matter where B and C lie on the the outside edge of the circle, these two relations will always be true:

∠CAD = ½ ∠CRD
∠BDA = ½ ∠BRA

And since triangles ∆CAR and ∆BDR are isosceles, we also have

∠CAR = ∠RCA
∠BDR = ∠RBD

Thus, we can fill in the original figure with more relations like so:
We know that x + y has to equal 90, and so ½(x + y) = 45. This is precisely value of ∠RBD + ∠RCA, so the correct answer is D.

*Bonus Question* How many words can you make from the letters A, B, C, D, and R?

(Q 156) GMAT/GRE Word Problem w/ Proportions

Source: http://gmatgremath.blogspot.com

Zach is going to mix three water-based solutions. Solution X is 40% alcohol, Solution Y is 30% sugar, and Solution Z is 30% alcohol and 10% sugar. If the final product is 10% alcohol and 23% sugar, what is the ratio of Solution X to Solution Z that Zach used?

(A) 2:3
(B) 1:2
(C) 1:1
(D) 2:1
(E) 4:3

Let's designate x as the proportion of Solution X, z as the proportion of Solution Z, and 1-x-z as the proportion of Solution Y. Remember, all the proportions (or percents, or probabilities) must add up to 1. The number we want to find is x/z.

To help illustrate things, this table shows the proportion of ingredients in each solution.

And this table illustrates how the final product is made from various proportions of the three solutions.
The final product is 67% water, 10% alcohol, and 23% sugar, which adds up to 1. The columns circled in red represent the system of equations we need to solve:
.4x + .3z = .1
.3(1-x-z) + .1z = .23,

which can be simplified to
4x + 3z = 1
3 - 3x - 3z + z = 2.3

With some algebra, you find the solution, which is x = .1 and z = .2. Thus, the ratio of x to z is 1/2. So the right answer is B.

(Q 155) GMAT/GRE 3 Statements

Source: http://gmatgremath.blogspot.com

Which of the following must be true for positive integers m and n greater than 1?

(i) (mn)! / (m+n)! is an integer
(ii) (m+n)! / (n!m!) is an integer
(iii) (m!)² / (n²)! is an integer, whenever m ≥ n

(A) i and ii
(B) i and iii
(C) ii and iii
(D) i, ii, and iii
(E) none

Starting with (i), we should recall the following property of factorials: a!/b! is always an integer if a ≥ b. Since m and n are integers greater than 1, it will always be true that mn ≥ m+n. Thus, (mn)!/(m+n)! is an integer. Therefore, we can eliminate choices C and E.

Moving on to (ii), this should remind you of the combination formula. Recall, if you have p objects, the number of ways to choose q objects is (p C q) = p!/[q!(p-q)!]. In this case, we have m+n objects and are choosing n. Thus, (m+n C n) = (m+n)!/[n!(m+n-n)!] = (m+n)!/n!m!. The combination formula always yields an integer. Thus, (ii) is true, and we can eliminate B.

Statement (iii) is trickier, but a good way to judge its veracity is simply to plug in numbers for m and n. We need to choose m at least as large as n, and both integers greater than 1, so let's try m = 3, and n = 2. Then,

(3!)² / (2²)! = 6²/4!
= 36/24 =1.5

So (iii) is not always true. Therefore, the correct answer is A.