CP #25

MGMAT CAT#2


The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?

		3
		14
		30
		42
		70

Solution

The greatest common factor (GCF) of two integers is the largest integer that divides both of them evenly (i.e. leaving no remainder).

One way to approach the problem is to consider what the GCFs stated in the question stem tell us about n:

The greatest common factor of n and 16 is 4. In other words, n and 16 both are evenly divisible by 4 (i.e., they have the prime factors 2 × 2), but have absolutely no other factors in common. Since 16 = 2 × 2 × 2 × 2, n must have exactly two prime factors of 2--no more, no less.

The greatest common factor of n and 45 is 3. In other words, n and 45 both are evenly divisible by 3, but have absolutely no other factors in common. Since 45 = 3 × 3 × 5, n must have exactly one prime factor of 3--no more, no less.  Also, n cannot have 5 as a prime factor.

So, n must include the prime factors 2, 2, and 3. Furthermore, n is not divisible by 8, 9, or 5.

The number 210 has the following prime factorization: 210 = 2 x 3 x 5 x 7. Thus n and 210 are both divisible by 2 and 3, and both might be divisible by 7, but it is not the case that both numbers are divisible by 4 or by 5:

(A) 3: missing a 2
(B) 14: missing a 3
(C) 30: n cannot be divisible by 5
(D) 42: Correct. n and 210 are both divisible by 2 and 3, and n could be divisible by 7.
(E) 70: n cannot be divisible by 5 and must be divisible by 3.

The correct answer is D.

CP #24

MGMAT CAT#2

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

Answer Explanation:
(1) SUFFICIENT: The factors of any number N can be sorted into pairs that multiply to give N. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. (For example, if N were 49, it would ahve the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. (If N is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that N is not a perfect square.

(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

The correct answer is D.

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I picked numbers to get to my answer, but made a silly mistake by omitting "1" as a distinct factor in Statement II.

And to the MGMAT explanation - hunh?!

CP #23

From GMATClub.com's m#05:


4. Two watermelons, A and B, are on sale. Watermelon A has a circumference of 6 inches; watermelon B, 5 inches. If the price of watermelon A is 1.5 times the price of watermelon B, which watermelon is a better buy?

(Assume that the two watermelons are spheres).

(C) 2008 GMAT Club - [t]m05#4[/t]

• A
• B
• Neither
• Both
• Impossible to determine

 

Approach 1:

Value = Volume / Price

The volume is proportional to radius^3 and the circumference is proportional to the radius. Increasing the circumference by 6/5 = 20% increases the volume by 1.2^3 = 1.728 (72.8%) and increases the price by 50%.

The increase in value of Watermelon A is 1.728 / 1.5 = 15.3% .

CP #22

GMATClub m#02 test:

Is ?

1. 
2. 

(C) 2008 GMAT Club - [t]m02#19[/t]

• Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
• Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
• BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
• EACH statement ALONE is sufficient
• Statements (1) and (2) TOGETHER are NOT sufficient

Statement (1) by itself is insufficient. Consider or .
Statement (2) by itself is insufficient. Consider or .
Statements (1) and (2) combined are sufficient. The absolute values of and are equal and , implying that is positive and is negative. Therefore, is false.

The correct answer is C.

CP #21

A train travels at an average speed of 200 km/hr without any stoppages. However, its average speed decreases to 120 km/hr on account of stoppages. On an average, how many minutes per hour does the train stop?

Options: 12, 18, 20, 24 minutes.


OA: 24 minutes

Solution:

You have to find out how many minuter PER HOUR does the train stop.

So, use T = 1 hour.

In 1 hour, the train travels 200 km.

If the distance travelled reduces to 120 km, how much time does it take the train to travel a distance of 120 km? The remaining time (1 hour - time taken to travel 120 km) will give us the stoppage time (per hour).

S = D / T

200 = 120 / T

T = 3/5 hour = 3/5 * 60 mins = 36 mins

So the train has taken 36 mins to travel 120 km in 1 hour. This means that the train stopped for how many minutes? 60 - 36 = 24 mins of stoppage time per hour.

CP #20

What is the number of integers from 1 to 1000 (inclusive) that are divisible by neither 11 nor by 35?

(C) 2008 GMAT Club - [t]m07#14[/t]

• 884
• 890
• 892
• 910
• 945

To count the number of integers from 1 to (inclusive) that are divisible by , find the value of . Use only the integer part of the resulting number. Based on the formula, the number of integers divisible by 11 is (even though the result could be rounded off to 91, use 90). In the same way, the number of the integers divisible by 35 is .
Subtract the number of integers that are divisible by both 11 and 35, so that they are not counted twice.
Therefore, .
.

The correct answer is A.

 

 

 

My answer was right. Wanted to keep this here to understand why they've used 1000/ 11*35. What I did was calculate 33*11 and 33*11*2 and saw that both are under 1000 and then counted 2. I should know this other method too.

CP

A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?

(C) 2008 GMAT Club - [t]m07#6[/t]

• 4
• 5
• 6
• 7
• 8
•  
•  
•  
• The amount of work in man-hours is: mh. After 60 days with 5 days of rain (55 net days) the crew of 10 had completed 550 mh of work, and had 550 mh more to go. Of the remaining 40 days, were raining. The last 550 mh were completed by 16 men: , and . Round the number of rainy days down to 5 as with 6 rainy days the crew would have worked less than 550 mh and not have completed the job.

CP#17

Question (from MGMAT Practice Test #1)
For positive integer k, is the expression (k + 2)(k² + 4k + 3) divisible by 4?

(1) k is divisible by 8.

(2)

k + 1 3

is an odd integer.

Solution
The quadratic expression k² + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:

If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.

If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.

The question might therefore be rephrased “Is k odd, OR is k + 2 divisible by 4?” Note that a ‘yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a ‘no’ to both conditions.

(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive ‘no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.

(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.

The correct answer is A.


Query
I think the answer is D. In Statement (2), for (k+1)/3 to be an odd INTEGER, k+1 has to be divisible by 3. Also, k must be even. Therefore, k could be 2,8,... And if k is an even integer, (k + 1)(k + 2)(k + 3) will not be divisible by 4 for any of these values of k. Therefore, Statement (2) also gives us the answer. Please tell me where I'm going wrong.

CP #16 - have not understood why not my way?

Question (from MGMAT Practice Test #1)
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

Solution
We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.

Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.

The correct answer is E. 

Remainder of today's CPs.

This is the link to Remainder questions. Sriharimurthy's explanations are really good. Good questions to repeat at some later stage! Have understood them for the time being. Should probably note down what I've understood here or elsewhere, so as to not go through the whole thread all over again. Phew. I'm tired now, have solved and understood quite a few challenging questions today!

To Do:
Combinatorics tutorial from MGMAT
Kaplan questions - P&C, Probability, and all the other topics.

CP #15

From GMATClub Math Book Remainders:

If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

OA: E

Using the concept,

If you take the decimal portion of the resulting number when you divide by "n", and multiply it to "n", you will get the remainder.
Eg. 8/5 = 1.6
0.6*5 = 3 = remainder

In our question,
0.12*t = R(remainder)
Since R has to be an integer, it must be a multiple of 0.12.
To make the calculation simpler, we multiply both sides by 100.
12*t = 100R

Now, put the value of different Rs in the equation and find which one is perfectly divisible by 12.

i.e. which one of

(A) 200
(B) 400
(C) 800
(D) 2000
(E) 4500

is perfectly divisible by 12 or 3*4.
All are divisible by 4, since they end in 00s.
Only (E) is divisible by 3 also (4+5=9).

Answer is therefore, (E) 45.

CP #14

From GMATClub:

10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?

(A) 144
(B) 131
(C) 115
(D) 90
(E) 45


OA: C
Chairmen shake hands 10*7=70 with business executives times. Business executives shake hands with each other 10 C 2 times or 45 times. The total is 115 .

Why 10C2?

10 business executives shakes hands with other 9 business executives in 10c2 ways = 45 ways
10c2 = 9+8+7+6+5+4+3+2+1 = 45

First executive shakes hands with remaining 9 executives
Second executive shakes hands with remaining 8 executives
Third executive shakes hands with remaining 7 executives
Fourth executive shakes hands with remaining 6 executives
Fifth executive shakes hands with remaining 5 executives
Sixth executive shakes hands with remaining 4 executives
Seventh executive shakes hands with remaining 3 executives
Eighth executive shakes hands with remaining 2 executives
Nineth executive shakes hands with remaining 1 executives
Tenth executive already shakes hands with all 9 executives.

CP #13

 From GMATClub's Probability chapter:


Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?
Solution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly 1/2 of all possible ways:
N = 1/2 * 5P2 = 10

Why half?

CP #12

Another one from GMATClub Math Book's Probability chapter.
Given that there are 5 married couples. If we select only 3 people out of 10, what is the probability that none of them are married to each other?

They have given 3 different ways to solve it. All have gone over my head. Help!

CP #11

This one's from the Probability chatper in GMATClub's Math Book.

If the probability of raining on any given day in Atlanta is 40%, what is the probability of raining on exactly 2 days in a 7-day period?

The way I look at it is this:
7 days, out of which on 2 it rains.
2 rain, 2 without rain = 0.4*0.4*0.6*0.6*0.6*0.6*0.6
= (0.4)^2 * (0.6)^5

However, the actual answer is
7C2 * (0.4)^2 * (0.6)^5

I don't understand why the 7C2 is required.

--

Went through the explanation in the chapter itself. It says that, it could rain on Day 1 and 2, or on Day 3 and 7. The sequence changes, although the number of days on which it rained remains the same. This must be factored in to get the correct answer. So other than the probability of it raining on those 2 and 5 days, we must also factor in the possibilites of different days on which it must rain.
At a preliminary level, I get it. But, I'll probably need a few more questions of the type or a different explanation to understand this thoroughly. Thoughts anyone?

CP #9

MGMAT CP 

Question

Two positive numbers differ by 12 and their reciprocals differ by 4/5. What is their product?

(A) 2/15
(B) 48/5
(C) 15
(D) 42
(E) 60

Solution

Don't be afraid to assign variables even when none are given in the problem. "Two positive numbers differ by 12" can be written as:
x – y = 12
And "their reciprocals differ by 4/5" can be written as:
1/y – 1/x = 4/5
(Note: Here, we've assigned x as the bigger of the two numbers and y as the smaller, so we've intuited that 1/y is the larger reciprocal and 1/x the smaller, and so arranged them in that order to write 1/y – 1/x = 4/5).
Now we have a system of two variables and two equations. Note that it is NOT necessary to solve for x and y, since we are being asked for the product, xy.
First, let's simplify the second equation by finding a common denominator for the terms on the left:
x/(xy) – y/(xy) = 4/5
(x – y)/(xy) = 4/5
Note that the denominator is xy, which is exactly the quantity we want to find.
Since we know from the first equation that x – y = 12, substitute 12:
12/(xy) = 4/5
60 = 4xy
15 = xy
The correct answer is (C) 15.
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It was a very simple question. I've highlighted where I made mistakes. Both were such stupid errors. Even if I hadn't figured the 2nd one, I would've solved it by an extremely long method.

CP #8

Here is the question I didn't get in my last PR test: 

Q. A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?

a) (1/3)^6 x (1/2)^3

b) (1/3)^6 x (1/2)

c) (1/3)^4

d) (1/3)^2 x (1/2)

e) (1/3)^2 x 5 

Solution:

I came up with an extremely simplified, but slightly longer solution, mainly because I didn't "get" the solutions posted completely. The method is still the same, but this is my explanation for why it is the way it is:

Suppose the shirts are - Y (yellow), G (green), W (white)

The shoes are - T (teal), V (violet)

And the pants are - R (red), L (lemon), A (aqua)

Yes, the man has a horrible sense of fashion. Let's call him Punk! 

Note: My explanation is based on a simple rule my math tutor taught me in school -- 'and' always translates into 'x' (multiplication) and 'or' always becomes '+' (addition). 

Now, the probability of him picking up a particular shirt, say Y = 1/3

This is the probability of each shirt, and so P(Y) = P(G) = P(W) = 1/3

Same for pants i.e. P(R) = P(L) = P(A) = 1/3
For shoes = P(T) = P(V) = 1/2

Day 1:

Mr. Punk can choose an outfit (Shirt & Shoes & Pants) by combining any of the above items.

P (choosing a shirt) = P(Y) or P(G) or P(W)

Similarly,

P (choosing a pair of shoes) = P(T) or P(V)

P (choosing a pair of pants) = P(R) or P(L) or P(A)

One of each item has to be picked up at the same time, so we will multiply the above. The equation is:

P (1Shirt, 1Shoe, 1Pant) = [ P(Y) + P(G) + P(W) ] x [ P(T) + P(V) ] x [ P(R) + P(L) + P(A) ]

The RHS (right hand side of the equation) can be put in words:

He must have picked either the Y 'or' G 'or' W shirt AND a T 'or' V shoe AND a R 'or' L 'or' A pant. I simply converted the 'or' to + & the AND to x.

P (1Shirt, 1Shoe, 1Pant) = (1/3 + 1/3 + 1/3) x (1/2 + 1/2) x (1/3 + 1/3 + 1/3)

= 3/3 x 2/2 x 3/3

= 1.

Most people can derive this by common sense, since the probability of him choosing n combination of outfits has to be 1, because although the combos may be different, the probability of him choosing 1 piece of clothing from each item is a 100% (he will not walk out half-naked!). Anyway, common sense is not so common, so I needed to explain this to myself.

Day 2:

Let us suppose that Mr. Punk picked out Y shirt, T shoes & R pants.

Here is where I had made a mistake. I had assumed that the items he could choose his day 2 outfits from had reduced. In reality, he once again had the same choice from 3 shirts, 2 shoes & 3 pants (yes, he doesn't send his clothes to the laundry after wearing them just once!)

Now the limitation of him choosing a different shirt & pants, and the same shoes, is what we have to look for i.e. what is the probability of that happening by chance, not by him deliberately choosing only from the remaining 2 shirts, etc.

The equation for day 2 will look like this:

P (1Shirt, 1Shoe, 1Pant) = [ P(G) + P(W) ] x [ P(T) ] x [ P(L) + P(A) ]

(since he's already worn the Y & R, and he has to choose only T again.)

P (1Shirt, 1Shoe, 1Pant) = (1/3 + 1/3) x (1/2) x (1/3 + 1/3)

= 2/3 x 1/2 x 2/3

= 2/9

Day 3:

Mr. Punk wore G shirt, T shoes & L pants on day 2.

For day 3, what is the probability of him wearing W shirt, T shoes & A pants?

P (W Shirt, T Shoe, A Pant) = P(W) x P(T) x P(A)

= 1/3 x 1/2 x 1/3

= 1/18 

Multiplying the final blue equations for days 1 AND 2 AND 3 (remember, AND => multiplication),

P (different shirt, pants, and same shoes, on the 3 days) = 1 x 2/9 x 1/18

= 1/81 or (1/3)^4 

PS. I've kept my earlier thoughts on the question...just in case I need them some day:

I made one MAIN mistake. I was thinking he had to choose a particular outfit, and I calculated the probability of that. That's why I was getting answer B I think. Anyway, after reading a couple of solutions posted on the web, I figured it out. We're looking for the probability of him not repeating that same outfit, and there IS replacement here! I had assumed that once shirt#1 is gone, he's left to choose between any of the remaining 2, when in fact he still has his 3 shirts to choose from, but now he can actually pick any 1 of the 2. This explanation is beginning to suck.

After some thought, I finally understood the solution. The solutions I found on GMAT forums helped.

CP #7

From GMATHacks again.

Commensalism is any relationship between two living things in which one benefits and the other is neither helped nor harmed. Oxpecker birds are commensal species that flock with the large mammals of the African Savannah. They feed on ticks, fleas, and flies that are attracted to the mammals' fur. Which of the following, if true, can most reasonably be inferred from the statements above?
(A)	Oxpecker birds are neither helped nor harmed by the large mammals of the African Savannah.
(B)	Ticks, fleas, and flies are commensal species in their relationship with both oxpecker birds and the large mammals of the African Savannah.
(C)	No species exist in a commensal relationship with oxpecker birds except for large mammals of the African Savannah.
(D)	In commensal relationships, the smaller of the species in the relationship usually benefits while the larger is neither helped nor harmed.
(E)	Preying on small creatures drawn to the fur of the large mammals of the African Savannah does not significantly affect those mammals.

-----------------------------------------------------------------

Answer: E

This is an inference question. The passage suggests that, since oxpecker birds are commensal species with large mammals, they benefit from the creatures that are attracted to the mammals' fur, but the mammals themselves are neither helped nor harmed by the relationship. Consider each choice, looking for a reasonable inference:

(A) This choice gets the commensal relationship exactly backwards.
(B) This is clearly wrong. If oxpecker birds feed on ticks, fleas, and flies, clearly the ticks, fleas, and flies are neither benefiting nor neutral in their relationship with the oxpecker birds.
(C) This choice is too extreme. The passage only describes this relationship; it doesn't tell us that it is exclusive.
(D) This is also too extreme. It may be true in some instances, but the three sentences of the passage don't provide enough evidence to reasonably deduce this.
(E) This is correct. It merely restates the definition of commensalism in terms of the role of the mammals in their relationship with oxpecker birds.

-----------------------------------------------------------------------------

I know it's not a difficult question, but I've made the same sort of mistake on earlier questions too. I had used POE to come down to A & E. But, by then I had lost track of the question I guess, and chose A. A Veritas Prep GMAT Tip of the Week talks about such mistakes.

CP #6

From GMATHacks.com

Bipedal dinosaurs' standing posture differs from virtually every visual depiction of them created before the 1970s, when scientists reevaluated their assumptions about tripod-style balance.
(A)	Bipedal dinosaurs' standing posture differs from
(B)	Bipedal dinosaurs stood in a posture that differs from
(C)	Bipedal dinosaurs exhibited standing postures that differ from those of
(D)	The standing posture of bipedal dinosaurs differs from
(E)	The characteristics of bipedal dinosaurs' standing posture differ from those of

------------------------------------------

Answer: C

The words "differ from" signal that there is a comparison. In this case, the comparison is between the actual posture of dinosaurs and the posture as depicted by certain images. As written, the sentence is incorrect: it compares the actual standing posture with "every visual depiction." We can't compare a posture with a picture as those are unlike things.

Choice (B) makes the same mistake, comparing a posture with pictures. (C) makes a proper comparison between "standing postures" and "those of." The word "those" takes the place of "standing postures," so we're comparing the actual standing postures with the standing postures as depicted. (D) makes the same mistake as (A) and (B). (E) is complicated but is yet another comparison error. It compares characteristics of standing posture with "those" (characteristics) of visual depictions. Those are different things, so (C) must be correct.

CP #5

Question #2 from Kaplan Problem Solving challenge.

The “connection” between any two positive integers a and b is the ratio of the smallest common multiple of a and b to the product of a and b.  For instance, the smallest common multiple of 8 and 12 is 24, and the product of 8 and 12 is 96, so the connection between 8 and 12 is 

The positive integer y is less than 20 and the connection between y and 6 is equal to . How many possible values of y are there?

	7
	8
	9
	10
	11

Answer Explantion:

If the connection between y and 6 is , then the least common multiple of y and 6 must equal the product 6y. The least common multiple of two numbers equals the product of the two numbers only when there are no common factors (other than 1). Since y is a positive integer less than 20, check all the integers from 1 to 19 to see which ones have no factors greater than 1 in common with 6:  1, 5, 7, 11, 13, 17, and 19. So there are 7 possible values for y.

 -------------------------------------------------------------------

In my own words (which I can actually understand!),

for the ratio to be 1:1, the LCM should be =6y (so that it cancels out with the product).

For this to be possible, y should NOT be a multiple of 6, and should not include 2,3 or any multiples of 2,3. The only numbers that are not multiples of 2,3 and 6 are 1,5,7,11,13,17,19, a total of 7 numbers!

Why I got it wrong:
I actually randomly guessed on this one cos I knew it would take me more thinking to get the right answer, and so I tried applying the pacing advice I read on a MGMAT forum today.