Thursday, September 09, 2010
A. 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase
rate = k*(A1^2)/B
If concentration of chemical B is increased by 100 percent then
rate = k*(A2^2)/2B
(A2/A1) ^ 2 = 2
A2/A1 = Square root (2)
(A2/A1) - 1 = Square root (2)-1 = 0.414
(A2-A1)/A1 = 0.414 = 40% approximately
members were to retire, and no other employee changes were to occur, what value of y would reduce the percent of
"long-term" employees in the company to 60%.
A) 200
B) 160
C) 112
D) 80
E) 56
Assume y=x
The number of people working more than 10 years = 70% of 800 = 560
Hence (560-x)/(800-x)=60%
Thus x=200
OR
The number of long term workers = 70% of 800 = 560
Now if y of the long term workers retired, then long term workers left are 560-y and the number of total employees
= 800-y
Thus (560-y)/(800-y) * 100 = 60
560-y = 480 - 0.6y
80 = 0.4y
y = 200 Ans
centimeters. What is the maximum possible length in centimeters of the shortest piece of wood?
A. 90
B. 100
C. 110
D. 130
E. 140
Shortest to Longest length ---- L1, L2, L3 = 140, L4, L5
L1 + L2 + 140 + L4 + L5 = 5 * 124
For L1 to be the maximum, L4 and L5 should be minimum
L1=L2
Thus, L1 + L2 = (5*124) - (140*3) = 620 - 420 = 200
Hence L1 = L2 = 100
(1) nx – n less than 0
(2) x–1 = –2
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). there are 12 even integers greater than X and less than Y
2). there are 24 integers greater than X and less than Y
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) 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.
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.
a) x^2 + y^2 > z^2
b) x + y > z
(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.
b positive ?
1). The slope of line k is negative.
2). a is less than b
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.
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.
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.
(1) x = even
(2) 13 is less than x is less than 17
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
(1) A(2) = 5
(2) A(1) - A(2) = 5
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.
(1) n is a multiple of 3.
(2) When n is divided by 2, the remainder is 1.
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.
1) Triangle A has sides whose lengths are consecutive integers
2) Triangle A is NOT a right triangle
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.
(1) x is divisible by one more positive integer than 3^4 is.
(2) x is the product of three different prime numbers
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.
(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.
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."
(1) 5^(k-1) greater than 3,000
(2) 5^(k-1) = 5^k - 500
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
decides to split the driving time equally with his friend George, instead of making the trip alone?
(1) The driving distance from Boston to New Orleans is 1500 miles.
(2) George’s driving speed is 1.5 times Edwin’s driving speed.
OE - The question asks for the percent decrease in Edwin’s travel time. To determine this, we need to be able to
find the ratio between, T1 (the travel time if Edwin drives alone) and T2 (the travel time if Edwin and George
drive together). Note that we do NOT need to determine specific values for T1 and T2; we only need to find the
ratio between them.
Percentage change is defined as follows: Difference/Original = (T1 - T2)/ T1 = 1 - (T2/ T1)
Ultimately, we can solve the percentage change equation above by simply determining the value of T2 /T1
Using the formula Rate × Time = Distance, we can write equations for each of the 2 possible trips
T1 = Travel time if Edwin drives alone
T2 = Travel time if Edwin and George drive together
E = Edwin’s Rate
G = George’s Rate
D = Distance of the trip
If Edwin travels alone: ET1 = D
If Edwin and George travel together: .5(E + G)T2 = D
(Since Edwin and George split the driving equally, the rate for the trip is equal to the average of Edwin and
George’s individual rates).
Since both trips cover the same distance (D), we can combine the 2 equations as follows:
ET1 = .5(E + G)T2
Then, we can isolate the ratio of the times (T2/T1) as follows:
E/ .5(E + G) = T2/ T1
Now we look at the statements to see if they can help us to solve for the ratio of the times.
Statement (1) gives us a value for D, the distance, which does not help us since D is not a variable in the ratio
equation above.
Statement (2) tells us that George’s rate is 1.5 times Edwin’s rate. Thus, G = 1.5E. We can substitute this
information into the ratio equation above:
E/ .5(E + G) = T2/ T1 ---> E/ .5(E + 1.5E) = T2/ T1 ---> E/ .5E + .75E = T2/T1
---> E/ 1.25E = T2/ T1 ---> 1/ 1.25 = T2/ T1 ---> .8 = T2/ T1
Thus, using this ratio we can see that Edwin’s travel time for the trip will be reduced as follows:
1 - (T2/T1) = 1 - .8 = .2 ---> 20%
Statement (2) alone is sufficient to answer the question.
The correct answer is B.
1) the base of the parallelogram is 10
2) one of the angles of the parallelogram is 45 degree
From (1) - For a parallelogram, Area = Base*Height => Height = 10 - There are infinite ways to draw a parallelogram
with 100 as area, as long as the height is 10 units.( parallelograms with varying slants from 1 to 179 degrees) --
insufficent
From (2) - one of the angles is 45 degrees - hence the opposite angle is 45 degrees too, and the two remaining
angles wil be 135 degrees each. But this doesn't tell us the measure of the sides to determine the perimeter -
insufficient
From(1) and (2) together
From (2) we know that there is only one of such parallelograms that has an angle 45 to the base.
Base = 10. One angle = 45 degrees
In Parallelogram opposite angles are equal ..
Hence there are 2*45 degrees
Sum of all angles = 360
Thus 2*45 + 2 x = 360 which gives each other angle as 135 degrees. Hence C
1) The product of the digits in n is 30
2) The sum of the digits in n is 10
From statement (1) : the factors of 30 are 1,2,3,5,6,10,15,30
Now because the number must be digits (single number) we do not need to consider 10,15 and 50
Now if the hundreds digit of the 3-digit number = any digit among the 1,2,3 digits ---> answer to the question is
clearly Yes.
But if the hundreds digit = 5 or 6 ---> answer to the question is No.
Hence (1) alone is insufficient.
From statement (2) : There are different combinations where the sum of digits can be equal to 10. e.g 541 and 145.
Hence (2) alone is insufficient.
Combining statements (1) and (2) we get :
If the hundreds digit= 5 , the second digit can only be 1,2,3 because if the digit is equal to 6 it violates
statement (2) Hence in this case answer to the question is Yes.
If hundreds digit = 6 then the number n must contain 5 so as to satisfy the first statement but then it will
violate statement (2) as the total of digits of n will exceed 10
Thus no 3-digit number exists with the hundreds digit equal to 6 satisfying both the statements together.
Hence the number n will always be less than 550 as it will be the combination of 1, 2, 3 and 5
(1) (y^2 - x^2) = 0
(2) xy/(x+y) = 0
Statement (1) insufficient -- y can take any value i.e can be +ve or -ve. We cannot assume y to be +ve. Therefore y
may or may not be equal to x
Statement (2) sufficient -- xy/(x+y) = 0 => xy = 0 => x = 0 or y = 0
In case x = 0, then y cannot be equal to 0. Hence y cannot be equal to x.
In case y = 0, then x cannot be equal to 0. Hence y cannot be equal to x.
Therefore sufficient.
Hence B
1) r is less than 4
2) q = 18
From Statement (1) -- q can be positive and r can be negative.
r can also be a real number but not necessarily an integer.
Hence insufficient.
From Statement (2) -- q = 18.
But r can be negative or can be positive.
r can also be a real number not an integer
Hence insufficient
Taking both statements (1) and (2) together -- Again r can be positive or negative. And again r can also be a real
number.
Hence insufficient
1) (-x)^3 = -x^3
2) (-x)^2 = -x^2
From statement (1):
if x = 0 both sides are equal
if x = 1 both sides are again equal {(-1)^3 = -1 & -1^3 = 1}
=> x = 0 or x = 1
Hence insufficient
From statement (2):
x can only be zero because the square of a number other than zero cannot be negative
{(-1)^2 = 1 which is not equal to -(1)^2)}
=> from above it is sufficient to say that x = 0
Hence sufficient
1). (x - 1)^2 less than and equal to 1
2). x^2 - 1 greater than 0
|x-1| less than 1 is only true when 0 less than x less than 1
From statement (1): (x-1)^2<=1
True when 0<=x<=2
If x=0.5, then |x-1| less than 1 is true
If x=2, then |x-1| less than 1 is not true
Hence insufficient
From statement (2): x^2>1 means x>1 and x<-1
True when x=1.5, but not when x=3
Hence insufficient
Statement (1) and (2) together: 1 is less than x is less than and equal to 2
Taking x=1.5 and x=2
Hence insufficient
1). a+b= -1
2). The graph intersects the y-axis at (0,-6)
From Statement (1) -- a+b = -1...no information about a and b ...hence insufficient
From Statement (2) -- If x = 0 the y = -6 thus ab = 6....insufficient
Taking statements (1) and (2) together: (x+a)*(x+b)=0
x^2+(a+b)x+ab=0
Hence x=-3, x=2
Thus the answer C

A student worked 20 days. For each of the amount shown (see attached table) in the first row of the table, second row gives the number of days the student earned that amount. Median amount of money earned per day for 20 days is?
A) 96
B) 84
C) 80
D) 70
E) 48
Median day = 20+1)/2 = 10.5 th -- money earned was 84 = Average value of 10th and 11th day in the sequence = Median amount of money Average value of 10th day = 84 Average value of 11th day = 84 Average value of 10th and 11th day = 84 ans
Strategies for Averages and Statistics
2. The median is the "middle" number in a group (when arranged in ascending or descending order) consisting of an odd number of numbers, and the average of the two middle numbers if there are an even number of numbers
3. For a set of consecutive integers, the median is the the average of the first and the last integer
4. Mode is the most frequently recurring number/numbers among the given set of numbers. It can be more than one
5. Range is the difference between the largest number and smallest number is a set
6. Calculation of Standard Deviation (SD):
1. Find the mean, \scriptstyle\overline{x}, of the values.
2. For each value xi calculate its deviation (\scriptstyle x_i - \overline{x}) from the mean.
3. Calculate the squares of these deviations.
4. Find the mean of the squared deviations. This quantity is the variance σ2.
5. Take the square root of the variance.
7. Variance is the square of the standard deviation
8. SD does not change when the same constant is added or subtracted to all the members of the set
9. If mean = maximum value it means that all values are equal and SD is 0
10. A set of numbers with range of zero means that all of the numbers are the same, hence the dispersion of the numbers from its mean is zero
11. For data with approximately the same mean, the greater the spread, the greater the SD.
12. SD is the square root of the average of the sum of square of the variation from the mean
13. The more uneven members are dispersed around their arithmetic average, the more their SD
14. You only need to know the difference between values and total number of values to compute SD
15. If we know all the numbers of the list, there is a definite SD, regardless of what it is, we can compute it and get an answer – this is helpful for DS questions
16. If the range is 0, then the SD must also be 0, because there is no variance
17. The SD of any list is not dependent on the average, but on the deviation of the numbers from the average. So just by knowing that two lists having different averages doesn't say anything about their standard deviation - different averages can have the same SD
18. The sum of the deviations of the elements from the mean must be 0
19.Closer the more values to the MEAN, lower the SD
20. If Range or SD of a list is 0, then the list will contain all identical elements
21. Standard Deviation is also useful when comparing the spread of two separate data sets that have approximately the same mean. The data set with the smaller Standard Deviation has a narrower spread of measurements around the mean and therefore usually has comparatively fewer high or low values.
In general, the more widely spread the values are, the larger the Standard Deviation is.
22. If you multiply all terms by x then SD =x times old SD and mean = x times old mean
23. For comparing the SD for two sets any information about mean ,median,mode and range are insufficient unless you can determine the individual terms from the given data
24. Symmetric about the mean means that the shape of the distribution on the right and left side of the curve are mirror-images of each other
25. For a given set of consecutive even numbers.. mean = median
26. When you have a set of consecutive numbers (integers, evens, odds, multiples), the mean is equal to the median
if you have the averages of all the FRACTIONS or PERCENTAGES of a group, then you'll be able to calculate the overall average of the group. This is a worthwhile fact to memorize for the data sufficiency problems.
- Note ----- Make sure you know that, when ALL numbers in a set are multiplied or divided by some number,** the mean and standard deviation are multiplied/divided by the same number
**This includes increasing or decreasing all the numbers in the set by some percentage (which can be accomplished by multiplication: e.g., 30% increase = multiplication by 1.3).
Strategies for Set Theory
u = union
n = intersection
1. For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
2. No of persons in exactly one set:
P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)
3. No of persons in exactly two of the sets: P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)
4. No of persons in exactly three of the sets: P(AnBnC)
5. No of persons in two or more sets: P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)
6. No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)
1. For three sets A, B, and C, P(AuBuC): (A+B+C+X+Y+Z+O)
2. Number of people in exactly one set: ( A+B+C)
3. Number of people in exactly two of the sets: (X+Y+Z)
4. Number of people in exactly three of the sets: O
5. Number of people in two or more sets: ( X+Y+Z+O)
6. Number of people only in set A: A
7. P(A): A+X+Y+O
8. P( AnB): X+O
Notes :
* In Datasufficiency problems, do not assume that overlap between sets (using double matrix problems) does not exists!!
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
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) 6
(B) 24
(C) 120
(D) 360
(e) 720
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
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.
A) 2951
B) 8125
C) 15600
D) 15302
E) 18278
Total ways to pick a 3 letter code = 26*26*26 = 26^3
Total ways to pick a 2 letter code = 26*26 = 26^2
Total ways to pick a 1 letter code = 26
Thus total stocks: 26^3 + 26^2 + 26 = 26(26^2 + 26 + 1) = 18278
A. 10/189
B. 15/196
C. 16/225
D. 25/144
E. 39/ 128
RULE: If denominator of a fraction has just the prime factors of 2 or 5 or both it is terminating otherwise not
128 = 2*2*2*2*2*2*2 => has only factors of 2..hence the ans E.
1). the median of the numbers in S is less than 5.
2). the median of the numbers in S is greater than 1
From statement (1): Median will be less than 5 only if n is located below 5
Thus the median will either be 1 if n less than 1 or n if 1 less than n less than 5
Hence in both cases n is less than 5 but it can also be n less than 0 ....insufficient
From statement (2): Median will be greater than 1 only if n is located above 1
Thus median will either be 5 if n greater than 5 or n if 1 less than n less than 5
Hence in both cases n greater than 1 but it can also be n greater than 7 ....insufficient
Taking statements (1) and (2) together: 1 less than n less than 5 which lies within the given interval 0 less than n less than 7 ...thus possible values for n are be 2, 3, or 4
Hence the answer C
1). 16 students at the school study both French and Japanese.
2). 10 percent of the students at the school who study Japanese also study French.
From statement (1):16 students study both French and Japanese, so 16/0.04=400 students study French. But we don't know how many the total number of students are so the number Japanese student can be at least 100 or more than 400....insufficient
From statement (2): 10% Japanese studying students = 4% French studying students
10/100 study Japanese = 4/100 study French
25 study Japanese = 10 study French
Hence study French > study Japanese ... thus more students at the school study French than study Japanese....sufficient
(1) q^4 is not a multiple of 128.
(2) q^2 has 27 factors, 7 of which are less than or equal to 10
OE:
From statement (1): Given q is an integer, thus q can be written as product of distinct prime factors, where the power of 2 must be a whole number(a non-negative integer). This implies that the power of 2 in the prime factorization of q^4 must be a multiple of 4. As 2^7 is not a factor of q^4, the highest power of 2 that could be a factor of q^4 is 2^4.
Hence q^4 is not a multiple of 64=2^6......sufficient
From statement (2): Given that q^2 has 27 factors. Thus q^2 could be of the form
(i) a^26, or
(ii)b^2*c^8 or
(iii) a^2*b^2*c^2 where a b and c are distinct prime numbers.
Because q^2 has 7 factors less than 11, (i) is impossible.
As for (ii) 2^8*3^2 has exactly 7 factors under 11 (1,2,3,4,6,8,9), in which case q^4 would be a multiple of 64. 3^8*2*2 is not a possibility for q^2 (it only has 6 factors under 1: 1,2,3,4,6,9)
Regarding (iii) if q^2=2^2*3^2*5^2, q^2 would have 8 factors under 11- 1,2,3,4,5,6,9,10 and q^2=2^2*3^2*7^2 would have 7 factors under 11: 1,2,3,4,6,7,9) In this case, q^4 would not be a multiple of 64. Thus (2) is insufficient
hence the answer A
A. $0.60
B. $0.70
C. $0.80
D. $0.90
E. $1.00
Day 1 = 0.10
Day 2 = 0.20
Day 3 = 0.20*2 = 0.40
Thus on 4th day the total fine is: 0.10 + 0.20 + 0.40 = 0.70
(1) n *triangle* 0 = n for all integers n.
(2) n *triangle* n = 0 for all integers n.
From statement (1): *triangle* can be both positive or negative as
n-0 = n
n+0 = n
Hence insufficient
From statement (2): *triangle* can only be negative in this case as
n-n = 0
Hence sufficient
1) (3r + 2 - s)(4r + 9 - s) = 0
2) (4r - 6 - s)(3r + 2 - s) = 0
Given that y = 3x+2 implies that does 3x+2-y = 0 contains the point (r,s) implies is (3r+2-s) = 0 ?
From statement (1): (3r+2-s)(4r+9-s) = 0 implies either (3r+2-s) = 0 or (4r+9-s) = 0.
Now when (3r+2-s)...the line passes through (r,s)
When (4r+9-s) = 0 ...we cannot determine that whether the line passes through (r,s) or not.
Hence insufficient
From statement (2): (4r-6-s)(3r+2-s) = 0 implies either (4r-6-s) = 0 or (3r+2-s) = 0
Now when (4r-6-s) = 0 ... we cannot determine that whether the line passes through (r,s) or not
When (3r+2-s) = 0..the line passes through (r,s)
Hence insufficient
Taking statement (1) and (2) together: (3r+2-s)(4r+9-s) = 0 and (4r-6-s)(3r+2-s) =0... We cannot have both 4r+9-s=0 and 4r-6-s=0 so it is (3r+2-s) = 0 ... only this equation makes both the equations to be 0
Hence sufficient
1. Slope of line joining pole A and B is 3/4 and slope of line joining poles B and C is -4/3
2. Length of line joining pole A and B is 12 and length of line joining B and C is 5
From statement (1): Given that the slope of line AB is 3/4 and slope of line BC is -4/3. This implies that the product of slopes = -1. Hence AB perpendicular BC and B is a right angle. Thus AC is a diameter which implies ABC form a semi-circle.
Hence sufficient
From statement (2): Given that length of line AB is 12 and length of BC is 5. However this does not imply that ABC is a right angled triangle. We can draw number of different triangles with the same given two sides but with different third side.
Hence insufficient
1). b=3a+3
2). b-a is an odd number
From statement (1): Given that b=3a+3
Thus a-3b=a-3(3a+3) = -8a-9 which may be even, odd, integer, non-integer, rational etc ... Hence insufficient
From statement (2): Given that b-a is an odd number implies b is of the form b=(2k+1)+a where k is an integer
Thus a-3b= a-3[(2k+1)+a] = -2a -6k-3 which may be even, odd, integer, non-integer, rational etc ..Hence insufficient
Taking statement (1) and (2) together: -8a-9=-2a-6k-3 for some integer k
or -6a=-6k+6=-6(k+1) implies a=k+1
Thus a is an integer, either odd or even
Now statement (2) tells us that b is also an integer and that exactly one of {a,b} is even
If a is even and b is odd, a-3b is odd
If b is even and a is odd a-3b is odd
Thus (1) and (2) combined tell us that a-3b is an odd number...hence sufficient
(1) The length of line segment of QR is equal to the length of line segment RS
(2) The length of line segment of ST is equal to the length of line segment TU
From statement (1): Length of line segment of QR is equal to the length of line segment RS ..this implies angle RQS = angle RSQ = p(say)
From statement (2): Length of line segment of ST is equal to the length of line segment TU .. this implies angle TUS = angle TSU = q(say)
Hence p+p+angle QRS = 180 --- eq(1) and q+q+angle UTS = 180 --- eq(2)
Thus, p+q+x = 180
Now because angle RPT = 90, QRS+UTS= 90
adding eq(1) and eq(2) we get:
2p+2q+QRS+UTS = 360
2p+2q+90=360
p+q = 270/2 = 135
Now x = 180-p-q..hence the answer C
x = 180 - (p+q) = 180 - 135 = 45
(1) Angle C= 90
(2) Angle B= 45
From statement (1): it is given that angle C = 90 degrees ...this implies that ABC is a right angle triangle with AB as the hypotenuse and DC as the median. We know that --- In all right triangles, the median on the hypotenuse is the half of the hypotenuse. Hence DC=5. The ans is A.
p^a * q^b * r^c * s^d = x, where x is a perfect square.
(1) 18 is a factor of ab and cd
(2) 4 is not a factor of ab and cd
OE: When a perfect square is broken down into its prime factors, those prime factors always come in "pairs." For example, the perfect square 225 (which is 15 squared) can be broken down into the prime factors 5 * 5 * 3 * 3. Notice that 225 is composed of a pair of 5's and a pair of 3's.
The problem states that x is a perfect square. The prime factors that build x are p, q, r, and s. In order for x to be a perfect square, these prime factors must come in pairs. This is possible if either of the following two cases hold:
Case One: The exponents a, b, c, and d are even. In the example 3^2 5^4 7^2 11^6, all the exponents are even so all the prime factors come in pairs.
Case Two: Any odd exponents are complemented by other odd exponents of the same prime. In the example 3^1 5^4 3^3 11^6, notice that 3^1 and 3^3 have odd exponents but they complement each other to create an even exponent (3^4), or "pairs" of 3's. Notice that this second case can only occur when p, q, r, and s are NOT distinct. (In this example, both p and r equal 3.)
Statement (1) tells us that 18 is a factor of both ab and cd. This does not give us any information about whether the exponents a, b, c, and d are even or not.
Statement (2) tells us that 4 is not a factor of ab and cd. This means that neither ab nor cd has two 2's as prime factors. From this, we can conclude that at least two of the exponents (a, b, c, and d) must be odd. As we know from Case 2 above, if paqbrcsd is a perfect square but the exponents are not all even, then the primes p, q, r and s must NOT be distinct.
The correct answer is B: Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(1) a and b share exactly one common factor
(2) a and b are both prime numbers
From statement (1): we know that a and b have only one common factor, and we also know that all positive integers share the common factor 1 only, so we know it must be 1...hence sufficient
From Statement (2): we know that a and b are both prime, this implies the greatest common factor will have to be 1 or if a = b could be the same prime number then the GCF would be a (=b). ...hence insufficient
NOTE: You cannot assume that a and b are different integers if the question stem does not states the same
1. K is parallel to the line with equation y=(1-m)x+(b+1)
2. K intersects the line with equation y=2x+3 at the point (2,7)
From statement (1): y=(1-m)x+(b+1) has the same slope as y=mx+b. (Parallel lines have same slope)
Thus 1-m = m
implies Slope of K=m=1/2 ---- Hence sufficient
From statement (2): just says line y=2x+3 is not parallel to K, these two lines can have any angle between them other than 0, 180, 360 degrees ---- hence insufficient
1). The distance from x to zero is equal to the distance from y to 1
2). The sum of the distance from x to zero and the distance from y to 1 is less than 1
From statement (1): |x-0|=|y-1|
x = y-1
x = 1-y ....hence insufficient
From statement (2): |x-0|+|y-1| less than 1
= x+y-1 less than 1
= x+1-y less than 1
= -x+y-1 less than 1
= -x+1-y less than 1 ....hence insufficient
Taking statements (1) and (2) together: still insufficient
(1) a^2+b^2>16
(2) a=|b|+5
The given curve will intersect the y-axis when x=0
Thus we get a^2 + (y-b)^2 = 16
<=> a^2 + y^2 + b^2 - 2yb = 16
<=> y^2 - 2yb + a^2 + b^2 -16 = 0
In order to have real roots b^2 - 4ac >= 0
=> 4b^2 - 4(1)(a^2 + b^2 -16) >=0
=> a^2 <=16
From statement (1): Given that a^2 + b^2 > 16
No information about a^2 <=16 --- hence insufficient
From statement (2): Given a = |b|+5
=> |b| is positive or zero.
=> a is atleast equal to 5 and value of a^2 is atleast 25.
But we know that a^2 <=16 ====> does not intersect the y axis ---- hence sufficient
Hence the answer B
(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project.
(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project.
From Statement 1): It is given that 25 percent of the projects at Company Z have 4 or more employees assigned to each project - but we donot know the percentage of projects who have employees less than 4 or in other words we do not have any information about the rest 75% projects ---- hence insufficient
From Statement 2): It is given that 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project - but we donot know the percentage of projects who have employees more than 2 or in other words we do not have any information about the rest 65% projects---- hence insufficient
Taking both the statements together:
25 percent of the projects at Company Z have employees 4 , 5, 6..
35 percent of the projects at Company Z have employees 2, 1, 0
=> 40% of projects have 3 employees = > median value is 3
(1-35)employees -- (36-75)employees -- (76-100)employees
2 or less than 2 ---------3, 3, 3, ------------ 4 or more than 4
1). If the selling price per coat had been twice as much, the store's gross profit on the 20 coats would have been 2400
2). If the store selling price per coat had been $2 more, the store's gross profit on the 20 coats would have been 440
Suppose the cost of each coat = x
Suppose sales price = y
=> Total cost = 20x and sales price = 20y
We are required to find the profit: 20(y-x).
From statement 1): It is given that 20(2y-x)=2400
Doesn't helps to find out 20(y-x) --- insufficient
From statement 2): It is given that 20(y+2-x) = 440,
=> 20(y-x) = 400 --- sufficient
Hence B
1) y/ x > y
2) x^3 > x^2
From statement (1): y/x > y
=> y > xy
=> y(1-x) > 0
y>0 when x is less than 1 or y less than 0> when x is greater than 1 ---- hence insufficient
From statement (2): x^3 > x^2
=> x^2(x-1) > 0
=> x^2 >0 and x>1
Since x>1, 1/x can not be greater than 1 ---- hence sufficient
Hence B
DS-2
1 the average hight of the n/3 talles people in the group is 6 feet and 2 1/2 inches and the average height of rest of the people in the group is 5 feet and 10 inches.
2 The sum of the lenghts of the people is 178 feet and 9 inches.
Looking at statement (2) first, we see that it is not sufficient, because the average (arithmetic mean) of a group of numbers is defined as (sum of data) / (# of data points). With statement (2), we only have the numerator of this expression (the # of people in the group is unknown), so we can't figure out the average.
Looking at statement (1) alone, we can set up the average as follows:
Average = (sum of data points) / (# of data points)
= [(n/3)(74.5) + (2n/3)(70)] / (n) <-- note that I used inches here, so I won't have to write in more fractions than necessary (trying to write fractions on this forum is not fun) = [(1/3)(74.5) + (2/3)(70)] / (n) There's no need to simplify further, because the 'n' is gone: you get one number. Therefore, this statement is sufficient. Answer = A Note that, if you have the averages of all the FRACTIONS or PERCENTAGES of a group, then you'll be able to calculate the overall average of the group. This is a worthwhile fact to memorize for the data sufficiency problems
Speed Time and Distance -1
A: 2.5
B: 2.4
C: 2.3
D: 2.2
E: 2.1
90/(V-3) = 1/2 + 90/(V+3)
90/(V-3) - 90/(V+3) = 1/2
[90(V+3) - 90(V-3)] / (V+3) (V-3) = 1/2
[90V + 270 - 90V +270] = [V^2 - 9]/2
540 *2 = V^2 - 9
1080 + 9 = V^2
1089 = V^2
V = 3*11 = 33
Downstream = 90/(v+3) = 90/36 = 10/4 = 2.5
Data Sufficiency - 1
Sunday, June 25, 2006
Remember......
2. At least 15 people in this world love you in some way.
3. The only reason anyone would ever hate you is because they want to be just like you.
4. A smile from you can bring happiness to anyone, even if they don't like you.
5. Every night, SOMEONE thinks about you before they go to sleep.
6. You mean the world to someone.
7. If not for you, someone may not be living.8. You are special and unique.
9. Someone that you don't even know exists loves you.
10. When you make the biggest mistake ever, something good comes from it.
11. When you think the world has turned its back on you, take a look: you most likely turned your back on the world.
12. When you think you have no chance of getting what you want, you probably won't get it, but if you believe in yourself, probably, sooner or later, you will get it.
13. Always remember the compliments you received. Forget about the rude remarks.
14. Always tell someone how you feel about them; you will feel much better when they know.
Unforgiven - A western at its classic best
Wednesday, June 21, 2006
Monday, June 19, 2006
Sunday, April 30, 2006
a paradise lost......
My solitude, my company.
The green grass brushes my feet,
wind rustles by my neck.
I see the passing skies,
the eirre silence mocking me
stars shine like glittering diamonds,
a celestial view, amazed at what I see
i see the lonely path ahead me,
the walk, long and strainiful .
I see this place with my eyes closed
so much beauty ..so serene
I know my hopes will be shortlived
but the memory isnt.
little wonders pass us all the day, a paradise lost...lost everyday.
Friday, 05 September 2003
Chained to the heart…
Flickering in the night,
Burning the midnight oil,
The flame burns on,
Just like the passion within.
I face the remorse of a burdened heart,
Just wising to let go.
Some of the things that aren’t justified,
Is better left alone.
Life has become a part of so many phases,
I seem to have lost that phase where I lived myself.
I drown myself in this eternity of make believe space,
Caring less for where I stand and what I do.
Saturday, October 18, 2003
…………feeling inside out
I feel repressed,
Feeling the pressure is killing me.
Makes me want to give up,
Give in to the arms of failure.
I try to give my best,
But always comes out as the worst.
My fears are the worst enemy,
Bringing me inside out.
The higher I try to get,
The harder I fall.
I can’t stand being judged,
Looked around by suspicion and contempt.
Life comes to halt after a full circle,
Begin a start from where it ends all…….
Wednesday, September 24, 2003
………..in the blink of an eye
in the blink of an eye.
What I believe in,
Comes against me.
My trust, my faith is what u broke,
Leaving me alone in this cruel world.
Time becomes a painful memory,
Trying to forget something is so hard to do.
Truth becomes a distortion,
The unreal seems much easier to accept.
Everything comes to an end,…in the blink of an eye.
Tuesday, September 23, 2003
Fallen glory…..
I seek my time,
That seems to have lost
Tune to myself,
Of what I am capable of.
I have lost my sheen,
And my ability to get in touch with myself,
I stay here fallen,
Exposing my weakness.
I wait to wake up,
Regaining consciousness to what has changed around me.
I reap harvest to my shameful deeds, A fallen glory is what I carry.
Wednesday, October 02, 2003
Saturday, January 28, 2006
Absolute Cricket moments
http://uprightvideos.blogspot.com/2006/01/shameful-moments-of-pakistani-cricket.html
Thats life for you.........
If you are jealous, she says it's bad
If you don't, she thinks you do not love her
If you attempt a romance, she says you didn't respect her
If you don't, she thinks you do not like her
If you are a minute late, she complains it's hard to wait
If she is late, she says that's a girl's way
If you visit another man, you're not putting in "quality time"
If she is visited by another woman, "oh it's natural, we are girls"
If you kiss her once in a while, she professes you are cold
If you kiss her often, she yells that you are taking advantage
If you fail to help her in crossing the street, you lack ethics
If you do, she thinks it's just one of men's tactics for seduction
If you stare at another woman, she accuses you of flirting
If she is stared by other men, she says that they are just admiring
If you talk, she wants you to listen
If you listen, she wants you to talk
Take the SUPERHERO Contest
My results were:
You are Hulk
| You are a wanderer with amazing strength. |
Saturday, January 21, 2006
Time Off from Work
Friend of mine is getting married,and asked me to come to hydrebad for that.well mite as well take that vacation for myself. Would love to go outside india though (preferably somewhere in Europe), a nice vacation out there would definitely cheer me up(cheer anybody up :) ).i Want to take some time off ....Wunder when would i get that in this hectic time schedule.
Sunday, January 01, 2006
Samuel Bakers classic - Platoons music
Another one of sad classical moment has been added to my collection.Adagio for strings -Samuel baker.The sound emits strong sentiments.Anguish,sadness,despair and loss,all embrolied in this melodious music.
Barber found initial inspiration in a passage from Vergil's Georgics describing how a rivulet gradually becomes a large river.The music made poular in the Oliver stones classic movie "Platton" that showed how the troops faced despair and challenges in the wartime Vietnam. The music exactly highlights the emotion of men at war and just touches your innermost soul. Get a download of it in the following location
http://www.modern-strings.de/sound/barber.htm
additional info
http://www.classical.net/music/comp.lst/works/barber/adagio.html