GMAT - Divisibility, Inequalities, Min Max Stats 4

B. √

11 310

*C

If x^2 = y^2, is true that x > 0 ?(1) x = 2y+1(2) y <= -1

C √

Is xy > 0?(1) x - y > -2(2) x - 2y < -6

E√

Is |a|=|ab|?(1) a ≠ b(2) a^10 - b^10 = 0

C Stats (AVG)

Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?1) the Model P televisions sold for $30 less than the Model Q televisions2) Either p=120 or q=120

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

C

The number of stamps that Kaye and Alberto had were in the ration of 5:3 respecctively. After Kaye gave Alberto 10 of her stamps, the ration of the number of Kaye had to the number of Alberto had was 7:5. As a result of the gift, Kaye had how many more stamps than AlbertoA. 20B. 30C. 40D. 60E. 90

C

What is the greatest prime factor of 6^8−3^8 ?A) 3B) 11C) 17D) 19E) 31

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)

D

How many positive 5-digit integers have the odd sum of their digits?A. 9*10^2B. 9*10^3C. 10^4D. 45*10^3E. 9*10^4

B

If M and N are positive integers greater than 1, does M have more unique prime factors than N?(1) 2N/M is an integer.(2) N^2/M is an integer.

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

So it is obvious that statement 2 is sufficient right away.But statement 1 is tricky to miss in a hurry.(1) x+y=2xx+y=2xy=xy=xorx−y=0x−y=0Hence St 2 is also sufficient. D

What is the value of x^2 - y^2?(1) x + y = 2x(2) x + y = 0

A*

If ab≠0, is |a−b|>|a+b|?1) ab<02) a>b

A *

If x is an integer, what is the value of x?(1) x^2 - 4x + 3 < 0(2) x^2 + 4x +3 > 0

1)z>2567..so there a re infinite values.Not Suff2)rounder off to 2600so z can range from 2550 to 2649(this is the main crux).Not suff1+2 same case as 2 leaving some numbers i.e 2568 to 2649E

What is the hundreds digit of the integer Z?1. 10z > 25,6702. z rounded to the nearest hundred is 2,600 Show Answer

The mean of a set is equal to the sum of terms divided by the number of terms in the set. Therefore,y + 3 = 10y = 7Given that y = 7, the terms of the set can now be ordered from least to greatest:6 - 4y, 6, y, y + 6, 2y, 6y à -22, 6, 7, 13, 14, 42The median of a set of six terms is the mean of the third and fourth terms (the two middle terms). The mean of the terms 7 and 13 is 10. Plug 7 for y into the answer choices and look for a match. Only (y + 3) gives a value of 10.The correct answer is B.

6, y, y + 6, 6 - 4y, 6y, 2yFor the set of terms shown above, if the mean of the set equals 10 then, in terms of y, the median must be y + 3 y y + 2

Stats - AVG. Statement 1: m = 52This is not enough to determine the ratio of m:f.Not sufficient. Statement 2: (9.8m + 9.1f)/(m + f) = 9.3--> 9.8m + 9.1f = 9.3m + 9.3f.--> 0.5m = 0.2f--> m/f = m : f = 2:5SufficientThe correct answer is B.

At a certain company, average (arithmetic mean) number of years of experience is 9.8 years for male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees and the number of the company's female employees?(1) There are 52 male employees at the company(2) The average number of years of experience for the company's male and female employees combined is 9.3 years.

The average to rise by $400, from $20,600 to $21,000, December donation should compensate those $400 for each of the first 11 months, so should have 11*400 more than $21,000: 11*400 + 21,000 = $25,400.Answer: D.

During the first 11 months of a recent year, a certain charitable organization received an average of $20,600 per month in donations. How much did the organization receive in donations during December of that year if the average amount of donations per month for the entire year was $21,000?A. $21,400B. $21,600C. $24,000D. $25,400E. $25,800

Solution: A. Note the given information, that b > a and both are positive. The only way, then, for cb to be less than ca, as statement 1 says, is for c to be negative. This is probably easiest to see via the inequality rule: If c were positive, then multiplying both sides by c in the given inequality b > a would keep the sign in the same direction (cb > ca). For that sign to flip, c must be negative. Statement 2 is a trap - clearly not sufficient on its own, it's designed to make you think "but what if c were a fraction". But recognize that it doesn't matter, as the algebra above proves. Statement 1 alone is sufficient, and the correct answer is A.

If 0 < a < b, is c < 0? cb < ca |c| > 1

E. *Any two digit integer nn can be expressed as n=10x+yn=10x+y, where xx and yy are digits, x>0x>0.Given: a=10x+ya=10x+y and b=10y+xb=10y+x. Q: a+b=11(x+y)=?a+b=11(x+y)=?(1) a−b=10x+y−10y−x=45a−b=10x+y−10y−x=45 --> x−y=5x−y=5. Multiple choices for x and y: (9,4), (8, 3), (7, 2), (6, 1). Not sufficient.(2) x−y=5x−y=5. Same info as above. Not sufficient.(1)+(2) No new info. Not sufficient.

If a and b are two-digit numbers that share the same digits, except in reverse order, then what is the sum of a and b?(1) a-b=45(2) The difference between the two digits in each number is 5.

Let us begin with B(n-2) * (n+2) = n^2 - 4since k<100 we have 96, 77, 60, 45, 32, 21, 12, 5 not sufficientfor A; from the above list we have 7x11, 7x3 in addition to possible numbers >100; hence not sufficientA & B together we have 77, 21 not sufficient E

If k = (n + 2)(n - 2), where n is an integer value greater than 2, what is the value of k?(1) k is the product of two primes(2) k < 100

*Is x^5 > x^4?Is x5>x4x5>x4? --> reduce by x^4: is x>1x>1?(1) x^3 > −x --> x(x2+1)>0x(x2+1)>0. Since x2+1x2+1 is always positive, then the second multiple, x, must also be positive. So, this statement implies that x > 0. Not sufficient.(2) 1/x < x --> multiply by x^2 (it has to be positive, so we can safely do that): x<x3x<x3 --> x(x2−1)>0x(x2−1)>0 --> (x+1)x(x−1)>0(x+1)x(x−1)>0 --> −1<x<0−1<x<0 or x>1x>1. Not sufficient.(1)+(2) Intersection of ranges from (1) and (2) is x>1x>1. Sufficient.Answer: C.

Is x^5 > x^4?(1) x^3 > −x(2) 1/x < x

S1 - r=s=1, therefore r+s=2; r-s=0; r x s=1; r/s = 1hence greatest is r+s.S2 - r-s is the least.When r=s=1, r+s is greatest.Suppose, when r=3,s=2r+s=5; r-s=1; r x s=6; r/s = 3/2 which gives r x s as the greatest.Since we get two different values, so B not sufficient.Hence answer is A.

Of the numbers r + s, r - s, r*s, and r/s, which is greatest?(1) r = s =1(2) r - s is the least of the numbers

ratio is 4:5:7, so if we have a common ratio as 'a', total becomes 4a+5a+7a = 16a..so either we require any of the values of three or common ratio 'a'.. lets see the statements- 1) The number of vans sold at the dealership last month was between 10 and 20.so 7a is between 10 and 20 , so ONLY value of 'a' can be 2..total sedan = 32suff2) The number of sedans sold at the dealership last month was less than 10now sedans are 4a, so it can have either 4 or 8 as its number and thus 'a' can be 1 or 2insuffA

The only types of vehicles sold at a certain dealership last month were sedans, trucks, and vans. If the ratio of the number of sedans to the number of trucks to the number of vans sold at the dealership last month was 4:5:7, respectively, what was the total number of vehicles sold at the dealership last month?1) The number of vans sold at the dealership last month was between 10 and 20.2) The number of sedans sold at the dealership last month was less than 10.

E

Is P + Q > 1/P + 1/ Q?(1) P < Q < 1(2) PQ < 1

E Stats - AVG

x, 3, 1, 12, 8If x is an integer, is the median of 5 numbers shown greater than the average of 5 numbers?(1) x > 6(2) x is greater than median of 5 numbersShow Answer

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.

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

%. let me set up the equations again and bring the answer for uBTW answer is 12%cost price =$60let actual retail price be R(the above price is not retail , its wholesale so don't confuse )Retail price > Whole sale price(always)so 40%discount on R =0.6R(selling price of th product )so question saysselling price -cost price=profit0.6R -60=25% of 60(whole sale price as per question)=15R=125question asks ( profit/actual retail price ) *10015/125 *100=12%

An electronics store purchased a CD player at a wholesale price of 60$ and then sold it at a 40 percent discount off the original retail price. If the store made a 25 percent profit on the wholesale price of the CD player, what was the store's profit as a percent of the original retail price?A. 10%B. 12%C. 15%D. 18%E. 20%

*stats. There are seven different dishes (4 main dishes and 3 side dishes), each with a different integer cost; we cannot use just two variables to represent the two types of dishes. Each of the four main dishes is more expensive than any of the side dishes. The seven dishes add up to $91. The question asks for the price of the most expensive side dish. This is a statistics question, so it will be useful to consider different cases. (1) SUFFICIENT: Since this problem involves 7 different integers, some of which must be larger than others, it's a good idea to test extreme possibilities. We're told that the most expensive main dish is $16 so try maximizing the cost of the other three: $15, $14, and $13. In this case, the total cost of the four main dishes is $58, leaving $33 to split among the three side dishes. The prices of the three side dishes must be different integers and all must cost less than $13. Again, try maximum values first: $12, $11 and $10 are the largest possible prices for the three side dishes and these indeed sum to $33. However, note that this maximized case is the only case that will work - otherwise the 3 side dishes will not sum to $33 (and the total will not sum to $91). If we back up and try something other than the maximum values for the four main dishes, the main dishes will then sum to less than $58 and the lowest price of these main dishes will be something less than $13. This will mean we'll need the sum of the three side dishes to be greater than $33, and we'll have to achieve that with numbers smaller than 12, 11, and 10. That's impossible! Thus, $12 is the price of the most expensive side dish. Conceptually, we could have started with the average of these 7 dishes that sum to $91, that is $91/7 or $13. With the greatest statistic at $16 (only $3 above the mean), we might have understood that the other 6 prices can't be that far away from the mean (the mean is the balancing point for all of the statistics in the set). The set of consecutive integers from $16 to $10 would have been the first to try. After that worked (13 is the mean of these 7 integers and 7 x 13 = 91), we could have proved that all other cases would decrease the sum to less than $91. (2) SUFFICIENT: Since this problem involves 7 different integers, some of which must be larger than others, it's a good idea to test extreme possibilities. We're told that the least expensive side dish costs $10, so this time try minimizing the cost of the other two: $11, and $12. The side dishes would then cost a total of $33, leaving $58 for the main dishes. Since the minimum values for each of the four side dishes, i.e. $13, $14, $15, and $16 already sum to $58 there is no way to choose any larger values for these prices, which also means that we also couldn't have chosen any larger values for the price of the side dishes. Thus, $12 is the price of the most expensive side dish. Conceptually, we could have started with the average of the 7 dishes that sum to $91, that is $91/7 or $13. With the smallest statistic at $10 (only $3 below the mean), we might have understood that the other 6 prices can't be that far away from the mean (the mean is the balancing point for all of the statistics in the set). The set of consecutive integers from $16 to $10 would have been the first to try. After that worked (13 is the mean of these 7 integers and 7 x 13 = 91), we could have proved that all other cases would increase the sum to greater than $91.Note that this is a C-trap. Knowing that the greatest and smallest values of the set of 7 different integers were only 6 apart would imply that they must be consecutive integers. However, it was possible to get the answer with each statement alone. The correct answer is (D).

At a restaurant, a group of friends ordered four main dishes and three side dishes at a total cost of $91. The prices of the seven items, in dollars, were all different integers, and every main dish cost more than every side dish. What was the price, in dollars, of the most expensive side dish?(1) The most expensive main dish cost $16.(2) The least expensive side dish cost $10.

*p and q are different two-digit prime numbers with the same digits, but in reversed order. What is the value of the larger of p and q?Given: p=prime=10x+yp=prime=10x+y and q=prime=10y+xq=prime=10y+x, for some non-zero digits xx and yy (any two digit number can be expressed as 10x+y10x+y but as both pp and qq are two-digit then xx and yy must both be non-zero digits).(1) p + q = 110 --> (10x+y)+(10y+x)=110(10x+y)+(10y+x)=110 --> x+y=10x+y=10. Now, if we were not told that pp and qq are primes than they could take many values: 91 and 19, 82 and 28, 73 and 37, 64 and 46, ... Thus the larger number could be: 91, 82, 73, or 64. But as we are told that both pp and qq are primes then they can only be 73 and 37, thus the larger number equals to 73. Sufficient.(2) p - q = 36 --> (10x+y)−(10y+x)=36(10x+y)−(10y+x)=36 --> x−y=4x−y=4. Again, if we were not told that pp and qq are primes than they could take many values: 95 and 59, 84 and 48, 73 and 37, 62 and 26, 51 and 15. Thus the larger number could be: 95, 84, 73, 62, or 51. But as we are told that both pp and qq are primes then they can only be 73 and 37, thus the larger number equals to 73. Sufficient.Answer: D.

p and q are different two-digit prime numbers with the same digits, but in reversed order. What is the value of the larger of p and q?(1) p + q = 110(2) p - q = 36 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 one ALONE is sufficient.EACH statement ALONE is sufficient.Statements (1) and (2) TOGETHER are NOT sufficient.

(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*. A first glance reveals that n must be an integer, so those terms 10n and 102n are powers of 10 (e.g. 1,000 or 100 or 10 or 110110 or 11001100, etc.). When powers of 10 are multiplied with decimals or integers, the problem is testing Decimals, specifically the decimal placement and magnitude of the resulting number. Ignoring decimals for a moment, notice that 63 is a multiple of 9, so simplifying in Statement (1) will likely pay off. Also, in Statement (1), the upper limit 0.5 is 101 times the lower limit of 0.05; the range of possibilities only spans one power of 10, so maybe there is only one solution for the exponent on the 10. (1) SUFFICIENT: Simplify mkmk first, using the equations from the question stem. mk=6.3 × 102n0.9 × 10n=6.30.9×102n10n=639×102n−n=7×10nmk=6.3 × 102n0.9 × 10n=6.30.9×102n10n=639×102n−n=7×10n n mk=7×10nmk=7×10n Valid case if between 0.05 and 0.5 0 7 too big -1 0.7 too big -2 0.07 valid -3 0.007 too small The only mkmk value between 0.05 and 0.5 is 0.07, so the only possible answer is n = -2 and this statement is sufficient. Eliminate choices (B), (C), and (E). (2) INSUFFICIENT: Because 0.02 is less than 0.9, the exponent n in the expression for k must be a negative number. When solving an inequality such as 0.9 × 10n < 0.02, algebra might be unnecessary work. Because n is restricted to integers, and further restricted by this statement to just negative integers, it is probably most efficient to Test Cases. If n = -1, k = 0.9 × 10-1 = 0.09 (too big...n can't be -1) If n = -2, k = 0.9 × 10-2 = 0.009 < 0.02 (valid case) If n = -3, k = 0.9 × 10-3 = 0.0009 < 0.02 (valid case) Since more than one value of n is possible, this statement is not sufficient. The correct answer is (A).

If n is an integer, m = 6.3 × 102n, and k = 0.9 × 10n, what is the value of n? (1) 0.05<mk<0.50.05<mk<0.5 (2) k < 0.02

Divisibility. C

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?(1) When x-y is divided by 5, the remainder is 1(2) When x+y is divided by 5, the remainder is 2

for this problem, i factored 2000 --> 2, 2, 2, 2, 5, 5, 5... or (5*3)(2*4).based on the factoring, x = 5 and y = 2... so xy = (5)(2) = 10the correct answer is d.

If x and y are positive integers, and x^3*y^4 = 2,000, which of the following is the value of xy?A. 2B. 4C. 8D. 10E. 20

We are given that (x^2)y = z^3, and we need to determine whether z^3 > 0. Recall that the values of z^3 and z have the same sign; thus, we need to determine whether z > 0.Statement One Alone:x(y^2) > 0Since y^2 is always nonnegative, x(y^2) > 0 means x > 0. However, since we cannot determine the sign of y, we cannot determine whether z^3 = (x^2)y is greater than 0.For example, if y > 0, then z^3 > 0; however, if y < 0, then z^3 < 0. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.Statement Two Alone:yz > 0Since yz > 0, we know that either both y and z are positive or both y and z are negative. However, since we cannot determine the sign of y, we cannot determine whether z^3 = (x^2)y is greater than 0.For example, if y > 0, then z > 0, and hence z^3 > 0; however, if y < 0, then z < 0, and hence z^3 < 0. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.Statements One and Two Together:Using the two statements together, we still cannot determine the sign of y and thus cannot determine whether z^3 > 0.Answer: E

If x^2y = z^3, is z^3 > 0 ?(1) xy^2 > 0(2) yz > 0

√ C. Using the two-case structure for absolute values, you have two possibilities as to the value of 2x+52x+5 . Either: 2x+5=3x-22x+5=3x-2 , in which case you can add two and subtract 2x2x from both sides to get: 7=x7=x (which doesn't match an answer choice). Or: 2x+5=−(3x-2)2x+5=−(3x-2) , in which case you need to distribute the negative across the parentheses, leaving: 2x+5=2-3x2x+5=2-3x . In this case, add 3x3x to both sides and subtract 5 from both sides to get: 5x=−35x=−3 x=−35x=−35 .As this matches answer choice C, C is correct.

If |2x + 5| = |3x − 2|, which of the following is a possible value of x?(a) -7(b) -19/5(c) -3/5(d) 3/5(e) 5

A

In the two-digit integer RS, R is the tens digit, and S is the units digit. What is the units digit of S^4?(1) RS is divisible by 2 but not by 5.(2) R = 5.

Per statement 1, m<5 and n<2 ---> no if m=-10, n = -5 but yes if m=1 and n =1 . Not sufficientPer statement 2, 1<m<3 and n2<25n2<25 ----> 1<m<3 and -5<n<5. Again you get 2 different answers for (m,n)=(2.9,4) or (1.5,-1). Thus not sufficient.Combining the 2 statements you get, 1<m<3 and -5<n<2, giving you -15<mn<6 and thus you get an unambiguous "no" and hence C is the correct answer. √

Is mn < 10?(1) m < 5 and n < 2(2) 1 < m < 3 and n^2<25

C

Is x < y ?(1) x^2 - y^2 < 0(2) -x - y < 0

1) The fisherman who caught the third-most fish caught 11 fish.for making the higher most catches the lowest posible, let everyone from thirdmost to lowest caught equal number of fishes, 11..so the two topmost will catch= 160-11*11=3939 will be caught by 2, which means atleast 1 caught>19Suff(2) The fisherman who caught the second-most fish caught 12 fishsame as (1), let all from second most to lowest catch equal number of fishes= 12*12=144so the highest will catch 160-144=16 >15SuffD

On a fishing expedition, a group of 13 fishermen caught a total of 160 fish. \ Did any one fisherman catch more than 15 fish?(1) The fisherman who caught the third-most fish caught 11 fish.(2) The fisherman who caught the second-most fish caught 12 fish

D. *As in the case of many—perhaps most—problems involving multiple absolute value signs, it is impractical to try to solve this algebraically. Instead, test some real numbers. Consider the effect of substituting different signs (and, where relevant, different sizes) of a and b into the expressions. Because the question asks which choice CANNOT have a negative value, try to find number combinations that will lead to a negative value. One possible way: if the answer consists of one expression minus another expression (as is the case in these answers), try to make the first expression either 0 or as small as possible. (A) If a = 1 and b = -1, then this answer choice yields |0| - |2| = -2. This choice can be negative so it is incorrect. In general, if a and b have opposite signs, then |a - b| will have a larger magnitude than |a + b|, making the quantity |a + b| - |a - b| negative. (B) If a = 1 and b = -1, then this choice yields |0| - |1| = -1. This choice can be negative so it is incorrect. (C) If a = 1 and b = -2, then this choice yields |0| - |-1| = 0 - 1 = -1. This choice can be negative so it is incorrect. (D) Correct. You might try individual numbers here, as you did for the others, but you would always get a 0 or positive result. After trying a couple but not finding a negative, move on to answer (E) to see whether you can disprove that one (in which case, this answer will be the only one left). Here's how to prove it definitively: Note that a2 must be 0 or positive, as must be b2. If either variable equals 0, then the value of the expression 2|ab| must also be 0, in which case the entire expression a2 + b2 - 2|ab| must equal either 0 or positive.If, on the other hand, neither variable equals zero, then the expression 2|ab| must be positive. Note that a2 can also be written |a|2. Rewrite this expression as:|a|2 + |b|2 - 2|ab|= |a|2 - 2|ab| + |b|2= (|a| - |b|)2This expression is a perfect square (everything inside the parentheses gets squared) and so must be either zero or positive. (E) This is the toughest one to disprove, because the little shortcut of making the first expression equal 0 doesn't work here. If a = b = 1/2, then this choice yields |1/4| - 1/2 - 1/2 = -3/4. This choice can be negative so it is incorrect. In general, if a and b are between 0 and 1, the expression |a3 + b3| - a - b will be negative. The correct answer is (D).

Which of the following expressions CANNOT have a negative value? |a + b| - |a - b||a + b| - |a||2a + b| - |a + b|a2 + b2 - 2|ab||a3 + b3| - a - b

If n<=150Then middle two numbers are 150 and 200, hence median is 175If n>=250Then middle two numbers are 200 and 250, hence median is 225These define the max and min median possible. If n is between 150 and 250, the median will be between 175 and 225In conclusion, median has to be between 175 & 225

150, 200, 250, nWhich of the following could be the median of the 4 integers listed above?I. 175II. 215III. 235

*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 ?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.

C*This problem is annoying because of the number of terms in the list; it's hard to wrap your head around 20 integers. Check the statements to see whether you can think through the problem using a smaller list, or whether it really is necessary to stick with a list of 20. In the case of both statements 1 and 2, the full size of the list doesn't matter; you can think the problem through using an easier list (say, 10 or even 5 numbers) that still represents the basic principles in question. (1) NOT SUFFICIENT: If the list consists of the numbers 2, 4, 6, 8, 10, then all of the values are different. If any value is increased by 1, the list will still consistent of five different values, so this scenario satisfies the statement. This list does not contain two consecutive integers, so the answer to the question is no. If, on the other hand, the list consists of four "1"s and a "2", then there are only 2 different values (1 and 2). If any of the 1's is increased, the result is 2, which is already in the list, so there are still two different values. If the 2 is increased, then the list will contain four 1's and a 3, and so the list will still contain only two distinct values. In this case, the original list does contain two consecutive integers (1 and 2), so the answer to the question is yes. Because there are two conflicting answers to the question (no and yes), this statement is not sufficient. (2) NOT SUFFICIENT: A list containing four 1's and a 2 contains two consecutive integers (1 and 2). If the list contains four 1's and a 3, then it doesn't contain any consecutive integers. Because there are two conflicting answers to the question, this statement is not sufficient. (1) AND (2) SUFFICIENT: Statement 2 indicates that at least one value occurs twice; call that value a. Statement 1 indicates that increasing any value in the list by 1 will not change the number of distinct values in the list. In this case, then, increasing one of the a values by one, to a + 1, will still leave you with a in the list (since there are at least two a values) as well as a + 1. The value of a + 1, then, must already have been in the original list; if it wasn't, then you would have just added a new value without getting rid of an old value, and statement 1 forbids this. For example, if the original list is {1, 1, 2, 4, 6}, then a = 1 and there are four distinct values in the list. Changing one of the 1's to 2 makes the list {1, 2, 2, 4, 6} and there are still four distinct values. This list does contain two consecutive integers (1 and 2). If the original list were {1, 1, 3, 5, 7}, then a = 1 and there are four distinct values in the list. Changing one of the 1's to 2 makes the list {1, 2, 3, 5, 7}, but now there are 5 distinct values in the list! This is not allowed, according to statement 1. As a result, whatever a is, a + 1 must also be in the original list. The original list must contain at least one pair of consecutive integers. The correct answer is (C).

A list contains twenty integers, not necessarily distinct. Does the list contain at least two consecutive integers? (1) If any single value in the list is increased by 1, the number of different values in the list does not change. (2) At least one value occurs more than once in the list.

Ratio - 90% of the exercise should be completed in 24min 18 sec (1458 sec)When she completes 40% of her exercise timer should show the time remaining for 60% of her exercise.If 90% requires 1458 sec then 60% requires --> (60 * 1458)/90 = (2/3)*(1458) = 2 * 486 = 972 secsRequired answer = 972/60 = 16 min 12 secAnswer: E

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?A. 10 min 48 secB. 14 min 52 secC. 14 min 58 secD. 16 min 6 secE. 16 min 12 sec

*The quadratic expression k2 + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.The question might therefore be rephrased "Is k odd, OR is k + 2 divisible by 4?" Note that a 'yes' to either of the conditions would suffice, but to answer 'no' to the question would require a 'no' to both conditions.(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive 'no' to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.The correct answer is A.

For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?(1) k is divisible by 8.(2) (k + 1)/3 is an odd integer

If an integer ends in 0, then that integer is divisible by 10; when that integer is divided by 10, the final 0 disappears. Therefore, the number of 0's at the end of an integer is exactly the number of times that integer is divisible by 10.Since 10 = 2 × 5, this number (the number of times the integer is divisible by 10) is the smaller of the powers to which 2 and 5 are raised in the integer's prime factorization. For instance, an integer whose prime factorization contains 28 × 57 and an integer whose prime factorization contains 27 × 58 will both end with seven 0's.In a factorial, such as 60!, there are many more 2's than 5's in the prime factorization: the factorial product has more numbers contributing 2's than 5's, and more of those numbers contribute repeated factors. Therefore, the question can be rephrased: How many 5's are in the prime factorization of (60!) ?60! is the product of all the integers from 1 to 60, inclusive. Each of the twelve multiples of 5 in this range - 5, 10, 15, ..., 60 - contributes a 5 to the prime factorization. Additionally, a second 5 is contributed by each of the two multiples of 25 (25 and 50). Therefore, the prime factorization of 60! contains fourteen 5's, and so 60! will end with fourteen 0's.The correct answer is C.

If 60! is written out as an integer, with how many consecutive 0's will that integer end? 6 12 14 42 56

*(1) The tens digit of N is 5.N = a5c. This implies that 5 is a factor of N, which means that the units digit of N (c) must be 5. c cannot be 0 since we are told that each digit of N is a factor of N itself and 0 is not a factor of any number. Thus we have that N = a55. Because N is odd, then a to be a factor of N, must be odd too. N can be:355;555;755;955.Only 555 satisfies all the conditions: each digit of N>200 is a factor of N itself . Sufficient.(2) The units digit of N is 5.N = ab5 --> more than one values of N are possible: 315, 515, 555. Not sufficient.Answer: A.

If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N? (1) The tens digit of N is 5. (2) The units digit of N is 5. 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 one ALONE is sufficient.EACH statement ALONE is sufficient.Statements (1) and (2) TOGETHER are NOT sufficient

Statement 1: This tells us only about a.If b were 1, then product ab = 17 and it would be prime.If b were 2 (or any other number), then product ab = 34 and it would not be prime.Hence insufficientStatement 2: This tells us only about bBut it says that b = 91 and this is not a prime.So if we multiply a non prime number with any other number, we will get a non prime number.Hence product ab is not prime.Hence SufficientOption B

If a and b are non-negative integers, is ab prime?1) a = 17 2) b = 91

√The problem can be solved using number theory or by testing smart numbers. Testing real numbers is often easier but can take longer. Number theory is often faster as long as the theory is fully understood; if not, number theory becomes impossible.(1) SUFFICIENT: For this statement, number theory is more straightforward. If b is 60% greater than a, then b = a + 0.60a, or b = 1.6a = (8/5)a. Because b is a positive integer, (8/5)a must also be a positive integer. Variable a, then, must be a multiple of 5 (in order to cancel out the 5 on the bottom of the fraction).Further, the numerator 8 indicates that b itself must be a multiple of 8 (because b equals 8 multiplied by a). Since a is a multiple of 5 and b is a multiple of 8, the product ab must be a multiple of 40. The remainder upon dividing ab by 40 will always be zero.(2) INSUFFICIENT: For this statement, the theory is quite complex. Instead, test scenarios (which still isn't easy - this is a hard statement!). Choose pairs of positive integers that fit this statement, then find the product ab and divide by 40 to find the remainder.Note: 40 = 2 × 2 × 2 × 5, so any values for a2b and ab2 must contain these factors. First, pick an easy pair for a and b, perhaps something that does multiply to 40 (see first example in the table below). Next, think how to get a different result. If a2b and ab2 are divisible by 40, then they need to contain all 4 terms (2, 2, 2, and 5); what can we pick so that a and b by themselves do NOT contain these 4 terms? (See below.) aba2bab2abab/4058(52)(2 × 2 × 2)(5)(2 × 2 × 2)24040/40 = 1 r.0210(22)(2 × 5)(2)(2 × 5)22020/40 = 0 r.20 As in the second example, if both a and b contain exactly one 2, then together ab does not contain enough 2's - but the terms a2b and ab2 do, because each one picks up an extra 2 when one of the variables is squared.These values provide two different values for the remainder (0 and 20), so this statement is insufficient.The correct answer is A.

If a and b are positive integers, what is the remainder when ab is divided by 40?(1) b is 60% greater than a.(2) Each of a^2b and ab^2 is divisible by 40.

√Start by rephrasing this Statistics question. A number that is divisible by 2 must be an even integer, and b + 5 is an integer whenever b is an integer. Our rephrased question is therefore: if a is even, is b an integer?(1) INSUFFICIENT: The median of two numbers is the average of those two numbers. Knowing that the average of a and b is not an integer does not help to determine whether b is an integer. A good way to test this statement is to Test Cases to prove insufficiency by showing that the median for a and b can be a non-integer using both integer ("Yes" to the question) and non-integer ("No" to the question) values of b. First, choose an integer value for b: let a = 2 and b = 3. The average of 2 and 3 is 2.5, a non-integer. This is a "Yes" to the original question: in this case, b is an integer. Next, deliberately try to choose a non-integer value of b that will also cause the median of a and b to be a non-integer. Let a = 2 and b = 3.5. The average is 2.75, a non-integer. This answers "No" to the original question: in this case, b is not an integer.(2) SUFFICIENT: Testing cases for this statement proves to be a much more arduous task due to the heightened constraints: three variable terms each with coefficients or added constants and the restriction that the average be even. Therefore it is more advisable to use the Average Formula and look for a pattern.Plug the givens into the average formula: , where n represents an even integer.Simplify this equation to getThis equation indicates that b must be an integer because every piece on the right-hand-side of the equation is an integer. must be an integer because n is an even integer. must be an integer because a is an even integer. And is equal to the integer 5. Therefore, b must be an integer, because adding or subtracting integers results in an integer.The correct answer is (B).

If a is divisible by 2, is b + 5 an integer?(1) The median of a and b is not an integer.(2) The average (arithmetic mean) of 3a, b, and b + 10 is an even integer.

*You can rephrase the question as follows:Is an integer?Is an integer?1002 can also be thought of as (102)2 or 104, which in turn can be broken down into primes: 2454 Thus, an alternative rephrase is as follows: Does the product abc contain four 2's and four 5's?(1) INSUFFICIENT: Simplifying the statement and analyzing it with regards to divisibility, you get ab = 100ab = 2252Thus the product ab contains two 2's and two 5's, but it is uncertain whether the inclusion of c in the product would add the two more 2's and two more 5's needed to satisfy the question.(2) SUFFICIENT: Simplifying the statement by breaking it down into primes, you get:c = (2252)b This means that c contains a set of two 2's and two 5's, b number of times. That is, if b = 1, c contains only two 2's and two 5's, however, if b = 2, c contains two sets for a total of four two's and four 5's, etc. Looking back at the given, a < b < c and all three are positive integers, thus the minimum value for b is 2. That means that c alone must have at least four 2's and four 5's and therefore so will the product abc.NOTE that this is a very tempting C-trap. Statement (1) provides information about a and b, and statement (2) provides information about c, appearing to complete the picture.The correct answer is (B).

If a, b, and c are positive integers such that a < b < c, is a% of b% of c an integer?(1) b = (a/100)^(-1)(2) c = 100^b

For n(n+1) to be a multiple of 4 either n or n+1 has to be a multiple of 4. So, n must be a multiple of 4 or 1 less than a multiple of 4:{1, 2, 3, 4,} {5, 6, 7, 8,} {9, 10, 11, 12,} ..., {97, 98, 99, 100}.As you can see half of the integers from 1 to 100, inclusive, guarantees divisibility by 4.Answer: C.

If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?A. 1/4B. 1/3C. 1/2D. 2/3E. 3/4

The value can be simplified to 12 . Given that x is divisible by 6, for the purpose of solving this problem x might be restated as 6y, where y may be any positive integer. The expression could then be further simplified to12 or24 Therefore each answer choice CAN be a solution if and only if there is aninteger y such that 24 equals that answer choice. The followingtable shows such an integer value of y for four of the possible answerchoices, which therefore CAN be a solution. y Solution 1 24 2 24 3 72 k 24k The answer choice that cannot be the value of is 24. For this expression to be a possible solution, y would have to equal 1/3, which is not a positive integer. Put another way, this solution would require that x = 2, which cannot be true because x is divisible by 6.The correct answer is B.

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of ? 24k 24 24 24 72

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.

1. 40 is the smallest integer that is divisible by 8 and n. ==> n=5,10,20,402. 60 is the smallest integer that is divisible by 12 and n.==> n=5,10,15,20,30,601+2).... n= 5,10,20Ans E

If n is a positive integer, what is the smallest positive integer that is divisible by 6, 15 and n?(1) 40 is the smallest integer that is divisible by 8 and n(2) 60 is the smallest integer that isterm-54 divisible by 12 and n.

D

If n is a positive integer, what is the units digit of the sum of the following series:1! + 2! + ... + n!? (The series includes every integer between 1 and n, inclusive)(1) n is divisible by 4.(2) n^2 + 1 is an odd integer.

A**If q, s, and t are all different numbers, is q < s < t ?(1) t - q = |t - s| + |s - q|Notice that the right hand side is positive (it's the sum of two absolute values, so two non-negative values, in fact, in our case two positive values, since we know that the variables are distinct). Thus the left hand side must also be positive, which means that t > q. So, we can have 3 cases for s:a. ---s---q-------t-------In this case s<q<ts<q<t:t−s>0t−s>0 and s−q<0s−q<0, which would mean that |t−s|=t−s|t−s|=t−s and |s−q|=−(s−q)|s−q|=−(s−q) (recall that |x| = x when x > 0 and x = -x when x <= 0).So, |t−s|+|s−q|=(t−s)−(s−q)=t−2s+q|t−s|+|s−q|=(t−s)−(s−q)=t−2s+q.So, in his case we'd have t−q=t−2s+qt−q=t−2s+q or q=sq=s. But we are told that q, s, and t are all different numbers, so this case is out.b. -------q---s---t-------In this case q<s<tq<s<t:t−s>0t−s>0 and s−q>0s−q>0, which would mean that |t−s|=t−s|t−s|=t−s and |s−q|=s−q|s−q|=s−q. So, |t−s|+|s−q|=(t−s)+(s−q)=t−q|t−s|+|s−q|=(t−s)+(s−q)=t−q.This matches the info given in the statement.c. -------q-------t---s---In this case q<t<sq<t<s:t−s<0t−s<0 and s−q>0s−q>0, which would mean that |t−s|=−(t−s)|t−s|=−(t−s) and |s−q|=s−q|s−q|=s−q. So, |t−s|+|s−q|=−(t−s)+(s−q)=−t+2s−q|t−s|+|s−q|=−(t−s)+(s−q)=−t+2s−q.So, in his case we'd have t−q=−t+2s−qt−q=−t+2s−q or t=st=s. But we are told that q, s, and t are all different numbers, so this case is out.Only q < s < t case is possible. Sufficient.(2) t > q. Not sufficient.Answer: A.

If q, s, and t are all different numbers, is q < s < t ?(1) t - q = |t - s| + |s - q|(2) t > q

If x represents the number of positive factors of integer y, is x odd?The question asks whether the number of factors of y is odd. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square. So, the question asks whether y is a perfect square.(1) y = n! where n is a positive integer greater than 1. Among factorials, only 0! and 1! are perfect squares. So, y is not. Sufficient.(2) y = m^2 − 1 where m is a positive integer greater than 1 --> y is 1 less, than a perfect square, so not a perfect square (two positive consecutive integers cannot both be prefect squares). Sufficient.Answer: D.

If x represents the number of positive factors of integer y, is x odd?(1) y = n! where n is a positive integer greater than 1(2) y = m^2 − 1 where m is a positive integer greater than 1

E. *Well, they don't tell us how many numbers there are in set S, so let's keep it as simple as possible. It's easier to find the median with an odd number of numbers, so we'll start with sets that have 3 elements (although we might have to try even-numbered sets, too, if we can't find good cases.)(1)Let's say the set is (10, 20, 30). The mean is 20. The median is also 20.Next, try (19, 20, 21). The mean is 20. The median is also 20.It seems like whenever we have one element larger than 20, and one element smaller than 20, the median is 20. That makes sense, because the third number in the set would have to be equal to 20. That number would be in the middle, so it would be the median.What if we try some sets of four numbers, then? There could be two numbers smaller than 20, and two numbers bigger than 20.(0, 0, 40, 40) - the mean is 20, and the median is 20 again (average of 0 and 20, the two middle numbers).BUT, how about (0, 19, 30, 31)? The mean is 20 (I figured this out by making sure the sum was 80) but the median definitely isn't 20.Insufficient.(2) This is definitely insufficient on its own, since we could test almost any set.(1+2) It's pretty easy to find a case where the median is equal to 20, and that fits both statements: (10,20,30) would work. What we need, then, is to find a case where the median isn't equal to 20. That is, we want a case similar to the one that worked for statement (1), above, except where all of the numbers are even.Let's use the same idea - a set that sums to 80 and has 4 elements, but that isn't 'evenly spaced'. One set that works here is (0, 18, 30, 32), which has a mean of 20 but a median of 24.Insufficient.The correct answer is E.

If the mean of set S is 20, what is the median of set S ?1. In set S there are as many numbers larger than 20 as there are numbers smaller than 20.2. All numbers in set S are even integers.

*(1) SUFFICIENT: Statement (1) can be rephrased as follows:-4x - 12y = 0-4x = 12yx = -3yIf x and y are non-zero integers, we can deduce that they must have opposite signs: one positive, and the other negative. Therefore, this last equation could be rephrased as|x| = 3|y|We don't know whether x or y is negative, but we do know that they have the opposite signs. Converting both variables to absolute value cancels the negative sign in the expression x = -3y.We are left with two equations and two unknowns, where the unknowns are |x| and |y|:|x| + |y| = 32|x| - 3|y| = 0Subtracting the second equation from the first yields4|y| = 32|y| = 8Substituting 8 for |y| in the original equation, we can easily determine that |x| = 24. Because we know that one of either x or y is negative and the other positive, xy must be the negative product of |x| and |y|, or -8(24) = -192.(2) INSUFFICIENT: Statement (2) also provides two equations with two unknowns:|x| + |y| = 32|x| - |y| = 16Solving these equations allows us to determine the values of |x| and |y|: |x| = 24 and |y| = 8. However, this gives no information about the sign of x or y. The product xy could either be -192 or 192.The correct answer is A.

If x and y are non-zero integers and |x| + |y| = 32, what is xy?(1) -4x - 12y = 0(2) |x| - |y| = 16

(1) x√<y√x<y. Since both sides of the inequality are positive (the square root from a positive number is positive), then we can safely square: x < y. Directly answers the question. Sufficient.(2) (x−3)2<(y−3)2(x−3)2<(y−3)2. If x=3x=3 and y≠3y≠3, the inequality will hold true: the left hand side will be 0, while the right hand side will be more than 0. Thus, if x=3x=3, y can be less than 3, giving a NO answer to the question, as well as more than 3, giving an YES answer to the question. Not sufficient.Answer: A.Hope it's clear.

If x and y are positive, is x < y?(1) x√<y√x<y(2) (x−3)2<(y−3)2

(1) -1 < x < 1. If x=1/2, then x=1/2, 1/x=2, and x^2=1/4, so 1/x has the greatest value but if x is negative, then both x and 1/x are negative and x^2=positive, so x^2 has the greatest value. Not sufficient.(2) x is negative. For negative x both x and 1/x are negative and x^2=positive, so x^2 has the greatest value. Sufficient.Answer: B. √

If x is a nonzero number, which of the three numbers x, 1/x and x^2 is greatest?(1) -1 < x < 1(2) x is negative

C***This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476If x is an integer, is |x| > 1?First of all: is |x|>1|x|>1 means is x<−1x<−1 (-2, -3, -4, ...) or x>1x>1 (2, 3, 4, ...), so for YES answer xx can be any integer but -1, 0, and 1.(1) (1 - 2x)(1 + x) < 0 --> rewrite as (2x−1)(x+1)>0(2x−1)(x+1)>0 (so that the coefficient of x^2 to be positive after expanding): roots are x=−1x=−1 and x=12x=12 --> ">>" sign means that the given inequality holds true for: x<−1x<−1 and x>12x>12. xx could still equal to 1, so not sufficient.(2) (1 - x)(1 + 2x) < 0 --> rewrite as (x−1)(2x+1)>0(x−1)(2x+1)>0: roots are x=−12x=−12 and x=1x=1 --> ">>" sign means that the given inequality holds true for: x<−12x<−12 and x>1x>1. xx could still equal to -1, so not sufficient.(1)+(2) Intersection of the ranges from (1) and (2) is x<−1x<−1 and x>1x>1. Sufficient.Answer: C.

If x is an integer, is |x|>1.(1) (1-2x)(1+x) < 0(2) (1-x)(1+2x) < 0

√The question "Is |x| less than 1?" can be rephrased in the following way. Case 1: If x > 0, then |x| = x. For instance, |5| = 5. So, if x > 0, then the question becomes "Is x less than 1?" Case 2: If x < 0, then |x| = -x. For instance, |-5| = -(-5) = 5. So, if x < 0, then the question becomes "Is -x less than 1?" This can be written as follows: -x < 1? or, by multiplying both sides by -1, we get x > -1? Putting these two cases together, we get the fully rephrased question:"Is -1 < x < 1 (and x not equal to 0)"? Another way to achieve this rephrasing is to interpret absolute value as distance from zero on the number line. Asking "Is |x| less than 1?" can then be reinterpreted as "Is x less than 1 unit away from zero on the number line?" or "Is -1 < x < 1?" (The fact that x does not equal zero is given in the question stem.) (1) INSUFFICIENT: If x > 0, this statement tells us that x > x/x or x > 1. If x < 0, thisstatement tells us that x > x/-x or x > -1. This is not enough to tell us if -1 < x < 1. (2) INSUFFICIENT: When x > 0, x > x which is not true (so x < 0). When x < 0, -x > x orx < 0. Statement (2) simply tells us that x is negative. This is not enough to tell us if -1 < x < 1. (1) AND (2) SUFFICIENT: If we know x < 0 (statement 2), we know that x > -1 (statement 1). This means that -1 < x < 0. This means that x is definitely between -1 and 1. The correct answer is C.

If x is not equal to 0, is |x| less than 1? (1) x|x| < x (2) |x| > x

E(***** )So you have basically 17!2−16!217!2−16!2, you can rewrite that as 16!216!2*(172172-1)= 16!216!2*(289-1)= 16!216!2*(288)now look at the answers, and what the stem tells you; you're told that ¡n!=n!^2, so 16!216!2 = ¡16!, and 288 is just 12^2 [144]*2,so 16!216!2*(288) = ¡16!(12^2)(2)E

If ¡n! = (n!)2, then ¡17! - ¡16! = ¡1! (¡16!)(¡4!)(2) (¡16!)(12)(2) 172 (¡16!)(122)(2)

Assume; g = number of girls, b = number of boysTotal number of students = b + gg / 4 = (b + g) /66g = 4(b + g) = 4b + 4g2g = 4b2/4 = b/gb/g = 1/2The answer is C. 1 to 2.

In a certain English class, 1/4 of the number of girls is equal to 1/6 of the total number of students. What is the ratio of the number of boys to the number of girls in the class?A. 1 to 4B. 1 to 3C. 1 to 2D. 2 to 3E. 2 to 1

C. *Your reasoning is not correct. Are you saying that all sets with even mean have even median? What about: {1, 1, 4} --> mean=2=evenmean=2=even and median=1=oddmedian=1=odd OR {0.6, 1.2, 4,2, } --> mean=2=evenmean=2=even and median=1.2≠integermedian=1.2≠integer.Is the median of set S even?(1) Set S is composed of consecutive odd integers --> set S is evenly spaced --> for every evenly spaced set mean=medianmean=median. But still insufficient.(2) The mean of set S is even. Insufficient as shown above.(1)+(2) From 1: mean=medianmean=median and from 2: mean=evenmean=even --> mean=median=evenmean=median=even. Sufficient.Answer: C.

Is the median of set S even?(1) Set S is composed of consecutive odd integers(2) The mean of set S is even.

Divisibility: The question asks whether x is divisible by the LCM of 2, 3, 4, 5, and 6, which is 60.(1) x = 10m, where m is a positive integer divisible by each integer from 2 through 5.m is divisible by the LCM of of 2, 3, 4, and 5, which is 60. Thus x = 10*(multiple of 60). Sufficient.(2) 10x = n, where n is a positive integer divisible by each integer from 2 through 9.n is divisible by the LCM of of 2, 3, 4, 5, 6, 7, 8, and 9, which is 2520 (9*7*8*5). Thus 10x = (multiple of 2520) --> x = (multiple of 252). If x = 252, then the answer is NO but if x is say, 252*60, then the answer is YES. Not sufficient.Answer: A.

Is the positive integer x divisible by each integer from 2 through 6 ?(1) x = 10m, where m is a positive integer divisible by each integer from 2 through 5(2) 10x = n, where n is a positive integer divisible by each integer from 2 through 9

D*Is x > 10^10 ?(1) x > 2^34 --> we should compare 234234 and 10101010 --> take the square root from both: we should compare 217217 and 105=100,000105=100,000. Now, 217=210∗27=1,024∗128>100,000217=210∗27=1,024∗128>100,000. Sufficient.OR: 234=(210)3.4=(1,024)3.4>(103)3.4=10(3∗3.4)=1010.2>1010234=(210)3.4=(1,024)3.4>(103)3.4=10(3∗3.4)=1010.2>1010.(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.Answer: D.

Is x > 1010 ?(1) x > 234(2) x = 235

Q: is x/3+3/x>2? Let's check when this statement holds true.x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.Nominator (x-3)^2 is NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.Let's move to the statements:(1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.Answer: C.

Is x/3 + 3/x > 2(1) x < 3(2) x > 1

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.

I'm going with C here.1) and 2) both give values for s and t, respectively. To keep the average at 40, we are able to guess the final unknown values in the set.T = {30, 35, 40, 45, 50}S = {25, 30, 40, 50, 55}There is a greater range with S, thereby showing it has a higher standard deviation.

List S and list T each contain 5 positive integers, and for each list the average (arithmetic mean) of the integers in the list is 40. If the integers 30, 40, and 50 are in both lists, is the standard deviation of the integers in list S greater than the standard deviation of the integers in list T ?(1) The integer 25 is in list S.(2) The integer 45 is in list T.

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