quantz work

8/7/2019 Quantz work

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Notes: MBA preparation

Quantitative Aptitude

Combinatory and Probability

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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?

a) 168.

b) 16.

c) 80.

d) 48.

e) 56.

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 oneshelf, one bed and one of the following: a chair or a closet?

a) 168.

b) 16.

c) 80.

d) 48.

e) 56.

3. Three people are to be seated on a bench. How many different sitting arrangements are possible

if Erik must sit next to Joe?

a) 2.b) 4.

c) 6.

d) 8.

e) 10.

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?

a) 648.

b) 504.

c) 576.

d) 810.e) 672.

5. Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if shedoesnt wear 2 specific shirts with 3 specific pants?

a) 41.

b) 66.


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c) 36.

d) 70.

e) 56.

6. A credit card number has 6 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. Howmany different credit card numbers exist?

a) 27.

b) 36.

c) 72.

d) 112.

e) 422.

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?

a) 2/5.

b) 3/5.

c) 3/10.d) 3/4

e) 2/3.

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

a) 8/125.

b) 1/30.

c) 2/5.

d) 1/720.

e) 3/10.

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?

a) 1/(XY).

b) X/Y.

c) Y/X.

d) 1/(X+Y).

e) 1/(X-Y).

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 shesright?

a) 1.

b) .

c) .

d) .

e) 1/3.


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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?

a) 1/4.b) 4/17.

c) 5/29.d) 4/13.

e) 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?

a) 9.

b) 8.

c) 7.

d) 12.

e) 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?

a) 1/5.

b) 1/2.

c) 1/3.

d) 1/4.

e) 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?

a) 1/24.

b) 1/2.

c) 1/3.

d) 2/3.

e) 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 isthe probability of drawing a white ball?

a) 1/5.

b) .c) 1/3.

d) 3/10.

e) .

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?


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a) 8/125.

b) 2/5.

c) 1/30.

d) 1/720.e) 13/48.

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

a) 1/9.

b) 1/12.

c) 5/36.

d) 1/6.

e) 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?

a) 1/8.b) 1/6.

c) 1/3.

d) 1/5.

e) .

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?

a) 1/8.

b) .

c) 1/5.

d) 3/8.e) 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?

a) 4.

b) 20.

c) 3.

d) 5.e) 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?


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a) 23.

b) 20.

c) 24.

d) 17.

e) 25.

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

a) 9/10

b) 16/20

c) 2/5

d) 3/5

e)

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?

a) 4/7b) 3/7

c) 35/128

d) 4/28

e) 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?

a)

b)

c)

d) 15/16e) 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?

a) 15/16

b) 11/16

c) 11/12

d) e) 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)?

a) 1/8

b)

c)

d) 3/8


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e) 5/8

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

a) 720b) 2,160

c) 2,400d) 4,320

e) 40,320

28. In how many ways can you sit 7 people on a bench if Suzan wont sit on the middle seat or on

either end?

a) 720

b) 1,720

c) 2,880

d) 5,040

e) 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?

a) 8b) 23

c) 29

d) 32

e) 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?

a) 23

b) 46

c) 57

d) 66

e) 67

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

a) 10/216

b) 12/216

c) 21/216

d) 23/216

e) 25/216


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32. How many diagonals does a polygon with 21 sides have, if one of its vertices does not

connect to any diagonal?

a) 21

b) 170c) 340

d) 357e) 420

33. How many diagonals does a polygon with 18 sides have if three of its vertices do not send

any diagonal?

a) 90

b) 126

c) 210

d) 264

e) 306

34. What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

a) 1/6b)

c)

d) 21/216

e) 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 additionalarea codes can be created if the company uses all 124 signs?

a) 246

b) 248

c) 492

d) 15,128

e) 30,256

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

a) 16

b) 64c) 128

d) 256

e) 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?


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a) 2520

b) 3150

c) 3360

d) 6000

e) 7500

38. A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three redhats 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?

a) 1/8

b)

c)

d) 3/8

e) 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?

a) 28

b) 44

c) 110d) 210

e) 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? (Remember that this does not ask about how many combinations

of games can be possible, its simply how many different games are possible).

a) 18

b) 19

c) 20

d) 21

e) None of the above

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

a) 2401b) 3584

c) 4096

d) 5040

e) 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?


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a) 1875

b) 2000

c) 2375

d) 2500e) 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?

a) 0.12

b) 0.23

c) 0.35

d) 0.45

e) 0.65

44. A six sided dice with faces numbered 1 thru 6 is rolled twice. What is the probability that theface with number 2 on it would not be facing upward on either roll?

A. 1/6

B. 2/3C. 25/36

D. 17/18

E. 35/36

The probability that face with no. 2 on it would not face upward on 2 rolls

= probability that the first roll does not have 2 facing upward * probability that the second roll

does not have 2 facing upward

= 5/6*5/6= 25/36 (The mistake I initially created was I took the probability of occurrence of 2 2s as 1/36

and just subtracted it from 1 to get 35/36. But this just takes into account that 2 does not face up

on either first or the second roll. We dont want it in either of the rolls).

How many different distinct ways can the letters in the wordVACATION be arranged?

A. 25,375B. 40,320C. 52,500D. 20,160E. 5,040

8!/2! = 20160 (As A appears twice)


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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.


The probability of picking the first jar is , 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.


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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.


The probability of the coin is independent on its previous outcomes and therefore the probabilityfor head or tail is always .

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.


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.


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.


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 = .

16. The best answer is C.The basic principle of this question is that one person cant 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 = 1/30.


When rolling two dice, there are 36 possible pairs of results (6 x 6).


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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: . The same

probability goes for the third child. Therefore the answer is x = .


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.


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:

1. Drawing blue-blue.

2. Drawing blue-red.

3. Drawing red-blue.

4. 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 = 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 = 6/20.

Red-red --> 3/5 x 2/4 = 6/20.Together = 18/20 = 9/10.


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 to rain so we have 4 days of that it will rain and 3 days of that it will


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not rain. We have 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:5. No even digits

6. One even digit.

7. Two even digits.

8. Three even digits.

9. Four even digits.

The probability of choosing an odd (or an even) digit is .

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).


Since 1 appears exactly three times, we can solve for the other four digits only. For every digit wecan 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 .

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.


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.


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 Suzanand the rest: 4*6! = 2880.


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.


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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.


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 thats 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.


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.


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 6options, (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.


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.


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Every letter must be chosen from 0 or 1 only. This means we have two options for every word

and 28 = 256 words total.


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 has3 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.


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= to get three red

hats. The same goes for three blue hats so + =1/2.

The probability to get 3 red or 3 blue can be expressed as follows:

(Prob to get 3 red + Prob to get 3 blue)

Prob to get 3 red = Probability to get 3 red * probability to get 1 blue

= Probability to get red * Probability to get red * Probability to get red *

Probability to get blue

Now, the mistake often created is this probability should take into account the following

combinations (R,R,R,B), (R,R,B,R), (R,B,R,R) and (B,R,R,R)

(This in short is 4C3)

So, the probability to get 3 red = 4 * (1/2) ^ 4

= 1/4

Similarly the probability to get 3 blue hats = 4*(1/2)^4 = 1/4

So, the total probability = + =


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.


The first digit has 7 possibilities (10 0,3 and 6). The other three digits have 8 possibilities each.

7*8*8*8= 3584.


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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.


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. (There is a 0.4 chance to win and 0.6

chance to lose. So, when 3 people win, 2 have to lose. Hence, the calculation is .4*.4*.4*.6*.6 =

0.02304, but this just accounts for the possibility that the first 3 win and the last 2 lose. However,

there can be 10 options for choosing this and hence the probability is 0.23

In New England, 84% of the houses have

a garage and 65% of the houses have a

garage and a back yard. What is the

probability that a house has a backyardgiven that it has a garage?

77%

109%

19%

None of

the above.

Probability = 0.65/0.84

= 77%

In a class of 30 students, there are 17 girls and

13 boys. Five are A students, and three of these

students are girls. If a student is chosen at

random, what is the probability of choosing a

girl or an A student?

None of

the above.


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Probability of choosing a girl = 17/30

Probability of choosing an A student = 2/30 (Because 3 are girls, so just consider 2 boys)

So total probability is 17+2/30 = 19/30

What is the probability that a card selected from a deck will be either an aceor a spade?

1. 2/522. 2/133. 7/264. 4/135. 17/52

Solution.Let A stand for a card being an ace, and S for it being a spade. Wehave to find p(A or S). Are A and S mutually exclusive? No. Are theyindependent? Why, yes, because spades have as many aces as any othersuit. Then,

p(A or S) = p(A) + p(S) - p(A) * p(S)With simple F/T we get:

p(A) = 4/52 = 1/13p(B) = 13/52 = 1/4

So,p(A or S) = 1/13 + 1/4 - 1/52 = 16/52 = 4/13

6 persons seat themselves at round table. What is the probability that 2 given persons are

adjacent?

(A) 1/5

(B) 2/5

(C) 1/10

(D) 1/7

(E) 2/15

I will go with B-2/5

6 people can be arranged in 5! ways.(total )

consider 2 persons as a single entity and then 5 people can be arranged in 4!*2 ways.

So answer is 4!*2/5! = 2/5

Q:There are 6 questions in a question paper? In how many ways can a student solve one or more

questions? The way to solve one or more questions can be described as = (way to solve 1 + wayto solve 2 + .... + way to solve all 6)

= 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6

= 63


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How many 5 letters word which consist of the letters D,I,G,I,T, are there,so that

the letter I are not next to each other?

a. 36b. 48

c.72d. 96

e.128

NUMBER OF COMBINATION WHEN 2I ARE NOT TOGETHER ARE

=TOTAL NUMBER OF COMBI-NUMBER OF COMBINATION WHEN 2I ARE TOGETHER

Taking both 'Is' together, we have 4 places to fill up with 4 letters. Hence, we have 4!

possibilities.

Total number of words can be 5!/2 (Divided by 2 as there are 2 'Is'.

So, the answer is 60-24 = 36

Five racers in a competition . No tie. How many possibilites A is ahead of B?

A 24

B 30

C 60

D 90

E 120

1st positiion - A is first ...that leaves 4*3*2*1 for theother positions

2nd position A is 2nd that leaves 3*1*3*2*1..... (note A is fixed in 2ndposition therefore permutation is 1)

3rd position A is 3rd that leaves 3*2*1*2*1

4th position A is the 4th position 3*2*1*1*1

5th position doesnt count cos a has to finish before B !! tada...add them up=60

2 couples and a single person are seated at random in a row of 5 chairs. What is the probability

that neither of the couples sit together in adjacent chairs.


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The total number of combinations to seat 5 people in 5 chairs = 5*4*3*2 = 120

Now, let us find ways to arrange ppl so that neither couples sit adjacent.

Let the first couple be c1 and c2, the second couple be c3 and c4 and the single person be s.

a) If s sits in the first chair, there are 4 possibilities for the second chair. There are 2

possibilities for the third chair (Not the partner of the person sitting in 2nd chair). There is1 possibility for the 4th chair and 1 possibility for the 5 th chair. So, in all, there are 4*2 = 8

ways. Again, due to symmetry, if s sits on the 5th chair, there are 8 possibilities.

b) If s sits on the second chair, there are 4 possibilities for the 1st chair. For the 3rd chair,there are 3 possibilities. 1 possibility each for the 4th and the 5th chair. In all, 4*3 = 12

possibilities. Again, due to symmetry, 12 possibilities if s sits on the 4th chair.

c) If s sits on the 3rd chair, there are 4 possibilities for the 1st chair. Only 2 possibilities forthe 2nd chair. 1 possibility each for the 4th and 5th chairs. So, 8 possibilities in all.

Summing up all the above possibilities = 8+8+12+12+8 = 48 possibilities.

Hence, the probability that no couples sit adjacent = 48/120 = 2/5

(This is based on the concept that s sits on the first chair OR on the second chair OR on the

third chair OR on the fourth chair OR on the fifth chair).

As a part of a game, 4 people each choose one number from 1 to 4. What is the likelihood

that all people will choose different numbers?

A, B, C and D are the persons. A can choose 1,2,3 and 4. B can choose 1,2,3 and 4 and so

on.

In all, there are 4^4 possibilities of number selections.

Out of these, the possibilities to have 4 distinct numbers = 4*3*2*1 (A has 4 selections, B has

3, C has 2 and D has 1) = 24

So, likelihood = 24/4^4 = 6/4^3 = 0.09 = 9%

Out of seven models, all of different heights, 5 models will be chosen for a photo shoot. If the

5 models stand in a line from shortest to the longest, and the 4th and 6th tallest models cannot

be adjacent, how many different arrangements of models is possible.

The number of ways to select 5 models out of 7 is 7C5 = 21.

Now, out of these 21 ways, the way to select models such that the 4th and 6th are adjacent to

each other are 12346, 12467, 23467, 13467 = 4 ways only.

So, when 4 and 6 cannot be adjacent, number of ways = 21-4 = 17

If 2 students are to be selected from a group of 12 students, how manypossible consequences are there?


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Number of consequences = 12C2 = 66 (Think of it as selecting

1,2 or 1,3 or 1,4 or 1,12, or 2,3 or 2,4 or 11,12) Adding

all these combinations, 11+10+..+2+1 = 66

Hence, the answer is 66.

If the question is to arrange these students, it would be 12P2

= 132 because an arrangement of 1,2 would be different from2,1

A Committee of 6 is chosen from 8 men and 5 women, so as to contain at least 2

men and 3 women. How many different committees could be formed if two of the

men refuse to serve together?

A- 3510

B- 2620

C- 1404

D- 700

E- 635

There are 2 ways of selecting atleast 2 men and atleast 3 women

select 2 men and 4 women or select 3 men and 3 women

selecting 2 men can be done in 3 ways

1. select 1st non-cooperating member and select 1 member from remaining 6(we are

excluding the 2nd non-cooperating member) = 1* 6c1 = 6

2. select 2nd non-cooperating member and select 1 member from remaining 6(we areexcluding the 1st non-cooperating member) = 1* 6c1 = 6

3. don't select any of the cooperating members = 6c2 = 15

same way do it for the selecting 3 men

finally you get

5c4(6+6+15) + 5c3(15+15+20)

answer is 635

OR

First let me provide the answer then explain

1) Select 3 men & 3 women = 8C3*5C3

2) Select 2 men & 4 women = 8C2*5C4

So Total combinations possible = 8C3*5C3 + 8C2*5C4


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3) Now from the above subtract the combinations where theose 2 men appear together.

In the first case (those 2 men appear together, we have to select only 1 other man and 3 more

women)

6C1 * 5C3

In the first case (those 2 men appear together, we only need to select 4 women)1 * 5C4

The Answer Is:(8C3 * 5C3) + (8C2*5C4) - [ 6C1 * 5C3 + 1 * 5C4 ]

= 560 + 140 - 65 = 635

If a committee of 3 people is to be selected from among 5 married couples sothat the committee does not include two people who are married to eachother, how many such committees are possible?

A. 20B. 40C. 50D. 80E. 120

Total ways to select 3 people = 10 c 3 = 120If among 3 people there 2 are married then no. of ways to select 3rd one outof rest 8 = 8c1 = 8since there are 5 couples total ways to do this is = 8*5 = 40But these cases are to be eliminated....so we are left with 120 - 40 = 80 cases..............Hence the answer............

2 similar examples below

1) Ten telegenic contestants with a variety of disorders are to bedivided into 2 groups for a competition, each of 5 members. Howmany combinations are possible?

Selecting 5 members out of 10, for group A = 10C5 = 252.Group B would have the rest of the members, and would have 1possibility. So, 252*1 = 252

Or, 10C5*5C5 = 252

2) Katie has 9 members that she must assign to 3 differentprojects. If 3 emloyees are assigned to each project and no oneis assigned to multiple ones, how many diff. Combinations arepossible?

Selecting 3 members for project A out of 9, = 9C3 = 84


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Selecting 3 members for project B out of 6 = 6C3 = 20

Selecting 3 members out of rem. 3 = 3C3 = 1

So, total combinations = 84*20 = 1680 (Same example as theabove one)

Lets permute:

Judges will select 5 finalists from 7 contenstants in a fashionshow. The judges will then rank the contenstatnts and awareprices to the 3 highest ranked contestants. How many differentarrangements of prize winners are possible?

= 7P5 = 7*6*5 = 210

3) Coach Miller is filling out the starting lineup for his indoor

soccer team. There are 10 boys on the team, and he mustassign 6 starters to the following positions: 1 goalkeeper, 2 ondefense, 2 in midfield, and 1 forward. Only 2 of the boys canplay goalkeeper, and they cannot play any other positions. Theother boys can each play any of the other positions. How manydifferent groupings are possible?

2C1*8C2*6C2*4C1 = 3360-------------------

How many ways the word "COMPUTER" can be arranged, where the vowelsshould occupy the even places?

3 vowels and 5 constn...

so 5*3*4*2*3*1*2*1... but remember because we have only 3 vowels andmore than one starting position for the first vowel then we must multiply thenumber of possibilities by 4

= 720*4 = 2880

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?

When first digit is 2,6 or 8, the combinations are 3*5*5*5*5When first digit is 4, the combination is 1*4*5*5*5*

Total = 2375

Alternatively,


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Total numbers = 4*5*5*5*5 = 2500

Numbers when 4 is at the first 2 digits = 1*1*5*5*5 = 125

Therefore, if 4 is not to appear more than once, 2500-125 = 2375

------------

If 6 people are to be divided to 3 different groups, each of which has 2people. How many such groups are possible?

- i get the method of 6C2 * 4C2 * 2C2 = 90

A certain roller coaster has 3 cars, and a passenger is equally likely to ride inany 1 of the 3 cars each time that passenger rides the roller coaster. If acertain passenger is to ride the roller coaster 3 times, what is the probabilitythat the passenger will ride in each of the 3 cars?

A-0 B-1/9 C-2/9 D-1/3 E-1

The probability to sit in a different car each time = (3*2*1)/(3*3*3) = 2/9

A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If thegardener isto select each of the bushes at random, one at a time, and plant them in arow, what is theprobability that the 2 rosebushes in the middle of the row will be the redrosebushes?A. 1/12B. 1/6C. 1/5D. 1/3E.

There are 2 ways to arrange the centre 2 red bushes. There are 2 ways toarrange the 2 white bushes at the sides. So, 4 arrangements. Totalarrangements would be 4*3*2 = 24

So, probability = 4/24 = 1/6

A photographer will arrange 6 people of 6 different heights for photograph byplacingthem in two rows of three so that each person in the first row is standing infront ofsomeone in the second row. The heights of the people within each row mustincreasefrom left to right, and each person in the second row must be taller than theperson


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standing in front of him or her. How many such arrangements of the 6 peoplearepossible?A. 5B. 6C. 9

D. 24E. 36

If a comttee of 3 people is to be selected from among 5 married couples sothat the comittee does not include tw people who are married to each other ,how many such committees are possible?

a) 20,b) 40), c) 50, d)80, e) 120

numbers of 3 people comttee from 10 people(5*2)

=10C3=10*9*8/6=120------------------1

numbers when couple are together

5*8C1=40----------------------------2

1-2

=80

ans is 80

How many different 6-letters sequence are there that consist of 1 A, 2 B's and3 C's?

a) 6,b) 60, c) 120,d) 360, e) 720

OA is B

6!/(1!*2!*3!)=60

There are 20 purple balls and 30 yellow balls in box A. There are 15 purpleballs and 35 yellow balls in box B. What is the probability that one ballselected randomly from the 2 box is purple?Reference key: 1/2*20/50+1/2*15/50=35/100

The probability to select either of the boxes is


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The probability to select a purple ball from box A is 20/50 and one purple ballfrom box B is 15/50

So, the probability is *20/50 + *15/50)

Dont forget to omit that selection of a box.

A couple want to have four babies, for each baby, 50% are male, 50% arefemale. Ask for the possibility of two boys and two girls.

The propobability of a boy or a girl is

The possibilities are BBGG, BGGB, BGBG, GGBB, GBBG, GBGB

So, 6/16 is the probability

i.e. 6/(1/2)^

what is the probability to get 3 heads and 2 tails on tossing a coin 5 times, inthe same sequence. (i.e. first 3 heads and then 2 tails)

the probability = 1/32 (Since only one combination (HHHTT)

the probability to find either head or tail in the first 3 tosses and the otherside in the last 2 would be

(HHHTT) or (TTHHH)

So, it is 2/32 = 1/16

9 people, including 3 couples, are to be seated in a row of 9 chairs.

What is the probability that

a. None of the Couples are sitting together

b. Only one couple is sitting together

c. All the couples are sitting together

a)1....couple 1 together.... 8!*2!2....couple 2 together.... 8!*2!3....couple 3 together.... 8!*2!


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4....couples 1 and 2 together.... 7!*2!*2!5....couples 1 and 3 together.... 7!*2!*2!6....couples 3 and 2 together.... 7!*2!*2!7....all couples together..6!*2!*2!*2!8....Atleast 1 couple together........ 1+2+3-4-5-6+7 = 3*8!*2-3*7!*4+6!*2*2*2= 3*2*7!*6 + 6!*8 = 6!*2 (3*7*6 - 4) = 6!*2*122total ways = 9!prob atleast one couple together = 6!*2*122 / 9*8*7*6! = 122*2/9*8*7 =61/126

prob that none of the couples is together = 1-61/126 = 65/126

b) only one couple sitting together = 8-4-5-6+2*7= 6!*2*122 - 3*7!*4+2*6!*8= 6!*2 (122-42+8) = 88 * 6! * 2req prob = 88 * 6! * 2/ 9! = 88*2/9*8*7 = 22/63

c) all couples sitting together = 6!*8/9! = 8/9*8*7 = 1/63

To verify my answers....exactly 2 couples are together = 4+5+6-3*7 = 3*4*7! - 3*6!*8= 3*4*6! *5 = 60*6!prob that exactly 2 couples are together = 60*6!/9! = 60/9*8*7 = 15/126

now .....prob of no couple together+exactly one couple together+exactly 2 couples

together+ all couples together = 165/126+22/63+1/63+15/126 = 65+44+2+15/126 = 126/126 = 1


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Notes: MBA preparation

Quantitative Aptitude

Compound Interest

Mean & Median

More MBA related downloads:http://howtoprepare4cat.blogspot.com/2008/09/study-materialqunatz.html
http://howtoprepare4cat.blogspot.com/2008/09/study-materialqunatz.htmlhttp://howtoprepare4cat.blogspot.com/2008/09/study-materialqunatz.html


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A total of $200,000 was deposited at a fixed annual interest rate which iscompounded quarterly. What is the interest of the first month?1) The interest in the second month is 1 percent more than first month2) The interest in the second month is $2 more than first monthReference key: D

Guys this is what I think shud be the solution............

First of all becoz' the interest is compounded quarterly it will be added to theprinciple only after 3 months...........

Let P = 200,000For the first month, the interest I1 = p*(r/100)*(1/12)For first 2 months , the interest I2 = p*(r/100)*(2/12)...... Here we take Pas the principle and not P+I1 becoz' any interest will be added to theprinciple only after the 3rd month and not before that as the rate iscompounded quarterly and not after every month.............

We have from option B , I2 = I1+2Solving this equation we can get the rate r.........and hence the interest

Now for option A...............wer have I2 = I1+I1*(1/100)Solving this also r can be obtained and hence the interest for the firstmonth....


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Hence the answer to this shud be D.....................------------------Ricardo deposits $1,000 in a bank account that pays 10%interest, compounded semiannually. Poonam deposits $1,000 ina bank account that pays 10% interest, compounded annually.If no more deposits are made, what is the difference betweenthe two account balances after 1 year?

A. $2.50B. $10C. $5D. $15E. $100

Interest for first 6 months(compounded semiannually) = amount X rate Xtime

(1000)(10/100)(6/12) = $50. So, amount + interest = $1000 + $50 =$1050Interest for remaining 6 months = (1050)(10/100)(6/12) = $52.50Amount after 1 yr in Ricardo's account = $1050 + $52.50 = $1102.50Poonam:Interest for the year (compounded annually) = (1000)(10/100)(1) = $100Total amount after 1 yr in Poonam's account = $1000 + $100 = $1100Therefore, difference = $1102.50 - $1100 = $2.50

A 2 year certificate of deposit is purchased for K dollars. If the> certificate earns interest at an annual rate of 6 percent compunded

> quarterly, which of the following represents the value, in dollars,> of teh certificate at the end of the 2 years?>> a) (1.06)2 K> b) (1.06)8 K> c) (1.015)2 k> d) (1.015)8 k> e) (1.03)4 k

S= P(1 +i/m)^nm, where P = principal, i = interest rate, n = # of years, m= # of compounding.Since the compounding is done quarterly, there will be 4 periods i.e m = 4

Therefore S = k(1 +0.06/4)^2*4=k(1.015)^8

D is the answer.

A 2-year certificate of deposit is purchased for k dollars. If the certificateearns interest at an annual rate of 6 percent compounded quarterly,which of the following represents the value, in dollars, of the certificateat the end of the 2 years?


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(A) (1.06)^2K(B) (1.06)^8 k(C) (1.015)^2K(D) (1.015)^8K(E) (1.03)^4K

Compound Interest

A = P ]1 + (r/n)](nt)

Original amount = PNumber of years = tNumber of times per year the interest is compounded = nAmount after t years = AAnnual interest rate in % = rInterest after t years = A - P

Amount of the CD after 2 yrs = k * (1 + 6/(100 * 4))^(2*4)

k(1.015)^8

Ans is D-------------

Feng invests his bonus check in a bank account that pays 20% interest,compounded annually. How many years will it take for the initial balance inthis account to double in value?

A. 2

B. 3C. 4D. 5E. 6

2=(1+.2)^t2=(1.2)^t1.2*1.2=1.441.44*1.44=2.0736so answer:C

A total of $1000 was invested for one year. Annual interest rate is r,compound interest is counted semiannually.If the total interest earned by$1000 for that year was $80.56, what is the value of r?

4


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.A sum of money was deposited at x percent compound semi-year interest. 6months later, the total of the money plus interest is $2021; 12 months later,the total of the money plus interest is $2082. x=? The numbers 2021, 2082are not sure.Reference key:1) a(1+x%)=2021 2) a(1+x%)2=20822)/1) = => x%=2082/2021-1=3%

A total of $200,000 was deposited at a fixed annual interest rate which iscompounded quarterly. What is the interest of the first month?1) The interest in the second month is 1 percent more than first month2) The interest in the second month is $2 more than first monthReference key: D

Someone plans to invest $10,000 in an account paying 3% annual interest

and compounded semi-annually. How much must he invest in anotheraccount paying 5% annual interests and compounded quarterly so that hisannual income from the 2 accounts in the first year are the same?Reference key: 9,812Let X be the amount he will invest, so,(1+0.05/4)^4*X=(1+0.03/2)^2*10,000

A total of $10,000 is deposited at the 7.5 percent annual interest rate,compounded monthly. What is the total value in the end of t years?Reference key: 10000*(1+0.075/12)^12t

A sum of money was deposited in a certain account for 2 years without anytransaction. What is the compounded annual interest rate?1) At the end of the second year, the amount in the account is 10.5 percentmore than the initial amount.2) The initial amount is $1,000.Reference key: ALet the initial amount be a and simple annual interest rate be r. Fromstatement 1, [a(1+r)^2-a]/a=10.5%.

Someone deposited a sum of money at annual compound rate ... 6 yearsbefore. There is no any transaction during the 6 years. How much did he

deposit at the beginning?1) At the end of the third year, the amount in the account was 16% morethan the initial amount.2) At the end of the sixth year, the amount in the account was ...Reference key: D


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A sum of $x has been invested in an account paying 8% compounded annualinterest for 5 years. What is the amount in the account now?Reference key: x*(1.08)^5

$ 10,000 was invested at the compounded annual rate r. r=?1) The total interest of the first 4 years is between a and b (a, b are specific

numbers)2) The total interest of the first 4 years is ... percent of the total interest ofthe first 2 yearsReference key: B

A total of $1000 was deposited at the 7 percent annual interest, compoundedmonthly. Without any transaction, at the end of t years, what would be thetotal amount in the account?Reference key: 1000*(1+7%/12)^12t

A sum of money was deposited in a certain account for 6 years without anytransaction. What is the compounded annual interest rate?1) At the end of the third year, the amount in the account is 16 percent morethan the initial amount.2) The initial amount is $1,000.Reference key: ALet the initial amount be a and simple annual interest rate be r. Fromstatement 1, [a(1+r)^3-a]/a=16%.

An investment has a rate of 7% per year compounded monthly. If a value $xis invested for one year, what is the total to be withdrawn?

Reference Key: B

MEAN AND MEDIAN


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Remember... its a RMS value ..... so easy to remember.

--------------

The mean of a list of numbers is m and the deviation (not sure here) is n. It

is known that 68% of the numbers are within m and n, what is thepercentage of the numbers that are less (or more) than m+n?Reference key: 84%=68%+(100%-68%)/2 [Or 16%]

Basics of SD

The mean of a sample of n values is x and the standarddeviation is s.Suppose we add a constant value a, to each observation so that the newdata is

What is the new mean and the new standarddeviationb) The new mean is x + a and the new standarddeviation is s.

The mean of a sample of n values is x and the standarddeviation is s.Suppose that the observations are multiplied by a constant value c, so thatthe new data is

What is the new mean and the new standarddeviation ?d) The new mean is cx and the new standarddeviation is cs.

The mean of 5 numbers is 6,is it deviation bigger than 10?

A. 4 numbers equal to 16.

B. one of the numbers is smaller than 4.

Answer is "A"

from (1), the numbers are 16,16,16,16,-34.

Once we know all the numbers, we can calculate the Deviation with the reqd.formula.


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The real question is not finding SD. It is to know whether we can find SD ornot?It doesn't matter whether SD is bigger than 10 or not??

So with choice (1), we can easily calculate SD.

Hence "A"

.If N is 3 times of the mean of 15 numbers, what is the ration of N to the 16numbers (including N)?Reference key: 1/6

A sequence has 600 numbers, what is the sum of numbers?1) The median is ...2) The mean is 110 percent of the median.Reference key: C

The average (arithmetic mean) of the 5 positive integers k, m, r, s, and t is16, and k < m < r < s < t. If t is 40, what is the greatest possible value ofthe median of the 5 integers?

16

18

19

20

22

Answer 18

How many numbers of 7 consecutive positive integers are divisible by 6?1) Their average is divisible by 62) Their median is divisible by 12Reference key: D

MY picK is D

k+k+1 ......k+6 )=6*7 =42

7k=21 k=3 so numbers from 3 to 9 suff one number divisible by 6

stat2 : 9,10, 11, 12 ,13, 14,15

so 12 divisible by 6


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hence d

Both the ranges of 2 lists are from 1 to 100, whose deviation is greater?1) List 1 has three 100 and two 50; List 2 has two 100 and three 50.2) The averages are the same.

Reference key: C

Why not E? We don't know how many elements in each set, so..i guess itshould be E.

set1: x1, x2, 50, 50, x6,... 100, 100, 100, ....xn

set2: y1, y2, y3, 50, 50, 50, ..., 100, 100, ...ym

even if their averages the same, we don't know elements.

Queen - apologies that this Question has not been answered sooner..... veryfew people are asking JJ questions,,,, anyway

The question itself is not very ambiguous ..its verging on misleading.. thereare too many interpretations to this questions AND remember that stat 1 andstat 2 AS WE HAVE SEEN MANY TIMES do not have to agree !!!!

Stat 2 : this tells us nothing about Stand dev

stat 1 : assuming that these are the only numbers in the list then great wecan anser it BUT it doesnt specifically say that these numbers represent(exhaustively) the list....

So combining stat 1 and 2 - we have an average of a set of numbers we dontknow for CERTAIN and stat 1 gives us some numbers.... useless unless wehave entire set.....SO ANS for me is E....

Formula for std dev = SQRT( [(x1-avg)^2+(x2-avg)^2+(x3-avg)^2]/n )

SOME EXPLAIN HOW TO GET C - thankyou !!!!

I'll go with C.

The total number of elements is given. Average is given as same for both(Stat -2). From this we can infer that the elements are spread out, more orless in a similar manner in both sets, on either side of the mean.

average = total sum / 100 ; Since denominator is constant, the numeratorwould be same for both.

Few numbers are given (Stat2). We can infer which set's SD is greater.


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Try with examples of smaller sets of numbers for proof.

Vam,

I guess your assumption that total number of elements is given is improper.Then have just given us the range.

From I we can not determine anything about standard deviation ( Standarddeviation is nothing but how much it has deviated from median)

From II alone we can not determine anything as median is different fromaverage.

If we combine, still it does not help us to find median.

I would go with E.

Yes Amit. If the set of numbers is given as constant and with the rest of thegiven conditions, then C would be correct. I am sure the actual question ontest would be much more clear and lucid.

Standard deviation is deviation from any measure of central tendency andnot just median. Average is fine for assessing SD or vice versa.

Vam - I am puzzled as to how you think average is used as a measure of stddev ???.... say 49 50 51 and 100 0 50... both same avergae but std devwidely different

perhaps I have misentrepreted what u've said ???????????????? PLeaseexplain

try subst the numbers and working it out ???

perhaps I have misentrepreted what u've said ????????????????

Yes it is misinterpretation- I wrote "Average is fine for assessing SD or


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vice versa.". Amit was harping on median alone. Average or mean is also avalid measure of central tendency used to estimate standard deviation.

try subst the numbers and working it out ???

Two sets of five numbers - {34567} and {12679} - with same average 5 butdifferent standard deviation. Therefore given the similar average and thenumber of elements of the two sets along with few elements - for ex - {67}and {79}, we can say which set would have greater Standard deviation.

Vam,

The list which willhave higher spread of data will have higher deviation.Heres since we don't know other elements of the list we can't determinewhich one will have higher std deviation.

I repeat - if we know the average and the number of elements for both sets(same for both), we can reasonably guess the relative extent of deviation ina particular set given few extreme numbers.

Given question - range is given. So the answer is E. The question is notframed correctly. Hence this ambiguity. (It shouldn't be a problem in the realtest as our fundamentals are fairly strong)

Set B has three positive integers with amedian of 9. If the largest possible range of thethree numbers is 19, given a certain mean, what isthat mean?(A) 22(B) 10(C) 9.6(D) 9

9 is the MEDIAN (another measure of central tendency).

only B is the correct answer.

9 is the MEDIAN (another measure of central tendency).

only B is the correct answer.

a, b, and care integers and a < b < c. S is the set of all integers from a to b,inclusive. Q is the set of all integers from b to c, inclusive. The median of set


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S is (3/4)b. The median of set Q is (7/8)c. IfR is the set of all integers froma to c, inclusive, what fraction ofcis the median of set R?

(A) 3/8(B) 1/2(C) 11/16(D) 5/7(E) 3/4

the answr shud b 11/16

Statistical Basics

The GMAT requires understanding of several basic statistical measures.Although some of the measures may be applied to large samples andpopulations, the GMAT focuses on the use of statistics for samples of limitedsize (a limited number of data points). The statistical measures which youmay encounter on the GMAT are explained and illustrated below.All the statistical measures used on the GMAT help characterize the centrallocation and distribution of the data. Consider the following two sets ofdata:

Data set 1 Data set 2

350400 400500 400 Median500 Median 450 is 425

500 500600 1500

2500 Total 3600 Total5 Number of points 6 Number of points

500 Arithmetic mean 600 Arithmetic meanMean : The arithmetic mean (or average) is the sum of the sample valuesdivided by the number of data points.Median : The median is the middle value of a group of numbers when theyare arranged in order of magnitude . For samples with an odd number ofdata points, the median is the middle number. For example, in data set 1,the median is the third of the five data points (500). For samples with aneven number of data points, the median is midway between the two middledata points. For example, in data set 2, the median is midway between 400and 450 (the third and fourth of the six data points) and equals 425 (theaverage of 400 and 450). For small samples, the median can be a bettermeasure of central tendency than the mean.Mode : The mode is the value that occurs most frequently. Since it ispossible that more than one value may have the same frequency in a set ofdata, there may be more than one mode; in fact, if no value is repeated,everyvalue is a mode. For data set 1 above, the mode is 500. For data set 2,


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the mode is 400. For small samples, the mode often indicates more aboutdata distribution than about central tendency.Range : The range is simply the largest value minus the smallest value. Fordata set 1, the range is 200 (600 C 400). For data set 2, the range is 1150(1500 C 350). For small samples, the range is a simple but useful measureof data distribution.Standarddeviation: The standarddeviation is a more sophisticated measureof data distribution. The standarddeviation can be described as the squareroot of the average squared deviation. Expressed mathematically, this is:

The deviation is the difference between the data value and the mean.Squaring this deviation makes the result positive, regardless of whether thedata point is above or below the mean. Dividing by the number of data pointsprovides an average of the squared deviation. Taking the square root givesthe standarddeviation the same units of measure as the data.The standarddeviation is a useful measure of data variability, even thoughits exact meaning may not be immediately obvious. Consider the standarddeviation an index of data variability. The more the data deviate from themean, the greater the standarddeviation will be. The greater the centraltendency C the closer data are grouped around the mean C the lower thestandarddeviation will be. The standarddeviation is a useful complement tothe range.The table below shows how the standarddeviation is computed for the twodata sets used in the previous discussion.

Data set 1 Data set 2

x x - avg (x Cavg) 2

x x - avg (x Cavg) 2

350 -250 62,500

400 -100 10,000 400 -200 40,000

500 0 0 400 -200 40,000

500 0 0 450 -150 22,500


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500 0 0 500 -100 10,000

600 100 10,000 1500 900 810,000

2500 Total 20,000 3600 Total 985,000

5 n 5 6 n 6

500 Avg. 4,000 600 Avg. 164,167

Std. deviation: 63 Std. deviation: 405

The table below presents a summary of the statistical measures for the twodata sets in the previous discussion:

Data set

1

Data set

2

350

400 400

500 400

500 450

500 500

600 1500

Mean 500 600

Median 500 425

Mode 500 400

Range 200 1150

Std. deviation 63 405

What the StandardDeviation IndicatesConsider the following three sets of data, which represent real estate sales

by a real estate office over a given time period (each sale is indicated by itsprice in thousands of dollars).The mean (average) for these sets of data are identical, but the data arespread very differently, as the histograms below show clearly.

Set A Set B Set C

450 450 450 500

500 500 550 550 550

250 450 450 450

450 450 450 550 1000

250 250 250 250 250

250 1000 1000 1000

Statistical measures other than the mean help characterize the datadistribution more fully. Note that Data Sets B and C have the same range, aswell as the same mean. The low median and mode for Set C may point to


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wider data dispersion than for Set B, but the most direct indication of datavariation is the standarddeviation.

Mean Median Mode(s) Range Std. Dev.

Set A 500 500 450, 500, 550 100 29

Set B 500 450 450 750 192

Set C 500 250 250 750 354

Here are details on the three data sets.

Set A Set B Set C

x xCavg x-av450 -50 2500450 -50 2500450 -50 2500500 0 0500 0 0500 0 0550 50 2500550 50 2500550 50 2500

Tot4500 7500n 9 9

Ave500 833Med500 SD: 28.9

Modes 450, 500, 550

Range 100

Set B xx -avg x - avg

250 -250 62500450 -50 2500450 -50 2500450 -50 2500450 -50 2500450 -50 2500450 -50 2500550 50 2500

1000 500 250000Total 4500 330000

n 9 9Average 500 36667Median 450 SD: 191.5

Mode 450Range 750

Set C x x-a250 -2250 -2250 -2250 -2250 -2250 -2

1000 51000 51000 5

Total 4500 n 9

Average 500Median 250 S

Mode 250Range 750

450 450 450 500

500 500 550 550 550

250 450 450 450

450 450 450 550 1000

250 250 250 25

250 1000 1000

So the standarddeviation is one of the statistical measures used tocharacterize the distribution and central tendency of a set of data. Thestandarddeviation is particularly good for measuring the amount of variationfrom the mean. On the GMAT, you probably will not need to calculate thestandarddeviation, but your are responsible for understanding what itmeans. The type of question that you might encounter is shown below:Q . If the average of 5 data points is 3.5, which new data point would resultin the smallest standarddeviation?

A. 2B. 2.5C. 3D. 3.5E. 4The correct answer to this question is D. To minimize the standarddeviation,one should choose the value closest to the present mean. Answer D allows usto choose a data point that equals the present mean, so it will add nothing tothe sum of the squared deviations. Since the number of data points will be


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one more than before, the standarddeviation will actually decrease slightly.There is no need to actually calculate the standarddeviation on this problem.Statistical Measures for Large Samples and PopulationsFor large samples and populations, the primary statistical measures used arethe mean and standarddeviation. The figure below shows a characteristicnormal distribution.

For a normal distribution (a sample or population which follows the typicalbell-shaped curve shown), 68% of the population lie within 1 standarddeviation of the mean. 95% of the population lie within 2 standarddeviationsof the mean, and 99.7% lie within 3 standarddeviations of the mean. Theother statistical measures (median, mode, and range) are subsumed by themean and standarddeviation. For a large population, the 50 th percentile(with a value equal to the mean) corresponds to the median for a smallsample. Likewise, the mean and 50 th percentile represent the mode of alarge normal distribution. For a large population, the range is not usedbecause even at the tails of the distribution, there is a finite probability

of finding a data point. Instead, one characterizes the probability using thenumber ofstandarddeviations away from the mean. The percentile scores onthe GMAT are derived in this way.

Set X has 5 numbers, which average is greater than their median. Set Y has7 numbers, which average is greater than their median also. If the 2 setshave no common number and are combined to a new set, is the average ofthe new set greater than its median?1) The average of Y is greater than the average of X2) The median of Y is greater than the median of X

If average of Y > X then we can't say nething abt their mediansIfmedian of Y > X then we can't say nething abt their averages

If both average and median of Y > X then

example 1::X:: 1 1 1 1 2


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Y:: 3 3 3 3 4 4 4

New Set = 1 1 1 1 2 3 3 3 3 4 4 4median = 3average = 30/12 = 2.5average < median

example 2::X:: 1 2 3 4 6Y:: 7 8 9 10 51 52 53

new set :: 1 2 3 4 6 7 8 9 10 51 52 53median = 7.5average = 206/12 = 17.17

average > median

Hence E...........

quantz work

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