Math Problems!........share It Here!

[twisted minds]

eto para sa inyo

 

sagutin nyo ha

pwedeng magdiscuss

isa ito sa mga problems nabinibigay ko sa mga students ko

 

si A bumaba sa escalator tumakbo nabilang nya 75 steps ang hinakbang nya

si B bumaba sa escalator naglakad nabilang nya 50 steps ang hinakbang nya

ang pagtakbo ni A is twice as fast sa paglakad ni B

 

ilan lahat ang steps ng escalator

 

hehe ... tricky ... A must be gong down the escalator going up ... and B must going down the escalator going down ... but still there are steps hidden inside so i really would not know how to calculate this

[dfgvan]

hehe ... tricky ... A must be gong down the escalator going up ... and B must going down the escalator going down ... but still there are steps hidden inside so i really would not know how to calculate this

 

 

actually it isn't

1. assume X to be the rate at which a step disappears. this will be constant as this will be the rate of the escalator

2. they both went down using the same escalator which was going down.

3. A = rate of A. B = rate of B.

 

V=D/T

both traveled the same distance

D is constant

 

 

answer 100 steps

[The_Blade]

Great answers up there. ^^^

 

OK more of Math problems.

 

Sabong problem (again) :upside:

One day Quin went to 5 cockpit arenas. Each arena has an entrance fee of 5pesos and a winning bet gets 5 times the money bet. Everytime Quin entered an arena he plays only once, bets all his money, and wins. Also he gives a 5peso tip to the guard of the arena he just played. At the end of the day, he has 555 pesos.

 

How much money did he originally have?

Answer needs to be accurate in decimal places.

 

 

orignal nomey = 7.6752

 

 

 

 

formula will be [(x-5) *5]-5 ---> reads as x (current money) minus 5 (entrace fee) then multiply the result to 5 (price you win) then subtract 5 from the result (tip to the guard).

 

so:

 

7.6752 - 5 = 2.6752; * 5 = 13.376; - 5 = 8.376 (first arena)

8,376 - 5 = 3.376; * 5 = 16.88; -5 = 11.88 (2nd arena)

11.88 - 5 = 6.88; * 5 = 34.4; -5 = 29.4 (3rd arena)

29.4 - 5 = 24.4; * 5 = 122; -5 = 117 (4th arena)

117 - 5 = 112 * 5 = 560; - 5 = 555 (5th arena)

 

this is actually done using manual steps (i worked backwards from 555 + 5 = 560; /5 = 112; + 5 = 117, and so on...), i'm still working for a formula for this...

 

 

 

let X0 be the original money.

let X5 the money on hand after the 5th arena =555

 

working the eqaution:

 

the money before the 5th arena is X4

 

(x4 - 5) * 5 - 5 =x5

5x4- 25-5 = x5=555

x4=x5/5 +30/5

x4= 555/5 + 30/5 = 117

 

x3 = x4/5 +30/5

 

working this out and expanding

 

we have

 

 

x0 = (x5+30)/5^5 + (30/ 5^4) +(30/5^3) +(30/5^2) + (30/5^1)

 

in general from the equation would be: where n=5

 

Xn-n = (xn+30)/5^n + (30/ 5^n-1) +(30/5^n-2) +(30/5^n-3) + (30/5^n-4)

 

 

proof:

x0 = (x5+30)/5^5 + (30/ 5^4) +(30/5^3) +(30/5^2) + (30/5^1)

 

X0 = 0.1872 +0.048+ 0.24+1.2+ 6= 7.6752

[destron]

Q1:

Two guys played checkers. They played 5 times. How is it possible that each of them won three games?

 

Q2:

How many times can you subtract 5 from 25?

[skitz]

Two guys played checkers. They played 5 times. How is it possible that each of them won three games?

 

Not specified wether they played against each other. It is possible that each guy won 5 games.

 

How many times can you subtract 5 from 25?

 

Infinite times. Question not limited to non-negative numbers. Another possible answer is ONE. After you subtract 5 from 25 one time, you can not subtract 5 from 25 anymore, you would be subtracting 5 from 20, then 5 from 15, etc.

Edited March 10, 2008 by skitz

[destron]

Two guys played checkers. They played 5 times. How is it possible that each of them won three games?

 

Not specified wether they played against each other. It is possible that each guy won 5 games.

 

How many times can you subtract 5 from 25?

 

Infinite times. Question not limited to non-negative numbers. Another possible answer is ONE. After you subtract 5 from 25 one time, you can not subtract 5 from 25 anymore, you would be subtracting 5 from 20, then 5 from 15, etc.

 

:thumbsupsmiley:

 

Next:

Q: Twelve members were present at a board meeting. Each member shook hands with all of the other members before and after the meeting. How many handshakes were given?

[taz31]

Ann purchased a house for $ 125176. She wants to sell it for a profit of exactly 13%. 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 5.7% (in addition to Ann's 13% profit). How much must the Agent sell the house for?

[bene_factor]

Let P be purchase price, c be commission rate, and t be target profit. Selling price S = P / (1 - c - t)

 

Reason?

 

We want S, after deducting the commission S x c, and the purchase price P, to be equal to the target profit S x t.

 

So we have S - Sc - P = St, and algebra follows.

[skitz]

:thumbsupsmiley:

 

Next:

Q: Twelve members were present at a board meeting. Each member shook hands with all of the other members before and after the meeting. How many handshakes were given?

Each member would shake hands 11 times to shake everyone's hand. So it 12 X 11 = 132 times before and 132 times after the meeting, for a grand total of 264.

 

 

^^ Ooops... I think I analyzed the problem wrong. It should be 11+10+9...+1 (so as not to count the same handshakes twice). So its (11+10+9+8...+2+1)x2 = 132

Edited March 15, 2008 by skitz

[The_Blade]

^^right, but how would you put it in a simple formula? say there are 1000 delegates.

Edited March 18, 2008 by The_Blade

[skitz]

^^ Darn! Took algebra too long ago... hehehehe... simple enough yung solution ko (for me) since I write code, it's all a matter of making it a loop (if 1000 delegates). But hey, I'd be interested with the "simplified" answer.

[The_Blade]

^^^ hehehe. ako rin tagal na. im trying to woRk it out... may naderived akong formula pero di ko ma prove ang sense kung tama ba.?galing sa factorial thingy...

 

like for example: Let N be the number of delagates so if N=12 your answer is 132. ang formula ko ay:

 

number of handshakes= [N!/ (N-2)! * 2! ] * 2 = N!/(N-2!) = 12!/ 10! = 12*11* 10!/10! = 12*11= 132

 

SO IF N=1000.

 

hANDSHAKES= 1000!/ 998! = 1000*999* 998!/998! = 1000*999= 999,000. is this right?

 

by the way going with your formula, where N = 12

 

then handshakes = (11+10+9 +........ +1)*2 = 66*2= 132 the terms inside the () can be expressed as

n+n-1+n-2+.......+n-(n-1). where n is the first term and n-(n-1) is the last term which is actually 1.

 

the sum of which can be expressed as

 

(first term +last term) * [(first term- last term +1)/2] = (n+1) * (n -1 +1 )/2 =(n+1)*(n/2)

 

check:

 

sum=(11+1)*(11/2) = 12* 5.5 = 66

 

therefore handshakes = 66*2 = 132

 

 

 

lets see if it applies if N=1000

 

n= 999

 

sum= (n+1)* (n/2) = (999+1) (999/2)= 1000*499.5 =499500 (first handshakes)

 

total handshakes = 499500 *2 = 999,000. seems right.

 

 

is there another formula kaya?

Edited March 18, 2008 by The_Blade

[destron]

Each member would shake hands 11 times to shake everyone's hand. So it 12 X 11 = 132 times before and 132 times after the meeting, for a grand total of 264.

 

 

^^ Ooops... I think I analyzed the problem wrong. It should be 11+10+9...+1 (so as not to count the same handshakes twice). So its (11+10+9+8...+2+1)x2 = 132

 

132 is correct.

 

^^right, but how would you put it in a simple formula? say there are 1000 delegates.

 

The sum formula for arithmetic progression can be used.

 

Sum = (n/2)(a1+an)

 

where

n = number of terms

a1 = first term

an = last term

 

Applying this to our problem, our number of terms(n) should always be the number of members(m) minus 1. And that the handshakes where done before and after, so our answer needs to be multiplied by 2. Multiplying the first equation by 2 and substituting n by (m-1) gives:

 

Handshakes = (m-1)(a1+an)

 

where

m = number of members

 

Handshakes = (12-1)(11+1) = 132

 

Still the formula can be simplified since we know that a1 always equal to (m-1), and (an) is always 1

 

Handshakes = (m-1)(m-1+1)

 

Handshakes = m(m-1)

 

(so if the members are 1000, total handshakes computes also to 999,000)

[destron]

More..

 

1. Simplify this expression: (99-9)(99-19)(99-29) .... (99-199)

2. If 5 is added to the numerator of a certain fraction and 2 is added to the denominator, the value of the fraction is tripled. What is the original fraction?

[The_Blade]

1. tricky Q.

(99-9)(99-19)(99-29) .... (99-199)= (99-9)(99-19)(99-29) .... (99-99) ..... (99-199)= 0

 

 

2. 1/2

 

1+5

2+2

 

=6/4

 

= 3/2