Manhattan GMAT Challenge - Crack GMAT with one Question Every Week

KevinFitzgerald

Answer to the August 14 problem
First, let's consider the different medal combinations that can be awarded to the 3 winners:

(1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.

(2) What if there is a 2-WAY tie?
--If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
--If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
--There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total).

(3) What if there is a 3-WAY tie?
--If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
--There are no other possible 3-WAY ties.

Thus, there are 4 possible medal combinations:

(1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G

Now let's determine how many different ways each combination can be distributed. We'll do this by considering four runners: Albert, Bob, Cami, and Dora.

COMBINATION 1: Gold, Silver, Bronze

Gold Medal: Any of the 4 runners can receive the gold medal: 4 possibilities

Silver Medal: There are only 3 runners who can receive the silver medal. Why? One of the runners has already been awarded the Gold Medal: 3 possibilities

Bronze Medal: There are only 2 runners who can receive the bronze medal. Why? Two of the runners have already been awarded the Gold and Silver medals: 2 possibilities

Therefore, there are 4 x 3 x 2 = 24 different victory circles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist.

COMBINATION 2: Gold, Gold, Silver.

Using the same reasoning as for Combination 1, we see that there are 24 different victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. However, it is important to realize that these 24 victory circles must be reduced due to "overcounting."

To illustrate this, consider one of the 24 possible Gold-Gold-Silver victory circles:

Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a SILVER.

Notice that this is the exact same victory circle as the following:

Bob is awarded a GOLD. Albert is awarded a GOLD. Cami is awarded a SILVER.

Each victory circle has been "overcounted" because we have been counting each different ordering of the two gold medals as a unique victory circle, when, in reality, the two different orderings consist of the exact same victory circle. Thus, the 24 victory circles must be cut in half; there are actually only 12 unique victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. (Note that we did not have to worry about "overcounting" in Combination 1, because each of those 24 possibilities was unique.)

COMBINATION 3: Gold, Silver, Silver.

Using the same reasoning as for Combination 2, we see that there are 24 possible victory circles, but only 12 unique victory circles that contain 1 GOLD medalist and 2 SILVER medalists.

COMBINATION 4: Gold, Gold, Gold.

Here, once again, there are 24 possible victory circles. However, because all three winners are gold medalists, there has been a lot of "overcounting!" How much overcounting?

Let's consider one of the 24 possible Gold-Gold-Gold victory circles:

Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a GOLD.

Notice that this victory circle is exactly the same as the following victory circles:

Albert-GOLD, Cami-GOLD, Bob-GOLD.
Bob-GOLD, Albert-GOLD, Cami-GOLD.
Bob-GOLD, Cami-GOLD, Albert-GOLD.
Cami-GOLD, Albert-GOLD, Bob-GOLD.
Cami-GOLD, Bob-GOLD, Albert-GOLD.

Each unique victory circle has actually been counted 6 times! Thus we must divide 24 by 6 to find the number of unique victory circles. There are actually only 24 / 6 = 4 unique victory circles that contain 3 GOLD medalists.

FINALLY, then, we have the following:

(Combination 1) 24 unique GOLD-SILVER-BRONZE victory circles.
(Combination 2) 12 unique GOLD-GOLD-SILVER victory circles.
(Combination 3) 12 unique GOLD-SILVER-SILVER victory circles.
(Combination 4) 4 unique GOLD-GOLD-GOLD victory circles.

Thus, there are 24 + 12 + 12 + 4 = 52 unique victory circles.

The correct answer is B.

KevinFitzgerald

Challenge Problem for August 21, 2006

A certain computer program randomly generates equations of lines in the form

. If point P is a point on a line generated by this program, what is the probability that the line does NOT pass through figure ABCD?

(A) 3/4
(B) 3/5
(C) 1/2
(D) 2/5
(E) 1/4

terrific

Answer should be 1/2. Reason: If we draw all lines passing thru P, we can see that only half wil be those which won't pass thru figure.

sandeep1007

The answer id D.

Because, from the first statement we get the values as 6,some number between 200 and 300, ...I donno whether there exist such numbers after this range....so minumum 2 values of n available with the first option.

Using the second option, the range is decreased to n<30 so there exists only one vale i.e. 6

So the value of n is 6.

The answer for this question is D.

Juss let me know if my explanation is wrong.

catdilse

Quote:

Originally Posted by KevinFitzgerald

Most of our students are trying to break the 700+ barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level Data Sufficiency problem (I'll post the solution next Monday).

Question
What is the positive integer n ?

(1) The sum of all of the positive factors of n that are less than n is equal to n
(2) n < 30

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

from statement no1 it is clear that 'n' is a perfect number,but we cant find the value of n.
now from statement no. 2 'n'<30 so n ican be 6 or 28 ,but still we dont have a unique answer.
so combining together is also not sufficient t find the answer hence E.

praanav

Answer is 1/2 (since the angle covered by line passing through the rectangle is 90, of the total possible of 180)

Quote:

Originally Posted by KevinFitzgerald

Challenge Problem for August 21, 2006

A certain computer program randomly generates equations of lines in the form

. If point P is a point on a line generated by this program, what is the probability that the line does NOT pass through figure ABCD?

(A) 3/4
(B) 3/5
(C) 1/2
(D) 2/5
(E) 1/4

nitinbhagnari

answer is 1/4 i.e A.

ssarkar

What's happening here??? Mods... please see...

MBAwanted

Quote:

Originally Posted by KevinFitzgerald

Most of our students are trying to break the 700+ barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level Data Sufficiency problem (I'll post the solution next Monday).

Question
What is the positive integer n ?

(1) The sum of all of the positive factors of n that are less than n is equal to n
(2) n < 30

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

both 6, 28 are eligible -- hence E