If q is a multiple of prime numbers, is q a multiple of r?
1) r is less than 4
2) q = 18
Answer: E
From Statement (1) -- q can be positive and r can be negative.
r can also be a real number but not necessarily an integer.
Hence insufficient.
From Statement (2) -- q = 18.
But r can be negative or can be positive.
r can also be a real number not an integer
Hence insufficient
Taking both statements (1) and (2) together -- Again r can be positive or negative. And again r can also be a real number.
Hence insufficient
Saturday, June 30, 2007
Tuesday, June 26, 2007
DS Question - 29
If x and y are integers, does IxI = y ?
(1) (y^2 - x^2) = 0
(2) xy/(x+y) = 0
Answer: B
Statement (1) insufficient -- y can take any value i.e can be +ve or -ve. We cannot assume y to be +ve. Therefore y may or may not be equal to x
Statement (2) sufficient -- xy/(x+y) = 0 => xy = 0 => x = 0 or y = 0
In case x = 0, then y cannot be equal to 0. Hence y cannot be equal to x.
In case y = 0, then x cannot be equal to 0. Hence y cannot be equal to x.
Therefore sufficient.
Hence B
(1) (y^2 - x^2) = 0
(2) xy/(x+y) = 0
Answer: B
Statement (1) insufficient -- y can take any value i.e can be +ve or -ve. We cannot assume y to be +ve. Therefore y may or may not be equal to x
Statement (2) sufficient -- xy/(x+y) = 0 => xy = 0 => x = 0 or y = 0
In case x = 0, then y cannot be equal to 0. Hence y cannot be equal to x.
In case y = 0, then x cannot be equal to 0. Hence y cannot be equal to x.
Therefore sufficient.
Hence B
DS Question - 28
Is the three-digit number n less than 550?
1) The product of the digits in n is 30
2) The sum of the digits in n is 10
Answer: C
From statement (1) : the factors of 30 are 1,2,3,5,6,10,15,30
Now because the number must be digits (single number) we do not need to consider 10,15 and 50
Now if the hundreds digit of the 3-digit number = any digit among the 1,2,3 digits ---> answer to the question is clearly Yes.
But if the hundreds digit = 5 or 6 ---> answer to the question is No.
Hence (1) alone is insufficient.
From statement (2) : There are different combinations where the sum of digits can be equal to 10. e.g 541 and 145.
Hence (2) alone is insufficient.
Combining statements (1) and (2) we get :
If the hundreds digit= 5 , the second digit can only be 1,2,3 because if the digit is equal to 6 it violates statement (2) Hence in this case answer to the question is Yes.
If hundreds digit = 6 then the number n must contain 5 so as to satisfy the first statement but then it will violate statement (2) as the total of digits of n will exceed 10
Thus no 3-digit number exists with the hundreds digit equal to 6 satisfying both the statements together.
Hence the number n will always be less than 550 as it will be the combination of 1, 2, 3 and 5
Hence ans is C
1) The product of the digits in n is 30
2) The sum of the digits in n is 10
Answer: C
From statement (1) : the factors of 30 are 1,2,3,5,6,10,15,30
Now because the number must be digits (single number) we do not need to consider 10,15 and 50
Now if the hundreds digit of the 3-digit number = any digit among the 1,2,3 digits ---> answer to the question is clearly Yes.
But if the hundreds digit = 5 or 6 ---> answer to the question is No.
Hence (1) alone is insufficient.
From statement (2) : There are different combinations where the sum of digits can be equal to 10. e.g 541 and 145.
Hence (2) alone is insufficient.
Combining statements (1) and (2) we get :
If the hundreds digit= 5 , the second digit can only be 1,2,3 because if the digit is equal to 6 it violates statement (2) Hence in this case answer to the question is Yes.
If hundreds digit = 6 then the number n must contain 5 so as to satisfy the first statement but then it will violate statement (2) as the total of digits of n will exceed 10
Thus no 3-digit number exists with the hundreds digit equal to 6 satisfying both the statements together.
Hence the number n will always be less than 550 as it will be the combination of 1, 2, 3 and 5
Hence ans is C
Problem Solving - 24
On his drive to work, Bob listens to one of three radio stations, A,B, or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns to B. If B is playing a song he likes, he listens to it; if not, he turns to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station , the probability is 0.30 that any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Bob will hear a song he likes?
A) 0.027
B) 0.090
C) 0.417
D) 0.657
E) 0.900
Answer: D
Probability Leo listens to A = (Probability that A is playing a song of Leo's choice ) = 0.3
Probability Leo listens to B = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is playing a song of Leo's choice ) = (1-0.3) * 0.3 = 0.7*0.3
Probobility Leo listens to C = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is not playing a song of Leo's choice ) * (Probability that C is playing a song of Leo's choice ) = 0.7 * 0.7 * 0.3
Total Probability = 0.3 + 0.7*0.3 + 0.7*0.7*0.3 = 0.657
A) 0.027
B) 0.090
C) 0.417
D) 0.657
E) 0.900
Answer: D
Probability Leo listens to A = (Probability that A is playing a song of Leo's choice ) = 0.3
Probability Leo listens to B = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is playing a song of Leo's choice ) = (1-0.3) * 0.3 = 0.7*0.3
Probobility Leo listens to C = (Probability that A is not playing a song of Leo's choice ) * (Probability that B is not playing a song of Leo's choice ) * (Probability that C is playing a song of Leo's choice ) = 0.7 * 0.7 * 0.3
Total Probability = 0.3 + 0.7*0.3 + 0.7*0.7*0.3 = 0.657
Problem Solving - 23
If x is an integer divisible by 15 but not divisible by 20, then x CANNOT be divisible by which of the following?
(A) 6
(B) 10
(C) 12
(D) 30
(E) 150
Answer: C
x is divisible by 15 -> x is divisible by 3 and 5
x is not divisible by 20 -> x is not divisible by 4
Now because x is divisible by 15, it has 3 and 5 as factors and because x is not divisible by 20, it definitely does not have 4 as a factor .
In the given options only 12 is the number with 4 as a factor.
Thus x will not be divisible by 12. Hence C is the answer.
(A) 6
(B) 10
(C) 12
(D) 30
(E) 150
Answer: C
x is divisible by 15 -> x is divisible by 3 and 5
x is not divisible by 20 -> x is not divisible by 4
Now because x is divisible by 15, it has 3 and 5 as factors and because x is not divisible by 20, it definitely does not have 4 as a factor .
In the given options only 12 is the number with 4 as a factor.
Thus x will not be divisible by 12. Hence C is the answer.
Problem Solving - 22
A certain restaurant offers 6 kinds of cheese and 2 kinds of fruit for it dessert platter. If each dessert platter contains an equal number of kinds of cheese and kinds of fruit, how many different dessert platters could the restaurant offer?
(A) 8
(B) 12
(C) 15
(D) 21
(E) 27
Answer: E
There are two kinds of platter ---- 1 cheese + 1 fruit
Total types of platter -- 6 * 2= 12 ---(1)
For 2 cheese + 2 fruits
Total -- 6C2 * 2C2 = 15 * 1 = 15 ---(2)
Hence adding (1) and (2) we get total = 12 + 15 = 27
Friday, June 01, 2007
DS Question - 27
The area of a parallelogram is 100. What is the perimeter of the parallelogram ?
1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree
Answer: C
From (1) - For a parallelogram, Area = Base*Height => Height = 10 - There are infinite ways to draw a parallelogram with 100 as area, as long as the height is 10 units.( parallelograms with varying slants from 1 to 179 degrees) -- insufficent
From (2) - one of the angles is 45 degrees - hence the opposite angle is 45 degrees too, and the two remaining angles wil be 135 degrees each. But this doesn't tell us the measure of the sides to determine the perimeter - insufficient
From(1) and (2) together
From (2) we know that there is only one of such parallelograms that has an angle 45 to the base.
Base = 10. One angle = 45 degrees
In Parallelogram opposite angles are equal ..
Hence there are 2*45 degrees
Sum of all angles = 360
Thus 2*45 + 2 x = 360 which gives each other angle as 135 degrees
Note: If we are able to draw the figure on given DS question it implies conditions given are sufficient.
1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree
Answer: C
From (1) - For a parallelogram, Area = Base*Height => Height = 10 - There are infinite ways to draw a parallelogram with 100 as area, as long as the height is 10 units.( parallelograms with varying slants from 1 to 179 degrees) -- insufficent
From (2) - one of the angles is 45 degrees - hence the opposite angle is 45 degrees too, and the two remaining angles wil be 135 degrees each. But this doesn't tell us the measure of the sides to determine the perimeter - insufficient
From(1) and (2) together
From (2) we know that there is only one of such parallelograms that has an angle 45 to the base.
Base = 10. One angle = 45 degrees
In Parallelogram opposite angles are equal ..
Hence there are 2*45 degrees
Sum of all angles = 360
Thus 2*45 + 2 x = 360 which gives each other angle as 135 degrees
Note: If we are able to draw the figure on given DS question it implies conditions given are sufficient.
Manhattan Challenge Problem of the week! - 05/28/07
Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
Answer: D
OE - In order to determine the probability that the World Series will last fewer than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.
Let's analyze one way this could happen:
Game 1 -----Game 2 -----Game 3 -----Game 4 -----Game 5 -----Game 6
T1 Wins-----T1 Wins-----T1 Wins-----T1 Loses-----T1 Loses-----T1 Loses
There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the two teams to split the first 6 games (3 wins for each).
There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 × 2 = 40 ways for the World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is
(1/2) ^ 7 = 1/ 128
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:
40 * 1/128 = 40/ 128 = .3125 = 31.255%
Thus the probability that the World Series will last fewer than 7 games is:
100% - 31.25% = 68.75%.
The correct answer is D.
(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%
Answer: D
OE - In order to determine the probability that the World Series will last fewer than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.
Let's analyze one way this could happen:
Game 1 -----Game 2 -----Game 3 -----Game 4 -----Game 5 -----Game 6
T1 Wins-----T1 Wins-----T1 Wins-----T1 Loses-----T1 Loses-----T1 Loses
There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the two teams to split the first 6 games (3 wins for each).
There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 × 2 = 40 ways for the World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is
(1/2) ^ 7 = 1/ 128
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:
40 * 1/128 = 40/ 128 = .3125 = 31.255%
Thus the probability that the World Series will last fewer than 7 games is:
100% - 31.25% = 68.75%.
The correct answer is D.
Manhattan Challenge Problem of the week! - 05/21/07
Edwin is planning to drive from Boston to New Orleans. By what percent would his travel time be reduced if Edwin decides to split the driving time equally with his friend George, instead of making the trip alone?
(1) The driving distance from Boston to New Orleans is 1500 miles.
(2) George’s driving speed is 1.5 times Edwin’s driving speed.
Answer: B
OE - The question asks for the percent decrease in Edwin’s travel time. To determine this, we need to be able to find the ratio between, T1 (the travel time if Edwin drives alone) and T2 (the travel time if Edwin and George drive together). Note that we do NOT need to determine specific values for T1 and T2; we only need to find the ratio between them.
Percentage change is defined as follows: Difference/Original = (T1 - T2)/ T1 = 1 - (T2/ T1)
Ultimately, we can solve the percentage change equation above by simply determining the value of T2 /T1
Using the formula Rate × Time = Distance, we can write equations for each of the 2 possible trips
T1 = Travel time if Edwin drives alone
T2 = Travel time if Edwin and George drive together
E = Edwin’s Rate
G = George’s Rate
D = Distance of the trip
If Edwin travels alone: ET1 = D
If Edwin and George travel together: .5(E + G)T2 = D
(Since Edwin and George split the driving equally, the rate for the trip is equal to the average of Edwin and George’s individual rates).
Since both trips cover the same distance (D), we can combine the 2 equations as follows:
ET1 = .5(E + G)T2
Then, we can isolate the ratio of the times (T2/T1) as follows:
E/ .5(E + G) = T2/ T1
Now we look at the statements to see if they can help us to solve for the ratio of the times.
Statement (1) gives us a value for D, the distance, which does not help us since D is not a variable in the ratio equation above.
Statement (2) tells us that George’s rate is 1.5 times Edwin’s rate. Thus, G = 1.5E. We can substitute this information into the ratio equation above:
E/ .5(E + G) = T2/ T1 ---> E/ .5(E + 1.5E) = T2/ T1 ---> E/ .5E + .75E = T2/T1
---> E/ 1.25E = T2/ T1 ---> 1/ 1.25 = T2/ T1 ---> .8 = T2/ T1
Thus, using this ratio we can see that Edwin’s travel time for the trip will be reduced as follows:
1 - (T2/T1) = 1 - .8 = .2 ---> 20%
Statement (2) alone is sufficient to answer the question.
The correct answer is B.
(1) The driving distance from Boston to New Orleans is 1500 miles.
(2) George’s driving speed is 1.5 times Edwin’s driving speed.
Answer: B
OE - The question asks for the percent decrease in Edwin’s travel time. To determine this, we need to be able to find the ratio between, T1 (the travel time if Edwin drives alone) and T2 (the travel time if Edwin and George drive together). Note that we do NOT need to determine specific values for T1 and T2; we only need to find the ratio between them.
Percentage change is defined as follows: Difference/Original = (T1 - T2)/ T1 = 1 - (T2/ T1)
Ultimately, we can solve the percentage change equation above by simply determining the value of T2 /T1
Using the formula Rate × Time = Distance, we can write equations for each of the 2 possible trips
T1 = Travel time if Edwin drives alone
T2 = Travel time if Edwin and George drive together
E = Edwin’s Rate
G = George’s Rate
D = Distance of the trip
If Edwin travels alone: ET1 = D
If Edwin and George travel together: .5(E + G)T2 = D
(Since Edwin and George split the driving equally, the rate for the trip is equal to the average of Edwin and George’s individual rates).
Since both trips cover the same distance (D), we can combine the 2 equations as follows:
ET1 = .5(E + G)T2
Then, we can isolate the ratio of the times (T2/T1) as follows:
E/ .5(E + G) = T2/ T1
Now we look at the statements to see if they can help us to solve for the ratio of the times.
Statement (1) gives us a value for D, the distance, which does not help us since D is not a variable in the ratio equation above.
Statement (2) tells us that George’s rate is 1.5 times Edwin’s rate. Thus, G = 1.5E. We can substitute this information into the ratio equation above:
E/ .5(E + G) = T2/ T1 ---> E/ .5(E + 1.5E) = T2/ T1 ---> E/ .5E + .75E = T2/T1
---> E/ 1.25E = T2/ T1 ---> 1/ 1.25 = T2/ T1 ---> .8 = T2/ T1
Thus, using this ratio we can see that Edwin’s travel time for the trip will be reduced as follows:
1 - (T2/T1) = 1 - .8 = .2 ---> 20%
Statement (2) alone is sufficient to answer the question.
The correct answer is B.
Manhattan Challenge Problem of the week! - 04/30/07
How many terminating zeroes does 200! have?
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
Answer : C
OE - To determine how many terminating zeroes a number has, we need to determine how many times the number can be divided evenly by 10. (For example, the number 404000 can be divided evenly by 10 three times, as follows:
404000/10 = 40400
40400/10 = 4040
4040/10 = 404
We can see that the number has three terminating zeroes because it is divisible by 10 three times.) Thus, to arrive at an answer, we need to count the factors of 10 in 200! Recall that
200! = 200 * 199 * 198 ........... * 4 * 3 * 2 * 1
Each factor of 10 consists of one prime factor of 2 and one prime factor of 5. Let’s start by counting the factors of 5 in 200!. Starting from 1, we get factors of 5 at 5, 10, 15, . . . , 190, 195, and 200, or every 5th number from 1 to 200. Thus, there are 200/5 or 40 numbers divisible by five from 1 to 200. Therefore, there are at least 40 factors of 5 in 200!.
We cannot stop counting here, because some of those multiples of 5 contribute more than just one factor of 5. Specifically, any multiple of 52 (or 25) and any multiple of 53 (or 125) contribute additional factors of 5. There are 8 multiples of 25 (namely, 25, 50, 75, 100, 125, 150, 175, 200). Each of these 8 numbers contributes one additional factor of 5 (in addition to the one we already counted) so we now have counted 40 + 8, or 48 factors of 5. Finally, 125 contributes a third additional factor of 5, so we now have 48 + 1 or 49 total factors of 5 in 200!.
Let us now examine the factors of 2 in 200!. Since every even number contributes at least one factor of 2, there are at least 100 factors of 2 in 200! (2, 4, 6, 8 . . .etc). Since we are only interested in the factors of 10 — a factor of 2 paired with a factor of 5 — and there are more factors of 2 than there are of 5, the number of factors of 10 is constrained by the number of factors of 5. Since there are only 49 factors of 5, each with an available factor of 2 to pair with, there are exactly 49 factors of 10 in 200!.
It follows that 200! has 49 terminating zeroes, and the correct answer is C.
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
Answer : C
OE - To determine how many terminating zeroes a number has, we need to determine how many times the number can be divided evenly by 10. (For example, the number 404000 can be divided evenly by 10 three times, as follows:
404000/10 = 40400
40400/10 = 4040
4040/10 = 404
We can see that the number has three terminating zeroes because it is divisible by 10 three times.) Thus, to arrive at an answer, we need to count the factors of 10 in 200! Recall that
200! = 200 * 199 * 198 ........... * 4 * 3 * 2 * 1
Each factor of 10 consists of one prime factor of 2 and one prime factor of 5. Let’s start by counting the factors of 5 in 200!. Starting from 1, we get factors of 5 at 5, 10, 15, . . . , 190, 195, and 200, or every 5th number from 1 to 200. Thus, there are 200/5 or 40 numbers divisible by five from 1 to 200. Therefore, there are at least 40 factors of 5 in 200!.
We cannot stop counting here, because some of those multiples of 5 contribute more than just one factor of 5. Specifically, any multiple of 52 (or 25) and any multiple of 53 (or 125) contribute additional factors of 5. There are 8 multiples of 25 (namely, 25, 50, 75, 100, 125, 150, 175, 200). Each of these 8 numbers contributes one additional factor of 5 (in addition to the one we already counted) so we now have counted 40 + 8, or 48 factors of 5. Finally, 125 contributes a third additional factor of 5, so we now have 48 + 1 or 49 total factors of 5 in 200!.
Let us now examine the factors of 2 in 200!. Since every even number contributes at least one factor of 2, there are at least 100 factors of 2 in 200! (2, 4, 6, 8 . . .etc). Since we are only interested in the factors of 10 — a factor of 2 paired with a factor of 5 — and there are more factors of 2 than there are of 5, the number of factors of 10 is constrained by the number of factors of 5. Since there are only 49 factors of 5, each with an available factor of 2 to pair with, there are exactly 49 factors of 10 in 200!.
It follows that 200! has 49 terminating zeroes, and the correct answer is C.
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