Math Questions August

FactorialThe factorial of a number is the product of all the positive integers from 1 upto the number. The factorial of a given integer n is usually written as n! and n! denotes the product of the first n natural number. -n! = n x (n – 1) x (n – 2) x ……… x 1n! = n (n – 1)0! = 1 as a rule.Note : Factorial is not defined for improper fractions or negative integers.

PermutationIf r objects are to be chosen from n, where n ≥ 1 and these r objects are to be arranged, and the order of arrangement is important, then such an arrangement is called a permutation of n objects taken r at, a time.Permutations is denoted by nPr or (n, r)

e.g., If it is required to seat 5 men and 4 women in a row such that women occupy the even places, in how many ways can this be done?In a row of 9 positions, there are four places, and exactly 4 women to occupy them, which is possible in 4! ways. The remaining S places can be filled up by 5 men in 5! ways,Total number of seating arrangements = 4! 5! = 24 x 120 = 2880

Important Permutation Rules:(i) The total number of arrangements of n things taken r at a time in which a particular thing always occurs.e.g., The number of ways in which 3 paintings can be arranged in an exhibition from a set of five, such that one is always included.number of ways 3. 5-1P3-1 = 3.4P2 = 36 or 3! (4C2) = 6.6 = 36

(ii) The total number of permutations of n distinct things taken r at a time in which a particular thing never occurs = n-1Pre.g., The number of ways in which 3 paintings from a set of five, can be displayed for a photo-shoot, such that one painting is never picked.= 5-1P3 = 4P3 ways = 24It can be observed thatrn-1Pr-1 + n-1Pr = nPr

(iii) The number of permutations of n different objects taken r at a time,

when repetitions are allowed, is nr. The f place can be filled by any one of the n objects in ‘n’ ways. Since repetition is allowed the second place can be filled in ‘n’ ways again. Thus, there are n x n x n r times ways = nr ways to fill first r positions.

Circular PermutationsSuppose four numbers 1, 2, 3, 4 are to be arranged in the form of a circle.The arrangement is read in anticlockwise direction, starting from any point as 1432, 4321, 3214 or 2143.These four usual permutation correspond to one circular permutation.Thus circular permutations are different only when the relative order of objects to be arranged is changed.Each circular permutation of n objects corresponds to n Linear permutations depending on where (of the n positions) we start.This can also be though of as keeping the position of one out of n objects fixed and arranging remaining n – 1 in (n – 1)! ways.CombinationsIf r objects are to be chosen from n, where r ≤ n and the order of selecting the r objects is not important then such a selection is called a combination of n objects taken r at a time and denoted by In a permutation the ordering of objects is important while in a combination it is immaterial. e.g.,AB and BA are 2 different Permutations but are the same combination.Usually (except in trivial cases) the number of permutations exceeds the number of combinations. Trivial cases are when r = 0 or 1.e.g., If there are 10 persons in a party, and if every two of them shake hands with each other, how many handshakes happen in the party?SoIution: When two persons shake hands it is counted as 1 handshake and not two hence here we have to consider only combinations.2 people can be selected from 10 in 10C2 ways.Hence, number of handshake = Combinatorial Identities:1. nCr = nCn – r2. nCo = nCn = 13. n+1Cr = nCr + nCr – 14. n+1Cr+1 = nCr+1 + n–1Cr + n–1Cr-15. nPr = r! nC6. The total no. of combinations of ‘n’ things taken some or all at a time nc = nC1 + nC2 + ……nCn = 2n – 1Important Combination Rules1. The number of combinations of ‘n’ things taken ‘r’ at a time in which p particular thin will always occur = n-pCr-p P things are

definitely selected in 1 way. The remaining r – p things can be selected from n – p things in n-pCr-p ways.

In how many ways can 7 letters be selected from the alphabet such that the vowels are always selected.Solution: There are 5 vowels a, e, i, o, u which are selected in 1 way then possible number of ways = 26-5C7-5 = 21C22.The number of combinations of ‘n’ things taken ‘r’ at a time in which ‘p’ particular things never occur is n-pCr (n – p ≥ r)p things are never to be selected.

Hence r things are to be selected from n - p in n–pCr ways It is clear that n - p ≥ r for this to be possible.e.g. In how many ways can 7 letters be selected from the alphabet such that the vowels are never selected.Solution: As vowels (a, e, i, o, u) are never selected. The 7 letters can be selected from (20 – 5), letters in = 26–5C7 = 21C73. The number of ways of dividing (partitioning) •n distinct things into r distinct groups, such that some groups can remain empty = rnOne object ran be put into r partitions in r ways\ objects can be partitioned in r x r x r .... n times = rn ways

Examplei) In how many ways can 11 identical white balls and 9 black bells be arranged in a row so that no two black balk are together?SolutionThe 11 white balls can be arranged in 1 way (all are identical)The 9 black balls can be arranged in the 12 places in 12P9 ways

ii) In how many ways can they be arranged if black balls were identical? (all other conditions remaining same)

SolutionThe 11 white balls can be arranged in 1 way.The 9 black balls can be arranged in the 12 places in 12C9 ways.Thus number of arrangements = 12C9iii) In how many ways can they be arranged if all the balls are different. (all other conditions remaining same)

SolutionThe 11 white balls can be arranged in 11! waysThe 9 black balls can be arranged in the 12 places in 12! ways.Total number of arrangements = 11!12!/3!

ExampleIn a multiple choice test there are 50 questions each having 4 options, which are equally likely.In how many ways can a student attempt the questions in the test?SolutionEach question can be attempted in 4 ways and not attempted in 1 way. \Each question can be attempted or unattempted in 5 ways.Thus 50 questions can be attempted or attempted in 550 ways.This will include the case when no questions are attempted.\The student can attempt the paper in (5 to the power of 50) – 1 ways.

ExampleHow many 4 digit numbers can be formed from the, digits 1, 5, 2, 4, 2, 9, 0, 4, 2i) with repetition of digits.ii) without repetition of digits.

i) In the given set 4 is repeated twice and 2 thrice\Number of distinct digits = 6The 4 digit number can be formed in 5.63 ways when repetition is allowed.Position I can be filled in 5 ways, (as it cannot have O)The remaining 3 positions can be filled in 6 ways each.Hence number of numbers = 5.63 = 1080ii). Position I can be filled in 5 ways.Position II can be filled in 5 ways (it can contain any of 5 digits except the one in position 1Thus number of such numbers = 5 x 5 x 4 x 3 = 300

Question: f(x) = -5x 197 -8x 248 +1; f(-1) = ?

(A)-16,251 (B)-12 (C)-2 (D)14 (E)16,251

1. To determine f(-1), we must first know (-1)197 and (-1)248. You should never multiply out lots of numbers.

2. Instead, look for a pattern:(-1)1=-1(-1)2=+1(-1)3=-1(-1)4=+1...(-1)(odd number)=-1(-1)(even number)=+1

3. This means (-1)197=-1 since 197 is odd and (-1)248=+1 since 248 is even.

4. This pattern simplifies the equation to:-5(-1) - 8(1) + 1, which equals 5 - 8 + 1, or -2.Thus, C is the correct answer.

Question: If (243 9 + 18z )(81 -18z )(27 15 - 9z ) = 1, what is the value of z?

(A)-27 (B)-10 (C)0 (D)10 (E)27

1. Rewrite the terms on the left-side with common prime bases.(2439 + 18z)(81-18z)(2715 – 9z)= (35(9 + 18z))(34(-18z))(33(15 – 9z))= (345 + 90z)(3-72z)(3(45 – 27z))

2. Simplify by combining terms with like bases.(345 + 90z)(3-72z)(3(45 – 27z))= 345 + 45 + 90z – 72z – 27z

= 390 -9z 3. As a result of this simplification, the equation is much easier to deal

with.390 -9z = 1

4. At this point, the problem may appear unsolvable as different bases and exponents exist. However, remember that any number raised to the 0th power is 1. Consequently, 1 can be rewritten as 30. The equation now reads:390 -9z = 1390 -9z = 1 = 30

5. Since the bases are the same, you can set the exponents equal to each other. The equation becomes:90 – 9z = 090 = 9zz = 10

Question:If x is a positive integer and z is a non-negative integer such that (2,066) z is a divisor of 3,176,793, what is the value of z x - x z ?

(A)-81 (B)-1 (C)0 (D)1 (E)It cannot be determined

1. An odd integer (e.g., 3,176,793) is not divisible by an even integer. For example, 3 is not divisible by 2 nor is 15 divisible by 2.(3,176,793/even integer) --> non integer

2. The only way (2,066)z can possibly be a divisor of 3,176,793 is if (2,066)z is an odd number. However, if z is any positive integer, (2,066)z will be an even number. (More specifically, it will have a units digit of 6). As a result, (2,066)z will not be a factor of 3,176,793 if z is a positive integer.

3. Since the question explicitly says that (2,066)z is a factor of 3,176,793, you know that, somehow, (2,066)z must be an odd number.

4. Remember that any number raised to the power of 0 will be 1.(any real number)0 = 1

5. The key to this problem is realizing that if z = 0, which is allowed because the question stem said z "is a non-negative number," (2,066)z will equal 1. Since the only way z can be a factor 3,176,793 is if z = 0, you know that z = 0.(3,176,793/(2,066)0) is the only way 2,066z is a factor of 3,176,793

6. You can now rewrite the question as follows:0(any positive integer) – (any positive integer)0

zx – xz = 0x – x0 = 0(pos int) – (pos int)0 7. Since 0 raised to any positive integer equals 0 and any positive

integer raised to 0 is 1, the question boils down to: 0 – 1 = -1.0(pos int) – (pos int)0 = 0 – 1 = -1

What is the units digit of (2) 3 (3) 3 (4) 3 (5) 7 (6) 2 (7) 2 ?

1. At first, this question looks daunting. Do we have to determine what all of these numbers are, raised to their respective powers, and then multiply them by one another? Recognize that if we just figure out what the units (ones) digit is for all of these numbers, we then just have to multiply the units digits by one another, which is much more manageable. This is because, to obtain the units digit of a product of two numbers, we only need to multiply the units digit of these two numbers by each other. Then the units digit of that product will be the units digit of the whole product. For example, the units digit of 38 * 9317 = 6, because 8 * 7 = 56, and the units digit of 56 is 6.

2. Thus, let us first determine the units digit of the numbers in this expression.

3. (2)3 = 2 * 2 * 2 = 8So the units digit of (2)3 is 8

4. (3)3 = 3 * 3 * 3= 9 * 3= 27So the units digit of (3)3 = 7

5. (4)3 = 4 * 4 * 4= 16 * 4= 64The units digit of (4)3 = 4

6. For (5)7, recognize that 5 to any power greater than 0 must have a 5 in the units digit. For example, 5 * 5 = 25; 25 * 5 = 125, etc.Therefore the units digit of (5)7 = 5

7. At this point, we can stop, if we recognize that 5 (i.e., the units digit of (5)7) * 4 (i.e., the units digit of (4)3) = 20. Remember that after determining all of the units digits, we were going to multiply them by one another to determine the units digit of the whole product. The fact that 5 * 4 = 20 is significant because, taking 0 as the units digits of this product, we then will multiply 0 by all of the other units digits we determine. This product, however, no matter what the other units digits are, must be 0, because any number multiplied by 0 is 0. Therefore, our units digit must be 0, answer choice (A).

8. At first, this question looks daunting. Do we have to determine what all of these numbers are, raised to their respective powers, and then multiply them by one another? Recognize that if we just figure out what the units (ones) digit is for all of these numbers, we then just have to multiply the units digits by one another, which is much more manageable. This is because, to obtain the units digit of a product of two numbers, we only need to multiply the units digit of these two numbers by each other. Then the units digit of that product will be the

units digit of the whole product. For example, the units digit of 38 * 9317 = 6, because 8 * 7 = 56, and the units digit of 56 is 6.





















29. At first, this question looks daunting. Do we have to determine what all of these numbers are, raised to their respective powers, and then multiply them by one another? Recognize that if we just figure out what the units (ones) digit is for all of these numbers, we then just have to multiply the units digits by one another, which is much more manageable. This is because, to obtain the units digit of a product of two numbers, we only need to multiply the units digit of these two numbers by each other. Then the units digit of that product will be the

units digit of the whole product. For example, the units digit of 38 * 9317 = 6, because 8 * 7 = 56, and the units digit of 56 is 6.














Question: If the sum of the consecutive integers from –35 to n inclusive is 150, what is the value of n?

(A)36 (B)39 (C)40 (D)60 (E)185

1. Consecutive integers are integers that follow in sequence, where the difference between two successive integers is 1. You are told that the sum of the consecutive integers from –35 to n inclusive is 150. The sum of all the consecutive numbers from –35 to –1 will be negative, yet the sum of the entire set is positive. This means that there must be positive integers in the set that go beyond +35 to result in a positive sum.

2. Consider that a number plus its opposite sum to zero:–5 + 5 = 0, –13 + 13 = 0, etc.Therefore:(–35) + (–34) + … + (–1) + 0 + 1 + … + 34 + 35 = 0

3. The next consecutive number in the set would be 36. Could this be n? If it were the sum of the set would be 0 + 36 = 36. So there must be more integers in the set. Move forward one number at a time, until the sum is 150:0 + 36 = 3636 + 37 = 7373 + 38 = 111111 + 39 = 150

4. Since the addition of 39 makes the sum 150, it must be the last consecutive integer in the set and the value of n. Therefore choice (B) is correct.

Question :Set S consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set S?

(A)1 (B)9 (C)14 (D)17 (E)It cannot be determined

Order the members of set S in ascending order. Leave out n since you do not yet know where it will fall.-27, -3, -2, 2, 3, 15, 17

1. At this point, you do not know where n will be. However, it is not essential to know the exact placement of n in order to find which answer choice could be the median of set S.

2. When n is included in set S, set S has an even number of terms. Consequently, the median will be the average of the middle two terms of set S.

3. It is important to reiterate that set S consists of "unique integers." Consequently, n cannot equal any of the existing values of S.

4. By logically examining set S, there are three options for the median:5. Option 1: n > 3 and Median = 2.5

1. If n > 3, the median will be 2.5-27, -3, -2, 2, 3, n, 15, 17-27, -3, -2, 2, 3, 15, n, 17-27, -3, -2, 2, 3, 15, 17, nIn each case, the median will be between 2 and 3 --> Median will be 2.5

6. Option 2: n < -3 and Median = 0 1. If n < -3, the median will be 0

-27, n, -3, -2, 2, 3,15, 17n, -27, -3, -2, 2, 3,15, 17In each case, the median will be between -2 and 2 --> Median will be 0

7. Option 3: Median Is (n+2)/2 1. If -2 < n < 2, n could be -1, 0, 1 (remember n must be a unique

integer, so it cannot be 2 or -2 since these numbers are already used).

2. The series is now ordered: -27, -3, -2, n, 2, 3, 15, 17Since the median will be the average of the middle two terms, we can simplify the series to:-2, n, 2, 3, which will simplify to:n, 2The median of the series will be the average of n and 2.Remember that n can equal -1, 0, or 1.If n = -1 --> Median will be 0.5If n = 0 --> Median will be 1If n = 1 --> Median will be 1.5

3. Since 1 is the only possible median that is included as an answer choice, it is the correct answer.

At a local appliance manufacturing facility, the workers received a 20% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked?

(A)83% (B)80% (C)20% (D)17% (E)12%

1. This question can be solved using either real numbers or algebra.2. Method 1: Use Real Numbers. 1. Assume that the worker in question, call him John, earned $50 by working

10 hours for $5 per hour.2. After the pay-raise, the total amount earned will be the same at $50, but

the hourly rate will increase by 20% to $5(1.2) = $6 per hour.3. Using the intuition that salary=(hours)(wage), you can find that

$50=(hours)($6) and therefore the number of hours worked after the change in pay will equal (50/6) = 8.33 hours

4. Since John worked 10 hours before the change in wage and he now works 8.33 hours, John reduced the number of hours he worked by (10-8.33)/10 = .1666666% = 17%

5. Note: It does not matter what numbers you use, as long as the original pay equals the post wage-change pay and the wage is increased by 20%.

3. Method 2: Use Algebra. 1. Assign variables to pieces of the problem:

Let p = the percentage of old hours worked after the raiseif p = 95%, the worker labors 95% of the old hours after the raise (i.e., 5% fewer hours after the raise)Let r = hourly rate that employees were paid before the raiseLet R = hourly rate that employees were paid after the raiseLet n = number of hours worked before pay-raiseLet N = number of hours worked after pay-raise

2. We know that the total pay before and after the pay-raise equal each other. The pay before was based on the rate r with hours n while the pay afterwards was based on rate R and hours N:rn = RN

3. The rate after the raise was 20% more than the rate before the raise:R = 1.20r Plug this value into the equation for R:rn = RNrn = 1.20rN

4. We are looking for the percentage of hours worked after the raise, so substitute N in terms of n and percent p.N = pn; the new number of hours equals the old number of hours multiplied by the percent of old hours worked after the raisern = (1.2r)(pn).

5. Divide both sides by r and n:1 = 1.2p

6. Divide by 1.2 to solve for p:p = 1/1.2 = 10/12 = 5/6 = .8333 = 83.33%.

7. Be careful: the question asks what percent the worker would reduce his hours, so subtract from 100% to yield 16.66%, or 17%. The correct answer is D.

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Let f(x) = x 2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses the y-axis at what y-coordinate?

(A)-4(B)-1(C)0(D)1(E)4

1. This problem requires working backward from solutions to a quadratic equation to the equation itself.Instead of factoring a quadratic equation to solve it (the process many questions require), this question requires you to go the other way—from solutions to factors.f(x) = x2 + bx + c = 0

2. The key to unlocking this problem is recognizing that if f(a) = 0 and f(x) is a quadratic equation in the form x2 + bx + c = 0, (x – a) is a factor of the equation.(x + d)(x + e) = x2 + dex + de = 0

3. If this past step does not make sense immediately, take a look at a few examples and allow them to convince you of this truth.x2 + 3x - 4 = 0(x - 1)(x+4)Solutions: f(1) = 0 or f(-4) = 0

x2 -9x + 20 = 0(x – 4)(x – 5)Solutions: f(4) = 0 or f(5) = 0

4. Following this same principle, you know that since f(1) = 0 and f(-4) = 0, (x – 1) and (x + 4) are factors.(x – 1)(x + 4) = 0

5. Multiply these factors together to form a quadratic equation.(x – 1)(x + 4) = 0x2 – x + 4x – 4 = 0x2 + 3x – 4 = 0

6. For any quadratic equation in the form ax2 + bx + c = 0, the y-intercept (i.e., the place where the equation crosses the y-axis) is c.Even if you did not know this, you could still find the y-axis. The line will cross the y-axis where x = 0.f(0) = (0)2 + 3(0) – 4 = -4

7. Y-Intercept is -4

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After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?

(A)8/27(B) 1/27(C)2/3(D)19/27(E)27/28

Pump Run # Still Left [as fraction of full tank] Removed [as fraction of full tank]

0 10Cumulative: 0

1 1 - (1/3) = 2/31(1/3) = 1/3 = 9/27Cumulative: 0 + (9/27) = 9/27

2 1 - (9/27) - (6/27) = 12/27 = 4/9(1/3)(2/3) = 2/9 = 6/27Cumulative: (9/27) + (6/27) = 15/27

3 1 - (9/27) - (6/27) - (4/27) = 8/27(1/3)(4/9) = 4/27Cumulative: (15/27) + (4/27) = 19/27

1. Each time the pump is used, 2/3 of the remaining gas stays and 1/3 of the remaining gas is removed.

2. The first time running the pump removed 1/3 of the full tank, leaving 2/3 of a full tank of gas. So, after running the pump once, 1/3 of the total gas in the tank has been removed.

3. The second time running the pump removed 1/3 of the 2/3 of a full tank that remained after running the pump for the first time. So, after running the

pump twice, (1/3) + (1/3 of the 2/3) of the remaining tank had been removed. Thus far, 5/9 of the total gas in the tank has been removed.

4. The third time running the pump removed 1/3 of the 4/9 of a full tank that remained after running the pump for the second time. So, after running the pump three times, 5/9 + 1/3 of 4/9 of the remaining tank had been removed. In total, 5/9 + 4/27 = 19/27 of the total gas in the tank at the beginning was removed after running the pump three times.

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Superior Security recently discovered a computer on its network had been hacked. The computer password contained 16 characters (including only numbers and letters). What is the probability that the password to Superior’s hacked computer was correctly guessed on the first try? (Assume that the hacker knew the password contained 16 characters, comprised only of numbers and both lower-case and upper-case letters.)

A)

52!/(16!36!)

B)

1/[62!/(16!46!)]

C)

1/[52!/(16!36!)]

D)

1/5216

E)

1/6216

1. The probability of guessing correctly on the first try is 1 divided by the number of pass permutations (i.e., unique arrangements of passwords). For example, if there were four possible password combinations, the probability of guessing the password correctly on the first try would be 1/4

2. For each of the 16 characters that form the password, the possible characters include: 26 lower case characters, 26 upper case characters, and 10 digits. So, for each character in the password, there are 62 possible characters.

Character 1 Character 2 Character 3 ... Character 15 Character 16

62 possibilities

62 possibilities

62 possibilities

62 possibilities

62 possibilities

62 possibilities

3. Since there are 16 characters in the password and each character could be one of 62 possibilities, there are a total of 6216 unique password permutations (i.e., there are 6216 different possible passwords, only one of which is the correct password). So, the probability of guessing the password on the first try is 1/6216.

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1. Divisibility

How many numbers can be formed by using 8,7,5,3,2 such that they are divisible by

125 ? Digits to be used exactly once.

(1) 20 (2) 4 (3) 2 (4) 1

any integer is divisible by 2^n, or 5^n. only when an integer formed by last n digits is divisible by 2^n, or 5^nin our case as 125=5^3 and n=3,the integers must end with 375, (125*3) or 875 (125*7) so possible numbers ending with 3752837582375and the same story with numbers ending with 8752387532875so i got 4 as an answer

Here is my approach:Since 1000 is divisible by 125 then, if the last three digits are divisible by 125 then the whole number is divisible by 125. Odd multiples of 125 which are less than 1000 are 125, 375, 625, 875. So we can use only 375 ( two numbers) and 875(two numbers) as the last three digits. So there are 4 numbers.

A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

(A)13 (B) 59 (C) 35(D)37(E)12

Correct Choice (D) Answer (37)

Explanatory AnswerLet the original number be ‘a’Let the divisor be ‘d’Let the quotient of the division of a by d be ‘x’

Therefore, we can write the relation as = x and the remainder is 24.i.e., a = dx + 24When twice the original number is divided by d, 2a is divided by d.We know that a = dx + 24. Therefore, 2a = 2dx + 48

The problem states that leaves a remainder of 11.2dx is perfectly divisible by d and will therefore, not leave a remainder.

The remainder of 11 was obtained by dividing 48 by d.When 48 is divided by 37, the remainder that one will obtain is 11.Hence, the divisor is 37.

How many zeros will be there in the value of 25!?

(A)25(B)8(C)6(D)5(E)2

Solution:25! is factorial 25 whose value = 25*24*23*22*…..*1When a number that has 5 as its factor is multiplied by another number that has 2 as its factor, the result will have ‘0’ in its units digit.In 25!, the numbers that have 5 as their factor are 5, 10, 15, 20, and 25. 25 is the square of 5 and hence has two 5’s in it.Therefore, 25! contains in it 6 fives.There are more than 6 even numbers in 25!. Hence, the limiting factor is the number of 5s. And hence, the number 25! will have 6 zeroes in it. Correct Answer Choice (C)

25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

How many integral divisors does the number 120 have?Solution:Express the number 120 as a product of powers of prime factors.In this case, 120 = 23 * 3 * 5The three prime factors are 2, 3 and 5.The powers of these prime factors are 3, 1 and 1 respectively.To find the number of factors / integral divisors that 120 has, increment the powers of the prime factors by 1 and then multiply them.

In this case, (3+1) * (1 + 1) * (1 + 1) = 4 * 2 *2 = 16.

The formula is 2n-1 where n is the number of powers of prime factors.

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How many keystrokes are needed to type numbers from 1 to 1000?

(A)3001(B)2893(C)2704(D)2890(E)None of these

Correct Choice (B) - Correct Answer (2893)Solution:

While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them require one keystroke. That is 9 key strokes.There are 90 two-digit numbers, from 10 to 99. Each of these numbers require 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers. There are 900 three-digit numbers, from 100 to 999. Each of these numbers require 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers.Then 1000 is a four-digit number which requires 4 keystrokes.Totally, therefore, one requires 9 + 180 + 2700 + 4 = 2893 keystrokes.

When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

(A) 11 (B) 17 (C) 13 (D) 23 (E)None of theseSolution:Let the divisor be d.When 242 is divided by the divisor, let the quotient be 'x' and we know that the remainder is 8.Therefore, 242 = xd + 8Similarly, let y be the quotient when 698 is divided by d.Then, 698 = yd + 9.242 + 698 = 940 = xd + yd + 8 + 9940 = xd + yd + 17As xd and yd are divisible by d, the remainder when 940 is divided by d should have been 17.owever, as the question states that the remainder is 4, it would be

possible only when leaves a remainder of 4.If the remainder obtained is 4 when 17 is divided by d, then d has to be 13. Correct Answer (C) Choice (13)

How many different positive integers exist between 10 6 (1000,000) and 10 7

(10,000,000), the sum of whose digits is equal to 2?

6 (B)7 (c)5 (D) 8 (E)18

Explanatory AnswerFind the number of such integers existing for a lower power of 10 and extrapolate the result of the present case.Between 10 and 100, that is 101 and 102, we have 2 numbers, 11 and 20 . Similarly, between 100 and 1000, that is 102 and 103, we have 3 numbers, 101, 110 and 200 . Similarly, between 1,000 and 10,000, that is 103 and 104, we have 4 numbers, 1001, 1010, 2,000 and 1100.

Similarly, between 10,000 and 100,000, that is 104 and 105, we have 5 numbers, 10001, 10010, 200,000, 10100 and 11000.Similarly, between 100,000 and 1000,000, that is 105 and 106, we have 6 numbers, 100001, 100010, 100100, 200,0000 and 110,000.

Similarly, between 1000,000 and 10,000,000 that is 106 and 107, we have 7 numbers, 1,000,001, 100,0010, 1000,100, 1001,000, 101,0000, 2,000,000 , and 11,000,000 .Therefore, between 106 and 107, one will have 7 integers whose sum will be equal to 2.

In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?

A) 1 : 2 (B) 2 : 1 (C) 1 : 3 (D) 3 : 1 (E) 2 : 3

Solution:Let there be 1 litre of the solution after mixing 20% methyl alcohol and 50% methyl alcohol.If the concentration of methyl alcohol in it is 40%, then 0.4 litres of the resultant mixture is methyl alcohol.Let x litres of the solution containing 20% methyl alcohol be mixed with (1 - x) litres of the solution containing 50% methyl alcohol to get 1 litre of the solution containing 40% methyl alcohol.X litres of 20% methyl alcohol solution will contain 20% of x = 0.2x litres of methyl alcohol in it.(1 - x) litres of 50% methyl alcohol solution will contain 50% of (1- x) = 0.5(1 - x) litres of methyl alcohol.The sum of these quantities of methyl alcohols added up to the total of 0.4 litres in the resultant mixture.Therefore, 0.2x + 0.5(1 - x) = 0.4 litres0.2x + 0.5 - 0.5x = 0.40.5 - 0.4 = 0.5x - 0.2x

x = litres

And 1 - x = 1 - litres.

So, the two solutions are mixed in the ratio of 1 : 2.

A, B and C, each of them working alone can complete a job in 6, 8 and 12 days respectively. If all three of them work together to complete a job and earn $ 2340, what will be C’s share of the earnings?A, B and C will share the amount of $ 2340 in the ratio of the amounts of work done by

them.As A takes 6 days to complete the job, if A works alone, A will be able to complete th

of the work in a day.Similarly, B will complete and C will complete of the work.So,

the ratio of the work done by A : B : C when they work together will be equal to Multiplying the numerator of all 3 fractions by 24, the LCM of 6, 8 and 12 will not change the

relative values of the three values.We get = 4 : 3 : 2.i.e., the ratio in which A: B : C will share $2340 will be 4 : 3 : 2.

Hence, C’s share will be = 520.

The difference between the value of a number increased by 12.5% and the value of the original number decreased by 25% is 30. What is the original number?

(A)60(B)80(C)40(D)120(E)160

Solution:Let the original number be x.Let A be the value obtained when x is increased by 12.5%.Therefore, A = x + 12.5% of xLet B be the value obtained when x is decreased by 25%.Therefore, B = x - 25% of x.The question states that A - B = 30i.e., x + 12.5% of x - (x - 25% of x) = 30x + 12.5% of x - x + 25% of x = 3037.5% of x = 30

x = = 80Correct Answer (B) Choice (80)In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election?A)240 000 (B)300,000( C)168,000 (D)36,000(E)24,000

Explanatory Answer

Let the percentage of the total votes secured by Party D be x%Then the percentage of total votes secured by Party R = (x – 12)%

As there are only two parties contesting in the election, the sum total of the votes secured by the two parties should total up to 100%

i.e., x + x – 12 = 1002x – 12 = 100or 2x = 112 or x = 56%.

If Party D got 56% of the votes, then Party got (56 – 12) = 44% of the total votes.

44% of the total votes = 132,000

i.e., = 132,000

ð T = = 300,000 votes.

The margin by which Party R lost the election = 12% of the total votes = 12% of 300,000 = 36,000.

If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

(A)3 (B) 1/3 (C) 2 (D) 9 (E) 1/9

Solution:

The sum of the first n terms of a Geometric Progression is given by , where ‘a’ is the first term of the G.P., ‘r’ is the common ratio and ‘n’ is the number of terms in the G.P.

Therefore, the sum of the first 6 terms of the G.P will be equal to

And sum of the first 3 terms of the G.P. will be equal to The ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1

i.e.

=> => r3 + 1 = 9 => r3 = 8 => r = 2 Correct Choice (C) Answer (2)

The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

(A)300 (B) 120 (C) 150 (D) 170 (E) 270Explanatory AnswerThe sum of the 4th and 12thterm = 20.Let t1 be the first term, t4 be the fourth term, and t12 be the 12thtermThen, t4 + t12 = 20≫ t1 + 3d + t1 + 11d = 20 2t1 + 14d = 20 t1 + 7d =120 t8 = 10.

The sum of the first 15 terms =In an arithmetic progression t1 + t15 = t2 + t14 = t3 + 13 =... = t8 + t8.

Therefore, the sum of the first 15 terms = = 150Correct Choice (C) Answer (150)

For what values of ‘k’will the pair of equations3x + 4y = 12 and kx + 12y = 30 not have a unique solution? (A)12 (B)9 (C)3 (D)7.5 (E)2.5Answer

A system of linear equations ax + by + c = 0 and dx + ey + g = 0 will have a unique solution if the two lines represented by the equations ax + by + c = 0 and dx + ey + g = 0 intersect at a point. That is, if they are not parallel lines.

ax + by + c = 0 and dx + ey + g = 0 will not be parallel lines if

In the above question, we need or k should not be equal 9 for the pair of equations to have a unique solution.

In other words, when k=9, the system of equation will not have any solution as the two lines represented by the equations will be parallel lines. Correct Choice (B) Answer (9)

3 4 x = 12

K 12 y 30

3*12-4k=036 = 4k or k=9

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In a certain game, a large container is filled with red, yellow, green, and blue

beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then

removed from the container. If the product of the point values of the removed

beads is 147,000, how many red beads were removed?

First I simplified 147,000 to 147. 147 divides by 3, since 1+4+7 is divisible by 3. 147 divided by 3 = 49, which in turn equals 7x7.

So 7x7x3 = 147.

Now we need to get 147 to equal 147,000. We do this by multiplying 147 x 1000. To get 1000, multiply (5x2)x(5x2)x(5x2).

So you have 7x7x3x5x2x5x2x5x2 = RxRxGxYxBxYxBxYxB

So there are: 2R 1G 3Y 3B I think the easiest way of thinking about it is factor out seven as many times as you can (each time you factor out a seven you're counting a red ball). 147000/7=21000, 21000/7=3000, and it's pretty easy to see (or check) that 3000 is some multiple of 2,3, and 5.

If the sequence x1, x2, x3, ... xn, ... is such that x1 = 3 and xn+1 = 2xn - 1 for n = 1,

then x20 - x19 equals which of the following?

A. 2^19 B. 2^20 C. 2^21 D. (2^20) - 1 E. (2^21) - 1

We have the sequence , , , …, … and for .

If you notice there is a specific pattern into it:

...

So, and .

Answer: A.

37. The probability that a certain coin will fall heads up is 50%. What is the probability that it

falls heads up on three of six tosses?(A)41 /26 (B)35/26 (C)1/26 (D)21 25 (E)5 24

How many even, three digit integers greater than 700 with distinct, non - zero digits are there? (A) 729 (B) 243 (C) 108 (D) 88 (E) 77 The best way to approach it is to split up the problem, and treat the 700's, 800's and 900's separately. To find the number of numbers beginning with a 7 that fit our criteria, we have 7 choices for our tens digit (excluding 7, because no repeats are allowed, 0, and whichever even number is used as the units digit) and 4 choices (all non-zero, even digits) for our ones digit, giving a total of 28 choices. The 900's work exactly the same way, because its an odd hundreds digit as well. For the 800's, we have 7 choices for the tens and only 3 for the ones (8 is no longer a choice for the ones column). That leaves us with at total of 21 legal numbers beginning with an 8. Add 28 + 28 + 21 = 77, so the correct answer is (E).

712 714 716 718 812 814 816 818 722 724 726 728 822 824 826 828 732 734 736 738 832 834 836 838 742 734 746 748 842 844 846 848 752 754 756 758 852 854 856 858 762 764 766 768 862 864 866 868 772 774 776 778 872 874 876 878 782 784 786 788 882 884 886 888 792 794 796 798 892 894 896 898

912 914 916 918 922 924 926 928 932 934 936 938 942 934 946 948 952 954 956 958 962 964 966 968 972 974 976 978 982 984 986 988 992 994 996 998

Set B has three positive integers with a median of 9. If the largest possible range of thethree numbers is 19, given a certain mean, what is that mean? (A) 22 (B) 10 (C) 9.6 (D) 9 (E) It cannot be determined from the information given

Plug in your answers here. Don't start with (C), because 9.6 is an annoying number to calculate with. Start with (B) instead. If the mean is 10 and the median is 9, what would the largest possible rangeof the three integers be? To find that, our three integers must fit into the equation

. The median,b , equals 9, So 21 = + c a . The range is defined as a c − , to make a c − as large as possible, given that 21 = + c a, we can set a=1 and 20= c That does give us a range of 19, so (B) is the correct answer

In a particular college dormitory with 130 people, 57 students watch E. R. and 48 watch Law & Order. If there are twice as many students who watch neither show as who watch both, how many students watch both?(A) 25 (B) 35 (C) 45 (D) 50 (E) 70

This question deals with the groups formula: Total = Group 1 + Group 2 - Both + Neither . Once you fill in the parts of the formula you are given, plug in your answers. Answer choice (A), the correct answer, gives you 130 = 57 + 48 - 25 + 50= 25 which works out correctly130- (57+48) =130- 105= 25What is the remainder when 332 is divided by 4? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4

There is always a pattern--all you have to do is find it. First of all, you can cross off answers (A), (C), and (E). For (A) or (E) to be correct, 332 would have to be divisible by 4, but it can't be (do you see why?). And for (C) to be correct, 332 would have to be even, but it won't be (again, can you explain why?). Now cycle through the first few exponents of 3 to find the pattern: 30

has a remainder of 1, 31 has a remainder of 3, 32 has a remainder of 1, and 33 has a remainder of 3. So the pattern that emerges is that all even exponents of 3 have a remainder of 1 when divided by 4, and all odd exponents have a remainder of 3. Since 32 is even, the correct answer is (B).

21. Chef Gundy is creating a new dessert that will be made from 3 ingredients. If he has 8 cookies

and one flavor of sorbet to choose from, what fraction of the possible arrangements will contain the

sorbet?

(A)1/ 84 (B)1/56 (C)1/9 (D)1/2-------------------------

Which of the following equations has a root in common with x2 - 6x + 5 = 0?x2 + 1 = 0 x2 - x - 2 = 0 x2 - 10x - 5 = 0 2x2 - 2 = 0 x2 - 2x - 3 = 0ExplanationSince +x2 - 6x + 5 = x2 -5x-x+5 =(x - 5)(x - 1), the roots of x2 - 6x + 5 = 0 are 1 and 5. When these two values are substituted in each of the five choices to determine whether or not they satisfythe equation, only in the fourth choice does a value satisfy the equation, namely, 2(1)2 - 2= 0. Thus the best answer is 2x2 - 2 = 0.If a , b and m are integers and a = b − 5 + 2 m , then which of the following integers must be odd?Ο a Ο b Ο a + 2b Ο ab Ο ab- 1Answer Suppose a= 3 or 4 and similarly b= 1 or 2 then apply conditions:ab − 1a, b and m are integers. Now, 2m − 5 is odd, which implies that a − b is odd. So, if a is odd, then bis even or vice versa. This eliminates options "a" and "b". The option "a + 2b" is eliminated since it may be odd or even depending on whether a is odd or even.

ab must be even because either a or b has to be even. Hence, (ab − 1) must be odd.If 1/ x < 1/ y , then (A) x < y (B) x > y (C) x < y if xy < 0 (D) x < y if xy > 0 (E) x > y if xy < 0 Answer x < y if xy < 0Multiply both sides of the inequality by xy; so, left-hand side = y and right-hand side = x. If xy > 0, then the inequality is unchanged; so, y < x. But, if xy < 0, then the inequality is reversed; so, y > x.If a 2 + 10 b 2 + c 2 = 2 b (3 a + c ), then which of the following is true? I. a = 3b II. a = b / 3 III. c = b IV. c = 3b V. c = a VI. c = a / 3Ο Only I Ο Only II Ο Only III Ο I, III and VI Ο II, IV and V Answer I, III and VIRearranging the given expression as follows: (a2 + 9b2 − 6ab + (b2 + c2 − 2bc) = 0 Thus, (a − 3b)2 + (b − c)2 = 0 Since the square of a quantity is always non-negative, the only possibility is that a = 3b and b=c.

---------A natural number is divided into two positive unequal parts such that the ratio of the original number to the larger (divided) part is equal to the ratio of the larger part to the smaller part. What is the value of this ratio?Ο (51/2 − 1) (B) (51/2 + 1) / 2 Ο (51/2 + 1) / 4 Ο (51/2 + 1) / (51/2 − 1) Ο (51/2 + 3) / (51/2 − 1) Answer (51/2 + 1) / 2Let N be the natural number and L be the larger of the two individual parts. Then, the given relation is N / L = L / N – L. ∴ N 2 – LN = L2 . Let the required ratio be R = N / L. Dividing the above equation by L2 gives the quadratic equation R2 − R − 1 = 0. The solution to the above quadratic equation is R = (1 + 51/2) / 2. Since N is a natural number, only the positive root is considered for R.

let the larger part be a .. the smaller part be b .. so the number is a+b.... .. according to the given condition (a+b)/a= a/b .... subtracting 1 from both the sides of the equation..

we have (a+b)/a - 1= a/b -1 � b/a= a/b - 1� b/a-a/b = -1�

let b/a = x � so we have x-1/x = -1 � x^2 - 1= -x x^2 + x - 1 = 0 ..

solving for x we have x = (5^1/2 - 1 )/2 .. but we want the ratio a/b � so a/b

= 1/x (since x =b/a) � so we have a/b = 2 / (5^1/2-1) .. multiplying both the numerator and denominator with 5^1/2 + 1 � we have ( 2 (5^1/2+1))/((5^1/2-1)*(5^1/2+1)) .. which is equal to � 2 * (5^1/2+1)/4 = (5^1/2 + 1)/2 .. so the answer is B �

Let x be the number, and let's divide x up so that:

x = y + z

and y > z. Because the question tells us that the ratio of x to y is equal to the ratio of y to z, we have the following:

(y+z)/y = y/z

so

1 + z/y = y/z

If we let r = y/z (notice that r is what the question is asking for), this becomes:

1 + 1/r = r 0 = r^2 - r- 1

You can use the quadratic formula (ruling out the negative solution, since r must be positive) to find that r = (1 + sqrt(5))/2.

A watch costs $100. Anakin bought the watch at a discount of x %. He then sold the watch to Luke at a discount of y %. How many dollars did Luke pay Anakin for the watch? Ο 100 − (x + y) Ο 100 − (x + y + (xy/100)) Ο 100 − (x + y − (xy/100))Ο 100 − (x − y + (xy/100)) Ο 100 − (x − y − (xy/100))

A train travels at the rate of 10 miles/hr for the first hour of a trip, at 20 miles/hr for the second hour, at 30 miles/hr for the third hour and so on. How many hours will it take the train to complete a 450-mile journey? Assume that the train makes no intermediate stops.Ο 8 Ο 8.5 Ο 9 Ο 9.5 Ο 10 Answer 9The speeds at which the train travels in successive hours are the terms of the arithmetic progression 10, 20, 30,... To find the time (say, t hours) taken by the train to complete the 450-mile journey, the sum of this sequence to the appropriate term is considered as follows. Time x Speed = Distance = 450 miles. 1 hour x (10 + 20 + 30 + ... to t terms) = 450 Dividing by 10, one obtains 1 + 2 + 3 + ... + t = 45 Now, 1 + 2 + 3 + ... + t = t (t + 1)/2 (a well-known result worth remembering). So, t (t + 1) = 90 giving t = 9.

Hans Solo, a navy admiral, has to choose 3 ships out of his fleet of 8 destroyer ships. In how many ways can this be achieved, assuming that he is not biased toward choosing any particular ship?Ο 24 Ο 48 Ο 56 Ο 120 Ο 336 Answer 56This problem requires calculation of the number of ways in which 8 different objects can be combined into groups of 3. The number of ways in which r objects can be chosen from n objects is given by the expression: nCr = n! / (r!(n − r)!), where ! stands for the factorial notation. Since n = 8 and r = 3, the required number of ways is 8! / (3! 5!) = 8 x 7 x 6 / (3 x 2 x 1) = 56. Note that 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1; 3! = 3 x 2 x 1; and 5! = 5 x 4 x 3 x 2 x 1.Joshua can paint a car in 40 minutes while Ariel can do the same job in 60 minutes. If they both work together, how long will they take to finish the job?

Ο 16 minutes Ο 18 minutes Ο 24 minutes Ο 30 minutes Ο 32 minutes Answer 24 minutesTo solve such problems, consider the rate at which a person or group does a particular job. Joshua does the job at the rate of (1/40)th of the job per minute. Ariel does the job at the rate of (1/60)th of the job per minute. When working together, their rates can be added. So, (1/40) + (1/60) = (3/120) + (2/120) = 5/120. Joshua and Ariel together do the job at the rate of (5/120)th of the job per minute. Thus, time taken to complete the job working together = 120/5 = 24 minutes.

The price of a pair of baseball gloves is increased by 15%. The next month, its price is further increased by 40%. This double increase in the price of the pair of baseball gloves would be equivalent to a single increase ofΟ 25% Ο 49% Ο 51% Ο 55% Ο 61%Answer Taking the original price of the pair of baseball gloves as $100 ($100 is chosen because it is easier to convert into a percentage later), New price after first increase = $115, i.e., 1.15 x 100 New price after second increase = $161, i.e., 1.4 x 115 ∴ Increase in price = $61 and so the increase is 61%

A motor-boat can travel upstream at a speed of 10 miles per hour and downstream at a speed of 14 miles per hour. If the boat travels at a constant speed, what is the speed at which the stream is flowing?Ο 1 mile per hour Ο 2 miles per hour Ο 3 miles per hour Ο 4 miles per hour Ο None of these Answer Let v be the speed at which the boat is moving (relative to the ground) and u be the speed at which the stream is flowing (also relative to the ground). When the boat travels upstream, its speed (relative to the ground) is reduced because the stream opposes the motion of the boat. So, the speed of the boat is v − u = 10. On the other hand, when the boat travels downstream, its speed (also relative to the ground) increases and is equal to v + u = 14. Subtracting v − u = 10 from v + u = 14 gives 2u = 4 or u = 12 miles per hour. So the ans: 2 miles per hour.

Four carpenters can individually complete a particular task in 3, 4, 5, and 7 hours, respectively. What is the maximum fraction of the task that can be completed in forty-five minutes if three of the carpenters work together at their respective rates?(A)11/30 (B) 47/80 (C) 3/5 (D)11/15 (E)5/6 Answer In one hour, the four carpenters can individually complete one-third, one-fourth, one-fifth and one-seventh of the task. For completion of the maximum fraction of the task, the rates of the three quickest carpenters must be added. Thus, (1/3) + (1/4) + (1/5) = (20 + 15 + 12)/60 = 47/60. In 1 hour, three carpenters together can complete (47/60)th of the task. In ¾ hr (i.e., 45 minutes), they can complete (47/80)th of the task. Note that (47/60) x ¾ = 47/80. Ans: 47/80A portion of $8000 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from each of the two portions is the same, then what is the total income from the two investments after one year?

Ο $280 Ο $300 Ο $320 Ο $360 Ο $480 Answer $300Let the $8000 amount be divided into two parts x and y such that x + y = 8000. The annual return on x is 5% and that on y is 3%; so, 5x / 100 = 3y / 100 or y = 5x / 3. On eliminating y, one obtains x + 5x / 3 = 8000 or 8x / 3 = 8000. Thus, x = 3000 and y = 5000. So, 5x / 100 = 150 and 3y / 100 = 150. ∴ Total income after one year is $150 + $150 = $300.

x and y are two distinct positive integers divisible by 5. Which of the following is necessarily divisible by 10?Ο x + y Ο x − y Ο xy Ο 2xy Ο x2 + y2 Answer 2xy is necessarily divisible by 10 because its factors are 2 and 5. The other quantities are not necessarily divisible by 10 as shown by the examples below. If x = 15 and y = 10, then x + y = 25. If x = 15 and y = 10, then x − y = 5. If x = 15 and y = 5, then xy = 75. If x = 15 and y = 10, then x2 + y2 = 225 + 100 = 325. So, ans : 2xyA number x is chosen at random from the set of integers that satisfy | x | < 9. What is the probability that (4 / x ) > x ? Ο 1 / 5 Ο 1 / 8 Ο 7 / 8 Ο 2 / 17 Ο 7 / 17Answer 7 / 17The values of x that satisfy |x| < 9 are −9 < x < 9. There are 17 integers (−8, −7, ..., −1, 0, 1, ..., 7, 8) in this interval. If both sides of (4 / x) > x are multiplied by a positive value of x, then x2 < 4. The only (positive) integer that satisfies −2 < x < 2 is 1. If both sides of (4 / x) > x are multiplied by a negative value of x, then x2 > 4. The six integers that satisfy x < −2 are −8, −7, −6, −5, −4 and −3. Since only 7 out of 17 integers satisfy (4 / x) > x, the probability is 7 / 17.The smallest number, which when decreased by 3 is divisible by 14, 18 and 30 isΟ 633 Ο 663 Ο 693 Ο 723 Ο 753 Answer 63314 = 2 x 7; 18 = 2 x 3 x 3; 30 = 2 x 3 x 5. So, Least Common Multiple (LCM) of 14, 18 and 30 = 2 x 3 x 3 x 5 x 7 = 630. Required number = 630 + 3 = 633.

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How many different integers can be expressed as the sum of two different integers from the set {-3, -2, 1, 4, 5, 9, 10}?

(A) 28 (B) 25 (C) 21 (D) 19 (E) 17If you look at the set of numbers, you will notice that the smallest integer you can form is -5 (from -3 + (-2)) and the largest integer you can form is 19 (from 9 + 10). So any integer that can be expressed as the sum of two numbers in {-3, -2, 1, 4, 5, 9, 10} will have to be between -5 and 19. Here is a table with the preliminary possibilities:

Now for each number in the table, see if you can express it as the sum of two different numbers from the set.

The numbers in yellow can be formed from adding two different numbers in the original set. There are 17 possibilities, so the answer is E.

Which of the following numbers is not a factor of 30^30?(A) 8000 (B) 3030 (C) 125 (D) 81 (E) 64First, let's find the prime decomposition of each number:30^30 = (2 x 3 x 5)^30 = (2^30)(3^30)(5^30), and8000 = 203 = (43)(53) = (26)(53)3030 = 303(10) = (3)(101)(2)(5)125 = 53

81 = 34

64 = 43 = 26

The original number 30^30 has a prime decomposition that consists only of 2's, 3's, and 5's. Any number that is a factor of 30^30 must also be comprised of 2's, 3's, and 5's only. But the number 3030 is has a prime factor of 101, so this cannot divide evenly into 30^30, thus it is not a factor of 30^30. So the correct answer is B.

http://3.bp.blogspot.com/_PzvixRyMFzs/SrKxi7m_yjI/AAAAAAAAAHQ/wYkGqNUNVQU/s1600-h/bbb1.bmp

http://2.bp.blogspot.com/_PzvixRyMFzs/SrKx9_12K1I/AAAAAAAAAHY/Rq62zg6i6KI/s1600-h/bbb2.bmp

Which of the following does not divide 210^21?(A) 9000 (B) 1000 (C) 770 (D) 625 (E) 343Notice that 210 = 2x3x5x7, so 210^21 = (2^21)(3^21)(5^21)(7^21).Also notice that9000 = 9(10^3) = (3^2)(2^3)(5^3)1000 = (2^3)(5^3)770 = 77(10) = (7)(11)(2)(5)625 = (25)(25) = 5^4343 = 7^3Every number except 770 is made of 2's, 3's, 5's, and/or 7's. But 770 contains a factor of 11 which is not in 210. So 770 will not divide evenly into 210^21, that is, 770 is not a factor of 210^21. Thus, the correct answer is C.

If a and b are prime numbers and a ² b (70)(72)(75)(77) = (49)(54)(55) (88)(625), what is a+b ? (A) 10 (B) 12 (C) 14 (D) 15 (E) 16The key to solving this problem is to factor each side as much as possible so that common terms ca be canceled. The left side isa²•b•7•10•8•9•25•3•7•11, or a²•b•7²•2⁴•5³•3³•11, and the right side is7•7•3•3•3•2•5•11•8•11•25•25, or 3³•7²•2⁴•11²•5⁵.When we set these expressions equal to each other and cancel common factors, we arrive ata²•b = 5²•11, so a = 5 and b = 11, and therefore the correct answer is E.

The positive integer m has half the number of distinct prime factors as the integer m+ 1 has. Which of the following could be the value of m ? (A) 24 (B) 29 (C) 64 (D) 69 (E) none of the aboveFor each of the answer choices A-D, we will count the number of distinct prime factors of m and m+1. If m has exactly one half the number of distinct prime factors as m+1 has, we will have the correct answer choice. Otherwise, the correct answer will be E.(A) m = 24, m+1 = 2524 has two distinct prime factors, 2 and 3. 25 has only one prime factor, 5. A is incorrect.(B) m = 29, m+1 = 3029 has one prime factor, 29. 30 has three distinct prime factors, 2, 3, and 5. B is incorrect.(C) m = 64, m+1 = 6564 has one prime factor, 2. 65 has two prime factors, 5 and 13. C is correct.(D) D is incorrect because 69 has two prime factors, 3 and 23, and 70 has three prime factors, 2, 5, and 7.

If y is a positive integer and y³ - y = 210, what is the value of 2y/(y-3)?(A) 5 (B) 4 (C) 7/2 (D) 5/2 (E) 2The equation y³ - y = 210 can be simplified on the left hand side by first factoring out the y:y(y² - 1) = 210. Then, if you notice another factorization, you will arrive at (y-1)(y)(y+1) = 210.This shows that 210 can be expressed as the product of 3 consecutive integers, so with a little guessing and checking, you can figure out which integers work. It turns out that (5)(6)(7) = 210, so y = 6.

Now, 2(6)/(6-3) = 12/3 = 4. So B is the correct answer.***Be careful that you don't mistakenly solve y² - y = 210 instead of y³ - y = 210. The equation y² - y = 210 has y = 15, and y = -14 as solutions.

What is the largest value of x such that 6^ x is a factor of 18!? (A) 16 (B) 12 (C) 8 (D) 4 (E) 3First, we should keep in mind that 6^x = (2^x)(3^x), and 18! = product of every integer between 1 and 18, including 1 and 18. Also, when we say 6^x is a factor of 18!, we mean 6^x divides evenly into 18!. 6^x will divide evenly into 18! so long as 6^x does not have more 2's and 3's in its prime decomposition than 18! has. Thus, we need to figure out how many 2's and 3's 18! is made of.This chart shows the 18 factors of 18! and the highest power of 2 that will go into each factor.

And this chart shows the the highest power of 3 that will go into 18!.

Since the highest power of 2 that divides 18! is 16, and the highest power of 3 that divides 18! is 8, then the highest power of 2x3 = 6 that can divide 18! is 8. (We must take the lower number.) So the answer is C.If y is an integer and y ³ - 4 y = 48, what is y ² + 4? (A) 29 (B) 24 (C) 20 (D) 16 (E) 13 Let's start by factoring the left side of y³ - 4y = 48. We have y³ - 4y = y(y² - 4) = y(y-2)(y+2). If we rearrange terms a little, it is (y-2)y(y+2) = 48 Notice that y-2, y, and y+2 must be either three consecutive even integers, or three consecutive odd integers. Since the product, 48, is even, these must be three consecutive even integers. So we need to find three consecutive even integers whose product is 48. If you experiment a little, you should come up with 2∙4∙6 = 48, and therefore, y = 4.4² + 4 = 20, so the correct answer is C.

Center Info) y ≠ x , and both y and x > 0

(Col A) (Column B)

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(y² + x²)/(y³ + x³) (y + x)/(y² + x²)Let's plug in small values for y and x. If we try y = 1 and x = 2, then Col A is 5/9 and Col B is 3/5. If you know your fractions, you can tell that Col B is larger with these numbers. Let's try another pair, y = 1 and x = 3. Then Col A is 10/28 and Col B is 4/10. Again, Col B is larger. If you're convinced that the pattern will hold for all x and y, go with B.To prove that Col B is always larger than Col A, you will have to work out the algebra. Start by multiplying both columns by (y³ + x³) and (y² + x²) to eliminate the fractions. Col A becomes (y² + x²)(y² + x²) and Col B becomes (y + x)(y³ + x³). When you foil both sides, you will get in Col A the expression y^4 + 2y²x² + x^4, and in Col B y^4 + y³x + yx³ + x^4.Subtract the 4th powers from both columns and then divide both columns by a factor of yx. Hopefully you will be able to produce 2yx in Col A, and y² + x² In Col B. Lastly, if you subtract 2yx from both column, you will get 0 in Col A and (y-x)² in Col B. The information in the center tells you that Col B will be larger. The first way is much easier!!! In any case, the answer is B.

(Col A) (Col B)x^2 - x^3 x - x^2

1 > x > 0Notice that the center information tells us that x, x^2, and x^3 are tiny numbers in this order: x > x^2 > x^3. That's because when you raise a small positive fraction or decimal to a positive integer power, you get an even smaller fraction or decimal. Now since x is positive, we can divide both columns by x and compare instead "x - x^2" in Col A, and "1 - x" in Col B. Remember, you can multiply and divide both columns by a variable only if the variable is restricted to positive values.

But that's not all we can factor out! Look that Col A can be factored into x(1 - x). We can see that 1 - x is also a positive number according to the center information, so we can divide both columns by 1 - x. Finally, we compare instead "x" in Col A ,and "1" in Col B.Thus, the answer is B.An alternative way to do this problem is to replace x with allowable numbers. Let's use numbers that are easy to work with, like x = .1. Then Col A becomes .01 - .001 = .0099, and Col B becomes .1 - .01 = .099. So Col B is larger.

Which of the following is true?(A) 3^10 > 2^15 > 5^6 (B) 2^15 > 5^6 > 3^10 (C) 5^6 > 3^10 > 2^15(D) 5^6 > 2^15 > 3^10 (E) 3^10 > 5^6 > 2^15We need to put the numbers 2^15, 3^10, and 5^6 in order. Let's compare them pairwise, starting with 2^15 and 3^10.If we raise 2^15 and 3^10 to the 1/5 power, we get(2^15)^(1/5) = 2^3(3^10)^(1/5) = 3^2.Since 3^2 = 9 and 2^3 = 8, it implies that 3^10 > 2^15. So we can cross off choices B and D.Now let's compare 5^6 and 2^15. If we raise both of these numbers to the 1/3 power, we get(5^6)^(1/3) = 5^2 (2^15)^(1/3) = 2^5.

Since 5^2 = 25 and 2^5 = 32, it implies that 2^15 > 5^6. So we can cross off C and E.The only choice that remains is A, so this is the correct order.

(Col A) The number of integers between 1 and 2000 (inclusive) that are both perfect squares and perfect cubes.

(Col B) The number of even integers between 1 and 2000 (inclusive) that are perfect 4th powers.

To analyze column A, we first note that any number that is both a perfect square and a perfect cube, must also be a perfect 6th power. This is because x^6 = (x²)³ = (x³)². Thus, we need to count the perfect 6th powers between 1 and 2000. 1^6 = 1, 2^6 = 64, and 3^6 = 27x27 = 729. So far that's three. If we try 4^6, we'll get 64x64, but 64x64 > 60x60 = 3600. That means 1, 64, and 729 are the only numbers between 1 and 2000 that are both perfect squares and perfect cubes.

To analyze column B, we need to look at the 4th powers of even numbers. 2^4 = 16, 4^4 = 256, and 6^4 = 36 x 36 = 1296. So far that's three. If we try 8^4, we'll get 8x8x8x8 = 64x64, but we know that's too big. Thus, 16, 256, and 1296 are the only even 4th powers between 1 and 2000. Since both columns are equal to 3, the answer is C.

a and b are distinct prime numbers, and the legs of a right triangle are a*sqrt(b) and b*sqrt(a). If c is the hypotenuse of the right triangle, which of the following must be true?

(i) c2 is composite (ii) c is not an integer (iii) c2 is odd (A) i only (B) ii only (C) i and ii (D) i and iii (E) ii and iii

Let's evaluate (i) since it occurs in three answer choices. Since we are dealing with a right triangle, we have [a*sqrt(b)]2 + [b*sqrt(a)]2 = c2. This simplifies to a2b + b2a = c2. The left hand side can be factored into (ab)(a+b), which implies that c2 is composite. So (i) is true and we can eliminate B and E.

Let's check (ii) now. If c2 = (ab)(a+b), and both a and b are prime, then we can deduce that c2 is not a perfect square, and thus c is not an integer.

To explain this more fully, notice that (ab)(a+b) = (a)(b)(a+b). We know that neither a nor b are perfect squares, and neither a nor b divides evenly into a+b. Then it is impossible for (ab)(a+b) to be a perfect square. Thus, (ii) is true and we can eliminate A and D. So the correct answer is C. Suppose a=3 and b=5 then A=3*√5 B=5*√3 C2= A2+B2

A cake recipe requires 3 eggs and 0.5 cups of milk. Up to one egg per cake can be replaced with 0.25 cups of milk. If Jessie has 21 eggs an 5 cups of milk, What is the maximum number of cakes she can make, assuming there are no bounds on the other ingredients?(A) 10 (B) 9 (C) 8 (D) 7 (E) 6

Jessie can make both 3-egg cakes and 2-egg cakes. A 3-egg cake requires 3 eggs and 0.5 cups of milk; a 2-egg cake requires and 0.75 cups of milk. We can determine the maximum number of cakes she can make by systematically counting the different combinations, however, this will be too time-consuming for the GMAT and GRE. A more efficient way is to solve it algebraically. Let's say A is the number

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of 3-egg cakes, and B is the number of 2-egg cakes.

One equation relating A and B is 3A + 2B ≤ 21. Another equation relating A and B is 0.5A + 0.75B ≤ 5, or if we multiply it by 4, 2A + 3B ≤ 20.

The total number of cakes is A + B; this is quantity for which we want to find an upper bound.Notice if we add the two inequalities, we get 5A + 5B ≤ 41, or A + B ≤ 8.2. So the total number of cakes is no more than 8, answer choice C. (In full detail, she can make five 3-egg cakes, and three 2-egg cakes.)

Theater A can hold 320 more people than Theater B. On the opening day of a play, Theater A sold twice as many tickets as Theater B. Theater A sold 90% of its seats, and Theater B sold 75%. What is the ratio of unsold seats in Theater A to unsold seats in Theater B?(A) ¼ (B) ½ (C) 2/3 (D) 5/6 (E) not enough information givenLet A and B be the capacities of Theaters A and B respectively. One equation we can derive is A = 320 + B.Also, if 90% of A's capacity is twice as much as 75% of B's capacity, then we have.9A = 2(.75B).9A = 1.5BA = (1.5/.9)BA = (5/3)BCombining these two equations gives us (5/3) B = 320 + B. Solving for B, we get(5/3)B = 320 + B(5/3)B - B = 320(5/3 - 1)B = 320(2/3)B = 320B = 320(3/2)B = 480. Substituting this value into one of the equations gives us A = 800.If Theater A sold 90% of its 800 seats, then there were 80 unsold seats. And if Theater B sold 75% of its 480 seats, then there were 120 unsold seats. So the ratio is 80/120 = 2/3, answer choice C.

At a company's factory, the cost of manufacturing q items is 500 + 2 q + 5sqrt( q ) dollars. Each item sells for 7 dollars. Assuming every item that is manufactured can be sold, how many items must be produces so that the company makes a profit of 700 dollars?

(A) 256 (B) 250 (C) 225 (D) 200 (E) 196

For any business, total profit equals total selling price minus total cost of production. The selling price for q items is 7q, and the cost of production is given by the expression 500 + 2q + 5sqrt(q). Thus, we can set up the equation

7 00 = Sale (7q) - Cost of production (500+2q + 5sqrt(q)), or7q-2q-5 sqrt (q)-500-700=0 or 5q - 5sqrt(q) - 1200 = 0.This equation can be divided by 5 to yieldq - sqrt(q) - 240 = 0.

It looks bad, but it's not that bad! There are several ways to find the solution. Notice that answer choices A, C, and E are square numbers. That means we can plug them

into the equation and test them. If we start with C, we get 225 - 15 - 240 = -20. So C is too small.If we try A, we get256 - 16 - 240 = 0 So A is the correct answer.

Boat X can cross the lake in 40 minutes, and Boat Y can cross in 30 minutes. Both boats start at opposite ends of the lake and travel toward each other, except Boat Y stalls and starts 20 minutes after Boat X starts moving. When they pass each other on the lake, what is the ratio of the distance that Boat X has traveled to the distance that Boat Y has traveled?

(A) 7:3 (B) 5:3 (C) 5:2 (D) 4:3 (E) 2:1

For simplicity, let's approach this problem by picking a number for the distance across lake. Since the relative speeds of the two boats do not depend on the lake's width, their relative traveling distances will not depend on the distance of the lake either. Let's choose 14 miles as the lake's distance across. And thus the rate for X is 14/40 miles per minute, and the rate for Y is 14/30 miles per minute.

Since X can traverse the entire distance in 40 minutes, then it can travel half the distance in 20 minutes. So by the time Y starts moving, X has already traveled 7 miles. The boats will pass each other somewhere in the next 7 miles.

Call dy the distance traveled by Y when the boats meet, dx the distance traveled by X, and ry and rx their respective rates. Then we have

dy = (ry)t and dx = (rx)t.

Since they take the same amount of time to meet when traveling those 7 miles, we can see that dy/dx = ry/rx.

This means the Y travels 4 miles, and X travels a total of 7 + 3 = 10 miles when the boats meet. Since 10:4 = 5:2, the correct answer is C.

A concert hall has a seating capacity of 1000 seats, which are either on the balcony, mezzanine, or main floor. There are twice as many main floor as balcony seats. For a particular sold-out show, the seat prices were $7 bal., $12 mezz., and $20 floor. The revenue from mezzanine seats was twice the revenue from balcony seats. How many balcony seats are there?

(A) 200 (B) 240 (C) 250 (D) 280 (E) 300

There are 3 unknown quantities in the problem, the number of balcony, mezzanine, and main floor seats. We can call them B, M, and F. To find the value of B, we need

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to find equations that relate the three variables.We knowB + M + F = 1000 (since the capacity is 1000)2B = F (since there are twice as many F seats as B seats)2(7B) = 12M (since the M revenue is twice the B revenue)Substituting the second equation into the first, we getB + M + 2B = 10003B + M = 1000M = 1000 - 3B

If we substitute M = 1000 - 3B into the last equation, we get2(7B) = 12(1000 - 3B)14B = 12000 - 36B50B = 12000B = 12000/50 = 240So there are 240 balcony seats, and the correct answer is (B).

Soraya's gold credit card has twice the spending limit of her silver card, and three times the spending limit of her bronze card. The balance on her bronze card is 1/3 of the spending limit on that card, the balance on her silver card that is 1/4 of the spending limit on that card, and the balance on her gold card is 1/5 of the spending limit on that card. If Soraya transfers all the balances to her gold card, approximately what percent of its limit will be unspent?

(A) 36% (B) 40% (C) 45% (D) 56% (E) 63%

Let's pick numbers for the spending limits of the three credit cards. Since (5)(4)(3) = 60, let's say the gold card has a limit of 360, the silver card a limit of 180, and the bronze card a limit of 120. Those numbers will work nicely with fractions like 1/3, 1/4, and 1/5.

The balance on the bronze card is (1/3)120 = 40. The balance on the silver card is (1/4)180 = 45. And the balance on the gold card is (1/5)360 = 72. If we transfer all those balances to the gold card, the gold card will have a balance of 40 + 45 + 72 = 157. The unspent limit will be 360 - 157 = 203.

The fraction of the limit that is unspent is 203/360. Notice that 203/360 is slightly greater than 200/360 = 5/9, and 5/9 = 55.5%. So the right answer must be close to 55.5%, which makes choice D correct. A dirty sock contains 15 marbles, and each marble is either red or blue . Cecile calculates that the probability of drawing two red marbles is exactly 1/5. How many blue marbles are in the sock?

(A) 8 (B) 9 (C) 10 (D) 11 (E) 12

This is a tricky probability question. Normally, you are given the number of items and asked to compute the probability. In this problem, you are given a probability and asked to find the number of items.

Worst case scenario, you have no idea how to do this problem on the GMAT or GRE. If you are going to guess randomly, the first choice you should eliminate is E, 12 blue marbles. This answer is too simplistic to be correct on this problem. 12 blue marbles and 3 red marbles would work only if the probability of drawing one red marble is 1/5, because 3/15 =1/5.

Another thing you can do if you get stuck: pick an answer choice and work backward to see if it is correct. Example, pretend there are 10 blue marbles and 5 red marbles (choice C). If there are 5 red marbles in the sock, then the probability of drawing one red and then another red is (5/15)(4/14) = (1/3)(2/7) = 2/21. This is less than 1/5, so you know that there must be more red marbles and fewer blue marbles.

All this preamble is to show you that there are several ways to approach GMAT and GRE math problems.

Let's solve this problem directly and say there are x red marbles. The probability of drawing two red marbles is [x/15][(x-1)/14] = (x²-x)/(15∙14).

Now solve (x²-x)/(15∙14) = 1/5. When you simplify, you'll get x² - x - 42 = 0, or (x-7)(x+6) = 0, so x = 7 red marbles. Therefore the correct answer is A, 8 blue marbles.

Seven children decide to play "game show," a game in which they select one person to be the host and divide the remaining children into two groups of three. The twins Jeff and Josh dislike each other and won't play on the same team. With that restriction, how many ways are there for the kids to distribute themselves?

(A) 140 (B) 120 (C) 84 (D) 50 (E) 35Let's first count the number of team combinations if Jeff is the game show host. Pick any remaining kid and fix him or her in one team. Then there are (5 ch 2) ways of picking the other team members for that kid. (5 ch 2) = 10 using the combination formula. So, if either Jeff or Josh is the host, there are 10 ways to divide the other kids into teams.

Now let's see how many distributions there are if neither Jeff nor Josh is the host. There are 5 ways to pick one kid to be the host and then place Jeff and Josh on different teams. There are now 4 kids left and (4 ch 2) = 6 ways to pick the remaining team members for Jeff. So there are 5x6 = 30 ways of arranging the teams if one of the nice kids is the host.

Lastly, since there are two arrangements with one of the twins as host and 30 arrangements with the twins on teams, there are 10 + 10 + 30 = 50 total ways to distribute the children. So D it is.

Alice drove 80 miles for two hours on the first leg of her trip, and then 70 miles for one hour on the second leg. Ben drove 150 miles for two and a half hours. Keeping Alice's drive time of 3 hours constant, how many miles per hour faster should Alice have driven during the first leg of her trip in order for her average speed to match Ben's average speed?

(A) 10 (B) 12 (C) 15 (D) 16 2/3 (E) 20

With 2 minutes per question on the GMAT and only 1.5 minutes per on the GRE, you have to consider if it's worth spending the time to set this one up as an algebra problem, or just plug and chug. Let's try the latter option. But first, let's calculate Alice's and Ben's average speeds. Alice drove a total of 150 miles in 3 hours, so her average speed is 50mph. Ben drove 150 in 2.5 hours, his avg is 60mph. And how fast was Alice driving on the first leg? 80/2=40, so 40mph.

If we plug in numbers, we want to pick easy ones, so let's ignore 16 2/3. Let's try 20. Supposing that Alice drove 20 mph faster on the first leg, she would have been going 60. But her speed on the second leg was 70, so already we know that's too much to make her average speed equal to Ben's. We need a number less than 20...

If we pick 15, then Alice drives 55mph for two hours on the first leg. How many miles does that equal? 55x2=110, so 110 miles. (It's OK that we changed Alice's mileage, we only need to keep her total drive time constant.) Add to that the next 70 miles and we get Alice's total mileage is 180 miles. Divide that by 3 hours and we get 60mph. Exactly Ben's average speed. The answer is C.\

Relevant formulas: Average speed = Total Distance/Total Time, speed x time = distance.

In a certain population, the percentage of people who have contracted an infectious disease after w weeks is (2 w )/(16 + 2 w ). After how many weeks will 80% of the population be infected?(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Since 80% = 4/5, we can set up the equation (2w)/(16 + 2w) = 4/5 and cross multiply. This will give us 5(2w) = 4(16 + 2w), or 5(2w) = 64 + 4 (2w). If we subtract 4(2w) from both sides we will get 2w = 64. This problem makes a case for having the powers of 2 memorized. If you know that 2^6 = 64, then you get w = 6, so the answer is C.

I've received several questions about how to count arrangements of objects in a circle recently. Here is a representative question about indistinguishable colored beads on a bracelet.

A bracelet is made of 8 beads, 5 yellow and 3 green. The yellow beads are indistinguishable from one another, and so are the greens. Considering that many arrangements of beads can be transformed into another by rotations and flips, how many fundamentally distinct arrangements of beads are there?(A) 5 (B) 8 (C) 24 (D) 28 (E) 56

First, let's look at a picture to understand how an arrangement can be transformed into another by rotations and flips. The picture below shows the same bracelet, but it has been rotated and flipped into different positions.

This picture suggests that counting the number of arrangements is more complicated than applying the usual combination or permutation formulas, and it is! In fact, this is a type of math problem that is usually solved using a theorem called Burnside's Lemma. Burnside's Lemma is something that you would only learn in college if you were a math or physics major, so what is a problem like this doing on the GRE or GMAT?

The GRE and GMAT wouldn't be very effective tests if they required you to know facts that only particular majors would know, so that means there must be an easier way to work out this particular problem. Let's try to find it.

For starters, we can eliminate some of the answer choices. (8 ch 3) = (8x7x6)/(3x2x1) = 56; we know this is too large because it counts rotations and flips of the same arrangement multiple times, so eliminate E. (8 ch 2) = (8x7)/(2x1) = 28, but that's only half of 56, still too big, so eliminate D. And C is 3x8 = 24, too large and too naive! Eliminate C too.

The choices left are 8 and 5. Since those numbers are small, we should try drawing the arrangements. It turns out that there are only 5 distinct arrangements of 3

http://en.wikipedia.org/wiki/Burnside's_lemma

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green and 5 yellow beads.

So the correct answer is A.

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Syvum TestsThe ratio of apples to oranges to bananas in a fruit-basket is 4 : 5 : 7. If the fruits in the basket are worth $96 and were purchased at an average rate of $1.50 per fruit, how many bananas are there in the fruit-basket?Ο 7 Ο 16 Ο 28 Ο 36 Ο 64 Number of fruits in the basket = 96/1.50 = 64 Sum of the parts of the ratio = 4 + 5 + 7 = 16 ∴ Number of bananas in the basket = (7/16) x 64 = 28In a right-angled triangle, the two unknown angles are in the ratio of 1:4. The smaller angle is (a)18o (b) 20o (c) 22o (d) 22.5o (e)30o The angles of a triangle add up to 180o. So the two unknown angles of the right-angled triangle add up to 90o. If the smaller angle is xo, then the other unknown angle is 4xo. So, x + 4x = 90 or x = 90/5 = 18.A certain integer n when divided by 5 yields a remainder of 4. Which of these cannot be an integer?(a) n / 9 (b) n / 10 (c) n / 14 (d) n / 19 (e) n / 24 Since n when divided by 5 yields a remainder of 4, n must be the sum of 4 and a multiple of 5, i.e., n could be 9 (4 + 5 x 1), 14 (4 + 5 x 2), 19 (4 + 5 x 3), 24 (4 + 5 x 4), etc. Thus, n has either a 4 or 9 in the units place. Integers (except multiples of 10 with a 0 in the units place) when divided by 10 give rise to decimal numbers and not integers. ∴ (n / 10) cannot be an integer (since no n is a multiple of 10).

Alternatively, substitute the following values for n in the choices given, to form integers: If n = 9 (4 + 5 x 1), then n / 9 = 1 (an integer) If n = 14 (4 + 5 x 2), then n / 14 = 1 (an integer) If n = 19 (4 + 5 x 3), then n / 19 = 1 (an integer) If n = 24 (4 + 5 x 4), then n / 24 = 1 (an integer)13. In how many different ways can the digits 2, 2, 7 and 9 be arranged, so that there is at least one 2 between 7 and 9 always?(a) 3 (b) 4 (c) 6 (d)9 (e) 12 There are three possibilities (2, 7, 9) for the first digit. If the first digit is 2, there are two arrangements: 2729 and 2927. If the first digit is 7, there are two arrangements: 7292 and 7229. If the first digit is 9, there are two arrangements: 9272 and 9227. ∴ There are 6 arrangements possible.

Which of the following cannot be the largest angle of a triangle?(a) 57 (b) 60 (c) 61 (d) 64 (e)73 An angle less than 60o cannot be the largest angle of any triangle as proven below. Let the angles of the triangle measure x, y and z (totaling 180o), with x denoting the largest angle.

Then, x = 180 − y − z with y < x and z < x. x is minimum when the other two angles are maximum (equal to x). So, xmin = 180 − xmin − xmin or xmin = 180/3 = 60. Thus, the minimum value of the largest angle in a triangle is 60o.

For any integer n greater than 0, n! denotes the product of all the integers from 1 to n , inclusive. How many multiples of 3 are there between 6! − 6 and 6! + 6, inclusive ?(a) One (b)Two (c) Three (d) Four (e) Five 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720. Since 714, 717, 720, 723 and 726 are divisible by 3, there are 5 multiples of 3 between 714 and 726. Note a number is divisible by 3 if the sum of its digits is divisible by 3.

A snail crawls at the rate of y inches per hour. How many minutes will it take the snail to crawl a distance of x inches ?(a) x / y (b) y / x (c) 60x / y (d)60y / x (e) y / (60x) 60x / ySpeed = Distance / Time Time = Distance / Speed = x / y hr = 60x / y minutes.

A bucket contains 24 liters of pure alcohol. Victor removes 8 liters of liquid from the bucket and replaces it with 8 liters of water. Then Sofia removes 6 liters of liquid from the bucket and replaces it with 6 liters of water. After Sofia replaces the liquid, what is the ratio of alcohol to water in the bucket? (GMAT/GRE)(A) 1:1 (B) 1:2 (C) 2:1 (D) 3:1 (E) 4:3For this GMAT or GRE ratio problem, we can determine the final ratio of alcohol to water (a part-to-part ratio) by determining the final concentration of alcohol (a part-to-whole ratio). Before Victor and Sofia come to the bucket, the concentration of alcohol is 100% = 1.

When Victor removes 1/3 of the alcohol and replaces it with water, the concentration is diluted to 2/3 of its original strength.

Then, when Sofia removes 1/4 of the liquid and replaces it with water, the solution is further diluted by a factor of 3/4. So the final concentration is (3/4)(2/3) = 2/4 = 1/2.

At the end of the process, the bucket is half alcohol and half water, i.e., equal parts alcohol and water. So the ratio is 1:1. The correct answer is A.The surface area of a cube increases by 300%. By what percent does its volume increase? (GMAT/GRE)(A) 520% (B) 600%(C) 625%(D) 700%(E) 732%Let's say the side length of the cube is originally 1. Before its surface area increases, its surface area is 6(1)² = 6, and its volume is 1³ = 1. When the surface area increases by 300%, it becomes 6 + 3(6) = 24.The surface area of a cube with side length s is 6s², so if 6s² = 24, then s = 2 is the new side length. The new volume is 2³ = 8. This is an increase of 700% over the original volume, so the correct answer is D.An equilateral triangle and a regular hexagon both have a perimeter of 12. What is the ratio of the area of the hexagon to the area of the triangle? (A) (3/2)sqrt(3) (B) (2/3)sqrt(3) (C) sqrt(3) (D) 2 (E) 3/2Each side of the hexagon has length 2, and each side of the triangle has length 4 since they are regular polygons. Here is a sketch drawn to scale

We could do this problem the normal way by finding their areas and computing the ratio. To find the area of the hexagon, you would need to break it up into equilateral triangles. This could be time consuming if you don't remember the properties of equilateral triangles. Luckily, there is an easier way that only requires the ability to count.Notice that you can partition each polygon into smaller equilateral triangles with side lengths of 2.

The larger triangle contains 4 such figures, and the hexagon contains 6. Thus, the ratio of their areas is 6/4, or 3/2. So the correct answer is EZach is going to mix three water-based solutions. Solution X is 40% alcohol, Solution Y is 30% sugar, and Solution Z is 30% alcohol and 10% sugar. If the final product is 10% alcohol and 23% sugar, what is the ratio of Solution X to Solution Z that Zach used?(A) 2:3 (B) 1:2 (C) 1:1 (D) 2:1 (E) 4:3 Let's designate x as the proportion of Solution X, z as the proportion of Solution Z, and 1-x-z as the proportion of Solution Y. Remember, all the proportions (or percents, or probabilities) must add up to 1. The number we want to find is x/z.To help illustrate things, this table shows the proportion of ingredients in each solution.

http://4.bp.blogspot.com/_PzvixRyMFzs/StDzpTmwfaI/AAAAAAAAALY/K2hblT13AUw/s1600-h/hextri1.bmp

http://2.bp.blogspot.com/_PzvixRyMFzs/StD06T7yGUI/AAAAAAAAALg/9hH56Q-369o/s1600-h/inprog.bmp

And this table illustrates how the final product is made from various proportions of the three solutions.

The final product is 67% water, 10% alcohol, and 23% sugar, which adds up to 1. The columns circled in red represent the system of equations we need to solve:.4x + .3z = .1.3(1-x-z) + .1z = .23,which can be simplified to4x + 3z = 13 - 3x - 3z + z = 2.3With some algebra, you find the solution, which is x = .1 and z = .2. Thus, the ratio of x to z is 1/2. So the right answer is B.

In the figure above, AB has length 60 and AD has length 30. What is the ratio of the area of shaded region to the area of triangle ABC?(A) 1:3 (B) 1:sqrt(3) (C) 1:4 (D) 1:6 (E) not enough informationThe answer is C; the shaded region takes up 1/4 of the area of the whole triangle. The picture below will help you see why:

http://4.bp.blogspot.com/_PzvixRyMFzs/SsrTODnvoYI/AAAAAAAAAKI/yC7vr1ZrPaI/s1600-h/tablee1.bmp

http://4.bp.blogspot.com/_PzvixRyMFzs/SsrTy6yw_iI/AAAAAAAAAKQ/9b83n8GR5sM/s1600-h/tablee2.bmp

http://2.bp.blogspot.com/_PzvixRyMFzs/SsvFFfvDDVI/AAAAAAAAAKY/pV8gjll8SeY/s1600-h/fig1.jpg

Because D is half way between A and B, the two blue triangles at the bottom have the same dimensions, and therefore, the same area. The light blue and yellow triangles also have equal area because they have the same dimensions. And finally, the yellow and pink triangles have the same dimensions and area. Thus, the large triangle can be divided into 4 equal triangles. That means the original shaded triangle has 1/4 the area of the big triangle.270 lbs of corn will feed 45 chickens for 30 days. For how many days can 75 chickens be fed with 300 lbs of corn?(A) 30 (B) 27 (C) 25 (D) 24 (E) 20We can start this GRE math problem by figuring out how much corn one chicken can eat in one day. If we know that 45 chickens can eat 270 lbs of corn in 30 days, then in one day, 45 chicken will eat 270/30 = 9 lbs of corn. And then one chicken will eat 9/45 = 1/5 of a pound of corn in one day.If we have 75 chickens, and each of them eats 1/5 lb of corn per day, then altogether they eat 15 lbs in one day. So 300 lbs of feed will last 300/15 = 20 days.So the correct answer is E.At Zach's store, the prices per pound of peanuts, raisins, almonds, and chocolate chips are $.30, $1.40, $.90, and $1.30 respectively. Zach is going to make trail mix that is (by weight) 40% peanuts, 30% raisins, 20% almonds, and 10% chocolate. How much should he charge per pound of trail mix?(A) $.63 (B) $.72 (C) $.85 (D) $.96 (E) $1.05This GRE problem is about computing a weighted average. To avoid any confusion with multiplying decimals, let's make the problem easier by finding the cost of 10 pounds of Zach's trail mix.

We would need 4 pounds of peanuts at $.30 per pound, or $1.203 pounds of raisins at $1.40 per, or $4.202 pounds of almonds at $.90 per, or $1.801 pound of chocolate chips at $1.30, or $1.30The total cost is $8.50 for 10 pounds, or $.85 per pound. So C.A circle of radius r is inscribed inside a triangle with angles 30, 60, and 90 degrees. In terms of r , what is the length of the shortest side of the triangle? (A) (2 + sqrt(3))r (B) 3r (C) (1 + sqrt(3))r (D) sqrt(3)r + 1 (E) r² + sqrt(3)

http://4.bp.blogspot.com/_PzvixRyMFzs/SsvFNv_YKvI/AAAAAAAAAKg/nwT-g9H7uSY/s1600-h/fig2.jpg

If you draw a line from the center of the circle to the top vertex, and two perpendicular lines from the center of the circle to the edges, you will get the figure above. As if by magic, the pink triangle is another 30-60-90 degree triangle. Recall that a triangle with angles 30-60-90 has side lengths in the ratio 1 : sqrt(3) : 2.

Then the shortest side has length r(1 + sqrt(3)), so C.Suppose that on test day you had no idea how to do this problem, it's still possible to get the correct answer by guessing intelligently. First, the correct answer has to be in the form nr where n is some number. Answer choices D and E can be eliminated since they aren't in the right form. Next, if you eyeball the figure, you notice that the shortest side of the triangle is only a bit longer than the diameter of the circle, which is 2r. That rules out A and B.Ms. Takahashi's class took a test and the average score of the entire class was 84. The average score of the boys was 90, and the average score of the girls was 81.

(Col A) The number of boys in the class(Col B) The number of girls in the class

From the number line, we can see that the class average is closer to the average of the girls, which means there must be more girls in the class than boys to skew the average in their direction.

If the number line picture is not convincing, you can prove it algebraically. Let B = # of boys and G = # of girls. Then the class average is (90B + 81G)/(B + G).If we set this equal to 84, we get (90B + 81G)/(B + G) = 84, or 90B + 81G = 84B + 84G, which can be simplified to 6B = 3G if we put like terms together.6B = 3G further simplifies to 2B = G, which means that there are actually twice as many girls as boys in the class. So the answer is that column B is larger.

http://4.bp.blogspot.com/_PzvixRyMFzs/SYZjYyrQzDI/AAAAAAAAAEI/MOio1jiIrGg/s1600-h/306090circinscr.jpg

http://4.bp.blogspot.com/_PzvixRyMFzs/SWPVHKOCptI/AAAAAAAAAEA/9nQjPSJosvo/s1600-h/numberline.jpg

If x and y are integers such that 12 ≤ x ≤ 30 and 12 ≤ y ≤ 30, then how many pairs of x and y are there such that the ratio of (5 + x ) to (7 + y ) equals the ratio of x to y ? (A) 1 (B) 2 (C) 3 (D) 4 (E) not enough informationThe question essentially asks how many solutions are there to the equation (5 + x)/(7 + y) = x/y where x and y are between 12 and 30. If we simplify the expression by cross multiplying, we get (5 + x)y = (7 + y)x, or 5y + xy = 7x + xy. If we subtract xy from both sides, we get 5y = 7x, or x/y = 5/7.There are exactly two pairs of x and y that satisfy the conditions of the question: x = 15 & y = 21, and x = 20 & y = 28. Therefore, the answer is B.Generally, on the GMAT and GRE math section, the answer "not enough information" is wrong in a multiple choice question. Not always, but if you have to guess, your chances are better if you eliminate that one.

(Q17) GMAT/GRE Proportions Dawn leaves a whole pizza in the break room at work and then leaves. Andy comes in the break room and eats half of it. Later Betty stops by and eats a third of what's left. Then Cate comes in and eats a fourth of what remains. When Dawn comes back, how much of the pizza is left for her?(A) 1/12 (B) 1/6 (C) 1/5 (D) ¼ (E) 2/5Since we have fractions with denominators 2, 3, and 4 in the problem, lets pretend that the pizza has 2x3x4 = 24 slices. This gives us a nice whole number to work with.If Andy eats half of the 24 slices, then there are 12 left. If Betty eats a third of those 12 slices, then there are 12 - 4 = 8 left. And if Cate eats a fourth of those 8 slices, there are 8 - 2 = 6 left. So Dawn gets 6 out of 24 slices, or 1/4. The correct answer is D.

http://gmatgremath.blogspot.com/2008/12/q17-gmatgre-proportions.html

10, 15, 25, 30, x

If d is the standard deviation of number in the list above for which of the following values of x would the value of the d be least.

a) 0 b) 5 c) 20 d) 30 e) 80

1) You assume a mean..say 25 in this case.2) You calc. SD = Sq.Rt.[{(mean - number)^2}/n]…where n=6=total no. in the series.3) Now take each and every choice in the ans option…You will see that the value of (mean – number)^2 is the least when ‘x’ is 20…and hence the SD.

See for yourself:

1. (25 – 0) = 25….25^2 = 625….too big! This will increase the numerator and hence the SD.2. (25 – 5)^2 = 4003. (25 – 20)^2 = 25….Clearly the smallest!4 & 5 are out of question.

Parleyconfer: to talk or negotiate, especially with an enemy noun (plural parleys) discussion: a talk or negotiation, especially between opposing military forces (dated)

synonym: parley(noun)

conference, meeting, discussion, confab (informal), negotiations, huddle (informal), deliberations (formal), consultation, talks

parley(verb)

confer, negotiate, talk, discuss, deliberate, consult

INVEIGH:

speak out angrily: to speak angrily in criticism of or protest at something

INVEIGLE:

1.

persuade: to charm or entice somebody into doing something that he or she would not otherwise have done inveigled me into making the trip

SYNONYM:

entice, persuade, trick, deceive, charm, con (informal), cajole, wheedle, influence, convince, beguile, allure, seduce

I would agree with your choice (One). Reasoning: the -33 rating would not be possible

without at least one rating of -2. Even if each respondent indicated -1, this would only total -

30. The trick is to consider the -26 as well. Could this be achieved without a -2? Yes. You'd

have to have a breakdown like this:

twenty-six "-1" ratings, followed by either:

four "0" ratings

or two "-1" and two "+1" ratings

If a sandbox in the shape of a right triangle has a hypotenuse of h feet, an area of a square feet, and one leg of length x feet, which of the following must be true?

(A)

(B)

(C)

(D)

(E)

We could solve this question using algebra, but since the algebra is relatively complicated, and since

there are variables in both the answer choices and the question prompt, it is a great candidate for the

plugging in numbers strategy.

The easiest right triangle to work with is a 3-4-5 right triangle. In this case, the hypotenuse is 5, so h =

5. The area of the triangle is , so a = 6. Finally, since

the triangle has legs of two different lengths, we could choose x = 4 or x = 3. We will choose x = 3,

because smaller numbers are generally easier to work with.

Now that we have chosen values for the variables, we plug those values into the answer choices in

order to identify the equation that must be true. Before we start, notice that answer choices B and E

have only addition (no subtraction). We know that each expression in is positive, and

we know that there is no way that three positive quantities can sum to zero, so there is no way the

equation in B can be true. The same goes for the expression in answer choice E.

Therefore, we will plug the values above into answers A, C, and D.

A: . Since , we can write this as

. Since this is not equal to zero, A is incorrect.

C: . Since , we can write this as

. C is a possible answer, but we cannot stop here, since it is

possible that answer D is also equal to zero.

D: . Since , we can write this as

. Since this is not equal to zero, D is incorrect.

The correct answer is C.

We can also solve the problem algebraically. Let y be the length of the second leg of the triangle.

Then, we can use the Pythagorean Theorem to write x2 + y2 = h2, and we can use the formula for area

of a triangle to write . In order to arrive at a statement that does not have the variable y in

it, we solve for y in the second equation and then substitute that value of y into the first equation.

Since , we can multiply by 2 and divide by x to find .

We then substitute this into the first equation. . This simplifies to . If we

subtract h2 from both sides, we get . This is answer C, so C is correct.

On his trip to Alaska, Joe traveled y percent of the total distance at an average speed of 50 miles per hour and the rest of the distance at an average speed of 70 miles per hour. What was Joe’s average speed in miles per hour for the whole trip in terms of y ? (A) 70(50 – y) (B) (y + 60)/4 (C) 10,000 – 199y (D) 17,500/(y + 250) (E) y/2 + 35Rather than doing the algebra, which can be more complicated than meets the eye on many GMAT Math questions, simply come up with a sample value for y. y = 50 is ideal because it will allow you to guesstimate and use the answers. In fact, if you use y = 50, you’ll have very little to do other than some arithmetic to work the answer choices. That’s because if y = 50, Joe is making half his trip at 50 mph and the other half at 70 mph, so you know Joe’s average speed is very close to 60 mph (the average of 50 mph and 70 mph). If you think Joe’s average speed is exactly 60 mph, you should realize that’s a little too easy to be true; on a problem that’s clearly somewhat involved (based on the answer choices), an answer that you could arrive at so immediately is probably incorrect. (In these situations, don’t second-guess yourself, but if you think a problem is really that simple, double-check by re-reading the question and making sure you know what you’re being asked.) OK, since exactly 60 mph is too easy but you know the answer is near 60 mph, you just have to figure out whether it would be a little above or a little below 60 mph. Ask yourself, “Does Joe spend more time going 50 mph or more time going 70 mph?” The answer is that Joe spends more time at the slower speed of 50 mph, so logic tells us that his average speed for the entire trip is slightly below 60 mph.

Now all you’ve got to do is the arithmetic to work the answer choices. With y = 50, you’re looking for an answer that’s slightly below 60:

http://www.beatthegmat.com/mba/2010/04/26/problems-with-problem-solving-guesstimate-and-use-the-answers

(A) 70(50 – 50) = 70(0) = 0 (B) (50 + 60)/4 = (110)/4 = 27.5 (C) 10,000 – 199(50) = 10,000 – 9,950 = 50 (D) 17,500/(50 + 250) = 17,500/(300) = 175/(3) = 58 1/3 (E) 50/2 + 35 = 25 + 35 = 60

(A) and (B) are way too low. (C) is also too low (the answer must be more than 50 mph if some percentage of the trip was traveled at 70 mph). When you get through calculating (D), it’s looking pretty good but you still need to check (E). (E) is exactly 60, which you know is NOT the answer, so you’re safe picking (D).

The correct answer is (D).

Our sample question shows that you’re often wasting your time on PS if you’re doing complicated algebra. Yes, there are some very advanced questions conquered best by algebra; however, when you see a variable(s), you should look to come up with a sample value for the variable(s) before you think about doing the algebra. After you’ve come up with a sample value, you’ll often find that you’re pretty much left with an arithmetic problem, in which case you’ll often be able to guesstimate and use the answers to get the credited response. The time to do algebra is if making up a sample value doesn’t immediately lead to progress toward a solution. For example, if you try making up a sample value but you have no idea what to do with that sample value in order to take the first step toward the solution, perhaps you need to translate the question from English to an algebraic expression. Still, for the vast majority of problems that involve variables, coming up with a sample value(s) is all you’ll need to do to get the correct answer, so this strategy, along with guesstimating and using the answers, is far more applicable and useful than brute-force solving with algebra (with which you’re more likely to make a mistake anyway, especially given the often deceptive complexity of the correct algebra and the time pressure you’re under on the Math section). Coming up with sample values for your variables will greatly increase your accuracy and your efficiency on PS!

By the way, this technique of using sample values rather than algebra is for testers in all score ranges. Even if you’re realistically aiming for 50 on Math, this technique will be an important part of your process. Even if you’re very good at algebra, it’s important to complement that ability with the ability to find shortcuts and out-game the test, and using sample values for your variables is one of the ways to do this. Try it and you’ll come to see the value of sample values—happy practice!

3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?(A) 1/8 (B) ¼ (C) 3/8 (D) ½ (E) 5/8 AnswerSince we are only dealing in fractions, let’s pick a simple number for the total number of students: 8. Thus, 3 students are in all 3 clubs. 4 students are in A, 5 students are in B, and 6 students (3/4 of 8 ) are in C. Since every student is in at least one club, then we know that the total number of club members, appropriately counting people in two or three clubs, must be 8.We can use the following idea: suppose there were no overlap in the club memberships at all. Then we could just add up the club memberships, and there would have to be 4 + 5 + 6 = 15 students. Now, we know that this is overcounting; we counted 3 students 3 times (for all 3 clubs), when we should have only counted them once. So we overcounted them twice, meaning that we should subtract 2×3 = 6 students from the 15, leaving 9. We still have an extra student in the count, since we really only have 8 people. Now, let’s think about the people in exactly 2 clubs: we counted those people twice, when we should have counted them just once. So we

http://www.beatthegmat.com/mba/2010/04/26/problems-with-problem-solving-guesstimate-and-use-the-answers

overcounted them once, and we should subtract their number once from the total. Representing this number by x, we get 9 – x = 8, so x = 1. The fraction of people in exactly 2 clubs is 1/8.An easier formula for this process that captures the same idea is this:Total people in at least one club= Total of separate memberships of each club,MINUS the people in exactly two clubs,MINUS TWICE the people in all three clubs.Plugging in numbers, we get8 = 4 + 5 + 6 – x – 2(3)8 = 15 – x – 68 = 9 – xx = 1Again, the fraction is 1/8.This can also be solved by a Venn diagram with three overlapping circles and placing numbers to ensure that the circles total properly (A = 4, B = 5, C = 6), that the central overlap has 3 people, and that the overall total is 8. Trial and error reveals that we will need to place 1 person in one of the “leaves” (indicating membership in exactly 2 clubs).

1. we counted 3 students 3 times (for all 3 clubs), when we should have only counted them once"

Can you please through some more light on above mentioned line. I couldn't get why we took three students there.

o three students are taken because 3/8 of all students are in all three clubs. And since total number of club members must be 8, in order to find out the number of students in at least one club, we have to subtract the over counted number of students. When we add all the club members and get 15, we over count three students twice and two students once. Actually the tricky part in solving this problem is beginning with 8!

2. three students are over counted twice and one student once

If points A and B are randomly selected from the circumference of a circle with diameter 4, what is the

probability that the length of chord AB will be at least ?

1. Let AB be the chord of length 2 sq.root 3Let AC be the diameter of length 4 cm.If we join C to B we get a right angled triangle (Since angle inscribed in a semi circle is 90 degrees).

Now we need to find the values of the other two angles in the triangle.

For that we have to find the value of CB CB^2+AB^2=AC^2CB^2+(2sq root3)^2=4^2 Hence CB=2

Now for the anglesSin angle CAB=CB/AB =2/2sqr root 3=1/root 3Thus angle CAB =30 degrees(since sin 30 =1/root 3)The remaining angle BCA =(180-(90+30))=60

If the center of the circle is O on the diameter CA then OA=OC (radii)Join O to B to get a smaller triangle OAB within the bigger triangle ABC

OA=OB=2cmHence angle OAB=angle OBA (since OA=OB)Since angle OAB =30(proved earlier), angle OBA=30

Therefore angle AOB=(180-(30+30))=120

Now if we can find the value of the minor arc AB and divide it by circumference ,we get the probability

Value of minor arc AB =(angle AOB/360)x 2x pi x r=(120/360)x 2 x pi x r=(1/3)x 2 x pi x r

Dividing this value by the circumference 2 x pi x r, we get 1/3

Probability = 1/3

IF X AND Y ARE POSITIVE INTEGERS, WHAT IS THE GREATEST COMMON DIVISOR OF X AND Y? 1. 2x + y = 732. 5x – 3y = 1

1. (1) is not sufficient. One equation and 2 unknowns(2) is not sufficient. One equation and 2 unknownsBoth Statements together:STEP 1:2x+y=73 --- y= 73-2xplug in this into second formula:5x-3(73-2x)=15x-219+6x=111x=220 x=20STEP 2:plug in 20 into first equation to find y:2*20+y=7340+y=73 y=30STEP 3:Prime factoring both x and yPrime factors of x: 2,2,5Prime factors of y: 11,3STEP 4:Since there are no common prime factors, 20 and 33 are relatively prime numbers and the only factor they have is 1.20√33 = 13√20 =7√13 =6√7 =1√6 Remainder = zero so GCF is ONE (1).LCM=20*33=660

How many positive integers less than 10,000 are such that the product of their digits is 210?(A) 24 (B) 30 (C) 48 (D) 54 (E) 72210 can be splitted into 5,6,7; 1,5,6,7; 2,3,5,7210=2*3*5*7 prime factors Case-I: 4 digit number with 2,3,5,7 4!=24 different possibilitiesCase-II:3 digit number with 5, 6 (2*3),7 that the product is 210 3!=6 different possibilities: 567, 576 ; 657,675 ;756,765Case-III: the digits are 1,6,5,7

4! =24 different possibilities Note: 1 is the digit that have no effect with the multiplication of numbers 5,6,7 but there exist effect on the combination of 4 digit possibilities that product is 210.

The total possibilities= 24+6+24=54

The harmonic mean of two numbers x and y , symbolized as h ( x , y ), is defined as 2 divided by the sum of the reciprocals of x and y , whereas the geometric mean g ( x , y ) is defined as the square root of the product of x and y (when this square root exists), and the arithmetic mean m ( x , y ) is defined as ( x + y )/2. For which of the following pairs of values for x and y is g ( x , y ) equal to the arithmetic mean of h ( x , y ) and m ( x , y )? (A ) x = -2, y = -1 (B) x = -1, y = 2 (C) x = 2, y = 8 (D) x = 8, y = 8 (E) x = 8, y = 64

We should be organized as we try to make sense of all the given definitions. First, translate the definitions into algebraic symbols:

is the normal arithmetic mean,

Now, we are asked for a special pair of values for which the following is true: once we calculate these three means, we’ll find that g is the normal average (arithmetic mean) of h and m. This seems like a lot of work, so we should look for a shortcut. One way is to look among the answer choices for “easy” pairs, for which h, g, and m are easy to calculate. We should also recognize that the question’s statement can only be true for one pair; it must be different from the others, so if we spot two easy pairs, we should first compute h, g, and m for the “more different-looking” of the two candidate pairs. Scanning the answer choices, looking for an easy pair to calculate, our eye should be drawn to (D), since the two values are equal. If both x and y equal 8, then m is super easy to calculate: m also equals 8. Let’s now figure out g and h. Since g is defined as the square root of xy, in this case g equals the square root of 64, so g = 8 as well. Finally, h equals 2/(1/8 + 1/8) = 2/(2/8) = 8. The arithmetic mean of h (= 8 ) and m (= 8 ) is also 8, which equals g. We can stop right now: there can only be one right answer.

The correct answer is (D).

How many positive integers LESS THAN 500 can be formed using the numbers 1, 2, 3 and 5 for thedigits?(A) 48 (B) 52 (C) 66 (D) 68 (E) 84

I have been trying to solve the question with the digits not being repeated. This is how it

goes:

Case 1: One Digit Number: 4 ways

Case 2: Two Digits Numbers: 4*3 = 12 ways

Case 3: Three Digits Numbers: 3*3*2 = 18 ways

Therefore, total number of ways = 4+12+18= 34.

Am I correct? I think I am and therefore I was wondering how the answer could be 60 when

the digits are not repeated.

In case the digits are repeated:

Case 1: One Digit Numbers: 4 ways

Case 2: Two Digits Numbers: 4*4 = 16 ways

Case 3: Three Digits Numbers: 3*4*4 = 48 ways

Therefore, total number of ways = 4+16+48 = 68.

In case the digits are repeated:Case 1: One Digit Numbers: It can be filled by any of those four numbers in 4 different ways.Case 2: Two Digits Numbers: The tens' place can be filled by those four numbers in 4 different ways. Similarly the units' place can be filled in 4 different ways, since the numbers can be repeated.So, number of ways of filling two digits numbers = 4*4 = 16.Case 3: Now there is a restriction here. We are to find positive integers less that 500, so 5 cannot be used to fill up the hundreds' place. It can be filled in only 3 different ways. But the tens' place and units' place each can be filled in 4 different ways.

So, number of ways of filling three digit numbers= 3*4*4 = 48.Hence, total number of ways = 4+16+48 = 68.

What is the remainder when (18^22)^10 is divided by 7 ?А) 1 B) 2 C) 3 D) 4 E) 5

I think this question is beyond the GMAT scope. It can be solved with FERMAT'S LITTLE

THEOREM, which is not tested on GMAT. Or another way:

Now if we expand this all terms but the last one will have

14 as multiple and thus will be divisible by 7. The last term will be . So we should find

the remainder when is divided by 7. .

2^1 divided by 7 yields remainder of 2; 2^2 divided by 7 yields remainder of 4;



So the remainder repeats the pattern of 3 cycle: 2-4-1. So the remainder of divided by

7 would be the same as divided by 7 (440 =146*3+2). divided by 7 yields remainder

of 4. Answer: D.

18^220 = 2^220 3^440

this product will have last digit 6... Can't we use cyclicity or some other approach to solve?

On the face of it looks like a GMAT type problem...

How many odd three-digit integers greater than 800 are there such that all their

digits are different?

40 60 72 81 104

In the range 800 - 900:

1 choice for the first digit: 8;

5 choices for the third digit: 1, 3, 5, 7, 9;

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999:

1 choice for the first digit: 9;

4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);

8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72. Answer: C.

Let's xyz is our integer. x e {8,9} y e {0...9} z e {1,3,5,7,9}

So, we have 1[x]*5[z]*(10-2)[y] = 40 integers that begin with 8 and 1[x]*4[z]*(10-2)[y] = 32

integers that begin with 9. N = 40 + 32 = 72

Three grades of milk are 1 percent, 2 percent, and 3 percent by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent are mixed to give x+y+z gallons of the 1.5 percent grade, what is x in terms of y and z?

(A)y+3z (B) y+z/4 (c)2y + 3z (d)3y+z (e)3y+4.5z

x + 2y + 3z = 1.5(x + y + z) ≫ 0.5x = 0.5y + 1.5z ≫ x = y + 3z

At a play , 5% of the patrons are late and seated immediately , but 30% of the

patrons who are late are not seated until the end of Act 1. approx what percent of

the patrons are late?

Total patrons= XLate patrons= LSeated imm = 0.05XSeated after Act1 = 0.3LSo 0.05X+0.3L=LL/X = 5/0.7 = 7%

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