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 [#permalink] New post 23 Jun 2005, 00:09
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By weight, liquid A makes up 7.0 percent of solution I and 14.5 percent of solution II. If 3 grams of solution I is mixed with 2 grams of solution II, then liquid A accounts for what percentage of the weight of the resulting solution?

We're told 3 grams of solution I is used, so liquid A = 7% of 3 grams = 0.21 grams
We're also told 2 grams of solution II is used, so liquid A = 14.5% of 2 grams = 0.29 grams

So the total weight of liquid A = 0.29 + 0.21 = 0.5 grams

The total weight of the solution = 5 grams. Therefore, A would account for 0.5/5 * 100% = 10% of the resulting solution.
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 [#permalink] New post 23 Jun 2005, 00:17
Can the positive integer P be expressed as the product of two integers, each of which is greater than 1 ?

1. 31 < P < 37
2. P is odd


From statement (1), P can either be 32, 33, 34, 35 or 36. For all 5 numbers, P can be expressed as the product of two integers, each of which is greater than 1. Thus, statement (1) alone is suffcient.

From statement (2), all we know is P is odd. If P is 1,3,5, or 7, then it cannot be expressed as the product of two integers, each of which is greater than 1. If P is an odd number from 9 and up, then it is possible to be expressed as the product of two integers, each of which is greater than 1. Thus, statement (2) alone is not suffcient.

The answer is therefore A.
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 [#permalink] New post 23 Jun 2005, 00:39
Is x/m(m^2 + n^2 + k^2) = xm + yn + zk

1) z/k=x/m
2) x/m=y/n


x/m(m^2 + n^2 + k^2) = xm + yn + zk can be simplified as xm + xn^2/m + xk^2/m. For the LHS to be equal to the RHS, xn^2/m = yn => y = xn/m and zk = xk^2/m => z = xk/m. So the question is really asking is y = xn/m and z = xk/m.

From statement (1), we have z/k = x/m => z = xk/m. This is not sufficient as we do not know what is y.

From statement (2), we have x/m = y/n => y = xn/m. This alone is not sufficient as we do not know what is z.

Using both statements, we have y = xn/m and z=xk/m and so we know the LHS = RHS.

Therefore, the answer is C.
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 [#permalink] New post 23 Jun 2005, 00:45
Is sqrt(x-5)^2 = 5-x?

1) -x|x| > 0
2) 5-x > 0


sqrt(x-5)^2 is really just (x-5). So what the question is asking is whether x-5 = 5-x => x = 10. If either statement tells us if x is 10 or any other particular value, then it is sufficient.

From statement (1), for -x|x| to be greater than zero, x must be negative. And if x is negative, then it can't be 10. Therefore, statement (1) is sufficient.

From statement (2), we have 5 - x > 0 => x < 5. We know x has to be less than 5, so x cannot be 10. Statement (2) alone is sufficient as well.

The answer is therefore D.
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 [#permalink] New post 23 Jun 2005, 01:00
Is k^2 + k - 2 > 0

1. k < 1
2. k > -1


Again, this is a yes/no D.S question.

From (1), we're told k is 0 and below. If k= 0, the inequality does not hold. If k = -1, the inequality will also not hold. However, when k = -3 and below, the inequality is true. Thus, statement (1) alone is not sufficient.

From (2), we're told k is 0 and above. For k=0, the inequality holds. If k = 1, then the inequality is false. Thus, statement (2) alone is not sufficient.

Using both statements, we have -1 < k < 1. Thus k can be 0, or a fraction such as -1/2 or 1/2. If k=0, the inequality is false. For k=1/2, the inequality is false. For k = -1/2, the inequality is also false. Thus using both statement, we know the inequality will not hold.

The answer is therefore C.
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 [#permalink] New post 23 Jun 2005, 19:25
Is sqrt(n) > 100

1. sqrt(n+1) > 100
2. sqrt(n-1) > 100


The question is really asking "Is n > 100,00?"

From statement (1), we have (n + 1) > 10000 => n > 9999. So n may or may not be larger than 10000. Statement (1) is therefore not sufficient.

From statement (2), we have (n - 1) > 10000 => n > 10001. So n is definitely larger than 10000. Statement (2) is therefore sufficient.

The answer is therefore B.
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 [#permalink] New post 26 Jun 2005, 21:29
Is 8x = 16 + 2x ?

1) -3x is greater than or equal to -9
2) 2x is greater than or equal to 6


Simplifying the equation in the question gives, 8x = 16 + 2x => 6x = 16 => x = 8/3. So the question is asking is x = 8/3.

From statement 1), -3x >= -9 => x <= 3. So statement (1) is clearly not sufficient since x can be greater than 3, or any value below 3.

From statement 2), 2x >= 6 => x >= 3. Statement (2) is sufficient since we know x is at least 3 and there is therefore no way that x can be 8/3.

The answer is therefore B.
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 [#permalink] New post 26 Jun 2005, 21:42
If 2001 sales increase x% from 2000, and 2002 sales decrease x% from 2001, and 2002 is 9% less than 2000. What is x?

If sales in 2000 is $y, then 2001 will be $(1+x/100)y
2002 will experience x% decrease from 2001, so 2002 sales will be: $(1-x/100)(1+x/100)y

The difference from 2000 will be:
sales in 2000 - sales in 2002
= y - (1-x/100)(1+x/100)y
= y - (1-x^2/10000)y
= y - y + yx^2/10000
=yx^2/10000

This difference expressed as a percentage of the sales in 2000 will be:
yx^2/10000y * 100% = x^2/100% = 9%, so x = 30%.
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 [#permalink] New post 27 Jun 2005, 19:12
Seed mixture X is 40% ryegrass and 60% bluegrass by weight; seed mixture Y is 25% ryegrass and 75% fescue. If a mixture of X and Y contains 30% ryegrass, what percentage of the weight of this mixture is X?

Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.
Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.

So we can now equate the parts of the ryegrass in the mixture as:

0.4X + 0.25(100-X) = 30
0.4X + 25 - 0.25X = 30
0.15X = 5
X = 5/0.15 = 500/15 = 100/3

So the weight of mixture X as a percentage of the weight of the mixture
= (weight of X/weight of mixture) * 100%
= (100/3)/100 * 100%
= 33%

** For most percentage problems, equating one value to 100 is very helpful for a quick solution.
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 [#permalink] New post 27 Jun 2005, 19:33
If both 5^2 and 3^3 are factors of n x 2^5 * 6^2 * 7^3, what is the smallest possible positive value of n ?

25
27
45
75
125


The questions tells us that the prime factorization, n * 2^5 * 6^2 * 7^3 is a multiple of both 5^2 and 3^3. n * 2^5 * 6^2 * 7^3 can be further broken down into n * 2^5 * (2*3)^2 * 7^3 = n * 2^5 * 2^2 * 3^2 * 7^3 = n * 2*7 * 3^2 * 7*3

To be a multiple of 3^3, we need another 3 and a 5^2 in the prime factorization above. So n will have to take on 25*3 = 75.

The answer is therefore D
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 [#permalink] New post 28 Jun 2005, 06:03
Are the integers a, b, and c consecutive?

(1) b - a = c - b.

(2) The average of {a, b, c} is equal to b

From statement (1), we have 2b = a + c. But this does not mean a,b and c are consecutive. We could have a = 1, b = 5, c = 9 and the equation still holds. Thus, statement (1) is not sufficient.

From statement (2), we have (a + b + c)/3 = b => a + b + c = 3b => a + c = 2b. This gets us back to what statement (1) gave, and so statement (2) is not sufficient.

There's really no point analysing for choice C since both statements mean the same thing.

The answer is therefore E.
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 [#permalink] New post 28 Jun 2005, 06:09
Everyday, except Sundays, Paul pays one dollar to get a lotto. Half of the time, he will wager an additional $2 on a lotto. one third of the time, he will win, with his average winnings being $15. What is his average weekly profit/loss in dollars?

We're told:

- Paul plays the lotto 6 days a week. He pays $1 to get a lotto on each of these days, so that's $6.
- For 3 of these days, he will wager an additional $2, so that's $6
- For every 3 days, there will be one day where he wager wins. So for 6 days, he should recoup 2 days worth of winnings - that's $30

So the weekly profit = $30 - $12 = $18
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 [#permalink] New post 28 Jun 2005, 06:14
What is the least value of the three digit integer y?

1) the sum of the three digits is 5
2) y is divisible by 5


From statement (1), we're told the sum of the three digits is 5. The least value will therefore be in the range of 100+. 104 should therefore be the smallest number. Statement (1) is sufficient.

From statement (2), were told y is divisible by 5. So the last digit should be either 0 or 5. So the number can either be 100 or 105. Since we're asked for the least value, therefore y has to be 100. Statement (2) is therefore sufficient.

The answer should be D.
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 [#permalink] New post 28 Jun 2005, 06:23
A screwdriver and a hammer currently have the same price . If the price of a screwdriver rises by 5% and the price of a hammer goes up by 3% ,how much more will it costs to buy 3 screwdrivers and 3 hammers?

3%
4%
5%
8%
24%


Since it's a percentage problem, one good way is to assume the screw-driver and hammer each cost $100 (we're told they cost the same).

The new cost of the screwdriver is therefore $105 and the new cost of a hammer is $103. The total cost of 3 screwdrivers and 3 hammers is therefore 3*(105) + 3*(103) = 315 + 309 = $624. This is $624 - $600 = $24 more than usual. The percentage increase in price is therefore 24/600 * 100% = 4%

The answer is therefore B.
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 [#permalink] New post 28 Jun 2005, 06:49
What is the average (arithmetic mean) of a, b, and c?

(1) (a + b) + (c + d) = 17
(2) d = 5


From statement (1), we have a + b + c + d = 17. Since we do not know the value of d, we can't compute the mean of a,b and c. Statement (1) is therefore insuffcient.

From statement (2), we have nothing but the value of d. So statement (2) alone is not sufficient.

Using both statements, we can obtain a value of a + b + c and therefore we can solve for the average of the three numbers.

The answer is therefore C.
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 [#permalink] New post 03 Jul 2005, 22:46
At Consolidated Foundaries, for a resolution to become policy, a quorum of at least half the 20 directors must pass the resolution by at least a two-third majority. At a meeting of the board of directors, did the resolution X pass or fail?

1) Ten directors voted for the resolution.
2) Seven directors voted against the resolution.


The question tells us at least 10 people must vote. And for the quorum to pass, the total # of votes in favor must be at least 2/3 the total # of votes.

From statement (1), all we know is that 10 directors voted in favor of the resolution. We know that voting took place, so at least 10 directors voted. If 10 directors voted, we have all directors voting for the resolution to pass. However, if 20 directors voted, then the resolution would fail since less than 2/3 majority voted in favor of the resolution.

From statement (2), we're told that 7 directors voted against the resolution. Since voting took place, at least 10 directors must have voted. So assuming 10 directors voted, only 3 directors voted in favor. Even if 20 directors voted, we would only have 13 in favor of the resolution, and this is still less than 2/3 majority. So statement (2) allows us to know that the resolution X failed to pass.

The answer is therefore (B).
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 [#permalink] New post 04 Jul 2005, 06:22
2^199 + 2^199 = 2^X

what is X?


The left hand side of the equation can be simplified to 2*(2^199) which is also equals to 2^(199+1) = 2^200.

Equating both left hand side and right hand side, we get X = 200.
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 [#permalink] New post 04 Jul 2005, 22:17
A certain company has a bonus plan that is based on yearly gross sales. For salespeople with yearly gross sales greater than $10,000 but less than $100,000, the bonus is calculated by taking the ten thousands digit, doubling it and multiplying it by 1000. If Peggy was the company’s top salesperson, and her yearly gross sales were between $10,000 and $100,000, was Peggy’s bonus greater than 15% of her yearly gross sales?

(1) No Salesperson had a yearly gross sales of less than $15,000 or greater than $50,000.

(2) Peggy’s bonus was $8,000.


We're told from the passage that for sales, S, in the range 10,000 < S < 100,000, the bonus, B, is calculated as B = 2 * (ten thousands digit) * 1000.

From statement (1), we're told that all the salesperson grossed between $15,000 to $50,000. So the maximum Peggy had to earn was $50,000, and the least would have to be more than $15,000. If it was $50,000, then her bouns would be 2 * (5) * 1000 = $10,000, and this amount is (10,000/50,000) * 100% = 20% of her gross sales. If it was, say, $16,000, then her bonus would be 2 * (1) * 1000 = 2000, and this amount is 2000/16000 * 100% = 12.5%. So we can't answer the question from the information given in statement (1).

From statement (2), we're told her bonus was $8000. Equating this value to the formula given for Bonues, we have: 2 * (x) * 1000 = 8000 => 4. So the ten thousands digit is 4. At 40,000, her bouns is 20%. Assuming she earned $50,000, her bouns would be 16% which is still greater than 15%. We know the maximum she can earn is $49,999, so her bonus as a percentage of her gross sales would be higher than 16%. Thus, statement (2) is sufficient for answering the quesiton.

The answer is B.
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A correction [#permalink] New post 07 Jul 2005, 11:00
The explanation on the bikers problem needs 1 correction. Sum of an AP is not n/2 + {2a = (n-1)d} as given in the explanation.
It should be n/2 * {2a+ (n-1)d}

If I am wrong, please correct me.
Thanks
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 [#permalink] New post 17 Jul 2005, 04:46
ywilfred wrote:
Is pq positive?

(1) (p + q)^2 > (p - q)^2
(2) (p - q)^2 is positive


From (1), We have (p + q)^2 > (p - q)^2. This can be simplified to:

(p^2 + 2pq + q^2) > (p^2 - 2pq + q^2)
4pq > 0
pq > 0.

So (A) alone is sufficient.

From (2), we have (p - q)^2 is positive, but this is not sufficient the square of any operation is always positive.

The answer is therefore A.


I disagree with this answer because it doesn't account for if either p or q is 0. If p is 0, pq is not positive. Therefore you can never tell if pq is positive or 0. The answer is E.
  [#permalink] 17 Jul 2005, 04:46
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