windriver

Stmt1: After reducing this asking price by 10 percent, the jewelry dealer sold the bracelet at a profit of $403. c is the cost price s is the selling price s = c + (40/100)c = 1.4c Now 10% of 1.4c is reduced =>new selling price = 1.4c - (10/100)(1.4c) = 1.26 c profit now is 1.26c-c = 0.26c = 403 => c = 1550 SUFF Stmt2: ----- The jewelry dealer sold the bracelet for $1,953. => 1.4c=1953 => c = 1395 SUFF Ans: D

Stmt1: a^2+b^2+2ab = 25 ab = (25-13)/2 = 6 SUFF Stmt2: a^2b^3 = 108 a^2*b^3=2^2*3^3 Since we are not told a and b should be integers, INSUFF Ans: A

Theater problem: ---------------- Stmt1:During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. Obviously insufficient because we don't know anything about the occupancy of the last 5 rows. INSUFF Stmt2: During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows. Again insufficient because we don't know anything about the occupancy of the first 10 rows. Stmt1 & Stmt2: We know the occupancy of rows 1-20 from Stmt1 and the occupancy of rows 11-25 from Stmt 2. But we don't know what the occupancy rate of the overlap between the two statements (i.e. rows 11-20) is. INSUFF Ans: E Probability problem: ------------------- Stmt1: More than ½ of the 10 employees are women. Number of women can be 6,7,8,9 or 10 6c2/10c2 0.5 INSUFF Stmt2: The probability that both representatives selected will be men is less than 1/10. mc2/10c2 => mc2 => mc2 => mc2 => m(m-1) => 2 => 7 7c2/10c2 0.5 INSUFF Stmt 1 and 2: ------------ INSUFF because from Stmt 2 the number of employees is clearly greater than 5, but that is still insufficient. Ans: E

But would someone be able to get through the combinations in under 2 min? Here's an example, (x+y+z)/(x+y*z). Here are the cases that can apply. x=positive integer, x=negative integer, x=positive fraction, x=negative, x=0. Same for y and z. For each number, you have 5 choices. Total permutations 125. Could someone do 125 computations in 2 min?

The remainder can be positive or negative. However, I think a positive is considered ( I am not 100% sure though). For example, what is the remainder of 10/7 ? Normally you would say 3. But, 10=(14-4). Now if (14-4) is divided by 7, then what is the remainder? 14/7 gives a remainder of 0 and -4/7 gives a remainder of -4. So, would that mean the remainder of 10/7 is -4? or is it 3? Both are correct, but we usually tend to pick the positive one.

Good question...a little confusing. I wish more such questions are posted. Stmt1: The number of voters who did not respond "favorable" for either candidate was 40. So, 40=(unsure+unfavorable) for either candidate or both candidates. So, those who are favorable for either candidate or both candidates=100-40=60 But, from the table, those who are favorable to A=40 and those who are favorable to B=30. So, of the 60, if 40 favor A, and 20 favor B, 10 of those who favor A must also favor B so that B becomes eligible for a share of 30 people. Therefore, 10 people favor both candidates. SUFF Stmt2:The number of voters who responded "unfavorable" for both candidates was 10. unfavorable to both = 10 (unsure + favorable) to both A and B or either A or B + unfavorable exclusively to A or B=90 Cant determine how many favor both A and B. Ans A.

I too find such problems time consuming. If we don't figure out there are multiple answers, we'll get it wrong. There is also no simple way to figure out if multiple answers exist other than trial and error. If anyone knows of a method, I'll be more than happy to know.

What is the best way to pick values. I usually cannot come up with the "right" numbers. For example, in the above problem, we will have to try "large -ve y", "small +ve x", "large -ve x", "small -ve y", "large -ve x", "large -ve y"...you get what I am saying. What is the best strategy to adopt when picking numbers. How do we identify what numbers to pick. Any ideas?

Problem: h=11q+r, r=? Stmt 1: h+10 = 11k => h=11k-10 => h=11k - 22 + 12mod11 => h = 11(k-2) + 12mod11 if q=k-2, then r=1 (12mod 11 =1). SUFF Stmt 2: h-1=11k => h = 11k + 1 if q = k, then r=1 SUFF Ans: D

Stmt1: lcm(x,1)/gcd(y,3) = 6 Cant say anything about the value of y. For example, x=18 and y=3 will satisfy this. Also, x=6 and y=1 will also satisfy this. Stmt 2: x=1, what is y? Can't say. Stmt1 & Stmt2: lcm(1,1)/gcd(y,3) = 6 => gcd(y,3)=1/6 I know of no positive integer that will give us a fraction for gcd with 3.

I am in the same boat as you are jadugar. Infact I've seen your other posts and I have great respect for you. If anything I should be the one who should say what you are saying :)

I don't like B because it says "horse feed restored market equilibrium". It's not horse feed that did that, it's the persuasion that did it. Therefore I think A is the answer.

A for this IMO. "As many X as Y" is the correct idiom. I am also having some difficulty with the GMATPrep question posted above. There are hopeful signs that we are shifting away from our heavy reliance on fossil fuels: more than ten times as much energy is generated through wind power now than it was in 1990. A) generated through wind power now than it was B) generated through wind power now as it was C) generated through wind power now as was the case D) now generated through wind power as it was E) now generated through wind power than was the case

I want to pick D between A and D because in A "have typically seen" could mean they can no longer see it if they visit again. In D, "have typically been able to see" means that they have seen it and there is a very high likelihood of seeing it again if they visited.

Here's my take Stmt1: avg of both classes = (82*37 + 40*74)/(37+40) = 77.84 Let class A be represented by a1, a2, a3.....a37 Let class B be represented by b1, b2, b3......b40 We know a19=80 We know b20+b21=2(78) Combining both classes, we can be sure that there would be 20 b's and 18 a's before you can get to b21 or a19. That makes it 38 students before you can even think of b21, or a19. All we care about is the 39th student in the combined student strength of 77 students. a19 we know is 80 which is greater than the avg of 77.84. How about b21? b21 is atleast equal to 78 (the median of class B). This again is greater than the avg of both classes. Therefore the median of the combined classes (atleast 78) is greater than the average of the combined classes (77.84) Answer A.

n can be either 3, or 5n+3. From the set of answer choices, 5n+3 is only solvable by E

How about a,b,-2,7,5? Here c will turn out to be -2 instead of 3. Unless there are additional constraints placed on the question, like all of them are positive integers, I don't think there is a unique answer to this problem. Once you plugin the constraint that all numbers in the series are positive, then you can get a=2.

It's actually quite simple. Think about it for a moment. 12 is 3*2^2. We know we have 14 numbers contributing a 3, 26 numbers contributing a 2. But we need a 2^2 contribution. That means that we have 26/2 = 13 numbers contributing 2^2 and 14 numbers contributing a 3. Since 13 is more limiting than 14, 13 has to be the answer.

a and b should be of similar sign. That's all it means. So the product is always positive.

C it should be. If the average is 15 and the least number is 2, then we can be sure that not all numbers in the set are 2. Therefore SD cannot be 0.

Something is not right with the question. You can get multiple answers as zukerman pointed out. Take any multiple of 4 >=32 and you will always get a remainder of 0. Any multiple of 4 less than 32 will yield a result that is the number you have chosen. Example, 16^4/32 remainder = 0. 16/32 remainder is 16.

12*11*10*9/2*3*4 = 495

Lets see. Look at the 3 conditions .x = y II. y = 1 III. x and y are prime integers. Should x=y? No. For 3x/y to be a prime that is greater than 2, you can have x=1, and y any one of 3/5, 1, 3/7, 3/11....... Should y=1. Again no as shown above Should x and y be prime integers. Again no as shown above. Ans should threfore be None

Lets see. Look at the 3 conditions .x = y II. y = 1 III. x and y are prime integers. Should x=y? No. For 3x/y to be a prime that is greater than 2, you can have x=1, and y any one of 3/5, 1, 3/7, 3/11....... Should y=1. Again no as shown above Should x and y be prime integers. Again no as shown above. Ans should threfore be None

1. Avg over 3 months = total price for three months/total number of dozens If a is the number of dozens sold in april, m is the number of dozens in May and j is the number of dozens sold in june a = (2/3)m, j=2a => total price = 1.26a + (3/2)a(1.2) + 1.08(2a) => total number of dozens = a + (3/2)a + 2a => avg over 3 months = (1.26a + (3/2)a(1.2) + 1.08(2a))/(a + (3/2)a + 2a) simplify to get 1.16 2. H+c=3.59, H+F=4.4, F=2C. Simplify to get F = 1.62 3. $10/wk, $30/month => 1year = 12 month price = 30 * 12 = 360 => 1year = 52 week price = 52 * 10 = 520 Savings = 520-360 = 160