GMAT - Divisibility, Inequalities, Min Max Stats 2

D√

Set J consists of 15 different integers. If the median of Set J is 15 and the range is 60, what is the smallest possible value contained within Set J?A. -60B. -48C. -45D. -38E. -32

A**

Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?A. 78B. 77 1/5C. 66 1/7D. 55 1/7E. 52

B

Given distinct positive integers 1, 11, 3, x, 2, and 9, which of the following could be the median?A. 3B. 5C. 7D. 8E. 9 Show Ans

Given: • The average (arithmetic mean) of the positive integers x, y, and z is 3• Also, x < y < z To find: • The greatest possible value of z Approach and Working:As the average is 3, • Then sum of all 3 = 3 * 3 = 9 Also, x < y < z and each are positive integers.To maximise z, we should take minimum value of x and y • Minimum possible x = 1• Minimum possible y = 2• Therefore, maximum possible z = 9 - (1 + 2) = 6 Hence, the correct answer is option B.Answer: B√

The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?A. 5B. 6C. 7D. 8E. 9

C

Three boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?(A) 1(B) 2(C) 3(D) 4(E) 5

D

Three straight metal rods have an average (arithmetic mean) length of 77 inches and the shortestterm-48 rod has a length of 65 inches. What is the maximum possible value of the median length, in inches, of the three rods?A. 71B. 77C. 80D. 83E. 89

*E

Which of the answer choices properly lists the following in increasing order from left to right (all roots are positive):2(13)2(13)5(16)5(16)10(110)10(110)30(115)

Answer: C.

a) mb) 10m/7c) 10m/7 - 9/7d) 5m/7 + 3/7e) 5m

Notice that we are told that x is a positive integer.The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order.Since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}. Now, since x≤5x≤5 then the maximum sum of the numbers is 5+5+6+8=24, so the maximum average is 24/4=6 (options C, D, and E are out). 3 cannot be the average because in this case the sum of the numbers must be 4*3=12, which is less than the sum of just two numbers 6 and 8 (remember here that since x is a positive integer it cannot decrease the sum to 12). So we are left only with option B (for x=3 the average is (3+5+6+8)/4=22/4=5.5).Answer: B.

8, 5, x, 6 The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?A. 3B. 5.5C. 6.25D. 7E. 7.5

D

A certain city with population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have?A. 10,700B. 10,800C. 10,900D. 11,000E. 11,100

*(1) The product of the greatest and smallest of the integers in the list is positive.Two cases:A. all integers in the list are positive: in this case product of all integers would be positive;ORB. all integers in the list are negative: now, if there is even number of integers, then product of all integers would be positive BUT if there is odd number of integers, then product of all integers would be negative.Not sufficient.(2) There is an even number of integers in the list.Clearly insufficient. {-2, 2} - answer NO; {2,4} - answer YES.(1)+(2) Now if we have scenario A (from 1) then answer is YES. If we have scenario B, then as there are even number of integers (from 2) the product of all integers still would be positive, so answer is still YES. Sufficient.Answer: C.

A certain list consist of several different integers. Is the product of all integers in the list positive?(1) The product of the greatest and smallest of the integers in the list is positive.(2) There is an even number of integers in the list.

Responding to a pm:5 pound bag - $13.8510 pound bag - $20.43 (the cost would be about $27 with two 5 pound bags so the 10 pound bag is much cheaper in terms of $/kg)25 pound bag - $32.25 ($32 is much less than $40, the cost of two 10 pound bags so a 25 pound bag is much cheaper in terms of $/kg)65 <= Quantity to Buy <= 80It is best to buy as many 25 pound bags as possible to make up the quantity. So let's buy 2 25 pounds bags to make up 50 pounds. Now we are left with 15 tp 30 pounds to fill. We can do this is following ways:1. Buy another 25 pound bag for $32.252. Buy two 10 pound bags for $40.86 (Ignore since it is more expensive than case 1)3. Buy one 10 pound and one 5 pound bag for about $34 (Ignore since it is more expensive than case 1)Hence, we should simply buy three 25 pound bags for $32.25*3 = $96.75Answer (B)_____________

A garden center sells a certain grass seed in 5-pound bags at $13.85 per bag, 10-pound bags at $20.43 per bag, and 25-pound bags $32.25 per bag. If a customer is to buy at least 65 pounds of the grass seed, but no more than 80 pounds, what is the least possible cost of the grass seed that the customer will buy?A) $94.03B) $96.75C) $98.78D) $102.07E) $105.3

*D

A list contains n distinct integers. Are all n integers consecutive?(1) The average (arithmetic mean) of the list with the lowest number removed is 1 more than the average (arithmetic mean) of the list with the highest number removed.(2) The positive difference between any two numbers in the list is always less than n

D

A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?(A) 62(B) 68(C) 75(D) 88(E) 100

A.√ The key to this problem is not to get hung up on the idea of standard deviation. Remember that the GMAT will never ask you to calculate standard deviation, but will instead ask you general, conceptual questions. So instead of focusing on remembering the formula for standard deviation, instead work on building the set described. You are told that Set A is made up of all 2-digit primes that are one less than a power of 2. So a good first step would be to write out all powers of 2 less than 100 since there are far fewer of those than there are 2-digit primes: {2, 4, 8, 16, 32, 64} Then eliminate the single-digit numbers to get: {16, 32, 64} And then subtract 1 from each: {15, 31, 63} Notice that 15 and 63 are both multiples of 3, and so they are not prime. This means that Set A contains only the number 31. Remember that if all items in a set are the same (even if there is only one term in the set), then the standard deviation must equal 0, answer choice (A).

If Set A is made up of all 2-digit primes that are one less than a power of 2, what is the standard deviation of set A? 0 15.1 17 20 23.7

*Let the set of 9 distinct integers in increasing order bea1a1 , a2a2 ... a9a9.From F.S 1, we know that a1∗a2∗....a4∗a5....a9a1∗a2∗....a4∗a5....a9 = a5a5 →→ a5(a1∗a2∗...a4....a9−1)a5(a1∗a2∗...a4....a9−1) = 0.Thus, either a5a5 = 0 OR (a1∗a2∗...∗a5....a9−1)(a1∗a2∗...∗a5....a9−1) = 0, the latter is not possible as no product of 8 distinct integers can ever equal 1.Thus, the median,a5a5 = 0 and not positive. Sufficient.From F. S 2, for -4,-3,-2,-1,0,1,2,3,4,the median is 0 and a NO for the question stem. Again, for the series -10,-3,-1,0,1,2,3,4,5, the median is 1, which is positive, and hence a YES for the question stem. Thus, Insufficient.A.

If list S contains nine distinct integers, at least one of which is negative, is the median of the integers in list S positive?(1) The product of the nine integers in list S is equal to the median of list S.(2) The sum of all nine integers in list S is equal to the median of list S.

If m, n, and p are integers, is m + n odd?(1) m = p^2 + 4p + 4. Not sufficient.(2) n = p^2 + 2m + 1. Not sufficient.(1)+(2) Add (1) and (2) m+n=p2+4p+4+p2+2m+1=2p2+4p+2m+5=2(p2+2p+m)+5=even+odd=oddm+n=p2+4p+4+p2+2m+1=2p2+4p+2m+5=2(p2+2p+m)+5=even+odd=odd. Sufficient.Answer: C.

If m, n, and p are integers, is m + n odd?(1) m = p^2 + 4p + 4(2) n = p^2 + 2m + 1

B√

If the average (arithmetic mean) of five distinct positive integers is 10, what is the least possible value of the greatest of the five numbers?(A) 11(B) 12(C) 24(D) 40(E) 46

D

If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?A. 50B. 52C. 49D. 48E. 44

E. √Many look at this question and assume that two numbers are less than 35 and two are greater than 35, but this is a good opportunity to "Play Devil's Advocate". Even if statements 1 and 2 are together, you have the possibility for these two different sets: {33, 33, 37, 37} --> two numbers below 35...this is the "obvious" set that the test wants you to see. {33, 33, 33, 41} --> three numbers below 35 but the average is still 35, so this proves that you can get two different answers On Data Sufficiency questions when you're picking numbers, it pays to try to get two different answers. The test often baits you by showing you one "obvious" scenario that leads to one answer, but astute test-takers are always looking to find the less-obvious situation that leads to another answer.

If the average of four numbers is 35, how many of the numbers are less than 35? (1) None of the numbers are exactly 35 (2) Two of the numbers are exactly 33

B. OA is b but your reasoning is not correct,First the question clearly says that x is positive and hence there is no question for x being -ve or + veSecondly the reason St 1 is no sufficient is that St says the unti digit ofx4x4 is 1, now if number x has 3 its unit digit then x4x4 will have unit digit as 1(34=81)1(34=81) and also when x has unit digit has 7 then also the x4x4 will have unit digit as 1at least two options hence not sufficient

If x is a positive integer, what is the units digit of x^2?(1) The units digit of x^4 is 1.(2) The units digit of x is 3.

E*We have a set: {1, 3, 8, 12, x} Question: is median>mean=x+1+3+8+125=x+245median>mean=x+1+3+8+125=x+245? Note that as we have odd (5) # of terms in the set then the median will be the middle term when arranged in ascending (or descending) order. So:if x≤3x≤3: {1, x, 3, 8, 12} then median=3median=3;if 3<x≤83<x≤8: {1, 3, x, 8, 12} then median=xmedian=x;if x≥8x≥8: {1, 3, 8, x, 12} then median=8median=8.(1) x>6x>6. If x=7x=7 then the median will be 7 as well: {1, 3, 7, 8, 12} and mean will be mean=7+245=6.2mean=7+245=6.2, so median=7>mean=6.2median=7>mean=6.2 and the answer is YES BUT if xx is very large number then the median will be 8: {1, 3, 8, 12, x=very large number} and mean will be more than median (for example if x=26x=26 then mean=26+245=10mean=26+245=10, so median=8<10=meanmedian=8<10=mean) and the answer will be NO. Not sufficient.(2) x is greater than the median of the 5 numbers --> so median=8median=8: now, if x=11x=11 then mean=11+245=7mean=11+245=7, so median=8>7=meanmedian=8>7=mean and the answer is YES. Again it's easy to get answer NO with very large xx. Not sufficient.(1)+(2) Again, x=11 and x=very large number give two diffrent answers to the question. Not sufficeint.Answer: E.

If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean) of the 5 numbers?(1) x > 6(2) x is greater than the median of the 5 numbers. x, 3, 1, 12, 8 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

A. Statement 1: On the number line, 0 is closer to x - 1 than to x.If zero is closer to x-1 than to x, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.If case #1 is true, we can see that x must be positiveIf case #3 is true, we can see that x must be positiveSince both possible cases yield the same answer to the target question, we can answer the target question with certainty.So, statement 1 is SUFFICIENTStatement 2: On the number line, 0 is closer to x than to x + 1.Recognize that x+1 will always be to the right of x.Also recognize that there are 3 possible ways to place x and x+1 with relation to zero.

Is the number x positive?(1) On the number line, 0 is closer to x - 1 than to x.(2) On the number line, 0 is closer to x than to x + 1.

A√

Last month 15 homes were sold in Town X. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?I. At least one of the homes was sold for more than $165,000.II. At least one of the homes was sold for more than $130,0000 and less than $150,000III. At least one of the homes was sold for less than $130,000.A. I onlyB. II onlyC. III onlyD. I and IIE. I and III

*Official Solution:List A consists of five distinct integers. Is the range of the numbers in list A greater than 8?(1) List A does not contain a multiple of 5. Clearly insufficient(2) No two numbers in list A are consecutive.In this case the minimum range would be if we consider a list with any five consecutive odd (or even) integers. So, if A is say {1, 3, 5, 7, 9} or {2, 4, 6, 8, 10} then the range would be 8, so less than 8. Of course, we can get a list with the range greater than 8, so this statement is also insufficient.(1)+(2) Without a multiple of 5, the minimum range would be if A is say {1, 3, 7, 9, 11} or {2, 4, 6, 8, 12} (so skipping a multiple of 5). So, the minimum range of the list is 10. Sufficient.Answer: C

List A consists of five distinct integers. Is the range of the numbers in list A greater than 8?(1) List A does not contain a multiple of 5.(2) No two numbers in list A are consecutive.

*For any evenly spaced set median = mean = the average of the first and the last terms.So the mean of S will be the average of the first and the last terms: mean = (x + x + 9*2)/2 = x+9, where x is the first term;The mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3;The difference will be (x + 9) - (x - 3) = 12.Answer: D.

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?(A) 2(B) 7(C) 8(D) 12(E) 22

C√

The average (arithmetic mean) of five positive integers a, b, c, d, e is 22 and a < b < c < d < e. If e is 40, what is the least possible value of the median of the five integers?A. 10B. 11C. 12D. 13E. 14

E*√

A set of 5 numbers has an average of 50. The largest element in the set is 5 greater than 3 times the smallest element in the set. If the median of the set equals the mean, what is the largest possible value in the set?(A) 85(B) 86(C) 88(D) 91(E) 92

D

A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?A. 33B. 35C. 38D. 48E. 75

Sum of the 30 integers = 150.Minimum possible sum of the 10 integers, each of which exceeds 5 = 10*6 = 60So, maximum possible sum of the remaining 20 integers, each of which doesn't exceed 5 = 150 - 60 = 90. Thus, the corresponding average = 90/20 = 4.5.FINAL ANSWER IS (D)

The average (arithmetic mean) of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average (arithmetic mean) of these 20 integers?A. 3B. 3.5C. 4D. 4.5E. 5

B

The average of four integers is 8. If the greatest of the four integers is 16, what is the minimum possible value of the least of the four integersA. -32B. -16C. -13D. 0E. 1

C

The average score in an examination of 10 students of a class is 60. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 40 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper isA) 87B) 95C) 99D) 100E) 103

E**The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?Given: sum(S)=sum(T)sum(S)=sum(T). Question: is t<st<s, where ss and tt are # of integers in lists S and T respectively.(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T --> sums<sumtsums<sumt --> cross multiply: sum∗t<sum∗ssum∗t<sum∗s. Now, if sum<0sum<0 then t>st>s (when reducing by negative flip the sign) but if sum>0sum>0 then t<st<s. not sufficient.(2) The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.(1)+(2):If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.Not sufficient.Answer: E.

The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T ?(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.(2) The median of the integers in S is greater than the median of the integers in T. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

Total weight of 7 packages = 225 * 7 = 1575 pounds.In a set of 7 numbers, 4th element will be the middle one and hence median. weight of 4th element =270 pounds. to maximise weight of lightest package, minimize the weight of others. 4th number is 270, so let the 5th 6th and 7th number of the set also be 270 pounds which is least possible weight they can have. total weight of 4th, 5th , 6th and 7th numbers = 270 *4 = 1080 pounds.so total weight remaining to be shared by 1st 2nd and 3rd number = 1575 - 1080= 495 pounds. let the 1st 2nd and 3rd element have the same value. so they can be 495/3 = 165 pounds each. so lightest package is 165 pounds in weight. B√

At a delivery store, seven packages have an average (arithmetic mean) weight of 225 pounds and a median weight of 270 pounds. What is the maximum possible weight, in pounds, of the lightest package?A. 25B. 165C. 195D. 225E. 270

B

Five pieces of wood have an average (arithmetic mean) length of 124 centimeters and a median length of 140 centimeters. What is the maximum possible length, in centimeters, of the shortest piece of wood?A. 90B. 100C. 110D. 130E. 140

Since the product of these values is 0, then the average must be 0. In order to then maximize the number of positive values, there are 24 values left. If all 24 were positive, then the average could not be 0, as the sum of the values needs to be 0. But if there were one negative value that was equal to -1 times the sum of the other 23 nonzero values, then the sum of values would be 0 and a total of 23 values could be positive. Therefore, the answer is D.

The average of 25 values is equal to product of these values. If only one of the numbers is 0, at most how many numbers can be greater than zero?A. 0B. 12C. 13D. 23E. 24

*A certain list consists of five different integers. Is the average (arithmetic mean) of the two greatest integers in the list greater than 70 ?Let's say these 5 unique integers are A, B, C, D, E (from min to max)So now we need to find if (D+E)/2 > 70 ---> D+E > 140(1) The median of the integers in the list is 70.Tell us that C = 70. To test that D+E > 140, I plug in the lowest possible values which are 71 and 72 (since they must be unique integers)71 + 72 > 140 ---> sufficient(2) The average of the integers in the list is 70.which means (A+B+C+D+E)/5 = 70 ---> A+B+C+D+E = 350To test whether D+E>140, we need to maximize the values A, B, and CIn this case the greatest possible values of A, B, and C are 68, 69, and 70 given that D and E must be greatest 2 integers.(Let's say if A, B, C are 69, 70, 71 ---> D+E = 350-69-70-71 = 140 which is not possible that D and E are greatest 2 integers)So we got the minimum values of D and E are 71 and 72. ---> Sufficient

A certain list consists of five different integers. Is the average (arithmetic mean) of the two greatest integers in the list greater than 70 ?(1) The median of the integers in the list is 70.(2) The average of the integers in the list is 70.

E*Six countries in a certain region sent a total of 75 representatives to an international congress, and no two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?Given: x1<x2<x3<x4<A<x6x1<x2<x3<x4<A<x6 and x1+x2+x3+x4+A+x6=75x1+x2+x3+x4+A+x6=75. Q: is A≥10A≥10(1) One of the six countries sent 41 representatives to the congress --> obviously x6=41x6=41 --> x1+x2+x3+x4+A=34x1+x2+x3+x4+A=34.Can A≥10A≥10? Yes. For example: x1=2x1=2, x2=3x2=3, x3=8x3=8, x4=10x4=10, A=11A=11 --> sum=34sum=34 (answer to the question YES);Can A<10A<10? Yes. For example: x1=4x1=4, x2=6x2=6, x3=7x3=7, x4=8x4=8, A=9A=9 --> sum=34sum=34 (answer to the question NO).(2) Country A sent fewer than 12 representatives to the congress --> A<12A<12.The same breakdown works here as well:Can 12>A≥1012>A≥10? Yes. For example: x1=2x1=2, x2=3x2=3, x3=8x3=8, x4=10x4=10, A=11A=11, x6=41x6=41 --> sum=75sum=75 (answer to the question YES);Can A<10A<10? Yes. For example: x1=4x1=4, x2=6x2=6, x3=7x3=7, x4=8x4=8, A=9A=9, x6=41x6=41 --> sum=75sum=75 (answer to the question NO).(1)+(2) The given examples fit in both statements and A in one is more than 10 and in another less than 10. Not sufficient.Answer: E.

Six countries in a certain region sent a total of 75 representatives to an international congress, and no two countries sent the same number of representatives. Of the six countries, if Country A sent the second greatest number of representatives, did Country A send at least 10 representatives?(1) One of the six countries sent 41 representatives to the congress.(2) Country A sent fewer than 12 representatives to the congress. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

A√ The relevant formula is Average = (SumofTerms)(NumberofTerms)(SumofTerms)(NumberofTerms) . Another way to look at the formula is Sum of Terms = Average * Number of Terms. For this set, the sum is 5 * 300, or 1500. To make the median as large as possible, set the three greatest numbers in the set to be 1997, 1998, and 1999, since the values are distinct. Thus the median would be 1997. Therefore the sum of the other 2 numbers is 1,500 - 5,994, which is -4,494. The correct answer is A.

The average of a set of five distinct integers is 300. If each number is less than 2,000, and the median of the set is the greatest possible value, what is the sum of the two smallest numbers? -4,494 -3,997 -3,494 -3,194 The answer cannot be determined from the information given