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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Since the thread is disrupted, and for good reasons, I just want to take the chance to thank you for posting up this sticky. These problems sound very familiar and I can absolutely picturing them appearing on GMAT, yet most of these problems are difficult enough for me to challenge myself to learn the underlying concept.

Anyways, this is definately a set well-put-together questions you have here. Thank you again. I'll delete this message in a couple days so you go on with your thread.

A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled

For this question, what I would do is to brute force the answer, instead of looking at the algebraic way of solving it. It's much faster, and doesn't require much manipulation with algebraic terms. But for the purpose of learning, we'll look at both ways.

Brute-Force:

1s - 4 ft 2s - 9 ft 3s - 14 ft 4s - 19 ft 5s - 24 ft 6s - 29 ft 7s - 34 ft 8s - 39 ft 9s - 44 ft 10s - 49 ft 11s - 54 ft

Total = 4 + 9 + 14 + 19 + 24 + 29 + 34 + 39 + 44 + 49 + 54 = 319 ft Pros of brute-forcing - Doesn't require much thought process. Just scribble and add

Cons of brute-forcing - Error Prone, since adding up under time pressure can lead to mistakes Algebraic Way:

We know for the first second, the rider travels 4 ft. Subsequent second, he travels 5 ft more than the previous second. From this, we know we're dealing with an arithmetic progression (A.P) where each term is (a+n) (a = first term, n = arithmetic difference).

The sum of an arithmetic progression, S, is defined as: S = n/2 + {2a + (n-1)d} -------(1) where n is the number of terms a is the first term d is the difference

S = n/2 (a+l) -----------------(2) where n is the number of terms a = first term l = last term

Since we only know n, a and d, we use equation (1).

S = 11/2 + {2(4) + (11-1)5} = 319 ft

Pros of using this method: -You can quickly work out the sum for a large number of terms (e.g. Sum of a A.P with 1000 terms)

Cons of using this method: - Need to be able to identify an A.P - Need to be able to recall the formula

Re: Formula is wrong [#permalink]
12 Aug 2005, 07:59

raki_sg wrote:

Please correct your formula S= n/2{2a + (n-1)d}

ywilfred wrote:

A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled

For this question, what I would do is to brute force the answer, instead of looking at the algebraic way of solving it. It's much faster, and doesn't require much manipulation with algebraic terms. But for the purpose of learning, we'll look at both ways.

Brute-Force:

1s - 4 ft 2s - 9 ft 3s - 14 ft 4s - 19 ft 5s - 24 ft 6s - 29 ft 7s - 34 ft 8s - 39 ft 9s - 44 ft 10s - 49 ft 11s - 54 ft

Total = 4 + 9 + 14 + 19 + 24 + 29 + 34 + 39 + 44 + 49 + 54 = 319 ft Pros of brute-forcing - Doesn't require much thought process. Just scribble and add

Cons of brute-forcing - Error Prone, since adding up under time pressure can lead to mistakes Algebraic Way:

We know for the first second, the rider travels 4 ft. Subsequent second, he travels 5 ft more than the previous second. From this, we know we're dealing with an arithmetic progression (A.P) where each term is (a+n) (a = first term, n = arithmetic difference).

The sum of an arithmetic progression, S, is defined as: S = n/2 + {2a + (n-1)d} -------(1) where n is the number of terms a is the first term d is the difference

S = n/2 (a+l) -----------------(2) where n is the number of terms a = first term l = last term

Since we only know n, a and d, we use equation (1).

S = 11/2 + {2(4) + (11-1)5} = 319 ft

Pros of using this method: -You can quickly work out the sum for a large number of terms (e.g. Sum of a A.P with 1000 terms)

Cons of using this method: - Need to be able to identify an A.P - Need to be able to recall the formula

Ah yes, thanks for pointing that out. I will make the change soon. Having problems with the internet connection....

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.

We do not have to consider so much because we know p times q must be greater than 0, and hence p or q cannot be 0 by themselves.

Town T has 20,000 residents, 60% of whom are female. what percentage of the residents were born in town T?

1. The number of female residents who were born in town t is twice the number of male residents who were NOT born in town T

2. The number of female residents who were NOT born in Town T is twice the number of female residents who were born in Town T

From the passage, we know T has 20,000 residents of which, 60% or (60/100)*20,000 = 12,000 are female, and 8,000 (20,000-12,000) are male.

We're asked what percentage of residents were born in town T. This would be found by (# of residents born in town T)/(# of residents in town T) * 100%. Since we already know the denominator, what the question is really asking is the number of residents born in town T.

From statement (1), we're told:

# of female residents born in town T = 2*(# of male residents not born in town T) We do now have the value for both LHS and RHS, and so statement (1) is not sufficient. Note: The question stem only gives us the total number of female and male residents, and never broke them down into those that were born in town T, and those that were born outside town T.

From statement (2), we're told:

# of female residents not born in town T = 2*(# of female residents born in town T)

Let's say the number of female residents who were born in town T is x, then the number of female residents who were not born in town T would be (12,000-x). We can now equate both LHS and RHS and solve for the number of female residents who were born in town T. However, this still does not answer the question as we're lacking a figure representing the number of male residents born in town T. So statement (2) alone is not sufficient.

Using both (1) and (2), we can now use the value we found in (2), to calculate the number of male residents who were not born in town T, and therefore find the number of male residents who were born in town T. This would allow us to answer the question, and so C is the answer.

I am not sure but look at this Men born in town =Mb Not born Mnb, Female Born Fb , not born = Fnb

Melting the three cheese balls and molding them together into one cheese ball will have one thing common. The volume of the big cheese ball will be the combined volume of the three cheese balls. To solve this, we need to use the formula for the volume of a sphere = 4/3*pi*(r^3)

For the cheese ball with dimeter of 1, V = 4/3 * pi * (1/2)^3 = pi/6 For the cheese ball with diameter of 2, V = 4/3 * pi * (1^3) = 4pi/3 For the cheese ball with diamater of 6, V = 4/3 * pi * (3^3) = 36pi

The total volume is therefore pi/6 + 4pi/3 + 36pi = 75pi/2

If the diamater of the big cheese cake = D, then the radius = D/2 So 4/3 * pi * (D/2)^3 = 75pi/2 D^3 = 225 D = 225^(1/3)

The answer is therefore B.

Nice explanation, but I would like to add one little shortcut; if you will.

4/3 pi r^3 is common everywhere so why even bother to go into detail calculation

The price of postage stamps has increased 5 cents per year every year since 1990. If 10 stamps were purchased every year from 1998 to 2002, the total cost would be $35. How much did a stamp cost in 1995?

A. 45 cents B. 35 cents C. 40 cents D. 48 cents E. 52 cents

We're told the stamps increased by 5 cents every year since 1990. So, this is a Arithmetic Sequence, where the frist term is the cost of a stamp in 1990, which we will cal a. The arithmetic difference is the increase in price, which is 5 cents. Thus, the nth term (or the nth year) would be defined as a + (n-1)d

For 1998, a stamp would cost a + (9-1)5 = (a + 40) cents (Note: use 9 as n since the 2008 is actually the ninth term)

For 1999, a stamp would cost a + (10-1)5 = (a + 45) cents For 2000, a stamp would cost a + (11-1)5 = (a + 50) cents For 2001, a stamp would cost a + (12-1)5 = (a + 55) cents for 2002, a stamp would cost a + (13-1)5 = (a + 60) cents

The cost of 10 stamps in 1998 would be (10a + 400) cents The cost of 10 stamps in 1999 would be (10a + 450) cents The cost of 10 stamps in 2000 would be (10a + 500) cents The cost of 10 stamps in 2001 would be (10a + 550) cents The cost of 10 stamps in 2002 would be (10a + 600) cents

Thus, the total cost of the stamps between 1998 to 2002 = (50a + 2500) cents = $35 = 3500 cents Solving for a gives us a = 20 cents

The cost of a stamp in 1995 = a + (6-1)5 = 20 + 25 = 45 cents

The answer is therefore A.

Another shortcut

1998 to 2002 (== 5 terms 98,99,00,01,02) and 10 stamps increased by 5 cents every year, so. 10x into 5 and 10(x+5) will result:

50x+500=3500 so x=60 cents in 1998 so 45 cents in 1995

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.

is not 8/3 less than 3. Of course is 8/3 is 2.666
I look at this way 6x=16
statement 1- x<=3 so if x=3 6x is not equal to 18 so answer is NO. for any value of x < or = 3 the answer si always going to be no and is sufficient. This is Yes or No question so either answer is sifficient.

Re: D is the answer [#permalink]
21 Oct 2005, 20:14

dsprob wrote:

for any value of x < or = 3 the answer si always going to be no and is sufficient. This is Yes or No question so either answer is sifficient.

IMO answer should be D

Not correct. x<=3 means x could equal 8/3 and the answer would be yes, but if x takes other values the answer would be no. 1) is thus insufficient. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

This equation S = n/2 + {2a + (n-1)d}
should be S = n/2 {2a + (n-1)d}

ywilfred wrote:

A bicycle rider coasts down the hill, travelling 4ft in the first second. In each succeeding second, he travels 5ft farther than in the preceeding second. If the rider reaches the bottom of the hill in 11 seconds, find the total distance travelled

For this question, what I would do is to brute force the answer, instead of looking at the algebraic way of solving it. It's much faster, and doesn't require much manipulation with algebraic terms. But for the purpose of learning, we'll look at both ways.

Brute-Force:

1s - 4 ft 2s - 9 ft 3s - 14 ft 4s - 19 ft 5s - 24 ft 6s - 29 ft 7s - 34 ft 8s - 39 ft 9s - 44 ft 10s - 49 ft 11s - 54 ft

Total = 4 + 9 + 14 + 19 + 24 + 29 + 34 + 39 + 44 + 49 + 54 = 319 ft Pros of brute-forcing - Doesn't require much thought process. Just scribble and add

Cons of brute-forcing - Error Prone, since adding up under time pressure can lead to mistakes Algebraic Way:

We know for the first second, the rider travels 4 ft. Subsequent second, he travels 5 ft more than the previous second. From this, we know we're dealing with an arithmetic progression (A.P) where each term is (a+n) (a = first term, n = arithmetic difference).

The sum of an arithmetic progression, S, is defined as: S = n/2 + {2a + (n-1)d} -------(1) where n is the number of terms a is the first term d is the difference

S = n/2 (a+l) -----------------(2) where n is the number of terms a = first term l = last term

Since we only know n, a and d, we use equation (1).

S = 11/2 + {2(4) + (11-1)5} = 319 ft

Pros of using this method: -You can quickly work out the sum for a large number of terms (e.g. Sum of a A.P with 1000 terms)

Cons of using this method: - Need to be able to identify an A.P - Need to be able to recall the formula

I'm confused. Does X represent the total amount of grams in mixture x? If so, does (100-x) represent total amount of grams in mixture y? And why do we use (100-x)?
Why isn't the mixture x and mixture y assumed to be 100g like mixture xy?

ywilfred wrote:

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:

It does look rather complicated and confusing doesn't it? It may help you a little bit by using a different letter. We know the definition of v*=v-v/3. We can say that x*=x-x/3.
Now let x=v*.
Substitute it back into the x* equation: x*=x-x/3=v*-v*/3.
Now we can substitute v*=v-v/3 into the equation:
x*=x-x/3=v*-v*/3=(v-v/3)-(v-v/3)/3
You can go from there and simplify the equation and get your answer. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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.

Hi,
I'm new here, but can someone please explain this one. From the actual problem, we know the smallest values of a 3-digit number is 100.

Statement 1 says the sum of the 3 digits is 5. This could be 104, 113, 203, etc. None of these values gives 100.
Statement 2 says y is divisible by 5. well 100 is divisible by 5, but so is 105, 110, 115, 120. Also Insufficient.

I would have chosen E.. and obviously got it wrong but can someone please clarify. Thanks in advance.

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.

Hi, I'm new here, but can someone please explain this one. From the actual problem, we know the smallest values of a 3-digit number is 100.

Statement 1 says the sum of the 3 digits is 5. This could be 104, 113, 203, etc. None of these values gives 100. Statement 2 says y is divisible by 5. well 100 is divisible by 5, but so is 105, 110, 115, 120. Also Insufficient.

I would have chosen E.. and obviously got it wrong but can someone please clarify. Thanks in advance.

The point is that only the least number was wanted. This is not likely a GMAT question though. From experience for answer D the two options should give you the same answer most of the time. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Re: Data Sufficency [#permalink]
19 May 2006, 06:29

andrecrompton wrote:

How much was spent for a certian amount of a fabric?

(1) The cost of the fabric was $15 per square yard.

(2) An additional 6 square yards of fabric would have cost an additional $90.

We want to know total cost. (1) gives unit price but not the amount bought. (2) gives again unit price for the additional 6 square yards. Both are insufficient, in my opinion, and combined is still insufficient. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

A leopard spots a deer from a distance of 200 meters. As the leopard starts chasing the deer, the deer also starts running. Given that the speed of the deer is 10 km/h and that of the leopard is 12 km/h, how far would have the deer run before it is caught?

A. 3 km B. 4 km C. 2 km D. 1 km E. 5 km

This question test the ability to recall the Distance-Speed-Time formula, and also to manipulate the question to formulate the equation for solving the question (by use of algebra). In such Distance-Speed-Time question (which involves A catching up with B), there's normally one quantity that's similar. Also, remember to make sure that your units match, otherwise your answer will not come out right.

We know from the passage: 1. The Leopard and the Deer are 200 meters apart = 0.2 km 2. THe speed of the Leopard = 12km/hr 3. The speed of the Deer = 10 km/hr

Let's assume that the Deer has travelled x km before it is caught. The Leopard would then have to travel (x+0.2)km. There is one quantity that is the same here, that is the time taken for the deer to travel that x km, and for the Leopard to travel (x+0.2km).

Since we know that Time = Distance/Speed, we can equate:

Time taken by Deer to travel x km = Time taken by Leopard to travel (x+0.2) km x/10 = (x+0.2)/12 12x = 10x + 2 2x = 2 x = 1 km

The answer is therefore, D.

The differential distance of 2 km is being covered in 1hr. So, 0.2 km would be covered in 0.1 hr. And the deer will travel 1 km in 0.1 hr @ 10km/hr.

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

Every student has a predefined notion about a MBA degree:- hefty packages, good job opportunities, improvement in position and salaries but how many really know the journey of becoming...