Mixing
Ratio. Number of sheeps and chickens
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At a farm, the ratio of the number of chickens to the number of sheep was 5:2. After the farmer sold 15 chickens, there was an equal number of chickens and sheep. How many chickens and sheep were there at the farm in the end?
My work:
Number of chickens = C
Number of Sheep = S
*We know (C/S) = (5/2)
*We know that the farmer sold 15 chickens and then had equal number of both animals.
So C - 15 = S
Then I plugged C - 15 = S into (C/S) = (5/2)
So I got C = 25. (So I'm guessing there where 25 chickens before?)
Since we know that there were 25 chickens before, I pluged C = 25 into C - 15 = S
So 25 - 15 = S
So, we know that there are 10 Chickens and 10 Sheep. So the answer should be 20 total?
If I'm correct, is there a better way to do this?
word-problem ratio
asked 2 hours ago
RukaiPlusPlus
283
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up vote
3
down vote
favorite
At a farm, the ratio of the number of chickens to the number of sheep was 5:2. After the farmer sold 15 chickens, there was an equal number of chickens and sheep. How many chickens and sheep were there at the farm in the end?
My work:
Number of chickens = C
Number of Sheep = S
*We know (C/S) = (5/2)
*We know that the farmer sold 15 chickens and then had equal number of both animals.
So C - 15 = S
Then I plugged C - 15 = S into (C/S) = (5/2)
So I got C = 25. (So I'm guessing there where 25 chickens before?)
Since we know that there were 25 chickens before, I pluged C = 25 into C - 15 = S
So 25 - 15 = S
So, we know that there are 10 Chickens and 10 Sheep. So the answer should be 20 total?
If I'm correct, is there a better way to do this?
word-problem ratio
asked 2 hours ago
RukaiPlusPlus
283
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
– fleablood
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
At a farm, the ratio of the number of chickens to the number of sheep was 5:2. After the farmer sold 15 chickens, there was an equal number of chickens and sheep. How many chickens and sheep were there at the farm in the end?
My work:
Number of chickens = C
Number of Sheep = S
*We know (C/S) = (5/2)
*We know that the farmer sold 15 chickens and then had equal number of both animals.
So C - 15 = S
Then I plugged C - 15 = S into (C/S) = (5/2)
So I got C = 25. (So I'm guessing there where 25 chickens before?)
Since we know that there were 25 chickens before, I pluged C = 25 into C - 15 = S
So 25 - 15 = S
So, we know that there are 10 Chickens and 10 Sheep. So the answer should be 20 total?
If I'm correct, is there a better way to do this?
word-problem ratio
asked 2 hours ago
RukaiPlusPlus
283
At a farm, the ratio of the number of chickens to the number of sheep was 5:2. After the farmer sold 15 chickens, there was an equal number of chickens and sheep. How many chickens and sheep were there at the farm in the end?
My work:
Number of chickens = C
Number of Sheep = S
*We know (C/S) = (5/2)
*We know that the farmer sold 15 chickens and then had equal number of both animals.
So C - 15 = S
Then I plugged C - 15 = S into (C/S) = (5/2)
So I got C = 25. (So I'm guessing there where 25 chickens before?)
Since we know that there were 25 chickens before, I pluged C = 25 into C - 15 = S
So 25 - 15 = S
So, we know that there are 10 Chickens and 10 Sheep. So the answer should be 20 total?
If I'm correct, is there a better way to do this?
word-problem ratio
word-problem ratio
asked 2 hours ago
RukaiPlusPlus
283
asked 2 hours ago
RukaiPlusPlus
283
asked 2 hours ago
RukaiPlusPlus
283
asked 2 hours ago
RukaiPlusPlus
283
asked 2 hours ago
RukaiPlusPlus
283
283
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
– fleablood
1 hour ago
add a comment |Â
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
– fleablood
1 hour ago
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
– fleablood
1 hour ago
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
– fleablood
1 hour ago
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
2
down vote
Initially you have $5$ units of chickens and $2$ units of sheeps.
After selling, we have $2$ units of chickens and $2$ units of sheeps.
So $3$ units is equal to $15$ animals and each unit represents $5$ animals.
In the end we have $4 times 5 = 20$ animals. $10$ chickens and $10$ sheeps.
answered 1 hour ago
Siong Thye Goh
84.6k1457106
add a comment |Â
up vote
1
down vote
The answer is right, the calculations look correct, and it is a very reasonable way to reach the answer.
Alternatively, you could've noted that the $15$ chickens sold were three fifths of all the chickens, and gone from there. It might have saved you a few seconds of calculation, in exchange for having to spend a few seconds writing down an explanation for why this is true.
answered 1 hour ago
Arthur
103k798179
add a comment |Â
up vote
1
down vote
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
Another way is $frac CS = frac 52 = frac 5a2a$ for some $a$ where $C = 5a$ and $S = 2a$.
Then $C - 15 = S$ means $5a - 15 = 2a$ so $3a = 15$ and $a =5$. So there were $25$ chickens and $10$ cows originally.
I'm not sure if that second way is better but captures the spirit of proportions more that the straight algebra.
answered 58 mins ago
fleablood
62.8k22679
add a comment |Â
up vote
1
down vote
$C$ number of chickens, $S$ number if sheep.
1) $C/S=5/2;$
2) $C-15 = S$;
1) $S= (2/5)C$
Combining with 2):
$C-15=(2/5)C;$
$5C -75 =2C;$
$3C = 75$; or $C=25.$
$C=25$ is the number of sheep before the transaction .
After:
$C_f= 25-15 =10$, the number of sheep after transaction .
$C_f=S=S_f=10.$
answered 26 mins ago
Peter Szilas
8,8482719
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up vote
0
down vote
You can check your answer by plugging your solution into the initial equation (think of systems of equations back in algebra).
We know without doubt, like you said, that $fracCS=frac52$ represents the ratio of chickens to sheeps, and that $C-15=S$ represents 15 chickens being sold and only then becoming equal to $S$,
therefore,
$ $
$we plug S=C-15 into fracCS as follows$
$$fracCC-15=frac52$$
$evaluate as follows$
$$(C-15)fracCC-15=frac52(C-15)$$
$$C=frac5c-752$$
$$2(C)=(frac5c-752)2$$
$once evaluated, you get$ $$2C=5c-75$$
$reword C-15=S as colorredCcolorred=colorred1colorred5colorred+colorredS then plug colorredC into the above equation to solve for colorblueS$
$$2(colorred1colorred5colorred+colorredS)=5(colorred1colorred5colorred+colorredS)-75$$
$$30+3S=75+5S-75$$
$$30+2S=5S$$
$$30=3S$$
$$colorblueScolorblue=colorblue1colorblue0 colorbluescolorbluehcolorblueecolorblueecolorbluepcolorblues$$
$Finally, plug this S into the original equality, fraccolorredCcolorblueS=frac52$
$$ fraccolorredCcolorblue1colorblue0=frac52$$
$$2C=colorblue5colorblue0$$
$$colorredCcolorred=colorred2colorred5 colorredCcolorredhcolorredicolorredccolorredkcolorredecolorredncolorreds$$
Therefore, you are correct to say 25 chickens and 10 sheeps in the end.
answered 43 mins ago
user581844
536
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Initially you have $5$ units of chickens and $2$ units of sheeps.
After selling, we have $2$ units of chickens and $2$ units of sheeps.
So $3$ units is equal to $15$ animals and each unit represents $5$ animals.
In the end we have $4 times 5 = 20$ animals. $10$ chickens and $10$ sheeps.
answered 1 hour ago
Siong Thye Goh
84.6k1457106
add a comment |Â
up vote
2
down vote
Initially you have $5$ units of chickens and $2$ units of sheeps.
After selling, we have $2$ units of chickens and $2$ units of sheeps.
So $3$ units is equal to $15$ animals and each unit represents $5$ animals.
In the end we have $4 times 5 = 20$ animals. $10$ chickens and $10$ sheeps.
answered 1 hour ago
Siong Thye Goh
84.6k1457106
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Initially you have $5$ units of chickens and $2$ units of sheeps.
After selling, we have $2$ units of chickens and $2$ units of sheeps.
So $3$ units is equal to $15$ animals and each unit represents $5$ animals.
In the end we have $4 times 5 = 20$ animals. $10$ chickens and $10$ sheeps.
answered 1 hour ago
Siong Thye Goh
84.6k1457106
Initially you have $5$ units of chickens and $2$ units of sheeps.
After selling, we have $2$ units of chickens and $2$ units of sheeps.
So $3$ units is equal to $15$ animals and each unit represents $5$ animals.
In the end we have $4 times 5 = 20$ animals. $10$ chickens and $10$ sheeps.
answered 1 hour ago
Siong Thye Goh
84.6k1457106
edited 1 hour ago
answered 1 hour ago
Siong Thye Goh
84.6k1457106
answered 1 hour ago
Siong Thye Goh
84.6k1457106
answered 1 hour ago
Siong Thye Goh
84.6k1457106
84.6k1457106
add a comment |Â
add a comment |Â
up vote
1
down vote
The answer is right, the calculations look correct, and it is a very reasonable way to reach the answer.
Alternatively, you could've noted that the $15$ chickens sold were three fifths of all the chickens, and gone from there. It might have saved you a few seconds of calculation, in exchange for having to spend a few seconds writing down an explanation for why this is true.
answered 1 hour ago
Arthur
103k798179
add a comment |Â
up vote
1
down vote
The answer is right, the calculations look correct, and it is a very reasonable way to reach the answer.
Alternatively, you could've noted that the $15$ chickens sold were three fifths of all the chickens, and gone from there. It might have saved you a few seconds of calculation, in exchange for having to spend a few seconds writing down an explanation for why this is true.
answered 1 hour ago
Arthur
103k798179
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The answer is right, the calculations look correct, and it is a very reasonable way to reach the answer.
Alternatively, you could've noted that the $15$ chickens sold were three fifths of all the chickens, and gone from there. It might have saved you a few seconds of calculation, in exchange for having to spend a few seconds writing down an explanation for why this is true.
answered 1 hour ago
Arthur
103k798179
The answer is right, the calculations look correct, and it is a very reasonable way to reach the answer.
Alternatively, you could've noted that the $15$ chickens sold were three fifths of all the chickens, and gone from there. It might have saved you a few seconds of calculation, in exchange for having to spend a few seconds writing down an explanation for why this is true.
answered 1 hour ago
Arthur
103k798179
answered 1 hour ago
Arthur
103k798179
answered 1 hour ago
Arthur
103k798179
answered 1 hour ago
Arthur
103k798179
103k798179
add a comment |Â
add a comment |Â
up vote
1
down vote
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
Another way is $frac CS = frac 52 = frac 5a2a$ for some $a$ where $C = 5a$ and $S = 2a$.
Then $C - 15 = S$ means $5a - 15 = 2a$ so $3a = 15$ and $a =5$. So there were $25$ chickens and $10$ cows originally.
I'm not sure if that second way is better but captures the spirit of proportions more that the straight algebra.
answered 58 mins ago
fleablood
62.8k22679
add a comment |Â
up vote
1
down vote
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
Another way is $frac CS = frac 52 = frac 5a2a$ for some $a$ where $C = 5a$ and $S = 2a$.
Then $C - 15 = S$ means $5a - 15 = 2a$ so $3a = 15$ and $a =5$. So there were $25$ chickens and $10$ cows originally.
I'm not sure if that second way is better but captures the spirit of proportions more that the straight algebra.
answered 58 mins ago
fleablood
62.8k22679
add a comment |Â
up vote
1
down vote
up vote
1
down vote
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
Another way is $frac CS = frac 52 = frac 5a2a$ for some $a$ where $C = 5a$ and $S = 2a$.
Then $C - 15 = S$ means $5a - 15 = 2a$ so $3a = 15$ and $a =5$. So there were $25$ chickens and $10$ cows originally.
I'm not sure if that second way is better but captures the spirit of proportions more that the straight algebra.
answered 58 mins ago
fleablood
62.8k22679
... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
Another way is $frac CS = frac 52 = frac 5a2a$ for some $a$ where $C = 5a$ and $S = 2a$.
Then $C - 15 = S$ means $5a - 15 = 2a$ so $3a = 15$ and $a =5$. So there were $25$ chickens and $10$ cows originally.
I'm not sure if that second way is better but captures the spirit of proportions more that the straight algebra.
answered 58 mins ago
fleablood
62.8k22679
answered 58 mins ago
fleablood
62.8k22679
answered 58 mins ago
fleablood
62.8k22679
answered 58 mins ago
fleablood
62.8k22679
62.8k22679
add a comment |Â
add a comment |Â
up vote
1
down vote
$C$ number of chickens, $S$ number if sheep.
1) $C/S=5/2;$
2) $C-15 = S$;
1) $S= (2/5)C$
Combining with 2):
$C-15=(2/5)C;$
$5C -75 =2C;$
$3C = 75$; or $C=25.$
$C=25$ is the number of sheep before the transaction .
After:
$C_f= 25-15 =10$, the number of sheep after transaction .
$C_f=S=S_f=10.$
answered 26 mins ago
Peter Szilas
8,8482719
add a comment |Â
up vote
1
down vote
$C$ number of chickens, $S$ number if sheep.
1) $C/S=5/2;$
2) $C-15 = S$;
1) $S= (2/5)C$
Combining with 2):
$C-15=(2/5)C;$
$5C -75 =2C;$
$3C = 75$; or $C=25.$
$C=25$ is the number of sheep before the transaction .
After:
$C_f= 25-15 =10$, the number of sheep after transaction .
$C_f=S=S_f=10.$
answered 26 mins ago
Peter Szilas
8,8482719
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$C$ number of chickens, $S$ number if sheep.
1) $C/S=5/2;$
2) $C-15 = S$;
1) $S= (2/5)C$
Combining with 2):
$C-15=(2/5)C;$
$5C -75 =2C;$
$3C = 75$; or $C=25.$
$C=25$ is the number of sheep before the transaction .
After:
$C_f= 25-15 =10$, the number of sheep after transaction .
$C_f=S=S_f=10.$
answered 26 mins ago
Peter Szilas
8,8482719
$C$ number of chickens, $S$ number if sheep.
1) $C/S=5/2;$
2) $C-15 = S$;
1) $S= (2/5)C$
Combining with 2):
$C-15=(2/5)C;$
$5C -75 =2C;$
$3C = 75$; or $C=25.$
$C=25$ is the number of sheep before the transaction .
After:
$C_f= 25-15 =10$, the number of sheep after transaction .
$C_f=S=S_f=10.$
answered 26 mins ago
Peter Szilas
8,8482719
edited 15 mins ago
answered 26 mins ago
Peter Szilas
8,8482719
answered 26 mins ago
Peter Szilas
8,8482719
answered 26 mins ago
Peter Szilas
8,8482719
8,8482719
add a comment |Â
add a comment |Â
up vote
0
down vote
You can check your answer by plugging your solution into the initial equation (think of systems of equations back in algebra).
We know without doubt, like you said, that $fracCS=frac52$ represents the ratio of chickens to sheeps, and that $C-15=S$ represents 15 chickens being sold and only then becoming equal to $S$,
therefore,
$ $
$we plug S=C-15 into fracCS as follows$
$$fracCC-15=frac52$$
$evaluate as follows$
$$(C-15)fracCC-15=frac52(C-15)$$
$$C=frac5c-752$$
$$2(C)=(frac5c-752)2$$
$once evaluated, you get$ $$2C=5c-75$$
$reword C-15=S as colorredCcolorred=colorred1colorred5colorred+colorredS then plug colorredC into the above equation to solve for colorblueS$
$$2(colorred1colorred5colorred+colorredS)=5(colorred1colorred5colorred+colorredS)-75$$
$$30+3S=75+5S-75$$
$$30+2S=5S$$
$$30=3S$$
$$colorblueScolorblue=colorblue1colorblue0 colorbluescolorbluehcolorblueecolorblueecolorbluepcolorblues$$
$Finally, plug this S into the original equality, fraccolorredCcolorblueS=frac52$
$$ fraccolorredCcolorblue1colorblue0=frac52$$
$$2C=colorblue5colorblue0$$
$$colorredCcolorred=colorred2colorred5 colorredCcolorredhcolorredicolorredccolorredkcolorredecolorredncolorreds$$
Therefore, you are correct to say 25 chickens and 10 sheeps in the end.
answered 43 mins ago
user581844
536
add a comment |Â
up vote
0
down vote
You can check your answer by plugging your solution into the initial equation (think of systems of equations back in algebra).
We know without doubt, like you said, that $fracCS=frac52$ represents the ratio of chickens to sheeps, and that $C-15=S$ represents 15 chickens being sold and only then becoming equal to $S$,
therefore,
$ $
$we plug S=C-15 into fracCS as follows$
$$fracCC-15=frac52$$
$evaluate as follows$
$$(C-15)fracCC-15=frac52(C-15)$$
$$C=frac5c-752$$
$$2(C)=(frac5c-752)2$$
$once evaluated, you get$ $$2C=5c-75$$
$reword C-15=S as colorredCcolorred=colorred1colorred5colorred+colorredS then plug colorredC into the above equation to solve for colorblueS$
$$2(colorred1colorred5colorred+colorredS)=5(colorred1colorred5colorred+colorredS)-75$$
$$30+3S=75+5S-75$$
$$30+2S=5S$$
$$30=3S$$
$$colorblueScolorblue=colorblue1colorblue0 colorbluescolorbluehcolorblueecolorblueecolorbluepcolorblues$$
$Finally, plug this S into the original equality, fraccolorredCcolorblueS=frac52$
$$ fraccolorredCcolorblue1colorblue0=frac52$$
$$2C=colorblue5colorblue0$$
$$colorredCcolorred=colorred2colorred5 colorredCcolorredhcolorredicolorredccolorredkcolorredecolorredncolorreds$$
Therefore, you are correct to say 25 chickens and 10 sheeps in the end.
answered 43 mins ago
user581844
536
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can check your answer by plugging your solution into the initial equation (think of systems of equations back in algebra).
We know without doubt, like you said, that $fracCS=frac52$ represents the ratio of chickens to sheeps, and that $C-15=S$ represents 15 chickens being sold and only then becoming equal to $S$,
therefore,
$ $
$we plug S=C-15 into fracCS as follows$
$$fracCC-15=frac52$$
$evaluate as follows$
$$(C-15)fracCC-15=frac52(C-15)$$
$$C=frac5c-752$$
$$2(C)=(frac5c-752)2$$
$once evaluated, you get$ $$2C=5c-75$$
$reword C-15=S as colorredCcolorred=colorred1colorred5colorred+colorredS then plug colorredC into the above equation to solve for colorblueS$
$$2(colorred1colorred5colorred+colorredS)=5(colorred1colorred5colorred+colorredS)-75$$
$$30+3S=75+5S-75$$
$$30+2S=5S$$
$$30=3S$$
$$colorblueScolorblue=colorblue1colorblue0 colorbluescolorbluehcolorblueecolorblueecolorbluepcolorblues$$
$Finally, plug this S into the original equality, fraccolorredCcolorblueS=frac52$
$$ fraccolorredCcolorblue1colorblue0=frac52$$
$$2C=colorblue5colorblue0$$
$$colorredCcolorred=colorred2colorred5 colorredCcolorredhcolorredicolorredccolorredkcolorredecolorredncolorreds$$
Therefore, you are correct to say 25 chickens and 10 sheeps in the end.
answered 43 mins ago
user581844
536
You can check your answer by plugging your solution into the initial equation (think of systems of equations back in algebra).
We know without doubt, like you said, that $fracCS=frac52$ represents the ratio of chickens to sheeps, and that $C-15=S$ represents 15 chickens being sold and only then becoming equal to $S$,
therefore,
$ $
$we plug S=C-15 into fracCS as follows$
$$fracCC-15=frac52$$
$evaluate as follows$
$$(C-15)fracCC-15=frac52(C-15)$$
$$C=frac5c-752$$
$$2(C)=(frac5c-752)2$$
$once evaluated, you get$ $$2C=5c-75$$
$reword C-15=S as colorredCcolorred=colorred1colorred5colorred+colorredS then plug colorredC into the above equation to solve for colorblueS$
$$2(colorred1colorred5colorred+colorredS)=5(colorred1colorred5colorred+colorredS)-75$$
$$30+3S=75+5S-75$$
$$30+2S=5S$$
$$30=3S$$
$$colorblueScolorblue=colorblue1colorblue0 colorbluescolorbluehcolorblueecolorblueecolorbluepcolorblues$$
$Finally, plug this S into the original equality, fraccolorredCcolorblueS=frac52$
$$ fraccolorredCcolorblue1colorblue0=frac52$$
$$2C=colorblue5colorblue0$$
$$colorredCcolorred=colorred2colorred5 colorredCcolorredhcolorredicolorredccolorredkcolorredecolorredncolorreds$$
Therefore, you are correct to say 25 chickens and 10 sheeps in the end.
answered 43 mins ago
user581844
536
edited 33 mins ago
answered 43 mins ago
user581844
536
answered 43 mins ago
user581844
536
answered 43 mins ago
user581844
536
536
add a comment |Â
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... or you could do $frac CS = frac 52$ so $frac 25S = frac 52$ so $S = 10$.
– fleablood
1 hour ago