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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 Number Sense for the GMAT appeared first on Magoosh GMAT Blog.
]]>1) Rank those three in order from smallest to biggest.
(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, I, II
2) Let P = 36000. Let Q equal the sum of all the factors of 36000, not including 36000 itself. Let R be the sum of all the prime numbers less than 36000. Rank the numbers P, Q, and R in numerical order from smallest to biggest.
(A) P, Q, R
(B) P, R, Q
(C) Q, P, R
(D) R, P, Q
(E) R, Q, P
3) Rank those three in order from smallest to biggest.
(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, II, I
Solutions for these number sense problems will come at the end of this blog article.
Many GMAT Quantitative Problems, like the foregoing pair, test number sense. What is number sense? Number sense is a good intuition for what happens to different kinds of numbers (positive, negative, fractions, etc.) when you perform various arithmetic operations on them.
Number sense is what allows some folks to “see” shortcuts such as estimation or visual solutions. For example, in any of the problems above, there’s absolutely no need to do any detailed calculations: in fact, folks with number sense can probably do all the math they need to do in their heads.
Of course, it would be near impossible to make anything like a complete list. The left-brain reductionist dreams of something like an exhaustive list one could study, but number sense is all about right-brain pattern matching. If you’re not familiar with the distinction of left/right hemisphere, see this GMAT post which touches on similar issues.
If you don’t have it, how do you get it? That’s not an easy question. There’s no magical shortcut to number sense, but here are some concrete suggestions.
1. Do only mental math. You shouldn’t be using a calculator to practice for the GMAT anyway. Try to do simpler math problems without even writing anything down. Furthermore, look for opportunities every day, in every situation, to do some simple math or simple estimation (e.g. there are about 20 cartons of milk on the grocery store’s shelf—about how much would it cost to buy all twenty?)
2. Look for patterns with numbers. Add & subtract & multiply & divide all kinds of numbers—positive integers, negative integers, positive fractions, negative fractions, and look for patterns. Number sense is all about pattern with numbers!! 🙂
3. This is a BIG one—in any GMAT practice problem that seemed (to you) to demand incredibly long calculations, but which had a very elegant solution of which you would have never dreamt—that problem & its solution are pure gold. In a journal, write down what insights were used to simplify the problem dramatically. Force yourself to articulate this, and return to this solution and to your notes on it often. Over time, you should develop an array of problems like this, and if you study those solutions, you probably will start to see patterns.
4. Similar to #3: search the two forums,GMAT Club and Beat the GMAT, for similarly difficult questions, and look for elegant solutions. That’s a great place to ask the experts (including yours truly) for more detailed explanations of their choices in the solution.
5. Here’s a variant on a game you can play, alone or with others who also want practice. Pick four single digit numbers at random—some repeats are allowed. You could roll a die four times, and use the results. Now, once you have those four numbers, your job is to use all four of them, each of them only once, and any arithmetic, to generate each number from 1 to 20. By “any arithmetic,” I mean any combination of: (a) add, subtract, multiply, divide; (b) exponents; and (c) parentheses & fractions
For example, if the four numbers I picked were {1, 2, 3, 4}, I could get 2 from
For any one number, you only need to come up with it in one way (although you can consider it a bonus to come up with multiple ways for a single number!) Here, I show three ways just to demonstrate the possibilities. A few examples for some of the higher numbers:
Notice that I used a variant of the expression for 13 to create an expression for 14. Also, if I changed the plus sign in the expression for 13 to a minus sign, I would get an expression for 11. Also, notice that if 1 is one of the four numbers, then if you don’t need it, you can simply multiply by it; furthermore, notice in the expression for 13 and in the second expression for 2 above, the exponent of 1 is a useful place to stash other numbers you don’t need!
As you practice, you will start to develop a sense of how expressions for one number can be tweaked to give you another number. Overall, using similar combinations, you have to get every number from 1-20 with these four, or with whatever four you pick. Actually, the set {1, 2, 3, 4} is a very good warm-up set. When you want more of a challenge, use {2,3,3,5}. 🙂
One of our Remote Test Prep Experts, Jeff Derrenberger, created an awesome web game based on this mental math game. Click on the banner to check it out!
Here’s a practice number sense problem. If you didn’t get anywhere with the practice problem, you may want to study the solution below carefully.
4) http://gmat.magoosh.com/questions/54
1) Notice that all three of these are close to fractions that equal 1/3. The fractions that equal 1/3 would be, respectively, 50/150, 110/330, and 300/900. First of all, only the second one has a higher numerator, so the second one is more than 1/3 and the other two are less than 1/3. Therefore, II is the greatest.
Now, from I and III, which is greater? Well, think about it this way. 50/150 = 300/900, because both of those equal 1/3. How much less than one third is each one of these? Well, 47/150 is 3/150 less than 50/150 = 1/3, and 299/900 is 1/900 less than 300/900 = 1/3. Well, clearly, 3/150 > 1/900 (the latter has a smaller numerator and a larger denominator!) Therefore, starting from 1/3, 47/150 goes down further than does 299/900. Therefore, 47/150, dropping down a larger distance, must be the minimum value. Therefore, the correct order is I, III, II.
Answer = (B)
2) We know that some of the factors of 36,000 are 18,000, 12,000, and 9,000. Right there, those three add up to 39,000 more than 36,000. Right there, we know that P < Q. We can eliminate (C) and (E).
Now, R is a little trickier. We don’t need to have detailed knowledge here. We know there are several prime numbers less than 100. Obviously, the density of prime numbers gets slightly less as we get bigger. Let’s assume, extremely conservatively, that when we get up into the 20 and 30 thousands, there is at least one prime number every thousand: one between 20K and 21K, one between 21K and 22K, all the way up to 36K. The 6 primes in the thirty thousands are all greater than 30K, so let’s estimate their sum as (30K)*6 = 180K. The 10 primes in the twenty thousands are all greater than 20K, so let’s estimate their sum as (20K)*10 = 200K. Right away, that’s 380K on an extremely conservative estimate–we didn’t even include any of the primes less than 20,000. There is no way that the sum of the factors of 36K, not including 36K itself, is going to be more than ten times 36K! Thus, R is much larger than Q, and the correct order is P < Q < R.
Answer = (A)
BTW, if your curious, according to Wolfram Alpha, the sum of the factors of 36000, not including 36000, is Q = 91,764, and the sum of all the prime number less than 36000 is R = 64,711,067.
3) Here, we have to “un-simplify” the square-roots to get a sense of their relative size.
From this, we see that II is less than I. From this alone, we can eliminate (A) & (B).
The trickier item on the list is III. Without a calculator, it would be nearly impossible compute an exact value for the fourth-root of 401. But consider this: the fourth root of a number is the number to the power of 1/4, and 1/4 = (1/2)*(1/2), so the fourth root is the square root of a square root. Now, of course, 401 is not itself a perfect square, but it is very close to a perfect square.
This demonstrates that III. is slightly larger than I. Therefore, the order from least to greatest is II, I, III.
Answer = (C)
Editor’s Note: This post was originally published in December, 2012, and has been updated for freshness, accuracy, and comprehensiveness.
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]]>The post Consecutive Integers and Multiples on the GMAT appeared first on Magoosh GMAT Blog.
]]>1) S is a set of n consecutive positive integers. Is the mean of the set a positive integer?
Statement #1: the range of S is an even integer
Statement #2: the median of S is a positive integer
2) If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?
I. 4
II. 6
III. 18
Explanations for these will come at the end of the article.
The word “consecutive” means “in a row; one after the other.” A set of three consecutive integers might mean {3, 4, 5} or {137, 138, 139} or {–25, –24, –23}. As long as the integers are in a row, it doesn’t matter whether they are big or small, positive or negative. In fact, the set {–1, 0, +1} contains one positive number and one negative number (of course, zero is neither positive nor negative).
a) Any set of n consecutive integers will contain exactly one number divisible by n. For example, any three integers in row must contain a multiple of 3; any 17 integers will contain one multiple of 17, etc. Now, you may look at the set {–1, 0, +1}, a set of three consecutive integers, and wonder: where is the multiple of 3? This is tricky. As it turns out, zero is a multiple of every integer, because (any integer) times zero equals zero.
b) In a set of three consecutive integers, we could have two evens and one odd, or two odds and one even, depending upon where we started. In a set of four consecutive integers, we would have to have two evens and two odds. More generally, if we have an odd number of consecutive integers, we could have more evens or more odds, depending on the starting value, but if we have an even number of consecutive integers, the evens and odds have to be evenly split.
c) If n is an odd number, then the sum of n consecutive integers is divisible by n. For example: for any three integers in a row, the sum is divisible by 3; for any 7 integers in a row, the sum is divisible by 7, etc.
Anyone can recognize that {6, 7, 8, 9, 10, 11} is a set of consecutive integers. When you are given plain old numbers, it’s easy to see whether they are consecutive integers. That’s too easy. The GMAT will not ask about that. Instead, the GMAT will give you algebraic representations of consecutive integers.
The following are examples of algebraic representations of consecutive integers:
{n, n + 1, n + 2, n + 3, n + 4, n + 5}
{n – 2, n – 1, n, n + 1}
{n + 12, n + 13, n + 14}
For all of those, if n equals any integer, then the set will be a set of consecutive integers. For simplicity, let’s pick n = 10. The first set becomes the set of integers from 10 to 15; the second, from 8 to 11; and the third, from 22 to 24. Even without knowing the value of n, we can apply the properties of consecutive integers to the set: for example, the second is a set of four consecutive integers, so it must have two evens and two odds; the third is a set of three consecutive integers, so the sum of those three numbers must be divisible by three.
If you had some insights reading this article, you may want to give the problems above a second look before reading the solutions below. Here’s another practice question from inside Magoosh:
5) http://gmat.magoosh.com/questions/943
If you have questions about anything unclear in this post, please let us know in the comments section.
1) First of all, for a set of consecutive integers, or for any set of evenly spaced numbers, the mean and the median are equal. If there’s an odd number of members of the list, then the median is the middle number. If there’s an even number of members of the list, then the median is the average of the two middle numbers. For example, the median of {1, 2, 3, 4, 5} is 3, a positive integer and member of the set. For consecutive integers, an even number of members would mean that the mean or median is the average of the two middle integers. For example, the median of {1,2, 3, 4} is the average of 2 and 3, that is, 2.5, not an integer. The only way the mean or median can be an integer is if the set of consecutive integers has an odd number of members.
Statement #1: If there are an even number of consecutive integers, then the evens and odds are balanced in the set, and the first and last number must be opposite: one must be even and the other must be odd. Thus, the range, the difference of (max) – (min) would be either (even) – (odd) or (odd) – (even), in either case, an odd number. If the range is odd, the number of consecutive integers is even.
If there are an odd number of consecutive integers, then the first and last numbers are either both even or both odd. The range would be either (even) – (even) or (odd) – (odd), in either case, an even number. If the range is even, the number of consecutive integers is odd. That must be the case here. As we have seen above, this means the mean or median is a positive integer. This statement, alone and by itself, is sufficient.
Statement #2: As we discussed above, the mean = the median. If the latter is a positive integer, so is the former. This statement, alone and by itself, is sufficient.
Both statement are separately sufficient. Answer = (D)
2) When we have a set of consecutive integers or consecutive multiples of the number, the range depends only on the size of the set, how many members, not where on the number line the set starts or ends. For example, any seven consecutive integers will have a range of 6, whether it’s 1 through 7 or 51 through 57. Thus, we can ignore the starting number, 255, which is just there to confuse us. We can pick any more convenient starting value.
Let’s start at a1 = 15 = 15*1. Then a2 = 15*2 = 30, and a3 = 15*3 = 45. Continuing in this pattern, the last number would be a23 = 15*23. Don’t multiply that yet. The range would be highest minus the lowest:
range = (a23) – (a1) = 15*23 – 15*1 = 15*(23 – 1) = 15*22
Now, use the doubling & halving trick. Half of 22 is 11, and twice 15 is 30, so
15*22 = 11*30 = 330
Answer = (B)
3) This is a tricky one. If you start plugging in values for n, you are sunk. The numbers are gigantic and unwieldy. This one begins with some clever factoring. Clearly, in the first factor, we can factor out a factor of n:
In the second one, we can us the Difference of Two Squares:
Now, put all that together, and rearrange the order:
Written in that order, we see this is a product of four consecutive integers. We absolutely know that two of the numbers are even and two are odd. If there are two even numbers, each one has a factor of two, so the product would have a factor of 2*2 = 4. The product must be divisible by 4. We know I must be true.
Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one. We have a multiple of 3, which has 3 as a factor, and at least one even number, which has 2 as a factor. They may be the same number or may not be: that doesn’t matter. As long as the entire product contains at least one factor of 3 and at least one factor of 2, then the product must be divisible by 6. We know II must be true.
To be divisible by 18=2*3*3, we need one factor of 2 and two factors of 3. We definitely have more than one factor of 2. The problem is: how many factors of three do we have? In a set of four consecutive integers, we could have two factors of three, as in {12, 13, 14, 15} or {7, 8, 9, 10}. BUT, we could also have a set of four consecutive integers with only one factor of three, as in {4, 5, 6, 7}. Thus, we could have two factors of three, and the product could be divisible by 18, but we are not sure. We cannot say that this must be true. Therefore, III cannot be among the answer.
Only I and II work. Answer = (C)
4) We know that odd integers are spaced two apart: if we have one odd integer, we can add 2 to get the next one. Starting from the given expression for the first member of the set, we can say:
This highest member, we can set equal to 7n
n = 2 or n = 5. Well, if n equals 2, then the first member would be 4, not an odd integer. That value doesn’t work with the problem. Therefore, we know n = 5. This means that first member is 25, and the whole set is {25, 27, 29, 31, 33, 35}. The median of this set is the average of the two middle numbers, 29 and 31; median = 30.
Answer = (D).
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]]>The post GMAT Quant: Difficult Units Digits Questions appeared first on Magoosh GMAT Blog.
]]>1) The units digit of is:
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
2) The units digit of is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
3) The units digit of is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
Admittedly, these problems are probably a notch harder than anything you are likely to see on the GMAT. If you understand these, you definitely will understand anything of this variety that the GMAT throws at you!
All of those problems above involve numbers with hundreds of decimal places. No one can calculate those answers without a calculator: in fact, no calculator would be sufficient to do the calculation, because no calculator can accommodate that many digits. If one needed the exact answer, one could always use that most extraordinary web computing tool, Wolfram Alpha. Of course, one will not have access to the web or a calculator or anything other than one’s owns wits when confronting a question such as this on the GMAT. How do we proceed?
It turns out, what appears as a ridiculously hard calculation is actually quite easier. No part of the calculation we are going to do will involve anything beyond single-digit arithmetic!
The units digits of large numbers are special: they form a kind of elite and exclusive club. The big idea: only units digits affect units digits. What do I mean by that? Well, first of all, suppose you add or subtract two large numbers —- the units digit of the sum or the difference will depend only on the units digits of the two input numbers. For example, 3 + 5 = 8 —- this means that any number ending in 3 plus any number ending in 5 will be a number ending in 8. If you remember your “column addition” processes from grade school, this one might make intuitive sense.
The one that can be a little harder for folks to swallow is multiplication. If you multiply two large numbers, the unit digit of the product will have the same units digit as the product of the units digit of the two factors. That’s a mouthful! In other words, let’s take 3*7 = 21, so a units digit of a 3, times a unit digit of a 7, equals a units digit of a 1. That means, we could take any large number ending in 3, times any large number ending in 7, and the product absolutely will have to have a units digit of 1. If this is new idea to you, I strongly recommend: sit down with a calculator and multiply ridiculously large numbers together, with all combinations of units digits, until you are 100% satisfied that this pattern works.
This part will be a recap of an earlier post on powers of units digits. When we raise to a power, of course, that’s iterated multiplication, so we just follow the multiplication rule above. As it turns out, a simple pattern will always emerge.
Suppose we were considering powers of 253 — first of all, only the units digit, 3, matters, for the units digit of the powers. Any number ending in three will have the same sequence of units digits for the powers.
= 3
= 9
9*3 = 27, so has a units digit of 7
7*3 = 21, so has a units digit of 1
1*3 = 3, so has a units digit of 3
3*3 = 9, so has a units digit of 9
9*3 = 27, so has a units digit of 7
7*3 = 21, so has a units digit of 1
Notice a pattern has emerged — 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, … It repeats like mathematical wallpaper. The pattern has a period of 4 — that is to say, it take four steps to repeat. This means, 3 to the power of any multiple of 4 has a units digit of 1: , , and all have a units digit of one. If I want to find power that’s not a multiple of 4, that’ easy: I just go to the nearest multiple of 4 and follow the wallpaper from there. For example, if I wanted —–
has a units digit of 1
has a units digit of 3
has a units digit of 9
As it happens, is a number that has 213 digits, but the units digit must be a 9.
The really expansive idea: everything I have just said about powers of 3 also applies to any larger number that happens to have a units digit of 3. Thus,
has a units digit of 9
That number has over a thousand digits (you don’t need to know how to figure that out!) but we know for sure that the units digit of this gargantuan number is 9.
If reading this article gave you any insights, you may want to give the questions above a second try. Here’s another problem, slightly easier and more GMAT-like, of this genre:
4) http://gmat.magoosh.com/questions/648
If there is anything you would like to say on this topic, or if you have any questions, please let me know in the comments section.
Finally, on a totally gratuitous note, here, in it’s the full thousand-plus-digit glory, is , courtesy of Wolfram Alpha:
= 6190880832531899190821690500833264378796558621138440416684569816896956548548008694
5501637120599244459832600741641042834529231770919235178215551804306285786976623543
72380800482154741117883167160446214356675850891464689434900627878445424968534812213
04515041521698298553935861735825956938171070649017829532068323699186643091574519875
641384511684642401730622089635116285314952964987090639595147866795944410916642166093
585848058327971863158257619930226042698661632905146850162960633520155118628867911625
239725418415604877699453370534194837316774432534349898272185517986005836675979188507
704257742239368667474408667895362250511057160490511029003928521905584001998500412272
300652930331121107733643816582958394189572596322595033481338694429893546070448926193
272806103607662243076062238492673013489615386273692928543218155895489937520257687664
947027647847750945509362588170852889875925160078794611182855714905968753089053225774
863189760920183769031795458368827168630624310066175033265292467587132663811805301906
7641362643313498166787213628751583911774745199740840719668395714479929
Notice, of course, that the units digit is 9.
1) First of all, all we need is the last digit of the base, not 137, but just 7. Here’s the power sequence of the units of 7
has a units digit of 7
has a units digit of 9 (e.g. 7*7 = 49)
has a units digit of 3 (e.g. 7*9 = 63)
has a units digit of 1 (e.g. 7*3 = 21)
has a units digit of 7
has a units digit of 9
has a units digit of 3
has a units digit of 1
etc.
The period is 4, so 7 to the power of any multiple of 4 has a units digit of 1
has a units digit of 1
has a units digit of 7
So the inner parenthesis is a number with a units digit of 7.
Now, for the outer exponent, we are following the same pattern — starting with a units digit of 7. The period is still 4.
has a units digit of 1
has a units digit of 7
has a units digit of 9
has a units digit of 3
So the unit digit of the final output is 3. Answer = B
BTW, this number is the great-granddaddy, the biggest number of all the big numbers mentioned in this post. The number clocks in with over 1300 digits!
2) We have to figure out each piece separately, and then add them. The first piece is remarkably easy — any power of anything ending in 5 always has a units digit of 5. So the first term has a units digit of 5. Done.
The second term takes a little more work. We can ignore the tens digit, and just treat this base as 3. Here is the units digit patter for the powers of 3.
has a units digit of 3
has a units digit of 9
has a units digit of 7 (e.g. 3*9 = 27)
has a units digit of 1 (e.g. 3*7 = 21)
has a units digit of 3
has a units digit of 9
has a units digit of 7
has a units digit of 1
The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.
has a units digit of 1
has a units digit of 3
has a units digit of 9
Therefore, the second term has a units digit of 9.
Of course 5 + 9 = 14, so something with a units digit of 5 plus something with a units digit of 9 will have a units digit of 4. Answer = B
3) We have to figure out each piece separately, and then multiply them. The powers of 4 are particularly easy.
has a units digit of 4
has a units digit of 6 (e.g. 4*4 = 16)
has a units digit of 4 (e.g. 4*6 = 24)
has a units digit of 6
has a units digit of 4
has a units digit of 6
Four to any odd power will have a units digit of 4. Thus, any number with a units digit of four, raised to an odd power, will also have a units digit of 4. The first factor, , has a units digit of 4.
Now, the base in the second factor ends in a 3 (we can ignore the tens digit). Here is the pattern for powers of three.
has a units digit of 3
has a units digit of 9
has a units digit of 7 (e.g. 3*9 = 27)
has a units digit of 1 (e.g. 3*7 = 21)
has a units digit of 3
has a units digit of 9
has a units digit of 7
has a units digit of 1
The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.
has a units digit of 1
has a units digit of 3
Thus, any number with a units digit of 7, when raised to the power of 37, will have a units digit of 3. The second factor, , has a units digit of 3.
Of course, 4*3 = 12, so any number with a units digit of 4 times any number with a units digit of 3 will yield a product with a units digit of 2.
Answer = A
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]]>The post GMAT Number Properties appeared first on Magoosh GMAT Blog.
]]>
Let’s first talk about the constituents of this realm. When I say the word “numbers”, what do you think? If the only thing that comes to mind is (1, 2, 3, 4, 5, …), then, with all due respect, you are thinking like a second-grader. Not surprisingly, such thinking will get you in trouble on the GMAT. If, when I say the word “numbers”, you immediately think not only of those numbers but also their negative counterparts, as well as zero, positive & negative fractions, positive & negative decimals, squareroots, pi, etc. etc., then you are thinking like a mathematician. One very simple trick for success on GMAT math problem: whenever you hear the word “number”, automatically force yourself to think exhaustively of every possibility.
Numbers live on what grade-school teachers call the “number line” and what mathematicians call the “real number line”, a continuous infinity of values. (The more you think about the idea of continuous infinity, the more it boggles the mind!) The real-number line is a real of perfect fairness —- the number -137/8 is just as much a legitimate member of the number line as is 4. If you have any prejudice that, for example (1, 2, 3, 4 ….) are the only “worthy” numbers and all the rest are somehow exceptions or illegal aliens or something like that, you have to retrain your mind to see the perfect fairness and equal worthiness of each and every number on the real number line.
It’s particular important if you are choosing numbers plugging in, say to test algebraic expression in a Data Sufficiency question, to remember to test numbers of all categories. Check out some more GMAT Data Sufficiency tips and hints.
In addition to knowing the infinite implications concerning the word “number”, two further terms you should know are “integers” (… -4, -3, -2, -1, 0, 1, 2, 3, 4, …) and “positive integers” (1, 2, 3, 4, …). On a rare occasion, the GMAT will also have a question concern “negative integers”, which are just the negative counterparts of the positive integers. Notice: while zero is an integer, it is neither positive nor negative. The word “integer” is etymologically related to the word “integrity”: both come from the Latin word for wholeness. Each integer is, as it were, a unique wholeness, a complete package in and of itself.
The “parts” between the “wholes” can be represented in two ways: fractions and decimals. Both can be either positive or negative. Mathematicians have both systems for these “partial” numbers because each system has its own advantages under certain circumstance. Technically, “fractions” and “decimals” don’t cover exactly the same turf, because, as I discuss in that post on decimals, fractions include only the rational numbers, whereas decimals include both rationals and irrational numbers, an infinitely bigger set. Thus, even though there’s an infinity of fractions between 0 and 1 as well as an infinity of decimals between 0 and 1, the infinity of decimals is infinitely bigger than the infinity of fractions (one of the many mind-boggling things about the continuous infinity of the real number line!) Irrational numbers include roots, which the GMAT loves to test.
The positive integers (1, 2, 3, 4, ….) are also called the “counting numbers”, and mathematicians call them the “natural numbers”, but “positive integers” seems to be the term the GMAT consistently uses. These numbers have a bunch of special properties just to themselves: primes, odd & even, multiples, factors, remainders, etc.
One of the hardest things the GMAT will ask you to do with positive integers is to count. This may sound paradoxical, as counting is one mathematical thing even the most inveterate math-phobe can do, but here, by “counting”, I mean problems like: “There are three boys in blue shirts, five boys in green shirts, and four girls in blue dresses. How many ways can they sit in seven seats if blah blah blah?” Those arrangements and sets and combinations —- that’s what the GMAT will have you count. Also, remember the trick of inclusive counting, which the GMAT loves to test.
Number properties is a huge topic: if you follow all those links, you will have a good understanding of the lay of the land. It’s also good to know the most common math mistakes on the GMAT, and the other most common GMAT questions. Finally, here’s a very easy practice question, to get you warmed up to thinking about this topic as it appear in test-format questions:
1) http://gmat.magoosh.com/questions/2621
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]]>The post GMAT Math: One is NOT a Prime Number appeared first on Magoosh GMAT Blog.
]]>
One definition of a prime number is: any number that has only two positive integer factors—itself and 1. The following are valid prime numbers.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ….
It is very good to know that 2 is the only even prime number. It is often handy to know the first eight or ten prime numbers.
Notice that 1 does not fit the fundamental definition for a prime number: it has only one positive integer factor, itself, not two. Therefore, it is not prime.
Now, to some folks, that rule will seem like a trivial technicality. We permanently excluded Fred from the elite country club because he owns one powder blue tie instead of two. We permanently excluded 1 from the elite set of prime numbers because it has only one positive factor instead of two. What kind of arbitrary rip-off is that? Who can we sue?
It turns out, as usual, mathematicians have deep reasons for the way they draw fine distinctions. Let’s think about this. One major rule of arithmetic is: each positive integer greater than one has a unique prime factorization. (This rule is so important, its official name is The Fundamental Theorem of Arithmetic!) The prime factorization of a number is like its DNA: we know its exact constituents, and thus can determine every single one of its factors. For example:
36 = 2*2*3*3
That is the unique prime factorization of 36, the only way to multiply prime numbers to get a product of 36. Now, pretend for a moment that 1 were a prime number: if we defined things that way, what would the consequences be? Instead of having a unique prime factorization, every number would have an infinite number of prime factorizations. For example, the first few prime factorizations of 36 would be:
36 = 2*2*3*3
36 = 1*2*2*3*3 (duh!)
36 = 1*1*2*2*3*3 (duh!)
36 = 1*1*1*2*2*3*3 (duh!) etc. (an infinite number of duh! statements)
Not only would we demolish a perfectly good rule of arithmetic, the Fundamental Theorem of Arithmetic, but in doing so, we also would gain an infinite number of absolutely useless statements. That’s a lose-lose trade-off! Mathematicians, sensing this lose-lose situation, choose to head it off at the pass simply by stating, by fiat, 1 is not a prime number. Choosing this particular definition renders irrelevant this troubling situation with an infinite number of duh! statements as well as numerous similar problematic situations. Mathematicians are crafty enough to realize they can avoid a whole boatload of problems just by making a single stipulation: 1 is not a prime number. This is the deep reason for the rule.
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]]>The post GMAT Math: The Many Meanings of Minus appeared first on Magoosh GMAT Blog.
]]>What does this symbol mean in math?
Technically, this symbol, typographically a dash, has three different meanings in mathematics, viz.:
a) a subtraction sign
b) a negative sign
c) an opposite sign
When the dash appears between two terms — between two numbers (5 – 3), between two variables (x – y), between a number and a variable (x – 2), etc. — then it indicates the operation of subtraction. This is, undoubtedly, the very first meaning folks associate with the dash, because folks learned this way back in grammar school math. Furthermore, as soon as kids understand money and spending money, essentially they understand something about subtraction, so rare is the kid who doesn’t get subtraction early on.
When the dash appears in front of a stand-alone number (–5, –2.7, , etc.), then it is a negative sign, denoting that the number in question is less than zero and to the left of zero on a standard number line. Folks tend to learn this idea relatively early on as well. Moreover, it’s easy to see how this meaning “blurs” into the subtraction-sign meaning, because after all, 8 + (–2) is just another way of saying 8 – 2. Both have a kind of “minus-making” meaning to them.
This is the one that can through folks into a tizzy. When you put the dash not in front of a stand-alone number but rather a stand-alone variable, then it is NOT a negative sign anymore. Rather, it is an opposite sign, which changes the sign of the variable to the opposite of whatever it was originally. If y is a positive, then –y is negative. BUT, if we know y is negative, then we know –y is positive.
Right there, that is precisely what wigs people out! Since before puberty, they were accustomed the dash having a universal “negatizing” effect, and yet, in this strange instance, when y is already negative, the dash in front of –y actually makes it positive. To some folks, this seems an unholy violation of everything they have ever learned about the sign! Technically, folks learn about the opposite sign somewhere in algebra, but it is seldom explained well there, setting folks up for this massive confusion when they encounter the opposite sign on, for example, the GMAT Quantitative section.
For example, the algebraic statement:
|y| = –y
is a sophisticated way of indicating that y ≤ 0. Conceivably, the GMAT could give you the former and expect you to deduce the latter.
It may be that the foregoing discussion gave you some insight into the practice question at the top of this pages. Take another look at that before reading the solution below. Also, here’s a free practice question with some positive/negative variable issues.
2) http://gmat.magoosh.com/questions/301
1) This is a difficult DS question. First of all, some folks might mistakenly think the prompt is already decided as it is written. After all, we know that the right side of the inequality,, is always positive. Some folks may mistakenly think that the left side of the inequality is always negative, but that would entail reading the dash incorrectly as a negative sign, rather than correctly as an opposite sign.
Statement #1 implies that y = ±1/5. (How do you know you have to take both the positive and negative square roots? See this GRE post.) Those two values imply different conclusions to the prompt. If y = +1/5, then –y/2 = –1/10, which is clearly less than . But, if y = –1/5, then –y/2 = +1/10, which happens to be greater than . The two different possible values imply different conclusions, which means no definitive answer to the prompt is possible. This statement, by itself, is insufficient.
Statement #2 is a fancy way of saying that y ≤ 0. If y = –5, then the inequality is false, but if y = –1/100, then –y/2 = +1/200, which is greater than , and the inequality is true. Two different possible values imply different conclusions, which means no definitive answer to the prompt is possible. This statement, by itself, is insufficient.
Statement combined: when we combine the restraints of both statements, we know that y can only have one value: y must equal –1/5, and this, by itself leads to a definite answer to the prompt. Thus, the combined statements are sufficient.
Answer = C.
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]]>First of all, consider this DS problem
1) Is P > Q?
Statement #1: P = 5*Q
Statement #2: P = Q + R
For folks who are not as comfortable with math, or are a bit rusty with math, may fall into any one of a number of traps here. Nothing is specified about any of these three variables, and on the GMAT, when nothing is specified, we have to assume that every category of number is possible. This means we have to consider
a) 1 (the great exception to many rules)
b) 0 (another great exception to many rules
c) positive integers (1, 2, 3, 4, etc.)
d) negative integers (-1, -2, -3, -4, etc.)
e) positive fractions (1/2, 1/10, etc.)
f) negative fractions (-1/2, -1/10, etc.)
For a mathematic statement in variables to be true, it must be true for every category of number. For example, the commutative law (a + b = b + a and c*d = d*c) works for every single number on the number line, all infinity of them. Statements like that are rare indeed.
In the problem above, if you fall into the trap of thinking only in terms of positive numbers, you may go astray. If we specified that P & Q & R must be positive integers, then each statements would be sufficient on its own, and the answer would be D. BUT, as it stands, those variables could be anything.
If P and Q are positive, then P = 5*Q implies P > Q, but if P and Q are negative, then P = 5*Q implies Q > P (for example, consider P = -10 and Q = -2). Statement #1 is not sufficient.
If R is positive, then adding R makes something bigger, but if R is negative, then adding R makes something smaller. Statement #2 is not sufficient.
Even if we combine both statements — if we pick three positive numbers (e.g. P = 5, Q = 1, and R = 4) we get a “yes” answer to the prompt; but, if we pick, say P = Q = R = 0, then the both math equations work, and the answer to the prompt would be “no.” Nothing is sufficient. Answer is E precisely because there are too many possibilities to determine anything.
Whenever you see variables, and no constraints are explicitly specified in the problem, you must consider all possible cases!
Here’s a challenging DS practice question:
1) http://gmat.magoosh.com/questions/966
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]]>The post GMAT Factorials appeared first on Magoosh GMAT Blog.
]]>First, consider these problems
3) Consider these three quantities
Rank these three quantities from least to greatest.
These are challenging problems, especially the third one! With a few simple insights about factorials, though, you will be able to manage all of these.
First of all, a few basics. The factorial is a function we can perform on any positive integer. The expression 5! (read “five factorial”) means the product of all the positive integers from that number down to one. In this particular case: 5! = 5*4*3*2*1 = 120. Here are the first ten factorials, just to give you a sense:
1! = 1
2! = 2*1 = 2
3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
5! = 5*4*3*2*1 = 120
6! = 6*5*4*3*2*1 = 720
7! = 7*6*5*4*3*2*1 = 5,040
8! = 8*7*6*5*4*3*2*1 = 40,320
9! = 9*8*7*6*5*4*3*2*1 = 362,880
10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800
It’s a good idea to have the first five memorized, simply because those come up frequently on the GMAT —the first five are pretty easy to figure out on your own anyway. Nobody expects you to have the last five here memorized. I give them here purely to give you a sense of how quickly the factorials grow. By the time we get to (10!), we are already over a million! Similarly, (13!) is more than a billion, and (15!) is more than a trillion. Holy schnikes!
Here are some big ideas to help you when you have to perform arithmetic operations with factorials.
Big Idea #1: every factorial is a factor of every higher factorial
The number 73! must be divisible by 72!, by 47!, by 12!, etc. The number 73! automatically has at least 72 known factors — all the factorials less than it!
Big Idea #2: you can “unpack” one factorial down to another.
Think about 8! —- we know:
8! = 8*7*6*5*4*3*2*1
Well, by the Associative Law, we can group factors in any groupings, so we could insert parenthesis wherever we like. In particular , if I include a selection of factors that goes all the way to the right, all the way to 1, that’s another factorial. Thus:
8! = 8*(7*6*5*4*3*2*1) = 8*(7!)
8! = 8*7*(6*5*4*3*2*1) = 8*7*(6!)
8! = 8*7*6*(5*4*3*2*1) = 8*7*6*(5!) etc.
I chose 8! because we can see all the factors, but clearly we could extend this idea to any factorial, no matter how large:
237! = 237*236*(235!)
That’s an “unpacking” of the first two factors of 237! (BTW, this term, “unpacking”, is my own creation: you will not see this term used anywhere else.)
Big Idea #3: when you divide two factorials, you “unpack” the larger one, and cancel it with the smaller one.
Example:
Big Idea #4: when you add/subtract two or more factorials, you unpack them all down to the lowest one, and factor out that common factor.
For more on “factoring out”, see the section “Distributing and factor out” in this post.
Examples:
(200!) – (199!) = 200*(199!) – (1)*(199!) = (200 – 1)*(199!) = 199*(199!)
(200!) + (199!) = 200*(199!) + (1)*(199!) = (200 + 1)*(199!) = 201*(199!)
Having read these rules, give those three practice problems another try before reading the solutions below. Here’s a fourth problem, with its own video explanation.
4) http://gmat.magoosh.com/questions/811
1) We are going to use the “factoring out” trick in both the numerator and the denominator:
Now, “unpack” that top factorial, to cancel the smaller one in the denominator:
Answer = D
2) “Unpack” the factorials in the numerator, so that everything is expressed as a product involving (89!) Then factor out that common factor.
Answer = D
3) Let’s consider the three expressions separately.
For expression I, we merely have to “unpack” the (49!) factor in the numerator.
So, expression I has a value of 49.
The expression II is tricky:
That’s a whole lot of factors in the numerator! That numerator is fantastically big: it has forty factors from 49 to 10 that are all greater than or equal to 10, so that means it’s automatically bigger than 10^40 (a number bigger than 1 trillion cubed — OMG!) That’s divided by 7! = 5040, so whatever this is, it’s way bigger than 49.
The expression III is a little easier than II:
That’s also a very big number. Each of the seven factors in the numerator is greater than 10, and 10^7 is ten million, so that numerator is more than 10,000,000. That’s divided by 7! = 5040. This is clearly bigger than 49. Notice, though, this has the same denominator as II, but many fewer factors in the numerator. Therefore II is much bigger than III.
Thus, from least to greatest, the order is I, III, II. Answer = B
Just as a note, if you are familiar with the idea of combinations, you may recognize expression II as a combinations number:
That would be the number of unique sets of 7 we could select from a pool of 49 unique items. There is no way you would be expected to calculate that number without a serious calculator or computer.
That’s a reasonably big number – just over 85 million. That’s a little more than a quarter the current population of the USA. For comparison, expression II is a real whopper:
This is a larger number than all humans and all other living things (animals & plants & all the way down to single-cell critters), including those alive now as well as those who have ever been alive on Earth. This number is larger than all the money in the world in pennies. This is more than the number of individual atoms comprising planet Earth and everything on Earth. This number is much much larger than the number of stars & planets & pulsars & quasars & black holes & whatever other star-like things in all galaxies & clusters in the visible Universe. The phrase “inconceivably big” does not even begin to capture how big this number is!
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]]>1) Given f(x) = 3x – 5, for what value of x does 2*[f(x)] – 1 = f(3x – 6)
If you find these questions completely incomprehensible, then you have found the right blog post.
The GMAT Quantitative section will ask an occasional question about function notation. Here is a basic catechism about functions and what you need to know about them for the GMAT.
A function is a rule, a “machine”, that takes an input and gives an output. When we are told the equation of a function, that equation makes explicit the rule this particular function is following. For example, for the function f(x) = 3x – 5, the rule is: whatever input x you give —- and that input could be any real number —- this function will multiply this input by 3 and then subtract five from the product: that difference is the output. If I put in an input of x = 2, then I get an output of 3(2) – 5 = 1, and the way we compactly write that fact with function notation is: f(2) = 1. In other words, an input of 2 gives an output of 1.
Notice — this is a very subtle issue. The x that appears in the equation of a function is a different sort of variable than the ordinary x of solve-for-x algebra. This x is what one might call a “formula variable”, like the a, b, and c in the quadratic formula. In other words, the x of function notation is not an x that is equal to only a single value; rather, it can be set equal to any value, any real number on the number line, when we want to plug that number into the function.
When we write f(x), many people new to function notation will misinterpret this as multiplication — as if there’s a thing “f” times the variable “x”. That is 100% incorrect. A function is a process through which the input goes. Cooking is a process through which food goes. Puberty is a process through which people go. A function is a process acting on the number, and the nature of that action is outside the categories of simple arithmetic actions (add, subtract, multiply, divide).
Relatedly, the parentheses of function notation are mathematically inviolable. Nothing may pass through these parentheses. Again, this can be anti-intuitive, because when parentheses are used in ordinary notation, you can distribute through parenthesis, factor out, etc. Because a function is a different category of mathematical object, its parentheses are of a different nature. Thus
If you can simply avoid these mistakes and respect at all times the inviolability of the function’s parentheses, you will already be in better shape than a sizable portion of GMAT test takers.
In the above section, I discussed ways that folks new to functions might misinterpret function notation. Now, I am going to discuss how functions are seen by people who really understand them. Suppose we have the function f(x) = 3x – 5. Here’s what a mathematician looking at this function sees:
Where folks new to function just see the letter x, mathematicians see a “box”, an empty slot, a space that is, in some ways, analogous to an artist’s blank canvas. Anything that get plugged into the box on the left needs to get plugged into the box on the right. We can plug in numbers — any of the continuous infinity of real numbers on the real number line. We can also plug in algebraic expressions: If I put (2x + 7) into the box on the left, I need to put that exact identical expression into the box on the right. I can even put whole functions — the same function or an entirely different function — into the boxes. In fact, the list of mathematical objects that can be plugged into a function extends into far more sophisticated mathematical objects (matrices, differential operators, etc.) that are well beyond the realm of GMAT Quant. The GMAT, though, will expect you to know what to do if they give you, say, the function f(x) = 3x – 5, and then ask you, say, to plug in the expression 2x + 7:
f(2x + 7) = 3*(2x + 7) – 5 = 6x + 21 – 5 = 6x + 16
If this is your first time encountering, or first time understanding, function notation, it is a worthwhile topic to practice, so that you are comfortable with it by test day. If you feel you have learned something from this, go back and try those two practice problems again before reading the solutions below. Also, here’s a practice question from our product:
3) https://gmat.magoosh.com/questions/147
1) We have the function f(x) = 3x – 5, and we want to some sophisticated algebra with it. Let’s look at the two sides of the prompt equation separately. The left side says: 2*[f(x)] – 1 —- this is saying: take f(x), which is equal to its equation, and multiply that by 2 and then subtract 1.
2*[f(x)] – 1 = 2*(3x – 5) – 1 = 6x – 10 – 1 = 6x – 11
The right side says f(3x – 6) — this means, take the algebraic expression (3x – 6) and plug it into the function, as discussed above in the section “How a mathematician things about a function.” This algebraic expression, (3x – 6), must take the place of x on both sides of the function equation.
f(3x – 6)= 3*[3x – 6] – 5 = 9x – 18 – 5 = 9x – 23
Now, set those two equal and solve for x:
9x – 23 = 6x – 11
9x = 6x – 11 + 23
9x = 6x + 12
9x – 6x = 12
3x = 12
x = 4
Answer = B
2) There are several ways to approach this problem. One quick way is to notice that if x = 2, f(2) = 2/3. That’s not the answer, but it gives us a shortcut. If f(k) = 2, then we see that f(f(k)) = f(2) = 2/3. So, really, finding the value of k that satisfies the prompt equation really simplifies to solving the equation f(k) = 2.
f(k) = k/(k + 1) = 2
Multiply both sides by the denominator.
k = 2*(k + 1)
k = 2k + 2
k – 2k = 2
–k = 2
Multiply both sides by –1.
k = –2
answer = A
To find out where functions sit 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
To find out where functions sit in the “big picture” of GRE Quant, and what other Quant concepts you should study, check out our post entitled:
What Kind of Math is on the GRE? Breakdown of Quant Concepts by Frequency
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