How do I calculate combining two ratios to create a new third ratio

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Can someone provide me with a formula that directs me in combining two known ratios to create a new third ratio?

Solution 1 contains 43mg ingredient A, and 3.7mg ingredient B for a ratio of about 12:1

Solution 2 contains 0.69mg ingredient A, and 48mg ingredient B for a ratio of about 1:70

Is there a formula I can use directing how much of each Solution to use to create a new solution with a ratio about 1:1?

ratio

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asked 9 hours ago

Tom

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up vote
1
down vote

favorite

Can someone provide me with a formula that directs me in combining two known ratios to create a new third ratio?

Solution 1 contains 43mg ingredient A, and 3.7mg ingredient B for a ratio of about 12:1

Solution 2 contains 0.69mg ingredient A, and 48mg ingredient B for a ratio of about 1:70

Is there a formula I can use directing how much of each Solution to use to create a new solution with a ratio about 1:1?

ratio

share|cite|improve this question

asked 9 hours ago

Tom

91

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Can someone provide me with a formula that directs me in combining two known ratios to create a new third ratio?

Solution 1 contains 43mg ingredient A, and 3.7mg ingredient B for a ratio of about 12:1

Solution 2 contains 0.69mg ingredient A, and 48mg ingredient B for a ratio of about 1:70

Is there a formula I can use directing how much of each Solution to use to create a new solution with a ratio about 1:1?

ratio

share|cite|improve this question

asked 9 hours ago

Tom

91

Can someone provide me with a formula that directs me in combining two known ratios to create a new third ratio?

Solution 1 contains 43mg ingredient A, and 3.7mg ingredient B for a ratio of about 12:1

Solution 2 contains 0.69mg ingredient A, and 48mg ingredient B for a ratio of about 1:70

Is there a formula I can use directing how much of each Solution to use to create a new solution with a ratio about 1:1?

ratio

share|cite|improve this question

asked 9 hours ago

Tom

91

share|cite|improve this question

share|cite|improve this question

asked 9 hours ago

Tom

91

asked 9 hours ago

Tom

91

asked 9 hours ago

Tom

91

91

add a comment | 

add a comment | 

3 Answers
3

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votes

up vote
2
down vote

Compute the amount of each ingredient. We might as well start with $46.7$ mg of solution $1$ and add $x$ mg of solution $2$. Then we have $43+frac 0.6948.69x$ of $A$ and $3.7+frac 4848.69x$ of $B$. Equate those and solve for $x$.

share|cite|improve this answer

answered 8 hours ago

Ross Millikan

275k21183348

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
  – Ross Millikan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! That's why you fixed $x$, I see !!
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

The first solution has $43A+3.7B$ and the second has $0.69A+48B$ we want a linear combination of these that the same amount of $A$ and $B$.
begineqnarray*
lambda(43A+3.7B)+mu(0.69A+48B)=(43 lambda+0.69 mu)A+(3.7 lambda+48 mu)B
endeqnarray*
So we need
begineqnarray*
43 lambda+0.69 mu&=&3.7 lambda+48 mu \
39.3 lambda&=&47.31 mu \
endeqnarray*
and $lambda=47.31,mu=39.3$ is an obvious solution.

share|cite|improve this answer

answered 8 hours ago

Donald Splutterwit

21.2k21143

add a comment | 

up vote
0
down vote

Let the amount of Solution $1$ be $alpha$ and of Solution $2$ be $beta$. Then:
$$43alpha+0.69beta=3.7alpha+48beta$$
This leads to:
$$39.3alpha=48.31betatoalpha=frac48313930beta$$
So use approximately $1.23$ units of Solution $1$ per unit of Solution $2$

share|cite|improve this answer

answered 6 hours ago

Rhys Hughes

3,8481227

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Compute the amount of each ingredient. We might as well start with $46.7$ mg of solution $1$ and add $x$ mg of solution $2$. Then we have $43+frac 0.6948.69x$ of $A$ and $3.7+frac 4848.69x$ of $B$. Equate those and solve for $x$.

share|cite|improve this answer

answered 8 hours ago

Ross Millikan

275k21183348

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
  – Ross Millikan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! That's why you fixed $x$, I see !!
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Compute the amount of each ingredient. We might as well start with $46.7$ mg of solution $1$ and add $x$ mg of solution $2$. Then we have $43+frac 0.6948.69x$ of $A$ and $3.7+frac 4848.69x$ of $B$. Equate those and solve for $x$.

share|cite|improve this answer

answered 8 hours ago

Ross Millikan

275k21183348

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
  – Ross Millikan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! That's why you fixed $x$, I see !!
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Compute the amount of each ingredient. We might as well start with $46.7$ mg of solution $1$ and add $x$ mg of solution $2$. Then we have $43+frac 0.6948.69x$ of $A$ and $3.7+frac 4848.69x$ of $B$. Equate those and solve for $x$.

share|cite|improve this answer

answered 8 hours ago

Ross Millikan

275k21183348

Compute the amount of each ingredient. We might as well start with $46.7$ mg of solution $1$ and add $x$ mg of solution $2$. Then we have $43+frac 0.6948.69x$ of $A$ and $3.7+frac 4848.69x$ of $B$. Equate those and solve for $x$.

share|cite|improve this answer

answered 8 hours ago

Ross Millikan

275k21183348

share|cite|improve this answer

share|cite|improve this answer

answered 8 hours ago

Ross Millikan

275k21183348

answered 8 hours ago

Ross Millikan

275k21183348

answered 8 hours ago

Ross Millikan

275k21183348

275k21183348

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
  – Ross Millikan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! That's why you fixed $x$, I see !!
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
  – Ross Millikan
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Got it !! That's why you fixed $x$, I see !!
  – Anik Bhowmick
  8 hours ago
  

  
  
  
  

If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
– Anik Bhowmick
8 hours ago

If you take $x$ mg of solution $1$ and $y$ mg of solution $2$, then we'll get simultaneous equations to solve. Am I right ?? @RossMillikan
– Anik Bhowmick
8 hours ago

Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
– Ross Millikan
8 hours ago

Yes. You could then set $x+y$ to the total amount of solution you want at the end. With my approach you just have one equation to solve, then need to scale up or down to get the total amount desired.
– Ross Millikan
8 hours ago

Got it !! That's why you fixed $x$, I see !!
– Anik Bhowmick
8 hours ago

Got it !! That's why you fixed $x$, I see !!
– Anik Bhowmick
8 hours ago

add a comment | 

up vote
1
down vote

The first solution has $43A+3.7B$ and the second has $0.69A+48B$ we want a linear combination of these that the same amount of $A$ and $B$.
begineqnarray*
lambda(43A+3.7B)+mu(0.69A+48B)=(43 lambda+0.69 mu)A+(3.7 lambda+48 mu)B
endeqnarray*
So we need
begineqnarray*
43 lambda+0.69 mu&=&3.7 lambda+48 mu \
39.3 lambda&=&47.31 mu \
endeqnarray*
and $lambda=47.31,mu=39.3$ is an obvious solution.

share|cite|improve this answer

answered 8 hours ago

Donald Splutterwit

21.2k21143

add a comment | 

up vote
1
down vote

The first solution has $43A+3.7B$ and the second has $0.69A+48B$ we want a linear combination of these that the same amount of $A$ and $B$.
begineqnarray*
lambda(43A+3.7B)+mu(0.69A+48B)=(43 lambda+0.69 mu)A+(3.7 lambda+48 mu)B
endeqnarray*
So we need
begineqnarray*
43 lambda+0.69 mu&=&3.7 lambda+48 mu \
39.3 lambda&=&47.31 mu \
endeqnarray*
and $lambda=47.31,mu=39.3$ is an obvious solution.

share|cite|improve this answer

answered 8 hours ago

Donald Splutterwit

21.2k21143

add a comment | 

up vote
1
down vote

up vote
1
down vote

The first solution has $43A+3.7B$ and the second has $0.69A+48B$ we want a linear combination of these that the same amount of $A$ and $B$.
begineqnarray*
lambda(43A+3.7B)+mu(0.69A+48B)=(43 lambda+0.69 mu)A+(3.7 lambda+48 mu)B
endeqnarray*
So we need
begineqnarray*
43 lambda+0.69 mu&=&3.7 lambda+48 mu \
39.3 lambda&=&47.31 mu \
endeqnarray*
and $lambda=47.31,mu=39.3$ is an obvious solution.

share|cite|improve this answer

answered 8 hours ago

Donald Splutterwit

21.2k21143

The first solution has $43A+3.7B$ and the second has $0.69A+48B$ we want a linear combination of these that the same amount of $A$ and $B$.
begineqnarray*
lambda(43A+3.7B)+mu(0.69A+48B)=(43 lambda+0.69 mu)A+(3.7 lambda+48 mu)B
endeqnarray*
So we need
begineqnarray*
43 lambda+0.69 mu&=&3.7 lambda+48 mu \
39.3 lambda&=&47.31 mu \
endeqnarray*
and $lambda=47.31,mu=39.3$ is an obvious solution.

share|cite|improve this answer

answered 8 hours ago

Donald Splutterwit

21.2k21143

share|cite|improve this answer

share|cite|improve this answer

answered 8 hours ago

Donald Splutterwit

21.2k21143

answered 8 hours ago

Donald Splutterwit

21.2k21143

answered 8 hours ago

Donald Splutterwit

21.2k21143

21.2k21143

add a comment | 

add a comment | 

up vote
0
down vote

Let the amount of Solution $1$ be $alpha$ and of Solution $2$ be $beta$. Then:
$$43alpha+0.69beta=3.7alpha+48beta$$
This leads to:
$$39.3alpha=48.31betatoalpha=frac48313930beta$$
So use approximately $1.23$ units of Solution $1$ per unit of Solution $2$

share|cite|improve this answer

answered 6 hours ago

Rhys Hughes

3,8481227

add a comment | 

up vote
0
down vote

Let the amount of Solution $1$ be $alpha$ and of Solution $2$ be $beta$. Then:
$$43alpha+0.69beta=3.7alpha+48beta$$
This leads to:
$$39.3alpha=48.31betatoalpha=frac48313930beta$$
So use approximately $1.23$ units of Solution $1$ per unit of Solution $2$

share|cite|improve this answer

answered 6 hours ago

Rhys Hughes

3,8481227

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let the amount of Solution $1$ be $alpha$ and of Solution $2$ be $beta$. Then:
$$43alpha+0.69beta=3.7alpha+48beta$$
This leads to:
$$39.3alpha=48.31betatoalpha=frac48313930beta$$
So use approximately $1.23$ units of Solution $1$ per unit of Solution $2$

share|cite|improve this answer

answered 6 hours ago

Rhys Hughes

3,8481227

Let the amount of Solution $1$ be $alpha$ and of Solution $2$ be $beta$. Then:
$$43alpha+0.69beta=3.7alpha+48beta$$
This leads to:
$$39.3alpha=48.31betatoalpha=frac48313930beta$$
So use approximately $1.23$ units of Solution $1$ per unit of Solution $2$

share|cite|improve this answer

answered 6 hours ago

Rhys Hughes

3,8481227

share|cite|improve this answer

share|cite|improve this answer

answered 6 hours ago

Rhys Hughes

3,8481227

answered 6 hours ago

Rhys Hughes

3,8481227

answered 6 hours ago

Rhys Hughes

3,8481227

3,8481227

add a comment | 

add a comment | 

 

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