The post Challenging GMAT Math Practice Questions appeared first on Magoosh GMAT Blog.
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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)
The post Challenging GMAT Math Practice Questions appeared first on Magoosh GMAT Blog.
]]>The post Matemática GMAT: Como Encontrar a Altura de um Triângulo? appeared first on Magoosh GMAT Blog.
]]>Primeiro de tudo, a “altura” de um triângulo é sua altitude. Qualquer triângulo possui três altitudes, portanto, três alturas. Veja, qualquer lado pode ser uma base. De qualquer vértice, você pode desenhar uma linha perpendicular ao oposto da base – esta é a altitude. Qualquer triângulo possui três altitudes e três bases. Você pode usar qualquer par de altitude-base para encontrar a área do triângulo, pela fórmula A = (1/2)bh.
Em cada um desses diagramas, o triângulo ABC é o mesmo. A linha verde é a altitude, a “altura”, e o lado com o quadrado perpendicular vermelho é a “base”. Todos os três lados do triângulo têm uma.
Dadas as alturas dos três lados do triângulo, a única forma de se encontrar a altura e a área dos lados individuais iria envolver trigonometria, o que está bem além do escopo do GMAT. Você está 100% ISENTO da responsabilidade de saber como fazer esse tipo de cálculo. Essa matemática é muito mais avançada do que é realmente necessário conhecer. Não se preocupe com essas coisas.
Na prática, se o problema do GMAT quer que você calcule a área de um triângulo, eles teriam que fornecer a altura. A única exceção seria em um triângulo retângulo – aqui, se uma das pernas é a base, a outra é a altitude, a altura, então é particularmente fácil encontrar a área de triângulos retângulos.
Se você não quiser saber nada sobre esse tópico pois definitivamente não precisa para o GMAT, pule esta seção!
a. Tecnicamente, se você sabe os três lados de um triângulo, é possível encontrar a área com algo chamado fórmula de Heron, mas isto também é algo além do que o GMAT espera de você. Mais do que precisa saber!
b. Se um dos ângulos de um triângulo é obtuso, então as altitudes para qualquer base adjacente ao seu ângulo obtuso está fora do triângulo. Super tecnicamente, uma altitude não é um segmento através de um vértice perpendicular à base oposta e sim, um segmento através de um vértice perpendicular à linha que contém a base oposta.
No diagrama acima, o triângulo DEF, uma das três altitudes é DG, que vai do vértice D à uma linha reta infinita que contém o lado EF. Isto é uma técnica que o GMAT não irá colocar no teste ou exigir de você. Novamente, mais do que precisa saber!
c. Se os três lados de um triângulo são todos inteiros perfeitamente positivos, então em toda a sua probabilidade, o valor matemático real das altitudes serão decimais bem feios. Muitas fontes de preparações do GMAT e professores no geral irão passar por cima disso, e, para facilitar a resolução dos problemas, irão fornecer um número perfeitamente positivo para a altitude também. Por exemplo, o real valor da altitude de C para AB no triângulo 6-7-8 do começo é:
Você tanto NÃO precisa saber encontrar esse número como também a maioria dos criadores das questões do GMAT vão poupar detalhes feios e apenas dizer, por exemplo, que a altitude = 5. Isto torna bem fácil o cálculo da área. Sim, tecnicamente, é uma mentira, mas uma que poupa os estudantes de vários decimais matemáticos feios que não requerem preocupação. Na verdade, professores de matemática de todos os níveis fazem isto o tempo todo – pequenas mentirinhas matemáticas, para poupar os estudantes de detalhes que não precisam saber.
Até onde eu sei, as próprias pessoas que escrevem o GMAT são caçadoras da verdade de todos os tipos, e nem ao menos fazem esses tipos de mentirinhas de “simplificação para estudantes”. É mais provável que contornem todo o problema, por exemplo, ao fazer todas as variáveis com comprimentos relevantes ou algo do tipo. Novamente, mais do que precisa saber!
É preciso saber geometria básica. Sim, há muita matemática além disso, e muito mais que poderia entender sobre triângulos e suas propriedades, mas você não é responsável por nada disto. É necessário conhecer a geometria básica de triângulos, incluindo a fórmula A = (1/2)b*h. Se o triângulo não é retângulo, você não tem responsabilidade nenhuma de saber como encontrar a altura – ela sempre será dada se precisar. Aqui está uma questão prática gratuita para você.
1) http://gmat.magoosh.com/questions/81
Esta postagem apareceu originalmente em inglês no Magoosh blog e foi traduzida por Jonas Lomonaco.
The post Matemática GMAT: Como Encontrar a Altura de um Triângulo? appeared first on Magoosh GMAT Blog.
]]>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 GMAT Practice Problems: Similar Geometric Figures appeared first on Magoosh GMAT Blog.
]]>1) In the figure, KLMN is a square, and angle KJN = 45°. Find the area of figure JKLMN.
2) In the diagram, HJLM is a square, and GH = 10. Find the area of trapezoid GHJK.
3) In the diagram, BD = 5, CD = 10, and AE = 40. What is the area of the shaded region?
4) In the figure above, angle PTQ = angle QRS = 70º, PT = 4, PR = 12, and the shaded region has an area of 48. What is the length of QW?
Answer will appear at the end of this blog.
Two geometric figures are similar if they have the same shape but are difference sizes. One is a smaller or larger version of the other. Similar figures always have all the same angles, and their sides are proportional. Some previous blog with relevant materials are
1) The GMAT’s Favorite Triangles: the 30-60-90 triangle and the 45-45-90 triangle
2) Similar Shapes, including scale factor
3) Scale Factors and Percent Increase/Decrease
You definitely need to know the two special right triangles in #1, and other blogs contain many time saving hints.
1) Since KN must be perpendicular to JM, we know that JNK must be a 45-45-90 triangle. The legs are equal, and the hypotenuse is the square root of 2 times larger than either leg. Let JN = KN = x. Then
In that work, we rationalized the denominator when we divided by the radical. This number for x is the side of the square, so square KLMN has an area which is this number squared.
That’s part of the area. Now, notice that triangle JKN would be equivalent to a square of the same size cut in half along the diagonal. This triangle must have exactly half the area of the square. Well, the square is 18, so the triangle must be 9, and together, they must have an area of 27.
Answer = (E)
2) This entire problem hinges on recognizing that triangles GHM and JKL are 30-60-90 triangles and using the properties of those triangles. First of all, notice that because HJLM is a square, it must be true that HM = JL. This means, the two triangles, GHM and JKL, must be congruent and have all the same sides & angles. GH = JK = 10.
Now, in a 30-60-90 triangle, the smaller leg is half the hypotenuse, so GM = LK = 5. The longer leg is the square root of 3 times the shorter leg, so
This is the side of the square, so we square this to get the area of square HJLM.
That’s part of the area. Now, we need the area of the triangles. One triangle is A = 0.5bh, so two triangles would simply be A = bh
Answer = (D)
3) In this problem, we need the crucial fact that the ratio of areas is proportional to scale factor squared.
First of all, side BD in BCD corresponds to side AE in ECA. AE:BD = 40:5 = 8, so the scale factor between the two triangles is 8. Every length in triangle ECA is 8 times bigger than the corresponding length in triangle BCD.
This means that the area of ECA is 8 squared, or 64 times bigger than the area of BCD. Well, it’s easy to calculate that BCD has an area of 25. This means that ECA must have area of
25*64 = 50*32 = 100*16 = 1600
Notice, we used the doubling and halving trick in that problem to simplify the multiplication. That’s the area of ECA. Well, the shaded area doesn’t include BCD, so
shaded = (triangle ECA) – (triangle BCD) = 1600 – 25 = 1575
Answer = (E)
4) We know the larger and smaller triangles are similar, because they share the angle at P, and one other angle in each is 70º. What’s tricky is that they have different orientations, so that side PT actually corresponds to side PR. We know PR:PT = 3, so that’s the scale fact. Every length in triangle PRS is three times more than the corresponding side in triangle PTQ. If lengths are multiplied by 3, area is multiplied by 3 squared, or 9. Let’s say that the area of triangle PTQ is A. Then the area of triangle PRS is 9A. The shaded area is the difference of the two triangle areas, or 8A. If 8A = 48, this means A = 6, and that’s the area of triangle PTQ.
Well, PT = 4 is a base of PTQ, and QW is a corresponding altitude: call its length h.
A = 0.5bh
6 = 0.5(4)h
6 = 2h
3 = h
The length of QW is 3.
Answer = (B)
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]]>1) The line shown passes through the point (A, 30). Which of the following is closest to the value of A?
2) Line A has a slope of and passes through the point (–2, 7). What is the x-intercept of Line A?
3) Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What is the value of Q?
4) A circle has a center of (1, 2) and passes through (1, –3). The circle passes through all of the following EXCEPT:
5) Points J (3, 1) and K (– 1, – 3) are two vertices of an isosceles triangle. If L is the third vertex and has a y-coordinate of 6, what is the x-coordinate of L?
6) A parabola has one x-intercept at (– 4, 0). If the vertex is at (2, 5), find the other x-intercept.
The coordinate plane is also known as the x-y plane and the Cartesian plane, so named after its discoverer, Mr. Rene Descartes (1596 – 1650), the mathematician & philosopher who also said “I think therefore I am.” Of course, it consists of two perpendicular numbers lines, the x- and y-axis, which define a grid that covers the entire infinite plane. Here are some previous blogs on this most remarkable mathematical object, the coordinate plane:
2) Lines & Slopes in the x-y plane
3) Midpoints and Parallel & Perpendicular Lines
5) Special Properties of the line y = x
6) an earlier set of Coordinate Geometry practice questions
7) a set of challenging Coordinate Geometry practice questions
If you get some insights from some of those blogs, you might give the problems above a second look before reading the solutions below.
1) The line appears to have a slope of ½ and y-intercept of 1, so its equation would be
y = 0.5x + 1
Plug in A for x and 30 for y.
30 = 0.5A + 1
29 = 0.5A
58 = A
Answer = (A)
2) Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the x-intercept. Let’s think about this in the vicinity of the x-intercept.
Obviously, (1, 2) is at a height of 2 above the x-axis. Let b be the distance from (1, 0) to the x-intercept. We know –h/b must equal the slope.
Now, we just have to add one to that to get the horizontal distance from the origin.
Answer = (B)
3) Think about this visually. From (–5, –1) to (2, 2), the line moves right 7 and up 3, so that’s a slope of 3/7.
From the point (2, 2) to the point (5, Q), there’s a horizontal distance of 3, an unknown vertical distance — call it h, so that h = Q – 2, or Q = h + 2. The ratio of this vertical and horizontal distance must equal the slope.
That’s h. Now, add 2 to get Q.
Answer = (C)
4) The radius is r = 5. First of all, 5 above the center, the circle goes through (1, 7) on the top, and 5 to the left & right of the center, the circle goes through (–4, 2) and (6, 2) on the same horizontal line as the center. That first point is choice (A).
That length of 5 can also be the hypotenuse of a 3-4-5 slope triangle, so starting from the center (1, 2), we could go over ±3 and up ±4, or over ±4 and up ±5. This means the circle must go through
right 3, up 4 = (4, 6)
right 4, up 3 = (5, 5) = option (E)
right 3, down 4 = (4, –2) = option (D)
right 4, down 3 = (5, –1)
left 3, up 4 = (–2, 6)
left 4, up 3 = (–3, 5) = option (D)
left 3, down 4 = (–2, –2)
left 4, down 3 = (–3, –1)
That’s all the points other than option (C). Notice that (–2, 6) and (4, 6) are on the circle, so another point between them, on the same horizontal line, (0, 6), could not be on the circle.
Answer = (C)
5) This one looks like it could involve a very complicated calculation, but there’s a very elegant way to do it. This involves special properties of the line y = – x. Notice that the coordinates of J & K have been switched, x for y and vice versa, and they have the opposite ± signs. This means that J and K are reflections of each other over the mirror line y = – x. As is always true of any reflection, every point on the mirror line is equidistant from a point and its reflection. Thus, we just need any point on the line y = – x, for example, (–6, 6).
Answer = (D)
6) The line of symmetry of a parabola always passes through the vertex, so the equation of the line of symmetry is the vertical line x = 2. The two x-intercepts are symmetrical around this line. The point (– 4, 0) is six units to the left of the symmetry line, so the other should be six units to right, at (8, 0).
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]]>1) In the diagram above, O is the center of the circle and angle AOB = 144º. What is the area of the circle?
Statement #1: The area of sector AOB is 40% of the area of the circle
Statement #2: Arc ACB has a length of .
2) In the diagram above, ABC is an equilateral triangle. D is the midpoint of AC. BD is a diameter of the circle. If AD = 4, what is the area of the circle?
3) In the diagram above, O is the center of the circle. What is the length of chord AC?
Statement #1: chord BC = 14
Statement #2: the circle has an area of
4) Points P, Q, R, S, and T all lie on the same line. The larger circle has center S and passes through P and T. The smaller circle has center R and passes through Q and S. What is the ratio of the area of the larger circle to the area of the smaller circle?
Statement #1: ST:PQ = 5/2
Statement #2: RT:PR = 13/7
5) A square and a circle intersect at more than one point. Does the square have more area than the circle?
Statement #1: there are exactly four intersection points
Statement #2: at least two of the intersection points are on vertices of the square
6) In the diagram above, all the points are on a line, and the number of each point indicates how many units that point is from zero. The points #1 – #6 are the centers of the six circles, and all circles pass through point zero. What is the total area of the shaded region?
7) In the diagram above, JKL is an equilateral triangle. Point M is the midpoint of segment JL, and M is the center of a circle that passes through points J and L. The shaded regions in the diagram indicate all the regions inside the circle that are outside the triangle. What fraction of the total area of the circle is outside the triangle?
8) In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?
Ah, the beauty of the circle, the most symmetrical of shapes! For more information about circles, see these blogs:
4) Inscribed & Circumscribed Circles
5) Circles, Arclength, and Sectors
If you have any insights while reading those blogs, you may want to give the problems above a second look. If you find any typos or anything on this page is unclear, please let us know in the comments section.
1) First of all, we should think about the prompt a bit. This angle, 144º, is what fraction of a full circle? Well, both 144 and 360 are divisible by 4: 144 ÷ 4 = 36 and 360 ÷ 4 = 90, so
This angle is 2/5 of the whole circle, so the arc is 2/5 of the whole circumference. Also, remember that if we can find anything about the whole circle, for example, the circumference, then we can find the radius, which would allow us to find the area.
Statement #1: This is a tautological statement. A tautological statement is a statement that, by definition, has to be true, and because of this, it contains no information. Statements such as “My car is a car” and “My employer employs me” are verbal tautologies: they contain no useful information. Much in the same way, we already know from the prompt that the angle takes up 2/5 of the circle, so of course the sector would take up 2/5, or 40%, of the area. This statement repeats information in the prompt, and contains no new information, so it doesn’t help us at all to figure out anything else. This statement, alone and by itself, is not sufficient.
Statement #2: We already know this arc is 2/5 of the whole circumference, so we could set up a proportion to find the circumference. From that, we could find the radius, and that would allow us to find the area. This statement, alone and by itself, is sufficient.
Answer = (B)
2) Because ABC is an equilateral triangle, right triangle ABD is a 30-60-90 triangle, one of the GMAT favorite triangles. From the ratios of that triangle:
Half of that is radius of the circle:
Use that to find the area:
Answer = (B)
3) The fact that AB is a diameter guarantees that angle C = 90º. If we had two sides of right triangle ABC, we could find the third using the Pythagorean Theorem.
Statement #1: this gives us only one side of a right triangle: not helpful. This statement, alone and by itself, is not sufficient.
Statement #2: this allows us to solve for the radius and, hence, the diameter, so we can determine side AB. Nevertheless, this gives us only one side of a right triangle: also not helpful. This statement, alone and by itself, is not sufficient.
Combined statements: We get the length of BC from the first statement, and the length of AB from the second. Now, we have two sides of the right triangle, so we can use the Pythagorean Theorem to solve for the third side, AC. Combined, the statements are sufficient.
Answer = (C)
4) Call the radius of the larger circle y, and y = PS = ST. Call the radius of the smaller circle x, and x = RS = RQ. If we took a ratio of the areas, the factors of would cancel and we would be left with the ratio (y/x) squared. If we could solve for this simpler ratio, y/x, then we could find the ratio of areas.
Statement #1: ST = y and PQ = y – 2x, so
This allows us to solve for the ratio y/x, which would allow us to find the ratio of areas. This statement, alone and by itself, is sufficient.
Statement #2: RT = y + x and PR = y – x, so
Cross-multiply.
This allows us to solve for the ratio y/x, which would allow us to find the ratio of areas. This statement, alone and by itself, is sufficient.
Answer = (D)
5) Statement #1: this information, with nothing more, could mean that the circle is either smaller or larger.
This statement, alone and by itself, is insufficient.
Statement #2: this information, with nothing more, could mean that the circle is either smaller or larger.
This statement, alone and by itself, is insufficient.
Combined Statements: One possibility is the circle that intersects the square four times by passing through all four vertices:
That circle is clearly bigger than the square. The circle absolutely cannot pass through exactly three vertices. If it pass through two vertices, it would have to intersect the side two more times. Possibilities include the following (point C is the center of the circle).
Notice that, as point C approaches the top side of the square, it gets closer and closer to the circle that has this top side as a diameter, equivalent to the first circle in the statement #1 diagram. That circle is clearly has less area than the square. Well, that circle won’t work here, because it intersects at only two points, but because point C could get closer and closer to the top side without touching it, which means the area of the circle in this diagram could get closer and closer to the area of the first circle in the statement #1 diagram. This means that we could make the circle in this diagram have less area than the square has.
Thus, even with the constraints of both statements, we can construct a circle that has an area that is either greater than or less than that of the square. Even with both statements, we cannot give a definitive answer to the prompt question. Both statements combined are insufficient.
6) First, let’s look at the outer “lobe,” the one between 10 and 12. The circle through point 12 has a center a 6 and radius of 6, so its area is . The circle through point 10 has a center a 5 and radius of 5, so its area is . If we subtract the latter from the former, we an area of for this lobe.
Now, let’s look the middle lobe, the one between 6 and 8. The circle through point 6 has a center a 3 and radius of 3, so its area is . The circle through point 8 has a center a 4 and radius of 4, so its area is . If we subtract the latter from the former, we an area of for this lobe.
Now, let’s look the smallest lobe, the one between 2 and 4. The circle through point 2 has a center a 1 and radius of 1, so its area is . The circle through point 4 has a center a 2 and radius of 2, so its area is . If we subtract the latter from the former, we an area of for this lobe.
Add the areas of the three separate lobes: .
Answer = (B)
7) This is a difficult one. For the sake of argument, let’s say that JM = ML = 1. Then the area of the total circle would be . That’s the denominator of our ratio.
Clearly, part of the shaded region is the semicircle beneath JL. That part has an area of . That’s the easy part.
The circle intersections sides JK and KL: call these points A & B.
Point A must be the midpoint of JK, and point B, the midpoint of KL. Triangle JAM must be an equilateral triangle, with sides equal to 1. First, we will compute the area of the sector:
The angle at M is 60º, so this is one sixth of the area of the circle.
To get the area of that small shaded part, known as a circular segment, we need to subtract the area of equilateral triangle JAM from the area of the sector. See this blog for the area of an equilateral triangle.
area of segment = (area of sector) – (area of triangle JAM)
We have two circular segments, so we double this to get both:
Now, add the area of the semicircle to get the area of the whole shaded region:
Divide this by the area of the circle, , to get the ratio:
Answer = (E)
8) This problem is based on a famous theorem of Hippocrates of Chios (c. 470 – c. 410 BCE), a predecessor of Euclid. The easiest way to see this is as follows.
Let AC = BC = 2S. This is the radius of the larger circle, which has an area of , so that the quarter circle:
must have an area of . Hold that thought.
Now, notice that triangle ABC is a 45-45-90 triangle, so that the length of the hypotenuse must be
and the radius of the smaller circle is
The area of that whole circle would be , and the area of the semicircle:
would be half of that, . Thus, the quarter circle of the larger circle and semicircle of the smaller circle have the same area. That’s a very big idea!
Now, look at the circular segment:
We don’t need to know that area: simply call it J.
If we subtract the circular segment, J, from the quarter circle of the bigger circle, we get the triangle:
– J = area of triangle ABC
If we subtract the circular segment, J, from the semicircle of the smaller circle, we get the lune:
– J = area of the lune
Because those two areas equal the same thing, they must equal each other.
area of triangle ABC = area of the lune
The two are equal, so their ratio is 1. Answer = (A)
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]]>
Hello! 🙂
This week, I’m solving a tough challenge question that you might have seen on our blog if you visit often! This challenge question draws on a whole host of geometry concepts and formulas, so it’s a great test for how much of geometry you really know.
And just in case the board isn’t clear, here’s a screenshot of the board with the problem solved:
If you have any questions or comments, please leave them for me below! 🙂
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]]>The post GMAT Geometry Practice Problems appeared first on Magoosh GMAT Blog.
]]>I. 3
II. 4
III. 5
2) Each circle in the diagram above has a radius of r = 6. What is the total area of the shaded regions?
3) Triangle STV has sides ST = TV = 17, and SV = 16. What is the area?
(A) 85
(B) 100
(C) 120
(D) 136
(E) 165
4) Given that ABCDE is a regular pentagon, what is the measure of ∠ACE?
(A) 24°
(B) 30°
(C) 36°
(D) 40°
(E) 45°
5) In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?
(A) 240
(B) 270
(C) 300
(D) 330
(E) 360
6) In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
7) In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?
8) In the diagram above, AB is parallel to EH, and BD is parallel to FH. Also, AB = BC, and EF = FH. If ∠EGC = 70°, then ∠D =
(A) 65°
(B) 70°
(C) 75°
(D) 80°
(E) 85°
Answers and explanations will come at the end of this article.
On the GMAT quantitative section, you need be familiar with the basics of Geometry: parallel lines & angles; all ordinary triangles, isosceles triangles, equilateral triangles, the two special triangles, as well as the Pythagorean Theorem and the triplets that satisfy it; quadrilaterals; polygons, including regular polygons; circles — their areas, their arcs, their tangent lines, and their inscribed polygons; and 3D solids. It’s important to know about similarity and scale factors. Some ideas in Geometry have far-reaching implications, while other situations can encourage naïve assumptions. If this topic is rusty, you may find some refreshers in those posts.
If you read some of those linked blogs, you may want to give the practice problems another look before reading the solutions below. Here’s another Geometry question, from inside the Magoosh product.
9) http://gmat.magoosh.com/questions/1024
If you having anything you would like to share about Geometry, or if you have a question about something appearing in this blog, please let us know in the Comments sections.
1) Here are diagrams showing all three cases:
All three are possible. Answer = (E)
2) It would be a huge mistake to do any advanced calculations to solve this. The three lower oval shapes can be inserted into the three hollow spaces adjacent to the upper shaded region, and together, the four shapes will neatly form a single circle. The circle has r = 6, so area = 36(pi). Answer = (B)
3) Let’s think about this triangle:
Triangle STV is isosceles, so the perpendicular line from vertex T is bisects base SV. Thus, SW = 8. Now, look at right triangle ATW: it has leg = 8 and hypotenuse = 17. It will save you a tremendous amount of calculations here if you already have memorized the 8-15-17 Pythagorean Triplet. Thus, TW = 15, and that’s the height. Area = (0.5)bh = (0.5)(16)(15) = 8*15 = 120. Answer = (C)
4) First of all, all the angles in any n-sided polygon add up to (n – 2)*180°. For any pentagon, that would be 3*180° = 540°. For a regular pentagon, the five angles are equal, so each one is 540/5 = 108°.
Now, look at isosceles triangle ABC, with an angle of 108° at B. The other two angles are equal: call each x. 108 + x + x = 180, which leads to x = 36°.
So, ∠BACA = 36° and ∠ECD = 36°, which means that ∠ACE = 36° Answer = (C)
5) Let’s begin by focusing on triangle FED. The angle ∠E spans a diameter, so ∠E = 90°. Thus, triangle FED is a right triangle with hypotenuse FD = 13 and leg ED = 5. It will save you a tremendous amount of calculations here if you already have memorized the 5-12-13 Pythagorean Triplet. Thus, FE = 12. Area = (0.5)bh = (0.5)(12)(5) = 30.
Because ED and GH are parallel, all the angles are equal, and the two triangles are similar. From ED = 5 to GH = 15 we scale up by a scale factor of k = 3. Lengths are multiplied by the scale factor, and areas are multiplied by the scale factor squared, k^2 = 9. 30*9 = 270 is the area of FGH. Answer = (B)
6) Step #1: Each circle has an area of
Step #2: One sector, one “slice of pie”, occupies 60°, one sixth of the circle.
Each sector has an area of
Step #3: Now look at the equilateral triangle.
This has a side of s = 6, so it’s area is
Step #4: A circular segment is the name for that little leftover piece, the part of the sector that’s beyond the triangle:
As may be clear, the (area of segment) = (area of sector) – (area of triangle), so the sector has an area of
Step #5: Now, notice that the shaded region in the diagram is two equilateral triangles minus two circular segments. That’s exactly as we’ll calculate it.
Answer = (D)
7) This is tricky. We are given the ratio of areas, and we want to know a ratio involving the radii. Let R be the radius of the bigger circle, and r be the radius of the smaller circle.
Take the square root and rationalize the denominator:
Well, CE = 2R, and BC = r, so CE/BC = 2R/r, which is twice this ratio
Answer = (D)
8) First of all, to solve this, we need to know the properties of parallel lines & angles. Since ∠EGC = 70°, we know that an angle formed by the same line intersecting both EH & AB will make the same angle: ∠A = 70°.
Now, we know that, because AB = BC, triangle ABC is isosceles, which means that ∠ACB = 70°. The three angles have to add up to 180°, so this tells us that ∠B = 40°.
At this point, we reach a very tricky move: both ∠B and ∠H are angles formed by the pairs of parallel lines — the sides of each one are parallel to the corresponding sides of the others. These means ∠B = ∠H = 40°.
Now, we know that, because EF = EH, triangle AFH is also isosceles, which means ∠GEF = 40°. The three angles have to add up to 180°, so this tells us that ∠F = 100°.
Finally, ∠F and ∠D are two angles on the same side of the same line between two parallel lines (“same side interior angles”). These angles must be supplementary, which means they have a sum of 180°. That means ∠D = 80°.
Answer = (D)
More on why ∠B and ∠H are congruent: For some people, this is intuitive, but for others this is a really trick one. Both of those angles are formed by two pairs of parallel lines, so they have to be congruent. Why? One way to think about this — suppose segment DEF were extended to the left of D, and suppose AB were extended above A, so that it intersected the extension of DEF — call that point Q. Then, we would have two triangles, HFE and BDQ — the pairs of angles at F & D and at E & Q would be equal, because the extended version of QDEF would a common transversal crossing both sets of parallel lines. Then, the two triangles would be similar, HFE ~ BDQ, by AA Similarity, and that would mean the angles at B & H would have to be congruent.
To find out where 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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]]>Hello! 🙂 Today, I talk through how to solve this tricky-looking geometry question. If you have any questions for us or suggestions for future video topics, let us know in the comments below. Enjoy!
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]]>1) In the diagram above, what is the measure of angle y?
Statement #1: x = 30°
Statement #2: line AB is parallel to line CD
2) In the diagram above, line m and line n are parallel. Given that angle a = 40° and angle c = 55°, what is the measure of angle b?
3) In the diagram above, angle measures in degrees are marked as shown, and segment BC is parallel to line AD. What is the measure of angle E?
Solutions will come at the end of this article.
Some of the most fundamental geometry facts have to do with the special properties of parallel lines. These facts include what angles are equal, and which angles have other mathematical relationships, if the lines are parallel. The special properties of parallel lines are also directly connected to one of the most famous theorems in Geometry, the 180°-Triangle Theorem:
The sum of all three angles in any triangle equals 180°.
This is a fact true, not just for certain triangles, but for every possible triangle. A more dramatic way to say this would be: God Himself would not be able to create a triangle in the plane the sum of whose angles is not 180°.
(Cool fact that is 110% irrelevant to the GMAT: in some alternate, non-Euclidean geometries, there are no parallel lines possible, and in these geometries, the 180°-Triangle Theorem does not hold. Consider the surface of the Earth, which is approximately spherical. Consider a triangle formed by three points: (1) the North Pole, (2) the intersection of the Prime Meridian and the Equator, and (3) the intersection of the 90° West meridian and the Equator. That’s a triangle with three right angles!!)
Important idea #1 in this context is: while there are a number of special geometry facts that are true for parallel lines, absolutely none of them are true for lines that are almost parallel. Almost parallel is absolutely worthless in geometry. This has an important implication for diagrams. Unless otherwise specified, all diagrams on the GMAT Problem Solving are drawn as accurately as possible. BUT, if two lines look parallel, you can’t assume they are parallel. Two lines that look parallel could be half a degree off from being truly parallel — that difference would not be visually apparent, but none of the special parallel-line facts would be true if the two lines are not exactly parallel. Your eyes can deceive you on this. You have to see, printed in black & white: the lines are parallel. Otherwise, you can’t assume anything.
OK, if we are guaranteed that the lines are parallel, and another line intersects these parallel lines, what do we know? This diagram summarizes everything you will need to know.
Notice that we could divide these eight angles into “big” angles (angles 1 & 4 & 5 & 8) and “small” angles (angles 2 & 3 & 6 & 7). Here’s what’s true:
1. All the big angles are equal
2. All the small angles are equal
3. Any big angle plus any small angle equals 180°
There are all kinds of fancy geometry names these angles had back in high school geometry — for example, angles 3 and 6 are “alternate interior angles” (does that bring back pre-prom memories?) —- but for the purpose of the GMAT, you don’t need to know any terms more technical than “big angles” and “small angles.” Keep it simple. J
It may that this refresher cleared up a few things for you. If you found the three questions at the beginning of this article challenging, then take another look at them before reading the solutions below. Here’s a slightly more challenging question along the same lines, for practice.
4) http://gmat.magoosh.com/questions/80
1) First of all, notice from the diagram: if the lines are parallel, then the two angles would be equal, x = y; but, if the lines are not parallel, we can conclude absolutely nothing about x & y. Furthermore, a visual assessment is not enough — yes, the lines look parallel, but that’s not a guarantee that they are parallel, and without this guarantee, we can do nothing.
Statement #1: Here, we know angle x, but we don’t know whether the lines are parallel, so we can conclude nothing. Alone & by itself, this statement is insufficient.
Statement #2: Now, we know the lines are parallel, but we don’t know the values of any variables. Alone & by itself, this statement is insufficient.
Combined Statements: Now, we know the lines are parallel, and we know x = 30°, so this means y = 30°. We now have definitive information that allows us to answer the prompt question. Together, the statements are sufficient.
Answer = C
2) Imagine we constructed a new line, parallel to lines m & n, through the vertex of the “crook” between the lines. This splits angle b into two smaller angles, b1 & b2.
Notice, by the parallel lines properties, b1 = a and b2 = c, so b = b1 + b2 = a + c. This means b = 40° + 55° = 95°. Answer = C
3) This is a very tricky one. First of all, in triangle ABC, the sum of the three angles must be 180°. We are given two angles, so we know the third angle, the angle at vertex C, must be 40°. Now, because segment BC is parallel to line AD, we know this angle at C, 40°, must be equal to angle EAD. Therefore, angle EAD = 40°. Now, we know two of the three angles in triangle EAD, and we know their sum must be 180° also, so the angle at E must be 45°. Answer = D
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