Tuesday, April 10, 2007

Manhattan Challenge Problem of the week! - 09/04/07

Stephanie, Regine, and Brian ran a 20 mile race. Stephanie and Regine's combined times exceeded Brian's time by exactly 2 hours. If nobody ran faster than 8 miles per hour, who could have won the race?

I. Stephanie II. Regine III. Brian

(A) I only
(B) II only
(C) III only
(D) I or II only
(E) I, II, or III

Answer: D

Let the race time of Stephanie = S
Let the race time of Regine = R
Let the race time of Brian = B

It is given that -- Stephanie and Regine's combined times exceed Brian's time by 2 hours.

Hence S + R = B + 2 ------- (1)

To win the race, an individual's time must be less than 1/3 of the the combined times of all the runners.

Therefore for Brian to win the race (=> that Brian would have the lowest time) his time would need to be less than 1/3 of the combined times for all the runners.

=> B is less than 1/3 of (S+R+B)

=>3B<> is less than (S+R+B)

=> 2B is less than (S+R)

Making use of (1) in above

2B<> is less than (S+R)

=> 2B<> is less than B+2


=> B<2



Thus to win the race Brian's time must be less than 2 hours. which is impossible as fastest Brian run is 8 miles/hr => that the least amount of time in which he can complete the 20 mile race is 2.5 hrs.

Hence Stephanie and Regine as possible winners. Since the question gives us same information about Stephanie and Regine, we cannot state either one as a possible winner.

Hence, the correct answer is D




Manhattan Challenge Problem of the week! - 02/04/07

Triangle A has one side of length x. If (x^8) ^ 1/2 = 81 , what is the perimeter of Triangle A?

1) Triangle A has sides whose lengths are consecutive integers
2) Triangle A is NOT a right triangle

(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 : C

OE - By simplifying the equation given in the question stem, we can solve for x as follows:

(x^8) ^ 1/2 = 81

x^4 = 81

x = 3


Thus, we know that one side of Triangle A has a length of 3.

Statement (1) tells us that Triangle A has sides whose lengths are consecutive integers. Given that one of the sides of Triangle A has a length of 3, this gives us the following possibilities: (1, 2, 3) OR (2, 3, 4) OR (3, 4, 5).

However, the first possibility is NOT a real triangle, since it does not meet the following condition, which is true for all triangles: The sum of the lengths of any two sides of a triangle must always be greater than the length of the third side. Since 1 + 2 is not greater than 3, it is impossible for a triangle to have side lengths of 1, 2 and 3.

Thus, Statement (1) leaves us with two possibilities. Either Triangle A has side lengths 2, 3, 4 and a perimeter of 9 OR Triangle A has side lengths 3, 4, 5 and a perimeter of 12. Since there are two possible answers, Statement (1) is not sufficient to answer the question.

Statement (2) tells us that Triangle A is NOT a right triangle. On its own, this is clearly not sufficient to answer the question, since there are many non-right triangles that can be constructed with a side of length 3.

Taking both statements together, we can determine the perimeter of Triangle A.

From Statement (1) we know that Triangle A must have side lengths of 2, 3, and 4 OR side lengths of 3, 4, and 5. Statement (2) tells us that Triangle A is not a right triangle; this eliminates the possibility that Triangle A has side lengths of 3, 4, and 5 since any triangle with these side lengths is a right triangle (this is one of the common Pythagorean triples). Thus, the only remaining possibility is that Triangle A has side lengths of 2, 3, and 4, which yields a perimeter of 9.

The correct answer is C: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

Thursday, March 29, 2007

DS Question - 22

n is a positive integer. What is the remainder when n is divided by 6?

(1) n is a multiple of 3.
(2) When n is divided by 2, the remainder is 1.

Answer: C

From statement (1) - n can be 3, 6, 9, 12...

Therefore:

(i) If n is odd, the reminder is 3 on dividing by 6.
(ii) If n is even, the reminder is 0 on dividing by 6.

Hence Insufficient.

From statement (2) - When n is divided by 2, the remainer is 1. Hence this is an odd integer as when an odd number is divided by 2, the remainder is 1.

n = 2*k + 1

(i) If n = 7 then n/6 = 6*1 + 1
(ii) If n = 9 then n/6 = 6*1 + 3

Hence insufficient.

taking statement (1) and (2) together - From Statement (1) we know that we need to know that whether n is odd or even in order to be able to conclude.Statement (2) gives us this information.

Hence Sufficient

Hence, C is the answer.

Wednesday, March 28, 2007

DS Question - 21

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Answer : B

It is given that 9x + 3y <= 20
It is required to prove that 12x + 12y <= 40 or 6x + 6y <= 20
Hence basically the question asks that whether three notebooks can be replaced by 3 pencils in the same price range or not....

Statement (1) - suggests that 2 notebooks can be exchanged for two pencils in the same price range --- well here third can always create some problem in the exchange...hence insufficient to answer the question asked...

Statement (2) - suggests that 4 notebooks can be replaced by 4 pencils in the same price range i.e 20 Swiss francs ----- Hence if 4 notebooks can be exchanged for 4 pencils in the same price range then definitely three notebooks can be exchanged for 3 pencils in the same price range.

Hence statement 2 is sufficient to answer the question.

Thus ans is B

OR

The solution to above problem can be found out graphically:
In the graph above :
Line AB with the axis covers the area representing the information given in the question --- 9x + 3y <= 20
Line ED with the axis covers the area representing the question asked - 6x + 6y <= 20
Statement (1) gives the area covered by line CD with the axis i.e 7x + 5y <= 20.
Clearly we can see the shaded area EFC (covered by line ED) lying outside the area covered by line CD. Hence statement (1) is insufficient to answer the question.
Statement (2) gives the area covered by line drawn from point F to the X - axis. Area given by statement (2) and the area given by line AB together cover the area covered by line ED.
Hence statement (2) is sufficient to answer the question.

Problem Solving - 18

If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is

A) 6
B) 12
C) 24
D) 36
E) 48

Answer : B

Focus primarily on the word "must" and after that shift your focus to the word "largest"

n^2 = 72q = 36*2q = 6^2*2q , where 2q must be the square of a number
q = 2p^2 , where p is a +ve integer
Hence n = 12p

As such 12 is the largest possible number that must divide n.

Other numbers that must divide n would be 1,2,3,4,6.

Apart from these the possible numbers that may divide n could be any numbers, depending on what p is

Manhattan Challenge Problem of the week! - 26/03/07

If 3x - 2y - z = 32 + z and (3x) ^ 1/2 - (2y + 2z) ^ 1/2 = 4 , what is the value of x + y + z ?

(A) 3
(B) 9
(C) 10
(D) 12
(E) 14

Answer: E


Each equation represents one of the elements in the common quadratic form:
Rewrite the given equation as follows:

3x - 2y - z = 32 + z
3x - (2y + 2z) = 32


Then, notice its relationship to the second given equation:

(3x) ^ 1/2 - (2y + 2z) ^ 1/2 = 4

The second equation is in the form a - b = 4,

while the first equation is in the form a ^ 2 - b ^ 2 = 32.

Since we know that and that a ^ 2 - b ^ 2 = 32 and that a - b = 4
we can solve for a + b , which must equal 8.

This gives us a third equation:

(3x) ^ 1/2 + (2y + 2z) ^ 1/2 = 8

Adding the second and third equations allows us to solve for x as follows:

(3x) ^ 1/2 - (2y + 2z) ^ 1/2 = 4
(3x) ^ 1/2 + (2y + 2z) ^1/2 = 8
_______________________

2{3 ^(1/2)} = 12
{3 ^(1/2)} = 6
3x = 36
x = 12

Plugging this value for x into the first equation allows us to solve for y + z as follows:

3x - 2y - 2z = 32
3(12) - 2y - 2z = 32
- 2y - 2z = -4
y + z = 2

The question asks for the value of x + y + z.

If x = 12 and y + z = 2, then x + y + z = 12 + 2 = 14.

The correct answer is E.

Friday, March 23, 2007

Probelm Solving - 17

On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)

Answer : C


Midpoint of the diagonal = (3,4)
Slope of the diagonal = -2/3
Slope of the other diagonal is the negative reciprocal of the first diagonal = 3/2

Let one of the vertices = (a,b).

The midpoint (diagonal) and the slope (diagonal) gives us the 1st equation = (y-4)/(x-3) = 3/2

The length of the diagonal = 2*sqrt13.

Calculate the length of the diagonal from the midpoint to the vertex (1/2 of diagonal) .

This gives us our 2nd equation = (x-3)^2 + (y-4)^2 = 13


Solving both the equations :

(x,y) = (1,1)

Thus, the distance is sqrt 2

DS Question - 20

Series of A(n) is such that A(n) = A(n-1) / n. How many elements of the series are bigger than 1/2 ?

(1) A(2) = 5
(2) A(1) - A(2) = 5

Answer : D

From statement (1) --- A(2) = 5 . It is given that A(n) = A(n-1) / n.
Hence A(2) = A(1) / 2
=> A(1) = 10
A(2) = 5
=A(3) = A(2) / 3 = 5 /3
A(4) = A(3) / 4 = 5 /( 3 * 4) = 5 /12

Hence statement (1) alone is sufficient to aswer the question.

From statement (2) --- A(1) - A(2) = 5
=> A(1) - A(1) /2 = 5
=> A(1) = 10
Hence A(2) = 5
A(3) = A(2) / 3 = 5 /3 ...

Hence statement (2) alone is sufficient to answer the question.

NOTE: Always remember sum of series is

S_n=x_1+x_2+\dots + x_n=\sum\limits_{i=1}^{n}x_i.
Further n cannot be equal to zero or negative as n = number of the term and not the value of the term.

Thursday, March 22, 2007

Problem Solving - 16

Six mobsters have arrived at the theater for the premiere of the film “Goodbuddies.” One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

(A) 6
(B) 24
(C) 120
(D) 360
(e) 720

Answer : D

Frankie wants to keep Joe in his sights, therefore Joe will always be ahead of F in the queue. (take care this doesnot implies that Joe and Frankie will be together)

1. Frankie is in last position in the queue, then Joe can be in any position from 1 to 5 = 5!

2. Frankie is in the 5th position in the queue then Joe can be in any of the positions 1 to 4 (positions ahead of Frankie) i.e 4 ways and rest of mobsters will be positioned in rest 4 places (4!) ways. = 4*4!

3. Frankie is in the 4th position then Joe can be take any place between 1 to 3 (positions ahead of Frankie) 3 ways and rest of the mobsters will be positioned in rest of 4 places i.e 4!ways. = 3*4!

4. Frankie is in the 3rd position then Joe can be in 1 or 2 places (i.e positions ahead of Frankie) 2 ways and rest of mobsters will be positioned in rest of the 4 places (4!) ways. = 2*4!

5. Frankie is in placed in the 2nd position then Joe can only be in 1st position i.e only position ahead him i.e 1 way and rest of mobsters will be placed in the rest of the 4 places (4!) ways. = 1*4!

Thus total no of ways = 5! + ( 4 * 4! ) + ( 3 * 4! ) + ( 2 * 4! ) + ( 1 * 4! ) = 360



Mnahattan Challenge Problem of the week! - 21/03/07

For a three-digit number xyz, where x, y, and z are the digits of the number, the function f(xyz) = 5x2y3z. If f(abc) = 3 * f(def), what is the value of abc - def?

(A) 1
(B) 2
(C) 3
(D) 9
(E) 27


Answer : A
It is given in the question that .
It is obvious that the digits b, c, e, f are integers from 0 to 9 whereas that the digits a and d are integers from 1 to 9 (can't be equal to 0 because they are in the hundreds place).
For the above statement to be true, must be equal to , and .
Hence .

As the only difference between abc and def is in the units digits, the difference between these three-digit numbers is equal to , or 1.

Thus the right answer is A.

Wednesday, March 21, 2007

DS Question - 19

Is x = square root (x^2) if

(1) x = even
(2) 13 is less than x is less than 17


Answer: B

From Statement (1) -- x = even . Always remember square root (x^2) = mod x.
E.g - squareroot (-4 ^ 2) = 4 ; suareroot (4^ 2) = 4. Hence 1 is insufficient.

From Statement (2) --
13 is less than x is less than 17 implies that x can be 14, 15 or 16 ..hence X IS POSITIVE...Hence sufficient.

Note : Is x = square root (x^2) is equivalent to stating Is x = mod x

Wednesday, March 14, 2007

Manhattan Challenge Problem of the week ! - 13/26/07

x years ago, Cory was one fifth as old as Tania. In x years, Tania will be twice as old as Cory. What is the ratio of Cory's current age to Tania's current age?

(A) 7:23
(B) 9:17
(C) 5:13
(D) 3:7
(E) 11:15

Answer : C

Use a chart to keep track of the ages in this problem:


x years ago

NOW

in x years

Corey

C - x

C

C + x

Tania

T - x

T

T + x

Then write algebraic expressions to represent the information given in the problem:

x years ago, Cory was one fifth as old as Tania

5(Cx) = Tx
5C – 5x = Tx
5CT = 4x


In x years, Tania will be twice as old as Cory

2(C + x) = T + x
2C + 2x = T + x
T – 2C = x

Substitute T - 2C in for x in the first equation and solve:

5C - T = 4(T - 2C)
5C - T = 4T - 8C
13C = 5T

C/T = 5/18

The correct answer is C.




Sunday, March 11, 2007

DS Question - 18

Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

1) The probability that the ball will both be white and have an even number painted on it is 0.

2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2.


Answer: E

The question is asking for P(W) or P(E)
=>P(W) or P(E) = P(W) + P(E) - P(W and E)

From statement (1) --- it is given that P(W and E) = 0 --- insufficient as we do not know individual probabilities.
From statement (2) --- it is given that P(W) - P(E) = 0.2 --- insufficient

Combining both statement (1) and (2) it is still insufficient as P(W) + P(E) cannot be calculated....P(W U E) = P(W) + P(E) - P( W and E)

NOTE : P(A and B) = P(A) * P(B) --- This is true only iff both A & B are Independent. Otherwise, it would be -- P(A and B) = P (A) * P(B/A) = P(B) * P(A/B).
From the question we do not know that if both A and B are independent.

Tuesday, March 06, 2007

DS Question - 17

In the xy-plane, the line k passes through the origin and through the point (a, b), where ab is not equal to 0. Is b positive ?

1). The slope of line k is negative.
2). a is less than b

Answer : C

It is given that the line passes through (0,0) and (a,b)
So it's slope = (b-0)/(a-0) = b/a

Statement (1) ---- Slope of line k is (-ve).
=> (b/a) less than 0 implies a and b are of opposite signs.
From the above we cannot conclude that b is less than 0.
Statement (2) ---- a is less than b ...again this alone cannot help us to infer that whether b is greater than 0 or less than 0 as b can take the value either way round.

Combining Statement (1) and (2) ---- we know that a and b are of opposite signs and that a is less than b.
Therefore, clearly a is negative and hence b is positive.

Hence the answer is C.


DS Question - 16

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.


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.



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.

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%.

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

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.

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 %....

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.

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.

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