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1) Let abcd be a general four-digit number and all the digits are non-zero. How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d?
(A) 6
(B) 7
(C) 24
(D) 36
(E) 42
2) Let abcd be a general four-digit number. How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two digit number cd?
(A) 4
(B) 6
(C) 12
(D) 24
(E) 36
3) There are 500 cars on a sales lot, all of which have either two doors or four doors. There are 165 two-door cars on the lot. There are 120 four-door cars that have a back-up camera. Eighteen percent of all the cars with back-up cameras have standard transmission. If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?
(A) 18
(B) 27
(C) 36
(D) 45
(E) 54
4) At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4^{th} or 5^{th} grade. There are 320 students in the 4^{th} grade, and there are 210 girls in the 5^{th} grade. Fifty percent of the 5^{th} graders and 40% of the 4^{th} graders take Mandarin Chinese. Ninety 5^{th} grade boys do not take Mandarin Chinese. The number of 4^{th} grade girls taking Mandarin Chinese is less than half of the number of 5^{th} grade girls taking Mandarin Chinese. Which of the following could be the number of 4^{th} grade boys in Mandarin Chinese?
(A) 10
(B) 40
(C) 70
(D) 100
(E) 130
5) A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube. These four are not touching each other. All outward faces are painted and all inward faces are not painted. These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square. The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom. How many individual faces will have to be painted to accommodate the requirements of this new design?
(A) 0
(B) 5
(C) 9
(D) 16
(E) 27
6) Twelve points are spaced evenly around a circle, lettered from A to L. Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices. What is the value of N?
(A) 48
(B) 52
(C) 60
(D) 72
(E) 120
7) Theresa is a basketball player practicing her free throws. On her first free throw, she has a 60% chance of making the basket. If she has just made a basket on her previous throw, she has a 80% of making the next basket. If she has just failed to make a basket on her previous throw, she has a 40% of making the next basket. What is the probability that, in five throws, she will make at least four baskets?
8) Suppose a “Secret Pair” number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other. For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not. How many “Secret Pair” numbers are there?
(A) 720
(B) 1440
(C) 1800
(D) 1944
(E) 2160
9) In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, – 6). What is the area of the part of this circle in the first quadrant?
10) In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84). Which of the following could be the slope of line L?
(A) 0
(B) 1/2
(C) 1/4
(D) 2/5
(E) 6/7
11) At the beginning of the year, an item had a price of A. At the end of January, the price was increased by 60%. At the end of February, the new price was decreased by 60%. At the end of March, the new price was increased by 60%. At the end of April, the new price was decreased by 60%. On May 1^{st}, the final price was approximately what percent of A?
(A) 41%
(B) 64%
(C) 100%
(D) 136%
(E) 159%
12) Suppose that, at current exchange rates, $1 (US) is equivalent to Q euros, and 1 euro is equivalent to 7Q Chinese Yuan. Suppose that K kilograms of Chinese steel, worth F Chinese Yuan per kilogram, sold to a German company that paid in euros, can be fashioned into N metal frames for chairs. These then are sold to an American company, where plastic seats & backs will be affixed to these frames. If the German company made a total net profit of P euros on this entire transaction, how much did the US company pay in dollars for each frame?
13) At the Zamenhof Language School, at least 70% of the students take English each year, at least 40% take German each year, and between 30% and 60% take Italian each year. Every student must take at least one of these three languages, and no student is allowed to take more than two languages in the same year. What is the possible percentage range for students taking both English and German in the same year?
(A) 0% to 70%
(B) 0% to 100%
(C) 10% to 70%
(D) 10% to 100%
(E) 40% to 70%
14) On any given day, the probability that Bob will have breakfast is more than 0.6. The probability that Bob will have breakfast and will have a sandwich for lunch is less than 0.5. The probability that Bob will have breakfast or will have a sandwich for lunch equals 0.7. Let P = the probability that, on any given day, Bob will have a sandwich for lunch. If all the statements are true, what possible range can be established for P?
(A) 0 < P < 0.6
(B) 0 ≤ P < 0.6
(C) 0 ≤ P ≤ 0.6
(D) 0 < P < 0.7
(E) 0 ≤ P < 0.7
(A) – 64
(B) – 7
(C) 38
(D) 88
(E) 128
Explanations for this problem are at the end of this article.
Here are twenty-eight other articles on this blog with free GMAT Quant practice questions. Some have easy questions, some have medium, and few have quite challenging questions.
1) GMAT Geometry: Is It a Square?
2) GMAT Shortcut: Adding to the Numerator and Denominator
3) GMAT Quant: Difficult Units Digits Questions
4) GMAT Quant: Coordinate Geometry Practice Questions
5) GMAT Data Sufficiency Practice Questions on Probability
6) GMAT Quant: Practice Problems with Percents
7) GMAT Quant: Arithmetic with Inequalities
8) Difficult GMAT Counting Problems
9) Difficult Numerical Reasoning Questions
10) Challenging Coordinate Geometry Practice Questions
11) GMAT Geometry Practice Problems
12) GMAT Practice Questions with Fractions and Decimals
13) Practice Problems on Powers and Roots
14) GMAT Practice Word Problems
15) GMAT Practice Problems: Sets
16) GMAT Practice Problems: Sequences
17) GMAT Practice Problems on Motion
18) Challenging GMAT Problems with Exponents and Roots
19) GMAT Practice Problems on Coordinate Geometry
20) GMAT Practice Problems: Similar Geometry Figures
20) GMAT Practice Problems: Variables in the Answer Choices
21) Counting Practice Problems for the GMAT
22) GMAT Math: Weighted Averages
23) GMAT Data Sufficiency: More Practice Questions
24) Intro to GMAT Word Problems, Part I
25) GMAT Data Sufficiency Geometry Practice Questions
26) GMAT Data Sufficiency Logic: Tautological Questions
27) GMAT Quant: Rates and Ratios
28) Absolute Value Inequalities
These are hard problems. When you read the solutions, don’t merely read them passively. Study the strategies used, and do what you can to retain them. Learn from your mistakes!
1) We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d. There are seven possibilities:
For each set, the sum-digit has to be in the one’s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits. Thus, for each item on that list, there are six different possible four-digit numbers. The total number of possible four-digit numbers would be 7*6 = 42. Answer = (E)
2) The fact that abcd is odd means that cd must be an odd number and that a & b both must be odd. That limits the choices significantly. We know that neither a nor b can equal 1, because any single digit number times 1 is another single digit number, and we need a two-digit product—there are no zeros in abcd. We also know that neither a nor b can equal 5, because any odd multiple of 5 ends in 5, and we would have a repeated digit: the requirement is that all four digits be distinct.
Therefore, for possible values for a & b, we are limited to three odd digits {3, 7, 9}. We can take three different pairs, and in each pair, we can swap the order of a & b. Possibilities:
Those six are the only possibilities for abcd.
Answer = (B)
3) Total number of cars = 500
2D cars total = 165, so
4D cars total = 335
120 4D cars have BUC
“Eighteen percent of all the cars with back-up cameras have standard transmission.”
18% = 18/100 = 9/50
This means that the number of cars with BUC must be a multiple of 50.
How many 2D cars can we add to 120 4D cars to get a multiple of 50? We could add 30, or 80, or 130, but after that, we would run out of 2D cars. These leaves three possibilities for the total number with BUC:
If a total of 150 have BUC, then 18% or 27 of them also have ST.
If a total of 200 have BUC, then 18% or 36 of them also have ST.
If a total of 250 have BUC, then 18% or 45 of them also have ST.
Then we are told: “40% of all the cars with both back-up cameras and standard transmission are two-door car.”
40% = 40/100 = 2/5
This means that number of cars with both back-up cameras and standard transmission must be divisible by 5. Of the three possibilities we have, only the third words.
Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)
18% or 45 of these also have ST.
40% of that is 18, the number of 2D cars with both BUC and ST.
Thus, the number of 4D cars with both BUC and ST would be
45 – 18 = 27
Answer = (B)
4) 700 student total
4G = total number of fourth graders
5G = total number of fifth graders
We are told 4G = 320, so 5G = 700 – 320 = 380
5GM, 5GF = fifth grade boys and girls, respectively
We are told 5GF = 210, so 5GM = 380 – 210 = 170
4GC, 5GC = total number of 4^{th} or 5^{th} graders, respectively taking Chinese
We are told
5GC = 0.5(5G) = 0.5(380) = 190
4GC = 0.4(4G) = 0.4(320) = 128
4GFM, 4GMC, 5GFC, 5GMC = 4^{th}/5^{th} grade boys & girls taking Chinese
We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do. Thus 5GMC = 80.
5GMC + 5GFC = 5GC
80 + 5GFC = 190
5GFC = 110
We are told:
4GFM < (0.5)(5GFC)
4GFM < (0.5)(100)
4GFM < 55
Thus, 4GFM could be as low as zero or as high as 54.
4GMC = 4GC – 4GFM
If 4GFM = 0, then 4GMC = 128 – 0 = 128
If 4GFM = 54, then 4GMC = 128 – 54 = 74
Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 ≤ N ≤ 128. Of the answer choices listed, the only one that works is 100.
Answer = (D)
5) The single cube has paint on all six sides. Each of the eight boxes in the 2 x 2 x 2 cube has paint on three sides (8 corner pieces). In the 3 x 3 x 3 cube, there are 8 corner pieces, 12 edge pieces (paint on two sides), 6 face pieces (paint on one side), and one interior piece (no paint). In the 4 x 4 x 4 cube, there are 8 corner pieces, 24 edge pieces, 24 face pieces, and 8 interior pieces. This chart summarizes what we have:
For the 10 x 10 flat square, we will need 4 corner pieces that have paint on three sides, 32 edge pieces that have paint on two sides (top & side), and 64 middle pieces that have paint on one side (the top).
We could use either the single total box or any of the 24 corner boxes for the four corners of the square. That leaves 21 of these, and 36 edge boxes, more than enough to cover the 32 edges of the square. The remaining ones, as well as all 30 face boxes, can be turned paint-side-up to fill in the center. The only boxes that will need to be painted, one side each, are the 9 interior boxes. Thus, we have 9 sides to paint.
Answer = (C)
6) Here’s a diagram.
First, let’s count the equilateral triangles. They are {AEI, BFJ, CGK, DHL}. There are only four of them.
Now, consider all possible isosceles triangles, excluding equilateral triangles, with point A as the vertex. We could have BAL, CAK, DAJ, and FAH. All four of those have a line of symmetry that is vertical (through A and G). Thus, we could make those same four triangles with any other point as the vertex, and we would never repeat the same triangle in the same orientation. That’s 4*12 = 48 of these triangles, plus the 4 equilaterals, is 52 total triangles.
Answer = (B)
7) There are five basic scenarios for this:
Case I: (make)(make)(make)(make)(any)
If she makes the first four, then it doesn’t matter if she makes or misses the fifth!
Case II: (miss)(make)(make)(make)(make)
Case III: (make)(miss)(make)(make)(make)
Case IV: (make)(make)(miss)(make)(make)
Case V: (make)(make)(make)(miss)(make)
Put in the probabilities:
Case I: (0.6)(0.8)(0.8)(0.8)
Case II: (0.4)(0.4)(0.8)(0.8)(0.8)
Case III: (0.6)(0.2)(0.4)(0.8)(0.8)
Case IV: (0.6)(0.8)(0.2)(0.4)(0.8)
Case V: (0.6)(0.8)(0.8)(0.2)(0.4)
Since all the answers are fractions, change all of those to fractions. Multiply the first by (5/5) so it has the same denominator as the other products.
Case I: (3/5)(4/5)(4/5)(4/5)(5/5) = 960/5^5
Case II: (2/5)(2/5)(4/5)(4/5)(4/5) = 256/5^5
Case III: (3/5)(1/5)(2/5)(4/5)(4/5) = 96/5^5
Case IV: (3/5)(4/5)(1/5)(2/5)(4/5) = 96/5^5
Case V: (3/5)(4/5)(4/5)(1/5)(2/5) = 96/5^5
Add the numerators. Since 96 = 100 – 4, 3*96 = 3(100 – 4) = 300 – 12 = 288.
288 + 256 + 960 = 1504
P = 1504/5^5
Answer = (E)
8) There are three cases: AABC, ABBC, and ABCC.
In case I, AABC, there are nine choices for A (because A can’t be zero), then 9 for B, then 8 for C. 9*9*8 = 81*8 = 648.
In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C. Again, 648.
In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C. Again, 648.
48*3 = (50 – 2)*3 = 150 – 6 = 144
3*648 = 3(600 + 48) = 1800 + 144 = 1948
Answer = (D)
9)
We know that the distance from A (0,6) to B (0, – 6) is 12, so triangle ABO is equilateral. This means that angle AOB is 60°. The entire circle has an area of
A 60° angle is 1/6 of the circle, so the area of sector AOB (the “slice of pizza” shape) is
The area of an equilateral triangle with side s is
Equilateral triangle AOB has s = 12, so the area is
If we subtract the equilateral triangle from the sector, we get everything to the right of the x-axis.
Again, that’s everything to the right of the x-axis, the parts of the circle that lie in Quadrants I & IV. We just want the part in Quadrant I, which would be exactly half of this.
Answer = (C)
10) One point is (50, 70) and one is (100, 89): the line has to pass above both of those. Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes about (100, 89)!
What is the slope from (50, 70) to (100, 90)? Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5. A line with a slope of 2/5 could pass just above these points.
Now, what about the third point? For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points. Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83). This means it would pass under the third point, (80, 84). A slope of 2/5 works for all three points.
We don’t have to do all the calculations, but none of the other slope values works.
Answer = (D)
11) The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.
Let’s say that the A = $100 at the beginning of the year.
End of January, 60% increase. New price = $160
End of February, 60% decrease: that’s a decrease of 60% of $160, so that only 40% of $160 is left.
10% of $160 = $16
40% of $160 = 4(16) = $64
That’s the price at the end of February.
End of March, a 60% increase: that’s a increase of 60% of $64.
10% of $64 = $6.40
60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40
Add that to the starting amount, $64:
New price = $64 + $38.40 = $102.40
End of April, 60% decrease: that’s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.
At this point, we are going to approximate a bit. Approximate $102.40 as $100, so 40% of that would be $40. The final price will be slightly more than $40.
Well, what is slightly more than $40, as a percent of the beginning of the year price of $100? That would be slightly more than 40%.
Answer = (A)
12) The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan. The German company must pay this amount.
Since 1 euro = (7Q) Chinese Yuan, then (1/(7Q)) euro = 1 Chinese Yuan, and (KF/7Q) euros = KF Chinese Yuan. That’s the amount that the Germans pay to the Chinese.
That is the German company’s outlay, in euros. Now, they make N metal chairs, and sell them, making a gross profit of P euros.
That must be the total revenue of the German company, in euros. This comes from the sale to the American company. Since $1 = Q euros, $(1/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.
That must be the total dollar amount that leaves the American company and goes to the German company. This comes from the sale of N metal frames for chairs, so each one must have been 1/N of that amount.
Answer = (A)
13) First, we will focus on the least, the lowest value. Suppose the minimum of 70% take English, and the minimum of 40% take German. Even if all 30% of the people not taking English take German, that still leaves another 10% of people taking German who also have to be taking English. Thus, 10% is the minimum of this region.
Now, the maximum. Both the German and English percents are “at least” percents, so either could be cranked up to 100%. The trouble is, though, that both can’t be 100%, because some folks have to take Italian, and nobody can take three languages at once. The minimum taking Italian is 30%. Let’s assume all 100% take German, and that everyone not taking Italian is taking English: that’s 70% taking English, all of whom also would be taking German. Thus, 70% is the maximum of this region.
Answer = (C)
14) Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch. The problem tells us that:
P(A) > 0.6
P(A and B) < 0.5
P(A or B) = 0.7
First, let’s establish the minimum value. If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7. All the requirements can be satisfied if P(B) = 0, so it’s possible to equal that minimum value.
Now, the maximum value. Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region. See the conceptual diagram.
The top line, 1, is the entire probability space. The second line, P(A or B) = 0.7, fixes the boundaries for A and B. P(A) is the purple arrow, extending from the right. P(B) is the green arrow extending from the left. The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.
Let’s say that P(A) is just slightly more than 0.6. That means the region outside of P(A), but inside of P(A or B) is slightly less than 1. That’s the part of P(B) that doesn’t overlap with P(A). Then, the overlap has to be less than 0.5. If we add something less than 1 to something less than 5, we get something less than 6. P(B) can’t equal 0.6, but it can any value arbitrarily close to 0.6.
Thus, 0 ≤ P(B) < 0.6.
Answer = (B)
15)
Answer = (E)
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]]>The post GMAT Tuesdays: Problem Solving – Figures Drawn as Accurately as Possible appeared first on Magoosh GMAT Blog.
]]>As you work through problem solving questions, you will see a lot of figures, charts, shapes, and lines! But can you trust them? What can you assume about those shapes? This week, I dive in and answer these questions! 😀
If you are curious to learn more about figures on the problem solving questions, I recommend reading this article.
And now for a closer look at this week’s board:
The post GMAT Tuesdays: Problem Solving – Figures Drawn as Accurately as Possible appeared first on Magoosh GMAT Blog.
]]>The post Challenging Coordinate Geometry Practice Questions appeared first on Magoosh GMAT Blog.
]]>1) Graph G has a line of symmetry of x = –5/2. Graph G passes through the point (3, 3). What is the x-coordinate of another point that must have a y-coordinate of 3?
(A) –8
(B) –7
(C) –5
(D) –4
(E) 2
2) In the figure above, the point on segment JK that is four times as far from K as it is from J is:
3) Which point is the reflection of the point (–7, 5) over the line y = –x?
(A) (–7, 5)
(B) (–5, 7)
(C) (5, –7)
(D) (7, –5)
(E) (7, 5)
4) In a coordinate system, P = (2, 7) and Q = (2, –3). Which could be the coordinates of R if PQR is an isosceles triangle?
I. (12, –3)
II. (–6, –9)
III. (–117, 2)
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III
5) Point W = (5, 3). Circle J has a center at point W and radius of r = 5. This circle intersects the y-axis at one intercept and the x-axis at two intercepts. What is the area of the triangle formed by these three intercepts?
(A) 7.5
(B) 12
(C) 15
(D) 24
(E) 30
6) Line M has a y-intercept of –4, and its slope must be an integer-multiple of 1/7. Given that Line M passes below (4, –1) and above (5, –6), how many possible slopes could Line M have?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
7) Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.)
(A) 25
(B) 33
(C) 36
(D) 41
(E) 58
8) In the x-y plane, point (p, q) is a lattice point if both p and q are integers. Circle C has a center at (–2, 1) and a radius of 6. Some points, such as the center (–2, 1), are inside the circle, but a point such as (4, 1) is on the circle but not in the circle. How many lattice points are in circle C?
(A) 36
(B) 72
(C) 89
(D) 96
(E) 109
Here are some blogs that you may find germane. If you look at one of these and have an “aha!”, then you might want to look over these problems again.
2) Special Properties of the line y = x
3) Distance between two points
4) Slopes
5) Midpoints and Parallel & Perpendicular lines
If you found this problems helpful or have a question about any after reading the TE, please let us know in the comments section.
1) The line of symmetry is x = –2.5. The point (3, 3) is 3 – (–2.5) = 5.5 to the right of this line of symmetry. It’s reflection must be 5.5 units to the left of the line of symmetry, so (–2.5) – (5.5) = –8 is the x-coordinate. Answer = (A)
2) Call the point P. Then, PK = 4*JP. Of course, JP + PK = JK, so JP + 4*JP = 5*JP = JK, and JP = (1/5)*JK. Point P should be one fifth of the segment away from J. This would be the point (–1, 3). Answer = (A)
3) The point (–7, 5) is in the second quadrant, relatively close to the line y = –x, so the reflection would have to be another point in the second quadrant, and the only answer choice in the second quadrant not equal to the original point is (B), (–5, 7). This question is very easy to solve by visual/spatial reasoning.
For those who want to know the “rule”: when a point is reflected over the line y = –x, the coordinates are reversed, and each one takes on its opposite sign. The –7 becomes +7, and the +5 becomes –5, and they switch places, which also results in (B).
4) Notice that points P & Q are separated by 10 units vertically.
I. R = (12, –3)
Then, point R is 10 units to the right of Q, so we get a big 45-45-90 Right Isosceles Triangle:
Option I works.
II. R = (–6, –9)
This is tricky. This left 8 and down 6 from Q, so the triangle formed by the x & y separations between Q and this R form a 6-8-10 right triangle, and the distance QR = 10, making this an isosceles triangle as well. This little 6-8-10 triangle on the QR connection is show in dashed lines:
Option II works.
III. R = (–117, 2)
It’s a mistake to do any calculations at all with this one. The line y = 2 is the perpendicular bisector of PQ:
This is an important Geometry idea: any point on the perpendicular bisector of a segment is automatically equidistant from the two endpoints of the segment. This means, we could pick absolutely any point on the line y = 2, call it R, and this point R would be equidistant from P & Q, making PQR an isosceles triangle. The point R = (–117, 2) is on this perpendicular bisector, so it is equidistant from P & Q, and PQR must therefore be isosceles.
Option III works.
Each of the three options works. Answer = (E).
5) Well, if we go 5 units to the left of (5, 3), we’re at (0, 3): that’s the single y-intercept of the circle.
Now, think about the two x-intercepts. Each one is a diagonal distance of r = 5 from (5, 3), and if we may a right triangle on either side, the vertical leg is the distance from (5, 3) straight down to the x-axis, which of course is 3.
Those two purple triangles must be 3-4-5 triangles, which means each one has a base of 4, and the distance between the two of them is 8. One is at (1, 0) and the other is at (9, 0).
Now, think about the triangle formed by these three intercepts. The base, from (1,0) to (9, 0) is 8, and while the third vertex is off-center, that doesn’t matter — the height is h = 3. A = (0.5)bh = (0.5)(8)(3) = 12. Answer = (B)
6) Well, for starters, zero is a multiple of every number, and a line with slope zero, the horizontal line y = –4 passes below (4, –1) and above (5, –6). That’s horizontal line is our starting point.
The point (4, –1) is over 4, up 3 from the y-intercept (0, –4). A line with a slope of +3/4 would go straight from (0, –4) to (4, –1). Thus, we need a slope that is less than +3/4. Notice that 3/4 = 21/28. Notice that 5/7 = 20/28, so this would be less than 3/4. Therefore, +1/7 through +5/7 will all slope up, obviously above (5, –6), and all will pass below (4, –1). That’s five upward sloping lines.
The point (5, –6) is over 5, down 2, from the y-intercept (0, –4). A line with a slope of –2/5 would go straight from (0, –4) to (5, –6). Thus, we need a slope that is more than –2/5; another way to say that is, we need a negative slope whose absolute value is less than +2/5. Well, 2/5 = 14/35, while 2/7 = 10/35 and 3/7 = 15/35, so (2/7) < (2/5) < (3, 7). The negatively sloping lines obviously pass below (4, –1), but only two of them, –1/7 and –2/7, pass above (5, –6).
That’s one horizontal line, five upward sloping lines, and two downward sloping lines, for a total of eight. Answer = (C).
7) First of all, Line Q 5y – 3x = 45 re-written in slope-intercept form is y = (3/5)x + 9. Line Q has a y-intercept of +9, so if Line S also has a y-intercept of +9, they would intersect on the y-axis, not in the second quadrant. Therefore, the y-intercept of Line S must be less than 9, and the highest value it could have is a y-intercept of 8.
Now, let’s think about the bottom end. Line Q has a y-intercept of +9 and an x-intercept of –15. Line S, perpendicular to Line Q, must have a slope of –5/3, the negative reciprocal of Line Q’s slope. If we start at the x-intercept of point Q, the point (–15, 0), we would follow the slope of –5/3 over 15 and down 25 to (0, –25). If Line S had a y-intercept of –25, it would intersect Line Q at (–15, 0), on the x-axis, and not in the second quadrant. Therefore, a y-intercept of –25 doesn’t work, and the lowest value that would work is one above it, y-intercept of –24.
Any y-intercept less than or equal to +8 and greater than or equal to –24 would work. That’s 8 positive y-intercepts, the intercept of zero at the origin, and 24 negative intercepts, for a total of 8 + 1 + 24 = 33. Answer = (B)
8) This is a tough question. First of all, obviously, the center a lattice point in the circle. If we move horizontally or vertically, the first 5 lattice points will be inside the circle, and the sixth one will be on the circle, so the points on the circle don’t count as being “in” the circle.
Now, we just are going to focus on one quadrant of the circle, the upper right quadrant. Suppose we go over one unit and up the circle.
Well, that’s a right triangle, with hypotenuse of r = 6, and horizontal leg of 1, so if the vertical leg is x, then
Well, the square root of 35 is bigger than 5, so five points in that column are inside the circle. Now, move two units to the right. Again, hypotenuse r = 6, short leg = 2, vertical leg = x, so
Still bigger than 5, so there five points in this second column. Now, move three units to the right. Again, hypotenuse r = 6, short leg = 3, vertical leg = x, so
Still bigger than 5, so there five points in this third column. Here’s where we are so far:
Clearly, by symmetry, we can reflect this array over the 45° line, to get more points in the circle:
Because that row at a height of 5 only is 3 points wide, we know that the column 5 units to the right can only be 3 points height. We have to check that highest point in the column 4 units to the right. Again, hypotenuse r = 6, short leg = 4, vertical leg = x, so
Clearly, this is bigger than 4, so that fourth column can be four points high.
That completes the quadrant. Within the quadrant, that’s 5 + 5 + 5 + 4 + 3 = 22 points. Multiply that by four, and that’s 88 points within the four quadrants. Add the 20 points on the vertical & horizontal segments, and the one point at the center, and we have 88 + 20 + 1 = 109. Answer = (E).
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]]>2) Special Properties of y = x
3) Distance between Two Points
4) Slopes
5) Midpoints and Parallel & Perpendicular lines
Here are five new practice problems on these topics.
1. The center of circle Q is on the y-axis, and the circle passes through points (0, 7) and (0, –1). Circle Q intersects the positive x-axis at (p, 0). What is the value of p?
2. In the diagram above, coordinates are given for three of the vertices of quadrilateral ABCD. Does quadrilateral ABCD have an area greater than 30?
Statement #1: point B has an x-coordinate of 4
Statement #2: quadrilateral ABCD is a parallelogram
3. In the x-y plane, point F = (3, –2). Point G is at (3, k), where k is an integer such that 5 ≤ k ≤ 40. If FG is to form the side of a square, how many different square can be created?
4. In the coordinate system above, which of the following is the equation of line p?
5. The graph above shows line H. Line J (not shown) does not pass through the first quadrant. Which of the following could be true?
I. line J is perpendicular to line H
II. line J is parallel to line H
III. line J intersect line H in the third quadrant
If would like to express anything, or have any question, please let us know in the comments section below!
1) The center of the circle must be halfway between (0, 7) and (0, –1), at the point C = (0, 3). We know the radius is 4. Now consider what this looks like:
Here C = (0, 3) is the center. From C to (0, 7) is a radius of 4, and from C to (0, –1) is also a radius of 4. Well, AC is another radius, so this also has a length AC = 4. Notice, now, that OCA is a right triangle. We know that OC = 3 and AC = 4
Answer = D
2) In this problem, the lower triangle ACD has a base of AC = 8, and a height, from the origin down to D, of 4. Therefore, the area of ACD = (1/2)(b)(h) = (1/2)(8)(4) = 16. We would need to know something about the upper triangle ABC to know the answer to the prompt question. We know the base of triangle ABC, AC = 8, but we don’t know anything about the height.
Statement #1: if we know the x-coordinate of point B, that doesn’t help us. We still know the base AC = 8, but we don’t know the height, only the vertical line along which point B will lie. Any height could be possible. This statement, alone and by itself, is insufficient.
Statement #2: the diagonal of any parallelogram (i.e. the line connecting two opposite vertices) divides it into two congruent triangles. Well, if ABCD is a parallelogram, then line AC is a diagonal, which means triangles ADC and ABD must be congruent and have equal area. This would allow us to calculate the total area and answer the prompt question. This statement, alone and by itself, is sufficient.
Answer = B
3) Idea #1: inclusive counting. From 5 to 40 inclusive, there are not 35, but 36 values.
Idea #2: The points F & G have the same x-coordinates, so FG must be a vertical segment.
There are 36 possible vertical segments. Any square with sides parallel to the x- & y-axes has two vertical sides and two horizontal sides. The vertical segment FG could be the right side or the left side of the square, so for any vertical segment there are two possible squares.
(36 segments) x (2 possible squares) = 72 squares
Answer = D
4) First of all, line p clearly has a negative slope. If the slope is negative, that means the x & y have opposite sign coefficients when written in slope-intercept form (i.e. y = mx + b). Thus, if we move the x to the opposite side, so that the x & y are on the same side, then they will have to have the same sign coefficients. The x & y coefficients could be both positive or both negative. The latter is not an option among the answer choices. We must have a plus-sign, so answers (C) & (D) are out right away.
Notice the x intercept is approximately (6, 0) —- it could be exactly equal to that, or approximately equal to that. Plug this in to the three remaining choices, and see what happens.
(A) 3(6) + 7(0) = 18 YES, exactly true
(B) 7(6) + 3(0) ≠ 18 no, not even close
(E) 3(6) + 7(0) = –18 no, not even close
Answer = A
5) If line J does not pass through first quadrant, then it must be a line with a negative slope and a negative y-intercept. Such a line could be perpendicular to line H:
Therefore, Statement I is possible.
Line H has a positive slope, and line J must have a negative slope, so there is absolutely no way for them to have the same slope. They absolutely cannot be parallel. Therefore, Statement II is impossible.
Both line H and line J pass through QIII, so there’s no reason they cannot intersect there. For example:
Therefore, Statement III is possible.
Answer = D
To find out where coordinate geometry sits in the “big picture” of GMAT Quant, and what other Quant concepts you should study, check out our post entitled:
What Kind of Math is on the GMAT? Breakdown of Quant Concepts by Frequency
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]]>1) Line A has the equation 3x + y = 7. Which of the following lines is perpendicular to Line A?
2) Line P has a positive slope and a positive y-intercept. Line Q also has a positive slope and a positive y-intercept. The two slopes and the two y-intercepts are four different numbers, none equal. Lines P & Q have a single intersection point — what is the full set of possible quadrants in which the intersection point could be?
3) The median of a triangle is the line from any vertex to the midpoint of the opposite side. Triangle ABC has vertices A = (0, 5), B = (–1, –1), and C = (5, 2). What is the slope of the median from A to the midpoint of BC?
In the previous post, I discuss the idea of slope, but I didn’t talk about the special cases of horizontal and vertical lines.
A horizontal line is all run and no rise: therefore, its slope, its rise/run, is zero. By contrast, a vertical line is all rise and no run — when we calculate its rise over run, we get a mathematical error, because we can’t divide by zero. This means, vertical lines have undefined slope; very roughly, we can think of vertical lines as having “infinite” slope, although technically, that way of interpreting vertical lines does not bear rigorous analysis of the idea of infinity —- which, of course, is well beyond anything on the GMAT.
Horizontal lines have a slope equal to zero. Vertical lines have an undefined slope: the value of their slope does not make any mathematical sense. I would strongly recommend against saying that either has “no slope”, because that is a term ripe for confusion: does “no slope” mean a slope with a value of zero or does it mean not being able to compute the value of the slope? I recommend that you absolutely banish the term “no slope” from your vocabulary.
Parallel lines in the x-y plane have equal slope. If the think about what the slope means geometrically, you will see that this has to be true. The GMAT expects you to know this.
The case with perpendicular lines is a little trickier. Let’s think about this. Consider any slanted line with a positive slope (i.e. moving up to the right) — then the line perpendicular to it will have to have a negative slope (i.e. moving down to the right, moving up to the left). Similarly, if the original line has a negative slope, the perpendicular would have to have a positive slope. Thus, the slope of a perpendicular line is always has a sign opposite of the sign of the original line. That’s part one of the idea.
Now, let’s think about if we take a “rise over run triangle” and rotate it 90°. Let’s ignore ± signs for a moment, and just think about absolute values.
So, we have two perpendicular lines, the original line AB and the perpendicular line AE, intersecting at point A. For each, we have constructed “rise over run” triangle for the slope. The slope of line AB = (original rise)/(original run) = BC/AC, and the slope of line AE = (new rise)/(new run) = AD/DE. But notice that these two triangles are congruent! Triangle ADE is just Triangle ACB rotated 90° around point A. Because the triangle are congruent, the corresponding lengths are equal: AC = AD and BC = DE. In other words, the rise of one is the run of the other! When we go from the original line to a line perpendicular to it, the rise and the run switch places in the slope fraction. When the numerator and denominator of a fraction switch places, that’s called a reciprocal. For example, 2/5 and 5/2 are reciprocals; 3/11 and 11/3 are reciprocals; 7 and 1/7 are reciprocals.
Now, put these two facts together — when line #1 has a slope, and line #2 is perpendicular to it, the slope of line #2 must have both the opposite sign from and must be a reciprocal of the original slop of line #1. More elegantly, perpendicular slopes are opposite reciprocals. The GMAT expects you to know this.
If one line has a slope = +4/7, then the perpendicular line has a slope = –7/4. If one line has a slope of –3, the perpendicular line has a slope = +1/3. If you know the slope of one line, you flip that fraction over and change its sign to the opposite to make the slope of the perpendicular line. If this is a new idea to you, or one which learned but forget a long time ago, then I strongly recommend you actually get some graph paper and actually graph sets of perpendicular lines and find their slopes, so you have a visual understanding of this idea.
Occasionally, the GMAT will ask to you to find the coordinates of the midpoint of a segment. You will be given the two endpoints of the segment — for example, (4, 1) and (10, 15). You find the coordinates of the midpoint by averaging the values they give you. The x-coordinate of the midpoint is the average of the x-coordinates of the end points; here, this average is (4 + 10)/2 = 7. Similarly, the y-coordinate of the midpoint is the average of the y-coordinates of the end points; here, this average is (1 + 15)/2 = 8. Therefore, the coordinates of the midpoint are (7, 8). If you can take an average —- twice —- then you can find the coordinates of a midpoint.
Any question about any points or lines in the x-y plane is a visual question. If you are not given a diagram, always sketch a diagram. You open up a whole new level of understanding when you add a visual approach.
If the two practice questions at the top gave you some difficulty when you first looked at them, now that you have read this article, take another look at them before you jump into the explanations below.
1) What’s tricky about this problem: we have to begin by solving the given equation for y, so that we know its slope.
The slope of the original line is m = –3, and the negative reciprocal of that is +1/3, so the perpendicular line must have a slope of +1/3. Among the answer choices, the only line with a slope of +1/3 is (C). Just so you have a visual for this, here’s a diagram of these line (whenever possible, always verify questions about the x-y plane visually, with at least a quick sketch!)
2) If you forget quadrants, take a look at this post. This is a tricky one, because you have to be careful to think about all cases. The first important point to realize is: any line with both a positive slope and a positive y-intercept goes through quadrants I & II & III, but never quadrant IV. Neither of these lines is ever in Quadrant IV, so they can’t intersect there. As it turns out, for different values of the slope, they can intersect in any of the other three quadrants. Here are visual examples.
Therefore, the answer = (D)
3) For your visual understanding, here’s a diagram of the situation.
First to find the midpoint of B & C — average the x-coordinates: (–1+ 5)/2 = 2; and average the y-coordinates: (–1 + 2)/2 = 1/2. Thus, the midpoint has coordinates (2, 1/2). We want the slope from A = (0, 5) to (2, 1/2). The rise is the change in the y-coordinates: 1/2 – 5 = -9/2. The run is the change in the x-coordinates: 2 – 0 = 2. Slope = rise/run = [–9/2]/2 = –9/4. Answer = (E)
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]]>1) What is the equation of the line that goes through (–2, 3) and (5, –4)?
2) The line y = 5x/3 + b goes through the point (7, –1). What is the value of b?
3) A line that passes through (–1, –4) and (3, k) has a slope = k. What is the value of k?
If these problems make your head spin, you have found the right post.
Slope is a measure of how steep a line is. There is very algebraic formula for the slope, and if you know that, that’s great! If you don’t know that formula, or used to know it and can’t remember it, I will say: fuhgeddaboudit! Here’s a much better way of thinking about slope. Slope is rise over run.
To calculate rise and run, first have to put the two points in order. It actually doesn’t matter which one we say is the first and which one, the second: all that matters is that we are consistent.
The rise is the vertical change — the change in y-coordinate (second point minus first). The run is the horizontal change — the change in the x-coordinate (again, second minus first). Once we have rise & run, divide them, rise divided by run, to find the slope.
For example, suppose our points are (–2, 4) and (5, 1). For the sake of argument, we’ll say that’s the order — (–2, 4) is the “first” and (5, 1) is the “second.” The rise is the change in height, the change in y-coordinate: 1 – 4 = –3 (notice, we had to do second minus first, which gave us a negative here!) The run is the horizontal change, the change in x-coordinate: 5 – (–2) = 5 + 2 = 7 (remember: subtracting a negative is the same as adding a positive!). Now, rise/run = –3/7 —- that’s the slope. Slope is definitely something you need to understand for the GMAT Quantitative section.
Whenever you find a slope, I strongly suggest doing a rough sketch, just to verify that the sign of the slope (positive or negative) and the value of the slope are approximately correct. Here’s a sketch of this particular calculation:
Your sketch, of course, does not need to be this precise. Even a rough sketch would verify that, yes, the slope should be negative. Again, I highly recommend performing this visual check every time you calculate slope.
Let’s get a bit philosophical for a moment. The technical name of the x-y plane is the Cartesian plane, named after its inventor, Mr. Rene Descartes. Although you may have met this sometime in middle school math and may now take it for granted, it is actually a brilliant mathematical device. It allowed for the unification of two ancient branches of mathematics: algebra and geometry. In more practical terms, every equation (an algebraic object) corresponds to a picture (a geometric object). That is a very deep idea.
A straight line is a very simple picture, and not surprisingly it has a very simple equation. There are a few different ways to write a line, but the most popular and easiest to understand is y = mx + b. The m is the slope of the line. The b is the y-intercept: where the line crosses the y-axis. For any given line, m & b are constants: for a given line, both m & b equal a fixed number. By contrast, x & y (sometimes call the “graphing variables”) do not equal just one thing. This is not the “x” of ordinary solve-for-x algebra. This is a very deep idea — x & y don’t equal any one pair of values; rather, every single point (x, y) on the line, the entire continuous infinity of points that make up that line — every single one of them satisfies the equation of the line. That is a powerful and often underappreciated mathematical idea.
Sometimes the GMAT will give you the equation of a line already in y = mx + b form. Sometimes, the GMAT give you the line in another form (e.g. 3x + 7y = 22), and you will have to do a little algebraic re-arranging —- essentially, solve for y —- to bring the equation into y = mx + b form. Sometimes, though, as in problem #2 above, they give you two points and ask you to find the equation of a line. Here the procedure. First, find the slope (as demonstrated above). Now, plug the slope in for “m” in the y = mx + b equation, and pick either point (it doesn’t matter which one) and plug those coordinates in for x & y in this equation. This will produce an equation in which everything has a numerical value except for “b” — that means, you can solve this equation for the value of b. Once you know m & b, you know the equation of the line.
After this introduction, go back and try those practice problems again before reading the solutions below. Here’s another relevant practice question:
4) http://gmat.magoosh.com/questions/821
In the next post, I will discuss midpoints and the issue of parallel & perpendicular lines. See also, the related posts on Distance in the Cartesian Plane, the Quadrants, and the special properties of the line y = x.
1) Here, we will follow the procedure we demonstrated in the last section. Call (–2, 3) the “first” point and (5, –4), the “second.” Rise = –4 – 3 = –7. Run = 5 – (–2) = 7. Slope = rise/run = –7/7 = –1. Visual check:
Yes, it makes sense that the slope is negative. We have the slope, so plug m = –1 and (x, y) = (–2, 3) into y = mx + b:
3 = (–1)*(–2) + b
3 = 2 + b
1 = b
So, plugging in m = –1 and b = 1, we get an equation y = –x + 1. Answer = A
2) Here, we already have the slope, so we just need to follow the second half of the “finding the equation” procedure. Plug (x, y) = (7, –1) into this equation:
Answer = E
3) This is a considerably more difficult one, which will involve some algebra. Let’s say that the “first” point is (–1, –4) and the “second,” (3, k). The rise = k + 4, which involves a variable. The run = 3 – (–1) = 4. There the slope is (k + 4)/4, and we can set this equal to k and solve for k.
k + 4 = 4k
4 = 3k
k = 4/3
Answer = C
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]]>First, try these practice questions.
1) What is the distance from (–7, 2) to (5, –3)?
2) Consider the three points in the x-y plane: P = (8, 4), Q = (6, 7), and R = (9, 0). Rank these three points from closest to the origin, (0, 0), to furthest from the origin.
3) Trapezoid JKLM in the x-y plane has coordinates J = (–2, –4), K = (–2, 1), L = (6, 7), and M = (6, –4). What is its perimeter?
The x-y plane is sometimes called the Cartesian plane, in honor of its creator, the multitalented philosopher and mathematician René Descartes (1596 – 1650). In the x-y plane, the x-coordinate of any point gives its horizontal position, how far right or left of the y-axis it is; the y-coordinate of any points gives its vertical position, how far above or below the x-axis it is. Notice that, on any horizontal line, all the y-coordinates would have to be the same, and on any vertical line, all the x-coordinates would have to be the same. Also notice: if any two points have the same x-coordinate, the segment that connects them must be vertical; and if any two points have the same y-coordinate, the segment that connects them must be horizontal. Those are very handy facts to know for the GMAT.
Big Idea #1: Horizontal & vertical distances are very easy to find!!
If two points have the same y-coordinate, then the segment connected them is horizontal. Simply subtract their x-coordinates to find the length of the segment between them.
If two points have the same x-coordinate, then the segment connected them is vertical. Simply subtract their y-coordinates to find the length of the segment between them.
BTW, whenever the GMAT gives you a geometric figure in the x-y plane and one or more of the sides are horizontal and/or vertical, the test is handing you a gift on a silver platter. Never fail to appreciate that.
Big Idea #2: For any oblique distance, use the Pythagorean Theorem
Perhaps in school you encountered something called the Distance Formula. If you already have that memorized and use it flawlessly, then good for you. If you don’t have it memorized, then I want you to blot out of your mind any trace of its existence. I don’t want you to memorize a needlessly complicated formula: instead, I want you to remember a simple procedure that accomplishes exactly the same thing.
The big idea is: use that most miraculously amazing theorem of mathematics, the Pythagorean Theorem. Combine this insight with Big Idea #1: we know horizontal & vertical distances are very easy to find, so construct a right triangle so that the distance you want to find is the hypotenuse, one leg is horizontal, and the other leg is vertical. Thus, you automatically get a triangle in which two of the sides are very simple to find, and the triangle is also automatically a right triangle, which guarantees that we can use the Pythagorean Theorem. For example, if we need to find the distance between A = (–1, 1) and B = (3, 3) —-
— then we simply draw a vertical segment going down from B, and a horizontal segment going to the right from A, and choose the third vertex, C, at the intersection:
Well, clearly AC = 4 and BC = 2, so then we are set up to find AB with the Pythagorean Theorem.
If you understand these two “big ideas”, then that’s all you need to know about distance in the x-y plane for the GMAT. If you had trouble with the practice questions, you may want to give them another look before reading the solutions below. Here’s another question from inside the Magoosh product:
4) http://gmat.magoosh.com/questions/820
1) Suppose we made a right triangle, as in the example above. The difference in the x-coordinates is (–7) – 5 = –12, so the horizontal leg would have a length of 12 (length has to be positive). The difference in the y-coordinates is 2 – (–3) = 5, so this is the length of the vertical leg. We recognize that we have a common Pythagorean triplet, so we don’t even need to perform a calculation: we know the length of the hypotenuse, and hence, the distance between the two points, is 13. Answer = D
2) Of course, the easiest distance to find is that of point R: point R is a distance of 9 from the origin. To find the distance to P, we construct a triangle with a horizontal leg of 8 and a vertical leg of 4.
Similarly,
Therefore P is the closest to the origin, and Q is the furthest. The order is P, R, Q. Answer = B
3) This one may look challenging initially, but all kinds of simplifying tricks. First of all, J & K have the same x-coordinate, so that’s a vertical segment, and the length is JK = 1 – (–4) = 5. L & M also have the same x-coordinate, so that’s also a vertical segment, and the length is LM = 7 – (–4) = 11. J & M have the same y-coordinate, so that’s also a horizontal segment, and the length is JM = 6 – (–2) = 8. Right there, we have the lengths of three of the four sides.
The last side to find is KL. From K = (–2, 1), we have to go over 8 and up 6 to get to L = (6, 7), so 6 and 8 the legs of the right triangle, and KL is the hypotenuse. We recognize another common Pythagorean triplet — (6, 8, 10), which is a multiple of the (3, 4, 5) triplet. Thus, without further calculations, we immediately know KL = 10.
Perimeter = JK + KL + LM + JM = 5 + 10 + 11 + 8 = 34
Answer = A
For those who would like a visual, here’s the graph of this trapezoid.
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]]>Fact: All lines with slopes of 1 make 45º angles with both the x- and y-axes.
Conversely, if a line makes a 45º angles with either the x- of y-axes, you know immediately its slope must be . This first fact is true, not only for y = x and y = –x, for all lines of the form y = mx + b in which m equals either 1 or –1. If the slope is anything other than , you would need trigonometry to figure out the angles, and that’s beyond the scope of GMAT math. The GMAT could expect you to know this one fact about these special lines, especially on Data Sufficiency.
Fact: Suppose we treat the line y = x as a mirror line. If you take any point (a, b) in the coordinate plane, and reflect it over the line y = x, the result is (b, a). It reverses the x- and y-coordinates!
The corollary of this is that if we compare any two points with reversed coordinates, say (2, 7) and (7, 2), we automatically know that each is the image of the other by reflection over the line y = x. Add now the geometry fact that a mirror line is the set of all points equidistant from the original point and its image. This means that the midpoint of the segment connect (2, 7) and (7, 2) must lie on the line y = x. In fact, any point on the line y = x will be equidistant from both (2, 7) and (7, 2). Without doing a single calculation, we know, for example, that the triangle formed by, say, (2, 7) and (7, 2) and (8, 8 ) must be an isosceles triangle. (See the diagram below.)
When we reflex over the line y = –x, the coordinate are reversed and made their opposite sign: e.g. (2, 7) reflect to (–7, –2), and (–5, 3) reflects to (–3, 5). The other conclusions, about equidistance, remain the same.
Fact: Any point (x, y) in the coordinate plane that is above the line y = x has the property that y > x. Any point (x, y) in the coordinate plane that is below the line y = x has the property that y < x.
Can you sense the veritable cornucopia of Data Sufficiency questions that could arise from this fact? If you every see a question about the coordinate plane asking whether y > x or y < x, chances are very good that the line y = x is hidden somewhere in the question.
1) Is the slope of Line 1 positive?
Statement #1: The angle between Line 1 and Line 2 is 40º.
Statement #2: Line 2 has a slope of 1.
(A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.
(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
(D) Each statement alone is sufficient to answer the question
(E) Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.
2) Point (P, Q) is in the coordinate plane. Is P > Q?
Statement #1: P is positive.
Statement #2: Point (P, Q) above is on the line y = x + 1
(A) Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.
(B) Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.
(C) Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.
(D) Each statement alone is sufficient to answer the question.
(E) Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.
3) A circle has a center at P = (–4, 4) and passes through the point (2, 3). Through which of the following must the circle also pass?
(A) (1, 1)
(B) (1, 7)
(C) (–1, 9)
(D) (–3, –2)
(E) (–9, 1)
1) A straightforward prompt.
Statement #1 is intriguing: it gives us a specific angle measure. This is tantalizing, but unfortunately, it is only the angle between Line 1 and Line 2, and that angle could be oriented in any direction. Therefore, we can draw no conclusion about the prompt from this statement alone. Statement #1, by itself, is insufficient.
Statement #2 is also tantalizing, because it’s numerically specific. But, unfortunately, this tells us a lot about Line 2 and zilch about Line one, so this statement is, by itself, is also insufficient.
Now, combine the statements. From statement #2, we know Line 2 has a slope of 1, which means the angle between Line 2 and the positive x-axis is 45º. We know, from statement #1, that Line #1 is 40º away from Line 2. We don’t know which way, above or below Line 2. If Line 1 is steeper than Line 2, it makes an angle of 45º + 40º = 85º with the positive x-axis. If Line 1 is less steep than Line 2, it makes an angle of 45º – 40º = 5º with the positive x-axis. Either way, its angle above the positive x-axis is between 0º and 90º, which means it has a positive slope. The combined statements allow us to give a definitive answer to the prompt question. Answer = C.
2) We see the x > y type question in the prompt, which makes us suspect that the line y = x will play an important part at some point.
Statement #1 just tells us P is positive, nothing else. The point (P, Q) = (4, 2) has the property that P > Q, but the point (P, Q) = (4, 5) has the property that P < Q. Clearly, just knowing P is positive does nothing to help us figure out whether P > Q. Statement #1, by itself, is wildly insufficient.
Statement #2 is intriguing. It discusses not the line y = x but the line y = x + 1. What is the relationship of those two lines? First of all, they are parallel: they have the same slope. The line y = x has a y-intercept of zero (it goes through the origin), while the line y = x + 1 has a y-intercept of 1. This means: any point on the line y = x + 1 must be above the line y = x. If (P, Q) is on y = x + 1, then it is above y = x, which automatically means Q > P. We can give a definite “no” answer to the question. By itself, Statement #2 is sufficient. Answer = B.
3) For this problem, there’s a long tedious way to slog through the problem, and there’s a slick elegant method that gets to the answer in a lightning fast manner.
The long slogging approach — first, calculate the distance from (–4, 4) to (2, 3). As it happens, that distance, the radius, equals . Then, we have to calculate the distance from (–4, 4) to each of the five answer choices, and find which one has also has a distance of —- all without a calculator. 🙁
The slick elegant approach is as follows. The point (–4, 4) is on the line y = –x, so it is equidistant from any point and that point’s reflection over the line y = –x. The reflection of (2, 3) over the line y = –x is (–3, –2). Since (–3, –2) is the same distance from (–4, 4) as is (2, 3), it must also be on the circle. Answer = D.
The post GMAT Math: Special Properties of the Line y = x appeared first on Magoosh GMAT Blog.
]]>The post Breakdown of GMAT Quant Concepts by Frequency appeared first on Magoosh GMAT Blog.
]]>Here are the samples of Official Material he used to figure out the GMAT Quant breakdown:
1.GMAT Official Guide (12th Edition) Problem Solving Practice Questions (Pg. 152-265)*
2. GMAT Official Guide (12th Edition) Data Sufficiency Practice Questions (Pg. 272-351)*
3. GMATPrep Test #1
4. GMATPrep Test #2
5. Released past exam from GMAC, test code 14, from the 90s/early 2000’s
Total | Percentage | ||
Arithmetic | Percents | 21 | 10.71% |
Arithmetic | Properties of Integers | 18 | 9.18% |
Arithmetic | Descriptive Statistics | 12 | 6.12% |
Arithmetic | Fractions | 11 | 5.61% |
Algebra | Linear Equations, Two Unknowns | 11 | 5.61% |
Algebra | Simplifying Algebraic Expressions | 9 | 4.59% |
Arithmetic | Powers & Roots of Numbers | 8 | 4.08% |
Arithmetic | Counting Methods | 7 | 3.57% |
Algebra | Linear Equations, One Unknown | 7 | 3.57% |
Algebra | Functions/Series | 7 | 3.57% |
Geometry | Coordinate Geometry | 7 | 3.57% |
Word Problems | Rate Problems | 7 | 3.57% |
Arithmetic | Ratio & Proportions | 6 | 3.06% |
Arithmetic | Decimals | 5 | 2.55% |
Arithmetic | Discrete Probability | 5 | 2.55% |
Algebra | Exponents | 5 | 2.55% |
Algebra | Inequalities | 5 | 2.55% |
Geometry | Quadrilaterals | 5 | 2.55% |
Geometry | Circles | 5 | 2.55% |
Word Problems | Interest Problems | 4 | 2.04% |
Algebra | Solving Quadratic Equations | 3 | 1.53% |
Geometry | Triangles | 3 | 1.53% |
Geometry | Rectangular Solids & Cylinders | 3 | 1.53% |
Word Problems | Work Problems | 3 | 1.53% |
Word Problems | Mixture Problems | 3 | 1.53% |
Geometry | Intersecting Angles and Lines | 2 | 1.02% |
Word Problems | Profit | 2 | 1.02% |
Word Problems | Sets | 2 | 1.02% |
Word Problems | Measurement Problems | 2 | 1.02% |
Word Problems | Data Interpretation | 2 | 1.02% |
Arithmetic | Real Numbers | 1 | 0.51% |
Arithmetic | Sets | 1 | 0.51% |
Algebra | Solving by Factoring | 1 | 0.51% |
Geometry | Polygons | 1 | 0.51% |
Word Problems | Discount | 1 | 0.51% |
Word Problems | Geometry Problems | 1 | 0.51% |
Algebra | Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 30 | 19.23% |
Arithmetic | Descriptive Statistics | 13 | 8.33% |
Arithmetic | Percents | 11 | 7.05% |
Algebra | Linear Equations, Two Unknowns | 11 | 7.05% |
Word Problems | Rate Problems | 9 | 5.77% |
Algebra | Inequalities | 8 | 5.13% |
Arithmetic | Sets | 7 | 4.49% |
Arithmetic | Ratio & Proportions | 6 | 3.85% |
Geometry | Triangles | 6 | 3.85% |
Geometry | Circles | 6 | 3.85% |
Arithmetic | Decimals | 5 | 3.21% |
Geometry | Coordinate Geometry | 5 | 3.21% |
Arithmetic | Fractions | 4 | 2.56% |
Algebra | Linear Equations, One Unknown | 4 | 2.56% |
Algebra | Exponents | 4 | 2.56% |
Geometry | Rectangular Solids & Cylinders | 4 | 2.56% |
Arithmetic | Discrete Probability | 3 | 1.92% |
Algebra | Functions/Series | 3 | 1.92% |
Word Problems | Interest Problems | 3 | 1.92% |
Geometry | Lines | 2 | 1.28% |
Word Problems | Work Problems | 2 | 1.28% |
Word Problems | Discount | 2 | 1.28% |
Word Problems | Profit | 2 | 1.28% |
Arithmetic | Real Numbers | 1 | 0.64% |
Arithmetic | Counting Methods | 1 | 0.64% |
Algebra | Simplifying Algebraic Expressions | 1 | 0.64% |
Algebra | Absolute Value | 1 | 0.64% |
Geometry | Quadrilaterals | 1 | 0.64% |
Word Problems | Measurement Problems | 1 | 0.64% |
Arithmetic | Powers & Roots of Numbers | 0 | 0.00% |
Algebra | Equations | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Sets | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 6 | 17.14% |
Arithmetic | Ratio & Proportions | 4 | 11.43% |
Arithmetic | Descriptive Statistics | 4 | 11.43% |
Algebra | Exponents | 3 | 8.57% |
Algebra | Functions/Series | 3 | 8.57% |
Arithmetic | Percents | 2 | 5.71% |
Arithmetic | Powers & Roots of Numbers | 2 | 5.71% |
Word Problems | Interest Problems | 2 | 5.71% |
Arithmetic | Fractions | 1 | 2.86% |
Arithmetic | Counting Methods | 1 | 2.86% |
Algebra | Equations | 1 | 2.86% |
Geometry | Intersecting Angles and Lines | 1 | 2.86% |
Geometry | Triangles | 1 | 2.86% |
Geometry | Circles | 1 | 2.86% |
Word Problems | Work Problems | 1 | 2.86% |
Word Problems | Sets | 1 | 2.86% |
Word Problems | Measurement Problems | 1 | 2.86% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Simplifying Algebraic Expressions | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Linear Equations, Two Unknowns | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Inequalities | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Geometry | Quadrilaterals | 0 | 0.00% |
Geometry | Rectangular Solids & Cylinders | 0 | 0.00% |
Geometry | Coordinate Geometry | 0 | 0.00% |
Word Problems | Rate Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Discount | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Descriptive Statistics | 6 | 18.75% |
Arithmetic | Properties of Integers | 4 | 12.50% |
Arithmetic | Ratio & Proportions | 2 | 6.25% |
Arithmetic | Percents | 2 | 6.25% |
Algebra | Simplifying Algebraic Expressions | 2 | 6.25% |
Algebra | Exponents | 2 | 6.25% |
Geometry | Coordinate Geometry | 2 | 6.25% |
Arithmetic | Fractions | 1 | 3.13% |
Arithmetic | Powers & Roots of Numbers | 1 | 3.13% |
Arithmetic | Counting Methods | 1 | 3.13% |
Algebra | Equations | 1 | 3.13% |
Algebra | Linear Equations, Two Unknowns | 1 | 3.13% |
Algebra | Inequalities | 1 | 3.13% |
Algebra | Functions/Series | 1 | 3.13% |
Geometry | Triangles | 1 | 3.13% |
Geometry | Circles | 1 | 3.13% |
Word Problems | Rate Problems | 1 | 3.13% |
Word Problems | Discount | 1 | 3.13% |
Word Problems | Measurement Problems | 1 | 3.13% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Geometry | Quadrilaterals | 0 | 0.00% |
Geometry | Rectangular Solids & Cylinders | 0 | 0.00% |
Word Problems | Work Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Interest Problems | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Sets | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 7 | 13.73% |
Arithmetic | Ratio & Proportions | 5 | 9.80% |
Arithmetic | Percents | 5 | 9.80% |
Arithmetic | Descriptive Statistics | 4 | 7.84% |
Word Problems | Sets | 4 | 7.84% |
Arithmetic | Powers & Roots of Numbers | 3 | 5.88% |
Algebra | Linear Equations, Two Unknowns | 3 | 5.88% |
Word Problems | Rate Problems | 3 | 5.88% |
Arithmetic | Fractions | 2 | 3.92% |
Arithmetic | Counting Methods | 2 | 3.92% |
Algebra | Exponents | 2 | 3.92% |
Algebra | Inequalities | 2 | 3.92% |
Geometry | Triangles | 2 | 3.92% |
Algebra | Simplifying Algebraic Expressions | 1 | 1.96% |
Algebra | Equations | 1 | 1.96% |
Algebra | Functions/Series | 1 | 1.96% |
Geometry | Quadrilaterals | 1 | 1.96% |
Geometry | Circles | 1 | 1.96% |
Geometry | Rectangular Solids & Cylinders | 1 | 1.96% |
Geometry | Coordinate Geometry | 1 | 1.96% |
Arithmetic | Decimals | 0 | 0.00% |
Arithmetic | Real Numbers | 0 | 0.00% |
Arithmetic | Sets | 0 | 0.00% |
Arithmetic | Discrete Probability | 0 | 0.00% |
Algebra | Linear Equations, One Unknown | 0 | 0.00% |
Algebra | Solving by Factoring | 0 | 0.00% |
Algebra | Solving Quadratic Equations | 0 | 0.00% |
Algebra | Absolute Value | 0 | 0.00% |
Geometry | Lines | 0 | 0.00% |
Geometry | Intersecting Angles and Lines | 0 | 0.00% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
Geometry | Polygons | 0 | 0.00% |
Word Problems | Work Problems | 0 | 0.00% |
Word Problems | Mixture Problems | 0 | 0.00% |
Word Problems | Interest Problems | 0 | 0.00% |
Word Problems | Discount | 0 | 0.00% |
Word Problems | Profit | 0 | 0.00% |
Word Problems | Geometry Problems | 0 | 0.00% |
Word Problems | Measurement Problems | 0 | 0.00% |
Word Problems | Data Interpretation | 0 | 0.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 65 | 13.83% |
Fractions | 19 | 4.04% | |
Decimals | 10 | 2.13% | |
Real Numbers | 2 | 0.43% | |
Ratio & Proportions | 23 | 4.89% | |
Percents | 41 | 8.72% | |
Powers & Roots of Numbers | 14 | 2.98% | |
Descriptive Statistics | 39 | 8.30% | |
Sets | 8 | 1.70% | |
Counting Methods | 12 | 2.55% | |
Discrete Probability | 8 | 1.70% | |
Algebra | Simplifying Algebraic Expressions | 13 | 2.77% |
Equations | 3 | 0.64% | |
Linear Equations, One Unknown | 11 | 2.34% | |
Linear Equations, Two Unknowns | 26 | 5.53% | |
Solving by Factoring | 1 | 0.21% | |
Solving Quadratic Equations | 3 | 0.64% | |
Exponents | 16 | 3.40% | |
Inequalities | 16 | 3.40% | |
Absolute Value | 1 | 0.21% | |
Functions/Series | 15 | 3.19% | |
Geometry | Lines | 2 | 0.43% |
Intersecting Angles and Lines | 3 | 0.64% | |
Perpendicular Lines | 0 | 0.00% | |
Parallel Lines | 0 | 0.00% | |
Polygons | 1 | 0.21% | |
Triangles | 13 | 2.77% | |
Quadrilaterals | 7 | 1.49% | |
Circles | 14 | 2.98% | |
Rectangular Solids & Cylinders | 8 | 1.70% | |
Coordinate Geometry | 15 | 3.19% | |
Word Problems | Rate Problems | 20 | 4.26% |
Work Problems | 6 | 1.28% | |
Mixture Problems | 3 | 0.64% | |
Interest Problems | 9 | 1.91% | |
Discount | 4 | 0.85% | |
Profit | 4 | 0.85% | |
Sets | 7 | 1.49% | |
Geometry Problems | 1 | 0.21% | |
Measurement Problems | 5 | 1.06% | |
Data Interpretation | 2 | 0.43% | |
Total | 470 | 100.00% |
Total | Percentage | ||
Arithmetic | Properties of Integers | 65 | 13.83% |
Arithmetic | Percents | 41 | 8.72% |
Arithmetic | Descriptive Statistics | 39 | 8.30% |
Algebra | Linear Equations, Two Unknowns | 26 | 5.53% |
Arithmetic | Ratio & Proportions | 23 | 4.89% |
Word Problems | Rate Problems | 20 | 4.26% |
Arithmetic | Fractions | 19 | 4.04% |
Algebra | Exponents | 16 | 3.40% |
Algebra | Inequalities | 16 | 3.40% |
Algebra | Functions/Series | 15 | 3.19% |
Geometry | Coordinate Geometry | 15 | 3.19% |
Arithmetic | Powers & Roots of Numbers | 14 | 2.98% |
Geometry | Circles | 14 | 2.98% |
Algebra | Simplifying Algebraic Expressions | 13 | 2.77% |
Geometry | Triangles | 13 | 2.77% |
Arithmetic | Counting Methods | 12 | 2.55% |
Algebra | Linear Equations, One Unknown | 11 | 2.34% |
Arithmetic | Decimals | 10 | 2.13% |
Word Problems | Interest Problems | 9 | 1.91% |
Arithmetic | Sets | 8 | 1.70% |
Arithmetic | Discrete Probability | 8 | 1.70% |
Geometry | Rectangular Solids & Cylinders | 8 | 1.70% |
Geometry | Quadrilaterals | 7 | 1.49% |
Word Problems | Sets | 7 | 1.49% |
Word Problems | Work Problems | 6 | 1.28% |
Word Problems | Measurement Problems | 5 | 1.06% |
Word Problems | Discount | 4 | 0.85% |
Word Problems | Profit | 4 | 0.85% |
Algebra | Equations | 3 | 0.64% |
Algebra | Solving Quadratic Equations | 3 | 0.64% |
Geometry | Intersecting Angles and Lines | 3 | 0.64% |
Word Problems | Mixture Problems | 3 | 0.64% |
Arithmetic | Real Numbers | 2 | 0.43% |
Geometry | Lines | 2 | 0.43% |
Word Problems | Data Interpretation | 2 | 0.43% |
Algebra | Solving by Factoring | 1 | 0.21% |
Algebra | Absolute Value | 1 | 0.21% |
Geometry | Polygons | 1 | 0.21% |
Word Problems | Geometry Problems | 1 | 0.21% |
Geometry | Perpendicular Lines | 0 | 0.00% |
Geometry | Parallel Lines | 0 | 0.00% |
The list of concepts tested on the quantitative section is from GMAC, which you can see on page 107 of the Official Guide (either 12th or 13th edition). It isn’t perfect– “Integer Properties” is a wide area of knowledge, whereas something like “Circles” is very specific.
NO. For the sake of simplicity and accuracy in reporting absolute frequency, we’ve only assigned each question to one concept. This means that even though GMAC lists “Perpendicular lines” as a topic tested on the GMAT, and we have 0 questions marked as pertaining to that topic, that certainly doesn’t mean the idea of perpendicular lines did not come up at all on all of the exams. It certainly appeared, but often in questions that were better categorized, overall, as “Coordinate Geometry”, or “Intersecting Angles and Lines”.
We hope this serves as a guideline for the relative frequency of math topics tested on the GMAT to help you decide how to focus your time! In Magoosh practice, you can set up customized practice sessions to focus on specific concepts, as well as review your performance on individual concepts to identify your weak spots using our Review tool.
Let us know whether you find this type of breakdown helpful, and whether you have any questions about any of the information above! 🙂
The post Breakdown of GMAT Quant Concepts by Frequency appeared first on Magoosh GMAT Blog.
]]>The post Quadrants on the GMAT: The Cartesian Plane appeared first on Magoosh GMAT Blog.
]]>The Cartesian Plane (a.k.a. the x-y plane) is a favorite GMAT topic, especially on GMAT Data Sufficiency. It’s a simple topic, but with just enough subtlety that the testmakers can spin endless questions from it.
The quadrants begin with I, where both x and y are positive, and rotate counterclockwise from there. Notice
OK, still relatively simple, but now what happens when you multiply or divide the x- and y-coordinates? Both (positive)*(positive) and (negative)*(negative) are positive, but if one factor is positive and one is negative, the product is negative. Similarly, (positive)/(positive) and (negative)/(negative) are positive, but if dividend is positive & divisor negative, or vice versa, the quotient is negative. Thus:
Already, that’s rich fodder for GMAT Data Sufficiency. Now, what happens when we add the x- and y-coordinates? (positive) + (positive) = (positive), and (negative) + (negative) = (negative), but (positive) + (negative) — hmmm — the sign of the sum depends on which has a larger absolute value. Thus:
In QI, x + y > 0
In QII, the sign of x + y is unclear (i.e. it depends on the values of x & y)
In QIII, x + y < 0
In QIV, the sign of x + y is unclear (i.e. it depends on the values of x & y)
Even worse: subtraction! We know a (positive) – (negative) must be positive, and a (negative) – (positive) must be negative, but if we have either a (positive) – (positive) or (negative) – (negative), then the sign of the difference depends on which has a larger absolute value. Thus:
In QI, the signs of (y – x) and (x – y) are unclear (i.e. it depends on the values of x & y)
In QII, y – x > 0 and x – y < 0
In QIII, the signs of (y – x) and (x – y) are unclear (i.e. it depends on the values of x & y)
In QIV, y – x < 0 and x – y > 0
Now, you probably have some appreciation of how many potential Data Sufficiency Questions the GMAT could concoct simply on the quadrants and the signs of x & y coordinates. A simple topic, but one worth thinking through thoroughly so you are ready on test day.
1) Is the point (x, y) in the fourth quadrant?
(1) xy > 0
(2) y > 0
2) http://gmat.magoosh.com/questions/1031
The question at that link will be followed by a video explaining the solution.
1) This DS question is a yes/no question. We don’t actually need to determine the quadrant of (x, y), only whether it’s in QIV.
Statement #1: xy > 0
This inequality means that the (x,y) is either in QI or QIII, as we saw above. We don’t know which, but in either case, we can definitively say: no, it’s not in QIV. Because the statement allows us to arrive at a definitive answer — and it doesn’t matter one peep whether it’s a yes or no answer, as long as it’s definitive — the statement is sufficient. We got a definitive no answer to the question, so Statement #1 is sufficient.
Statement #2: y > 0
This inequality means that the point (x,y) is north of the x-axis, in either QI or QII. We don’t know which, but in either case, we can definitively say: no, it’s not in QIV. Again, definitive answer to the question, so Statement #2 is sufficient.
Both statements sufficient: answer = D.
The solution & explanation to question #2 are available at the link above.
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]]>