So, I just found out I can post questions here instead of emailing the instructor, I have looked through previous posts, but wasn't still quite sure about these.
I can't understand the solutions for these. It seems they take short-cuts??
[Series of Question 2 - From Guide #3]
P. 59 - 4
P. 84 - 18 (Why 20*20?)
P. 203 - 3
P. 218 - 5
P. 220 - 9
P. 224 - 18
P. 225 - 19
I have posted more for different books.
Thanks.
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- sendalot
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Re: Series of Question 2 - From Guide #3
These are great questions, and a lot of them are really hard. Also, a lot of them require pictures that I can't embed in the answers, so be sure to keep your book in front of you while you're reading these over. There are quite a few questions here, so it may be necessary to split each one of these into its own post some time in the future, but for now, I'll put all the answers here.
Guide 3, Page 59 #4
The longest side of an isosceles right triangle is 20*rad(2). What is the area of the triangle?
An isosceles right triangle is a 45-45-90 triangle, because if it has two sides that are the same length, it must have two angles that are the same size. A 45-45-90 triangle's sides are in the ratio of 1:1:rad(2), so if the longest side of the triangle (the hypotenuse) is 20*rad(2), the size of the two legs are both 20.
The area of a triangle is equal to its 1/2 times its base times its height, and in this case, the base and height are both 20. Thus, the area of the triangle is A = 1/2 (20)(20) = 200.
Guide 3, Page 84 #18
The lengths of the two shorter legs of a right triangle add up to 40 units. What is the maximum possible area of the triangle?
This is a hard problem, and I like the explanation of this problem given on page 90.
The rule for finding the largest possible area of a space with limited perimeter is to find the "roundest" space that can be enclosed. We could choose the lengths of our sides to be 39 and 1, and have two legs that added up to a length of 40, but this would be a very pointy-looking triangle that won't contain much area. As the sides get closer to equal, our triangle becomes more round, and it is the roundest when our two legs are of equal length, 20 and 20.
The area of a right triangle with legs 20 and 20 is (1/2)*20*20 = 200.
Guide 3, Page 203 #3
Polygon ABCD is a square. BD is the diagonal, and the length of BD is x.
Quantity A: .6x
Quantity B: The length of side CD.
Because we know that ABCD is a square, let's call its side length s. Since BD is a diagonal of the square, triangle ABD is an isosceles right triangle, which means that its sides are in a ratio of 1:1:rad(2).
Since the sides of the square are of length s and the diagonal is length x, s:x = 1:rad(2), which means s*rad(2) = x.
Now, let's look at the answer choices. Quantity A is .6x. We've just shown that x = s*rad(2), so Quantity A is .6*rad(2)*s. rad(2) is about 1.4, so Quantity A is close to .84*s. Quantity B is the length of side CD, which is just a side of the square, s.
Since s is positive, .84s < s, so the Quantity B will always be larger.
Guide 3, Page 218 #5
See the guide itself for the very complicated picture. F is the midpoint of AE, D is the midpoint of CE. Which of the following must be true? Choose all that apply.
A) FD is parallel to AC
B) Area of Triangle DEF = Area of Triangle CDF
C) Area of triangle ABF < Area of Triangle DEF
D) Angle AFB = Angle BFC
E) Area of triangle ACE = AC x BF
This is a challenging problem, and I'd suggest rereading the answer on page 229 several times, because of how tricky each one of these statements is to prove or disprove.
A is true. Because F and D are midpoints of their respective sides, triangle AEC and FED are similar--they have the exact same angles, but the sides of FED are exactly 1/2 of the size of the sides in AEC. Angles DFE and CAE are "corresponding angles", and because they are equal, AC and DF must be parallel.
B is true. One way to see this is to rotate the triangle so that its base is the line segment CE is the base. To find the height of the triangle, we would need to drop an altitude from the highest point of both triangles, which is point F, to the base, which is at the same level since CD and DE are collinear.
An alternate way is to use trig (which is not necessary for the GRE, but can come in handy!), and use the more general formula for the area of a triangle: A = (1/2)ab*sin(C), where C is the angle between sides a and b. Let CD = DE = a, DF = b. The area of CDF is 1/2 ab*sin(angle CDF), and the area of triangle DEF is 1/2 ab*sin(angle FDE). Angle FDE = 180 - angle CDF, and it is a property of the sine function that sin(180 - x) = sin(x).
Thus, the area of triangle CDF is 1/2 ab sin(CDF) and the area of triangle FDE = 1/2 ab sin(CDF), which are the exact same thing.
We can use this exact argument to show that the area of triangle ACF is equal to the area of triangle FCE. Because angle ABF is right, BF is the height of triangle ACF. The area of triangle ACF is then 1/2 (base) x (height) = 1/2 AC x BF.
Since the Area(ACF) + Area(FCE) = Area(ACE), and Area(ACF) = Area(FCE), 2 x Area(ACF) = Area(ACE). Area(ACF) = 1/2 AC x BF, so Area(ACE) = 2 x 1/2 x AC x BF = AC x BF. Therefore E is true.
C and D are false. It is hard to prove this in words--there is a picture in the answers that demonstrates that C and D both aren't necessarily true, but here's an example that satisfies the conditions in the problem. Let ACE be an isosceles right triangle with hypotenuse AE. In this case, B will be the midpoint of AC, and ABF will have the exact same area as DEF. This disproves C.
To prove D is false, let ACE be an equilateral triangle. Angles ACF and CFE will be 90 degrees, because the median of an equilateral triangle is also its altitude. Angle CFD will be 30 degrees, which means angle BFC will be 60. Angle AFB will therefore be 30 degrees, disproving D.
Guide 3, Page 220 #9
See picture. In the figure above, circle O has radius 8, and AD is parallel to BC. If the length of arc AYD is twice the length of minor arc BXC,what is the length of arc BXC?.
For this problem, we need to follow the information we are given. BC is parallel to AD , so angle BCA and CAD are alternate interior angles, so they must be equal. Angle BCA = 45 degrees.
Angles BCA and CAD are inscribed angles, which means they are both equal to 1/2 of the arc that they cut off. Thus, Arc AB = Arc CD = 90 degrees.
We can divide the entire circle's circumference into arcs AB, BXC, CD, and DYA with no overlap.
360 degrees = AB + BXC + CD + DYA
360 = 90 + BXC + 90 + DYA
Arc DYA = 2 * Arc BXC, so we have
360 = 180 + BXC + 2*BXC
180 = 3*BXC
60 = BXC
Now that we know arc BXC is 60 degrees, we can compute its length using the formula Arc Length = Circumference * (arc degrees/360).
Circumference = 2pi*radius = 2pi*8 = 16pi.
Length of BXC = 16pi * 60/360
Length of BXC = 16pi * 1/6
Length of BXC = 8pi/3
The correct answer is B.
Guide 3, Page 224 #18
See picture. A and B are circles with identical radii, such that the center of circle A is on the circumference of circle B, and vice versa. What is the area of the overlap between circle A and circle B?
This is a classic super-challenging problem, and unfortunately, the easiest way to solve it is exactly how it's written out in the answers on page 236. I strongly suggest you reread that a couple of times, and I'll try to give you some intuition behind how to do this problem in general.
The overlap area we're looking at is pretty close to a sector of a circle, but with circular arcs instead of lines for edges. The first thing we should do is to try to find the size of this arc. If we put points X and Y at the intersection points between circles A and B, and draw lines AX, AY, BX, BY, and AB, we have that triangles ABX and ABY are equilateral. Why? Because all of the line segments we just drew are radii of circle A or circle B, and A and B have the same radius.
Angle YAX = angle XAB + angle BAY, and since triangles ABX and ABY are equilateral, angle XAB = angle YAB = 60 degrees.
Angle YAX = 60 + 60 = 120 degrees.
Now, let's approximate the area of the shaded region by the area of the sector of circle A that corresponds to central angle YAX The area of the sector is equal to the area of the circle * (central angle/360).
The area of the circle is pi*x^2, so the area of the sector is (120/360)*pi*x^2 = 1/3 * pi * x^2.
Now let's pause for a moment. Even if you had absolutely no idea of how to compute the area of the circular pieces we chopped out when dealing with the sector, we know that the shaded area is going to be a little bigger than the area of the sector, which was 1/3*pi*x^2.
We also know that the shaded area is less than half the area of either circle, 1/2*pi*x^2. To see this, split circle A into a left half and a right half. There are points in the right half of the circle that are not in the shaded region.
So right here, we can *eliminate* any answer choice that is less than 1/3*pi*x^2, and any answer that is bigger than 1/2 * pi * x^2. Answer choices A and B are too small, while answer choices D and E are too large. Therefore answer choice C is correct.[\b]
[b]Guide 3, Page 225 #19
In the xy-plane, line n is a line that passes through the origin. Which of the following statements individually provides sufficient evidence to determine whether the slope of line n is greater than 1? Indicate all such statements.
A) Line n does not pass through any point (a,b) where a and b are positive and a > b.
B) Line m is perpendicular to line n and has a slope of -1.
C) Line n passes through the point (c,d+1) where c and d are consecutive integers and c > d.
.
If line n passes through the origin, its y-intercept must be zero, so the equation for line n will be y = kx.
A is true. For logic problems like these, it will help us to sometimes prove the contrapositive of a statement. For a statement If p, then q, the contrapositive is if not q, then not p. A statement is true if and only if its contrapositive is true. Here's an example of a statement:
If it is raining outside, the ground is wet.
This is true, if and only if its contrapositive is true. The contrapositive is:
If the ground is not wet, it is not raining outside.
Instead of showing that if line n does not pass through any point (a,b) where a and b are positive and a > b, then line n will not have a slope greater than 1, we will prove the contrapositive. If line n has a slope greater than one, it will pass through a point (a,b) where a and b are positive and a > b.
Let's look at the point on line n where x = 1. The equation of line n is y = kx, so this point is (k, 1). But since we know that the slope of the line is greater than 1, k must be greater than 1. If we let a = k and b = 1, then line n passes through a point (a,b) where a > b and a and b are both positive. We have proven the contrapositive of the statement in answer choice A, so answer choice A must be true.
B is true. If line m is perpendicular to line n, then it has the opposite reciprocal slope. If line m's slope is -1, that means that line n's slope is -(1/-1) = 1, which is sufficient to prove that the slope of the line is not greater than 1.
C is true. If c and d are consecutive integers and c > d, then c = d+1. If line n passes through point (c,d+1), it passes through the point (c, c).
Then line n goes through points (0,0) and (c,c), which means that the slope of line n is equal to (c-0)/(c-0) = c/c = 1, which is sufficient to prove that the slope is not greater than 1.
Whew! That's a lot to chew on. Keep working these tough problems--eventually they'll come naturally to you.
Michael
Guide 3, Page 59 #4
The longest side of an isosceles right triangle is 20*rad(2). What is the area of the triangle?
An isosceles right triangle is a 45-45-90 triangle, because if it has two sides that are the same length, it must have two angles that are the same size. A 45-45-90 triangle's sides are in the ratio of 1:1:rad(2), so if the longest side of the triangle (the hypotenuse) is 20*rad(2), the size of the two legs are both 20.
The area of a triangle is equal to its 1/2 times its base times its height, and in this case, the base and height are both 20. Thus, the area of the triangle is A = 1/2 (20)(20) = 200.
Guide 3, Page 84 #18
The lengths of the two shorter legs of a right triangle add up to 40 units. What is the maximum possible area of the triangle?
This is a hard problem, and I like the explanation of this problem given on page 90.
The rule for finding the largest possible area of a space with limited perimeter is to find the "roundest" space that can be enclosed. We could choose the lengths of our sides to be 39 and 1, and have two legs that added up to a length of 40, but this would be a very pointy-looking triangle that won't contain much area. As the sides get closer to equal, our triangle becomes more round, and it is the roundest when our two legs are of equal length, 20 and 20.
The area of a right triangle with legs 20 and 20 is (1/2)*20*20 = 200.
Guide 3, Page 203 #3
Polygon ABCD is a square. BD is the diagonal, and the length of BD is x.
Quantity A: .6x
Quantity B: The length of side CD.
Because we know that ABCD is a square, let's call its side length s. Since BD is a diagonal of the square, triangle ABD is an isosceles right triangle, which means that its sides are in a ratio of 1:1:rad(2).
Since the sides of the square are of length s and the diagonal is length x, s:x = 1:rad(2), which means s*rad(2) = x.
Now, let's look at the answer choices. Quantity A is .6x. We've just shown that x = s*rad(2), so Quantity A is .6*rad(2)*s. rad(2) is about 1.4, so Quantity A is close to .84*s. Quantity B is the length of side CD, which is just a side of the square, s.
Since s is positive, .84s < s, so the Quantity B will always be larger.
Guide 3, Page 218 #5
See the guide itself for the very complicated picture. F is the midpoint of AE, D is the midpoint of CE. Which of the following must be true? Choose all that apply.
A) FD is parallel to AC
B) Area of Triangle DEF = Area of Triangle CDF
C) Area of triangle ABF < Area of Triangle DEF
D) Angle AFB = Angle BFC
E) Area of triangle ACE = AC x BF
This is a challenging problem, and I'd suggest rereading the answer on page 229 several times, because of how tricky each one of these statements is to prove or disprove.
A is true. Because F and D are midpoints of their respective sides, triangle AEC and FED are similar--they have the exact same angles, but the sides of FED are exactly 1/2 of the size of the sides in AEC. Angles DFE and CAE are "corresponding angles", and because they are equal, AC and DF must be parallel.
B is true. One way to see this is to rotate the triangle so that its base is the line segment CE is the base. To find the height of the triangle, we would need to drop an altitude from the highest point of both triangles, which is point F, to the base, which is at the same level since CD and DE are collinear.
An alternate way is to use trig (which is not necessary for the GRE, but can come in handy!), and use the more general formula for the area of a triangle: A = (1/2)ab*sin(C), where C is the angle between sides a and b. Let CD = DE = a, DF = b. The area of CDF is 1/2 ab*sin(angle CDF), and the area of triangle DEF is 1/2 ab*sin(angle FDE). Angle FDE = 180 - angle CDF, and it is a property of the sine function that sin(180 - x) = sin(x).
Thus, the area of triangle CDF is 1/2 ab sin(CDF) and the area of triangle FDE = 1/2 ab sin(CDF), which are the exact same thing.
We can use this exact argument to show that the area of triangle ACF is equal to the area of triangle FCE. Because angle ABF is right, BF is the height of triangle ACF. The area of triangle ACF is then 1/2 (base) x (height) = 1/2 AC x BF.
Since the Area(ACF) + Area(FCE) = Area(ACE), and Area(ACF) = Area(FCE), 2 x Area(ACF) = Area(ACE). Area(ACF) = 1/2 AC x BF, so Area(ACE) = 2 x 1/2 x AC x BF = AC x BF. Therefore E is true.
C and D are false. It is hard to prove this in words--there is a picture in the answers that demonstrates that C and D both aren't necessarily true, but here's an example that satisfies the conditions in the problem. Let ACE be an isosceles right triangle with hypotenuse AE. In this case, B will be the midpoint of AC, and ABF will have the exact same area as DEF. This disproves C.
To prove D is false, let ACE be an equilateral triangle. Angles ACF and CFE will be 90 degrees, because the median of an equilateral triangle is also its altitude. Angle CFD will be 30 degrees, which means angle BFC will be 60. Angle AFB will therefore be 30 degrees, disproving D.
Guide 3, Page 220 #9
See picture. In the figure above, circle O has radius 8, and AD is parallel to BC. If the length of arc AYD is twice the length of minor arc BXC,what is the length of arc BXC?.
For this problem, we need to follow the information we are given. BC is parallel to AD , so angle BCA and CAD are alternate interior angles, so they must be equal. Angle BCA = 45 degrees.
Angles BCA and CAD are inscribed angles, which means they are both equal to 1/2 of the arc that they cut off. Thus, Arc AB = Arc CD = 90 degrees.
We can divide the entire circle's circumference into arcs AB, BXC, CD, and DYA with no overlap.
360 degrees = AB + BXC + CD + DYA
360 = 90 + BXC + 90 + DYA
Arc DYA = 2 * Arc BXC, so we have
360 = 180 + BXC + 2*BXC
180 = 3*BXC
60 = BXC
Now that we know arc BXC is 60 degrees, we can compute its length using the formula Arc Length = Circumference * (arc degrees/360).
Circumference = 2pi*radius = 2pi*8 = 16pi.
Length of BXC = 16pi * 60/360
Length of BXC = 16pi * 1/6
Length of BXC = 8pi/3
The correct answer is B.
Guide 3, Page 224 #18
See picture. A and B are circles with identical radii, such that the center of circle A is on the circumference of circle B, and vice versa. What is the area of the overlap between circle A and circle B?
This is a classic super-challenging problem, and unfortunately, the easiest way to solve it is exactly how it's written out in the answers on page 236. I strongly suggest you reread that a couple of times, and I'll try to give you some intuition behind how to do this problem in general.
The overlap area we're looking at is pretty close to a sector of a circle, but with circular arcs instead of lines for edges. The first thing we should do is to try to find the size of this arc. If we put points X and Y at the intersection points between circles A and B, and draw lines AX, AY, BX, BY, and AB, we have that triangles ABX and ABY are equilateral. Why? Because all of the line segments we just drew are radii of circle A or circle B, and A and B have the same radius.
Angle YAX = angle XAB + angle BAY, and since triangles ABX and ABY are equilateral, angle XAB = angle YAB = 60 degrees.
Angle YAX = 60 + 60 = 120 degrees.
Now, let's approximate the area of the shaded region by the area of the sector of circle A that corresponds to central angle YAX The area of the sector is equal to the area of the circle * (central angle/360).
The area of the circle is pi*x^2, so the area of the sector is (120/360)*pi*x^2 = 1/3 * pi * x^2.
Now let's pause for a moment. Even if you had absolutely no idea of how to compute the area of the circular pieces we chopped out when dealing with the sector, we know that the shaded area is going to be a little bigger than the area of the sector, which was 1/3*pi*x^2.
We also know that the shaded area is less than half the area of either circle, 1/2*pi*x^2. To see this, split circle A into a left half and a right half. There are points in the right half of the circle that are not in the shaded region.
So right here, we can *eliminate* any answer choice that is less than 1/3*pi*x^2, and any answer that is bigger than 1/2 * pi * x^2. Answer choices A and B are too small, while answer choices D and E are too large. Therefore answer choice C is correct.[\b]
[b]Guide 3, Page 225 #19
In the xy-plane, line n is a line that passes through the origin. Which of the following statements individually provides sufficient evidence to determine whether the slope of line n is greater than 1? Indicate all such statements.
A) Line n does not pass through any point (a,b) where a and b are positive and a > b.
B) Line m is perpendicular to line n and has a slope of -1.
C) Line n passes through the point (c,d+1) where c and d are consecutive integers and c > d.
.
If line n passes through the origin, its y-intercept must be zero, so the equation for line n will be y = kx.
A is true. For logic problems like these, it will help us to sometimes prove the contrapositive of a statement. For a statement If p, then q, the contrapositive is if not q, then not p. A statement is true if and only if its contrapositive is true. Here's an example of a statement:
If it is raining outside, the ground is wet.
This is true, if and only if its contrapositive is true. The contrapositive is:
If the ground is not wet, it is not raining outside.
Instead of showing that if line n does not pass through any point (a,b) where a and b are positive and a > b, then line n will not have a slope greater than 1, we will prove the contrapositive. If line n has a slope greater than one, it will pass through a point (a,b) where a and b are positive and a > b.
Let's look at the point on line n where x = 1. The equation of line n is y = kx, so this point is (k, 1). But since we know that the slope of the line is greater than 1, k must be greater than 1. If we let a = k and b = 1, then line n passes through a point (a,b) where a > b and a and b are both positive. We have proven the contrapositive of the statement in answer choice A, so answer choice A must be true.
B is true. If line m is perpendicular to line n, then it has the opposite reciprocal slope. If line m's slope is -1, that means that line n's slope is -(1/-1) = 1, which is sufficient to prove that the slope of the line is not greater than 1.
C is true. If c and d are consecutive integers and c > d, then c = d+1. If line n passes through point (c,d+1), it passes through the point (c, c).
Then line n goes through points (0,0) and (c,c), which means that the slope of line n is equal to (c-0)/(c-0) = c/c = 1, which is sufficient to prove that the slope is not greater than 1.
Whew! That's a lot to chew on. Keep working these tough problems--eventually they'll come naturally to you.
Michael
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Re: Series of Question 2 - From Guide #3
Hello Sir,
I wanted to get your attention to the Guide 3, problem 19. The only option that are true(according to strategy guide) is, B & C. Was it a typo from your end? if not, could you please justify, Why you have included option A in the answer choice?
I wanted to get your attention to the Guide 3, problem 19. The only option that are true(according to strategy guide) is, B & C. Was it a typo from your end? if not, could you please justify, Why you have included option A in the answer choice?
- tommywallach
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Re: Series of Question 2 - From Guide #3
Hey Video,
This thread is kinda silly, in that it's full of questions but no one can know which ones without clicking the link, which makes it pretty useless. Would you mind posting a new thread for your question, with the actual question # in the subject line. Thanks!
-t
This thread is kinda silly, in that it's full of questions but no one can know which ones without clicking the link, which makes it pretty useless. Would you mind posting a new thread for your question, with the actual question # in the subject line. Thanks!
-t
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