destron Posted March 11, 2008 Share Posted March 11, 2008 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? Quote Link to comment
taz31 Posted March 12, 2008 Share Posted March 12, 2008 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? Quote Link to comment
bene_factor Posted March 12, 2008 Share Posted March 12, 2008 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. Quote Link to comment
skitz Posted March 14, 2008 Share Posted March 14, 2008 (edited) :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 Quote Link to comment
The_Blade Posted March 18, 2008 Share Posted March 18, 2008 (edited) ^^right, but how would you put it in a simple formula? say there are 1000 delegates. Edited March 18, 2008 by The_Blade Quote Link to comment
skitz Posted March 18, 2008 Share Posted March 18, 2008 ^^ 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. Quote Link to comment
The_Blade Posted March 18, 2008 Share Posted March 18, 2008 (edited) ^^^ 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 Quote Link to comment
destron Posted March 27, 2008 Share Posted March 27, 2008 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) wheren = number of termsa1 = first terman = 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) wherem = 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) Quote Link to comment
destron Posted March 28, 2008 Share Posted March 28, 2008 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? Quote Link to comment
The_Blade Posted March 29, 2008 Share Posted March 29, 2008 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+52+2 =6/4 = 3/2 Quote Link to comment
destron Posted March 31, 2008 Share Posted March 31, 2008 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+52+2 =6/4 = 3/2 Correct. Heres another one: A five digit number is represented by ABCDE (where A is the ten thousandth number, B is the thousandth number, C is the hundredths, D is the tenths, and E is the ones.) Find the value of this number to make this expression true, and show how it is derived: ABCDEx 4_______EDCBA Note: A given letter always represent the same digit, no two letters represent the same number. Quote Link to comment
The_Blade Posted April 3, 2008 Share Posted April 3, 2008 ^^^ answer is 21978 21978 X 4 = 87912 i just use trial and error starting from multiples of 4. also used my experience with numbers 123456789. do you know that if you multiply 123456789 by 2, 4 & 8 you always get the same unique numbers in differnt orders? try it. Quote Link to comment
Queen Darkeinjel Posted July 10, 2008 Share Posted July 10, 2008 i taught math in elementary for just a very short amount of time... here are three problems that they were so unable to answer... they said it was too hard... so the questions goes like this: the only rule... you are not allowed to repeat any numbers... 1. the highest 6 digit numeral has the digit in the hundred thousands place two times the digit in the thousands place. the digit in the hundreds place is three times the digit in the ten thousands place. if the digit in the hundreds place is 9, what is the numeral? 2. the lowest 5 digit numeral has the digit 4 in the tens place. if the digit in the thousands place is two times bigger than the digit in the tens place what is the numeral? 3. what is the largest 6 digit numeral with the digit 9 in the tens place and 8 in the ones place? thanks for answering... have fun!!! Quote Link to comment
bene_factor Posted July 10, 2008 Share Posted July 10, 2008 132 is correct. The sum formula for arithmetic progression can be used. Sum = (n/2)(a1+an) wheren = number of termsa1 = first terman = 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) wherem = 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) No arithmetic series or recursion necessary. If there are m members, and you asked each of them after the first round of handshakes how many people they shook hands with, they would each say (m-1). So in all that gives you m(m-1). But then each handshake would end up being double-counted, we have to divide the answer by 2. But then there is another round so we multiply by 2 again. Quote Link to comment
The_Blade Posted August 29, 2008 Share Posted August 29, 2008 i taught math in elementary for just a very short amount of time... here are three problems that they were so unable to answer... they said it was too hard... so the questions goes like this: the only rule... you are not allowed to repeat any numbers... 1. the highest 6 digit numeral has the digit in the hundred thousands place two times the digit in the thousands place. the digit in the hundreds place is three times the digit in the ten thousands place. if the digit in the hundreds place is 9, what is the numeral? 2. the lowest 5 digit numeral has the digit 4 in the tens place. if the digit in the thousands place is two times bigger than the digit in the tens place what is the numeral? 3. what is the largest 6 digit numeral with the digit 9 in the tens place and 8 in the ones place? thanks for answering... have fun!!! 1.) 834976 Let ABCDEF the numeral D=9D=3B=> B=D/3=3A=2C since 9 is already used. next biggest digit is 8 therefore A=8 => C= A/2= 4E=7; F=6 by CS same method applies... 2.) 182433.) 765498 haha bagal ko mag type, mabilis pa download ng maria ozawa... Quote Link to comment
Queen Darkeinjel Posted September 5, 2008 Share Posted September 5, 2008 1.) 834976 Let ABCDEF the numeral D=9D=3B=> B=D/3=3A=2C since 9 is already used. next biggest digit is 8 therefore A=8 => C= A/2= 4E=7; F=6 by CS same method applies... 2.) 182433.) 765498 haha bagal ko mag type, mabilis pa download ng maria ozawa... really good...my students had a hard time answering that one...logic right more than math??? Quote Link to comment
bottomfeeding Posted September 23, 2008 Share Posted September 23, 2008 hello po. i'm a civil engineer by profession I hope I can contribute something to this forum. taas na po kasing mag backread e tanong ko lang po kung anong klaseng math po pinaguusapan natin dito? basic math? advanced? o specialized like strength of materials, thermodynamics, and the like. Quote Link to comment
bottomfeeding Posted September 23, 2008 Share Posted September 23, 2008 ito na lang po muna for starters...simple problems muna... the logarithm of 125 to the base of 5 is x. what is the value of x? Quote Link to comment
bottomfeeding Posted September 23, 2008 Share Posted September 23, 2008 "Small minds discuss persons. Average minds discuss events. Great minds discuss ideas. Really great minds discuss mathematics." Quote Link to comment
The_Blade Posted October 4, 2008 Share Posted October 4, 2008 hello po. i'm a civil engineer by profession I hope I can contribute something to this forum. taas na po kasing mag backread e tanong ko lang po kung anong klaseng math po pinaguusapan natin dito? basic math? advanced? o specialized like strength of materials, thermodynamics, and the like. welcome engr! how do you relate the winspeed and pressure? when pagasa says bagyong lunengneg is travelling 200KPH, how much pressure will it induce to a building of 200m high within 5km radius? how much pressure will it induce to a 20ftX50ft bill board in EDSA within 5 km radius? or how much pressure will it induced to GMA antenna 50ft high? Quote Link to comment
bottomfeeding Posted October 8, 2008 Share Posted October 8, 2008 welcome engr! how do you relate the winspeed and pressure? when pagasa says bagyong lunengneg is travelling 200KPH, how much pressure will it induce to a building of 200m high within 5km radius? how much pressure will it induce to a 20ftX50ft bill board in EDSA within 5 km radius? or how much pressure will it induced to GMA antenna 50ft high? naku mahirap to a. Pressure = ½ x (density of air) x (wind speed)2 x (drag coeff.). 1. given: density = 1.25 kg/m3 windspeed = 200kph or 55.55mps drag coeff. = assuming that the building is shaped like a cube = 1 building height = 200m Pressure = ½ x (density of air) x (wind speed)2 x (drag coeff.) = ½ x 1.25 kg/m3 x (55.55mps)2 x 1 P= 34.72 N/m2 or 34.72 Pa Load generated by the wind on the building = 34.72Pa x 200m = 6944 N/m 2. given: density = 1.25 kg/m3 windspeed = 200kph or 55.55mps drag coeff. = 1 area = 20ftX50ft = 6.1mX15.24m = 92.164m2 Pressure = ½ x (density of air) x (wind speed)2 x (drag coeff.) = ½ x 1.25 kg/m3 x (55.55mps)2 x 1 P= 34.72 N/m2 or 34.72 Pa Load generated by the wind on the billboard = 34.72Pa x 92.164m2 = 3200 N 3. Medyo mahirap to. I don't even know if i can answer this. I haven't seen the GMA antenna. The way i see it kelangan ng aerodynamics ito as the antenna would act as an object passing through a fluid. I'll try to research muna dito. Sa lahat po ng babasa I don't claim to know everything and that what i posted here is absolutely correct. It's just correct as far as I know. So any corrections po ay welcome. More power to us!!! Quote Link to comment
bottomfeeding Posted October 8, 2008 Share Posted October 8, 2008 Edit ko lang po kasi di ko raise to the 2nd power tung velocity. welcome engr! how do you relate the winspeed and pressure? when pagasa says bagyong lunengneg is travelling 200KPH, how much pressure will it induce to a building of 200m high within 5km radius? how much pressure will it induce to a 20ftX50ft bill board in EDSA within 5 km radius? or how much pressure will it induced to GMA antenna 50ft high? naku mahirap to a. Pressure = ½ x (density of air) x (wind speed)2 x (drag coeff.). 1. given: density = 1.25 kg/m3 windspeed = 200kph or 55.55mps drag coeff. = assuming that the building is shaped like a cube = 1 building height = 200m Pressure = ½ x (density of air) x (wind speed)2 x (drag coeff.) = ½ x 1.25 kg/m3 x (55.55mps)2 x 1 P= 1928.62 N/m2 or 1928.62 Pa Load generated by the wind on the building = 1928.62Pa x 200m = 385.73 KN/m 2. given: density = 1.25 kg/m3 windspeed = 200kph or 55.55mps drag coeff. = 1 area = 20ftX50ft = 6.1mX15.24m = 92.164m2 Pressure = ½ x (density of air) x (wind speed)2 x (drag coeff.) = ½ x 1.25 kg/m3 x (55.55mps)2 x 1 P= 1928.62 N/m2 or 1928.62 Pa Load generated by the wind on the billboard = 1928.62Pa x 92.164m2 = 177.75 KN 3. Medyo mahirap to. I don't even know if i can answer this. I haven't seen the GMA antenna. The way i see it kelangan ng aerodynamics ito as the antenna would act as an object passing through a fluid. I'll try to research muna dito. Sa lahat po ng babasa I don't claim to know everything and that what i posted here is absolutely correct. It's just correct as far as I know. So any corrections po ay welcome. More power to us!!! Quote Link to comment
The_Blade Posted October 9, 2008 Share Posted October 9, 2008 @ bottomfeeding... good answers i am impressed. we are looking for wind engrs baka pwede ka? Quote Link to comment
Guest xerxes Posted October 11, 2008 Share Posted October 11, 2008 really good...my students had a hard time answering that one...logic right more than math??? sabi mo elementary math yung tinuturo mo? anong grade? :thumbsdownsmiley: :thumbsdownsmiley: Quote Link to comment
Guest xerxes Posted October 11, 2008 Share Posted October 11, 2008 Sa DE, pinakamahirap na siguro ang PDE. Dyan na papasok ang eigenfunction decomposition, orthogonal functions, bessel at legendre functions etc.In "practical" matters, any version ng electromag. Mapa transmission line theory o simpleng maxwell equations.Masakit din sa ulo ang discrete transforms. solution of partial de for me is very difficult particularly when applied ot real problems like heat tra :thumbsdownsmiley: Quote Link to comment
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