IS x^4 + y^4 > z^4 ?
a) x^2 + y^2 > z^2
b) x + y > z
Answer - E
(1) x^2 + y^2 > z^2, when squared, gives the stated equation.
From this we cannot conclude definitively whether x^4 + y^4 > z^4 because the equation contains (2x^2 * y^2).
If this is removed, then x^4 + y^4 may or may not be > z^4.
Thus insufficient..
e.g - 2+3+4 > 5 --- if we remove 1 number from the left hand side of the inequality then the inequality may or may not hold true..
OR
Statement (1) ---- x^2 + y^2 > z^2
Let x = {(2) ^ 1/2}, y = {(3) ^ 1/2}, z = {(4) ^1/2}
Hence 2+3>4 but at the same time 4+9<16
Let x = 2, y= 4, z=3
4+16>9
And 16 + 256 > 81
Thus (1) is insufficient.
Statement (2) ---- x+y> z
Let x=2, y=3, z=6
Hence 2+3<6
Now let x=2, y=6, z=3
Then 2+6>3
Both 1 and 2 together:insufficient
Hence answer E.
Tuesday, March 06, 2007
Manhattan Challenge Problem of the week! - 7 march 07
The function f(n) = the number of factors of n. If p and q are positive integers and f(pq) = 4, what is the value of p?
(1) p + q is an odd integer
(2) q is less than p
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Answer - E
The product pq factors: {1, p, q, and pq}. If pq has no additional factors, then positive integers p and q must be prime. Pay heed to the fact that neither p nor q can be equal to 1; otherwise, pq would not have 4 distinct factors.
Statement (1) -- the sum of p and q is an odd integer. Therefore either p is even and q is odd, or p is odd and q is even. Since we know that both p and q are prime, either p or q must be equal to 2 (the only even prime number). But we do not know which of the two integers is equal to 2.
Statement (2) -- q is less than p -- gives us no information about the value of p
Taking both statements together, we can conclude that, because 2 is the smallest prime number, q must equal 2. But, we cannot determine the value of p.
(1) p + q is an odd integer
(2) q is less than p
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Answer - E
The product pq factors: {1, p, q, and pq}. If pq has no additional factors, then positive integers p and q must be prime. Pay heed to the fact that neither p nor q can be equal to 1; otherwise, pq would not have 4 distinct factors.
Statement (1) -- the sum of p and q is an odd integer. Therefore either p is even and q is odd, or p is odd and q is even. Since we know that both p and q are prime, either p or q must be equal to 2 (the only even prime number). But we do not know which of the two integers is equal to 2.
Statement (2) -- q is less than p -- gives us no information about the value of p
Taking both statements together, we can conclude that, because 2 is the smallest prime number, q must equal 2. But, we cannot determine the value of p.
Friday, March 02, 2007
DS Question - 15
A certain clothing manufacturer makes only two types of men's blazer: cashmere and mohair. Each cashmere blazer requires 4 hours of cutting and 6 hours of sewing. Each mohair blazer requires 4 hours of cutting and 2 hours of sewing. The profit on each cashmere blazer is $40 and the profit on each mohair blazer is $35. How many of each type of blazer should the manufacturer produce each week in order to maximize its potential weekly profit on blazers?
1) The company can afford a maximum of 200 hours of cutting per week and 200 hours of sewing per week.
2) The wholesale price of cashmere cloth is twice that of mohair cloth.
(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.
Answer - A
First, let c be the number of cashmere blazers produced in any given week and let m be the no of mohair blazers produced in any given week.Let p be the total profit on blazers for any given week.Since the profit on cashmere blazers is $40 per blazer and the profit on mohair blazers is $35 per blazer, we can form the equation p = 40c + 35m. In order to know the maximum potential value of p, we need to know the maximum values of c and m.
Statement (1) tells us that the maximum number of cutting hours per week is 200 and that the maximum number of sewing hours per week is 200.
Since it takes 4 hours of cutting to produce a cashmere blazer and 4 hours of cutting to produce a mohair blazer, we can construct the following inequality: 4c + 4m < = 200.
Since it takes 6 hours of sewing to produce a cashmere blazer and 2 hours of sewing to produce a mohair blazer, we can construct the following inequality: 6c + 2m < = 200 .
In order to maximize the number of blazers produced, the company should use all available cutting and sewing time. So we can construct the following equations:
4c + 4m = 200
6c + 2m = 200
Since both equations equal 200, we can set them equal to each other and solve:
4c + 4m = 6c + 2m -->
2m = 2c -->
m = c -->
4m + 4(m) = 200 -->
8m = 200 -->
m = 25 -->
m = c -->
c = 25
So when m = 25 and c = 25, all available cutting and sewing time will be used. If p = 40c + 35m, the profit in this scenario will be 40(25) + 35(25) or $1,875. Is this the maximum potential profit?
Since the profit margin on cashmere is higher, might it be possible that producing only cashmere blazers would be more profitable than producing both types? If no mohair blazers are made, then the largest number of cashmere blazers that could be made will be the value of c that satisfies 6c = 200 (remember, it takes 6 hours of sewing to make a cashmere blazer). So c could have a maximum value of 33 (the company cannot sell 1/3 of a blazer). So producing only cashmere blazers would net a potential profit of 40(33) or $1,320. This is less than $1,875, so it would not maximize profit.
Since mohair blazers take less time to produce, perhaps producing only mohair blazers would yield a higher profit. If no cashmere blazers are produced, then the largest number of mohair blazers that could be made will be the value of m that satisfies 4m = 200 (remember, it takes 4 hours of cutting to produce a mohair blazer). So m would have a maximum value of 50 in this scenario and the profit would be 35(50) or $1,750. This is less than $1,875, so it would not maximize profit.
So producing only one type of blazer will not maximize potential profit, and producing both types of blazer maximizes potential profit when m and c both equal 25.
Statement (1) is sufficient.
Statement (2) tells us that the wholesale cost of cashmere cloth is twice that of mohair cloth. This information is irrelevant because the cost of the materials is already taken into account by the profit margins of $40 and $35 given in the question stem.
Statement (2) is insufficient.
The answer is A: Statement (1) alone is sufficient, but statement (2) alone is not.
1) The company can afford a maximum of 200 hours of cutting per week and 200 hours of sewing per week.
2) The wholesale price of cashmere cloth is twice that of mohair cloth.
(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.
Answer - A
First, let c be the number of cashmere blazers produced in any given week and let m be the no of mohair blazers produced in any given week.Let p be the total profit on blazers for any given week.Since the profit on cashmere blazers is $40 per blazer and the profit on mohair blazers is $35 per blazer, we can form the equation p = 40c + 35m. In order to know the maximum potential value of p, we need to know the maximum values of c and m.
Statement (1) tells us that the maximum number of cutting hours per week is 200 and that the maximum number of sewing hours per week is 200.
Since it takes 4 hours of cutting to produce a cashmere blazer and 4 hours of cutting to produce a mohair blazer, we can construct the following inequality: 4c + 4m < = 200.
Since it takes 6 hours of sewing to produce a cashmere blazer and 2 hours of sewing to produce a mohair blazer, we can construct the following inequality: 6c + 2m < = 200 .
In order to maximize the number of blazers produced, the company should use all available cutting and sewing time. So we can construct the following equations:
4c + 4m = 200
6c + 2m = 200
Since both equations equal 200, we can set them equal to each other and solve:
4c + 4m = 6c + 2m -->
2m = 2c -->
m = c -->
4m + 4(m) = 200 -->
8m = 200 -->
m = 25 -->
m = c -->
c = 25
So when m = 25 and c = 25, all available cutting and sewing time will be used. If p = 40c + 35m, the profit in this scenario will be 40(25) + 35(25) or $1,875. Is this the maximum potential profit?
Since the profit margin on cashmere is higher, might it be possible that producing only cashmere blazers would be more profitable than producing both types? If no mohair blazers are made, then the largest number of cashmere blazers that could be made will be the value of c that satisfies 6c = 200 (remember, it takes 6 hours of sewing to make a cashmere blazer). So c could have a maximum value of 33 (the company cannot sell 1/3 of a blazer). So producing only cashmere blazers would net a potential profit of 40(33) or $1,320. This is less than $1,875, so it would not maximize profit.
Since mohair blazers take less time to produce, perhaps producing only mohair blazers would yield a higher profit. If no cashmere blazers are produced, then the largest number of mohair blazers that could be made will be the value of m that satisfies 4m = 200 (remember, it takes 4 hours of cutting to produce a mohair blazer). So m would have a maximum value of 50 in this scenario and the profit would be 35(50) or $1,750. This is less than $1,875, so it would not maximize profit.
So producing only one type of blazer will not maximize potential profit, and producing both types of blazer maximizes potential profit when m and c both equal 25.
Statement (1) is sufficient.
Statement (2) tells us that the wholesale cost of cashmere cloth is twice that of mohair cloth. This information is irrelevant because the cost of the materials is already taken into account by the profit margins of $40 and $35 given in the question stem.
Statement (2) is insufficient.
The answer is A: Statement (1) alone is sufficient, but statement (2) alone is not.
Problem Solving - 15
During a behavioral experiment in a psychology class, each student is asked to compute his or her lucky number by raising 7 to the power of the student's favorite day of the week (numbered 1 through 7 for Monday through Sunday respectively), multiplying the result by 3, and adding this to the doubled age of the student in years, rounded to the nearest year. If a class consists of 28 students, what is the probability that the median lucky number in the class will be a non-integer?
(A) 0%
(B) 10%
(C) 20%
(D) 30%
(E) 40%
Answer - A
Since any power of 7 is odd, the product of this power and 3 will always be odd. Adding this odd number to the doubled age of the student (an even number, since it is the product of 2 and some integer) will always yield an odd integer. Therefore, all lucky numbers in the class will be odd.
The results of the experiment will yield a set of 28 odd integers, whose median will be the average of the 14th and 15th greatest integers in the set. Since both of these integers will be odd, their sum will always be even and their average will always be an integer. Therefore, the probability that the median lucky number will be a non-integer is 0%.
(A) 0%
(B) 10%
(C) 20%
(D) 30%
(E) 40%
Answer - A
Since any power of 7 is odd, the product of this power and 3 will always be odd. Adding this odd number to the doubled age of the student (an even number, since it is the product of 2 and some integer) will always yield an odd integer. Therefore, all lucky numbers in the class will be odd.
The results of the experiment will yield a set of 28 odd integers, whose median will be the average of the 14th and 15th greatest integers in the set. Since both of these integers will be odd, their sum will always be even and their average will always be an integer. Therefore, the probability that the median lucky number will be a non-integer is 0%.
Manhattan Challenge Problem of the week ! - 02/26/07
What is xy?
(1)
(2)
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Answer : A
Simplifying the original expression yields: 
Therefore: xy = 0 or y — x = 0. Our two solutions are: xy = 0 or y = x.
Statement (1) says y > x so y cannot be equal to x. Therefore, xy = 0. Statement (1) is sufficient.
Statement (2) says x < x =" y" xy =" 0.">
The correct answer is A.
Monday, February 26, 2007
DS Question - 14
How many odd integers are greater than integer X and less than the integer Y?
1). there are 12 even integers greater than X and less than Y
2). there are 24 integers greater than X and less than Y
Answer -- B
From statement 1) -- We cannot determine about both X, Y being even or odd. ... hence insufficient
From statement 2) -- There are 24 integers between X and Y
Let it start with an even integer ...if so then it will end with an odd integer or if it starts with an odd integer then it will end with an even integer ...hence there will always be 12 odd and 12 even integers in the total of 24 consecutive integers
e.g consider X=2, Y=27 thus the series will be is 2 ... 26, 27
X=3, Y=28, thus the series will be 3, ... 27, 28
In each case, the total number of odd integers is the same ... hence sufficient
1). there are 12 even integers greater than X and less than Y
2). there are 24 integers greater than X and less than Y
Answer -- B
From statement 1) -- We cannot determine about both X, Y being even or odd. ... hence insufficient
From statement 2) -- There are 24 integers between X and Y
Let it start with an even integer ...if so then it will end with an odd integer or if it starts with an odd integer then it will end with an even integer ...hence there will always be 12 odd and 12 even integers in the total of 24 consecutive integers
e.g consider X=2, Y=27 thus the series will be is 2 ... 26, 27
X=3, Y=28, thus the series will be 3, ... 27, 28
In each case, the total number of odd integers is the same ... hence sufficient
Problem solving - 14
Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
Answer - A -- generally answer for such questions is the sum of the series of integers from 1 to n+1 where n = number of items to be distributed.
In this ques n = 5, thus the answer is 21.
For n = 6 it will be 28. For n = 7 it will be 36.....
However here is the explanation based on the counting method...
1) 1 person gets all 5 donuts -
Possibility: 3.
2). 2 persons get all 5 donuts -
The one without donuts - possibility - 3.
The other 2 persons have 4 ways.
Hence in total -- 3*4 = 12.
3). All have donuts -
The way to divide can only be - 1, 2, 2; 1, 3, 1
But this arrangement can get interchanged midst 3 persons, hence it becomes --
(3!)/ (2) + (3!)/ (2) = 6.
(3!)/ (2) is needed because donuts are all same without difference.
3! is counting the same arrangement twice.
Hence the answer - 3 + 12 + 6 = 21.
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
Answer - A -- generally answer for such questions is the sum of the series of integers from 1 to n+1 where n = number of items to be distributed.
In this ques n = 5, thus the answer is 21.
For n = 6 it will be 28. For n = 7 it will be 36.....
However here is the explanation based on the counting method...
1) 1 person gets all 5 donuts -
Possibility: 3.
2). 2 persons get all 5 donuts -
The one without donuts - possibility - 3.
The other 2 persons have 4 ways.
Hence in total -- 3*4 = 12.
3). All have donuts -
The way to divide can only be - 1, 2, 2; 1, 3, 1
But this arrangement can get interchanged midst 3 persons, hence it becomes --
(3!)/ (2) + (3!)/ (2) = 6.
(3!)/ (2) is needed because donuts are all same without difference.
3! is counting the same arrangement twice.
Hence the answer - 3 + 12 + 6 = 21.
Thursday, February 22, 2007
Problem solving - 13
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%
Answer - C
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%
OR
we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....
A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%
Answer - C
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%
OR
we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....
Problem solving - 12
If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?
A. 10
B. 12
C. 14
D. 16
E. 18
Answer -- C
P = 1*2*3*.....*30
if u factorize P, what is the power of 3 ?
So we have to find out how many 3's are there in he product.
As is known 3,6,9,12,15,18,21,24,27,30 is total count of 10 numbers
but
9 = 3*3 thus 1 extra 3
18 = 3*3*2 thus 1 extra 3
27 = 3*3*3 thus 2 extra 3
Thus in total 10 + 1 + 1 + 2 = 14 3's are there in product.
Hence k is 14.
A. 10
B. 12
C. 14
D. 16
E. 18
Answer -- C
P = 1*2*3*.....*30
if u factorize P, what is the power of 3 ?
So we have to find out how many 3's are there in he product.
As is known 3,6,9,12,15,18,21,24,27,30 is total count of 10 numbers
but
9 = 3*3 thus 1 extra 3
18 = 3*3*2 thus 1 extra 3
27 = 3*3*3 thus 2 extra 3
Thus in total 10 + 1 + 1 + 2 = 14 3's are there in product.
Hence k is 14.
Problem solving - 11
A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13?
(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30
Answer -- B
The first 30 natural nos are 1,2,3.....28,29,30.
There are 10 multiples of 3 in the above range and there are 2 multiples of 13 in the same range.
Hence there are total 12 i.e (10+2) nos which can be either a multiple of 3 or 13.
Total numbers = 30.
So the probability is 12/30 = 2/5.
(A) 17/30
(B) 2/5
(C) 7/15
(D) 4/15
(E) 11/30
Answer -- B
The first 30 natural nos are 1,2,3.....28,29,30.
There are 10 multiples of 3 in the above range and there are 2 multiples of 13 in the same range.
Hence there are total 12 i.e (10+2) nos which can be either a multiple of 3 or 13.
Total numbers = 30.
So the probability is 12/30 = 2/5.
Problem solving - 10
A infinite sequence is 1, 11, 111, 1111, 11111, ..., what is the tens digit of the sum of the first 40 terms?
A) 1
B) 2
C) 3
D) 4
E) 9
Answer -- C
The sum of the units is 40. Then add 4 to the tens , whose sum is 39 and 3 is 43 which gives 3 as tens digit
A) 1
B) 2
C) 3
D) 4
E) 9
Answer -- C
The sum of the units is 40. Then add 4 to the tens , whose sum is 39 and 3 is 43 which gives 3 as tens digit
Thursday, January 25, 2007
Manhattan Challenge Problem of the week ! - Jan 8
If x = 3 ^ 21 and y = 6 ^ 55, what is the remainder when xy is divided by 10?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
Answer : The correct answer is E.
Since every multiple of 10 must end in zero, the remainder from dividing xy by 10 will be equal to the units’ digit of xy. In other words, the units’ digit will reflect by how much this number is greater than the nearest multiple of 10 and, thus, will be equal to the remainder from dividing by 10. Therefore, we can rephrase the question: “What is the units’ digit of xy?”
Next, let’s look for a pattern in the units’ digit of 3 ^ 21. Remember that the GMAT will not expect you to do sophisticated computations; therefore, if the exponent seems too large to compute, look for a shortcut by recognizing a pattern in the units' digits of the exponent:
3 ^ 1 = 3
3 ^ 2 = 9
3 ^ 3 = 27
3 ^ 4 = 81
3 ^ 5 = 243
As you can see, the pattern repeats every 4 terms, yielding the units digits of 3, 9, 7, and 1. Therefore, the exponents 3 ^ 1, 3 ^ 5, 3 ^ 9, 3 ^ 13, 3 ^ 17, and 3 ^ 21 will end in 3, and the units’ digit of 3 ^ 21 is 3.
Next, let’s determine the units’ digit of 6 ^ 55 by recognizing the pattern:
6 ^ 1 = 6
6 ^ 2 = 36
6 ^ 3 = 256
6 ^ 4 = 1,296
As shown above, all positive integer exponents of 6 have a units’ digit of 6. Therefore, the units' digit of 6 ^ 55 will also be 6.
Finally, since the units’ digit of 3 ^ 21 is 3 and the units’ digit of 6 ^ 55 is 6, the units' digit of 3 ^ 21 × 6 ^ 55 will be equal to 8, since 3 × 6 = 18.
Therefore, when this product is divided by 10, the remainder will be 8.
(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
Answer : The correct answer is E.
Since every multiple of 10 must end in zero, the remainder from dividing xy by 10 will be equal to the units’ digit of xy. In other words, the units’ digit will reflect by how much this number is greater than the nearest multiple of 10 and, thus, will be equal to the remainder from dividing by 10. Therefore, we can rephrase the question: “What is the units’ digit of xy?”
Next, let’s look for a pattern in the units’ digit of 3 ^ 21. Remember that the GMAT will not expect you to do sophisticated computations; therefore, if the exponent seems too large to compute, look for a shortcut by recognizing a pattern in the units' digits of the exponent:
3 ^ 1 = 3
3 ^ 2 = 9
3 ^ 3 = 27
3 ^ 4 = 81
3 ^ 5 = 243
As you can see, the pattern repeats every 4 terms, yielding the units digits of 3, 9, 7, and 1. Therefore, the exponents 3 ^ 1, 3 ^ 5, 3 ^ 9, 3 ^ 13, 3 ^ 17, and 3 ^ 21 will end in 3, and the units’ digit of 3 ^ 21 is 3.
Next, let’s determine the units’ digit of 6 ^ 55 by recognizing the pattern:
6 ^ 1 = 6
6 ^ 2 = 36
6 ^ 3 = 256
6 ^ 4 = 1,296
As shown above, all positive integer exponents of 6 have a units’ digit of 6. Therefore, the units' digit of 6 ^ 55 will also be 6.
Finally, since the units’ digit of 3 ^ 21 is 3 and the units’ digit of 6 ^ 55 is 6, the units' digit of 3 ^ 21 × 6 ^ 55 will be equal to 8, since 3 × 6 = 18.
Therefore, when this product is divided by 10, the remainder will be 8.
Manhattan Challenge Problem of the week ! - Jan 15
Sn = 2Sn-1 + 4 and Qn = 4Qn-1 + 8 for all n > 1. If S5 = Q4 and S7 = 316, what is the first value of n for which Qn is an integer?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Answer : C is the correct answer.
If S7 = 316, then 316 = 2S6 + 4, which means that S6=156.
We can then solve for S5:
156 = 2S5 + 4, so S5 = 76
Since S5 = Q4, we know that Q4 = 76 and we can now solve for previous Qn’s to find the first n value for which Qn is an integer.
To find Q3: 76 = 4Q3 + 8, so Q3 = 17
To find Q2: 17 = 4Q2 + 8, so Q2 = 9/2
It is clear that Q1 will also not be an integer so there is no need to continue.
Q3 (n = 3) is the first integer value.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Answer : C is the correct answer.
If S7 = 316, then 316 = 2S6 + 4, which means that S6=156.
We can then solve for S5:
156 = 2S5 + 4, so S5 = 76
Since S5 = Q4, we know that Q4 = 76 and we can now solve for previous Qn’s to find the first n value for which Qn is an integer.
To find Q3: 76 = 4Q3 + 8, so Q3 = 17
To find Q2: 17 = 4Q2 + 8, so Q2 = 9/2
It is clear that Q1 will also not be an integer so there is no need to continue.
Q3 (n = 3) is the first integer value.
Tuesday, January 23, 2007
Manhattan Challenge Problem of the week ! - Jan 1
"Smurfs, Elves and Fairies"
One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?
(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7
Answer : D is the correct answer.
The combined rate of individuals working together is equal to the sum of all the individual working rates.
Let s = rate of a smurf, e = rate of an elf, and f = rate of a fairy. A rate is expressed in terms of treehouses/hour. So for instance, the first equation below says that a smurf and an elf working together can build 1 treehouse per 2 hours, for a rate of 1/2 treehouse per hour.
1) s + e = 1/2
2) s + 2 f = 1/2
3) e + f = 1/4
The three equations can be combined by solving the first one for s in terms of e, and the third equation for f in terms of e, and then by substituting both new equations into the middle equation.
1) s = 1/2 – e
2) (1/2 – e) + 2 (1/4 – e) = 1/2
3) f = 1/4 – e
Now, we simply solve equation 2 for e:
(1/2 – e) + 2 (1/4 – e) = 1/2
2/4 – e + 2/4 – 2 e = 2/4
4/4 – 3e = 2/4
-3e = -2/4
e = 2/12
e = 1/6
Once we know e, we can solve for s and f:
s = 1/2 – e
s = 1/2 – 1/6
s = 3/6 – 1/6
s = 2/6s = 1/3
f = 1/4 – e
f = 1/4 – 1/6
f = 3/12 – 2/12
f = 1/12
We add up their individual rates to get a combined rate:
e + s + f
=1/6 + 1/3 + 1/12
=2/12 + 4/12 + 1/12
= 7/12
Remembering that a rate is expressed in terms of treehouses/hour, this indicates that a smurf, an elf, and a fairy, working together, can produce 7 treehouses per 12 hours. Since we want to know the number of hours per treehouse, we must take the reciprocal of the rate. Therefore we conclude that it takes them 12 hours per 7 treehouses, which is equivalent to 12/7 of an hour per treehouse.
The correct answer is D.
One smurf and one elf can build a treehouse together in two hours, but the smurf would need the help of two fairies in order to complete the same job in the same amount of time. If one elf and one fairy worked together, it would take them four hours to build the treehouse. Assuming that work rates for smurfs, elves, and fairies remain constant, how many hours would it take one smurf, one elf, and one fairy, working together, to build the treehouse?
(A) 5/7
(B) 1
(C) 10/7
(D) 12/7
(E) 22/7
Answer : D is the correct answer.
The combined rate of individuals working together is equal to the sum of all the individual working rates.
Let s = rate of a smurf, e = rate of an elf, and f = rate of a fairy. A rate is expressed in terms of treehouses/hour. So for instance, the first equation below says that a smurf and an elf working together can build 1 treehouse per 2 hours, for a rate of 1/2 treehouse per hour.
1) s + e = 1/2
2) s + 2 f = 1/2
3) e + f = 1/4
The three equations can be combined by solving the first one for s in terms of e, and the third equation for f in terms of e, and then by substituting both new equations into the middle equation.
1) s = 1/2 – e
2) (1/2 – e) + 2 (1/4 – e) = 1/2
3) f = 1/4 – e
Now, we simply solve equation 2 for e:
(1/2 – e) + 2 (1/4 – e) = 1/2
2/4 – e + 2/4 – 2 e = 2/4
4/4 – 3e = 2/4
-3e = -2/4
e = 2/12
e = 1/6
Once we know e, we can solve for s and f:
s = 1/2 – e
s = 1/2 – 1/6
s = 3/6 – 1/6
s = 2/6s = 1/3
f = 1/4 – e
f = 1/4 – 1/6
f = 3/12 – 2/12
f = 1/12
We add up their individual rates to get a combined rate:
e + s + f
=1/6 + 1/3 + 1/12
=2/12 + 4/12 + 1/12
= 7/12
Remembering that a rate is expressed in terms of treehouses/hour, this indicates that a smurf, an elf, and a fairy, working together, can produce 7 treehouses per 12 hours. Since we want to know the number of hours per treehouse, we must take the reciprocal of the rate. Therefore we conclude that it takes them 12 hours per 7 treehouses, which is equivalent to 12/7 of an hour per treehouse.
The correct answer is D.
Manhattan Challenge Problem of the Week ! - dec 18
Chicago Rain
If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4
Answer: Correct ans is C. For OE click on the link below.
Manhattan Challenge Problem
If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?
(A) 1/32
(B) 2/25
(C) 5/16
(D) 8/25
(E) 3/4
Answer: Correct ans is C. For OE click on the link below.
Manhattan Challenge Problem
Manhattan Challenge Problem of the week ! - dec 25
Is n less than 1 ?
(1) nx – n less than 0
(2) x–1 = –2
Answer : C
From stat(1): (n^x) – n less than 0 -- no information about x --- insufficient
From stat(2): x^–1 = –2 -- no information about n --- insufficient
Taking both the statements together we get: From stat 2 we get x = -1/ 2
Substituting value of x in stat 1 we get (n^-1/2) -n less than 0
(n^3/2)>1
=> n will be greater than 1 ---- sufficient
Hence C
(1) nx – n less than 0
(2) x–1 = –2
Answer : C
From stat(1): (n^x) – n less than 0 -- no information about x --- insufficient
From stat(2): x^–1 = –2 -- no information about n --- insufficient
Taking both the statements together we get: From stat 2 we get x = -1/ 2
Substituting value of x in stat 1 we get (n^-1/2) -n less than 0
(n^3/2)>1
=> n will be greater than 1 ---- sufficient
Hence C
Thursday, September 28, 2006
Problem Solving - 9
9). AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined
Answer - D
Explanation -- AB + CD = AAA
AB and CD are two digit numbers, hence AAA must be 111
Therefore 1B + CD = 111
B can assume any value between 3 and 9
If B = 3, then CD = 111-13 = 98 and C = 9
If B = 9, then CD = 111-19 = 92 and C = 9
Hence for all values of B between 3 & 9 ---- C = 9
Thus the answer is D i.e (C = 9)
(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined
Answer - D
Explanation -- AB + CD = AAA
AB and CD are two digit numbers, hence AAA must be 111
Therefore 1B + CD = 111
B can assume any value between 3 and 9
If B = 3, then CD = 111-13 = 98 and C = 9
If B = 9, then CD = 111-19 = 92 and C = 9
Hence for all values of B between 3 & 9 ---- C = 9
Thus the answer is D i.e (C = 9)
Tuesday, September 05, 2006
Manhattan Challenge Problem of the week! - sept 4
The Power of Absolutes
If a and b are integers and a is not equal to b, is ab > 0?
(1) a^b > 0
(2) a^b is a non-zero integer
(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.
If a and b are integers and a is not equal to b, is ab > 0?
(1) a^b > 0
(2) a^b is a non-zero integer
(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.
Wednesday, August 30, 2006
Manhattan Challenge Problem of the week ! - Aug 28
High Functioning
The vertical position of an object can be approximated at any given time by the function: p(t) = rt – 5t2 + b where p(t) is the vertical position in meters, t is the time in seconds, and r and b are constants. After 2 seconds, the position of an object is 41 meters, and after 5 seconds the position is 26 meters. What is the position of the object, in meters, after 4 seconds?
(A) 24
(B) 26
(C) 39
(D) 41
(E) 45
Answer - D , for OE click on link below.
Manhattan challenge problem
The vertical position of an object can be approximated at any given time by the function: p(t) = rt – 5t2 + b where p(t) is the vertical position in meters, t is the time in seconds, and r and b are constants. After 2 seconds, the position of an object is 41 meters, and after 5 seconds the position is 26 meters. What is the position of the object, in meters, after 4 seconds?
(A) 24
(B) 26
(C) 39
(D) 41
(E) 45
Answer - D , for OE click on link below.
Manhattan challenge problem
Manhattan Challenge Problem of the week ! - Aug 7
Question
If j and k are positive integers where k > j, what is the value of the remainder when k is divided by j?
(1) There exists a positive integer m such that k = jm + 5.
(2) j > 5
(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.
For the answer click on the link below -
What remains to be seen ?
If j and k are positive integers where k > j, what is the value of the remainder when k is divided by j?
(1) There exists a positive integer m such that k = jm + 5.
(2) j > 5
(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.
For the answer click on the link below -
What remains to be seen ?
Manhattan Challenge Problem of the week ! - Aug 14
Question
A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
For answer click on the link below -
Rush paint job
A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
For answer click on the link below -
Rush paint job
Manhatten Challenge Problem of the week ! - Aug 21
Question
What is the ratio of 2x to 3y?
(1) The ratio of x^2 to y^2 is equal to 36/25.
(2) The ratio of x^5 to y^5 is greater than 1.
(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.
For the answer click on link below -
Ratios and exponents
What is the ratio of 2x to 3y?
(1) The ratio of x^2 to y^2 is equal to 36/25.
(2) The ratio of x^5 to y^5 is greater than 1.
(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.
For the answer click on link below -
Ratios and exponents
Monday, July 31, 2006
Manhattan Challenge Problem of the week ! july 24
Is the positive integer x odd?
(1) x = y2 + 4y + 6, where y is a positive integer.
(2) x = 9z2 + 7z - 10, where z is a positive integer.
(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.
Answer - The correct answer is B.
For complete solution click on the link below.
Manhattan challenge problem
(1) x = y2 + 4y + 6, where y is a positive integer.
(2) x = 9z2 + 7z - 10, where z is a positive integer.
(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.
Answer - The correct answer is B.
For complete solution click on the link below.
Manhattan challenge problem
Saturday, July 29, 2006
Maths - Prime Factors
Methods and tricks to solve questions related to Prime factors.
1). Counting the Number of Factors -- If you factor a number into its prime power factors, then the total number of factors is found by adding one to all the exponents and multiplying those results together.
Example: The total number of factors of 108 are --
108 = (2^2) * (3^3) thus we add 1 to the exponents and multiply the results together.
(2+1)*(3+1) = 3*4 = 12.
Hence total number of factors of 108 are 12.
Verifying the above
The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108 , which are total 12 in number.
2). Factoring Factorials - First write the number you are factoring as a product of two or more numbers. For example, suppose we want to factor the number 30.
We know that 30 = 5 x 6,
so we write for the first step of our factor tree:
30
/ \
5 6
Next we factor the factors, if possible. In other words, for each number in the product from the first step, we try to write it down as a product of even smaller numbers.
For instance, in our example, we would try to factor both 5 and 6. As we noted above, 5 is prime, so we can't factor it further.
We are done with that branch of the factor tree.
However, 6 is not prime. 6 = 2 x 3, so we can extend our factor tree as follows:
30
/ \
5 6
Now further
6
/ \
2 3
Now, we continue this process until all of the branches of the tree end in prime numbers.
In our example, we are done after two steps, since 5, 2, and 3 are all prime numbers.
The factors of the number are the numbers at the end of the different branches on the factor tree. To figure out what power each prime factor is raised to, count the number of times the prime factor appears in the factor tree.
In our example, each of the factors appears only once, so the prime factorization of 30 is: 30 = 2 x 3 x 5.
1). Counting the Number of Factors -- If you factor a number into its prime power factors, then the total number of factors is found by adding one to all the exponents and multiplying those results together.
Example: The total number of factors of 108 are --
108 = (2^2) * (3^3) thus we add 1 to the exponents and multiply the results together.
(2+1)*(3+1) = 3*4 = 12.
Hence total number of factors of 108 are 12.
Verifying the above
The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108 , which are total 12 in number.
2). Factoring Factorials - First write the number you are factoring as a product of two or more numbers. For example, suppose we want to factor the number 30.
We know that 30 = 5 x 6,
so we write for the first step of our factor tree:
30
/ \
5 6
Next we factor the factors, if possible. In other words, for each number in the product from the first step, we try to write it down as a product of even smaller numbers.
For instance, in our example, we would try to factor both 5 and 6. As we noted above, 5 is prime, so we can't factor it further.
We are done with that branch of the factor tree.
However, 6 is not prime. 6 = 2 x 3, so we can extend our factor tree as follows:
30
/ \
5 6
Now further
6
/ \
2 3
Now, we continue this process until all of the branches of the tree end in prime numbers.
In our example, we are done after two steps, since 5, 2, and 3 are all prime numbers.
The factors of the number are the numbers at the end of the different branches on the factor tree. To figure out what power each prime factor is raised to, count the number of times the prime factor appears in the factor tree.
In our example, each of the factors appears only once, so the prime factorization of 30 is: 30 = 2 x 3 x 5.
Thursday, July 27, 2006
Problem Solving - 8
How many factors of 2940 are NOT factors of 112?
(A) 22
(B) 30
(C) 32
(D) 34
(E) None of these
Answer - B
Prime factors of 2940 = (2 ^ 2) * 3 * 5 * (7 ^ 7)
Hence total number of factors of 2940
= (2 + 1) * ( 1 + 1) * (1 + 1) * ( 2 + 1) = 36
(by the method of counting number of factors)
Prime factors of 112 = (2 ^ 4) * 7
Therefore common factors for the above two numbers are
2^2 * 7
Total number of factors for (2 ^ 2) * 7 = (2 + 1) * ( 1 + 1) = 6
Hence total factors of 2940 which are not factors of 112 are
36 - 6 = 30, Hence B is the answer.
(A) 22
(B) 30
(C) 32
(D) 34
(E) None of these
Answer - B
Prime factors of 2940 = (2 ^ 2) * 3 * 5 * (7 ^ 7)
Hence total number of factors of 2940
= (2 + 1) * ( 1 + 1) * (1 + 1) * ( 2 + 1) = 36
(by the method of counting number of factors)
Prime factors of 112 = (2 ^ 4) * 7
Therefore common factors for the above two numbers are
2^2 * 7
Total number of factors for (2 ^ 2) * 7 = (2 + 1) * ( 1 + 1) = 6
Hence total factors of 2940 which are not factors of 112 are
36 - 6 = 30, Hence B is the answer.
Thursday, July 20, 2006
Problem Solving - 7
Eight points lie on the circumfrence of a circle.What is the positive diff b/w th no. of triangles and the no. of quadrilaterals that can be formed by connecting these points?
a. 8
b. 14
c. 56
d. 70
e. 1,344
Answer - B
Since the points are on the circumference of the circle thus no 3 or 4 points will be colinear, thus all points can be used to draw a triangle or quadilateral
Hence 8C4 = 70 and 8C3 = 56 (we make use of combination here because here the order of points do not matter)
Taking the difference we get 70 - 56 = 14.
a. 8
b. 14
c. 56
d. 70
e. 1,344
Answer - B
Since the points are on the circumference of the circle thus no 3 or 4 points will be colinear, thus all points can be used to draw a triangle or quadilateral
Hence 8C4 = 70 and 8C3 = 56 (we make use of combination here because here the order of points do not matter)
Taking the difference we get 70 - 56 = 14.
DS Question 13
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.
Answer - B
Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of
p1 ^ q1 * p2 ^ q2 * ....... pk ^qk
where p1 is less than p2 , p2 is less than p3 and so on.
Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of
2 * p1^q1 * p2^q2 * .... * pk ^ qk
but if n is even then 2n can be factorised as
p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1
Hence , now the total number of factors is based on choice of prime powers for each prime in term.
Thus if n is odd, then 2n must have twice as many factors as n.
Alternatively in simpler language
If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8
But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10
Thus if n is odd the 2nd condition is satisfied. Hence ans is B
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.
Answer - B
Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of
p1 ^ q1 * p2 ^ q2 * ....... pk ^qk
where p1 is less than p2 , p2 is less than p3 and so on.
Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of
2 * p1^q1 * p2^q2 * .... * pk ^ qk
but if n is even then 2n can be factorised as
p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1
Hence , now the total number of factors is based on choice of prime powers for each prime in term.
Thus if n is odd, then 2n must have twice as many factors as n.
Alternatively in simpler language
If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8
But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10
Thus if n is odd the 2nd condition is satisfied. Hence ans is B
Monday, July 17, 2006
Manhattan Challenge problem of the Week!
Dig those Digits - Challenge problem of the week
Click on the link below to see the question and its solution.
Manhattan challenge problem
Click on the link below to see the question and its solution.
Manhattan challenge problem
Wednesday, July 05, 2006
DS Question - 12
Does x + y = xy?
(1) x is neither a positive integer nor a negative integer
(2) y is neither a positive integer nor a negative integer
Answer - E - x and y can be either 0 or fractions.. which makes both conditions insufficient.
(1) x is neither a positive integer nor a negative integer
(2) y is neither a positive integer nor a negative integer
Answer - E - x and y can be either 0 or fractions.. which makes both conditions insufficient.
Manhattan Challenge Problem of the Week - July 5
Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compunded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month?
(A) 333
(B) 383
(C) 402
(D) 433
(E) 483
Answer - C
For solution click on the link below
Manhattan challenge problem
(A) 333
(B) 383
(C) 402
(D) 433
(E) 483
Answer - C
For solution click on the link below
Manhattan challenge problem
Problem Solving - 6
For every positive integer n, the function h(n) is defined to be the product of all of the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
A. between 2 and 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
Answer - E
h(n)=2*4*6....100 =2(1*2*3*........50)
h(n) + 1 = 2(1*2*...50) + 1
Prime No ={2,3,........}
h(n) contains multiples of all the primes uptil 50 .... Hence E
OR
Let X = 2*4*6*.....*100
then X contains multiples of all the primes uptil 50
eg.
17 * 2 = 34 is in X.
37 * 2 = 74 is in X
Hence all the primes between 2 - 50 will be a factor of X => that none of them will be a factor of (X+1), hence the smallest prime should be at least greater than 50
Hence E.
A. between 2 and 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
Answer - E
h(n)=2*4*6....100 =2(1*2*3*........50)
h(n) + 1 = 2(1*2*...50) + 1
Prime No ={2,3,........}
h(n) contains multiples of all the primes uptil 50 .... Hence E
OR
Let X = 2*4*6*.....*100
then X contains multiples of all the primes uptil 50
eg.
17 * 2 = 34 is in X.
37 * 2 = 74 is in X
Hence all the primes between 2 - 50 will be a factor of X => that none of them will be a factor of (X+1), hence the smallest prime should be at least greater than 50
Hence E.
DS Question - 11
a, b, c all are positive, is a/b > (a+c)/(b+c) ?
1. a > c
2. a > b
Answer - B
statement (1). - insufficient as we do not know the relation between a and b
statement (2). - a > b gives answer irrespective of whether ( c <> b) that, a / b > (a + c) / (b + c)
a/b > [a+c]/[b+c]
= a(b+c) > b(a+c)
= ab+ac > ab+bc
= ac > bc
=a > b
1. a > c
2. a > b
Answer - B
statement (1). - insufficient as we do not know the relation between a and b
statement (2). - a > b gives answer irrespective of whether ( c <> b) that, a / b > (a + c) / (b + c)
a/b > [a+c]/[b+c]
= a(b+c) > b(a+c)
= ab+ac > ab+bc
= ac > bc
=a > b
Tuesday, July 04, 2006
DS Question - 10
When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15 ?
(1) n- 2 is divisible by 5.
(2) t is divisible by 3.
Answer - C
Statement (1) alone - no sufficient information about t
Statement (2)alone - no sufficient information about n
From the ques --- n = a*3 + 2
t = b*5 + 3
Using Statement (1) --- n = p*5 + 2
Thus n = 15*k + 2
Using Statement (2) --- t = q*3 + 3
Thus t = 15*j + 3
nt = 15*k(15*j + 3) + 30*j + 6
=> always leaves a remainder of 6 when divided by 15.
(1) n- 2 is divisible by 5.
(2) t is divisible by 3.
Answer - C
Statement (1) alone - no sufficient information about t
Statement (2)alone - no sufficient information about n
From the ques --- n = a*3 + 2
t = b*5 + 3
Using Statement (1) --- n = p*5 + 2
Thus n = 15*k + 2
Using Statement (2) --- t = q*3 + 3
Thus t = 15*j + 3
nt = 15*k(15*j + 3) + 30*j + 6
=> always leaves a remainder of 6 when divided by 15.
Problem Solving - 5
If |x| > 3, which of the following must be true?
A). x>3
B). x<3
C). x=3
D). x is not equal to 3
E). x<-3
Answer - D
A,B,E - Ruled out - a modulus sign in the question=>cannot have a single signed (can always find a solution with the opposite sign )
When x is positive ----- x >3
When x is negative ---- x <-3
Thus x can never be equal to 3.
A). x>3
B). x<3
C). x=3
D). x is not equal to 3
E). x<-3
Answer - D
A,B,E - Ruled out - a modulus sign in the question=>cannot have a single signed (can always find a solution with the opposite sign )
When x is positive ----- x >3
When x is negative ---- x <-3
Thus x can never be equal to 3.
DS Question - 9
Find an integer S.
(1) S=3
(2) S^3=3^S
Answer - B
Statement (1) - insufficient ---- S=3 or –3
Statement (2) - sufficient ---- S = 3
X^Y = Y^X only if X and Y are positive integers of equal value
=> X and Y have to be positive, otherwise one side will give a negative value when X or Y is odd while other side will give a decimal value
Given that Y is equal to 3, X has to be equal to 3.
(1) S=3
(2) S^3=3^S
Answer - B
Statement (1) - insufficient ---- S=3 or –3
Statement (2) - sufficient ---- S = 3
X^Y = Y^X only if X and Y are positive integers of equal value
=> X and Y have to be positive, otherwise one side will give a negative value when X or Y is odd while other side will give a decimal value
Given that Y is equal to 3, X has to be equal to 3.
Monday, July 03, 2006
Problem Solving - 4
What is the sum of all three digit positive integers, the sum of whose digits is a multiple of 3?
a) 900
b) 165150
c) 165000
d) 329199
e) 328900
Answer - B
Sum of n numbers = n/2(2a+(n-1)d)
We want to add the numbers from 102 to 999 that are divisible by 3.
We divide by 3 to get 34 to 333, or 300 numbers in total.
Here n = 300, a = 34, d= 1
Sum = 1/2*300*(333+34) = 150 * 367 = 55,050
Multiplying by 3 again gives 165,150 .
TIP
To add the first n integers from 1 to n
add together 1 + n = n+1
then 2 + (n-1) = n + 1 until you get to n/2 (depending on whether n is odd or even)
Ultimately we have n/2 pairs that all add up to (n+1)
So the sum is 1/2 * n * (n+1)
If you don't start at 1, or have a gap of more than 1 --- slightly adjust the formula.
a) 900
b) 165150
c) 165000
d) 329199
e) 328900
Answer - B
Sum of n numbers = n/2(2a+(n-1)d)
We want to add the numbers from 102 to 999 that are divisible by 3.
We divide by 3 to get 34 to 333, or 300 numbers in total.
Here n = 300, a = 34, d= 1
Sum = 1/2*300*(333+34) = 150 * 367 = 55,050
Multiplying by 3 again gives 165,150 .
TIP
To add the first n integers from 1 to n
add together 1 + n = n+1
then 2 + (n-1) = n + 1 until you get to n/2 (depending on whether n is odd or even)
Ultimately we have n/2 pairs that all add up to (n+1)
So the sum is 1/2 * n * (n+1)
If you don't start at 1, or have a gap of more than 1 --- slightly adjust the formula.
DS Question - 8
If X and Y are nonzero integers, what is the remainder when X is divided by Y?
(1) When X is divided by 2y, the remainder is 4
(2) When X+Y is divided by y, the remainder is 4
Answer - B
Statement (1) ---- x = (2y)M +4
if y > 4 the remainder of x/y will be 4
if y= 1 or 2 or 4 there will be no remainder,
if y=3, remainder will be 1,
so it depends on value of y...
so insufficient
Statement (2) -----(x+y)/y=M+4
It is clear that y/y=1 therefore (x/y) has to be 4 - Thus
sufficient
(1) When X is divided by 2y, the remainder is 4
(2) When X+Y is divided by y, the remainder is 4
Answer - B
Statement (1) ---- x = (2y)M +4
if y > 4 the remainder of x/y will be 4
if y= 1 or 2 or 4 there will be no remainder,
if y=3, remainder will be 1,
so it depends on value of y...
so insufficient
Statement (2) -----(x+y)/y=M+4
It is clear that y/y=1 therefore (x/y) has to be 4 - Thus
sufficient
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