Problem Solving Flashcards
Qs. 1 There are these 8 numbers in a set 9.4,9.9,9.9,9.9,10.0,10.2,10.2,10.5 The mean and standard deviation of the 8 numbers are 10.0 and 0.3 respectively, what percent of the 8 numbers are within 1 standard deviation of the mean?
a) 90%
b) 85%
c) 80%
d) 75%
e) 70%
Answer: D mean = 10 sd = 0.3 "one standard deviation" = 1*(sd) = 0.3. 1 stdev range would be mean plus/ minus std Dev Lower range:10 - 0.3 = 9.7 Higher range:10 + 0.3 = 10.3 6 numbers out of 8 fall within the range Then 6/8 or 3/4 are within range 3/4 = 75% Hence answer is D
Qs. 2: The fourth grade at school X is made up of 300 students who have a total weight of 21,600 pounds.
If the weight of these four graders has a normal distribution and standard deviation equals 12 pounds, approximately what percentage of the fourth graders weighs more than 84 pounds?
a 12% b 16% c 36% d 48% e 60%
Answer: B
Mean = 21600/300 = 72
Mean + 1sigma (std dev) = 72 + 12 = 84
+/- 1 covers 68.2% (34.1% + 34.1%) of the normal distribution. Therefore, remain distribution of data 100% - 68.2% = 31.8%.
Since we care only about data points above 84 lbs => 31.8 / 2 = 15.9% = 16%
Qs. 3 A teacher prepares a test. She gives 5 objective type question out of which 4 have to be answered. Find the total ways in which they can be answered if the first 2 questions have 3 choice and last 3 have 4 choice.
Answer:816
Options available
- Get first one wrong. X3444= 192
- Get 2nd one wrong 3X444 = 192
- Get 3rd one wrong 33X44= 144
- Get 4th wrong 334X4=144
- Get 5th wrong 3344X=144
Hence 144+144+144+192+192= 816
Qs. 4 The letters of the word PROMISE have to be arranged so that no two vowels come together. Find the number of arrangements.
word PROMISE has vowels O,I,E
if you chose two vowels then you will have following combinations
OI - it can also also be arranged as IO
IE - it can also be arranged as EI
OE - it can also be arranged as EO
Consider one combination OI
Now you have PRMSE and OI
Following are some of the combinations that can be obtained
ESMRPOI
ESMRPIO
and
SMRPEIO - > Here all the 3 vowels are together
So just accounting for two vowels at a time should cover everything.
Thus for each vowel pair we have invalid combinations as 2 * 5! -> for OI and IO 2 * 5! -> for IE and EI 2 * 5! -> for EO and OE this is same as 2 * 3C2 * 5! = 720
Total combinations are 7! = 5040
so valid combinations are 7! - 720 = 4320
Qs. 5 Four questions repeated outcomes
Qs. 1 Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)
Qs. 2 Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.)
Qs. 3 There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
Qs. 4 There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
Repeated Outcomes [#permalink] Wed Mar 09, 2005 9:29 am
2 This post received
KUDOS
Repeated Outcomes
If there’re only k possible outcomes for each object, total possible outcomes for n objects is k^n.
Explanation: For each object, there are k outcomes. So the total number of outcomes would be kkk …*k, for n times.
Example:
Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total number of fruits = 9
For each fruit there are 2 possible outcomes: included, not included
Total outcome = 2^9-1
(The minus one is to take out the one possible outcome where nothing is in the basket.)
Example:
Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.)
Total outcome = (2^3-1)(2^4-1)(2^2-1)
Example:
There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
P = (good outcomes)/(total possible outcomes).
Let the 3 reports be A, B, and C.
Each report must be assigned to a secretary.
Total possible outcomes:
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 3. (Any of the 3 secretaries.)
Number of options for C = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
333 = 27.
Good outcomes:
A good outcome occurs when each report is assigned to a DIFFERENT secretary.
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 2. (Either of the 2 remaining secretaries.)
Number of options for C = 1. (Only 1 secretary left.).
To combined these options, we multiply:
321 = 6.
Thus:
(good outcomes)/(total possible outcomes) = 6/27 = 2/9.
Example:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
P = (good outcomes)/(total possible outcomes).
Let the 4 reports be A, B, C and D.
Each report must be assigned to a secretary.
Total possible outcomes:
Number of options for A = 3. (Any of the 3 secretaries.)
Number of options for B = 3. (Any of the 3 secretaries.)
Number of options for C = 3. (Any of the 3 secretaries.)
Number of options of D = 3. (Any of the 3 secretaries.)
To combine these options, we multiply:
333*3 = 81.
Good outcomes:
For each secretary to be assigned at least 1 report, exactly 1 secretary must receive a PAIR of reports, while the other 2 secretaries receive 1 report each.
Number of pairs that can be formed from the 4 reports = 4C2 = (43)/(21) = 6.
Number of ways to assign this pair = 3. (Any of the 3 secretaries.)
Number of ways to assign the next report = 2. (Either of the 2 remaining secretaries.)
Number of ways to assign the last report = 1. (Only 1 secretary left.)
To combine these options, we multiply:
632*1 = 36.
Thus:
(good outcomes)/(total possible outcomes) = 36/81= 4/9.
Qs. 6 Three questions on probability of repeated experiments.
Qs. 1 When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses?
Qs. 2 The probability of raining is 0.3 and not raining is 0.7. What is the probability of getting three days of rains among seven days?
Qs. 3 The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?
Binomial Distribution
In each individual test, the probability of A happening is p and not happening is 1-p. What is the probability of A happening exactly k times in n repeated tests?
Formula:
C(n,k) * p^k * (1-p) ^ n-k
Example:
When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses?
C(4, 2) * (1/2) ^2 * (1/2) ^2 = 6/16 = 3/8
Example:
The probability of raining is 0.3 and not raining is 0.7. What is the probability of getting three days of rains among seven days?
C(7, 3)0.3^30.7^4
Example:
The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?
Probability that at least two babies are boys
= Probability that two babies are boys + probability that three babies are boys + … + Probability that ten babies are boys
(also) = 1- Probability that non are boys - probability that only one is a boy
Choose the easier route
P=1-C(10,0)0.5^00.5^10-C(10,1)0.5^10.5^9
=1-0.5^10-100.5^10
=1-110.5^10
Qs.8 Of the 10 employees at a certain company, 5 had annual salaries of $20,000, 4 had annual salaries of $25,000, and 1 had an annual salary of $30,000. If a bonus equal to 10 percent of annual salary was given to each employee, what was the total amount of he bonuses? (A) $230,000 (B) $75,000 (C) $30,000 (D) $23,000 (E) $7,500
Answer: D
10% of 20k = 2k
For 5 employees = 5*2k = 10k
10% of 25k = 2.5k
For 4 employees = 4*2.5k = 10k
10% of 30k = 3k
For 1 employee = 3k
Add all = 10 + 10 + 3 = 23k
Qs. 9 A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?
A. 15/28 B. 1/4 C. 9/16 D. 1/32 E. 1/16
Answer : A
Prob (not blue) * Prob ( 2nd not blue) = 6/8*5/7 = 15/28
Qs. 10 In a group of 30 students, 25 are taking mathematics, 22 English, and 19 history. Every
student is taking at least one of the courses. The greatest number of students who COULD be
taking all three courses is x. The least number of students who COULD be taking all three
courses is y. What is the value of x + y?
a. 17
b. 19
c. 22
d. 23
e. 24
E
T=M+E+H-(ME+MH+EH)-2MEH
30=25+22+19-(ME+MH+EH)-2MEH
(ME+MH+EH) -2MEH = 36
To maximize MEH, the (ME+MH+EH) must be minimum. Hence (ME+MH+EH) must be zero
Putting it in the above equation MEH = 18
Similarly for Minimum value (ME+MH+EH) must be maximum
Sine M=25 the maximum value of EH would be 30-25= 5
Similarly MH would be 8
And ME would be 11
Therefore MEH = 6
Thus x+y would be 18+6= 24
Qs. 11 a company has 15 disrtibution centres and uses color coding to identify each center.either a single color or a pair of two different colors is chose to represent each center uniquely.what is the mnimum number of colors needed for the coding and the order of the colors doesnot matter?
1 Colour codes: 4 2 Colour codes: we need to choose 2 colors from 4. This can be accomplished in 4c2 ways 4c2= 4x3x2x1/(2x(4-2))= 6 Similarly 3 Colour codes: 4c3 =4 For 4 Colour codes: 4c4=1 Hence 4+6+4+1=15
Therefore 4 is the correct answer.
Qs. 12 At a certain financial institution, 30% of the clients use both credit card and a cheque book, but 40% of the clients who use the credit card do not use the cheque book. What percentage of the members of the institution use the credit card?
A. 35 B. 40 C. 50 D. 65 E. 75
Pleas help me and give me a nice explanation, of course an explanation which takes less than 2 min. to do.
Good post? |
IMO C.
Let the total clients = 100
No. of clients using both credit card and cheque book = 30.
Let total no of clients using credit card = x
So, no. of clients using credit card but no cheque book = 0.4x
According to question,
0.4x+30 = x
x= 50 (which is the total number of clients using credit card)
Since we had assumed total no. of client to be 100, this value of x is the percentage answer.
You can aslo do it by making table as:
CC no CC Total
CB 30
no CB 0.4x
Total x 100
It took little less than a min to solve this. I hope this was helpful.
Qs. 13 Could the answer be an integer if x is an integer greater than 1?
a) x^(10) + x^(–10)
b) x^(1/6) + x^(1/2)
Sufficient or not sufficient?
Answer is A only
Good post? |
PLZ solve this question ……..
One variable is equal to 40% more than X. The same variable is equal to 20% less than Y. By how much percentage is Y greater than X?
Good post? |
Originally Posted by sanaulcse
# One variable is equal to 40% more than X. The same variable is equal to 20% less than Y. By how much percentage is Y greater than X?
Let x be 10
So the other variable ( Z ) be 14
Now it is given 14 is 20% less than Y
So the other variable Z is 80% of Y
Hence Y = 17.5
Now we have all the 3 variables , X , Y and Z
X = 10
Y = 17.5
Z = 14
So let’s find out how much is Y greater than X ,
Y - X is 7.5
So percentage is Y greater than X is -
( 7.5 / 10 ) * 100 => 75%
Hope this helps…
If the sequence X 1 (X one), X 2 (X two), X 3 (X three), …..Xn is such that X 1= 3 and X n+1 = 2 (Xn)-1 for n=1 , then X 20 - X 19 =
A) 2 ^ 19 B) 2 ^ 20 C) 2 ^ 21 D) (2 ^ 20) -1 E) (2 ^ 21) -1
Can some one please explain how this problem can be solved without making the lengthy calculations.
Good post? |
Xn+1 = (2Xn) - 1 or Xn = (Xn-1) - 1
X1 = 3 = 2^1 + 1 X2 = 2*(3) - 1 = 5 = 2^2 + 1 X3 = 2*5 - 1 = 9 = 2^3 + 1
So, it follows that the nth term of the sequence is obviously Xn = 2^n + 1
Therefore, X^20 - X^19 = 2^20 + 1 - 2^19 - 1 = 2^20 - 2^19 = 2^19 * (2-1) = 2^19.
(Choice A is the answer)
If x < 0 , SQRT ( -x *|x|)
1) -x
2) -1
3) 1
4) x
5) sqrt(x)
SPOILER: Official Answer A. I guess its D. Am I missing something here If x < 0 , SQRT ( -x *|x|) A) -x B) -1 C) 1 D) x E) sqrt(x) A quick way to solve this question is to plug in a value for x. Since xs say that x= -5
So, the sqrt[-x (|x|)] = sqrt[5 (|-5|)] = sqrt[5(5)] = sqrt[25] = 5
Answer choice A is -x, so if x= -5, then -x = 5. Since our output above is 5, answer choice A is correct.
Good post? |
DS- Multiplication by zero
Official Guide Quant Review - 2nd Edition - DS-Question 46
What is the value of x^2 - y^2 ? (x square minus y square)
(1) x - y = y + 2
(2) x - y = 1 / (x+y)
Statement 2 tells us that x-y = 1/(x+y). From this, we can conclude that x+y does not equal zero. If x+y did equal zero, then 1/(x+y) would be undefined in which case it couldn’t equal some other value. So, knowing that 1/(x+y) equals some other value, it’s safe to conclude that x-y does not equal zero.
Good post? |
GMAT Prep Set theory Problem
Of the 200 members of a certain association, each member who speaks german also speaks english, and 70 of the members speak only spanish. if no member speaks all three languages, how many of the members speak two of the three languages?
(1) 60 of the members speak only english.
(2) 20 of the members do not speak any of the three languages.
more specifically:
there are eight subsets: none E only G only S only ES EG GS EGS
let’s fill in the list with the information that we already have from the problem:
none = 20 (from statement 2) E only = 60 (from statement 1) G only = 0 (because they all speak english too) S only = 70 (given) ES = \_\_\_\_\_\_\_ EG = \_\_\_\_\_\_\_ GS = \_\_\_\_\_\_\_ EGS = 0 (given)
the only blanks combine to give the desired quantity. we can’t find the values of the individual blanks, but we don’t care; all that matters is their sum, which is easily found by subtracting 20, 60, and 70 (as well as the two 0’s, if you want) from the total of 200. there’s no need to perform this calculation, because it’s data sufficiency and we know there’s going to be a unique numerical answer.
ans = c
Good post? | Easy data sufficiency question that's confusing. OG 12th ed Is X an integer? 1. X/2 is an integer 2.2x is an integer.
The answer is A.
In the game Cako, a player is awarded one tick for every third Alb captured, and one click for every fourth Berk captured. The total score is equal to the product of clicks and ticks. If a player has a score of 77, how many Albs did he capture?
(1) The difference between Albs captured and Berks captured is 7.
(2) The number of Albs captured is divisible by 5.
.Actually, this problem is testing knowledge of modular arithmetic - that is arithmetic with remainders.
Consider how many ticks you get for each Alb captured
Alb Tick 0 0 1 0 2 0 3 1 4 1 5 1 6 2
The point here is that you do not get fractional ticks (T) for each Alb captured - rather you get one for every third Alb (A) and one click (C) for every fourth Berk (B).
So from the question stem, we know the T + C = 77
Now consider 1). We are told that A - B = 7. It is tempting to see this as sufficient. Turn Alb and Berk into ticks and clicks and you might get A/3 + B/4 = 77. You have two equations in two variables. If you solve, you get A = 135 and B = 128. This gives you 45 ticks and 32 clicks. Now when you look at 2) and see that Alb is divisible by four, there appears to be a conflict between the two statements.
The problem with the above is that the equation you should have sent up above is Q (A/3) + Q (B/4) =77 where Q(x/y) is the integer quotient, excluding the remainder. So eg Q(10/5) =2 and Q(13/5) =2. Now, (1) does not yield a unique solution. A =135, B=128 and A =136, B=129 and A=137 and B=130 all yield T =45 C = 32. Looking just at the Alb for this example, see 135/3 = 45 r0, 136/3 = 45 r1, 137/3 = 45 r2.
Now consider 2. It’s clearly insufficient by itself. Consider 1 and 2 together. Of the three possible solutions we found in 1), only A=136, B=129 has A divisible by 4. So 1) and 2) are sufficient together and the answer is C.
I hope you add this problem back into the database - after a couple of problems like this, the real exam will seem easy. ;)
Good post? |
800 level Difficult Exponent question.
How can i solve this Data Sufficiency problem, anyone pull me out from the dark……..
If m is a positive integer, is the value p+q at least twice the value of 3^m+4^m ?
1) p=3^(m+1) and q=2^(2m+1)
2) m=4
First, let’s rewrite the target question as “Is p+q > 2(3^m + 4^m) ?”
Statement 1: If we replace p and q with their respective values, we can reword the target question as:
Is 3^(m+1)+2^(2m+1) > 2(3^m + 4^m) ? (do we have sufficient information to answer this question? let’s find out)
To simplify the right-hand-side, first recognize that 4^m = (2^2)^m = 2^2m
So, we can reword the target question as: Is 3^(m+1)+2^(2m+1) > 2(3^m + 2^2m) ?
If we expand the right-hand-side, we get: Is 3^(m+1)+2^(2m+1) > (2)3^m + 2^(2m+1)?
At this point, we can subtract 2^(2m+1) from both sides to get: Is 3^(m+1) > (2)3^m?
Now, if we divide both sides by 3^m, we get: Is 3 > 2?
Yes, 3 is greater than 2.
Since we can answer the reworded target question with certainty, statement 1 is sufficient.
Statement 2: Since we have no information about p and q, we cannot answer the target questions. So, statement 2 is not sufficient and the answer is
SPOILER: A
.
Cheers,
Brent
Good post? |
Triangle Problem
ABC and PQR are both triangles. Is the area of triangle ABC greater than the area of triangle PQR?
1.Sides AB=PQ, BC=QR, and AC= PR
2. Angles A=P, B=Q, and C=R
A. Together the statements are sufficient, but alone neither statement is sufficient
B. Together the statements are not sufficient
C. Both statements are sufficient alone
D. Statement 1 alone is sufficient, but statement 2 alone is not sufficient
E. Statement 2 alone is sufficient, but statement 1 alone is not sufficient
Lolz, this is a funny one
The answer should be (A)
Using 1
If the 3 sides of a triangle are equal, then by SSS (side,side,side) the 2 triangles are congruent
we do not need to know much about the angles
Using 2
If the 3 angles have the same angle, then by AAA (angle, angle, angle) the 2 triangles are similar, but the area may be different
Of the students in the class, 55% of the females and 35% of the males passed the exam. Did more than half the students pass the exam?
1) More than half the students in the class are females
2) The number of female students is 20 more than the the number of male students.
SPOILER: Official Answer - B
Let M - number of males, F - number of females
Need to compare (M+F)/2 and 0,55F+0,35M given that F = M+20
(M + M + 20)/2 compare with 0,55(M +20) + 0,35M
(2M+20)/2 compare with 0,55M+0,35M+11
M+10
Average
A student finds the average of 10 positive integers. Each integers contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this the average becomes 1.8 less than the previous one.What was the difference of the two digits b and a?
A. 8 B. 6 C. 2 D. 4 E. None of the above
Originally Posted by shanssv
A student finds the average of 10 positive integers. Each integers contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this the average becomes 1.8 less than the previous one.What was the difference of the two digits ba and ab?
Answer: C. 2
Let Sc be the sum of 10 current numbers then the average (Current) is Avc = Sc / 10
Similalrly let Sp be the sum of 10 previous numbers then the average (previous) is Avp = Sp / 10
But as per the questions Avc = Avp - 1.8 then Sc/10 = Sp/10 - 1.8 therefore Sc/10 = (Sp - 18)/10 therefore Sc = Sp -18 or Sc -Sp = 18
Now Let A be the sum of all other numbers except ba then Sc = A + (b=10a)
Similaly Sp = A + (a + 10b)
Put in above equation A + (b + 10a) - A - (a + 10b) = 18 hence 9(b-a) = 18 therefore b - a = 2 Hence C is teh correct answer
on Average Three math classes: X, Y, and Z, take an [COLOR=blue ! important][COLOR=blue ! important]algebra[/COLOR][/COLOR] test. The average score in class X is 83. The average score in class Y is 76. The average score in class Z is 85. The average score of all students in classes X and Y together is 79. The average score of all students in classes Y and Z together is 81. What is the average for all the three classes? 81 81.5 82 84.5 Ans : B
Can someone help me on this one?
This is a problem on weighted ave.
x83+y76/(x+y)= 79
y76+z85/(y+z) = 81
We need to find
x83+y76+z*85/(x+y+z)
Good post? | PS Questions (GMAT Guru's Erin and 800Bob Please Help) Part 2 9. 1 – [2 –(3 – [4 – 5] + 6) + 7] = (A) –2 (B) 0 (C) 1 (D) 2 (E) 16
SPOILER: D
11. If 0.497 mark has the value of one dollar, what is the value to the nearest dollar of 350 marks? (A) $174 (B) $176 (C) $524 (D) $696 (E) $704
SPOILER: D
12. A right cylindrical container with radius 2 meters and height 1 meter is filled to capacity with oil. How many empty right cylindrical cans, each with radius 1/2 meter and height 4 meters, can be filled to capacity with the oil in this container? (A) 1 (B) 2 (C) 4 (D) 8 (E) 16
SPOILER: C
16. A family made a down payment of $75 and borrowed the balance on a set of encyclopedias that cost $400. The balance with interest was paid in 23 monthly payments of $16 each and a final payment of $9. The amount of interest paid was what percent of the amount borrowed? (A) 6% (B) 12% (C) 14% (D) 16% (E) 20%
SPOILER: D
12. An instructor scored a student’s test of 50 questions by subtracting 2 times the number of incorrect answers from the number of correct answers. If the student answered all of the questions and received a score of 38, how many questions did that student answer correctly? (A) 19 (B) 38 (C) 41 (D) 44 (E) 46
SPOILER: D
1) D….Apply PEMDAS or BODMAS…ans:2
2) E….350/0.497 ~704
3) C….2^21=n(1/2)^2*4…n=4
4) D ..amnt borrowed=400-75=325
Total amnt paid=75+23*16+9=452
amont paid as interest=452-400=52
->52/325=16
5) I think there is something missing in the question..it is not mentioned the marks awarded for the correct answer.
If we assume it to be 1 than ans is 38+(12/2)=44…D
Average Question A teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 80, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?
A.74 B.75 C.76 D.77 E.78
Good post? |
b (654)+(806)+(77*5)/15
average score The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test? A. 2x - 8 B. x - 4 C. 8 - 2x D. 16 - x E. 8 - x/2
IMO A.
If the score for 5 students was 8 –> the total score for these 5 students is 40
(40 + 5y) / 10 = x
solve for Y and it is answer A. 2x - 8
an easy one An instructor scored a student’s test of 50 questions by subtracting 2 times the number of incorrect answers from the number of correct answers. If the student answered all of the questions and received a score of 38, how many questions did that student answer correctly? (A) 19 (B) 38 (C) 41 (D) 44 (E) 46
for me the answer is e correct=x incorrect=50-x x-2(x-50)=38 solve for x=46 correct
Good post? |
mean-median
Class A and B took a same test. For class A, median score is 80, average score is 82; for class B, median score is 78, average score is 74. Combining A and B, is the average greater than the median?
1). A has 37 students and B has 40 students.
2). A and B have 77 students.
Good post? |
Stmt 1:
A = 37, B = 40
=> median is the 38th element
A has 18 students below 80(median)
B has 20 students below 78 (median = (20 th element +21st element/2)
=> median of the combined set can be 80 or atleast > 78 as 38 students are below 80
Avg = 3782 + 4074/77 => avg < 78 (only if equal number of students are there, the avg will be 78)
Median > avg
Sufficient
Stmt 2:
No info on how many students are in A & B individually.
Insufficient
Ans A.
Good post? |
Averages
A student was able to score 600 in 12 tests.He scored less than or equal to 80% of his average score per test in the four of these tests.If he did not score more than 60 in any of the test,wht is the minimum number of tests in which he should have scored more than 50
A.8 B.4 C.3 D.2 E.7
Good post? |
i think i would agree with mastermind.
as per what karmaholic says:
__________________________________________________ ____________
Marks obtained in four tests = 160 (4*40) [40 = 80% of 50 and 50 is the average score]
Total Marks = 600. So the student needs another 440 marks from remaining 8 tests.
Now the max he can get in these tests is 60.
__________________________________________________ _____________
now we need the min number of tests in which he has to score max ie 60
if
no of tests with max score =2 then 2x60 + 6x50 = 420 .. option D incorrect
no of tests with max score =3 then 3x60 + 5x50 = 430 .. option C incorrect
no of tests with max score =5 then 4x60 + 4x50 = 440 .. correct option B
Here r 2 more,
1) A farmer has a field that measures 1000 ft wide by 2000 ft long. There is an untillable strip 20 ft wide on the inside edge of the field, and a 30 ft wide untillable strip bisects the field into two squares (approximate). Approximately what percentage of the field is tillable?
A) 98%
B) 93%
C) 91%
D) 90%
E) 88%
2) Elsa has a pitcher containing x ounces of root beer. If she pours y ounces of root beer into each of z glasses, how much root beer will remain in the pitcher?
A) x /(y + z)
B) xy - z
C) x /(yz)
D) x - yz
E) x/(y - z)
Vinay
B
D
hi Vinay,
“On the inside edge” means that the border of the land is untillable:
(1000 x 20) + (1000 x 20) + (2000 x20) + (2000 x 20) = 120,000 ft squared
Then there’s another untillable part that cuts the field into two squares:
30 x 1000 = 30,000 ft. squared
Total untillable land = 30,000 + 120,000 = 150,000 ft. squared
Total land = 2000 x 1000 = 2,000,000 ft. sq
% tillable land = (Total land - untillable land) / Total land =
(2000,000 - 150,000) / 2000,000 x 100% = 92.5%
Hence, 93%
Btw, Vinay, are these questions from the 800score tests?
Tina
Here are five difficult combinations/permutations problems.
If you can get all 5 right, you will win a free subscription to magoosh GRE!
EDIT BY MODERATOR
I will be taking answers until Fri.
- A committee of three must be formed from 5 women and 5 men. What is the probability that the committee will be exclusive to one gender?
(A) 1/60 (B) 1/120 (C) 1/8 (D) 1/6 (E) 1/3
- A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)?
(A) 1/55 (B) 1/66 (C) 2/17 (D) 1/110 (E) 2/55
- A homework assignment calls for students to write 5 sentences using a total of 10 vocabulary words. If each sentence must use two words and no words can be used more than once, then how many different ways can a student select the words?
(A) 10!/5! (B) 10!/32 (C) 5! x 5! (D) 2! x 5! (E) 10!
- Team S is to comprise of n debaters chosen from x people? Team R is comprised of n + 1 debaters chosen from x+1 people.
Column A
Number of unique team S
Column B
Number of unique Team R
- A lunar mission is made up of x astronauts and is formed from a total of 12 astronauts. A day before the launch the commander of the program decides to add p astronauts to the mission. If the total number of possible lunar missions remain unchanged after the commander’s decision, then which of the following cannot be the value of p?
(A) x (B) x + 3 (C) 3 (D) 6 (E) 8
Not sure if these are right, but here are my answers:
1. = (5C3+5C3) / 10C3 = 20 / 120 = 1/6 => ANSWER D
- A-L = 12 letters
20 desirable possibilities [10 for forward 3-letter sequences (like ABC) 10 for backward 3-letter sequences (like FED)]
121110 = total possibilities
=> 20/(121110) = 1/66
=> ANSWER B
3. = 10P2 * 8P2 * 6P2 * 4P2 * 2P2 = 10*9*8*7*6*5*4*3*2 = 10! => ANSWER E
4. = x! / (n!)(x-n)! v (x+1)! / (n+1)! (x-n)! = 1 v (x+1) / (n+1) = n+1 v (x+1) = n v x => Answer depends on values of n and x => ANSWER D
- 12! / (x!)(12-x)! = 12! / (x+p)!(12-x-p)!
=> x(12-x) = (x+p)(12-x-p), where x and p are integers
When p = 3, we get:
6x = 27 => x = 9/2 which is not an integer, therefore p != 3
=> ANSWER C
Good post? |
Ds
| | x | - | y | | = | x | - | y | ?
(1) x ^ 2 = y ^ 2
(2) x + y = 0
I dont have the Official Answer … please explain
Answer would be (D)
(A)X ^2 = Y^2 —> |X| = |Y|
once you get that the question can be answered
(B) X+ Y=0 –> X= -Y —> |X| = |Y|
again main question can be answered.
Og11 Ps 180
In a nationwide poll, N people were interviewed. If 1/4 of them answered “yes” to question 1, and of those, 1/3 answered “yes” to question 2, which of the following expressions represents the number of people interviewed who did not answer “yes” to both questions?
A. N/7 B. 6n/7 C. 5N/12 D. 7N/12 E. 11N/12
Official Answer: E
I am confused with this answer because the question specifically asks for “people interviewed who did not answer ‘yes’ to BOTH questions. 11N/12 factors in people who said “yes” to first and “no” to second, and “no” to first and “yes” to second. Shouldn’t the answer be N/2 (3N/4 * 2/3)?
chosters, Official Answer is right.
Total people interviewd=N
of people said “yes” to question 1 = N/4
In the question it states that 1/3 of people who answered “yes” to question 1 also answered “yes” question 2.
What it means is people answered “yes” to question 1 and 2 = 1/3*N/4=N/12
So the number of people interviewed who did not answer “yes” to both questions = Total people interviwed- number of people answered “yes” to both questions=N-N/12
=11N/12
Hence E.
Let me know if you have any questions.
Good post? |
venn diag
If 75% of a class answered the first question on a certain test correctly, 55% answered the second question correctly, and 20% answered neither of the questions correctly, what percent answered both correctly ?
Good post? |
Answer 50%
80 = 75 +55 - p(a int b)
1
budhi
Within my grasp!
Join Date Apr 2005 Posts 180 Rep Power 8
Good post? | Teacher /Class Ratio ? Each of the 32 schools has 2 classes. The number of the teachers is 37. One class has one teacher; one teacher teaches at least one class and at most 3 classes. What are the greatest number and the least number of the teachers who teach 3 classes? A: 0,13 B: 0,14 C: 0,15 D: 0,16, E: 0,17
SPOILER: Official Answer C [0,13]
Please explain.
Total of 64 classes
Total of 37 teachers
A teacher has to teach atleast 1 class
If each teacher were to teach 2 classes, then there would need to be 37*2 = 74 classes but there are only 64 so least number of teachers who teach 3 classes = 0
Now there are 64-37 = 27 more classes than there are teachers. The greatest number of teachers who can teach 3 classes = 26/2 = 13 (and one teacher will teach 2 classes)
So solution set {greatest, least} = {13,0}
Permutation & Combination
In an IPL cricket competition, there are 9 teams. How many matches were held and in how many ways can each team play each other?
Please help.
You are trying to find the number of ways that each team can play with each other.
This is about combinations not permutations. Order does not matter.
So out of 9 teams, you want to pick 2, such that the order does not matter. That's 9nCr2 That's 9! / (2! * (9-2)!) = 9! / (2! *7!) = 8*9 / 2 = 72/2= 36
Good post? |
Difficult Permutation Problem
I’m having a lot of trouble making my answer look like the Official Answer. Can someone help?
Here’s the problem:
In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?
Official Answer to follow after some discussion
Originally Posted by choked
I’m having a lot of trouble making my answer look like the Official Answer. Can someone help?
Here’s the problem:
In how many different ways can 12 people be seated at two tables with one seating 7 and the other seating 5?
Official Answer to follow after some discussion
It depends on whether or the each seat at a table is unique. What I mean is that if we consider the tables to have chair1, chair2, chair3, etc… we get a different answer than if we just think of the tables as having 5 or 7 chairs. In other words, does seating the people in the same order but shifting everyone over to the right or the left count as the same arrangement or not?
If we’re only concerned about which people are next to each other and not which particular chairs they occupy, then we have 12c5 (or 12c7) ways to divide them into groups of 7 and 5. Then we have 6! ways to arrange the people at the larger table and 4! ways to arrange the people at the smaller table.
This gives us (12c5)6!4!, which will be a huge number.
If we care about which chairs they occupy, then the 6! and 4! become 5! and 7!. They would cancel out with the denominator in 12c5, so we would just get 12!.
These calculations seem right to me for these 2 different cases. What do you think?
Permutation question!
Hi,
Could anyone of you please help me explain how can we solve and visualize the below two scenarios:
1) How to arrange 7 people in 5 chairs.
2) How to arrange 5 people in 7 chairs.
Thanks.
Originally Posted by vd17
Hi,
Could anyone of you please help me explain how can we solve and visualize the below two scenarios:
1) How to arrange 7 people in 5 chairs.
2) How to arrange 5 people in 7 chairs.
Thanks.
A lot of people prefer to just apply the permutation formula here, but the truth of the matter is that you don’t need to know the permutation formula for the GMAT. In fact, there are very few true permutation questions on the GMAT. Instead, you should have a solid grasp of the Fundamental Counting Principle (FCP).
1) How to arrange 7 people in 5 chairs Take the task of sitting people and break it into stages Stage 1: seat someone in chair #1 Stage 2: seat someone in chair #2 Stage 3: seat someone in chair #3 Stage 4: seat someone in chair #4 Stage 5: seat someone in chair #5
Stage 1 can be accomplished in 7 ways (7 people to choose from)
Stage 2 can be accomplished in 6 ways (6 people to choose from, once the first person was seated)
Stage 3 can be accomplished in 5 ways
Stage 4 can be accomplished in 4 ways
Stage 5 can be accomplished in 3 ways
Applying the FCP, the total number of ways to complete all 5 stages, and occupy all 5 chairs = 7x6x5x4x3
2) How to arrange 5 people in 7 chairs.
We need to take a different approach here, since some chairs will not be occupied. However, every person will get a seat.
So, take the task of sitting people and break it into stages Stage 1: seat person #1 in a chair Stage 2: seat person #2 in a chair Stage 3: seat person #3 in a chair Stage 4: seat person #4 in a chair Stage 5: seat person #5 in a chair
Stage 1 can be accomplished in 7 ways (7 chairs to choose from)
Stage 2 can be accomplished in 6 ways (6 chairs to choose from, once the first chair was occupied)
Stage 3 can be accomplished in 5 ways
Stage 4 can be accomplished in 4 ways
Stage 5 can be accomplished in 3 ways
Applying the FCP, the total number of ways to complete all 5 stages, and seat all 5 people = 7x6x5x4x3
Cheers,
Brent
1
davidformba
Trying to make mom and pop proud
Join Date Jun 2004 Location USA Posts 24 Rep Power 9
Good post? |
Can someone explain to me the key difference as to why the Combination and permutation formulas are different by n!
Combination=
C(n,r)=n!/(n-r)!
Permutation=
P(n,r)=n!/n!(n-r)!
I get that Permutation doesn’t care about the order. But I’m looking for a core, simple explanation behind the formulas so that the concept is understood better by me. I have the formulas down, but wonder if I’d do better in knowing the bigger picture rather than taking the formula and looking for the plug in numbers. Any comment would be appreciated.
Good post? |
Hi Dave,
First of all, you got the Combination and Permutation mixed up - but I guess that was the whole point of your post
I came up with this mnemonic device to keep these straight:
P=Prizes (permutations)
C=Committee (combinations)
Here’s the full story:
[/SIZE]Permutations (“Prizes”)[/SIZE]
Let’s say you have 5 people who are running a race, and you want to know in how many ways three prizes (gold, silver, and bronze medal) could be awarded.
Any of the 5 could win the gold, any of the remaining 4 could win silver, and any of the other 3 could win bronze.
So you get 5 x 4 x 3 possibilities.
You can get this result using the permutation formula:
P(5,3) = 5! / (5-3)! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 60
Notice that the effect of the denominator in the formula is just to cancel out the last two terms from the numerator. That’s why I prefer to think of it as simply an incomplete factorial where you multiply out only as many terms as you have selections. In this case, instead of the complete factorial of 5 x 4 x 3 x 2 x 1, you only have three prizes to be awarded, so you stop after the third term.
[/SIZE]Combinations (“Committee”)[/SIZE]
Now think of the same 5 people, but your task this time is to form a committee of 3 (with no special roles, just equal members). You could do your selection in the same way as above, but you would find that some of these permutations give you the same committee. For example, it doesn’t matter whether your selection is person A, then C, then D or whether it is C, then D, then A. So the straightforward selection process we used to award prizes for the race needs to be adapted a bit. We need to divide by the number of possible ways in which a particular committee could have been picked. This is simply the factorial of the number of committee members, in this case 3!
For example, consider these 6 permutations:
ACD
ADC
CAD
CDA
DAC
DCA
These selections all result in the same committee, so they are all equivalent to a single combination.
That’s why the combination formula includes the division by n!, so
C(5,3) = P(5,3) / 3! = (5 x 4 x 3) / (3 x 2) = 10
Hope that clears it up. I completely agree with you that understanding the concept behind it is a much more solid approach than just memorizing the formulas. Once you understand the concept, you can reconstruct the formulas very quickly, even if you forget them under pressure. Also, if you can “map” the problem to one of the two scenarios above, you should have no problem picking the correct formula to use.
Good post? |
Permutation & combination 1
How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number?
(1) 520 (2) 370 (3) 345 (4) None
For a 4 digit num to be div by 4, the last two digits shuld be divisible by 4.
we shall have 14 such combinations.
The remaining digits are 6.
(i) For the combinations which include 0 (there wuld be 4 combinations - 20, 40, 60, 04)-> there are 6 ways to choose the first digit and 5 ways to choose the second digit to make it 4 digit num.
(ii) For the remaining combinations (14-4 = 10)-> there are only 5 ways to choose the first digit (coz 0 would make it 3 digit num) and 5 ways to choose the second digit to make it 4 digit num.
so,
(i) 304 = 120
(ii) 2510 = 250
total = 370
ans . (2)
It took some time to get this answer… is there a way to solve this in 3 to 4 steps..!!??
In how many ways can 5 letters go into 5 envelopes such that:
- No letter goes into its corresponding envelope?
- Exactly 1 letter goes into its corresponding envelope?
A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):
Number of derangements = n! (1/2! - 1/3! + 1/4! + … + ((-1)^n)/n!)
Given 5 letters A, B, C, D and E:
Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44.
Total possible arrangements = 5! = 120.
P(no letter is in the correct position) = 44/120 = 11/30.
2nd Part
A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):
Number of derangements = n! (1/2! - 1/3! + 1/4! + … + ((-1)^n)/n!)
Let the correct ordering of the 5 letters be A-B-C-D-E.
P(A is correctly placed):
P(A is in the correct position) = 1/5. (Of the 5 positions, only 1 is correct.)
P(B, C, D and E are all incorrectly placed):
Total number of derangements = 4! (1/2! - 1/3! + 1/4!) = 12-4+1 = 9.
Total possible arrangements = 4! = 24.
P(no letter is in the correct position) = 9/24 = 3/8.
Since we want both events to happen, we multiply the probabilities:
1/5 * 3/8.
Since the correctly placed letter could be A, B, C, D or E – yielding 5 options for the correctly placed letter – the result above must be multiplied by 5:
5 * 1/5 * 3/8 = 3/8.
Permutation Question
I need some help with digesting the answer to this question:
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, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?
The answer:
Ignoring Frankie’s requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie’s requirement, the six mobsters could be arranged in 720/2 = 360 different ways.
My thoughts:
I understand the 6! There are 720 arrangements without the constraint. However, I don't understand why we divide by 2. I would think that since Frankie wants to stand behind Joey, the limitations would be: Slot 1: Joey, Frankie has 5 choices Slot 2: Joey, Frankie has 4 choices Slot 3: Joey, Frankie has 3 choices and so on.
Thanks in advance
Hi,
I would approach this problem this way:
if there are 6 people to be arranged in 6 places, then number of ways they can be arranged would be 6! ways.
Since Frankie has to be always behind Joe not necessarily immediately behind him:
So let us fix Joe’s position & count the number of options for frankie & remaining people:
A : if Joe at 1st position then Frankie has 5 slots behind him.
B
C
D
E
F
Say for eg: he occupies 3rd slot then remaining 4 persons can be arranged in total 4! ways. This means no of ways this option will have is
a) 1(for Joe) X 5 (for Frankin) X 4!(for remaining 4)
Second possibility: A B Joe's position C D E F
Now Joe is at second position so franlin has 4 slots (if he has to remain behind him)
b) This means no of ways this option will have is
1(for Joe) X 4 (for Frankin) X 4!(for remaining 4)
Similarly
c) 1(for Joe) X 3 (for Frankin) X 4!(for remaining 4)
d) 1(for Joe) X 2 (for Frankin) X 4!(for remaining 4)
e) 1(for Joe) X 1 (for Frankin) X 4!(for remaining 4)
Add all the options from a to e: You will get 360.
Hope this helps.
Cheers
Kundan
Good post? |
Permutation & combination 2
In how many ways can four prizes each having 1st, 2nd , 3rd positions be given to 3 boys , if each boy is eligible to recieve more than one prize?
(1) 12P3 (2) 36 * 36 (3) 64 (4) 12C4 *3!
Good post? |
Originally Posted by targetsep08
yes Official Answer is (2). can someone explain.
Prize one [1st, 2nd and 3rd] can be distribute among three boy in 3P3=6 ways.
Prize two, three and four also can be distribute in 6 ways
So total way is =6666=3636
Engineers - Permutation/Combination problem
Ben needs to form a committee of 3 from a group of 8 engineers to study design imporvements for a product.
If two of the engineers are too inexperienced to serve together on the committe, how many different committees can Ben form?
20 30 50 56 336
Good post? |
There are 8C3 combination. Out of these there are 6 combination where both the inexperienced engineers will be together.
8C3 - 6 = 50
Permutation & Combination - IMS MaxGMAT (PS and DS)
I was doing the IMS GMAT guide for Perm & Comb and came across a couple of confusing ones. someone por favor, el señor shed some light
Problem Solving
- There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
(a) 20!/4!
(b) 20!/5(4!)
(c) 20!/(4!)5
(d) 20!
(e) 5!
Data Sufficiency
1. How many ways m digit numbers can be formed using n distinct digits?
I. m=3 and n =5
II. n=2m
- what is the value of nCr if the value of nC3=56?
I. n - r = 3
II. r = 3
Will post the Official Answer later
I was doing the IMS GMAT guide for Perm & Comb and came across a couple of confusing ones. someone por favor, el señor shed some light
Problem Solving
- There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?
(a) 20!/4!
(b) 20!/5(4!)
(c) 20!/(4!)5
(d) 20!
(e) 5!
Official Answer : (C) 20!/(4!)5. Can someone explain in detail?
Data Sufficiency
1. How many ways m digit numbers can be formed using n distinct digits?
I. m=3 and n =5
II. n=2m
Official Answer: A
- what is the value of nCr if the value of nC3=56?
I. n - r = 3
II. r = 3
Official Answer: D. Can Some explain in detail?
Will post the Official Answer later
Permutation/combination
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A.20 B.40 C.50 D.80 E.120
Good post? |
solution: D
this is the repeated question
Answer is (10 * 8 * 6) / 3! = 80
permutation sum unable to understand
.*Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.
2) in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats
ans:
(4! * 4! * 2) - (7 * 3! * 3! *2)
isn’t 7 supposed to be 4…??
anyone help!!
i agree with mauni it should be (4! * 4! * 2) - (4 * 3! * 3! *2). can some body plaese explain? thanks.
numbers greater than a thousand (permutation)
How many numbers greater than a thousand can be made using the following digits without repetition: 1, 0, 3, 4, and 5?
a) 96^2
b) 96x2
c) 576
d) (24)^2
e) 24x96
Official Answer: B <— highlight to reveal
Originally Posted by mbafan
I disagree with the Official Answer. However, my answer does not fit any of those above.
If 0 is chosen for the first number, that’s OK > then the number automatically defaults to a 4-digit number. The lowest 4-digit number
is 01345 or 1345.
You cannot do that. For example, choose 0 as your first digit. Then you only have 1 option for that. Then you choose the other 4 digits from 4321 [notice when you do this, you have effectively eliminated the possibility of a 4 digit>1000 that includes a 0. So you’ve under-counted.
And then you do the 5 digits by excluding the numbers that started with 0.
So 4432*1–All these just tells you that you have counted 4 digit numbers greater than 1000 that do no include 0, and all 5 digit numbers.
[add those two pieces, 96+24=120, which is what you have] [The only 4 digit numbers you can create by 0 being the first digit are those that do not include 0’s. And if the number starts with anything other than a 0, then that number is a 5 digit number]
comments please.
Good post? |
Permutation and combination
How many different combinations of outcomes can you make by rolling three standard dice if the order of the dice doesnot matter.
A.24 B.30 C.56 D.120 E.216
SPOILER: C
Originally Posted by ACETARGET
How many different combinations of outcomes can you make by rolling three standard dice if the order of the dice doesnot matter.
A.24 B.30 C.56 D.120 E.216
The key here is WITHOUT ORDERING, as I had mentioned in my previous post.. such problems can be classified in 4 categories!
This one falls in WITH REPETITION, W/O ORDER
i.e. (1, 1, 1) is allowed
(1, 2, 3) is allowed but not (1, 3, 2)
Thus answer is:
C(6+3-1, 3) = C(8,3) = 56
Good post? |
solve it
A talent contest has 8 contestants. Judges must award prizes for 1st, Second and third places . If there are no ties a) How many different ways can the prizes be awarded, and b) how many different groups of 3 people can get prizes?
Originally Posted by shootout
A talent contest has 8 contestants. Judges must award prizes for 1st, Second and third places . If there are no ties a) How many different ways can the prizes be awarded, and b) how many different groups of 3 people can get prizes?
These problems can be classified into 4 categories:
R - Repetition O - Order 1) With R, With O 2) With R, Without O 3) Without R, With O 4) Without R, Without O
a) Falls in without repetition, with order:
ex:
123, 132 are countable but not 122, 222 - (assuming same person cannot get more than one prizes)
Then the applied rule is permutation:
876 (8P3)
b) Falls in Without R, Without O ex: 123, 134 allowed, but not 123, 132 (w/o order) Then the applied rule is combination (8C3) = 8!/(3!*5!) = 42
c) If a person can get more than one prize, then it falls into with R, with O.
Then the rule applied is:
(8^3) = 8 * 8 * 8
i dont remember the 4th one, but can derive it if required
circular permutation
Eight men and two women are to be seated around a table. In how many ways can they be arranged if the two women are not sit directly opposite one another.
Please explain.
Good post? |
Hi,
8 Men and 2 Women can be seated around the table in (10-1)! ways.
No of ways you can arrange 2 women around a table sitting opposite to each other = 10
No of ways we can arrange 8 men = 8!
totoal no of ways you can arrange 8 men and 2 women where the 2 women sitting opposite to each other = 10 * 8!
total no of ways you can arrange 8 men and 2 women around a table, where the 2 women are not sitting opposite to each other = 9! - 10 * 8!
Thanks,
Sharath
Reply Reply With Quote
dinner party permutation
A dinner party consisting of 5 couples, sit around a rectangular table, with ladies and gentlemen altering. The host and hostess each occupy one end of the table and their guests are arranged four on each side. Find the number of ways the party can be seated.
Please explain.
Good post? |
Wherever the host sits, he has to be surrounded by two female guests, and so the gender arrangement gets completely defined. Because this is a rectangular table, he only has 2 options of where to sit (had this been a circular table, he’d have 10). So wherever he sits, there are only 4 places where female guests can sit and 4 places where male guests can sit. Since order is important, we use factorials:
24!4!=1152 as cesar82 said.
The other possibility is if the couples have to sit together, but the problem didn’t specify that explicitly so it seems you have to go with the solution above.
1.There are 9 beads in a bag. 3 beads are red, 3 beads are blue, and 3 beads are black. If two beads are chosen at random, what is the probability that they are both blue?
A. 1/81
B. 1/12
C. 2/9
D. 1/3
E. 1/4
2 A letter is randomly selected from the word Mississippi. What is the probability that the letter will be an s?
A. 1/11
B. 3/10
C. 4/11
D. 1/4
E. 1/3
- A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?
A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16
- A fair coin is tossed, and a fair six-sided die is rolled. What is the probability that the coin come up heads and the die will come up 1 or 2?
A. 1/2
B. 1/4
C. 1/6
D. 1/12
E. 1/3
- A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
A. 21/50
B. 3/13
C. 47/50
D. 14/15
E. 1/5
- A fair, six-sided die is rolled. What is the probability that the number will be odd?
A. 1/4
B. 1/2
C. 1/3
D. 1/6
E. 1/5
- A letter is randomly select from the word studious. What is the probability that the letter be a U?
A. 1/8
B. 1/4
C. 1/3
D. 1/2
E. 3/8
- A bag contains 2 red beads, 2 blue beads, and 2 green beads. Sara randomly draws a bead from the bag, and then Victor randomly draws a bead from the bag. What is the probability that Sara will draw a red marble and Victor will draw a blue marble?
A. 2/13
B. 1/5
C. 1/3
D. 1/10
E. 2/15
- If two fair, six-sided dice are rolled, what is the probability that the sum of the numbers will be 5?
A. 1/6
B. 1/4
C. 1/36
D. 1/18
E. 1/9
- If four fair coins are tossed, what is the probability of all four coming up heads?
A. 1/4
B. 1/6
C. 1/8
D. 1/16
E. 1/32
- The probability that a certain event will occur is 5/9. What is the probability that the event will NOT occur?
A. 5/9
B. 4/9
C. 2/9
D. 1/4
E. 1/2
- A certain bag contains red, blue, yellow, and green marbles. If a marble is randomly drawn from the bag, the probability of drawing a blue marble is .2, the probability of drawing a red marble is .3, and the probability of drawing a yellow marble is .1. What is the probability of drawing a green marble?
A. .5
B. .6
C. .2
D. .4
E. .3
- A bag contains 3 red marbles, 3 blue marbles, and 3 green marbles. If a marble is randomly drawn from the bag and a fair, six-sided die is tossed, what is the probability of obtaining a red marble and a 6?
A. 1/15
B. 1/6
C. 1/3
D. 1/4
E. 1/18
- A fair, six-sided die is rolled. What is the probability of obtaining a 3 or an odd number?
A. 1/6
B. 1/5
C. 1/4
D. 2/3
E. 1/2
- At a certain business school, 400 students are members of the sailing club, the wine club, or both. If 200 students are members of the wine club and 50 students are members of both clubs, what is the probability that a student chosen at random is a member of the sailing club?
A. 1/2
B. 5/8
C. 1/4
D. 3/8
E. 3/5
- A bag contains six marbles: two red, two blue, and two green. If two marbles are drawn at random, what is the probability that they are the same color?
A. 1/3
B. 1/2
C. 1/8
D. 1/4
E. 1/5
- There are five students in a study group: two finance majors and three accounting majors. If two students are chosen at random, what is the probability that they are both accounting students?
A. 3/10
B. 2/5
C. 1/5
D. 3/5
E. 4/5
- Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?
A. 5/7
B. 23/24
C. 6/7
D. 34/35
E. 8/13
- A bag has six red marbles and six blue marbles. If two marbles are drawn randomly from the bag, what is the probability that they will both be red?
A. 1/2
B. 11/12
C. 5/12
D. 5/22
E. 1/3
Good post? | 1. B [ 3/9 * 2/8 = 1/12 ] 2. C [ 4/11 ] 3. ? [ 1-2/8 * 1/7 = 27/28] 4. C [ 1/2 * 2/6 = 1/6 ] 5. D [ 1-3/10*2/9 = 14/15 ] 6. B [ 3/6 = 1/2 ] 7. B [ 2/8 = 1/4 ] 8. E [ 2/6 * 2/5 = 2/15 ] 9. E [ 4/36 = 1/9 ] 10. D [ 1/2*1/2*1/2*1/2 = 1/16 ] 11. B [ 1-5/9 = 4/9 ] 12. D [ 1-0.2-0.3-0.1 = 0.4 ] 13. E [ 3/9*1/6 = 1/18 ] 14. E [ 3/6 = 1/2 ] 15. B only wine=200-50=150 only sailing = 400-50-150=200 sailing=200+50=250 probability=250/400=5/8 16. E [ 3 * 2/6 * 1/5 = 1/5 ] 17. A [ 3/5 * 2/4 = 3/10 ] 18. D [ 1- 3/7 * 2/6 * 1/5 = 34/35 ] 19. D [ 6/12 * 5/11 = 5/22 ]
Good post? |
A few different questions
If you join all the vertices of a heptagon, how many quadrilaterals will you get?
72 36 25 35 120
RA:35
35
Good post? |
1.
For a quadrilateral, we need 4 vertices out of the 7 of a heptagon. Hence, 7C4 = 35 quadrilaterals can be formed.
A teacher prepares a test. She gives 5 objective type questions out of which 4 have to be answered. Find the total ways in which they can be answered if the first 2 questions have 3 choices and the last 3 have 4 choices.
255 816 192 100 144
RA: 816
816
2.
Once can select to answer the first 2 and 2 from the last three in 3 ways. These can be answered in 3344 = 144 ways, yielding a total of 3144 = 432 ways.
Or one may choose to answer the last 3 and 1 of the first two. This can be done in 2 ways. Each way would result in 444*3 = 192 ways, giving a total of 384 ways.
Adding the two together would give a total of 816 ways.
In how many ways can 15 students be seated in a row such that the 2 most talkative children never sit together? 14!*14! 15*14! 14! 14!*13 15!
RA: 14!*13
14!*13
3.
First, we arrange the remaining 13 students in 13! ways. This gives 14 places where the 2 children can be placed. The first child can be put in any of the 14 places and the second child can be put in any of the 13 places. This gives a total of 13! * 14 * 13 = 14! * 13 ways.
