Manhattan Challenge Problem of the week! - 04/23/07

Which of the following sets includes ALL of the solutions of x that will satisfy the equation:

Answer: OA - C

OE - One way to solve equations with absolute values is to solve for x over a series of intervals. In each interval of x, the sign of the expressions within each pair of absolute value indicators does not change.

In the equation Ix-2I - Ix-3I = Ix-5I , there are 4 intervals of interest:

x less than 2 - In this interval, the value inside each of the three absolute value expressions is negative.

2 less than x less than 3 - In this interval, the value inside the first absolute value expression is positive, while the value inside the other two absolute value expressions is negative.

3 less than x less than 5 - In this interval, the value inside the first two absolute value expressions is positive, while the value inside the last absolute value expression is negative.

5 less than x - In this interval, the value inside each of the three absolute value expressions is positive.

Use each interval for x to rewrite the equation so that it can be evaluated without absolute value signs.

For the first interval, x less than 2, we can solve the equation by rewriting each of the expressions inside the absolute value signs as negative (and thereby remove the absolute value signs):

- x + 2 - ( -x + 3 ) = - x + 5

- x + 2 + x - 3 = - x + 5

x = 6

Notice that the solution x = 6 is NOT a valid solution since it lies outside the interval x less than 2. (Remember, we are solving the equation for x SUCH THAT x is within the interval of interest).

For the second interval 2 less than x less than 3, we can solve the equation by rewriting the expression inside the first absolute value sign as positive and by rewriting the expressions inside the other absolute values signs as negative:

x - 2 - ( -x + 3 ) = - x + 5

x - 2 + x - 3 = - x + 5

3x = 10

x = 10/3

Notice, again, that the solution x = 10/3 is NOT a valid solution since it lies outside the interval 2 less than x less than 3

For the third interval 3 less than x less than 5, we can solve the equation by rewriting the expressions inside the first two absolute value signs as positive and by rewriting the expression inside the last absolute value sign as negative:

x - 2 - ( x - 3 ) = - x + 5

x - 2 - x + 3 = - x + 5

x = 4

The solution x = 4 is a valid solution since it lies within the interval 3 less than x less than 5

Finally, for the fourth interval 5 less than x, we can solve the equation by rewriting each of the expressions inside the absolute value signs as positive:

x - 2 - ( x - 3 ) = x - 5

x - 2 - x + 3 = x - 5

x = 6

The solution x = 6 is a valid solution since it lies within the interval 5 less than x.

We conclude that the only two solutions of the original equation are x = 4 and x = 6. Only answer choice C contains all of the solutions, both 4 and 6, as part of its set.

Therefore, C is the correct answer.

Manhattan Challenge Problem of the week! - 16/04/07

x and y are positive integers. , what is the value of xy?

(A) 48
(B) 36
(C) 24
(D) 18
(E) 12

Answer: OA - E

5^x - 5^y = 2^(y-1) * 5^(x-1)
Divide both sides by 5^(x-1)....we get
[5^x/ 5^(x-1)] - [5^y/ 5^(x-1)] = 2^(y-1)

Simplifying the first tem we get

5 - [5^x/ 5^(x-1)] = 2^(y-1)

Add [5^y/ 5^(x-1)] to both sides...we get

5 = 2^(y-1) + [5^y/ 5^(x-1)]

We know that 2^(y-1) and [5^y/ 5^(x-1)] are positive and each of them is less than 5

We can now list all the possibilities, by testing small integer values for y. We only have to test a few values because we know that 2^(y-1) must be less than 5, which means that y can only be 1, 2 or 3. (If y is 4, then 2^(y-1) would be 8, which is greater than 5.)

If y = 1, then 2^(y-1) = 2^(1-1) = 2^0 , which means that [5^y/ 5^(x-1)] must equal 4 (remember, the sum of the two terms must be equal to 5). However, since x and y are positive integers, there is no way to make [5^y/ 5^(x-1)] equal to 4.

If y = 2, then 2^(y-1) = 2^(2-1) = 2^1 = 2, which means that [5^y/ 5^(x-1)] must equal 3. However, since x and y are positive integers, there is no way to make [5^y/ 5^(x-1)] equal to 3.

If y = 3, then 2^(y-1) = 2^(3-1) = 2^2 = 4 , which means that cmust equal 1. This is possible. Using y = 3, we can solve for x as follows:

[5^y/ 5^(x-1)] = 1
=> [5^3/ 5^(x-1)] = 1
=> 5^3 = 5^(x-1)
=> 3 = x-1
=> x = 4

The only possible solution is y = 3 and x = 4. Therefore, xy = (4)(3) = 12. The correct answer is E.

Problem Solving - 21

Mrs. Napueta has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Napueta been given if there are 120 ways that she could distribute the vouchers?

(A) 13

(B) 14

(C) 15

(D) 16

(E) more than 16


Answer: C

Formula - Number of ways of distributing n identical things among r persons when each person may get any number of things = C(n+r-1, r-1)

Hence

Let total number of vouchers = n

All 4 nephews get atleast 2 vouchers, thus remaining vouchers to be distributed among 4 nephews where each can get 0 or more vouchers = n-8

Using the formula above we get = C(n-8+4-1, 3) = (n-8+4-1)C3 = 120 = (n-5)(n-6)(n-7) = 120 * 3! = 720

Hence n = 15

DS Question - 26

Is 5^k less than 1,000?

(1) 5^(k-1) greater than 3,000

(2) 5^(k-1) = 5^k - 500


Answer: D

From statement (1) - 5^(k-1) is greater than 3000

=> 5^k/5 is greater than 3000
=> 5^k is greater than 15000

Hence sufficient

From statement (2) - 5^(k-1) = 5^k - 500

=> 5^(k-1) = 5^k - 500
=> 5^k - 5^(k-1) = 500
=> 5^k(1- 5^-1) = 500
=> 5^k(4/5) = 500
=> 5^k = 2500/4 = 625
Therefore 5^k is less than 1000

Hence sufficient

DS Question - 25

Is 5^k less than 1,000?

(1) 5^(k-1) is less than 3000
(2) 5^(k-1) = 5^k - 500

Answer: B

From statement (1) - It is given that 5^(k-1) less than 3000
=> 5^k/5 less than 3000
=> 5^k less than 15000
=> 5^k could be either more than 1,000 or less than 1,000.
Hence insufficient

From statement (2) - It is given that 5^(k-1) = 5^k - 500
=> 5^k/5 = (5^k) - 500
=> 5^k = 5*(5^k) - 2500
=> (5-1)*5^k = 2500
=> 4*5^k = 2500
=> 5^k = 2500/4 = 625

Hence 5^k less than 1,000.

Hence sufficient.

DS Question - 24

What is the 999th term of the series S ?

(1) The first 4 four terms of S are (1 + 1)² , (2 + 1)² , (3 + 1)² , and (4 + 1)².
(2) For every x, the xth term of S is (x + 1)².

Answer: B

From statement (1) - It is given that the 1st 4 four terms of S are (1 + 1)² , (2 + 1)² , (3 + 1)² , and (4 + 1)². We only know the first four terms. We cannot assume that terms following the first 4 terms will be like -- nth term = (n+1)^2.
There can be a sequence that follows a different rule for any term in it

From statement (2) - It is given that for every x, the xth term of S is (x + 1)². Hence sufficient as it implies "for any number x that refers to the term in S."

Problem Solving - 20

What is the area of the trapezoid pictured to the right?

(A) 22.5

(B) 24

(C) 27

(D) 45

(E) 54


Answer: A

Since the triangle is a right isosceles triangle, the other side of the triangle (the height of the trapezoid) must also be 3. The top base if the trapezoid must be 6 since it is the opposite side of a rectangle.

The area of a trapezoid = 1/2 (sum of parallel sides) × (distance between the parallel sides)
= 1/2 (9 + 6) (3) = 22.5

The area can also be found here by breaking up the figure into a rectangle (area of 6 × 3 = 18) and a triangle (area of ½ × 3 × 3 = 4.5) and adding these two areas.

The correct answer is A.

Problem Solving - 19

How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500


Answer: Given OA - A is incorrect ... ans is 2688 which is not an option in the choices...

When 2nd and 3rd digit gets repeated:

The first digit will be a non zero even (2, 4, 6, 8) = 4 ways
3rd digit is a non even prime = (3, 5, 7) = 3 ways
2nd digit is a REPEAT of that prime: 1 way
the fourth digit has not been used: 8 ways
the fifth digit has not been used: 7 ways

Hence 4*3*8*7 = 672 ways

Now, the non repeating case:

1st digit will be a non zero even (2, 4, 6, 8) = 4 ways
3rd digit (3, 5, 7) = 3 ways
2nd (no repeat and odd) = 4 ways
4th digit = 7
5th digit = 6

Hence 4*3*4*7*6 = 2016

Total number of ways = 2016 + 672 = 2688 ways


Note:
1). 1 is neither a prime nor a composite number.
2). 2 is the only even prime number.

DS Question - 23

If x is a positive integer, then how many factors does x have ?

(1) x is divisible by one more positive integer than 3^4 is.

(2) x is the product of three different prime numbers


Answer: D

From statement (1) - x has 5 factors 1, 3, 9, 27, 81. Adding one more factor from statement (1) makes total number of factors to be six.
Thus x has 6 factors.
Hence sufficient

From statement (2) - x = abc where a, b, c are prime numbers. Thus in all x has 8 factors - 1, a, b, c, ab, bc, ac, abc.
Thus x has 8 factors.
Hence sufficient


Note: Here each condition gives a different answer but still satifies the sufficient conditions.

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.

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.

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.

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

Further n cannot be equal to zero or negative as n = number of the term and not the value of the term.

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) = 5^x2^y3^z. 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.

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