there are 10 balls of different colours

for minimum amount : the first ball will be gumball
for the first 10 balls there is 1 gumball,ie the first ball of the 10

for the 2nd 10 balls there is 1 gumball,ie the first ball of the 10

for the 3rd 10 balls we only need the first one which would be the gumball.
hence we have 10+10+1=21

one quater is $0.25

three gumball cost one quater which is $0.25

we can get 7 three balls in 21 i.e

21/3=7

the total amount is now 7*$0.25=$1.75

@edusegzy
Please help me find the eigen values and eigen vectors of the matrix A.

3 5 -2 0 7 -4 1

0 -8 2 -3 0 9 -7
where A =
6 1 -8 3 0 4 -6

1 9 -1 -6 0 7 0

-2 -5 -9 7 3 0 8

SOLUTION: for all i know a matrix has eigen values and eigen vectors if it is a square matrix.This is a rectangular matrix of order (5X7).
anyone that has anything contrary should post it

edusegzy


Great work buddy. Where did you write the GRE? How difficult or should I say, challenging was the quant section? How did your friend

prepare for the GRE verbal? Men, you just need to help in my preparation.In fact, can you give an epical account of how you prepared for the test

and your impression of the difficulty index? The GRE is a linchpin in my application as it would really help me in determining where to apply. I've

got 34 days left. I need tips man. Those scores you quoted,1520, 1600/1600, they are really sources of inspiration!Kind of dream scores, though

achievable of course! I'm highly motivated. Ask you friend to bare the secrets to such an atronomical score . I really need to score high.

Actually, I'm not studying in Nigeria. I'm currently in my senior year in Russia. Hopefully shall be a Ph.D student come fall'08. I'll keep you

posted as time passes. Shall be applying to top 10 in my field and probably 2-3 safe schools--it'll all come down to the upshot

of the GRE exams. After the GRE, I'll have a ready list then.

Hope to read an account of your GRE experience and those of other GRE overachievers you've come across. They'll be greatly invigorating!

Oops ( scratches his head)!!! Been a while, dude. I need to dust off my linear Algebra Text from the shelf.

My cerebration on eigen values is quite seedy and hazy. Be right back!

you got the answer. Eigenvalues/vectors only exist for square matrices.

Well, let's move on.

A candidate who gets 20% in an exam fails by 30 marks and the other candidate who gets 32% gets 42 marks more than the pass mark. Find

the pass mark in percentage.

a) 20% b) 25% c) 30% d) 40%

Your answer is 25%

i.e where x = total marks in absolute terms and P is pass mark in absolutes

0.2x = P - 30
0.32x = P + 42

so 0.12x = 72

thus x = 600

substituting above, 120 = P - 30

therefore P = 150

in %, P = 150/600 = 25%

Eyohimself, edusegzy, you're very much welcome!


I took the GRE a longtime ago. My firm belief is that any self-respecting math enthusiast should hit at least the 98th percentile range in that exam.

The key is lots of practice, i.e., daily practice. I used the GRE Big Book, back then. Though the exam is a CBT (computer-based test), the Big Book (some paper-based past questions) was quite helpful.

The verbal is another story. Learning some Latin, Greek, German, French, Italian, Spanish and Japanese will definitely help. The English language has borrowed significantly from these languages and many English words derive their origins from them. The idea of cramming words is only a temporary fix. Reading a wide array of subject matter will definitely help.

Let's keep the questions and answers flowing!

@ RichyBlack,

Was that the only book you used for preparation?

I've got the smaller version of the BIG BOOK with 9 test.

I learnt the bigger version one has 27 .

The math there is too simple. That can't be actual GRE

questions. The real GRE is more tasking there wrt quant.

The verbal is at par with the real GRE. The BigBook boots confidence a lot.

What was your quant score like?

Eyohimself:



A bag contains 5 one dollar notes, 4 fifty cent coins, and 3 twenty cent coins. A boy is asked to take 2 coins at random at random from the bag as his


birthday gift, what is the probability of the boy drawing the maximum possible amount?

a) 5/33 b) 11/66 c) 1/6 d) 5/12

It seems there is a problem with this question. You see the problem?

Eyohimself:


@ RichyBlack,

Was that the only book you used for preparation?

I've got the smaller version of the BIG BOOK with 9 test.

I learnt the bigger version one has 27 .

The math there is too simple. That can't be actual GRE

questions. The real GRE is more tasking there wrt quant.

The verbal is at par with the real GRE. The BigBook boots confidence a lot.

What was your quant score like?

@Eyohimself
So, it seems you don't have the "bigger version", i.e. the Big Book. The (smaller) version that you have is is just for light practice. Get the big book, do the tests in exam-like conditions as best as possible, review the ones you missed and learn from them, then see the results after 4 weeks. Do a test at least every other day. The ETS (diskettes) and Kaplan CDs are better for simulating exam conditions - timing, CBT environment, scoring, etc. Ultimately, your preparation for the exam goes beyond those practice tests. Your long-term preparation (at least from the early days in primary school) will also matter. My quant score was (800/800, 99th percentile). Remember also that gaining admission to US or Canadian schools (for those schools that use the GRE) is a process that considers more than just your GRE scores. Other things that will help are:
1. High undergrad GPA (1st class or 2:1 will make things easier)
2. Strong letters of reference
3. Excellent statement of purpose and or personal statement
4. Prior research in field of interest

Are you preparing to take the GRE?

@ RichyBlack,

Yeah, I didn't notice it. There was a mistake and it has been corrected.

A bag contains 5 one dollar coins, 4 fifty cent coins, and 3 twenty cent coins. A boy

is asked to take 2 coins at random from the bag as his birthday gift, what is the probability of the boy drawing the maximum possible amount?

a) 5/33 b) 11/66 c) 1/6 d) 5/12

@RichyBlack,

Kudos man for that great math score. The scaled percentile of 99 shows that you were

high up there in the test-tester's pool. I can infer that you've got a knack for math.

Yeah, I'm about to write the GRE in a month's time, precisely in 30 days time. There's an area in math that

I never understood when I was in high school and I'm still finding it difficult getting even now.

Thats the area of combination and permutation. Questions from this area flusters me utterly. The reasoning

behind such questions always go beyond my cerebration unnoticed. Maybe you can help here. Doffing my cap for you

in acknowledgment of your stock math predilections, can I count on you in this regard to put me through the underpinnings

of this concept of recondite attribute as I seem to be continously wallowing in this combinatorial and

permutational impasse? Other areas are quite surmountable. They are more like taking candy from a baby!

Most questions I'll be posting will be from this area and I hope I'll get useful explanations from the math prodigies on nairaland.

Meanwhile, I've corrected the question I earlier posted. Hope to get solutions. Others with questions should bring them here so we

tender solutions and compare approaches. I feel this is going to be helpful as it strengthens and expands one's problem solving abilities. Let's keep

posting questions and proferring solutions.

edusegzy,

Where you at ? Your estimable inputs are highly solicited.

Straighthen out of your state of doldrums and come assist in

keeping us going sprightly. Hope to read your post soonest.

eyohimself,
Thou art highly motivated.

.
i was working on my accomodation hence i was not chanced to be on the net.But i'm back to rub minds with the great eyohimself,richybalcky et al

Eyohimself:

@ RichyBlack,

Yeah, I didn't notice it. There was a mistake and it has been corrected.

A bag contains 5 one dollar coins, 4 fifty cent coins, and 3 twenty cent coins. A boy

is asked to take 2 coins at random from the bag as his birthday gift, what is the probability of the boy drawing the maximum possible amount?

a) 5/33 b) 11/66 c) 1/6 d) 5/12

the correct answer should be A(5/33)


EXPLANATION

if the boy is to draw the maximum possible amount , then the two coins must be one dollar coins each.
to draw the first coin the probability would be 5/12
to drawthe second the probability would be 4/11

hence the probability of drawing the two would be
5/12*4/11=5/33

I WOULD COME UP WITH MORE EXPLANATIONS

@ edusegzy, your answer is correct.

thanks davidylan

Just curious, is _[b]davidylan [/b]= PTH = davidylan?

Eyohimself:

@ RichyBlack,

Yeah, I didn't notice it. There was a mistake and it has been corrected.

A bag contains 5 one dollar coins, 4 fifty cent coins, and 3 twenty cent coins. A boy

is asked to take 2 coins at random from the bag as his birthday gift, what is the probability of the boy drawing the maximum possible amount?

a) 5/33 b) 11/66 c) 1/6 d) 5/12

@Eyohimself
Without that correction, there would have been no correct answer from the options.

As edusegzy showed, and _davidylan seconded, 5/33 is the correct answer.

Eyohimself:

@RichyBlack,

Kudos man for that great math score. The scaled percentile of 99 shows that you were

high up there in the test-tester's pool. I can infer that you've got a knack for math.

Yeah, I'm about to write the GRE in a month's time, precisely in 30 days time. There's an area in math that

I never understood when I was in high school and I'm still finding it difficult getting even now.

Thats the area of combination and permutation. Questions from this area flusters me utterly. The reasoning

behind such questions always go beyond my cerebration unnoticed. Maybe you can help here. Doffing my cap for you

in acknowledgment of your stock math predilections, can I count on you in this regard to put me through the underpinnings

of this concept of recondite attribute as I seem to be continously wallowing in this combinatorial and

permutational impasse? Other areas are quite surmountable. They are more like taking candy from a baby!

Most questions I'll be posting will be from this area and I hope I'll get useful explanations from the math prodigies on nairaland.

Meanwhile, I've corrected the question I earlier posted. Hope to get solutions. Others with questions should bring them here so we

tender solutions and compare approaches. I feel this is going to be helpful as it strengthens and expands one's problem solving abilities. Let's keep

posting questions and proferring solutions.

As regards combination and permutation, see them as just sophisticated counting processes. While combination emphasizes counting the number of subsets (having clearly defined properties) in a given set, permutation generally deals with counting the number of arrangements/orderings of a given set of elements (with clearly defined restrictions).

Bring the questions you have, and I'll be hoping that there is enough ammunition on this thread to subdue them .

Meanwhile, GOODLUCK on the GRE!

edusegzy,

Welcome back!!! Your solution was right. Well check out this question:

An identification code read from left to right consists of 2 digits, a dash, 3 digits, a dash, and then 4 digits. Each digit can be any number from 0 through 9.

A) The number of different identification codes possible.

B) 10^9

RichyBlacK:

Just curious, is _[b]davidylan [/b]= PTH = davidylan?

Thinkest thou not that he that thou searchest for, indeed i am he!

_davidylan:

Thinkest thou not that he that thou searchest for, indeed i am he!

Shakespeare-speak! Aight.

there are 9 spaces to be filled and each space can be filled in 10 different ways hence the number of different codes to be generated would be 10^9

i.e 10 X 10 X10X10X10X10X10 X10X10=10^9

U guys are doing a good job proferring solutions.

edusegzy, kudos man!

The number of 5 digit telephone numbers having atleast one of their digits repeated is
A) 90000
B) 100000
C) 30240
D) 69760
E) 69000



How would you solve this?

Eyohimself:

The number of 5 digit telephone numbers having atleast one of their digits repeated is
A) 90000
B) 100000
C) 30240
D) 69760
E) 69000



How would you solve this?

Let G = set of all 5-digit telephone numbers
Let P = set of all 5-digit telephone numbers with no digits repeated
Let Q = set of all 5-digit telephone numbers with at least one digit repeated

It is easy to see that:
1. G = P u Q
2. P n Q = null set (or P and Q are mutually exclusive)

|G| = 10^5 = 100,000
|P| = 10*9*8*7*6 = 30,240
|Q| = |G| - |P| = 100,000 - 30,240 = 69,760 (D)

5 Black socks and 5 Green socks are in a dark room.What is the minimum number of socks you must take out in order to get a complete pair?
a.) 2 b.) 3 c.) 4 d.) 5

hollandis:

5 Black socks and 5 Green socks are in a dark room.What is the minimum number of socks you must take out in order to get a complete pair?
a.) 2 b.) 3 c.) 4 d.) 5

if you take 2 it is possible for you to have one green and one black .That way you dont have a complete pair.

if you take out three there is no way you wouldn't have a complete pair.
it is either a black and two green or a green and two black, or three green, or three black and that way you have a complete pair in each of the situations.

so the minimum number of sucks you must take out is 3(answer :b)

yeah, you are right edusegzy. The answer is surely three.