GMAT - Divisibility, Inequalities, Min Max Stats 5

√We can solve this problem using Straightforward Math and our knowledge of the rules of exponents which tells us that (xy)n = xnyn. First we know that (50)50 = (2 × 25)50 and using the rules of exponents we know that (2 × 25)50 = 250× 2550. So (50)50 - (25)50 = 250× 2550 - (25)50 = 2550 (250 - 1) or Answer Choice (E).

(50)^50 - (25)^50 =?A. (25)^50B. (2^50) - 1C. (25^50) - 1D. 2^50 (25^50 - 1)E. 25^50 (2^50 - 1)

√*Given list consists of 10 evenly spaced integers. Mean=(First+Last)/2=13 and Sum=(Mean)*(# of terms)=130.Given that list M is obtained by removing 2 integers from the list shown.To determine the standard deviation of list M we must know which 2 integers were removed.(1) The average (arithmetic mean) of the numbers in list M is equal to the average of the numbers in the list shown --> the mean of list M is also 13. Thus the sum of the integers in list M is 13*8=104, which means that the sum of the 2 integers removed is 130-104=26. The 2 integers removed could be: (4, 22), (6, 20), ..., (12, 14). Not sufficient.(2) List M does not contain 22. We know only one of the numbers removed. Not sufficient.(1)+(2) From (1) we know that the the sum of the 2 integers removed is 26 and from (2) we know that one of the integers removed is 22. Therefore the second integer removed is 26-22=4. List M consists of the following 8 integers: {6, 8, 10, 12, 14, 16, 18, 20}. So, we can determine its standard deviation. Sufficient.Answer: C.

4, 6, 8, 10, 12, 14, 16, 18, 20, 22List M (not shown) consists of 8 different integers, each of which is in the list shown. What is the standard deviation of the numbers in list M ?(1) The average (arithmetic mean) of the numbers in list M is equal to the average of the numbers in the list shown.(2) List M does not contain 22.

*Solution:Question Stem Analysis:We need to determine the median of a list consisting of 6 numbers, including one unknown value.Statement One Alone:Since we know x > 7, arranging the numbers from smallest to largest, we have either:4, 5, 6, 7, x, 9 or 4, 5, 6, 7, 9, xWe see that either way the median will be (6 + 7)/2 = 6.5. Statement alone is sufficient.Statement Two Alone:We can't determine a unique value for the median, given the information in statement two. For example, if x = 8, then the median and mean are both 6.5. However, if x = 2, then the median and mean are both 5.5. Statement two alone is not sufficient.Answer: A

7, 9, 6, 4, 5, xIf x is a number in the list above, what is the median of the list?(1) x > 7(2) The median of the list equals the arithmetic mean of the list.

√A certain bookcase has 2 shelves of books. On the upper shelf, the book with the greatest number of pages has 400 pages. On the lower shelf, the book with the least number of pages has 475 pages. What is the median number of pages for all of the books on the 2 shelves?Notice that we are told that all books on the upper shelf have fewer pages than all books on the lower shelf.(1) There are 25 books on the upper shelf. No info about the lower shelf. Not sufficient.(2) There are 24 books on the lower shelf. No info about the upper shelf. Not sufficient.(1)+(2) There are total of 25+24=49 books, thus the median number of pages for all of the books will be the book with 25th largest number of pages. Now, since on the upper shelf which has 25 books, the book with the greatest number of pages has 400 pages, then median=400 pages (recall that all books from the lower shelf have more pages than all books on the upper shelf, so if we order all books in ascending number of pages, then we'll have 25 books from the upper shelf and then 24 books from the lower shelf). Sufficient.Answer: C.

A certain bookcase has 2 shelves of books. On the upper shelf, the book with the greatest number of pages has 400 pages. On the lower shelf, the book with the least number of pages has 475 pages. What is the median number of pages for all of the books on the 2 shelves?(1) There are 25 books on the upper shelf.(2) There are 24 books on the lower shelf.

√Given:1. A certain company consists of three divisions, A, B, and C.2. Of the employees in the three divisions, the employees in Division C have the greatest average (arithmetic mean) annual salary.Asked: Is the average annual salary of the employees in the three divisions combined less than $55,000 ?(1) The average annual salary of the employees in Divisions A and B combined is $45,000.Since the average annual salary of the employees in Divisions C is unknownNOT SUFFICIENT(2) The average annual salary of the employees in Division C is $55,000.Since, of the employees in the three divisions, the employees in Division C have the greatest average (arithmetic mean) annual salary.The average annual salary of the employees in Divisions A and B combined is less than $55,000.The average annual salary of the employees in the three divisions combined is less than $55,000.SUFFICIENTIMO B____________

A certain company consists of three divisions, A, B, and C. Of the employees in the three divisions, the employees in Division C have the greatest average (arithmetic mean) annual salary. Is the average annual salary of the employees in the three divisions combined less than $55,000 ?(1) The average annual salary of the employees in Divisions A and B combined is $45,000.(2) The average annual salary of the employees in Division C is $55,000.

*(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.

*A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?Say the three numbers are x, y, and x, where x < y < z. The median would be y and the average would be (x + y + z)/3. So, the question asks whether y = (x + y + z)/3, or whether 2y = x + z.(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median --> the range is the difference between the largest and the smallest numbers of the set, so in our case z - x. We are given that z - x = 2(z - y) --> 2y = x + z. Sufficient.(2) The sum of the 3 numbers is equal to 3 times one of the numbers --> the sum cannot be 3 times smallest numbers or 3 times largest number, thus x + y + z = 3y --> x + z = 2y. Sufficient.Answer: D.

A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median.(2) The sum of the 3 numbers is equal to 3 times one of the numbers.

√400 different numbers... Median = average of 200th and 201st....(1) Of the numbers in the list, 280 are less than the average.here 280 are less, so ofcourse 200th and 201st will also be less than average...ans YESSuff(2) Of the numbers in the list, 30 percent are greater than or equal to the average30% of 400 = 120 so ONLY 120 are greater than average, same as aboveSuffD

A certain list consists of 400 different numbers. Is the average (arithmetic mean) of the numbers in the list greater than the median of the numbers in the list?(1) Of the numbers in the list, 280 are less than the average.(2) Of the numbers in the list, 30 percent are greater than or equal to the average.

*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.

√An integer grater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?Given: n>20n>20 --> two digit integer can be written as follows: n=10b+a>20n=10b+a>20 --> 2≤b≤92≤b≤9, 0≤a≤90≤a≤9.(1) The tens digit of n is a factor of the units digit of n. This statement implies that a=kba=kb, (0≤k≤40≤k≤4) --> n=10b+a=10b+kb=b(10+k)n=10b+a=10b+kb=b(10+k) --> as b≥2b≥2, nn will always be composite and factor of bb. Sufficient(2) The tens digit of n is 2. This statement implies that b=2b=2, but nn can be for instance composite 25 or prime 29. Not sufficient.Answer: A.

An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?(1) The tens digit of n is a factor of the units digit of n.(2) The tens digit of n is 2

√A certain list, L, contains a total of n numbers, not necessarily distinct, that are arranged in increasing order. If L1L1 is the list consisting of the first n1n1 numbers in L and L2L2 is the list consisting of the last n2n2 numbers in L, is 17 a mode for L ?Say the list is a, b, c, c, d, e, f, h, h, h, k, l..L1L1 could be just a, and L2L2 could be h,h,k,l ORL1L1 could be a,b,c,c,d and L2L2 could be c,c,d..h,h,k,l(1) 17 is a mode for L1L1 and 17 is a mode for L2L2.Since the list is in increasing order and mode is 17 in both cases, the list L1 and L2 either have all numbers within themselves OR only some other 17 are left.For example 1,3,5,17,17,17,17,17,20,25....L1 could be 1,3,5,17,17 and L2 could be 3,5,17,17,17,17,17,20,25OR L1 could be 1,3,5,17,17 and L2 could be 17,17,20,25So the mode of L will be 17 in each case.Suff(2) n1+n2=nn1+n2=nWe do not know the numbers.InsuffA

A certain list, L, contains a total of n numbers, not necessarily distinct, that are arranged in increasing order. If L1L1 is the list consisting of the first n1n1 numbers in L and L2L2 is the list consisting of the last n2n2 numbers in L, is 17 a mode for L ?(1) 17 is a mode for L1L1 and 17 is a mode for L2L2.(2) n1+n2=n

*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

*A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?Basically the question asks whether nn (# of students) is a multiple of mm (# of classrooms), or whether nm=integernm=integer, because if it is then we would be able to assign students to classrooms so that each classroom has the same number of students assigned to it.Given: 3<m<13<n3<m<13<n.(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it --> 3nm=integer3nm=integer, from this we cannot say whether nm=integernm=integer. For example nn indeed might be a multiple of mm (n=14n=14 and m=7m=7) but also it as well might not be (n=14n=14 and m=6m=6). Not sufficient.(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it --> 13nm=integer13nm=integer, now as given that 3<m<133<m<13 then 13 (prime number) is not a multiple of mm, so 13nm13nm to be an integer the nn must be multiple of mm. Sufficient.Answer: B.

A school administrator will assign each student in a group of N students to one of M classrooms. If 3<M<13<N, is it possible to assign each of the N students to one of the M classrooms so that each classroom has the same number of students assigned to it?(1) It is possible to assign each of 3N students to one of M classrooms so that each classroom has the same number of students assigned to it.(2) It is possible to assign each of 13N students to one of M classrooms so that each classroom has the same number of students assigned to it

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Absolute Values

√Let's go through the numbers 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 and count how many meet our criteria. Approach Strategically: The integers that will have a remainder of two or less when divided by 4 implies integers with remainders of 2 as well as 1 and 0. These integers are multiples of 4 (remainder of 0), and integers that are 1 and 2 greater than multiples of 4. In the given range there are 8 such numbers: 16, 17, 18, 20, 21, 22, 24 and 25 so the correct answer is Answer Choice (E).

How many of the integers between 15 and 25, inclusive, have a remainders of 2 or less when divided by 4?45678

√*At a certain company, a test was given to a group of men and women seeking for promotions. If the average (artihmetic mean) score for the group was 80, was the average score for the women more than 85?(1) The average score for the men was less than 75. Since the average score for the group was 80 then the average score for women must be more than 80, but we don't know whether it's more than 85. Not sufficient.(2) The group consisted of more men than women. Clearly insufficient.(1)+(2) From (1) we have that the distance between the average score for the men and the average score for the whole group is more than 5 and from (2) we have that there are more men than women. Now, in order to compensate that difference and to make the average for the whole group 80 the average of the smaller group (women) must be further from 80 than the average for the larger group (men), so the average score for the women must be more than 85. Sufficient.Answer: C.

At a certain company, a test was given to a group of men and women seeking for promotions. If the average (arithmetic mean) score for the group was 80, was the average score for the women more than 85?(1) The average score for the men was less than 75.(2) The group consisted of more men than women

*To find the range we should know:A. Balance before he started depositing, initial balance - we know that there was initial balance because for may balance was 2600 and maximum amount he could deposited for this period (from January till May) is: 5 months*120=600;B. In which month Carl stopped depositing $120 and started withdrawing $50.We have:APRIL___MAY__JUNE---?----$2,600----?---(1) April balance < 2625 --> he deposited in May;Because if he didn't then April balance would have been $2,600+50=$,2650 and we know that in April balance was<2625:APRIL____MAY__JUNE2,480----$2,600----?---Notice that we can find the initial balance based on this info: $2,600=x(initial balance)+5months*120 --> x+600=2600 --> x=2000.Though this statement is still insufficient as we still don't know in which month Carl stopped depositing and started withdrawing.(2) June balance < 2675 --> he didn't deposited in June --> he withdrew in June.Because if he deposited, then in June deposit would have been May balance +$120: $2,600+$120=$2,720>$2,675.APRIL___MAY____JUNE---?----$2,600---$2,550---But again this statement is still insufficient as we still don't know when he started withdrawing, all we know it was not after June.(1)+(2) We know that initial balance was $2,000 and that Carl deposited in May and started withdrawing in June:APRIL_____MAY____JUNE$2,480----$2,600---$2,550---Thus we can find the range. Sufficient.Answer: C.________________

Beginning in January of last year, Carl made deposits of $120 into his account on the 15th of each month for several consecutive months and then made withdrawals of $50 from the account on the 15th of each of the remaining months of last year. There were no other transactions in the account last year. If the closing balance of Carl's account for May of last year was $2,600, what was the range of the monthly closing balances of Carl's account last year?(1) Last year the closing balance of Carl's account for April was less than $2,625.(2) Last year the closing balance of Carl's account for June was less than $2,675.

*Does the integer k have a factor p such that 1<p<k?Question basically asks whether kk is a prime number. If it is, then it won't have a factor pp such that 1<p<k1<p<k (definition of a prime number).(1) k>4!k>4! --> kk is more than some number (4!=244!=24). kk may or may not be a prime. Not sufficient.(2) 13!+2≤k≤13!+1313!+2≤k≤13!+13 --> kk can not be a prime. For instance if k=13!+8=8∗(2∗3∗4∗5∗6∗7∗9∗10∗11∗12∗13+1)k=13!+8=8∗(2∗3∗4∗5∗6∗7∗9∗10∗11∗12∗13+1), then kk is a multiple of 8, so not a prime. Same for all other numbers in this range. So, k=13!+xk=13!+x, where 2≤x≤132≤x≤13 will definitely be a multiple of xx (as we would be able to factor out xx out of 13!+x13!+x, the same way as we did for 8). Sufficient.Answer: B.

Does the integer k have a factor p such that 1<p<k?(1) k > 4!(2) 13!+2≤k≤13!+13

√For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2. This implies that the set is even;y spaced. In any evenly spaced set the mean (average) is equal to the median. Sufficient.(2) The range of the n numbers in the set is 2(n- 1). This is completely useless. Not sufficient.Answer: A.P.S. The range is the difference between the smallest and largest elements of the set.

For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.(2) The range of the n numbers in the set is 2(n- 1).

√In simple words, 2-height is just the number of 2s in a positive integer x.So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5If x = 15, its 2-height is 0 because there are no 2s in 15.and so on.So, an even number will have a 2-height of at least 1.An odd number will have a 2-height of 0."is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"(1) k > mk could have more 2s than m or it could have fewer 2s than m.For example, if k = 4 and m = 3, k has two 2s while m has none.If k = 11 and m = 8, k has no 2s while m has 3.Not sufficient.(2) m/k is an even integer.When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.Answer (B)

For any positive integer x, the 2-height of x is defined to be the greatest nonnegative integer n such that 2^n is a factor of x. If k and m are positive integers, is the 2-height of k greater than the 2-height of m?(1) k > m(2) k/m is an even integer.

√ak=(1+1k+1)ak=(1+1k+1)ak=k+2k+1)ak=k+2k+1)a1a2...ana1a2...an= 32∗43∗54∗.........∗n+2n+132∗43∗54∗.........∗n+2n+1a1a2...ana1a2...an=n+22n+22a1a2...ana1a2...an= n2+1n2+1n2+1n2+1 is an integer, if n2n2 is an integerSo basically question stem is whether n is a multiple of 2 or not.Statement 1- n+1 is a multiple of 3if n+1 is 3, n is 2 (even)if n+1 is 6, n is 5 (odd)InsufficientStatement 2- n is multiple of 2 or evenSufficient

For each positive integer k, let ak=(1+1k+1)ak=(1+1k+1). Is the product a1a2...ana1a2...an an integer?(1) n + 1 is a multiple of 3.(2) n is a multiple of 2.

√"x rounded to the nearest hundred is 500" means 450≤x<550450≤x<550 (inequality you've written) BUT it IS NOT the same as "500 is the multiple of 100 that is closest to x", which should be 450<x<550450<x<550 as 450 is equidistant from 400 and 500 and we cannot say that 450 rounded to nearest multiple of 100 is 500, it's 500 OR 400. That's why endpoints must be excluded.If 500 is the multiple of 100 that is closest to X and 400 is the multiple of 100 closest to Y, then which multiple of 100 closest to X + Y ?"500 is the multiple of 100 closest to X" --> 450<x<550450<x<550;"400 is the multiple of 100 closest to Y" --> 350<y<450350<y<450.(1) x<500 --> 450<x<500450<x<500 --> add this inequality to inequality with y --> 800<x+y<950800<x+y<950. If x+y=810 then closest multiple of 100 is 800 BUT if x+y=860 then closest multiple of 100 is 900. Not sufficient.(2) y<400 --> 350<y<400350<y<400 --> add this inequality to inequality with x --> 800<x+y<950800<x+y<950. The same here: if x+y=810 then closest multiple of 100 is 800 BUT if x+y=860 then closest multiple of 100 is 900. Not sufficient.(1)+(2) Sum 450<x<500450<x<500 and 350<x<400350<x<400 --> 800<x+y<900800<x+y<900 --> and again if x+y=810 then closest multiple of 100 is 800 BUT if x+y=860 then closest multiple of 100 is 900. Not sufficient.Answer: E.

If 500 is the multiple of 100 that is closest to x and 400 is the multiple of 100 that is closest to y, which multiple of 100 is closest to x + y ?(1) x < 500(2) y < 400

√If S is a set of four numbers w, x, y, and z, is the range of the numbers in S greater than 2 ?(1) w - z > 2. The range of a set is the difference between the largest and the smallest elements of the set, since the difference of some particular two numbers are already more than 2 then the the range must also be more than 2. Sufficient.(2) z is the least number in S --> just says that from four unknowns z is the smallest one (obviously one of the unknowns would be the smallest one, what difference does it make to know that it's z?). Not sufficient.Answer: A.

If S is a set of four numbers w, x, y, and z, is the range of the numbers in S greater than 2 ?(1) w - z > 2(2) z is the least number in S.

√No, B is not correct. It's straight E: if p=3 then every term in S is divisible by p but if p=11 then some terms in S are divisible by p (for example 99 and 9999) and some are not (for example 9 and 999). Not sufficient.Answer: E.

If S is the infinite sequence: S1=9, S2=99, S3=999, ..., SK = 10^K-1, ..., is every term in S divisible by the prime number p?(1) p is greater than 2.(2) At least one term in sequence S is divisible by p.

√We need to first list the factors of 64: 1, 2, 4, 8, 16, 32, 64. Which of the factors are multiples of 8? The only multiples of 8 in this set are 8, 16, 32 and 64, so there are 4 elements in set S or Answer Choice (C).

If S is the set of integers that are both positive factors of 64 and also multiples of 8, how many elements are there in set S? 23468

*If k is an integer greater than 1, is k equal to 2^r for some positive integer r?Given: k=integer>1k=integer>1, question is k=2rk=2r.Basically we are asked to determine whether kk has only 2 as prime factor in its prime factorization.(1) k is divisible by 2^6 --> 26∗p=k26∗p=k, if pp is a power of 2 then the answer is YES and if pp is the integer other than 2 in any power (eg 3, 5, 12...) then the answer is NO.(2) k is not divisible by any odd integers greater then 1. Hence kk has only power of 2 in its prime factorization. Sufficient.Answer: B.

If k is an integer greater than 1, is k equal to 2^r for some positive integer r?(1) k is divisible by 2^6.(2) k is not divisible by any odd integer greater than 1.

*(1) 10^d is a factor of f --> k∗10d=30!k∗10d=30!.First we should find out how many zeros 30!30! has, it's called trailing zeros. It can be determined by the power of 55 in the number 30!30! --> 305+3025=6+1=7305+3025=6+1=7 --> 30!30! has 77 zeros.k∗10d=n∗107k∗10d=n∗107, (where nn is the product of other multiples of 30!) --> it tells us only that max possible value of dd is 77. Not sufficient.Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that 10d10d is factor of this number, but 10d10d can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically dd can be any integer from 1 to 7, inclusive (if d>7d>7 then 10d10d won't be a factor of 30! as 30! has only 7 zeros in the end). So we cannot determine single numerical value of dd from this statement. Hence this statement is not sufficient.(2) d>6 Not Sufficient.(1)+(2) From (2) d>6d>6 and from (1) dmax=7dmax=7 --> d=7d=7.Answer: C.

If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?(1) 10^d is a factor of f(2) d > 6

*Note that k, m, r, s, and t are positive integers, thus neither of them can be zero.Given: 0<k<m<r<s<t=400<k<m<r<s<t=40 and average=k+m+r+s+405=16average=k+m+r+s+405=16. Question: medianmax=?medianmax=? As median of 5 (odd) numbers is the middle number (when arranged in ascending or descending order) then the question basically asks to find the value of rmaxrmax.average=k+m+r+s+405=16average=k+m+r+s+405=16 --> k+m+r+s+40=16∗5=80k+m+r+s+40=16∗5=80 --> k+m+r+s=40k+m+r+s=40. Now, we want to maximize rr:General rule for such kind of problems:to maximize one quantity, minimize the others;to minimize one quantity, maximize the others.So to maximize rr we should minimize kk and mm, as kk and mm must be distinct positive integers then the least values for them are 1 and 2 respectively --> 1+2+r+s=401+2+r+s=40 --> r+s=37r+s=37 --> rmax=18rmax=18 and s=19s=19 (as rr and ss also must be distinct positive integers and r<sr<s). So, rmax=18rmax=18Answer: B.

If k, l, m, and n are non-prime integers, is the average (arithmetic mean) of k, l, and m greater than the average of k, l, m, and n? (1) 10 > k > l > m > n > 0 (2) l = 2n and k = m

*I get A.You can set this up by adding the fractions with a common denominator. You get(2k/12) + (3m/12) = t/12so now we know that2k + 3m = tAnd we want to know if t and 12 share a factor other than 1.1 says that k is a multiple of 3. There is a rule that when you add together numbers that are all a multiple of the same number, the sum is also a multiple of that number. So with statement 1 we know that 2k is a multiple of 3, and it's given that 3m is also a multiple of 3, so t must be a multiple of 3. Since 12 is also a multiple of 3, both t and 12 share at least 3 as a common factor.2 says that m is a multiple of 3. Using the same rule, we don't know anything about 2k, so we really don't know what t is a multiple of at all. It could be even or odd, and it doesn't have to have any of the factors that 12 has. So it's not enough information.

If k, m, and t are positive integers and k/6 + m/4 = t/12 , do t and 12 have a common factor greater than 1 ?(1) k is a multiple of 3.(2) m is a multiple of 3.

*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.

*Statement 1- Let's say m = (2*a*b*...), which a,b,... are prime numbers.- Let's say n = (2*x*y*...), which x,y,... are prime numbers.- (m+n)4(m+n)4 = (2∗a∗b∗...)+(2∗x∗y∗...)4(2∗a∗b∗...)+(2∗x∗y∗...)4 = 2((a∗b∗...)+(x∗y∗...))42((a∗b∗...)+(x∗y∗...))4- Here we can see that a,b,x,y can be anything prime numbers.- #1 If (a*b) + (x*y) = EVEN, YES, m+n4m+n4 divisible by 4.- #2 If (a*b) + (x*y) = ODD, NO, m+n4m+n4 cannot divisible by 4.- Hence, INSUFFICIENT.Statement 2- Each of m and n has only one maximum one factor of 2 so it cannot divisible by 4.- Try to plug simple number : m = 3 and n = 5, divisible by 4, YES.- Try to plug another random number : m = 5 and n = 9, NOT divisible by 4, NO.- Hence, INSUFFICIENT.Both statement- Go back to this equation :- (m+n)4(m+n)4 = (2∗a∗b∗...)+(2∗x∗y∗...)4(2∗a∗b∗...)+(2∗x∗y∗...)4 = 2((a∗b∗...)+(x∗y∗...))42((a∗b∗...)+(x∗y∗...))4- If m & n divisible by 2 but not by 4, so a,b,x,y MUST BE ODD.- (ODD*ODD) + (ODD*ODD) = ODD + ODD = EVEN.- 2 * ANY even number MUST BE DIVISIBLE BY 4.- Hence, SUFFICIENT.

If m and n are positive integers, is m + n divisible by 4 ?(1) m and n are each divisible by 2.(2) Neither m nor n is divisible by 4.

*TIP:On the GMAT we can often see such statement: kk is halfway between mm and nn on the number line. Remember this statement can ALWAYS be expressed as:m+n2=km+n2=k.Also on the GMAT when we see the distance between x and y, this can be expressed as |x−y||x−y|.BACK TO THE QUESTION.If m and r are two numbers on a number line, what is the value of r?(1) The distance between r and zero is 3 times the distance between m and zero --> |r−0|=3|m−0||r−0|=3|m−0| --> |r|=3|m||r|=3|m| --> r=3mr=3m OR r=−3mr=−3m. Clearly insufficient.(2) 12 is halfway between m and r --> r+m2=12r+m2=12 --> r+m=24r+m=24. Clearly insufficient.(1)+(2) r=3mr=3m OR r=−3mr=−3m and r+m=24r+m=24.r=3mr=3m --> r+m=3m+m=24r+m=3m+m=24 --> m=6m=6 and r=18r=18ORr=−3mr=−3m --> r+m=−3m+m=24r+m=−3m+m=24 --> m=−12m=−12 and r=36r=36Two different values for rr. Not sufficient.Answer: E.

If m and r are two numbers on a number line, what is the value of r?(1) The distance between r and 0 is 3 times the distance between m and 0.(2) 12 is halfway between m and r.

√If n and t are positive integers, what is the greatest prime factor of nt?(1) The greatest common factor of n and t is 5 --> if n and t does not have any prime greater than 5 then the greatest prime factor of nt will be 5 (example: n=5 and t=5 or n=10 and t=15) BUT if n and/or t have some primes more than 5 then the greatest prime factor of nt will be more than 5 (example: n=35 and t=5 --> the greatest prime of nt is 7 or n=5 and t=55 --> the greatest prime of nt is 11)(2) The least common multiple of n and t is 105 --> the least common multiple of two integers contains all common primes of these integers, thus 105 has all the primes which appear in both n and t --> the greatest prime factor of 105 is 7, hence it's the the greatest prime factor of nt (no greater factor can "appear" in nt if it's not in either of them). Sufficient.Answer: B.

If n and t are positive integers, what is the greatest prime factor of nt?(1) The greatest common factor of n and t is 5(2) The least common multiple of n and t is 105

√If n is a positive integer, is n^3 - n divisible by 4 ?n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible by 4.(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.Answer: A.

If n is a positive integer, is n^3 - n divisible by 4 ?(1) n = 2k + 1, where k is an integer(2) n^2 + n is divisible by 6

√If r and s are positive integers, is r/s an integer?(1) Every factor of s is also a factor of r. If every factor of s is also factor of r, then in fraction r/s, s will just be reduced and we get an integer. Sufficient.(2) Every prime factor of s is also a prime factor of r. The powers of prime factors of s could be higher than powers of prime factors of r. eg 25/125=1/5 not an integer. Not sufficient.Answer: A.

If r and s are positive integers, is r/s an integer?(1) Every factor of s is also a factor of r.(2) Every prime factor of s is also a prime factor of

√*If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?Odd consecutive integers is an evenly spaced set. For any evenly spaced set the mean equals to the average of the first and the last terms, so in our case mean=10=x1+xn2mean=10=x1+xn2 --> x1+xn=20x1+xn=20. Question: x1=?x1=?(1) The range of the n integers is 14 --> the range of a set is the difference between the largest and smallest elements of a set, so xn−x1=14xn−x1=14 --> solving for x1x1 --> x1=3x1=3. Sufficient.(2) The greatest of the n integers is 17 --> xn=17xn=17 --> x1+17=20x1+17=20 --> x1=3x1=3. Sufficient.Answer: D.

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?(1) The range of the n integers is 14(2) The greatest of the n integers is 17

√#(a,b)# = a+b / G.C.F (a,b)Find value of #(6,x)# = ? X+ve Intst1:The greatest common factor of 6 and x is 1/7 the sum of 6 and x#(6,x)# = 6+x / 6+x/7 => 7sufficientst2: x is a positive odd integer not equal to 1 -> InsufficientAnswer A

If the expression #(a,b)# is defined to be the sum of a and b divided by the greatest common factor of a and b, what is the value of #(6,x)#, for positive integer x ?(1) The greatest common factor of 6 and x is 1/7 the sum of 6 and x(2) x is a positive odd integer not equal to 1

√If the integer n is greater than 1, is n equal to 2?(1) Any prime satisfies this statement (2, 3, 5, ...), thus n may or may not equal to 2. Not sufficient.(2) n has at least following factors: 1 and n (the number itself). As given that "the difference of any two distinct positive factors of n is odd", then must be true that: n−1=oddn−1=odd --> n=odd+1=evenn=odd+1=even. Can nn be even number more than 2? No, because if n=even>2n=even>2 it obviously has 2 as a factor and again as "the difference of any two distinct positive factors of n is odd", then n-2 must be odd, but n−2=even−2=even≠oddn−2=even−2=even≠odd. Hence n=2n=2. Sufficient.OR: "the difference of any two distinct positive factors of n is odd" means that number must have only one odd factor and only one even factor. (If odd factors, (or even factors) >1, then the difference of the pair of two odd factors (or even factors) will be even not odd). Only number to have only one odd and only one even factor is 2. Sufficient.Answer: B.

If the integer n is greater than 1, is n equal to 2?(1) n has exactly two positive factors.(2) The difference of any two distinct positive factors of n is odd.

*If x and y are integers great than 1, is x a multiple of y?(1) 3y2+7y=x3y2+7y=x --> y(3y+7)=xy(3y+7)=x --> as 3y+7=integer3y+7=integer, then y∗integer=xy∗integer=x --> xx is a multiple of yy. Sufficient.(2) x2−xx2−x is a multiple of yy --> x(x−1)x(x−1) is a multiple of yy --> xx can be multiple of yy (x=2x=2 and y=2y=2) OR x−1x−1 can be multiple of yy (x=3x=3 and y=2y=2) or their product can be multiple of yy (x=3x=3 and y=6y=6). Not sufficient.Answer: A.

If x and y are integers greater than 1, is x a multiple of y?(1) 3y2+7y=x3y2+7y=x(2) x2−xx2−x is a multiple of y

*If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?Given: x=8y+12x=8y+12.(1) x = 12u, where u is an integer --> x=12ux=12u --> 12u=8y+1212u=8y+12 --> 3(u−1)=2y3(u−1)=2y --> the only thing we know from this is that 3 is a factor of yy. Is it GCD of xx and yy? Not clear: if x=36x=36, then y=3y=3 and GCD(x,y)=3GCD(x,y)=3 but if x=60x=60, then y=6y=6 and GCD(x,y)=6GCD(x,y)=6 --> two different answers. Not sufficient.(2) y = 12z, where z is an integer --> y=12zy=12z --> x=8∗12z+12x=8∗12z+12 --> x=12(8z+1)x=12(8z+1). So, we have y=12zy=12z and x=12(8z+1)x=12(8z+1). Now, as zz and 8z+18z+1 do not share any common factor but 1 (8z and 8z+1 are consecutive integers and consecutive integers do not share any common factor 1. As 8z has all factors of z then z and 8z+1 also do not share any common factor but 1). Thus, 12 must be GCD of xx and yy. Sufficient.Answer: B

If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?(1) x = 12u, where u is an integer(2) y = 12z, where z is an integer

√*(1) GCD(x,y)=10GCD(x,y)=10 --> if x=10x=10 and y=10y=10, then xy=100xy=100, which is not divisible by 8 BUT if x=10x=10 and y=20y=20, then xy=200xy=200 which is divisible by 8. Two different answers. Not sufficient.(2) LCM(x,y)=100LCM(x,y)=100 --> if x=1x=1 and y=100y=100, then xy=100xy=100, which is not divisible by 8 BUT if x=4x=4 and y=50y=50, then xy=200xy=200 which is divisible by 8. Two different answers. Not sufficient.(1)+(2) The most important property of LCM and GCD is: for any positive integers xx and yy, x∗y=GCD(x,y)∗LCM(x,y)=10∗100=1000x∗y=GCD(x,y)∗LCM(x,y)=10∗100=1000 --> 1000 is divisible by 8. Sufficient.

If x and y are positive integers, is xy a multiple of 8 ?(1) The greatest common divisor of x and y is 10.(2) The least common multiple of x and y is 100.

√Target question: What is the value of x?Given: x is a positive integer greater than 1Statement 1: 2x is a common factor of 18 and 24Positive factors of 18: 1, 2, 3, 6, 9, 18Positive factors of 24: 1, 2, 3, 4, 6, 8, 12, 24COMMON factors of 18 and 24: 1, 2, 3, 6So, 2x = 1, 2x = 2, 2x = 3 or 2x = 6This means, x = 0.5, x = 1, x = 1.5 or x = 3Since we're told that x is a positive integer greater than 1, we can see that x MUST equal 3Since we can answer the target question with certainty, statement 1 is SUFFICIENTStatement 2: x is a factor of 6So, x could equal 1, 2, 3, or 6Since we're told that x is a positive integer greater than 1, we can see that x COULD equal 2, 3, or 6Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENTAnswer: A

If x is a positive integer greater than 1, what is the value of x?(1) 2x is a common factor of 18 and 24(2) x is a factor of 6

(..****......)......If x is a positive integer, how many positive integers less than x are divisors of x ?(1) x^2 is divisible by exactly 4 positive integers less than x^2.x2x2 will be divisible by exactly 5 integers including x2x2, only when x is a perfect square of a prime number..for example x=a2..x2=a4x=a2..x2=a4... Number of factors = (4+1)=5(4+1)=5So, x or a2a2 will have (2+1) or 3 factors, and our answer is 2 as we have to exclude x itself.Suff(2) 2x is divisible by exactly 3 positive integers less than 2x.This can have two cases for 4 factors including 2x itself..a) x is any other prime number other than 2...21x121x1 will have factors as (1+1)(1+1)=2*2=4.. x will have (1+1) or 2 factors and just 1 factor less than x itself.b) x is also in terms of 2 or x=22x=22, so 2x=232x=23...factors =(3+1) or 4, and x or 2222 will have (2+1) or 3 factors, and our answer is 2 as we have to exclude x itself.InsuffA

If x is a positive integer, how many positive integers less than x are divisors of x ?(1) x^2 is divisible by exactly 4 positive integers less than x^2.(2) 2x is divisible by exactly 3 positive integers less than 2x.

√Hey, guys, is it rule? LCM(a,b,c)=LCM(LCM(a,b),c) ?If yes, we can use it:1) LCM(x,6,9) = LCM(30,9)=902) LCM(x,6,9) = LCM(45,6)=90D

If x is a positive integer, what is the least common multiple of x, 6 and 9?(1) The least common multiple of x and 6 is 30.(2) The least common multiple of x and 9 is 45.

√First of all: 1/4x - 5 <=0 should be written as 1/4*x - 5 <=0.1. If x is to be selected at random from T, what is the probability that 14∗x−5≤014∗x−5≤0?14∗x−5≤014∗x−5≤0 --> is x≤20x≤20?(1) T is a set of 8 integers. Clearly insufficient.(2) T is contained in the set of integers from 1 to 25, inclusive. Though the wording is a little bit strange but it means that set T is a subset of a set of integers from 1 to 25, inclusive. Set T can be {1,5,7} or {21,22,25}... Also insufficient.(1)+(2) T can be set of 8 integers, which are ALL less than or equal to 20 and in this case P(x≤20)=1P(x≤20)=1 or T can be set of 8 integers which are NOT ALL all less than or equal to 20 and in this case P(x≤20)<1P(x≤20)<1. Not sufficient.Answer: E.

If x is to be selected at random from T, what is the probability that 1/4∗x−5≤0(1) T is a set of 8 integers.(2) T is contained in the set of integers from 1 to 25, inclusive.

√If x + y >0, is x > |y|?x>|y|x>|y| means:A. x>−yx>−y, if y≤0y≤0;B. x>yx>y, if y>0y>0.So we should check whether above two inequalities are true.First inequality is given to be true in the stem (x>-y), so we should check whether x>yx>y is true.(1) x > y. Sufficient.(2) y < 0 --> |y|=−y|y|=−y. Question becomes is x>−yx>−y. This given to be true in the stem. Sufficient.Answer: D.

If x+y > 0, is x> |y|i. x>yii.y<0

**If x, y and d are integers and d is odd, are both x and y divisible by d?Question asks whether both x and y are multiples of d.(1) x+y is divisible by d --> if x=y=d=1x=y=d=1 then the answer is YES but if x=2x=2, y=4y=4 and d=3d=3 then the answer is NO. Not sufficient.(2) x-y is divisible by d --> if x=y=d=1x=y=d=1 then the answer is YES but if x=4x=4, y=1y=1 and d=3d=3 then the answer is NO. Not sufficient.(1)+(2) From (1) x+y=dqx+y=dq and from (2) x−y=dpx−y=dp --> sum those two equations: 2x=d(q+p)2x=d(q+p) --> x=d(q+p)2x=d(q+p)2. Now, since xx is an integer and dd is an odd number then q+p2=integerq+p2=integer (d/2 cannot be an integer because d is odd) --> x=d∗integerx=d∗integer --> xx is a multiple of dd. From (1) {multiple of d}+y={multiple of d} --> hence yy must also be a multiple of dd. Sufficient.Answer: C.If integers aa and bb are both multiples of some integer k>1k>1 (divisible by kk), then their sum and difference will also be a multiple of kk (divisible by kk):Example: a=6a=6 and b=9b=9, both divisible by 3 ---> a+b=15a+b=15 and a−b=−3a−b=−3, again both divisible by 3.If out of integers aa and bb one is a multiple of some integer k>1k>1 and another is not, then their sum and difference will NOT be a multiple of kk (divisible by kk):Example: a=6a=6, divisible by 3 and b=5b=5, not divisible by 3 ---> a+b=11a+b=11 and a−b=1a−b=1, neither is divisible by 3.If integers aa and bb both are NOT multiples of some integer k>1k>1 (divisible by kk), then their sum and difference may or may not be a multiple of kk (divisible by kk):Example: a=5a=5 and b=4b=4, neither is divisible by 3 ---> a+b=9a+b=9, is divisible by 3 and a−b=1a−b=1, is not divisible by 3;OR: a=6a=6 and b=3b=3, neither is divisible by 5 ---> a+b=9a+b=9 and a−b=3a−b=3, neither is divisible by 5;OR: a=2a=2 and b=2b=2, neither is divisible by 4 ---> a+b=4a+b=4 and a−b=0a−b=0, both are divisible by 4.

If x, y, and d are integers and d is odd, are both x and y divisible by d ?(1) x + y is divisible by d.(2) x − y is divisible by d.

√If x≠0x≠0, is |x|<1|x|<1?Is |x|<1|x|<1? --> is -1<x<11<x<1 (x≠0x≠0)?(1) x2<1x2<1 --> −1<x<1−1<x<1. Sufficient.(2) |x|<1x|x|<1x --> since LHS (|x|) is an absolute value which is always non-negative then RHS (1/x), must be positive (as |x|<1x|x|<1x), so 1x>01x>0 --> x>0x>0.Now, if x>0x>0 then |x|=x|x|=x and we have: x<1xx<1x --> since x>0x>0 then we can safely multiply both parts by it: x2<1x2<1 --> −1<x<1−1<x<1, but as x>0x>0, the final range is 0<x<10<x<1. Sufficient.Answer D.

If x≠0, is |x| <1?(1) x2<1(2) |x| < 1/x

*If y≥0y≥0, what is the value of x?(1) |x−3|≥y|x−3|≥y. As given that yy is non negative value then |x−3||x−3| is more than (or equal to) some non negative value, (we could say the same ourselves as absolute value in our case (|x−3||x−3|) is never negative). So we can not determine single numerical value of xx. Not sufficient.Or another way: to check |x−3|≥y≥0|x−3|≥y≥0 is sufficient or not just plug numbers:A. x=5x=5, y=1>0y=1>0, and B. x=8x=8, y=2>0y=2>0: you'll see that both fits in |x−3|>=y|x−3|>=y, y≥0y≥0.Or another way:|x−3|≥y|x−3|≥y means that:x−3≥y≥0x−3≥y≥0 when x−3>0x−3>0 --> x>3x>3OR (not and)−x+3≥y≥0−x+3≥y≥0 when x−3<0x−3<0 --> x<3x<3Generally speaking |x−3|≥y≥0|x−3|≥y≥0 means that |x−3||x−3|, an absolute value, is not negative. So, there's no way you'll get a unique value for xx. INSUFFICIENT.(2) |x−3|≤−y|x−3|≤−y. Now, as |x−3||x−3| is never negative (0≤|x−3|0≤|x−3|) then 0≤−y0≤−y --> y≤0y≤0 BUT stem says that y≥0y≥0 thus y=0y=0. |x−3|≤0|x−3|≤0 --> |x−3|=0=y|x−3|=0=y (as absolute value, in our case |x-3|, can not be less than zero) --> x−3=0x−3=0 --> x=3x=3. SUFFICIENTIn other words:−y−y is zero or less, and the absolute value (|x−3||x−3|) must be at zero or below this value. But absolute value (in this case |x−3||x−3|) can not be less than zero, so it must be 00.Answer: B.

If y≥0y≥0, what is the value of x?(1) |x−3|≥y|x−3|≥y(2) |x−3|≤−y|x−3|≤−yShow Answer

√*This is not a good question, (well at least strange enough) as neither of statement is needed to answer the question, stem is enough to do so. This is the only question from the official source where the statements aren't needed to answer the question. I doubt that such a question will occur on real test but if it ever happens then the answer would be D.If zy<xy<0zy<xy<0 is |x−z|+|x|=|z||x−z|+|x|=|z|Look at the inequality zy<xy<0zy<xy<0:We can have two cases:A. If y<0y<0 --> when reducing we should flip signs and we'll get: z>x>0z>x>0.In this case: as z>xz>x --> |x−z|=−x+z|x−z|=−x+z; as x>0x>0 and z>0z>0 --> |x|=x|x|=x and |z|=z|z|=z.Hence in this case |x−z|+|x|=|z||x−z|+|x|=|z| will expand as follows: −x+z+x=z−x+z+x=z --> 0=00=0, which is true.And:B. If y>0y>0 --> when reducing we'll get: z<x<0z<x<0.In this case: as z<xz<x --> |x−z|=x−z|x−z|=x−z; as x<0x<0 and z<0z<0 --> |x|=−x|x|=−x and |z|=−z|z|=−z.Hence in this case |x−z|+|x|=|z||x−z|+|x|=|z| will expand as follows: x−z−x=−zx−z−x=−z --> 0=00=0, which is true.So knowing that zy<xy<0zy<xy<0 is true, we can conclude that |x−z|+|x|=|z||x−z|+|x|=|z| will also be true. Answer should be D even not considering the statements themselves.As for the statements:Statement (1) says that z<xz<x, hence we have case B.Statement (2) says that y>0y>0, again we have case B.Answer: D.

If zy < xy < 0, is |x - z | + |x| = |z|?(1) z < x(2) y > 0

√Statement 1: greatest common divisor of M and N is 6. So, M and N are multiple of 6. But, an exact value of N cannot be determined. Insufficient!Statement 2: LCM of M and N is 36. M can be 9 and N can be 12 or M can be 12 and N can be 18. Multiple possible answer. Insufficient!Combining 1&2, M and N are multiple of 6 and LCM is 36. So the only possible values on M and N can be 12 and 18 respectively.Sufficient!Answer C!

M and N are integers such that 6<M<N.What is the value of N?(1) The greatest common divisor of M and N is 6(2) The least common multiple of M and N is 36OG Q 2017 New Question (Book Question: 297)

*In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?Let's consider the temperatures in April in ascending order: t1≤t2...≤t15≤t16≤...≤t29≤t30t1≤t2...≤t15≤t16≤...≤t29≤t30. The median temperature would be the average of 15th and 16th greatest temperatures. The question asks: is this value less than the average temperature, (which would be the sum of temperatures divided by 30).(1) In City X last April, the sum of the 30 daily high temperatures was 2,160° --> gives the average temperature, but not info about the median. Not sufficient.(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature --> each temperature (term) from the 18 lowest ones (from t1t1 to t18t18) given to be less than average, hence the median (average of 15th and 16th terms) would also be less than the average. Sufficient.Answer: B.

In City X last April, was the average (arithmetic mean) daily high temperature greater than the median daily high temperature?(1) In City X last April, the sum of the 30 daily high temperatures was 2,160°.(2) In City X last April, 60 percent of the daily high temperatures were less than the average daily high temperature. Show Answer

*(1) For month M, total revenue from sales was $3,500.We are not in a position to comment since we have no clue about any sales figure or no of days etc(2) For month J, total revenue from sales was $6,000.M and j each alone cannot give us information about the meanWhen 1 and 2 is combined still it isn't sufficient since individual days will acoount for different sales therefore we cannot deterministically determine and nail down the sameTherefore IMO insufficient hence E

In a retail store, the average (arithmetic mean) sale for month M was d dollars. Was the average (arithmetic mean) sale for month J at least 20 percent higher than that for month M?(1) For month M, total revenue from sales was $3,500.(2) For month J, total revenue from sales was $6,000.

√Statement 1- |x|+|y|=1(1,0),(0,1), (-1,0) and (0,-1) satisfy equationCase 1-If co-ordinates of P (-1,0) and that of Q is (0,1)Slope= 1−00−(−1)1−00−(−1)= 1Case 2-If co-ordinates of P (-1,0) and that of Q is (1,0)Slope= 0−01−(−1)0−01−(−1)= 0InsufficientStatement 2- |x+y|=1Again (1,0),(0,1), (-1,0) and (0,-1) satisfy the equationWe can make the same casesCase 1-If co-ordinates of P (-1,0) and that of Q is (0,1)Slope= 1−00−(−1)1−00−(−1)= 1Case 2-If co-ordinates of P (-1,0) and that of Q is (1,0)Slope= 0−01−(−1)0−01−(−1)= 0InsufficientCombining both equations gives us nothing new, as those 4 points satisfies both equations.InsufficientE

In the standard (x,y) coordinate plane, what is the slope of the line containing the distinct points P and Q ?(1) Both P and Q lie on the graph of |x| + |y| = 1.(2) Both P and Q lie on the graph of |x + y| = 1.

√*Is (x−3)2−−−−−−−√=3−x(x−3)2=3−x?Remember: x2−−√=|x|x2=|x|. Why?Couple of things:The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0.So x2−−√≥0x2≥0. But what does x2−−√x2 equal to?Let's consider following examples:If x=5x=5 --> x2−−√=25−−√=5=x=positivex2=25=5=x=positive;If x=−5x=−5 --> x2−−√=25−−√=5=−x=positivex2=25=5=−x=positive.So we got that:x2−−√=xx2=x, if x≥0x≥0;x2−−√=−xx2=−x, if x<0x<0.What function does exactly the same thing? The absolute value function! That is why x2−−√=|x|x2=|x|Back to the original question:So (x−3)2−−−−−−−√=|x−3|(x−3)2=|x−3| and the question becomes is: |x−3|=3−x|x−3|=3−x?When x>3x>3, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.When x≤3x≤3, then LHS=|x−3|=−x+3=3−x=RHSLHS=|x−3|=−x+3=3−x=RHS, hence in this case equation holds true.Basically question asks is x≤3x≤3?(1) x≠3x≠3. Clearly insufficient.(2) −x|x|>0−x|x|>0, basically this inequality implies that x<0x<0, hence x<3x<3. Sufficient.Answer: B.

Is (x−3)2−−−−−−−√=3−x(x−3)2=3−x?(1) x≠3x≠3(2) −x|x|>0

√*For me (D) sqrt( (x-5)^2) = 5 - x ?<=> |x-5| = 5-x ?<=> |5-x| = 5-x ?This is true if 5-x >= 0 <=> x =< 5So we end up with checking if x =< 5 ?Stat 1-x |x| > 0<=> x < 0 < 5SUFF.Stat 25 - x > 0<=> x < 5SUFF.

Is (x−5)2−−−−−−−√=5−x(x−5)2=5−x?(1) −x|x|>0−x|x|>0(2) 5−x>0

√*Is 1/(a - b) > b - a ?(1) a < b --> we can rewrite this as: a−b<0a−b<0 so LHS is negative, also we can rewrite it as: b−a>0b−a>0 so RHS is positive --> negative<positive. Sufficient.(2) 1 < |a - b| --> if a−b=2a−b=2 (or which is the same b−a=−2b−a=−2) then LHS>0 and RHS<0 and in this case the answer will be YES if a−b=−2a−b=−2 (or which is the same b−a=2b−a=2) then LHS<0 and RHS>0 and in this case the answer will be NO. Not sufficient.Answer: A.

Is 1/(a - b) > b - a ?(1) a < b(2) 1 < |a - b|

*If a number is the square of an integer, then in the prime factorization of that number, all of the exponents will be even. For example, (3^8)(5^6) is the square of an integer (it is the square of (3^4)(5^3) ) whereas (3^7)(5^6) is not.Statement 1 tells us that in the prime factorization of (2^2)(n), all of the exponents are even. The prime factorization of n is identical to that of (2^2)(n) except that in n, the exponent on the '2' is smaller by two, and subtracting two doesn't change an even number to an odd number. So if the exponents in (2^2)(n) are even, so are the exponents in n, and Statement 1 is sufficient.Statement 2 tells us that in the prime factorization of n^3, all of the exponents are even. The exponents in the prime factorization n^3 are all 3 times as big as the exponents in the prime factorization of n. Multiplying by 3 doesn't change evenness or oddness, so if all the exponents in n^3 are even, so are the exponents in n, and Statement 2 is also sufficient.The answer is D.

Is the positive integer n the square of an integer?1) 4n is the square of an integer2) n^3 is the square of an integer

√Is the range of the integers 6, 3, y, 4, 5, and x greater than 9?Given integers are: {3, 4, 5, 6, x, y}(1) y > 3x. If x=1x=1 and y=4y=4 then the range=6-1=5<9 but if x=100x=100 then the range>9. Not sufficient.(2) y > x > 3. If x=4x=4 and y=5y=5 then the range=6-3=3<9 but if x=100x=100 then the range>9. Not sufficient.(1)+(2) From x>3x>3 we have that the least value of xx is 4, and from y>3x=12y>3x=12 we have that the least value of yy is 13, hence the least value of the range is 13-3=10>9. Sufficient.Answer: C._________________

Is the range of the integers 6, 3, y, 4, 5, and x greater than 9?(1) y > 3x(2) y > x > 3

√Statement (1): We are given a value for x. It's hard to determine whether that value is greater or smaller than 1010 without a calculator, but there is no need to calculate - just to know that we could. Wherever 322 falls on the number line, we'll know for certain whether it's less than 1010. Therefore, Statement (1) is Sufficient to answer the question. Eliminate choices (B), (C), and (E). Statement (2): We are given the statement that x > 320. If we had a calculator, we could calculate 320, but unlike in Statement (1), knowing that we could calculate isn't enough. If 320 is greater than 1010, then the answer to the question is a clear "no," but if 320 is less than 1010, then we don't know whether x < 1010. We need to somehow relate 320 to 1010. At first this seems difficult, as 3 is not among the factors of 10. Since we aren't able to relate the bases, we'll have to relate the exponents. Let's re-express 320, aiming for an exponent of 10: 320 = 32(10) Exponent rules tell us ( xa)b = xab. There's no reason why we can't use this rule "backwards": x > 320 x > 32(10) x > (32)10 x > 910 910 is less than 1010, so x could be less than 1010. But since there's no upper limit on x, it could easily be greater. Since we do not have a definite answer, Statement (2) is Insufficient. Eliminate choice (D). Therefore, Answer Choice (A) is correct.

Is x < 1010? (1) x = 322 (2) x > 320

*Good question. +1.Is x> 1?(1) (x+1) (|x| - 1) > 0. Consider two cases:If x>0x>0 then |x|=x|x|=x so (x+1)(|x|−1)>0(x+1)(|x|−1)>0 becomes (x+1)(x−1)>0(x+1)(x−1)>0 --> x2−1>0x2−1>0 --> x2>1x2>1 --> x<−1x<−1 or x>1x>1. Since we consider range when x>0x>0 then we have x>1x>1 for this case;If x≤0x≤0 then |x|=−x|x|=−x so (x+1)(|x|−1)>0(x+1)(|x|−1)>0 becomes (x+1)(−x−1)>0(x+1)(−x−1)>0 --> −(x+1)(x+1)>0−(x+1)(x+1)>0 --> −(x+1)2>0−(x+1)2>0 --> (x+1)2<0(x+1)2<0. Now, since the square of a number cannot be negative then for this range given equation has no solution.So, we have that (x+1)(|x|−1)>0(x+1)(|x|−1)>0 holds true only when x>1x>1. Sufficient.(2) |x| < 5 --> −5<x<5−5<x<5. Not sufficient.Answer: A.

Is x > 1?(1) (x+1)(|x|−1)>0(x+1)(|x|−1)>0(2) |x|<5

*|a-c| + |a| = |c||a-c| = |c| - |a|to satisfy "|a-c| = |c| - |a|", there are 3 options:1. a = c2. a = 03. |c| > |a| (c>a in magnitude) but have similar signs.statement 1: ab > bceliminates option 1 and 2, but leaves us with option 3 which can't be known as the sign of b is unknown.statement 2: ab < 0gives no information except that a and b ≠0≠0 and that a and b has different signs.by combining statement 1 and 2: (see the testing table)0>ab>bc while a and b have different signs, so a and c must have similar signsso if b>0, then both a and c are negative and c>a in magnitude (|c| > |a|)so if b<0, then both a and c are positive and c>a in magnitude (|c| > |a|)these criteria satisfies the tested equationso C

Is |a-c| + |a| = |c|?(1) ab > bc(2) ab < 0

√*Is |x-y|>|x|-|y|?Probably the best way to solve this problem is plug-in method. Though there are two properties worth to remember:1. Always true: |x+y|≤|x|+|y||x+y|≤|x|+|y|, note that "=" sign holds for xy≥0xy≥0 (or simply when xx and yy have the same sign);2. Always true: |x−y|≥|x|−|y||x−y|≥|x|−|y|, note that "=" sign holds for xy>0xy>0 (so when xx and yy have the same sign) and |x|>|y||x|>|y| (simultaneously). (Our case)So, the question basically asks whether we can exclude "=" scenario from the second property.(1) y < x --> we can not determine the signs of xx and yy. Not sufficient.(2) xy < 0 --> "=" scenario is excluded from the second property, thus |x−y|>|x|−|y||x−y|>|x|−|y|. Sufficient.Answer: B.

Is |x-y|>|x|-|y| ?(1) y < x(2) xy < 0

√Is |x| < 1 ?Is |x| < 1 --> is -1 < x < 1.(1) x/|x| < x.Two cases:A. x<0x<0 --> x−x<xx−x<x --> −1<x−1<x. Since we consider the case when x<0x<0, then we'd have −1<x<0−1<x<0. Answer YES.B. x>0x>0 --> xx<xxx<x --> 1<x1<x. Answer NO.Not sufficient.(2) x/|x| < 1. This simply means that x is a negative number: if x is positive then x/|x| = 1, if x is negative x/|x| = -1 < 1. Not sufficient.(1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient.Answer: C.

Is |x| < 1 ?(1) x/|x| < x(2) x/|x| < 1

√K is a set of numbers such that(i) if x is in K, then -x is in K, and(ii) if each of x and y is in K, then xy is in K.Is 12 in K?(1) 2 is in K --> according to (i) -2 is n K --> according to (ii) -2*2=-4 is in K --> according to (i) -(-4)=4 is in K and so on. Thus we know that 2, -2, -4, 4, 8, -8, 16, -16, ... are in K, so basically powers of 2 and their negative pairs. Is 12 in K? We don't know. Not sufficient.(2) 3 is in K --> according to (i) -3 is n K --> according to (ii) -3*3=-9 is in K --> according to (i) -(-9)=9 is in K and so on. Thus we know that 3, -3, -9, 9, 27, -27, 81, -81, ... are in K, so basically powers of 3 and their negative pairs. Is 12 in K? We don't know. Not sufficient.(1)+(2) From (1) 4 is in K and from (2) 3 is in K, hence according to (ii) 4*3=12 must also be in K. Sufficient.Answer: C.

K is a set of numbers such that(i) if x is in K, then -x is in K, and(ii) if each of x and y is in K, then xy is in K.Is 12 in K?(1) 2 is in K.(2) 3 is in K.

*Last year the average (arithmetic mean) salary of the 10 employees of Company X was $42,800. What is the average salary of the same 10 employees this year?Note that it's almost always better to express the average in terms of the sum.Given: total salary is $42,800*10.Question: new total salary=?(1) For 8 of the 10 employees, this year's salary is 15 percent greater than last year's salary. Clearly insufficient.(2) For 2 of the 10 employees, this year's salary is the same as last year's salary. Clearly insufficient.(1)+(2) Consider two possible cases:2 lowest salaries didn't change and 8 highest salaries increased by 15%;2 highest salaries didn't change and 8 lowest salaries increased by 15%;Ask yourself, would new total salary be the same for both cases? No, because the increase in amount for the first case will be greater than the increase in amount for the second case. Hence even taken together statements are not sufficient.Answer: E.

Last year the average (arithmetic mean) salary of the 10 employees of Company X was $42,800. What is the average salary of the same 10 employees this year?(1) For 8 of the 10 employees, this year's salary is 15 percent greater than last year's salary.(2) For 2 of the 10 employees, this year's salary is the same as last year's salary.

*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.

*Question basically ask on which day did Jane make the payment of $300.Or algebraically: average daily balance=x∗600+(30−x)∗30030average daily balance=x∗600+(30−x)∗30030 --> for the first xx days the balance will be $600 and then for (30−x)(30−x) days the balance will be $600-$300=$300 (meaning that Jane made a payment of $300 on the (x+1)st day). So all we need to know is on which day did Jane make the payment of $300, the value of x+1.(1) Jane's payment was credited on the 21st day of the billing cycle --> directly gives the value of x+1x+1 --> x+1=21x+1=21 --> x=20x=20: for 20 days the balance was $600 and for the rest 10 days (from 21st) it was $300 --> we can find the average daily balance. Sufficient.(2) The average daily balance through the 25th day of the billing cycle was $540 --> average25 days=540=x∗600+(25−x)∗30025average25 days=540=x∗600+(25−x)∗30025 --> we can find xx --> we can find the average daily balance. Sufficient.Answer: D.

On Jane's credit card account, the average daily balance for 30-day billing cycle is the average (arithmetic mean) of the daily balances at the end of each of the 30 days. At the beginning of a certain 30-day billing cycle, Jane's credit card account had a balance of $600. Jane made a payment of $300 on the account during the billing cycle. If no other amounts were added to or subtracted from the account during the billing cycle, what was the average daily balance on Jane's account for the billing cycle?(1) Jane's payment was credited on the 21st day of the billing cycle.(2) The average daily balance through the 25th day of the billing cycle was $540.

*(1) s>0s>0, clearly not sufficient.(2) The distance between tt and rr is the same as the distance between tt and -ss: |t−r|=|t+s||t−r|=|t+s|.t−rt−r is always positive as rr is to the left of the tt, hence |t−r|=t−r|t−r|=t−r;BUT t+st+s can be positive (when t>−st>−s, meaning tt is to the right of -ss) or negative (when t<−st<−s, meaning tt is to the left of -ss, note that even in this case ss would be to the left of tt and relative position of the points shown on the diagram still will be the same). So we get either |t+s|=t+s|t+s|=t+s OR |t+s|=−t−s|t+s|=−t−s.In another words: t+st+s is the sum of two numbers from which one tt, is greater than ss. Their sum clearly can be positive as well as negative. Knowing that one is greater than another doesn't help to determine the sign of their sum.Hence:t−r=t+st−r=t+s --> −r=s−r=s;ORt−r=−t−st−r=−t−s --> 2t=r−s2t=r−s.So the only thing we can determine from (2) is: t−r=|t+s|t−r=|t+s|Not sufficient.(1)+(2) s>0s>0 and t−r=|t+s|t−r=|t+s|. s>0s>0 --> t>0t>0 (as tt is to the right of ss) hence t+s>0t+s>0. Hence |t+s|=t+s|t+s|=t+s. --> t−r=t+st−r=t+s --> −r=s−r=s. Sufficient.Answer: C.

On the number line shown, is zero halfway between r and s?(1) s is to the right of zero(2) The distance between t and r is the same as the distance between t and -s

*We are given that on the number line, point R has coordinate r and point T has coordinate t, and we need to determine whether t < 0.Statement One Alone:-1 < r < 0Statement one tells us that r is a negative proper fraction. However, without knowing anything about t, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.Statement Two Alone:The distance between R and T is equal to r^2.The distance between two values on the number line is the absolute value of the difference between the two values. Thus statement two gives us the equation |r - t| = r^2. However, without knowing anything about r and t, we can't determine whether t is less than zero. For instance, r could be 2 and t could be -2; or r could be -2 and t could be 2. In each of the cases, |r - t| = 4 = r^2; but in one case t > 0 and in the other t < 0. Statement two is not sufficient to answer the question. We can eliminate answer choice B.Statements One and Two Together:Using statements one and two, we know that r is a negative proper fraction and |r - t| = r^2. Thus, r - t = r^2 OR r - t = -r^2. Solving each of these for t, we get: t = r - r^2 OR t = r + r^2.Since r is a negative proper fraction, no matter what the value of r is, t will always be a negative number. For instance, if r = -1/2, then r^2 = 1/4 and t will either be -1/2 - 1/4 = -3/4 or -1/2 + 1/4 = -1/4.The reason why t cannot be positive is that when we square r (a negative proper fraction), the value of r^2 (though positive) will be less than the absolute value of r. Recall that t = r - r^2 or t = r + r^2. When a positive proper fraction with a smaller absolute value is added to (or subtracted from) a negative proper fraction with a larger absolute value, the sum (or difference) will always be less than zero.Answer: C

On the number line, point R has coordinate r and point T has coordinate t. Is t < 0?(1) -1 < r < 0(2) The distance between R and T is equal to r^2

*On the number line, the distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?The distance between x and y is greater than the distance between xx and zz, means that we can have one of the following four scenarios:A. y--------z--x (YES case);B. x--z--------y (YES case);C. y--------x--z (NO case);D. z--x--------y (NO case)'The question asks whether we have scenarios A or B (zz lies between xx and yy).(1) xyz<0xyz<0 --> either all three are negative or any two are positive and the third one is negative. If we place zero between yy and zz in case A (making yy negative and xx, zz positive), then the answer would be YES but if we place zero between yy and xx in case C, then the answer would be NO. Not sufficient.(2) xy<0xy<0 --> xx and yy have opposite signs. The same here: We can place zero between yy and xx in case A and the answer would be YES but we can also place zero between yy and xx in case C and the answer would be NO. Not sufficient.(1)+(2) Both case A (answer YES) and case C (answer NO) satisfy the statements. Not sufficient.A. y----0----z--x (YES case) --> xyz<0xyz<0 and xy<0xy<0;C. y----0----x--z (NO case) --> xyz<0xyz<0 and xy<0xy<0.Answer: E.

On the number line, the distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?(1) xyz < 0(2) xy < 0

√S is a set of integers such thati) if a is in S, then -a is in S, andii) if each of a and b is in S, then ab is in S.Is -4 in S?(1) 1 is in S --> according to (i) -1 is in S. Is -4 in S? We don't know. Not sufficient.(2) 2 is in S --> according to (i) -2 is in S --> according to (ii) -2*2=-4 is in S. Sufficient.Answer: B.

S is a set of integers such thati) if a is in S, then -a is in S, andii) if each of a and b is in S, then ab is in S.Is -4 in S?(1) 1 is in S.(2) 2 is in S.

*As the question stem doesn't make it clear where to start, we'll first try a few examples.This is an Alternative approach.(1) OK, but how much is the maximum? Could be 80, 85, 90, anything above the mean.Insufficient.(2) Still no useful information.... though finding specific examples is time consuming (as the equation for std is involved), we can think of the std as an expression with 10 variables representing the 10 different scores. So the maximum could, as above, be any number, such as 50, 30, 90, etc...Insufficient.Combined: Still not helpful, as we now have two equations for 10 variables (our 10 scores): std = 5 and mean = 75. So we cannot solve.(E) is our answer.

The 10 students in a history class recently took an examination. What was the maximum score on the examination?(1) The mean of the scores was 75.(2) The standard deviation of the scores was 5.

√Solution:We need to determine whether the mean of 6 integers is at least 10. Notice that the sum of the original 5 integers is 9 x 5 = 45. If we let the additional integer be n, the question becomes(45 + n) / 6 ≥ 10 ?45 + n ≥ 60 ?n ≥ 15 ?Therefore, if we can determine whether the additional integer is at least 15, then we can determine whether the mean of the 6 integers is at least 10.Statement One Alone:Knowing that the additional integer is at least 14 does not mean it's at least 15. If n = 15, then the mean of the 6 integers is 10; however, if n = 14, then the mean is less than 10. Statement one alone is not sufficient.Statement Two Alone:Knowing that the additional integer is a multiple of 5 does not mean it's at least 15. If n = 15, then the mean of the 6 integers is 10; however, if n = 10, then the mean is less than 10. Statement two alone is not sufficient.Statements One and Two Together:Knowing that the additional integer is at least 14 and a multiple of 5 allows us to say it's at least 15 since 15 is the smallest integer that satisfies both conditions. Therefore, we can say that indeed, the mean of the 6 integers is at least 10.Answer: C

The arithmetic mean of a collection of 5 positive integers, not necessarily distinct, is 9. One additional positive integer is included in the collection and the arithmetic mean of the 6 integers is computed. Is the arithmetic mean of the 6 integers at least 10 ? 1. The additional integer is at least 14. 2. The additional integer is a multiple of 5.

√*The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?Given: 2<m<p2<m<p and pm≠integerpm≠integer. p=xm+rp=xm+r. q: r=?r=?(1) The greatest common factor of m and p is 2 --> pp and mm are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as pm≠integerpm≠integer), then the remainder must be more than 1. Sufficient.(2) The least common multiple of m and p is 30 --> if m=5m=5 and p=6p=6, then remainder=1=1 and thus the answer to the question will be NO. BUT if m=10m=10 and p=15p=15, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.Answer: A.

The integers m and p are such that 2<m<p, and m is not a factor of p. If r is the remainder when p is divided by m, is r>1?(1) The greatest common factor of m and p is 2.(2) The least common multiple of m and p is 30.

√Finding the Number of Factors of an Integer:First make prime factorization of an integer n=ap∗bq∗crn=ap∗bq∗cr, where aa, bb, and cc are prime factors of nn and pp, qq, and rr are their powers.The number of factors of nn will be expressed by the formula (p+1)(q+1)(r+1)(p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.Example: Finding the number of all factors of 450: 450=21∗32∗52450=21∗32∗52Total number of factors of 450 including 1 and 450 itself is (1+1)∗(2+1)∗(2+1)=2∗3∗3=18(1+1)∗(2+1)∗(2+1)=2∗3∗3=18 factors.Back to the original question:The positive integer k has exactly two positive prime factors, 3 and 7. If k has a total of 6 positive factors, including 1 and k, what is the value of K?"k has exactly two positive prime factors 3 and 7" --> k=3m∗7nk=3m∗7n, where m=integer≥1m=integer≥1 and n=integer≥1n=integer≥1;"k has a total of 6 positive factors including 1 and k" --> (m+1)(n+1)=6(m+1)(n+1)=6. Note here that neither mm nor nn can be more than 2 as in this case (m+1)(n+1)(m+1)(n+1) will be more than 6.So, there are only two values of kk possible:1. if m=1m=1 and n=2n=2 --> k=31∗72=3∗49k=31∗72=3∗49;2. if m=2m=2 and n=1n=1 --> k=32∗71=9∗7k=32∗71=9∗7.(1) 3^2 is a factor of k --> we have the second case, hence k=32∗71=9∗7k=32∗71=9∗7. Sufficient.(2) 7^2 is NOT a factor of k --> we have the second case, hence k=32∗71=9∗7k=32∗71=9∗7. Sufficient.Answer: D.

The positive integer k has exactly two positive prime factors, 3 and 7. If k has a total of 6 positive factors, including 1 and k, what is the value of K?(1) 3^2 is a factor of k(2) 7^2 is NOT a factor of k

√The range of the numbers in set S is x, and the range of the numbers in set Tis y. If all of the numbers in set T are also in set S, is x greater than y?(1) Set S consists of 7 numbers. Nothing about set T. Not sufficient.(2) Set T consists of 6 numbers. Nothing about set S. Not sufficient.(1)+(2) It's quite easy to get two different answers. For example, if S={1, 2, 3, 4, 5, 6, 7} and T={1, 2, 3, 4, 5, 6}, then x=6>5=y, but if S={1, 2, 3, 4, 5, 6, 7} and T={1, 2, 3, 4, 5, 7}, then x=6=y. not sufficient.Answer: E.

The range of the numbers in set S is x, and the range of the numbers in set T is y. If all of the numbers in set T are also in set S, is x greater than y?(1) Set S consists of 7 numbers.(2) Set T consists of 6 numbers.Show Answer

√*We have three prices: a, b and c. (a+b+c)/3=120The median price would be: the second biggest.a<=b<=c --> median price b.(1) One of the prices is 110, less than average of 120. It's possible 110 to be a or b price, so insufficient.(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120Answer B.

Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was $120,000. What was the median price of the three houses?(1) The price of Tom's house was $110,000.(2) The price of Jane's house was $120,000.

√1. We don't know individual values. So insufficient.2. We know only the ranges. Value can be anything. So insufficient1+2 salaries are 0,6300,12600,18900,25200,315006300,12600,18900,25200,31500,37800So any values possible. So insufficientHence E

What is the average (arithmetic mean) annual salary of the 6 employees of a toy company?(1) If the 6 annual salaries were ordered from least to greatest, each annual salary would be $6,300 greater than the preceding annual salary.(2) The range of the 6 annual salaries is $31,500.

√What is the median number of employees assigned per project for the projects at Company Z?(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project. Not sufficient on its own.(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project. Not sufficient on its own.(1)+(2) Since 35% of of the projects have 2 or fewer (≤2≤2)employees and 25% of the projects have 4 or more (≥4≥4) employees, then 100%-(25%+35%)=40% of the projects have exactly 3 employees assigned to each of them. So, the median number of employees assigned per project is 3. Sufficient.Answer: C.To elaborate more: consider there are 100 projects: {p1, p2, ..., p100}{p1, p2, ..., p100}. The values of p1p1 to p35p35 will be 0, 1, or 2; the values of p36p36 to p75p75 will be exactly 3; the values of p76p76 to p100p100 will be 4 or more. Median=p50+p512=3+32=3Median=p50+p512=3+32=3.For example list can be: {2, 2, 2, ..., (p35=2), (p36=3), 3, ..., (p75=3), (p76=4), 4, ..., (p100=4)}{2, 2, 2, ..., (p35=2), (p36=3), 3, ..., (p75=3), (p76=4), 4, ..., (p100=4)};OR:{0, 0, 1, 1, 1, 2, 2, ..., (p35=2), (p36=3), 3, ..., (p75=3), (p76=4), 5, 7, 27, ..., (p100=10000)}{0, 0, 1, 1, 1, 2, 2, ..., (p35=2), (p36=3), 3, ..., (p75=3), (p76=4), 5, 7, 27, ..., (p100=10000)} (of course there are a lot of other breakdowns).In any case median=3.

What is the median number of employees assigned per project for the projects at Company Z?(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project.(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project

√There are 9 consecutive integers..(1) The median of the integers in the list is greater than 0.The nine numbers could be anything..2, 4, 6, 8, 10, 12, 14, 16, 18....Median = 10102, 104, 106, 108, 110, 112, 114, 116, 118....Median = 110Insufficient(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is -4.Let the least negative integer be xx.The least positive integer will be 2, if the set contains positive integers.But let us take there are NO positive integers, then x+0=−4.....x=−4x+0=−4.....x=−4sequence would be -4, -2, 0, 2, 4, 6, 8, 10, 12 --- But this contains positive integers too, while we took the least positive integer as none.So x+2=−4...x=−6x+2=−4...x=−6, and sequence is -6, -4, -2, 0, 2, 4, 6, 8, 10---Median = 2.Sufficient B

What is the median of the nine consecutive even integers in a certain list?(1) The median of the integers in the list is greater than 0.(2) Of the integers in the list, the sum of the least of the negative integers and the least of the positive integers is -4.

√What is the ratio of the average (arithmetic mean) height of students in class X to the average height of students in class Y?(1) The average height of the students in class X is 120 centimeters.(2) The average height of the students in class X and class Y combined is 126 centimeters.Average height of students in class X: xaverage=sum of the heights in Xnxaverage=sum of the heights in Xn, where nn is the # of students in class X;Average height of students in class Y: yaverage=sum of the heights in Ymyaverage=sum of the heights in Ym, where mm is the # of students in class Y.Question: xaverageyaverage=?xaverageyaverage=?Each statement alone is clearly insufficient.From (1): xaverage=120xaverage=120;From (2): 126=xaverage∗n+yaverage∗mn+m126=xaverage∗n+yaverage∗mn+m;Above information is not sufficient to calculate yaverageyaverage, so we can not get the ratio. If you try different values for mm and nn you'll get different values of yaverageyaverageAnswer: E.

What is the ratio of the average (arithmetic mean) height of students in class X to the average height of students in class Y?(1) The average height of the students in class X is 120 centimeters.(2) The average height of the students in class X and class Y combined is 126 centimeters.

*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)

*√Let the hardcover books sold be h, so softcover books sold become 2h.For revenue we require the average prices too, so let the average price of softcover and hardcover be x and y.We are looking for : Is 2hx>hy? Or Is 2x>y?(1) The average (arithmetic mean) price of the hardcover books sold at the store yesterday was $10 more than the average price of the softcover books sold at the store yesterday.y=x+10.So the question becomes: Is 2x>x+10? OR Is x>10?We don't know whether x is greater than 10 or not.Insufficient(2) The average price of the softcover and hardcover books sold at the store yesterday was greater than $14.Total price = h*y+2h*xTotal books = h+2h = 3hAverage = hy+2hx3h=y+2x3>14.........y+2x>42hy+2hx3h=y+2x3>14.........y+2x>42We cannot say anything about relation of x and y.InsufficientCombinedy+2x>42 and y=x+10x+10+2x>42........3x>32........x>10.67>10x+10+2x>42........3x>32........x>10.67>10SufficientC

Yesterday Bookstore B sold twice as many softcover books as hardcover books. Was Bookstore B's revenue from the sale of softcover books yesterday greater than its revenue from the sale of hardcover books yesterday?(1) The average (arithmetic mean) price of the hardcover books sold at the store yesterday was $10 more than the average price of the softcover books sold at the store yesterday.(2) The average price of the softcover and hardcover books sold at the store yesterday was greater than $14.

√We need to find (x+z)/2Statement 1. x-y = y-z . this means z, y, x are in Arithmetic Progression (increasing order also)..And x+z = 2y or (x+z)/2 = y. But we don't know the value of y, so insufficient.Statement 2. Clearly insufficient.Combining the two statements: Let the three AP terms: z = z, y = z+d, x = z+2dNow from second statement we are given that x^2 - y^2 = zor (z+2d)^2 - (z+d)^2 = zSolving we get 2zd + 3d^2 = z.. But this is NOT sufficient to find the value of z/d (thus that of y too) so the question cannot be answered.Hence E answer

√If x, y, and z are positive numbers, what is the value of the average (arithmetic mean) of x and z ?(1) x - y = y - z(2) x^2 - y^2 = z