Math help with 7th grade
- Thread starter GABRIELLA
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Rule #1 - Name Stuff!
C = # of Chickens
R = # of Rabits
Common Knowledge is required. Chickens have 2 legs eash and rabits have 4 legs each.
Translate the two hints using the definitions.
1045 Animals
C + R = 1045
360 Legs
??? That can't be right. One leg each and we're already up to 1045! 3600 legs maybe? Maybe I can guess what's for dinner, tonight!
Fix the problem statement and give it a go.
C = # of Chickens
R = # of Rabits
Common Knowledge is required. Chickens have 2 legs eash and rabits have 4 legs each.
Translate the two hints using the definitions.
1045 Animals
C + R = 1045
360 Legs
??? That can't be right. One leg each and we're already up to 1045! 3600 legs maybe? Maybe I can guess what's for dinner, tonight!
Fix the problem statement and give it a go.
Hello, Gabriella!
I assume your finger slipped when you typed "104".
Now follow tkhunny's game plan.
\(\displaystyle C\) = number of chickens.
\(\displaystyle R\) = number of rabbits.
There were 104 animals: .\(\displaystyle C + R \:=\:104\) .[1]
Chickens are bipeds. .There are: \(\displaystyle 2C\) chicken legs.
Rabbits are quadrupeds. .There are: \(\displaystyle 4R\) rabbit legs.
Total number of legs is 360. . \(\displaystyle 2C + 4R \:=\:360\) .[2]
Solve this system of equations: . \(\displaystyle \begin{array}{ccc}C + R &=& 104 \\ 2C + 4R &=& 360 \end{array}\)
I assume your finger slipped when you typed "104".
Now follow tkhunny's game plan.
\(\displaystyle C\) = number of chickens.
\(\displaystyle R\) = number of rabbits.
There were 104 animals: .\(\displaystyle C + R \:=\:104\) .[1]
Chickens are bipeds. .There are: \(\displaystyle 2C\) chicken legs.
Rabbits are quadrupeds. .There are: \(\displaystyle 4R\) rabbit legs.
Total number of legs is 360. . \(\displaystyle 2C + 4R \:=\:360\) .[2]
Solve this system of equations: . \(\displaystyle \begin{array}{ccc}C + R &=& 104 \\ 2C + 4R &=& 360 \end{array}\)
Because this question was posted in the ARITHMETIC section, and the topic indicates that it is a 7th grade problem, I am going to use an arithmetic-only approach.
A very valid problem-solving technique is the "guess and check" method, where one makes a GUESS at what a possible solution might be and CHECKS to see if that possible solution is, indeed, correct. If the solution isn't correct, the "guess" can be adjusted and the process repeated.
I'll assume that you meant "there were 104 animals," since it isn't possible for 1045 animals to have 360 legs unless a whole bunch of them don't have any legs at all.
Take a guess.....suppose there are 52 chickens and 52 rabbits (so that we have a total of 104 animals). Let's check this guess. How many legs would there be? Well, each chicken has two legs, so 52 chickens would have 52*2, or 104 legs. And each rabbit has four legs, so 52 rabbits would have 52*4, or 208 legs. The total number of legs would be 104 + 208, or 312. Hmmm....not enough legs! What does this tell us? We need MORE rabbits and fewer chickens! So we can make a better guess.
What if we have 60 rabbits? Since there needs to be 104 animals, that means there'd be 104 - 60, or 44 chickens. Let's check this guess. If there are 60 rabbits, that would be 60*4, or 240 legs. And if there are 44 chickens, that would be 44*2, or 88 legs. 240 + 88 is 328 legs. Still not enough!
Ok...you can continue this process. You know that we still need more rabbits, so your next guess logically should have a higher number of rabbits, and a smaller number of chickens. It shouldn't take you too long to come up with the correct numbers.
If you have been studying algebra in your seventh grade class, of course you may wish to use an algebraic approach as suggested by some of the other posters.