Montarvillois Posted November 24, 2004 Share Posted November 24, 2004 OK, I know there is no good section to port this but I need help in solving this little puzzle for one of my son's homeworks A soccer team is composed of 40% boys and 60% girls, after their last recruitment campaign, the boys recruited 3 more and with these three additions, the number of boys and girls is equal. What would be the answer and what method would I use to solve it ? Apologies to the mods, I really need help from one of MW geniuses. Thanks Link to comment Share on other sites More sharing options...
RichterX Posted November 24, 2004 Share Posted November 24, 2004 there were originaly 15 members in the team, 9 girls and 6 boys Method? send me a PM Link to comment Share on other sites More sharing options...
RichterX Posted November 24, 2004 Share Posted November 24, 2004 (edited) Okay I was working on the method cause I calculated in my head Girls = 60% of team Boys = 40 % of the team 40% of the team + 3 = 60% 3boys = 20% 20% * 5 = 100% = total percentage of the team 3 * 5 = 15 =total members of team Calculating numbers of boys and girls 15/100 = 0.15 0.15 * 40% of boys = 6boys 0.15 * 60% of girls = 9girls Edited November 24, 2004 by RichterX Link to comment Share on other sites More sharing options...
RicePiece Posted November 24, 2004 Share Posted November 24, 2004 wicked ... i can put my bachelor of mathematics degree to some good use... let X = the number of boys let Y = the number of girls we know X / (X+Y) = .4 Solve for X to get X = (2/3)Y by multiplying both sides by the denominator (X+Y). we know X + 3 = Y so plug in the value of X we derived above to get (2/3)Y + 3 = Y and solve to get: Y = 9 plug Y=9 back into X + 3 = Y to get X + 3 = 9 to get X =6 So X = 6, Y = 9. check: X / (X+Y) = 6/15 = .4 and Y / (X+Y) = .6 and X +3 = Y Link to comment Share on other sites More sharing options...
Montarvillois Posted November 24, 2004 Author Share Posted November 24, 2004 AARRrgggh !!!! Where was I in school ???? this thing is killing me. Link to comment Share on other sites More sharing options...
Montarvillois Posted November 24, 2004 Author Share Posted November 24, 2004 My son (11 year old) says he understands, I don't know what kills me most; knowing he got it and I don't or not getting it period Thanks for your help guys ! Link to comment Share on other sites More sharing options...
Lynx7725 Posted November 24, 2004 Share Posted November 24, 2004 Solve by algebra. Let B be the number of boys before recruitment, and G be the number of girls before recruitment. Thus: (1) G = B+3 (2) G = 0.6(B+G) Substituting (1) into (2), B+3 = 0.6(B+B+3) <-- (3) Resolving (3) B+3 = 0.6(B+B+3) 10B+30 = 6(2B+3) 10B+30 = 12B + 18 2B = 12 B=6 <-- (3) Substituting (3) back into (1) G=9 Link to comment Share on other sites More sharing options...
Lynx7725 Posted November 24, 2004 Share Posted November 24, 2004 (edited) Double Post, pls delete, thanks. Edited November 24, 2004 by Lynx7725 Link to comment Share on other sites More sharing options...
JsARCLIGHT Posted November 24, 2004 Share Posted November 24, 2004 AARRrgggh !!!!Where was I in school ???? this thing is killing me. You were sitting in the back next to me drawing in the margins of your math book waiting for gym class so you could move around some more. You are not the only one who forgot all basic math. Link to comment Share on other sites More sharing options...
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