# Math Help - 3rd Grade Math Help

1. ## 3rd Grade Math Help

Max went to the fair with 5 friends.some paid $2 each to ride the bumper cars.the rest paid$3 each to ride the roller coaster.they spent $16 in all.how many rode each ride ? if angel plants the same number of pepper plants in each of 5 rows, he will have none left over.if he plants 1 in the first row and then 1 more in each row than he did in the previous row, he will plant peppers in only 4 rows.how many plants does he have ? paul has 30 postcards.he has 4 times as many large postcards as small postcards.how many small postcards does he have ? how many large postcards ? can anyone help ? i've helped my daughter with about 20 of these questions, but now my brain has died 2. Max went to the fair with 5 friends.some paid$2 each to ride the bumper cars.the rest paid $3 each to ride the roller coaster.they spent$16 in all.how many rode each ride ?

I have 2 questions, one of them should reveal a problem in the problem:

1. Are you sure there are 5 friends?

2. Are you sure they spent $16 dollars? Because, as the problem is, I'm getting -1 friends rode the bumper cars and 6 people rode the roller coaster, so obviously something is wrong, here's my work: $2a + 3b = 16$ a = Friends that rode bumper cars b = Friends that rode the roller coaster $a + b = 5$ $a = 5 - b$ $2(5 - b) + 3b = 16$ $10 - 2b + 3b = 16$ $10 + b = 16$ $b = 6$ $a + 6 = 5$ $a = -1$ ??? 3. 1st: Max + friends = 6 people, suppose all ride roller coaster: 6x$3=18$suppose one rides bumpers and the rest ride roller: 5x$3+1x$2=$17
suppose two ride bumpers and four ride roller: 4x$3+2x$2=$16 jackpot, so the combination of four on roller, two on bumpers works. The problem isn't very well posed because there are other possible answers, e.g. max rides the bumpers 8 times while his friends watch...I think my answer is fairer 2nd: look at it the other way round, the question tells you how many seeds he has: 1+2+3+4 because he can complete this and no more so he has 10 seeds, what number of seeds in each of the 5 rows if they are evenly spaced...? btw, this looks obvious if you draw it on paper, you get a triangle and a rectangle made out of dots as different ways of laying the same number of seeds. 3rd: you have two unknowns S=number of small, L=number of large. you also have two bits of information L+S=30 (total is large and small) and L=4xS (4 times as many large as small). If you substitute L=4xS into other equation you get one equation with one unknown: 4xS+S=30 so (4+1)xS=30 so s=30/5=6 so there are 6 small cards, using L=4xS you get L=4x6=24. 6+24=30 This last problem is important because it is the beginnings of seeing how to relate what is known and unknown and what information is given in equations. This IS maths 4. Originally Posted by Aryth $a + b = 5$ here is your mistake, there are 6 people, not 5. there is Max, plus his 5 friends EDIT: Oh, ath3na got it... and an hour ago too! 5. Originally Posted by Aryth ... $2a + 3b = 16$ a = Friends that rode bumper cars b = Friends that rode the roller coaster $a + b = 5$ ... you had the right idea with your working but made one of those mistakes we all make...5+1=6! Incidentally you get the same results as my fairest result because of a tacit assumption which is that a+b=6 (implying that they all get a go). The actual conditions of the problem don't seem to imply that this is necessary, e.g if between them they have 5 rides on the bumpers and 2 rides on the roller coaster the problem is also satisfied. All possible solutions can be enumerated by rearranging and imposing 'reality conditions'... a = 8 - (3b)/2, any (a,b) pair which satisfies this is allowed so long as: 1) a>0 2) a,b are integers (whole numbers) if a >0 then 3b/2 < 8 so b<16/3 therefore b=<5. also 3b must be a multiple of 2 and 3 is odd so b must be even. so the only solutions are going to be those where b is an even number less than or equal to 5: b=0,2,4 give a=8,5,2 respectively (0/2=0 and so divides cleanly by two so can be counted as even) so the complete set of possible solutions must be: (a,b)=(8,0), (5,2), (2,4) and these rides can be distributed however you like between max and his friends and still be a solution to the problem...if there were only 5 friends and we weren't supposed to consider max then we would get the same solution set but would distribute them among 5 people rather than 6...there is no stipulation that it is fair, but leaving some (a,b) pair wouldn't be a solution to the problem. Judging by the fact that you (Aryth) have the most beautiful equation ever written as your posting signature I bet you got all of the above by looking at the question again or at least a correct answer as soon as you realised it probably meant 6 people... but I thought i'd give a complete and hopefully reasonably readable coverage just in case other people come across this ...just realised that Jhevon found the original fault and replied before me, but i'm gonna post anyways because I wrote a bit more 6. Honestly, I thought back as if I were a 3rd grader and typed all that out. Of course, I hit the one error I would have hit back then... Anyway. I assumed they all bought 1 ticket for either$2 or \$3, though combinations and repeats are possible, I think they are highly unlikely due to the increased complexity per combination or repeat of a ride.

So, simply put:

$2a + 3b = 16$

$a + b = 6$

$2(6 - b) + 3b = 16$

$b = 4$

$a + 4 = 6$

$a = 2$

Due to the simplicity, I believe that this would be, as you had put previously, a third grade solution to an ambiguous problem.