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Flashcards in Problem Solving Deck (144):
1

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

2

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%

3

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

1. Get first one wrong. X*3*4*4*4= 192
2. Get 2nd one wrong 3*X*4*4*4 = 192
3. Get 3rd one wrong 3*3*X*4*4= 144
4. Get 4th wrong 3*3*4*X*4=144
5. Get 5th wrong 3*3*4*4*X=144

Hence 144+144+144+192+192= 816

4

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

5

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 k*k*k ...*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:
3*3*3 = 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:
3*2*1 = 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:
3*3*3*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 = (4*3)/(2*1) = 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:
6*3*2*1 = 36.

Thus:
(good outcomes)/(total possible outcomes) = 36/81= 4/9.

6

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)*0.3^3*0.7^4

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)*0.5^0*0.5^10-C(10,1)*0.5^1*0.5^9
=1-0.5^10-10*0.5^10
=1-11*0.5^10

7

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

8

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

9

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

10

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.

11

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.

12

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

13


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...

14

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)

15

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.

16


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.

17


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

18


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.

19

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. ;)

20


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

21


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

22

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

23

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

24

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.

x*83+y*76/(x+y)= 79

y*76+z*85/(y+z) = 81

We need to find
x*83+y*76+z*85/(x+y+z)

25


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....2^2*1=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

26

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 (65*4)+(80*6)+(77*5)/15

27

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

28

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

29


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 = 37*82 + 40*74/77 => 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.

30


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

31

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

32

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.




1. 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

2. 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


3. 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!

4. 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


5. 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

2.
A-L = 12 letters
20 desirable possibilities [10 for forward 3-letter sequences (like ABC) 10 for backward 3-letter sequences (like FED)]

12*11*10 = total possibilities

=> 20/(12*11*10) = 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

5.
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

33


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.

34

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.

35


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)

36

#1
budhi
Within my grasp!


Join Date
Apr 2005
Posts
180
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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}

37

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

38


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?

39

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

40

#1
davidformba
Trying to make mom and pop proud

Join Date
Jun 2004
Location
USA
Posts
24
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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.

41


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) 30*4 = 120
(ii) 25*10 = 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..!!??

42


In how many ways can 5 letters go into 5 envelopes such that:

1. No letter goes into its corresponding envelope?
2. 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.

43

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

44


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 =6*6*6*6=36*36

45

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

46

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
1. 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

2. 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
1. 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

2. 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

47

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

48

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.

49

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 4*3*2*1 [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 4*4*3*2*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.

50


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

51


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:
8*7*6 (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

52

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

53

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:

2*4!*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.

54

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

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

4. 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

5. 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

6. 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

7. 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

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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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 ]

55


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.

56

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 3*3*4*4 = 144 ways, yielding a total of 3*144 = 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 4*4*4*3 = 192 ways, giving a total of 384 ways.

Adding the two together would give a total of 816 ways.

57

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.

58

How many five digit numbers can be formed using the digits 0, 2, 3, 4 and 5, when repetiotion is allowed such that the number formed is divisile by 2 or 5 or both?

100
150
3125
1500
125

RA: 1500

My guess was: 4*5*5*5*4 = 2000; 4(0 doesnt work at the beginning)*5*5*5*4(0, 2,4,5 are divisble by 2 or 5 ...)




1500

4.

I agree that the answer should be 2000, as you have explained.

59

How many heptagons can be drawn by joining the vertices of a polygon with 10 sides?
562
120
105
400
282

RA: 120

120


5.

Similar to first question. The answer is 10C7 = 120 ways.

60

Ten different letters of an alphabet are given. 2 of these letters followed by 2 digits are used to number the products of a company. In how many ways can the products be numbered?

52040
8100
5040
1000
4000

RA:1000

Found on GRE ...

1000


6.

I think the answer should be 10*10*10*10 = 10,000 ways as it does not mention anything about repetition not being allowed.

61

Question 1:

If Henry were to add 5 gallons of water to a tank that is already ¾ full of water, the tank would be 7/8 full. How many gallons of water would the tank hold if it were full?
(A) 25
(B) 40
(C) 64
(D) 80
(E) 96

3/4x + 5 = 7/8x
1/8x = 5
x = 40 (B)

62


Good post? |
Question 2:

The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

f(v) = f(m)/3^2 => 3^y / 3^2 => difference between m and v is y-2. Since y is the tens digit, the difference will be 20. Therefore (d) is the answer.

63

Question 3:

At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

To determine # of apples and oranges in first selection: (40(x) + 60 (10-x)) / 10 = 56
This leads to x = 2, therefore 2 apples and 8 oranges.

To determine how many to return: (80 + 60(x)) / (2+x) = 52
This leads to x = 3...Therefore she must return 8-3 = 5 oranges... E

64

Clintonn...thanks for this set....would you by any chance have the Official Answer to all the questions. Plus, there seems to be a typo in 28, is it just my copy?

I completely hit a brick wall on 17...:

17. Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24

any suggestions on how to go about solving this one?

If f(v) = f(432) = (2^4)(3^3)(5^2), f(m)= 3^2((2^4)(3^3)(5^2))

Therefore f(m) = (2^4)(3^5)(5^2)

Therefore m=452 and v=432...452-432=20

65

1. If Henry were to add 5 gallons of water to a tank that is already ¾ full of water, the tank would be 7/8 full. How many gallons of water would the tank hold if it were full?
(A) 25
(B) 40
(C) 64
(D) 80
(E) 96

Answer IMO B
7/8x-3/4x = 1/8x = 5
Therefore x = 8*5 = 4

66

2. The function f is defined for each positive three-digit integer n by f(n) = 2x3y5z , where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m)=9f(v), them m-v=?
(A) 8
(B) 9
(C) 18
(D) 20
(E) 80

Answer : IMO D
Let m = 100A + 10B +C
V = 100D + 10E + F

Then f(m) = 2a3b5c
f(v) = 2d3e5f

multiplying f(v) by 9 = 9f(v) = 32*2d*3e*5f
comparing powers of f(m) we get a=d, b=e+2 and c=f

putting these in the original equation we get m-v = 100A +10(e+2) +C – 100A -10e –C = 20
Hence answer is D

67

3. At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

IMO B
(0.4x+0.6y)/10 =5.6 where x = number of apples and y = number of oranges
Also x+y = 10. from the two equation we can calculate that x = 2 and y = 8
Now the average price after A oranges are put back (i.e A = number of oranges to be put back)
(0.4x+0.6y-0.6A)/(10-A) = 0.52
Solving for A we get A = 2

68

4. Professor Vasquez gave a quiz to two classes. Was the range of scores fro the first class equal to the range of scores for the second class?
(1) In each class, the number of students taking the quiz was 26, and the lowest score in each class was 70.
(2) IN each class, the average (arithmetic mean) score on the quiz was 85

IMO E
Statement I is tells us that the lowest limit is same but the upper limit can be different therefore the range (i.e maximum – minimum can be different). Hence it is insufficient.
Statement II gives us the mean, does not tell us about the maximum or minimum limits therefore it is also insufficient
Together we cannot calculate the upper limit therefore the combined equations are also insufficient.

69

5. If x and y are positive integers, what is the value of x?
(1) 3x5y=1,125
(2) y=3

IMO A
Statement I: 1,125 = 53*32 therefore x = 2 and y = 3 as bases are same powers must be same too. Hence equation I is sufficient.
Statement II: value of y is given. We cannot calculate x from this equation therefore the statement is insufficient.

70

6. If |d-9| = 2d, then d=
(A) -9
(B) -3
(C) 1
(D) 3
(E) 9

IMO D
Since multiple of d is equal to the absolute value of an number therefore it must be positive. Rule out A and B. Plug in values of C, D and E. Only D satisfies the equation therefore D must be the answer

71

7. There are 11 women and 9 men in a certain club. If the club is to select a committee of 2 women and 2 men, how many different such committees are possible?
(A) 120
(B) 720
(C) 1060
(D) 1520
(E) 1980

IMO E
11C2*9C2 = 1980

72

8.


In the figure shown above, two identical squares are inscribed in the rectangle. If the perimeter of the rectangle is 18 by squareroot2, then what is the perimeter of each square?
(A) 8 by squareroot2
(B) 12
(C) 12 by squareroot2
(D) 16
(E) 18

IMO B
Let x be the length of each side of the square. Draw two lines as shown in the figure below. Let us look at the green line first. The length of the green line = root 2*x (property of an isosles triangle…) therefore from the figure we can infer that the width of the rectangle = root 2*x. Next look at the blue line. Use same principle but this time we have two squares so length of rectangle = 2*root 2*x.x

Now the perimeter of the rectangle is 2L+2W= 18 * root 2. Put in values of Length and width. We get x = 3. Thereofre perimeter of each square = 4*3 = 12 Hence answer is B.

73

9. Is the measure of one of the interior angles of quadrilateral ABCD equal to 60 degrees?
(1) Two of the interior angles of ABCD are right angles.
The degree measure of angle ABC is twice the degree measure of angle BCD.

IMO C
Statement I : since the sum of all the angles of a quadrilateral is 360 and we know that two of the angles are 90 degrees therefore the sum of the other two angles would be 180. (i.e 360-2*90= A + B) where A and B are the other two angles. Therefore there would be an angle that is atleat greater than 60 degrees. (i.e if one angle is 10 degrees then the other angle would have to be 170 degrees). However it is uncertain whether the angle would be equal to 60 or not. Therefore statement I is insufficient.
Statement II : A = 2 B. On its own the equation is insufficient.
Combined Statement I and II : 360-2*90=A+B
Therefore 3B =180 hence B = 60 Therefore the one of the angle is equal to 60 hence C should be the answer

74

10. $10,000 is deposited in a certain account that pays r percent annual interest compounded annually, the amount D(t), in dollars, that the deposit will grow to in t years in given by D(t)=10,000(1+(r/100))t What amount will the deposit grow to in 3 years?

(1) D(t) = 11,000
(2) r=10

IMO D
Both equations satisfy as D(t) = 11,000 tells us the amount the deposit will grow in three years. I am assuming that there is a misprint as t = 3. Also in statement II r is given there we can calculate how much will the deposit grow in three years by pluggin in the values in the equation.

75

11. If a and b are integers, is b even?

(1) 3a + 4b is even
(2) 3a + 5b is even



IMO C
Statement 1 tells us that a is even. Since a is even in statement 2 therefore b must also be even.

76

12. Working alone at its constant rate, machine K took 3 hours to produce ¼ of the units produced last Friday. Then machine M started working and the two machines, working simultaneously at their respective constant rates, took 6 hours to produce the rest of the unites produced last Friday. How many hours would it have taken machine M, working along at its constant rate, to produce all of the units produced last Friday?
(A) 8
(B) 12
(C) 16
(D) 24
(E) 30

IMO D
Machine K took 3 hours to do ¼ of the job. Therefore it would take it 4*3 = 12 hours to do the complete job on its own
Similarly both Machine K and M took 6 hours to do ¾ of the job therefore it would take 6*4/3 = 8 hours to do the complete job.
Use the formula
1/k + 1/m = 1/T = where k = hours that would be taken by machine K if it had operated alone. m= hours that would be taken by machine M if it had operated alone. T = hours that would be taken by both machines if they operated together
1/12+1/m = 1/8 solve for m . M = 24

77

15. At a certain bookstore, the regular price of each book is 20 percent less than its list price. If during a sale the price of each book at the store was 15 percent less than its regular price, then the price of a book during the sale was what percent less than its list price?
(A) 30%
(B) 32%
(C) 35%
(D) 38%
(E) 40%

IMO B
Let price list price be 100. Then regular price = 80
85% of regular price = 68
100-68 = 32 Hence B is the answer

78

16. What is the value of the ratio of x to y2?
(1) the rate of x2 to y is 5/3
(2) the ratio of x to 1 is 5/3

IMO C
From statement I x*x/y=5/3
From statement II : x = 5/3 Combining the two equation put in first equation
X/Y * 5/3 = 5/3 therefore X/Y = 1 Hence X = Y = 5/3
Now X/Y^2 = 1* 1/(5/3) = 3/5

G

79

17. Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24

IMO E
19+6-1 = 24. We subtract 1 because 19 inches less and 6 inches greater recounts the length of red stick.


G

80

18. If K is a positive integer, is K the square of an integer?
(1) K is divisible by 4
(2) K is divisible by exactly 4 different prime numbers

IMO B
Since K is a product of 4 different prime number. It cannot be the square of an integer. For example 2*3*5*7 = 210 (not the square of an integer)

G

81

20. Bob, Mia, and Lou each own a number of shares of stock in a company. How many shares does Lou own?
(1) Bob and Mia together own a total of 800 shares.
(2) Bob and Lou together own a total of 500 shares.

IMO E
B+M = 800, B –L = 500. Three variables and two equations insolvable.

G

82

21. In the xy-plane, what is the x-intercept of a certain line?
(1) The line passes through the point (0,2)
(2) The y-intercept of the line is 2

IMO E. Both equations give only the y intercept of the equation

G

83

23. (0.8)-5 / (0.4)-4=

(A) 3/32
(B) 5/64
(C) 1/2
(D) 1
(E) 2

IMO B

(4/5)-5 / (2/5)-4 = 5/64

G

84

24. By what percent was the price of a certain television set discounted for a sale?

(1) The price of the television set before it was discounted for the sale was 25 percent greater than the discounted price
(2) The price of the television set was discounted by $60 for the sale

IMO C

G

85

25. For a certain race, 3 teams were allowed to enter 3 members each. A team earned
6-n points whenever one of its members finished in nth place, where 1<=5. There were no ties, disqualifications, or withdrawals. If no team earned more than 6 points, what is the leas
t possible score a team could have earned?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

IMO D
Least points that a member can have is 6-5=1 when n=5.
Least possible score a team could earn is when all 3 members finish 5th. So, pts = 1+1+1 = 3

G

86

26. Is x < y?
(1) The ratio of x to y is 7/9
(2) xy>0

IMO E
Statement I: x = 7/9y. now X would be less than y if both are positive. However if both x and y are negative x would be greater than y therefore statement I is not sufficient.


G

87

27. For each landscaping job that takes more than 4 hours, a certain contractor charges a total of r dollars for the first 4 hours plus 0.2r dollars for each additional hour or fraction of an hour, where r>100. Did a particular landscaping job take more than 10 hours?

(1) The contractor charges a total of $288 for the job
(2) The contractor charges a total of 2.4r dollars for the job.

IMO C
Equation = r + 0.2r(H-4) = total charge where r= dollars for the first 4 hours and H = Total number of hours worked. Note that there are three variables i.e r , H and total charges
Statement I gives us total charges. Therefore it is not sufficient
Statement II also gives us total charges but in terms of r

Combining the two equation we get the 2.4r = 288. Calculate r. and put in the equation
r + 0.2r(H-4) = 288. We can calculate H from the equation therefore the two equations combined gives us the answer. Hence C is the right answer.

G

88

28. In the rectangular coordinate system, points (4,0) and (-4, 0) both lie on circle C. What is the maximum possible value of the radius of C?

(A) 2
(B) 4
(C) 8
(D) 16
(E) There is no finite maximum value

IMO E
We don’t know whether the two points represent the end points of diameter or a cord. Therefore E should be the answer

G

89

30. A train traveled from station X to station Y at a constant speed of 88 feet per second. Is the distance that the train traveled from station X to station Y greater than 40 miles? (1 mile = 5,280 feet)
(1) It took less than 45 minutes for the train to travel from station X to station Y
(2) It took more than 42 minutes for the train to travel from station X to station Y

IMO B
In one minute the train travels =88*60=5280 feet= 1mile per hour
Statement I says that it took less than 45 minutes of travel. So we cannot calculate anything from this information.
Statement 2 says that it look more than 42 minutes. Hence a train traveling at 1 mile per hour would have traveled more than 40 miles in 42 minutes. Hence the statement I is sufficient.

G

90

31. There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125

IMO A
It is a permutation problem with repeated instances
5!/(3!*1!*1!) = 20

32. missing


G

91

33. (0.8) -5 / (0.4) -4 =

(A) 3/32
(B) 5/64
(C) ½
(D) 1
(E) 2

IMO B
Explanation given above.

G

92

34. A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?

(A) 1: 2
(B) 4: 5
(C) 1: 1
(D) 3: 2
(E) 5: 3

IMO A
Let X be the revenue from cars in 1996 and Y be the revenue from trucks in 1996.
Therefore 0.89X +1.07Y = 1.01(X+Y)
Solving gives X/Y = ½ so A should be the answer


G

93

35. What is the median number of employees assigned per project for the projects at Company Z?

(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.
IMO C
Statement I tells us that 75% of the projects have less than 4 employess. So the median can be 3,2,1
Statement II tells us that 65% of the projects have more than 2 employees assigned to each job
Combining the two equation we can find out that mid 40% of the projects have 3 employees assigned to it. Hence C

G

94

36. If x2 + y2 =1, is x + y =1?
(1) xy =o
(2) y = 0

IMO B
Statement I has no use
In statement II. Put y = 0 in the equation therefore x^2 = 1 therefore x = 1. put y = 0 and x = 1 in x + y = 1. it satisfies the equation therefore statement II is sufficient.

G

95

37. Of the following, which is greatest?

(A) (1/5) -2
(B) (1/3) -2
(C) 3-2
(D) 5-2
(E) 23

IMO A
(1/5)^(-2) = 5^2
(1/3)^(-2) = 3^2
3^ (2) = 1/(3^2)
5^ (-2) = 1/(5^2)
2^3 = 2^3

G

96

Combinatory and probability


1. In a workshop there are 4 kinds of beds, 3 kinds of closets, 2 kinds of shelves and 7 kinds of chairs. In how many ways can a person decorate his room if he wants to buy in the workshop one shelf, one bed and one of the following: a chair or a closet?

168.
16.
80.
48.
56.

G

97

2. In a workshop there are 4 kinds of beds, 3 kinds of closets, 2 kinds of shelves and 7 kinds of chairs. In how many ways can a person decorate his room if he wants to buy in the workshop one shelf, one bed and one of the following: a chair or a closet?

168.
16.
80.
48.
56.

G

98

3. Three people are to be seated on a bench. How many different sitting arrangements are possible if Erik must sit next to Joe?

2.
4.
6.
8.
10.

G

99

4. How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

648.
504.
576.
810.
672.


G

100

5. Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants?

41.
66.
36.
70.
56.

G

101

6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

27.
36.
72.
112.
422.

G

102


7. In jar A there are 3 white balls and 2 green ones, in jar B there is one white ball and three green ones. A jar is randomly picked, what is the probability of picking up a white ball out of jar A?

2/5.
3/5.
3/10.
3/4
2/3.

G

103

8. Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

8/125.
1/30.
2/5.
1/720.
3/10.

G

104

9. The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?

1/(XY).
X/Y.
Y/X.
1/(X+Y).
1/(X-Y).

G

105

(165)^2 - (164)^2

A) 1
B) 2
C) 4
D) 325
E) 329

E

5*5 = 25
4*4 = 16

Unit digit of 5 minus unit digit of 6 should result in a unit digit of 9. Option e is the only one with unit digit of 9.

106

If the temperature rises 25 percent from d degrees to 60 degrees then 60-d=

A) 12
B) 30
C) 35
D) 48
E) 85

A

60/d=25/100
d=6000/125
d=48
60-d= 12
Hence A

107

a=[(1/10)^2]
B=1/5
C=root (1/100)

The values of a,b and c are shown above. Which of the following is correct

A) a

B

A=1/100
B=1/5
C= 1/10

Hence a is the least as the denominator is the highest followed by c and then b

108

10. Danny, Doris and Dolly flipped a coin 5 times and each time the coin landed on “heads”. Dolly bet that on the sixth time the coin will land on “tails”, what is the probability that she’s right?

1.
½.
¾.
¼.
1/3.

11. In a deck of cards there are 52 cards numbered from 1 to 13. There are 4 cards of each number in the deck. If you insert 12 more cards with the number 10 on them and you shuffle the deck really good, what is the probability to pull out a card with a number 10 on it?

1/4.
4/17.
5/29.
4/13.
1/3.

12. There are 18 balls in a jar. You take out 3 blue balls without putting them back inside, and now the probability of pulling out a blue ball is 1/5. How many blue balls were there in the beginning?

9.
8.
7.
12.
6.

13. In a box there are A green balls, 3A + 6 red balls and 2 yellow ones.
If there are no other colors, what is the probability of taking out a green or a yellow ball?

1/5.
1/2.
1/3.
1/4.
2/3.

14. The probability of Sam passing the exam is 1/4. The probability of Sam passing the exam and Michael passing the driving test is 1/6.
What is the probability of Michael passing his driving test?

1/24.
1/2.
1/3.
2/3.
2/5

15. In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?

1/5.
½.
1/3.
3/10.
¼.

16. Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary. What is the probability that only girls will be elected?

8/125.
2/5.
1/30.
1/720.
13/48.

17. Two dice are rolled. What is the probability the sum will be greater than 10?

1/9.
1/12.
5/36.
1/6.
1/5.

18. The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?

1/8.
1/6.
1/3.
1/5.
¼.

19. On one side of a coin there is the number 0 and on the other side the number 1. What is the probability that the sum of three coin tosses will be 2?

1/8.
½.
1/5.
3/8.
1/3.

20. In a flower shop, there are 5 different types of flowers. Two of the flowers are blue, two are red and one is yellow. In how many different combinations of different colors can a 3-flower garland be made?

4.
20.
3.
5.
6.



21. In a jar there are balls in different colors: blue, red, green and yellow.
The probability of drawing a blue ball is 1/8.
The probability of drawing a red ball is 1/5.
The probability of drawing a green ball is 1/10.
If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

23.
20.
24.
17.
25.

22. In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

9/10
16/20
2/5
3/5
½

23. In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

4/7
3/7
35/128
4/28
28/135

24. A Four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

¼
½
¾
15/16
1/16

25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?

15/16
11/16
11/12
½
5/8


26. What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

1/8
¼
½
3/8
5/8

27. In how many ways can you sit 8 people on a bench if 3 of them must sit together?

720
2,160
2,400
4,320
40,320

28. In how many ways can you sit 7 people on a bench if Suzan won’t sit on the middle seat or on either end?

720
1,720
2,880
5,040
10,080

29. In a jar there are 15 white balls, 25 red balls, 10 blue balls and 20 green balls. How many balls must be taken out in order to make sure we took out 8 of the same color?

8
23
29
32
53

30. In a jar there are 21 white balls, 24 green balls and 32 blue balls. How many balls must be taken out in order to make sure we have 23 balls of the same color?

23
46
57
66
67






31. What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?

10/216
12/216
21/216
23/216
25/216


32. How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

21
170
340
357
420


33. How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal? Use the formula: number of diagonals: n (n-3)/2 where n is the number of sides. Each vertex sends of n-3 diagonals.

90
126
210
264
306

34. What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously? Use the formula: number of diagonals: n (n-3)/2 where n is the number of sides. Each vertex sends of n-3 diagonals.

1/6
¼
½
21/216
32/216

35. The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?

246
248
492
15,128
30,256

36. How many 8-letter words can be created using computer language (0/1 only)?

16
64
128
256
512

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

2520
3150
3360
6000
7500

38. A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one? (Tricky)

1/8
¼
½
3/8
7/12

39. Ruth wants to choose 4 books to take with her on a camping trip. If Ruth has a total of 11 books to choose from, how many different book quartets are possible?


28
44
110
210
330

40. A computer game has five difficulty levels. In each level you can choose among four different scenarios except for the first level, where you can choose among three scenarios only. How many different games are possible?

18
19
20
21
None of the above



41. How many four-digit numbers that do not contain the digits 3 or 6 are there?


2401
3584
4096
5040
7200

42. How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?


1875
2000
2375
2500
3875

43. In a department store prize box, 40% of the notes give the winner a dreamy vacation; the other notes are blank. What is the approximate probability that 3 out of 5 people that draw the notes one after the other, and immediately return their note into the box get a dreamy vacation?

0.12
0.23
0.35
0.45
0.65


44.The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?
a) 1/8.

b) 1/6.

c) 1/3.

d) 1/5.

e) E) ¼





45. A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?

46. A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?

47. There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

48. Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, and then drew cards one at a time randomly from the box, without returning the cards he had already drawn to the box. In order to ensure that the sum of all cards he drew was even, how many cards did Jerome have to draw?

49. From (1, 2, 3, 4, 5, 6), one number is picked out and replaced and one number is picked out again. If the sum of the 2 numbers is 8, what is the probability that the 2 numbers included the number 5?

50. Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

51. In how many ways can 11 # signs and 8* signs be arranged in a row so that no two * signs come together?

52. 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




53. 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

54. There are 5 red marbles, 3 blue marbles, and 2 green marbles. If a user chooses two marbles, what is the probability that the two marbles will be a different color?

55. 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?



















Explanations:

1. The best answer is C.
You must multiply your options to every item. (2 shelves) x (4 beds) x (3 closets + 7 chairs) = 80 possibilities.

2. The best answer is C.
You must multiply your options to every item. (2 shelves) x (4 beds) x (3 closets + 7 chairs) = 80 possibilities.


3. The best answer is B.
Treat the two who must sit together as one person. You have two possible sitting arrangements. Then remember that the two that sit together can switch places. So you have two times two arrangements and a total of four.

4. The best answer is C.
For the first digit you have 9 options (from 1 to 9 with out 0), for the second number you have 9 options as well (0 to 9 minus the first digit that was already used) and for the third digit you have 8 options left.
So the number of possibilities is 9 x 9 x 8 = 648.


5. The best answer is D.
There are (8 x 9) 72 possibilities of shirts + pants. (2 x 3) 6 Of the combinations are not allowed. Therefore, only (72 – 6) 66 combinations are possible.


6. The best answer is A.
First digit is 1, the second is 2, the third can be (7,8,9), the forth can be (3,6,9), the fifth and the sixth are dependent with one another. The fifth one is 3 times bigger than the sixth one, therefore there are only 3 options there: (1,3), (2,6), (3,9).
All together there are: 1 x 1 x 3 x 3 x 3 = 27 options.


7. The best answer is C.
The probability of picking the first jar is ½, the probability of picking up a white ball out of jar A
Is 3/(3+2) = 3/5. The probability of both events is 1/2 x 3/5 = 3/10.


8. The best answer is B.
The probability for the first one to be black is: 4/(4+6) = 2/5.
The probability for the second one to be black is: 3/(3+6) = 1/3.
The probability for the third one to be black is: 2/(2+6) = 1/4.
The probability for all three events is (2/5) x (1/3) x (1/4) = 1/30.




9. The best answer is B.
Let Z be the probability of breaking the jar, therefore the probability of both events happening is Z x (1/X) = (1/Y). Z = X/Y.


10. The best answer is B.
The probability of the coin is independent on its previous outcomes and therefore the probability for “head” or “tail” is always ½.


11. The best answer is A.
The total number of cards in the new deck is 12 +52 = 64.
There are (4 + 12 = 16) cards with the number 10.
The probability of drawing a 10 numbered card is 16/64 = 1/4.


12. The best answer is E.
After taking out 3 balls there are 15 left. 15/5 = 3 blue balls is the number of left after we took out 3 therefore there were 6 in the beginning.


13. The best answer is D.
The number of green and yellow balls in the box is A+2.
The total number of balls is 4A +8.
The probability of taking out a green or a yellow ball is (A+2)/(4A+8)=1/4.


14. The best answer is D.
Indicate A as the probability of Michael passing the driving test.
The probability of Sam passing the test is 1/4, the probability of both events happening together is 1/6 so: 1/4 x A = 1/6 therefore A = 2/3.


15. The best answer is B.
Indicate A as the probability of drawing a white ball from the jar.
The probability of drawing a red ball is 1/5.
The probability of drawing both events is 1/10 so, 1/5 x A = 1/10.
Therefore A = ½.


16. The best answer is C.
The basic principle of this question is that one person can’t be elected to more than one part, therefore when picking a person for a job the “inventory” of remaining people is growing smaller.
The probability of picking a girl for the first job is 4/10 = 2/5.
The probability of picking a girl for the second job is (4-1)/(10-1) = 3/9.
The probability of picking a girl for the third job is (3-1)/(9-1) = 1/4.
The probability of all three events happening is: 2/5 x 3/9 x ¼ = 1/30.


17. The best answer is B.
When rolling two dice, there are 36 possible pairs of results (6 x 6).
A sum greater than 10 can only be achieved with the following combinations: (6,6), (5,6), (6,5).
Therefore the probability is 3/36 = 1/12.


18. The best answer is E.
The gender of the first-born is insignificant since we want all children to be of the same gender no matter if they are all boys or girls.
The probability for the second child to be of the same gender as the first is: ½. The same probability goes for the third child. Therefore the answer is ½ x ½ = ¼.


19. The best answer is D.
The coin is tossed three times therefore there are 8 possible outcomes
(2 x 2 x 2). We are interested only in the three following outcomes:
(0,1,1), (1,0,1), (1,1,0).
The probability requested is 3/8.


20. The best answer is A.
We want to make a 3-flower garlands, each should have three colors of flowers in it.
There are two different types of blue and two different types of red.
The options are (2 blue) x (2 red) x (1 yellow) = 4 options.


21. The best answer is A.
If 1/8 is the probability of drawing a blue ball then there are 40/8 = 5 blue balls in the jar. And with the same principle there are 8 red balls and 4 green ones. 40 – 5 – 8 – 4 = 23 balls (yellow is the only color left).


22. The best answer is A.
Since we want to draw at least one red ball we have four different possibilities:
Drawing blue-blue.
Drawing blue-red.
Drawing red-blue.
Drawing red-red.
There are two ways to solve this question:
One minus the probability of getting no red ball (blue-blue):
1-2/5 x ¼ = 1-2/20 = 18/20 = 9/10/
Or summing up all three good options:
Red-blue --> 3/5 x 2/4 = 6/20.
Blue-red --> 2/5 x ¾ = 6/20.
Red-red --> 3/5 x 2/4 = 6/20.
Together = 18/20 = 9/10.



23. The best answer is C.
We have 7!/(4!*3!) = 35 different possibilities for 4 days of rain out of 7 consecutive days (choosing 4 out of seven). Every one of these 35 possibilities has the following probability: every day has the chance of ½ to rain so we have 4 days of ½ that it will rain and 3 days of ½ that it will not rain. We have ½ to the power of 7 = 1/128 as the probability of every single event. The total is 35 x 1/128 = 35/128.


24. The best answer is D.
For every digit we can choose out of 8 digits (10 total minus 1 and 4). There are four different options:
No even digits
One even digit.
Two even digits.
Three even digits.
Four even digits.
The probability of choosing an odd (or an even) digit is ½.
One minus the option of no even digits: 1- (1/2)4= 15/16.
You can also sum up all of the other options (2-5).


25. The best answer is B.
Since 1 appears exactly three times, we can solve for the other four digits only. For every digit we can choose out of 8 digits only (without 1 and 0). Since we have 4 prime digits (2, 3, 5, 7) and 4 non-prime digits (4, 6, 8, 9), the probability of choosing a prime digit is ½.
We need at least two prime digits:
One minus (the probability of having no prime digits + having one prime digit):
There are 4 options of one prime digit, each with a probability of (1/2)4.
There is only one option of no prime digit with a probability of (1/2)4.
So: [1- ((1/2)4+(1/2)4*4)] = 11/16.


26. The best answer is D.
There are three different arrangements of a boy and two girls:(boy, girl, girl), (girl, boy, girl), (girl, girl, boy). Each has a probability of (1/2)3. The total is 3*(1/2)3=3/8.


27. The best answer is D.
Treat the three that sit together as one person for the time being. Now, you have only 6 people (5 and the three that act as one) on 6 places: 6!=720. Now, you have to remember that the three that sit together can also change places among themselves: 3! = 6. So, The total number of possibilities is 6!*3!= 4320.


28. The best answer is C.
First, check Suzan: she has 4 seats left (7 minus the one in the middle and the two ends), After Suzan sits down, the rest still have 6 places for 6 people or 6! Options to sit. The total is Suzan and the rest: 4*6! = 2880.

29. The best answer is C.
The worst case is that we take out seven balls of each color and still do not have 8 of the same color. The next ball we take out will become the eighth ball of some color and our mission is accomplished.
Since we have 4 different colors: 4*7(of each) +1=29 balls total.
Of course you could take out 8 of the same color immediately, however we need to make sure it happens, and we need to consider the worst-case scenario.


30. The best answer is D.
The worst case would be to take out 21 white balls, 22 green and 22 blue balls and still not having 23 of the same color. Take one more ball out and you get 23 of either the green or the blue balls. Notice that you cannot get 23 white balls since there are only 21, however, you must consider them since they might be taken out also.
The total is: 21+22+22+1= 66.


31. The best answer is E.
Start checking from the smaller or bigger numbers on the dice. We will check from bigger numbers working downwards: start with 6, it has the following options: (6,5,1), (6,4,2), (6,3,3). Now pass on to 5: (5,5,2), (5,4,3). Now 4: (4,4,4). And that’s it, these are all number combinations that are possible, if you go on to 3, you will notice that you need to use 4, 5 or 6, that you have already considered (the same goes for 2 and 1). Now analyze every option: 6,5,1 has 6 options (6,5,1), (6,1,5), (5,1,6), (5,6,1), (1,6,5), (1,5,6). So do (6,4,2) and (5,4,3). Options (6,3,3) and (5,5,2) have 3 options each: (5,5,2), (5,2,5) and (2,5,5). The same goes for (6,3,3). The last option (4,4,4) has only one option. The total is 3*6+2*3+1=18+6+1 = 25 out of 216 (63) options.


32. The best answer is B.
We have 20 vertices linking to 17 others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170. The vertex that does not connect to any diagonal is just not counted.


33. The best answer is A.
We have 15 Vertices that send diagonals to 12 each (not to itself and not to the two adjacent vertices). 15*12=180. Divide it by 2 since any diagonal links 2 vertices = 90. The three vertices that do not send a diagonal also do not receive any since the same diagonal is sent and received. Thus they are not counted.


34. The best answer is A.
The options for a sum of 14: (6,4,4) has 3 options, (6,5,3) has 6 options, (6,6,2) has 3 options, (5,5,4) has 3 options. We have 15 options to get 14.
The options for a sum of 8: (6,1,1) has 3 options, (5,2,1) has 6 options, (4,3,1) has 6 options, (4,2,2) has 3 options, (3,3,2) has 3 options. We have 21 options to get 8.
Total: 21+15= 36/216 = 1/6.


35. The best answer is C.
The phone company already created 122*122 area codes, now it can create 124*124.
1242-1222=(124+122)(124-122) = 246*2 = 492 additional codes.
There are other ways to solve this question. However this way is usually the fastest.


36. The best answer is D.
Every letter must be chosen from 0 or 1 only. This means we have two options for every word and 28 = 256 words total.


37. The best answer is A.
The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6=2520.


38. The best answer is C.
Getting three red out of 4 that are taken out has 4 options (4!/(3!*1!)) each option has a probability of (1/2)4 since drawing a red or blue has a 50% chance. 4*1/16= ¼ to get three red hats. The same goes for three blue hats so ¼+¼ =1/2.


39. The best answer is E.
Choosing 4 out of 11 books is: 11!/(4!*7!) = 330 possibilities.


40. The best answer is .
On four levels there are 4 scenarios = 16 different games. The first level has 3 different scenarios. The total is 19 scenarios.


41. The best answer is B.
The first digit has 7 possibilities (10 – 0,3 and 6). The other three digits have 8 possibilities each. 7*8*8*8= 3584.


42. The best answer is C.
Not considering the fact that 4 cannot appear more than once, we have a total of 4*5*5*5*5=2500. Now we deduct the possibilities where 4 does appear more than once (in this case it can appear only twice on the two leftmost even digits). In order to do so, we put 4 in the first and second leftmost digits. The rest of the digits are odd: 5*5*5=125. 2500-125=2375.


43. The best answer is B.
The chance of winning is 0.4 and it stays that way for all people since they return their note. The number of different options to choose 3 winners out of 5 is 5!/(3!*2!) = 10. Each option has a chance of 0.4*0.4*0.4*0.6*0.6 = 0.02304 * 10 = 0.2304.

44. The best answer is E.
First child could be B or G, similarly 2nd and third could be B or G. Hence total number of ways =2*2*2 =8.

Favorable number of ways = all B or all G (i.e. BBB or GGG)
=2
hence P(E)= Favorable number of ways/Total number of ways
=2/8
=1/4

45. Number of ways 2 can be selected without taking the three in question. : 7C2
Number of ways 2 can be selected out of 10 = 10C2
Prob. = 7C2/10C2 = 21/45 = 7/15

46. No. of ways to pick 3 individuals out of 5 couples such that a couple is included is 5C1*8C1 = 5*8 = 40 (order of choice here doesn't matter)

Total no. of ways to pick 3 folks from ten = 10C3 = 10*9*8/(3*2) = 120

Therefore ways to select 3 individuals out of 5 couples such that no couple is included = 120-40 = 80.

47. Total no of ways: 3^4
1st sec can get a report in 4c1 ways = 4
2nd sec .........................3c1 way = 3
3rd sec...........................2c1 way = 2
last can be distributed to any of the three = 3

So probability is = 4*3*2*3/ 3^4 = 8/9

48. Lets assume the worst-case scenario.

Jerome draws an odd. Then an even. This gives an odd number. Next he draws another even. Now we have an odd again. Then he draws an even. Again the sum is odd.
So, to sum it up, 

We have 10 odd and 10 evens. 
If his first draw is odd, then the next 10 are even, we still have an odd sum. The tie breaker will be the 12th card, which now has to be odd since all evens have been exhausted. So if the first card drawn is odd, then we must DRAW 12 CARDS.

If the first is even, then the second is odd, again we have an odd number. Now we only have 9 evens lefts, we must exhaust all of them to get an odd one. So again, 12 cards.

So the answer is 12. The 12th draw ensures an even sum.

49. It is already given that sum is 8. 
So total number of events is 5 i.e. (2,6) (6,2) (3,5) (5,3) (4,4)
Events in which 5 is included 2.
So probability = 2/5.

50. Since order is important, this is a permutation problem, not a combination one
# of sequences possible = 8P3 = 336

36 unassigned => # of participants = 336-36 = 300

51. 1st we place the 11#, now there are 10 places between them and 2 on the extreme left and extreme right of them, total places = 10 + 2 = 12

If * is placed at any of these 12 places, no 2 *'s will be together
so the number ways will be 12C8 = 12C4 = 495

52. The probability will be 1- probability that no blue marbles are selected
1- 3C2 / 10C2 = 14/15
Another method:
1 Blue - 1 Red + 1 Red - 1 blue + Both Blue = 7/10*3/9 + 3/10*7/9 + 7/10*6/9= 14/15. Since we are not replacing the marbles, order matters and so 1 Blue- 1 Red is not the same as 1 Red- 1 Blue.

53. 1 st card that is not blue = 6/8
2nd card that is not blue = 5/ 7

= 30/56 = 15/ 28

Another method: 1 - P(at least 1 blue card)

54. 1 -(probability of same color)

1 -(5C2/10C2 + 3C2/10C2 + 2C2/10C2)
=1 - (10+3+1)/45
= 1 - 14/45 = 31/45

55. Probability that they are the same color = 1- probability that are NOT the same color.

Probability that they are not the same color = Probability of (1R-1B + 1B-1G + 1G-1R + 1B-1R + 1G-1B + 1R-1G) = 1- 24 /30 = 1/5 (**Order of color matters)

G

109

If a, b, and c are not equal to 0 or 1 and if a^x = b, b^y = c, and c^z = a, then xyz =
(A) 0
(B) 1
(C) 2
(D) a
(E) abc

This question is pretty challenging for me.please help me out here.

B

If a^x = b then a = b^(1/x) similarly b= c^(1/y) and c=a^(1/z)

Put value of c in equation 2
b=[a^(1/z)]^(1/y)
Therefore b= a^(1/(ya))

But b= a^x then
A^x=a^(1/(ya))
Taking root x on both sides
A=a^(1/(xyz))

Since bases are same powers would have to be same
1/(xyz) = 1
Xyz=1

Hence B is the correct answer

110

1) If (x - 2y)(x + 2y) = 5 and (2x - y)(2x + y) = 35, then x^2-y^2 / x^2+y^2

(A) -8/5
(B) -4/5
(C) 0
(D) 4/5
(E) 7/5

D

For question 1), you can solve as following :

After expanding the two equation given in the problem, you get the following :

x^2 - 4y^2 = 5 .....(i)

4x^2 - y^2 = 35.....(ii)

Add the two equations and you get following :

5x^2 - 5y^2 = 40

Therefore, x^2 - y^2 = 8.... (a)

Now subtract (i) from (ii)

3x^2 + 3y^2 = 30

Therefore, x^2 + y^2 = 10.....(b)

Divide (a) by (b) and you get = 8/10 = 4/5

Answer is (d)

111

A computer is programmed to generate a list of multiples of Prime numbers 2, 3 and 5 as shown below

Program 1: list multiples of 2
Program 2: list multiples of 3
Program 3: list multiples of 5

How many integers between 1 and 100 will appear on all three of theists of the program produced above

A) none
B) 1
C) 3
D) 5
E) an infinite number of integers

C

A number will have to be a multiple of 2,3 and 5

5*2*3= 30

Hence 30, 60 and 90 lie between 1 to 100 therefore 3 numbers

112

2)If l/m+n=m/n+1=n/l+m=k,where k is a real number, then which one of the following does k equal?
(A) 1/3
(B) 1/2
(C) 1
(D) 2
(E) 3

B

For question 2)

L/m+n = k

therefore, L/k = m+n

Similarly, m/k = n+L

n/k = L+m

Add all three equations :

1/k (m+n+L) = 2(m+n+L)

Therefore k = 1/2

Answer = B

113

A photography dealer ordered 60 model X cameras to be sold for $250 each, which represent a 20% markup over the dealer's initial cost for each camera. Of the cameras ordered , 6 were never sold and were returned to manufacturer for a refund of 50% of the dealer's initial cost.what was the dealer's aprrox profit or loss as a percent of dealers initial cost of 60 cameras.
a) 7%loss
b) 13%loss
c) 7% profit
D) 13%profit
e) 15 % profit

Was able to solve it but took me around 4 minutes. Is there any shortcut??

This was my way of doing it. I took just more than 2 minutes but this was due to the final step which required lengthy calculations.

The dealer plans to cell the cameras for USD 250 which is 20% more than dealer's intial cost.

Let the initial cost be "a"

Therefore, a + 0.2a = 250

Solve for "a" , a = 208.

Therefore initial cost = 208 * 60 = $ 12480.

The dealer sold 54 cameras at $ 250 = $ 13500

The dealer sold 6 cameras at $ 104 (50% refund on initial cost) = $ 624

Total selling price = $ 14124.

Profit = (14124 - 12480)/ 12480 = 13 % profit.

114

On level farmland, two runners leave at
the same time from the intersection of
two country roads. One runner jogs due
north at a constant rate of 8 miles per
hour while the second runner jogs due
east at a constant rate that is 4 miles per
hour faster than the first runner's rate.
How far apart, to the nearest mile, will
they be after 1/2 hour ?
(A) 6
(B) 7
(C) 8
(D) 12
(E) 14

B

they will have gone 4 and 6 miles, respectively. Just do the pythagorean theorem :


4^2 + 6 ^ 2 = 52,

The square root of this is closest to 7.

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115

A square playground has the same
area as a rectangular playground that
is 30 meters longer but 20 meters
narrower. What is the length, in
meters, of a side of the square
playground?
(A) 10 square root 5
(B) 10 square root 6
(C) 25
(D) 50
(E) 60

E

x^2 = (x +30)(x-20)

x^2 = x^2 - 10x - 600

when you solve this you will get -60 -> the magnitude of this answer matches and honestly the sign does not matter for this question. Really the wording should be flipped though.

116

Michelle deposited a certain sum of money in a savings account on July 1st, 2007. She earns an 8% interest compounded semiannually. The sum of money in the account on December 31st, 2009 is approximately what percent of the initial deposit?

117%
120%
121%
135%
140%

Can some one show an easy way to solve this? Thanks.

C

117

Michelle deposited a certain sum of money in a savings account on July 1st, 2007. She earns an 8% interest compounded semiannually. The sum of money in the account on December 31st, 2009 is approximately what percent of the initial deposit?

117%
120%
121%
135%
140%

Can some one show an easy way to solve this? Thanks.

C

Simple interest over 5 periods would be 20% (4 x 5). You know you are going to be more than this because of compounding but not almost double or double as is in D and E.

118

I've seen a post about this question, but no one ever actually answered the question.

For every positive even integer n, the function h(n) equals the product of all even integers from 2 to n, inclusive. If P is the smallest prime factor of h(100)+1, then P is...

1) Between 2 and 10
2) Between 10 and 20
3) Between 20 and 30
4) Between 30 and 40
5) >40

3rd question on my 1st GMAT Prep I would know how to do this if it were the summation of all even integers from 2 to n, but I'm not sure how to deal with the product of all even integers. Anyone?

E

The previous explanation is correct. I will try to elaborate in more detail, and hopefully that will help.

Let's break it down piece by piece.....

h(100) = multiply every even number from 2 to 100

h(100) =2x4x6x8....etc

Can we simplify this? Yes.

h(100) = 2 * (2*2) * (2*3) * (2*4)....all the way up to (2*50)

I think the first "trick" of this question is that they are trying to get you to misapply the distributive property of addition.

If the question asked 2+4+6+8, you could solve by figuring out (2*1)+(2*2)+.... and you could pull out all your 2's, so you have (2*(1+2+3+4)....) - and then find the sum of all the numbers from 1 to 50).

However, this is not that kind of question. It is asking you to MULTIPLY every even number from 2 to 100.

So we can't distribute. It is literally (2*1)*(2*2)*(2*3)......so we are multiplying the number 2 times every single number between 1 and 50 in a composite way. Why is it not between 1 and 100, you ask........because its only the even numbers between 1 and 100, so there are only 50 numbers.

This is the same as saying multiply 2, 50 times, and then multiply all the numbers from 1 to 50. Translated into math, this is 2^50 * 50!.

So, h(100) = 2^50 * 50!

Ok - now we have a probem. We have simplified our equation, but we still are working with an incalculable number for GMAT purposes. So we can't use our standard divisibility rules in any predicable way.

This is where the solution using prime numbers comes into play.

Let's play around with a smaller version of h(100) for practice. Let's try h(6).

h(6) = 2x4x6 = 48

h(6) + 1 = 49

What is the SMALLEST PRIME NUMBER THAT DIVIDES h(6)?

Well, 49 = 7x7 - so 7 is the smallest prime number that divides h(6)

Let's try h(4).

h(4) = 2x4 = 8

8+1 = 9

What's the smallest prime number that divides 9?

Well, it's 3.

Hopefully you are noticing a pattern. The smallest prime number that divides h(n)+1 is always larger than the largest prime number in h(n).

So, before we even get to the formal explanation, we see that rule. If we extend that rule to h(100) +1 - then the largest prime number that divides h(100)+1 is larger than the largest prime in the set H(100). We can easily see from above that 47 is the largest prime number in h(100). So our answer has to be greater than 47.....so we know that it must be E.

Now, a more formal "proof" for this question is along the lines of what someone explained before.

We know h(100) = 2^50 * 50!, as explained earlier.

We know from that if a number is divisible by a prime number, and then you add 1 to the number in question, it CANNOT be divisible by that same prime. For example. if 10 is divisible by 5, can 11 be divisible by 5? Absolutely not.

Take this logic, and apply it to the problem. We are looking for the largest PRIME divisor. So by the above logic, if h(100) is divisible by a certain prime divisor, h(100) + 1 CANNOT be divisible by that same prime divisor.

We know that h(100) = 50! * 2^50, and 50! contains every prime from 1 to 50. Therefore, h(100) must be divisible by every prime from 1 to 50. (If you don't get the part about being divisible by primes, then you need to study this for the GMAT. While this type of question in particular will not come up often, you need to know prime factorization and its applications to do well on GMAT Math).

Since h(100) is divisible by every prime from to 1 to 50, H(100) + 1 CANNOT be divisible by ANY of those primes. Therefore, our prime in question must be greater than 50.

119

In an election, candidate Smith won 52% of the total vote in Counties A and B. He won 61% of the vote in County A. If the ratio of people who voted in County A to County B is 3:1, what percent of the vote did candidate Smith win in County B ?

25%
27%
34%
43%
49%

Need to know alternate methods of solving this problem in less than 2 min!!

source: https://www.freequestionaday.com/gmat

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A

GMAT Kolaveri wrote:
In an election, candidate Smith won 52% of the total vote in Counties A and B. He won 61% of the vote in County A. If the ratio of people who voted in County A to County B is 3:1, what percent of the vote did candidate Smith win in County B ?

25%
27%
34%
43%
49%

Need to know alternate methods of solving this problem in less than 2 min!!

source: https://www.freequestionaday.com/gmat
Let the number of voters in A = 3.
Let the number of voters in B = 1.
The total number of voters in A and B = 3+1 = 4.

Treat the percentages as averages.
Sum in A and B = (number)(average) = 4*52 = 208.
Sum in A = (number)(average) = 3*61 = 183.
Sum in B = 208-183 = 25.
Average in B = sum/number = 25/1 = 25.

The correct answer is A.

120

Working alone, Printers X, Y, and Z can do a certain printing job, in 12, 15, and 18 hours respectively. What is the ratio of the time it takes printer x to do the job tot he time it takes printers Y and Z working together at their individual rates?
4/11
1/2
15/22
22/15
11/4
OG 22/15


The first thing I noticed was that it would take the printer working by himself longer so I eliminated answers 1 2 3. However I dont know how to do the math correctly to pick between the oteher two. I guessed correctly as 11/4 seemed to be too much time difference.

D


avada wrote:
Working alone, Printers X, Y, and Z can do a certain printing job, in 12, 15, and 18 hours respectively. What is the ratio of the time it takes printer x to do the job tot he time it takes printers Y and Z working together at their individual rates?
4/11
1/2
15/22
22/15
11/4
The time ratio for X:Y:Z = 12:15:18.
To make the math easier, divide all of the times by 3:
X:Y:Z = 4:5:6.

Let the job = the LCM of 4, 5, and 6 = 60.

X's rate = w/t = 60/4 = 15 pages per hour.
Y's rate = w/t = 60/5 = 12 pages per hour.
Z's rate = w/t = 60/6 = 10 pages per hour.
Combined rate for Y+Z = 12+10 = 22 pages per hour.

Rate and time are RECIPROCALS.
Since (rate for X alone) : (rate for Y+Z) = 15:22, (time for X alone) : (time for Y+Z) = 22:15.

The correct answer is D.

121

OG Diagnostic Qs. 1:
Last month a certain music club offered a discount to preferred customers. After the first compact disc purchased, preferred customers paid $3.99 for each additional compact disc purchased. If a preferred customer purchased a total of 6 compact discs and paid $15.95 for the first compact disc, then the dollar amount that the customer paid for the 6 compact discs is equivalent to which of the following? (A) 5(4.00) + 15.90 (B) 5(4.00) + 15.95 (C) 5(4.00) + 16.00 (D) 5(4.00 – 0.01) + 15.90 (E) 5(4.00 – 0.05) + 15.95

G

122

The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?(A) 150(B) 175 (C) 200(D) 225(E) 300

G

123

The average (arithmetic mean) of the integers from 200 to 400, inclusive, is how much greater than the average of the integers from 50 to 100, inclusive?(A) 150(B) 175 (C) 200(D) 225(E) 3003. The sequence a1, a2, a3,...,an,... is such thatan = for all n ≥ 3. If a3 = 4 anda5 = 20, what is the value of a6 ?(A) 12(B) 16(C) 20(D) 24(E) 28

G

124

Among a group of 2,500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from the 2,500 people, what is the probability that the person selected will be one who invests in municipal bonds but NOT in oil stocks?(A) (B) 725(C) 720(D) 2150(E) 2750

G

125

A closed cylindrical tank contains 36π cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?(A) 2(B) 3(C) 4(D) 6 (E) 9

G

126

6. A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap?(A) 15(B) 20(C) 30(D) 40(E) 45

G

127

A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?(A) 1720(B) 180(C) 110(D) 19(E) 15

G

128

If a certain toy store’s revenue in November was 25 of its revenue in December and its revenue in January was 14 of its revenue in November, then the store’s revenue in December was how many times the average (arithmetic mean) of its revenues in November and January

G

129

9. A researcher computed the mean, the median, and the standard deviation for a set of performance scores. If 5 were to be added to each score, which of these three statistics would change?(A) The mean only(B) The median only(C) The standard deviation only(D) The mean and the median(E) The mean and the standard deviation

G

130

In the figure shown, what is the value of v + x + y + z + w ?(A) 45(B) 90(C) 180(D) 270(E) 360

G

131

11. Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 90 (B) 82 (C) 80 (D) 45 (E) 36

G

132

12. Positive integer y is 50 percent of 50 percent of positive integer x, and y percent of x equals 100. What is the value of x ? (A) (B) (C) (D) (E) 50 100 200 1,000 2,000

G

133

13. If s and t are positive integers such that s t= 64.12, which of the following could be the remainder when s is divided by t ? (A) (B) (C) 2 4 8 (D) 20 (E) 45

G

134

14. Of the 84 parents who attended a meeting at a school, 35 volunteered to supervise children during the school picnic and 11 volunteered both to supervise children during the picnic and to bring refreshments to the picnic. If the number of parents who volunteered to bring refreshments was 1.5 times the number of parents who neither volunteered to supervise children during the picnic nor volunteered to bring refreshments, how many of the parents volunteered to bring refreshments? (A) 25 (B) 36 (C) 38 (D) 42 (E) 45

G

135

The product of all the prime numbers less than 20 is closest to which of the following powers of 10 ? (A) 109 (B) 108 (C) 107 (D) 106 (E) 105

G

136

16. If (A) 3221 −=+ xx, then 4x2 = 1 (B) 4 (C) 2 − 2x (D) 4x − 2 (E) 6x − 1 17. If n = (A) (B) (C) (D) (E) 16 81, 2, 7, and k ? (A) One (B) Two (C) Three (D) Four (E) Five

G

137

what is the value of ?1 91 44 92 39 2 18. If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have? (A) Four (B) Five (C) Six (D) Seven (E) Eight

G

138

19. If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths

G

139

20. A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?(A) (B) 1:1 (C) :112(D) (E) 2:1

G

140

John deposited $10,000 to open a new savings account that earned 4 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in John’s account 6 months after the account was opened?(A) $10,100(B) $10,101(C) $10,200(D) $10,201(E) $10,400

G

141

A container in the shape of a right circular cylinder is 12 full of water. If the volume of water in the container is 36 cubic inches and the height of the container is 9 inches, what is the diameter of the base of the cylinder, in inches?(A) 169π(B) 4π(C) 12π(D) 2π(E) 42π

G

142

23. If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6, then xy must be a multiple of which of the following? I. 8 II. 12 III. 18(A) II only(B) I and II only(C) I and III only(D) II and III only(E) I, II, and III

G

143

24. Aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron jog so that he spends a total of t hours jogging and walking?(A) xty(B) x + txy(C) xy tx + y(D) x + y + txy(E) y + t–txy

G

144

etters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

Example:
From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?

This is a case of "with replacement":
Total outcomes: 9^4
Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2
You can calculate the probability from here.

Note that there is a special thing with letter problems, ie. the repeating letters. In the above examples the repeating letters don't matter. However they would matter if the question is asked differently.

Example:
There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?

Solution:
Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.
Consonants: S, T, R, L

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)

After you've got the four letters you need to order them to get different outcomes. So the first case (with different vowels) would be C(3,2)*C(4,2)*P(4,4)=432, and the second case (with same vowels) would be C(1,1)*C(4,2)*P(4,4)/2=72. The final outcome would be 504.

etters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

Example:
From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?

This is a case of "with replacement":
Total outcomes: 9^4
Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2
You can calculate the probability from here.

Note that there is a special thing with letter problems, ie. the repeating letters. In the above examples the repeating letters don't matter. However they would matter if the question is asked differently.

Example:
There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?

Solution:
Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.
Consonants: S, T, R, L

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)

After you've got the four letters you need to order them to get different outcomes. So the first case (with different vowels) would be C(3,2)*C(4,2)*P(4,4)=432, and the second case (with same vowels) would be C(1,1)*C(4,2)*P(4,4)/2=72. The final outcome would be 504.