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# Help with math problem for my son please?

Topic: Steps in math problem solving
June 19, 2019 / By Darin
Question: This is a bonus question that he has and it's worth 50 points. I want him to do it on his own, but I need to understand it in order to help him if he gets stuck. I'm TERRIBLE at math! Can someone help me so I can help him please! The problem is as follows: Santa Claus was having difficulties. This year he decided to take as many reindeer as he could find and let a workshop elf ride each one. When he did a count, however, he found he had a problem. There were 4 more elves than there were reindeer, but there were 24 more reindeer legs than elf legs. How many elves and reindeer were there, providing there were no more than 20 of each? That's the problem..and he has to show his work. So..an example of how you would draw this out would be very much appreciated! Like I said..I want him to do it.but I want to be able to help him if I can. Please..serious answers only. I'm not in the mood to wade through the lame answers that people put on just to get points. Can someone please help me? He's in 7th grade.

Azriel | 7 days ago
I think i figured this answer out ill show you step by step: 1.Firstly they have said that there are no more than 20 of each so keep that in mind 2.Then you the fact that there are 4 more elves than reindeer and that there were 24 more reindeer legs than elves 3.Now a reindeer has 4 legs and an elf has 2. 4.Therefore do the following: Number of Elves/Number of Elf Legs/Number of Reindeer/Number of Reindeer Legs/Total Leg Difference Take Differences of 4 between the Number of Elves and The Number if Reindeer When: 18 Elves/(18*2)=36/14 Reindeer/(14*4)=56/(56-36)=20 That Means when there are 18 elves there should be 14 reindeer but the leg difference is 20 so this assumption was wrong. Repeat this with 19 Elves and 15 Reindeer..you get a leg difference of 22 which is also wrong But with 20 Elves and 16 reindeer you get a leg difference of 24 which is true to the question because there are 4 more elves than reindeer..they both are not more than 20 each and their are 24 more reindeer legs than elf legs. Hope this solves your problem!
👍 182 | 👎 7
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We found more questions related to the topic: Steps in math problem solving

Originally Answered: Omg.will u help me with my 8th graders math problem? It has algebra, fractions and is a story problem?
1miles / 8 minutes ( which is the speed of the person) * 1 minute = 1/8 this is how far you run in 1 minute 1 miles / 8 minutes (speed) * 11 minutes = 11/8

Willa
Answer: There are 16 reindeer and 20 elves. Work: r = # of reindeer e = # of elves there were 4 more elves than there were reindeer therefore r + 4 = e there were 24 more reideer legs than elf legs therefore 4r (because 1 reindeer has 4 legs so the number of reindeer times 4 legs each would give us the total number of reindeer legs) = 2e (because each elf has 2 legs) + 24 (because there were 24 more reindeer legs than elf legs) this equation clearer: 4r = 2e + 24 you have to then substitue the first equation into the second equation: e = r + 4 4r = 2e + 24 ---> 4r = 2(r+4) +24 ---> you then solve this equation for r so you have to distribute the 2: 4r = 2r + 8 + 24 then subtract both sides by 2r: 2r = 8 + 24 then add the 8 and 24 together: 2r = 32 then divide by 2: r = 16 therefore you have 16 reindeer. you take this value of r and plug it back into the first equation (e = r + 4) : e = 16 + 4 therefore e = 20 this also fits under the conditions that there are no more than 20 of each! i hope you understand this...let me know if you dont
👍 70 | 👎 4

Sharmain
If you let the number of reindeer be represented by the letter "x" and the number of elves be represented by "y", then you can say that y = x + 4 since there are 4 more elves than reindeer. Also, if we have "y" elves it follows that there will be double this number of elf legs.....that is "2y" By the same logic, "x" reindeer will have "4x" legs. Since there are 24 more reindeer legs than elf legs, 4x = 2y + 24 But since y = x + 4 4x = 2(x + 4) + 24 4x = 2x + 8 + 24 4x = 2x + 32 2x = 32 x = 16 Therfore, the number of reindeer is 16 (since x represents the number of reindeer) and the number of elves is 20, since there are 4 more elves than reindeer. Hope this helps.
👍 64 | 👎 1

Page
I found the answer! there are 16 reindeer because 16*4(there are 16 and they have 4 legs)equals 64. And 4 more elves would be 20. 20*2 because the y have 2 legs is 40. 64-40=24 more reindeer legs. so,16 Reindeer and 20 Elves
👍 58 | 👎 -2

Mamie
okay, I just guessed and checked and I think that the answer is 16 reindeer and 20 elves. I got this by taking a random number for the reindeer,less than 20 ( I chose 16) multiplying it by four (reindeer have 4 legs) which equaled 64. then I subtracted 24 (because there were 24 more reindeer legs than elf legs) that equaled 40 I then divided that by 2 (because elves have two legs) to find the number of elves. Which ended up with 16 and 20 and 16 is four less than 20! Get it? =] p.s. A lot of these other answers are wrong because they are assuming that their are only 24 reindeer legs, when actually it says that "there are 24 MORE reindeer legs than elf legs" Not 24 reindeer legs total.
👍 52 | 👎 -5

Kimberlyn
What grade is he in? That's hard. And I'm in 8th grade and in Algebra 2. Let's see... Wait... I think it's basically guess and check. Let's try it. So there's 16 reindeer, meaning 64 legs. There are 20 elves, meaning 40 legs. Yep, that's a 24 leg difference. Dang... I've been sitting here for a good 1/2 hour trying to figure that out... Oh well, at least I finished it. I think it was written that way to confuse people into thinking more then they needed to. Hope that helped!
👍 46 | 👎 -8

Originally Answered: (math homework) Pre-calculus math problem?
C(x) = 0.43x2 +18.3x +15.6 R(x) =xp =x{-0.02x+129 = -0.02x^2 +129x R(x) - C(x) >0 -0.02x^2 +129 x - 0.43 x^2 -18.3x -15.6 >0 -0.45 x^2 +110.7x -15.6> 0 0.45 x^2 -110.7 x +15.6 < 0 x < 110.7 +/- {sqrt (110.7)^2 - 4*(0.45)(15.6)}}/0.9 x < [110.7 + 110.5698421813109/0.9 < 245.8553802014565555556 x < 245 jackets....................................

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