The post Challenging GMAT Math Practice Questions appeared first on Magoosh GMAT Blog.

]]>

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:

- a) {1, 2, 3}, sum = 6
- b) {1, 2, 4}, sum = 7
- c) {1, 2, 5}, sum = 8
- d) {1, 3, 4}, sum = 8
- e) {1, 2, 6}, sum = 9
- f) {1, 3, 5}, sum = 9
- g) {2, 3, 4}, sum = 9

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:

- use {3, 7}, product = 21, abcd could be 3721 or 7321
- use {3, 9}, product = 27, abcd could be 3927 or 9327
- use {7, 9}, product = 63, abcd could be 7963 or 9763

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 GMAT Practice Problems: Sequences appeared first on Magoosh GMAT Blog.

]]>14, 23, 32, 41, 50, 59, …

1) In the sequence above, each term is 9 more than the previous term. What is the 40^{th} term of the sequence?

- (A) 360

- (B) 365

- (C) 369

- (D) 374

- (E) 383

2) What is the difference between the fourth and third terms of the sequence defined by

- (A) 18

- (B) 23

- (C) 47

- (D) 65

- (E) 83

3) Which of the following could be true of at least some of the terms of the sequence defined by

I. divisible by 15

II. divisible by 18

III. divisible by 27

- (A) I only

- (B) II only

- (C) I and II only

- (D) I and III only

- (E) I, II, III

4) Let S be the set of all positive integers that, when divided by 8, have a remainder of 5. What is the 76^{th} number in this set?

- (A) 605

- (B) 608

- (C) 613

- (D) 616

- (E) 621

5) Let T be a sequence of the form . If and , find .

- (A) 37

- (B) 38

- (C) 39

- (D) 40

- (E) 41

6) What is the sum of all the multiples of 20 from 160 to 840?

- (A) 14,000

- (B) 17,500

- (C) 18,060

- (D) 28,000

- (E) 35,000

7) A sequence is defined by for n > 2, and it has the starting values of and . Find the value of .

- (A) 25

- (B) 32

- (C) 36

- (D) 93

- (E) 279

9) In the set of positive integers from 1 to 500, what is the sum of all the odd multiples of 5?

- (A) 10,000

- (B) 12,500

- (C) 17,500

- (D) 22,500

- (E) 25,000

10) If =, , , , and , what is the value of ?

- (A) 7

- (B) 11

- (C) 19

- (D) 42

- (E) 130

Solutions will follow this article.

Sequences are a tricky topic on the GMAT Quant section. In that linked article, I discuss sequence notation, which is a variant of function notation, and I discuss recursive sequences, that is, sequences in which each term is determined by the previous term or terms.

An **arithmetic sequence** is a sequence in which we add some fixed amount to each term to get the next term. Another way to say that is that, if we subtract any term from the following term, the difference will always be the same: this difference is called the **common difference** of the arithmetic sequence. Let d be the common difference. Then, in algebraic form, the terms of the arithmetic sequence would be

Recall that, say, for the 3^{rd} term, the little subscript 3 is the **index**, that is, the position on the list. For this particular sequence, every term equals the first term plus a factor times **d**, and that factor is always one less than the index. This means we can write the general term as:

That is a very important formula, although, as always, don’t just memorize it; instead, remember the logic of the argument that leads up to it.

This is an important formula because any evenly spaced list is an arithmetic sequence. The consecutive multiples of any factor form an arithmetic sequence.

The sum of the first n terms in an arithmetic sequence is given by the formula

That’s the sum of the first term and the last term, times half the number of items on the list. You can also thing of that as the average of the first & last terms times the number of items on the list. One special case is the sum of the first n integer, given by

If you had some insights while reading that first article on sequences or the section on arithmetic sequences, then take another look at the problems above before looking at the solutions below. If you found this article helpful, or if you have an alternative solution for solving any of these problems, please let us know in the comments section below!

1) This is an arithmetic sequence, with

= 14 and d = 9

Using the formula for the nth term, we find that:

= 14 + 9*39 = 14 + 351 = 365.

Answer = **(B)**.

2) This is general sequence, with an explicitly defined nth term.

Answer = **(C)**.

3) First of all, if n = 8, then

= (16 – 1)(16 + 3) = 15*19

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then

= (24 – 1)(24 + 3) = 23*27

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

Answer = **(D)**.

4) Think about the first few numbers on this list:

5, 13, 21, 29, 37, 45, …

Notice that 5 is the first number in S, because when 5 is divided by 8, the quotient is zero and the remainder is 5.

This is an arithmetic sequence with

= 5 and d = 8

Using the general formula for the nth term of an arithmetic sequence, we have

= 5 + 8*(75) = 5 + 4*(150) = 5 + 600 = 605

Answer = **(A)**.

5) The formula in the first sentence tells us that this is an arithmetic sequence. The first term and the common difference are unknown, but we can generate two equations from the values of the two terms given.

Subtract the first equation from the second, and we get 16d = 48, which means d = 3. From the value of third term, we can see that first term must equal 11. Therefore,

Answer = **(B)**.

6) This is arithmetic sequence, and we have the first and last terms already. How many terms are there? Well, 160 = 20*(8) and 840 = 20*(42); we have to use inclusive counting to see that there are 42 – 8 + 1 = 35 terms.

Answer = **(B)**.

7) This is a recursive sequence, and we have to find it term by term.

Answer = **(E)**.

8) This is a recursive sequence, so we have to work backwards term by term.

Answer = **(A)**.

9) Let’s think about the terms in this sequence:

5, 15, 25, 35, …., 485, 495

The first term is 5 and the last is 495. There are 100 multiples of 5 from 1 to 500, so there are 50 odd multiples and 50 even multiple. The sum is:

Answer = **(B)**.

10) This recursive sequence is probably more difficult than anything that the GMAT is going to throw you, but solving this problem is not too bad. As with any recursive sequence, we have to go term by term.

Answer = **(C)**.

The post GMAT Practice Problems: Sequences appeared first on Magoosh GMAT Blog.

]]>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**

The post GMAT Quant: Difficult Units Digits Questions appeared first on Magoosh GMAT Blog.

]]>The post GMAT Quant: Must Be True Problems appeared first on Magoosh GMAT Blog.

]]>1) If A is a number, which of the following must be true for any A?

2) If F and G are integers, with F < G, which of the following must be true?

3) If J is an integer, which of the following must be true?

4) If p and q are two different odd prime numbers, such that p < q, then which of the following must be true?

For many math problems, one has to focus on finding the one right answer. That approach, with these problems, is highly problematic. If you are super-talented with math, fluent in Algebra, then perhaps you can scan the list of choices and immediately spot the correct answer. If you are talented enough to do that, then good for you!

For most folks, that approach is simply not feasible. The much more efficient solution involves elimination. You see, if any statement “must be true”, then it will be true for any value you pick. It takes incredible skill (or fantastic luck) to pick a single value that eliminates four answer choices and leaves only one. If that happens by chance, great, but don’t focus on that. Focus on speed and efficiency. Pick one value, eliminate what you can, then pick another, eliminate more, etc. Whittle down the choices until you are left with one. That’s the most efficient approach for these problems.

You have to pick values to plug in. At the start of picking, pick very easy values — with a few easy choices, you should be able to eliminate two or three answer choices, making your overall job much easier.

If you picked the same values for all four questions above, that’s a problem. Those questions were specifically written so that the allowable numbers would be different in different questions.

When you hear “A is a number”, of what do you think? Remember that the general word “number”, like the general term “human being”, includes all types. Just as the individuals under the term “human being” constitute a bewildering variety, so do the citizens of the realm of “numbers.” Numbers include positive & negative, wholes & fractions & decimals, square-roots, pi, etc. etc. It’s a realm of perfect democracy — for example, -(pi)/5 is just as much a number as is 8. If, when a GMAT problem says “number”, your mind defaults to {1, 2, 3, 4, ..}, the “counting numbers”, then with all due respect, that’s really a third-grade way of thinking about the word “numbers”, and the GMAT will viciously punish that kind of thinking. You always must be aware of all possibilities for any category.

numbers = everything, positive/negative, zero, fractions, decimals, everything on the continuous infinity of the number line.

integers = positive and negative whole numbers = { … -3, -2, -1, 0, 1, 2, 3, …}

positive integers = the counting numbers, 1, 2, 3, 4, ..}; this list does NOT include zero, because zero is neither positive nor negative.

Remember that “odd integers” includes both positive and negatives, and “even integers” includes positives & zero & negatives. Prime numbers are, by definition, a subset of positive integers. There are no negative primes, and neither zero nor one is a prime number. It’s a very good idea to have the first few prime numbers memorized:

Primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, …)

Notice that 2 is the only even prime number (all other even numbers are divisible by 2, and hence are not prime). After 2, the primes are an irregularly spaced set of odd numbers that continues to infinity. In fact, let’s talk a moment about their pattern

In this section, I am going to dip into very advanced math, far far beyond the GMAT realm. In modern mathematics, most work around the “pattern of prime numbers” revolves around the Riemann Hypothesis, the single hardest un-answered question in all of mathematics. The great mathematician Bernhard Riemann proposed this grand question as a conjecture in an 1859 paper, and it has utterly baffled the most brilliant mathematical minds on the planet every since. Forget about answering the question — you need close to a Ph.D. in mathematics just to understand what the question is asking and what it has to do with prime numbers! (A brief graph of the central player of the Riemann Hypothesis, the Riemann Zeta Function, is shown below.)

All of which is a long-winded way of saying: there is no easy algebraic formula that always produces prime numbers. A simple pattern for prime numbers absolutely does not exist. Thus, in a “must be true” question, any answer of the form “[simple algebra expression] is a prime number” cannot possibly be true all the time. That absolutely has to be a wrong answer.

If you had any big “aha!” while reading this article, you may want to give those problems another glance before reading the solutions below. If you have any further questions or observations, please let us know in the comments section below.

1) First of all, notice that A is a “number” — could be positive or negative or zero, whole or fraction or negative. Let’s first try a very easy choice: A = 0. Then, for the five answer choices, we get:

(A) undefined

(B) false

(C) true

(D) true

(E) undefined

Remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers. We are down to (C) & (D). Try another relatively simple choice, A = -2.

(C) 1 = 1, true

(D) +2 ≠ -2, false

Remember that if A is negative, a negative squared is a positive, and the square-root sign always has a positive output. Therefore, the left side of the (D) equation is always positive, and can’t be true if A is negative.

Answer = (C).

2) For this problem, both F & G are integers — they could be positive or negative or zero, but they must be whole numbers. Here’s a suggestion — one good choice would be to pick a positive number with smaller absolute value and a negative number with a larger absolute value: say, F = -4 and G = 1. Let’s true those in the answers:

(A) 16 < 1 false

(B) -64 < 1 true

(C) 4 < 1 false

(D) 4 < 1 false

(E) 1-16 = -15 > 0 false

With one magical choice, we were able to eliminate four answer choices. Keep in mind, the combination small positive/big negative is a good one, and so is pairing a negative number with 0 (e.g. F = -4, G = 0), because zero is greater than any negative number.

Answer = (B)

3) Here, J is an integer, so it can be positive or negative or zero, but not a fraction or a decimal. Zero is always a special case among integers, so let’s start there. J = 0: then,

(A) undefined error

(B) 0 > 0 false

(C) 0 > 0 false

(D) 1 > 0 true

(E) -1 > 0 false

Here, the special case was enough to eliminate four answers. Again, remember, if some choice leads to an “undefined” value, a divide-by-zero error, then it is not “true” for that value. Error is just as good as false for eliminating answers.

Answer = (D)

4) Here, p and q are odd prime numbers. First of all, eliminate (A), because there’s no simple algebraic rule that generates prime numbers. The easiest choices for numbers are p = 3 and q = 5. Try this:

(B) 3 + 5 = 8 is divisible by 4 — true

(C) 5 – 3 = 2 is not divisible by 4 — false

(D) 5 + 3 + 1 = 9 = (5^2) – (4^2) — true

(E) (3^2) + (5^2) = 34

This last number requires comment. Every odd number is the difference between two perfect squares, because adjacent squares . When we subtract two even squares or two odd squares, the number is always divisible by 4. Thus, 34 cannot be the difference of any pair of squares. Thus, on the basis of this choice, we can eliminate (E). We are left with (B) & (D).

Try p = 3 and q = 7.

(B) 3 + 7 = 10 is not divisible by 4 — false

(D) 3 + 7 + 1 = (6^2) – (5^2) — true

In fact, since p & q are both odd, (p + q + 1) must be odd, and as stated above, any odd number is the difference of two perfect squares. That’s why (D) must be true.

Answer = (D)

The post GMAT Quant: Must Be True Problems appeared first on Magoosh GMAT Blog.

]]>The post Sequences on the GMAT appeared first on Magoosh GMAT Blog.

]]>A sequence is a list of numbers that follow some mathematical patterns. More formally, a sequence is a function whose inputs are limited to the positive integers. Terms are denoted by a letter for the whole sequence, and in the subscript, the **index**, which is the place on the list. Here are some examples of common sequences. First, (A) and (B) are **arithmetic sequences**:

A) , , , , , …

B) , , , , , …

In an arithmetic sequence, the same “**common difference**” is added or subtracted each time. In (A), we add 3 to each term to get the next. In (B), we subtract 4 to get each new term. If you graphed the terms of an arithmetic sequence against the indices, the dots would follow a straight line in the x-y plane.

The next two, sequences (C) and (D), are **geometric sequences**:

C) , , , , , …

D) , , , , , …

In a geometric sequence, the same “**common ratio**” is multiplied or divided each time. In (C), we multiply each term by 3 to get the next. In (D), we divide by 2 to get each new term. If you graphed the terms of an geometric sequence against the indices, the dots would follow an exponential function in the x-y plane.

Notice also — (D) begins with a “zeroth” term. The first term of a sequence can have an index of either 1 or 0: both appear on the GMAT.

There are other exotic sequences in mathematics. Here are two more

E), , , , , , , , …

F), , , , , , , , …

Both of these are famous in mathematics. The first is the Fibonacci sequence, which I will discuss further below. The second, known in math as the Partition Function (http://en.wikipedia.org/wiki/Partition_(number_theory)) gets into way way more difficult math that you need to know for the GMAT.

A explicit series is a series in which the general rule for finding each term can be stated, either verbally or mathematically. Sometimes the GMAT will give you the general rule for a sequence in algebraic form:

- (A) 4

- (B) 6

- (C) 8

- (D) 12

- (E) 18

If the rule is given algebraically, all we have to do is plug in the index number to find the value of each term. Here, we plug in n = 3 to find the third term, and plug in n = 4 to find the fourth term. The third and fourth terms are:

so the difference between them is 12 – 6 = 6, which is answer **B**. I will give another practice question of this genre at the end.

The explicit description might also be given verbal. For example

2) In each term of a sequence, 9 is added to get the next term. If the first term is 2, what is the eighty-first term?

- (A) 632

- (B) 695

- (C) 713

- (D) 722

- (E) 731

Notice, the rule for the whole sequence is articulated: we know it is an arithmetic sequence with a common difference of 9. You can try this question now. I will show the solution at the end of this article.

In mathematics, a recursive definition is one where you need the result at end of each step in order to proceed to the next step. A sequence has a recursive definition if the only way to calculate the third term is first to calculate the second term, and the only way to calculate the fourth term is first to calculate the third term, etc. etc. Clearly, if we were asked for, say, the sixth term of a recursive sequence, we would have to calculate each and every term along the way: there’s no “shortcut” we can use to shoot directly to the sixth term.

Recursive defined sequence on the GMAT are almost always given in algebraic form. Here’s an example of a question involving one.

- (A) 4

- (B) 11

- (C) 20

- (D) 31

- (E) 116

Whenever you see the nth term equal to some expression involving an (n – 1)th term, that’s a recursively defined sequence. The idea is: to find the n = 3 term, you would have to plug the n – 1 = 2 term into the formula; to find the n = 4 term, you would have to plug the n – 1 = 3 term into the formula; etc. I will discuss a full solution to this one below.

Sometimes, a sequence is recursively defined not simply in terms of the previous term, but in terms of the previous two terms. The most famous sequence defined this way is the **Fibonacci Sequence**:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

This is famous because these numbers are found all over Nature: the number of seeds in almost any fruit, the number of petals on any flower, even the number of digits on one human hand or foot. Some theorists feel Fibonacci Numbers even give insight into the fluctuations of the stock market: http://www.goldennumber.net/fibonacci-stock-market-analysis/

The Fibonacci series has a recursive definition that can be written algebraically as:

Think about what that says. If I want the n = 5 term, for example, I have to take the sum of the n – 1 = 4 term with the n – 2 = 3 term. To find any term, I have to add the two previous terms. You start with two 1’s, follow that rule, and you generate the Fibonacci sequence.

Whenever you see the nth term expressed as some combination of the (n-1)th and (n-2)th terms, that’s a recursively defined sequence in which each term is defined in terms of the previous two terms. You will always be given two “starter” terms to such a sequence. Here’s a sample problem involving one.

- (A) 1

- (B) 7

- (C) 22

- (D) 155

- (E) 721

I will give a complete solution to this below.

If you understand how to plug into the algebraic sequences, and how to extend the patterns and find terms of the verbal defined sequences, you will be able to handle everything the GMAT throws at you involving sequence.

As promised, another question of the first type discussed.

- (A) 1/2

- (B) 2/7

- (C) 5/8

- (D) 7/10

- (E) 7/12

Practice question #1 was explained in the text of the article, immediately following the question, and the answer (**B**) was stated there.

2) Let’s think about the first few terms of this sequence.

To find each new term, we add 9. The result is, each term equals 2 plus a sum of 9’s. How many nines? Well, the second term has one 9, the third term has two 9’s, the fourth term has three 9’s, so it’s clear the nth term would have a sum of (n – 1) 9’s. Thus, the eighty-first term would be 2 plus eighty 9’s. Well, eighty 9’s = 80*9 = 720, so the eighty-first term = 722, answer **D**.

3) This is a recursively defined sequence, so we have to find the values of each term up until the one we want, the 5th term. Fortunately, the “starter” term we are given is the 2nd term, so we have a bit of a head start. We don’t know why the question started us with the second term, instead of the first, term, but it works to our advantage, so we won’t complain. To find the next term, use the recursion formula:

Notice, there’s no shortcut with recursive sequences: we have to find the third term = 4 and then use that to find the fourth term = 11, then used that to find the fifth term = 116. There’s no way to go directly to the fifth term with calculating each one of the previous terms. The fifth term = 116, answer **E**.

4) This is a recursively defined sequence, defined in terms of the previous two terms, so we have to find the values of each term up until the one we want, the 6th term. Again, we have to find the terms one by one, using the recursion formula.

So, the sixth term = 22, answer = **C**.

5) Ah, an explicitly defined sequence! After recursive sequences, this is much easier. All we have to do to find the 7th term is to plug n = 7 into the formula.

The seventh term is 1/2, Answer = **A**.

6) http://gmat.magoosh.com/questions/925

7) http://gmat.magoosh.com/questions/807

To find out where sequences 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

The post Sequences on the GMAT appeared first on Magoosh GMAT Blog.

]]>