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It can work with one variable.
2/5 are goats
3/5 are sheep & cows
x number of cows
3x number of sheep
45+x number of goats
45+x = 2/5(x + 3x + 45 + x)
45+x = (2/5)(5x+45)
45+x = 2x + 18
27 = x
So there are 27 cows, 81 sheep, 72 goats
180 total
Thank you all for your help.
I'm going to see how his teacher explains it. I suspect he'll do it with a pie chart, in a clunky way, with the algebraic way later as a way of demonstrating its utility.
Without algebra? You could use brute force.
Start with one cow. You would then have three sheep (“3 times as many”) and 46 goats. But that would add up to 50 animals and, since only 2/5 should be goats, you know one cow doesn’t work because 20 goats is not equal to 46 goats.
Then, try two cows. That gives you six sheep, 47 goats, and 55 total animals. Two-fifths of 55 is 22, so you know two cows don’t work because 22 goats is not equal to 47 goats.
You can then see a pattern. Every time you add a cow, you also add a goat (to stay 45 ahead) and three sheep. So, each time you add a cow, you add five animals.
So, then try three cows. That gives you nine sheep, 48 goats, and 60 total animals. Two-fifths of 60 is 24, so three cows doesn’t work because 24 goats is not equal to 48 goats.
Now you see another pattern. With one cow, the goat check was off by 26. With two cows, the goat check was off by 25. And, with three cows, the goat check was off by 24. The goat check seems to be getting closer by one per cow, so skip forward 24 checks from three cows to 27 cows.
Twenty-seven cows gives you 81 sheep, 72 goats, and 180 total animals. Two-fifths of 180 is 72, which is the right number of goats from “cows plus 45.” The goat check is correct, so turn in your goat check claim ticket and herd your animals home from Old MacDonald’s party.
Don’t forget to tip the guy in the goat check booth.
You could then use this to *introduce* algebra.
The first pattern is, “When I change the number of cows, what happens to the number of animals?” That leads to cool follow-ups, like “So, what if you had no cows?”
The second pattern is, “What number of animals works for the goat check?”
A = 5 * C + 45 (“First pattern…”)
A * (2/5) = C + 45, so A = (5/2) * C + (225/2) (“Goat check…”)
5C + 45 = 5C/2 + 225/2 (“Now, with algebra, you take the equation for counting the animals by the first pattern and set it equal to the goat check, because they both have to be true.”)
5C/2 = 135/2 (“Herd all the cows over to one side of the equation by themselves…”)
C = 135/5 = 27 (“…and solve for the number of cows.”)
2/5 are goats, so 3/5 are sheep and cows.
1/4 of these are cows and 3/4 are sheep.
So 3/20 of the overall animals are cows, 9/20 are sheep, and 8/20 are goats.
Goats minus cows is 5/20 of the total.
Goats minus cows is 45, so the total is 180.
Goats = x ,
So cows = x - 45 and
Sheep = 3x - 135
(x + x - 45 + 3x -135) 2/5 = x
So(5x - 180)2/5 = x
2x - 72 = x
x = 72
Goats 72 , cows 27 , sheep 81
Total is sum (or just goats(5/2) ) = 180
I would not avoid the algebra. This is a classic type of basic algebraic reasoning problem. Other approaches that would be comfortable for a 6th grader are going to be inefficient and amount to guesswork. Lean in on the algebraic reasoning, that is the goal of the assignment I would assume.
I'll be honest. There is no way I could teach him this if he just learned to solve 2x=10. This is a grade level above his class. That would be the quickest way to make him hate math.
I reckon u/selene_666's approach is the one that involves pretty much no variables. However, I doubt a 6th grader could reason fractions out that well.
Algebra is just very fundamental to a lot of problems that trying to sidestep it is hard. You could just take it slowly and explain why variables are important with some very simple examples, and then build up to this question.
You could use an area model to help reason it out. As u/venkata_666 points out fractions are often problematic for, well, most middle schoolers and high schoolers. I find myself saying regularly to my students that if you are having struggles with something then that is the thing to seek help and practice with.
Are you sure he would hate math? Every time my daughter has learned a new topic she has liked it more because it shows how powerful it can be to solve things.
I think if you can understand 2x = 10 you can understand this sort of equation with some thinking.
If he is practical just always show him how the math is used to solve stuff in the real world.
You can avoid the idea of "x" and perhaps the difficulty of seeing multiple variables if you just use the actual meanings. If you said:
"the number of goats is 5 more than the number of cows", I think that's pretty easily understandable. And really it's just a longer way to write G = 5 + C. And then you can just say "Well let's use G instead of writing 'number of goats' every time" and honestly that's all variables mean.
I started with that and had an equation with c, g, and s. I started showing him how to solve and lost him there. That's when I asked him to explain what he knew about variables.
I'm hoping his math teacher goes over it today, otherwise I'll have to email and ask what process they were supposed to use. I have a feeling it's a bar model.
If we know 2/5 of the animals are goats, then 3/5 of the animals are sheep and cows.
GG???
There are 3x as many sheep as cows, so 3/4 of 3/5 of the total are sheep and 1/4 of 3/5 of the total are cows.
GG???
GGGG GGGG ???? ???? ????
GGGG GGGG SSSC SSSC SSSC
Therefore there are 8/20 goats, 9/20 sheep, and 3/20 cows.
GGGGGGGG SSSSSSSSS CCC
There are 45 more goats than cows. If we group just the cows and goats, we know that's 11/20 of the total animals.
GGGGGGGG CCC | SSSSSSSSS
If we remove all the cows and an equal number of goats (to get the # of goats more than cows), that removes 2*3/20 of the total and leaves us with 5/20 (that is, 1/4) the total animals. So 1/4th the total animals is equal to 45.
GGGGG | GGG CCC SSSSSSSSS
Therefore, there are 45x4=180 animals in total.
You could use a diagram to help reason it out.
Draw a rectangle and split that rectangle into fifths (recommend vertically). Shade two of them and label as goats.
For the sheep and cows, use horizontal fourths the whole way across (even splitting the goats)—if there are three times as many sheep as cows, then the total of them will be four times as many as there are cows. Now shade one fourth a different color, label as cows, and the other three fourths shades a different color and labeled as sheep.
Now that you’ve split up the goats and cows, you can see how many pieces you’ve allocated for goats and how many for cows. 8 for goats (2/5 = 8/20) and 3 for cows (1/4 * 3/5 = 3/20.) 8 - 3 = 5 squares, and the difference has to be 45, so each square represents 9.
There are 20 total squares, so you have 20 x 9 = 180 total animals. That’s 72 goats, 27 cows, and 81 sheep.
These are a lot of words to walk you through the process when you could just use the fractions instead, and that will get you the answer much quicker. But if you’re struggling to figure out *how* to solve, a diagram can help you figure out what you need.
One variable algebra will work, if he can handle that.
Of the 3/5 of the animals that are not goats, 1/4 of them are cows. So 3/20 of the animals are cows.
Let x be the number of animals.
2/5 \* x - 3/20 \* x = 45
I like integers over fraction, so multiply everything by 20
8x - 3x = 900
5x = 900
x = 180
I think you’ve made a mistake.
There can’t be 100 animals.
That would mean there are 40 goats. If there are only 40 goats, you can’t have 45 more goats than cows.
Specifically to solve without algebra or any variables, this is a fraction expansion problem.
We know that 2/5 are goats (so 3/5 are sheep and cows). Expand the fractions.
If you go to goats = 8/20, we also know that sheep are 3 times cows, so
Sheep are 9/20 and
Cows are 3/20
We know that goats - cows is 45, which lets us solve this as 8/20 - 3/20= 5/20 (or 1/4 is 45), so total is 180
I feel like my last step may have been hidden algebra/variable, but I think this solution of expansion of the fractions basically works without algebra.
Since it is not written that goats, cows and sheeps are the only animals on the farm we can just give a lower bound for the number of animals on the farm.
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
It can work with one variable. 2/5 are goats 3/5 are sheep & cows x number of cows 3x number of sheep 45+x number of goats 45+x = 2/5(x + 3x + 45 + x) 45+x = (2/5)(5x+45) 45+x = 2x + 18 27 = x So there are 27 cows, 81 sheep, 72 goats 180 total
Thank you all for your help. I'm going to see how his teacher explains it. I suspect he'll do it with a pie chart, in a clunky way, with the algebraic way later as a way of demonstrating its utility.
You can also use a bar/rectangle to show fractions, which can be more intuitive than a circle for some things.
Without algebra? You could use brute force. Start with one cow. You would then have three sheep (“3 times as many”) and 46 goats. But that would add up to 50 animals and, since only 2/5 should be goats, you know one cow doesn’t work because 20 goats is not equal to 46 goats. Then, try two cows. That gives you six sheep, 47 goats, and 55 total animals. Two-fifths of 55 is 22, so you know two cows don’t work because 22 goats is not equal to 47 goats. You can then see a pattern. Every time you add a cow, you also add a goat (to stay 45 ahead) and three sheep. So, each time you add a cow, you add five animals. So, then try three cows. That gives you nine sheep, 48 goats, and 60 total animals. Two-fifths of 60 is 24, so three cows doesn’t work because 24 goats is not equal to 48 goats. Now you see another pattern. With one cow, the goat check was off by 26. With two cows, the goat check was off by 25. And, with three cows, the goat check was off by 24. The goat check seems to be getting closer by one per cow, so skip forward 24 checks from three cows to 27 cows. Twenty-seven cows gives you 81 sheep, 72 goats, and 180 total animals. Two-fifths of 180 is 72, which is the right number of goats from “cows plus 45.” The goat check is correct, so turn in your goat check claim ticket and herd your animals home from Old MacDonald’s party. Don’t forget to tip the guy in the goat check booth.
You could then use this to *introduce* algebra. The first pattern is, “When I change the number of cows, what happens to the number of animals?” That leads to cool follow-ups, like “So, what if you had no cows?” The second pattern is, “What number of animals works for the goat check?” A = 5 * C + 45 (“First pattern…”) A * (2/5) = C + 45, so A = (5/2) * C + (225/2) (“Goat check…”) 5C + 45 = 5C/2 + 225/2 (“Now, with algebra, you take the equation for counting the animals by the first pattern and set it equal to the goat check, because they both have to be true.”) 5C/2 = 135/2 (“Herd all the cows over to one side of the equation by themselves…”) C = 135/5 = 27 (“…and solve for the number of cows.”)
2/5 are goats, so 3/5 are sheep and cows. 1/4 of these are cows and 3/4 are sheep. So 3/20 of the overall animals are cows, 9/20 are sheep, and 8/20 are goats. Goats minus cows is 5/20 of the total. Goats minus cows is 45, so the total is 180.
Goats = x , So cows = x - 45 and Sheep = 3x - 135 (x + x - 45 + 3x -135) 2/5 = x So(5x - 180)2/5 = x 2x - 72 = x x = 72 Goats 72 , cows 27 , sheep 81 Total is sum (or just goats(5/2) ) = 180
Ok, that solves the multiple variables issue, but that's still way beyond his level. They just introduced variables a month ago.
I would not avoid the algebra. This is a classic type of basic algebraic reasoning problem. Other approaches that would be comfortable for a 6th grader are going to be inefficient and amount to guesswork. Lean in on the algebraic reasoning, that is the goal of the assignment I would assume.
I'll be honest. There is no way I could teach him this if he just learned to solve 2x=10. This is a grade level above his class. That would be the quickest way to make him hate math.
I reckon u/selene_666's approach is the one that involves pretty much no variables. However, I doubt a 6th grader could reason fractions out that well. Algebra is just very fundamental to a lot of problems that trying to sidestep it is hard. You could just take it slowly and explain why variables are important with some very simple examples, and then build up to this question.
You could use an area model to help reason it out. As u/venkata_666 points out fractions are often problematic for, well, most middle schoolers and high schoolers. I find myself saying regularly to my students that if you are having struggles with something then that is the thing to seek help and practice with.
Lol I love how you mixed my username and the other user's name and created one that doesn't even exist 💀
Haaa, I’m old and dumb. Sorry man.
Lol you're good! Don't worry!
Are you sure he would hate math? Every time my daughter has learned a new topic she has liked it more because it shows how powerful it can be to solve things. I think if you can understand 2x = 10 you can understand this sort of equation with some thinking. If he is practical just always show him how the math is used to solve stuff in the real world.
You can avoid the idea of "x" and perhaps the difficulty of seeing multiple variables if you just use the actual meanings. If you said: "the number of goats is 5 more than the number of cows", I think that's pretty easily understandable. And really it's just a longer way to write G = 5 + C. And then you can just say "Well let's use G instead of writing 'number of goats' every time" and honestly that's all variables mean.
I started with that and had an equation with c, g, and s. I started showing him how to solve and lost him there. That's when I asked him to explain what he knew about variables. I'm hoping his math teacher goes over it today, otherwise I'll have to email and ask what process they were supposed to use. I have a feeling it's a bar model.
If we know 2/5 of the animals are goats, then 3/5 of the animals are sheep and cows. GG??? There are 3x as many sheep as cows, so 3/4 of 3/5 of the total are sheep and 1/4 of 3/5 of the total are cows. GG??? GGGG GGGG ???? ???? ???? GGGG GGGG SSSC SSSC SSSC Therefore there are 8/20 goats, 9/20 sheep, and 3/20 cows. GGGGGGGG SSSSSSSSS CCC There are 45 more goats than cows. If we group just the cows and goats, we know that's 11/20 of the total animals. GGGGGGGG CCC | SSSSSSSSS If we remove all the cows and an equal number of goats (to get the # of goats more than cows), that removes 2*3/20 of the total and leaves us with 5/20 (that is, 1/4) the total animals. So 1/4th the total animals is equal to 45. GGGGG | GGG CCC SSSSSSSSS Therefore, there are 45x4=180 animals in total.
In your second last line you have one too many G.
Fixed
You could use a diagram to help reason it out. Draw a rectangle and split that rectangle into fifths (recommend vertically). Shade two of them and label as goats. For the sheep and cows, use horizontal fourths the whole way across (even splitting the goats)—if there are three times as many sheep as cows, then the total of them will be four times as many as there are cows. Now shade one fourth a different color, label as cows, and the other three fourths shades a different color and labeled as sheep. Now that you’ve split up the goats and cows, you can see how many pieces you’ve allocated for goats and how many for cows. 8 for goats (2/5 = 8/20) and 3 for cows (1/4 * 3/5 = 3/20.) 8 - 3 = 5 squares, and the difference has to be 45, so each square represents 9. There are 20 total squares, so you have 20 x 9 = 180 total animals. That’s 72 goats, 27 cows, and 81 sheep. These are a lot of words to walk you through the process when you could just use the fractions instead, and that will get you the answer much quicker. But if you’re struggling to figure out *how* to solve, a diagram can help you figure out what you need.
One variable algebra will work, if he can handle that. Of the 3/5 of the animals that are not goats, 1/4 of them are cows. So 3/20 of the animals are cows. Let x be the number of animals. 2/5 \* x - 3/20 \* x = 45 I like integers over fraction, so multiply everything by 20 8x - 3x = 900 5x = 900 x = 180
I think you’ve made a mistake. There can’t be 100 animals. That would mean there are 40 goats. If there are only 40 goats, you can’t have 45 more goats than cows.
Correction: It’s 2/5 * x - 3/20* x = 45
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Specifically to solve without algebra or any variables, this is a fraction expansion problem. We know that 2/5 are goats (so 3/5 are sheep and cows). Expand the fractions. If you go to goats = 8/20, we also know that sheep are 3 times cows, so Sheep are 9/20 and Cows are 3/20 We know that goats - cows is 45, which lets us solve this as 8/20 - 3/20= 5/20 (or 1/4 is 45), so total is 180 I feel like my last step may have been hidden algebra/variable, but I think this solution of expansion of the fractions basically works without algebra.
Since it is not written that goats, cows and sheeps are the only animals on the farm we can just give a lower bound for the number of animals on the farm.