shamit1982

1) The team has played 2/3 of its games. They have played 20 games. That means 10 games remain. Total of 30 games. 3/4*30 = 22.5. So in order to have 3/4 of the games won they must win at least 23. Therefore they need 6 wins, leaving room for only 4 losses. D 2) Its basically a triangle problem. The hypotunous = 10, Karen to the cafe is x, and dan to cafe is x-2. x^2+(x-2)^2 = 10^2 Solve for x --> I ended up with something like this: (x-8)(x+6) = 0 so x=8 or x=-6. C

5c3 * (1/2)^5 = 5/16 5c3 is the number of ways that have it raining on exactly 3 days. (1/2)^5 is the probability of having 3 rain days and 2 non-rain days.

I think this makes use of the rule that if a number is divisible by 3 the sum of its digits must add up to a number divisible by 3. So, you know for sure that 3m is divisble by 3. You also know that m and 3m have the same digit sum. Therefore m is divisible by 3 as well. I would have went with 3. I don't understand how they have both 9 and 3 as answer choices. If m is a multiple of 9, doesn't that also make it a multiple of 3?

Qn 1: I agree with cippo that you can solve for how many stamps using B with trial and error. Qn3: Math error on my part lsv is correct. a+b = 190....so avg = 95 Qn4: The question asks for the slope of I, not just the slope of k. I think you need both a and b to solve this.

Thanks, that makes sense. I was reading the question incorrectly. I thought r represented the length of the piece of colored glass for some reason. Have to be more careful...

Good question. If we assume replacement then A is suff: n/3n * n/3n * n/3n = n^3/(27*n^3) = 1/27 If the balls are not replaced: A - Insuff B - Insuff but c will work since you have all the variables. To me, the question sounds like its saying that there is no replacment. Since it almost sounds like all 3 were drawn at once. OA?

Please take a look at the problem in the attachment. Maybe I'm missing something, but I think the real answer is not present in the options.

1) A If .15x + .29y = 4.40. The only combination I got to work was y=10 and x=10. Since we are working with a 9 in the hundreths digit, and the total sum has a 0, y has to be a multiple of 5. It can only be 0, 5, 10, 15. Anything else would be over 4.40. The only one that can work is 10. Forcing x=10. 2) D R/B = 3/8 R = (3/8)B --> R = (3/8)*72 = 27 3) 115 x+y+20 / 3 = x+y+20+30/4 + 10 -> solve for x+y = 230 x+y / 2 = 115 4) C Each individual statement was insuff, but together you can figure out slope of k = 3/4. Then solve for (3/4)*I = -1. I = -4/3.

K/A = 5/3 (K-10)/(A+10) = 7/5 K = 5A/3 ((5A/3)-10)/(A+10) = 7/5 Solve for A: A = 90 K/90 = 5/3 ---> K = 150 Answer asks for (K-10) - (A+10) = 40 C

If p and n are positive and p > n, what is the remainder when p^2 - n^2 is divided by 15? 1) The remainder when p+n is divided by 5 is 1. 2) The remainder when p-n is divided by 3 is 1. Not really sure how to approch something like this. Suggestions would be appreciated.

I think the two quotes I highlighted above are contradicting. The mode of the first seven numbers is 135, that means no other number repeated more times then it in the first 7 values. IMO It has to be C.

Yeah, I missed this one as well. Its a tricky one, but I think it makes sense. We know x - y = 1/2 and x/y > 1 One of the restrictions that the second condition poses is |x| > |y|. Another restriction is that x and y must be the same sign. If both values are positive: x has to be greater then y, so this satisfies both conditions. i.e. x=1 y=1/2 If both values are negative: The only way to get x-y = 1/2 is to have |x| i.e. x = -1 y = -1.5 Thats why its C.

The stem says "At least 100 students at a certain high school study Japanese." Therefore you really don't know how many study Japanese, and that is why 1 is insufficient.

I think thankont is correct about Q17. My mistake.

If n and t are positive integers, is n a factor of t? (1) n = 3^(n-z) (2) t = 3^n IMO E. If we knew that 0 Q17: If Meg is 1st bob can be in 4C1 positions Meg is 2nd bob can be in 3C1 positions etc So I'd say 4C1*3C1*2C1*1C1 = 24 (A) OA's please.

The probabilty of 1 being defective and the other one not should be the following: (n/10)*((10-n)/9) + ((10-n)/10)*(n/9) = 7/15 --> 2*(n*(10-n))/90 = 7/15 (Thats where the 2 comes from Swiss_boy) Simplify --> -2n^2 + 20n -42 = 0 (Can use quadratic equation here or...) Div by -2 n^2 - 10n + 21 = 0 factor: (n - 3)(n - 7) = 0 Since the stem already says that n

IMO Put the equation into something more familiar (Y = mx+b) ax + by + c = 0 (Solve for y) --> y = (a/-b)x + (c/-b) 1) a = 4 (4/-b) = 2/3 ---> solve for b --> b = -6 Suff 2) c = -6 y = (a/-b)x + (6/b) Too many unknows Insuff Answer must be A

1) (2^k)*(5^(k-1)) = 2*2^(k-1)*5^(k-1) = 2*(2*5)^(k-1) = 2*(10)^(K-1) 2) This one took me alittle over 2 min to figure out, but here is my approach anyway: Determine that 10^5 = 10000, so 9999 = 10^5-1 and 9999^2 = (10^5-1)^2 = 10^10 - 2*10^5 + 1 Therefore, 9999^2-1^2 = 10^10 - 2*10^5 + 1 - 1 = 10^5(10^5 -2) 3) This one i just plugged in the answer choices, and it took very little time. Hope that helps.

TBAY I was confused by that as well. However, if you continue with the route you were taking, I think you can reach the same answer. Basically, you want to know: (? represents >, a ? [a+ x/100(a)] - y/100[a+ x/100(a)] expand --> a ? a(1+ x/100 - y/100 - xy/10000) divide by a (have to assume a is +ve) --> 1 ? 1 + x/100 - y/100 - xy/10000 sub 1, and mult by 100 --> 0 ? x - y - xy/100 xy/100 ? x - y now you know that ? has to be Hope that helps.

The OA is B. Thanks for the help. For some reason I was under a fog and thought the only 2^6 could equal 64. Was forgetting about 4^3 and 8^2. Thats why I thought it was D, but now I see my mistake. Thanks! :tup:

If someone could explain the answer to the problem below it would be appreciated. This question is from GMATPrep, so please look away if you are planning to take the practice tests later. If the integers a and n are greater then 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a? 1) a^n = 64 2) n = 6 Just to keep things interesting, I will give out OA after a few replies. Thanks!

During a certain month, Dave watched n two-hour movies and r ninety-minute movies. Had he spent the same amount of time watching television, he could have watched 60 thirty-minute television shows. Each of the following is a possible ratio of n to r EXCEPT 3:16 1:2 2:3 (Correct answer) 9:8 3:1