Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 22 May 2015, 21:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Beginners Forum (Re-worked questions)

Author Message
TAGS:
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [1] , given: 0

1
KUDOS
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.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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%. GMAT Club Legend Joined: 07 Jul 2004 Posts: 5077 Location: Singapore Followers: 22 Kudos [?]: 186 [0], given: 0 [#permalink] 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. GMAT Club Legend Joined: 07 Jul 2004 Posts: 5077 Location: Singapore Followers: 22 Kudos [?]: 186 [0], given: 0 [#permalink] 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 GMAT Club Legend Joined: 07 Jul 2004 Posts: 5077 Location: Singapore Followers: 22 Kudos [?]: 186 [0], given: 0 [#permalink] 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. GMAT Club Legend Joined: 07 Jul 2004 Posts: 5077 Location: Singapore Followers: 22 Kudos [?]: 186 [0], given: 0 [#permalink] 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 GMAT Club Legend Joined: 07 Jul 2004 Posts: 5077 Location: Singapore Followers: 22 Kudos [?]: 186 [0], given: 0 [#permalink] 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. GMAT Club Legend Joined: 07 Jul 2004 Posts: 5077 Location: Singapore Followers: 22 Kudos [?]: 186 [0], given: 0 [#permalink] 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%

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5077
Location: Singapore
Followers: 22

Kudos [?]: 186 [0], given: 0

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 companyâ€™s top salesperson, and her yearly gross sales were between $10,000 and$100,000, was Peggyâ€™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) Peggyâ€™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.

Intern
Joined: 07 Jul 2005
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 0

A correction [#permalink]  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
Intern
Joined: 27 Jun 2005
Posts: 18
Followers: 0

Kudos [?]: 2 [0], given: 0

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.

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.

Go to page   Previous    1   2   3   4   5    Next  [ 97 posts ]

Similar topics Replies Last post
Similar
Topics:
Question about error logs from beginner 2 22 Sep 2010, 09:48
Beginner 1 18 Jul 2008, 01:03
Beginner 2 18 Jul 2008, 01:03
Here's another one I picked off the Beginners forum. It's a 7 23 Jun 2005, 00:25
I took this from the Beginners forum, and I'm posting it 3 23 Jun 2005, 00:22
Display posts from previous: Sort by