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The number of 4 digit even numbers that can be formed using 0,1,2,3,4,5,6 without repetition is
A) 120
B) 300
C) 420
D) 20
E) 42


RichyBlack, edusegzy, hollandis and others, pls explain how

to solve this. I'm utterly confused. Would I ever understand permutation

and combination?

Eyohimself:

The number of 4 digit even numbers that can be formed using 0,1,2,3,4,5,6 without repetition is
A) 120
B) 300
C) 420
D) 20
E) 42


RichyBlack, edusegzy, hollandis and others, please explain how

to solve this. I'm utterly confused. Would I ever understand permutation

and combination?

@Eyohimself
Let D = {0,1,2,3,4,5,6}, i.e. the set of digits given.
Let PQRS represent a 4-digit number formed using the digits from D such that no digit is repeated.
Let E be the set of even numbers represented by PQRS
Let O be the set of odd numbers represented by PQRS
Let A be the set of all numbers represented by PQRS

It doesn't take too much contemplation to see that:
1. A = E u O
2. E n O = null set (simply showing the mutual exclusivity of the set of odd and even numbers)

We want to find |E|, i.e. the cardinality of E

This, fortunately, can be done (easily) by direct construction, i.e. by just counting them.

Elements in E will have 4 digits (PQRS) and S will be either 0, 2, 4, or 6.
Let us fix S = 0 (we'll do this later for 2, 4, and 6)
So, there are:
6 choices for P
5 choices for Q
4 choices for R
Therefore, there are 6*5*4 = 120 numbers in E with S=0
Extending this to S = 2, S = 4, and S = 6, we see that there are 120+120+120+120 = 480 numbers in E.
Answer = 480 (correct answer not among options - maybe a typo).

Check:
=====
We can continue the construction for O.
The number of elements in O = 120 + 120 + 120 = 360.
For A, counting the number of elements is as follows:
P has 7 possibilities
Q has 6 possibilities
R has 5 possibilities
S has 4 possibilities

So, total possibilities, or number of elements in A = 7*6*5*4 = 840.

But, |E| + |O| = 480 + 360 = 840 = |A|.

Check complete
============

If you need further explanations, please let me know.

@RichtBlack,

Why can't I think this was when I see problems like

this? All other math topics are like second nature to me

except this one. Your logic is great RichyBlack. I now

understand how to solve problems like this. But there's an

issue here which I think will need some clarifications from

you.

Now look at this:

RichyBlacK:


Elements in E will have 4 digits (PQRS) and S will be either 0, 2, 4, or 6.
Let us fix S = 0 (we'll do this later for 2, 4, and 6)
So, there are:
6 choices for P
5 choices for Q
4 choices for R
Therefore, there are 6*5*4 = 120 numbers in E with S=0
Extending this to S = 2, S = 4, and S = 6, we see that there are 120+120+120+120 = 480 numbers in E.
Answer = 480 (correct answer not among options - maybe a typo).

Well, for S=0, I understood how you assigned 6 choices to P,

5 choices to Q,

and 4 choices to R.

Now, for S= 2 , 4 , 6 :

P I think can't have 6 choices. Why?

0 cannot start a number.

With S=2,

There will be 5 choices for P ( i.e, 1, 3, 4, 5, 6) excluding 0 because 0

cannot be the firs[/b]t digit of a 4-digit number.

Q will also have 5 choices ( i.e, 0, 3, 4, 5, 6) assuming the next digit assigned to Q was [b]1.

R will have 4 choices then and S which is already = 2 will have just 1 [/b]choice.

=> There'll be 5*5*4= [b]100 numbers in E with S=2.

Same applies for S=4 and S=6.

Extending the results,

We find that there will be 120 + 100 + 100 +100 = 420 numbers in E.

Answer = 420-- option C.

Thanks man, RichyBlack for showing me the method to use.

The problem I've got with questions like this is where to start from.

I hope you'll continue to throw more light on questions I'll be posting.

If I can get to understand this topic, I'll attribute it to this wonderful

forum, edusegzy and you in particular for your insightful and vivid

solutions. You kinda a math prodigy I think.

Eyohimself:


Now look at this:



Well, for S=0, I understood how you assigned 6 choices to P,

5 choices to Q,

and 4 choices to R.

Now, for S= 2 , 4 , 6 :

P I think can't have 6 choices. Why?

0 cannot start a number

Yes, you're right!

In the representation of natural numbers, zero is typically not used as the first digit. This added restriction reduces the sizes of each set discussed.

|E| = 120+100+100+100 = 420
|O| = 100+100+100 = 300
|A| = 6*6*5*4 = 720 = 420+300 = |E| + |O|.

Thanks for pointing that out

Eyohimself:


Extending the results,

We find that there will be 120 + 100 + 100 +100 = 420 numbers in E.

Answer = 420-- option C.

Thanks man, RichyBlack for showing me the method to use.

The problem I've got with questions like this is where to start from.

I hope you'll continue to throw more light on questions I'll be posting.

If I can get to understand this topic, I'll attribute it to this wonderful

forum, edusegzy and you in particular for your insightful and vivid

solutions. You kind of a math prodigy I think.

@Eyohimself

Try to identify the possible sets (or events) in the problem and their relationships, e.g. mutual exclusivity, independence, etc.

Further identify the particular set the problem is interested in.

Then use permutation, combination, etc. to count the elements in the set, if it's a counting problem (like the one just discussed).

As you solve more of these problems, you'll definitely get better at it.

richy blacky:

great is thy insight brother.

Shells at the Shore
question

image with 3 buckets of 10 shells each and 4 loose shells image with 2 buckets of 10 shells each and 3 loose shells

image with 3 buckets of 10 shells each and 0 loose shells image with 2 buckets of 10 shells each and 1 loose shell

Jan, Kate, Mike, and Sam have lots of shells and many pails. They put groups of 10 shells in each pail.

* Kate has 2 pails and 3 more shells left over.

* Sam has 1 more pail than Kate, but no shells left over.

* Jan has 4 more shells than Sam.

* Kate has 2 more shells than Mike.

How many shells does each have?

Shells at the Shore
question


Jan, Kate, Mike, and Sam have lots of shells and many pails. They put groups of 10 shells in each pail.

* Kate has 2 pails and 3 more shells left over.

* Sam has 1 more pail than Kate, but no shells left over.

* Jan has 4 more shells than Sam.

* Kate has 2 more shells than Mike.

How many shells does each have?

The wording of the question is kind of vaguely expressed.

But I'll do it the way I understood.

If Kate has 2 pails and 3 more shells leftover,

=> in the 2 pails there'll be 20 shells.

In all, Kate has 20 + 3 = 23 shells then.

Sam has 1 more pail than Kate,

=> Sam has 3 pails with a total of 30 shells.

Jan has 30 + 4 = 34 shells.

Kate has 2 more shells than Mike.

Kate had 23 shells, => Mike will have 23 - 2 = 21 shells.

In otherwords,

Kate has 23 shells, Sam has 30 shells, Jan has 34 shells and Mike has 21 shells.

I'm confused with the wording of the first proviso:

Kate had (2 pails and 3 more shells) left over or

he had 2 pails and (3 more shells) left over?

The rationale to my solution is based on the latter.

[s]Can we solve this?

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. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.[/s]

Can we solve this?

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. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.

Eyohimself:

Can we solve this?

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. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.

Let's call the last 4 digits ABCD.
A has 3 possibilities 7,8,9
B has 3 possibilities 3,6,9
C and D form a pair (C,D) with 3 choices (3,1), (6,2), (9,3)

Total possibilities 3*3*3 = 27 (answer = A)

nice one richy blacky

i worked it earlier and was getting 81 when i saw your solution i saw where i was getting it wrong

THanks man

Nice one there RichyBlack.

I'm interestd in how fast this problem could be solved and to see if I can find a faster approach.




The jewels in a crown consist of diamonds, rubies and emeralds. If the ratio of diamonds to rubies is 5/6

and the ratio of rubies to emeralds is 8/3, what is the least number of jewels that could be in the tiara?

a) 16

b) 22

c) 40

d) 53

e) 67

Eyohimself:

I'm interestd in how fast this problem could be solved and to see if I can find a faster approach.




The jewels in a crown consist of diamonds, rubies and emeralds. If the ratio of diamonds to rubies is 5/6

and the ratio of rubies to emeralds is 8/3, what is the least number of jewels that could be in the tiara?

a) 16

b) 22

c) 40

d) 53

e) 67

Let
R be the number of rubies
D be the number of diamonds
E be the number of emeralds

So, D/R = 5/6 and R/E = 8/3 (given)
Choose any of these as a unit. Let's choose E.

So,
R = (8/3)E
D = (5/6)R = (20/9)E

Let R+D+E = X
X must be a whole number.

Therefore,
R+D+E = (8/3)E + (20/9)E + E = (53/9)E = X

So, the question reduces to: What is X if E is the least value that makes X a whole number?
E = 9, and X = 53 (Answer = D)

edusegzy:

nice one richy blacky

i worked it earlier and was getting 81 when i saw your solution i saw where i was getting it wrong

THanks man

Yep, decoupling the pair is a common mistake. You're welcome.

Eyohimself:

Nice one there RichyBlack.

You're welcome man.

eyohimself:
Ann purchased a house for $ 170270. She wants to sell it for a profit of exactly 11%. However, she does not want to sell it herself. She wants a Real Estate Agent to sell it for her. The Agent must make a commission of exactly 5.5% (in addition to Ann's 11% profit). How much must the Agent sell the house for? Give your answer to the nearest dollar.
RIchyblacky:
are you in nigeria?

edusegzy:


RIchyblacky:
are you in nigeria?

Yep, I reside in Apapa, however, I'm currently visiting "the land of the free and home of the brave".

Determine the point of intersection (if there is one) of the lines :
(x-1)/3=(y-3)/2=z+2 and x=(7-y)/3=(z+7)/2 try this mathematics scholar

Pocket Change


Your friend has a dollar bill and needs change for the vending machine. In your pocket, you have 7 pennies, 5 nickels, 8 dimes, and 3 quarters.

How many different ways can you make change for a dollar?

Solar System Scales


Kate and Shireen have completed their model of the solar system. The model looks great hanging in the classroom. But before they can get a grade for their project, their teacher has one more assignment for them.

“We know,” she says, “that you had to use one scale for planet sizes and a much smaller scale for showing distances of the planets from the Sun. If you had used the larger scale for the distances, your model would not have fit in the classroom, because the distances are so immense. Can you show the class just why the model wouldn’t have fit? If you had used the same scale for distance that you used for planet sizes, how far from your model Sun would the dwarf planet Pluto have to be placed?”

Well, for the planet sizes the girls had used a pea measuring 0.5 cm in diameter for their model of Mercury. The real Mercury has a diameter of 4800 km. The real Pluto is nearly 5900 × 106 kilometers from the Sun.

How far, in miles, from their model Sun would Kate and Shireen have had to place a model of Pluto?


i ve sent some series of questions and i am expecting the answer

edusegzy:

eyohimself:
Ann purchased a house for $ 170270. She wants to sell it for a profit of exactly 11%. However, she does not want to sell it herself. She wants a Real Estate Agent to sell it for her. The Agent must make a commission of exactly 5.5% (in addition to Ann's 11% profit). How much must the Agent sell the house for? Give your answer to the nearest dollar.

He must sell it for :


0.11*170270 + 0.055*170270 + 170270 = $198 364.55 ( $198 365--to the nearest $)

( Assuming they are making the gains on the cost price).

ajisafejt:

Determine the point of intersection (if there is one) of the lines :
(x-1)/3=(y-3)/2=z+2 and x=(7-y)/3=(z+7)/2 try this mathematics scholar

The solution stems down to solving these multiple equations simultaneously.

Now,(x-1)/3=(y-3)/2=z+2 => x-1= 3z + 6 => x=3z + 7, (A)

x=(7-y)/3=(z+7)/2 => 2x = z + 7 , (B)

Solving A and B simultaneously we have x= 14/5 and z = -7/5 .

Now letz solve for y.

(x-1)/3=(y-3)/2=z+2 => 2x- 2 = 3y - 9 => 3y = 2x + 7

Therefore, 3y = 2* 14/5 + 7 = 63/5 . From here, y = 63/15.

The lines intersect at the point ( 14/5 , 21/5 , 7/5 ) .

ajisafejt:

Pocket Change


Your friend has a dollar bill and needs change for the vending machine. In your pocket, you have 7 pennies, 5 nickels, 8 dimes, and 3 quarters.

How many different ways can you make change for a dollar?

Let d = dimes, n = nickels, q = quarters, p = pennies.

quarter = 25 cent coin
dime = a 10 cent coin
nickel = 5 cent coin
penny = 1 cent coin

Lets start with d. Using d, the number of combination pairs possible are:

(d,n) (d,q,n) (d,q,p) (d,n,p)

Without d, the number of possible combination pairs are:

(n,p,q) (n,q)

This implies the total number of ways would be 11.

RichyBlacK:

Let
R be the number of rubies
D be the number of diamonds
E be the number of emeralds

So, D/R = 5/6 and R/E = 8/3 (given)
Choose any of these as a unit. Let's choose E.

So,
R = (8/3)E
D = (5/6)R = (20/9)E

Let R+D+E = X
X must be a whole number.

Therefore,
R+D+E = (8/3)E + (20/9)E + E = (53/9)E = X

So, the question reduces to: What is X if E is the least value that makes X a whole number?
E = 9, and X = 53 (Answer = D)

I like your solution- it is well reasoned out. Well this was how I solved mine when I simulated a test condition.

Frankly speaking, if I had the slightest idea of your method, thats what I would have used. I just gave my first

PowerPrep Practice test and this was one of my questions. On this question, I spent about 2 mins. I wanted to see

other approaches.This was how I solved it:

Given:

D/R = 5/6 and R/E = 8/3 . The question reduces to finding D:R:E in the lowest form.

D/E= D/R * R/E = 5/6 * 8/3 = 20/9

=> D : E = 20 : 9

To find the common R for both ratio, we take the HCF of the two R's i.e 6 and 8.

HCF (6 and 8 ) = 24.

=> D:R:E= 20:24:9

The least number becomes 20 + 24 + 9 = 53 ( the ratio as you can see is in its lowest term)

my ogas in d house can u pls solve dis maths question for me


if 18,000 was earned from investing in 2 accounts A & B at15% and 20% respectively at 3:2 ratio what is d amount invested in B.

THE ANSWER IS NEEDED URGENTLY.Tanks

gracee:



if 18,000 was earned from investing in 2 accounts A & B at15% and 20% respectively at 3:2 ratio what is d amount invested in B.

THE ANSWER IS NEEDED URGENTLY.Tanks

Given:

A:B=3:2 => A/B = 3/2 => A= (3/2)B.

Lets assume amount invested in account A and account B are A and B respectively.

For account A,

at 15% ineterest rate, I'll see in my account when it matures an earning of:

0.15*A

For B at 20%,you'll see:

0.2*B

Now these two earnings total 18 000.

=> 0.2B + 0.15A = 18 000

=> 0.2B + 0.15*1.5B = 18 000

Multiply through by 1000:

200B + 225B = 18 000 000

=> B= (18 000 000)/425 = 42 352. 94

=> The amount invested in B was approximately 42 352

i must really say that u guys are doing a fantastic job oput there well i have just one question
PLs can some one Differentiate X^x for me
thanks

lilrukevwe:

i must really say that u guys are doing a fantastic job oput there well i have just one question
PLs can some one Differentiate X^x for me
thanks

I'm assuming that you meant x^x (or X^X). If not, then you have to be clear with respect to what, because X is not the same as x.

Let y = x^x
Let log() represent the natural log function
Let exp() represent the exponential function

So, log(y) = x*log(x)

Let log(y) = P
So, y = exp(P) and P = x*log(x)

We know that dy/dx = dy/dp * dp/dx
We also know that dy/dp = exp(P)
And that dp/dx = x*(1/x) + log(x) = 1 + log(x)

Therefore, dy/dx = exp(P)*(1 + log(x)) = exp(x*log(x))*(1 + log(x)) = x^x*(1 + log(x))