Word problem : income tax and mixture

[NotaMathPerson]

Hello everyone!

I need help on setting the equations for these problems.View attachment 5643

For prob 30. I could not solve it because there's no given amount in lbs of the mixture.
For prob 28 I really had no idea what it is asking.

Please bear with the monetary unit used those problems. These problems are from very old book.
Thanks!

 

[MarkFL]

Using this monetary system, there are 12 pence (d) in a shilling (s) and 20 shillings, or 240 pence, in a pound (£).

Let's look at problem 29 first, and use the pound as the monetary unit Let's let $A$ be the income tax raised on incomes below 100 and $B$ be the income tax raised on incomes above 100. We will then let $C$ be the total incomes below 100 and $D$ be the total incomes above 100. So we may write:

\(\displaystyle A=\frac{7}{240}C\)

\(\displaystyle B=\frac{1}{20}D\)

\(\displaystyle A+B=18750\)

\(\displaystyle C+D=500000\)

Since we are being asked for $A$, let's substitute for $C$ and $D$ in the last equation using the first 2 equations:

\(\displaystyle A+B=18750\)

\(\displaystyle \frac{240}{7}A+20B=500000\)

Now, let's multiply the first equation by -140 and second equation by 7:

\(\displaystyle -140A-140B=-2625000\)

\(\displaystyle 240A+140B=3500000\)

Now, adding these two equations, what do you find?

 

[NotaMathPerson]

Using this monetary system, there are 12 pence (d) in a shilling (s) and 20 shillings, or 240 pence, in a pound (£).

Let's look at problem 29 first, and use the pound as the monetary unit Let's let $A$ be the income tax raised on incomes below 100 and $B$ be the income tax raised on incomes above 100. We will then let $C$ be the total incomes below 100 and $D$ be the total incomes above 100. So we may write:

\(\displaystyle A=\frac{7}{240}C\)

\(\displaystyle B=\frac{1}{20}D\)

\(\displaystyle A+B=18750\)

\(\displaystyle C+D=500000\)

Since we are being asked for $A$, let's substitute for $C$ and $D$ in the last equation using the first 2 equations:

\(\displaystyle A+B=18750\)

\(\displaystyle \frac{240}{7}A+20B=500000\)

Now, let's multiply the first equation by -140 and second equation by 7:

\(\displaystyle -140A-140B=-2625000\)

\(\displaystyle 240A+140B=3500000\)

Now, adding these two equations, what do you find?

I edited it.

Hello! solving for A =8750 pounds

How about the prob 30?

 

[MarkFL]

Let's let $i$ be the number of pounds of the inferior tea needed and $S$ be the number of pounds of the superior. Thus, the total amount spent $T_S$ per pound on the two teas (in shillings) is:

\(\displaystyle T_S=\frac{3I+5S}{I+S}\)

The total amount he is to receive per pound is 3.4, and that is to be equal to 110% of the amount spent per pound:

\(\displaystyle \frac{11}{10}\cdot\frac{3I+5S}{I+S}=3.4\)

Can you now find $I$ in terms of $S$?

 

[NotaMathPerson]

Hello!

As I was studying your solution for prob 29, I came up with a method that uses only one variable

Let $x=$ income tax for all incomes below 100 pounds.
$18750-x=$ income tax for all incomes above 100 pounds

$\frac{240}{7}x+20(18750-x)=500000$

Solving for x I get $x=8750$ pounds which agrees with the first method.

And also is it okay if I don't call the unknown as "income tax raised on incomes below..." and just call it plainly as "income tax below..."? Because I think that part of the statement gave me trouble solving this problem not to mention english is not my primary language.

And for problem 30 yes I was able to come up for an equation for I. But I am still at a loss as to where you get 3.4? And the 110%?
Can you make me understand first what's really going on in the problem? Thank you and I really appreciate your help!

 

[MarkFL]

...And also is it okay if I don't call the unknown as "income tax raised on incomes below..." and just call it plainly as "income tax below..."? Because I think that part of the statement gave me trouble solving this problem not to mention english is not my primary language...

Yes, you can label your variables however you like, as long as it makes sense to you and each variable represents a relevant quantity.

And for problem 30 yes I was able to come up for an equation for I. But I am still at a loss as to where you get 3.4? And the 110%?
Can you make me understand first what's really going on in the problem? Thank you and I really appreciate your help!

There are 20 pence per shilling, so 3s 8d is 3.4s. The problem states that the seller is to gain 10% per pound sold, so we take the amount paid and increase if by 10% to get the selling price. To increase something by 10% we multiply it by 11/10.

 

[NotaMathPerson]

Yes, you can label your variables however you like, as long as it makes sense to you and each variable represents a relevant quantity.
There are 20 pence per shilling, so 3s 8d is 3.4s. The problem states that the seller is to gain 10% per pound sold, so we take the amount paid and increase if by 10% to get the selling price. To increase something by 10% we multiply it by 11/10.

Hello! Sorry, I can't still follow. Can you show me how it is done using only one variable? From there, I may be able to do it.

And also fo 3.4 do you mean 3.67? Because 8d is 8 pence and there is 12 pence in 1 shilling so 8/12 =2/3s.

Thanks!

 

[MarkFL]

Yes, sorry 12 pence in a shilling (there are 20 shillings per pound), so adjust the equation accordingly...the method is the same.

 

[NotaMathPerson]

Let's let $i$ be the number of pounds of the inferior tea needed and $S$ be the number of pounds of the superior. Thus, the total amount spent $T_S$ per pound on the two teas (in shillings) is:

\(\displaystyle T_S=\frac{3I+5S}{I+S}\)

The total amount he is to receive per pound is 3.4, and that is to be equal to 110% of the amount spent per pound:

\(\displaystyle \frac{11}{10}\cdot\frac{3I+5S}{I+S}=3.4\)

Can you now find $I$ in terms of $S$?

Hello.

Solving for I, $I=25s$ where will I plug this in?

 

[MarkFL]

Hello.

Solving for I, $I=25s$ where will I plug this in?

You should get:

\(\displaystyle I=5S\)

And this tells you that there needs to be 5 times as much of the inferior tea in the mixture as there is the superior tea. Thus, for every pound of superior tea used in the mix, 5 pounds of inferior tea much be used.

 

[NotaMathPerson]

Let's let $i$ be the number of pounds of the inferior tea needed and $S$ be the number of pounds of the superior. Thus, the total amount spent $T_S$ per pound on the two teas (in shillings) is:

\(\displaystyle T_S=\frac{3I+5S}{I+S}\)

The total amount he is to receive per pound is 3.4, and that is to be equal to 110% of the amount spent per pound:

\(\displaystyle \frac{11}{10}\cdot\frac{3I+5S}{I+S}=3.4\)

Can you now find $I$ in terms of $S$?

Why didnt you multiply the rhs by 1.1?

 

[MarkFL]

Why didnt you multiply the rhs by 1.1?

We are told what the selling price/lb. is to be, so why would we increase that by 10%? We are told that this given selling price/lb. will be 10% greater than the price paid/lb. So, we increase the price paid by 10%, and then set it equal to the given selling price.

 

[NotaMathPerson]

We are told what the selling price/lb. is to be, so why would we increase that by 10%? We are told that this given selling price/lb. will be 10% greater than the price paid/lb. So, we increase the price paid by 10%, and then set it equal to the given selling price.

Is 3s. 8d the price paid/lb of the mixture?

 

[MarkFL]

Is 3s. 8d the price paid/lb of the mixture?

That's the price per/lb. the mixture is to be sold.

 

[NotaMathPerson]

That's the price per/lb. the mixture is to be sold.

Hello! Now I see. The original cost per/lb of the mixture was not given. I thought it was 3s. 8d.
In 3s. 8d , included are the original cost per/lb of the mixture plus 10% of original cost per/lb. Am I correct on this?

And also, can you help me set up an equation where only one variable is present? Because this exercise is from a topic of problems leading with one vriable equation. Thanks a lot.

 

[MarkFL]

Hello! Now I see. The original cost per/lb of the mixture was not given. I thought it was 3s. 8d.
In 3s. 8d , included are the original cost per/lb of the mixture plus 10% of original cost per/lb. Am I correct on this?

And also, can you help me set up an equation where only one variable is present? Because this exercise is from a topic of problems leading with one vriable equation. Thanks a lot.

Yes, the price/lb. paid for the mixture must be computed by taking the toal amount paid, and dividing by the combined weight, resulting in a weighted average.

There are two unknowns here...that aren't related by a given total weight. Since we are being asked to find how many pounds of the inferior tea must be mixed with each pound of the superior (which we'll call $k$), we could go about it this way:

Let:

\(\displaystyle S\) = the weight of the superior tea purchased.

\(\displaystyle kS\) = the weight of the inferior tea purchased.

Then we may state the total price paid per pound as:

\(\displaystyle \frac{3kS+5S}{kS+S}=\frac{3k+5}{k+1}\)

Now we are down to 1 variable...

Then, marking this up by %10 and equating it to the stated selling price/lb.:

\(\displaystyle \frac{11}{10}\cdot\frac{3k+5}{k+1}=\frac{11}{3}\)

\(\displaystyle 9k+15=10k+10\)

\(\displaystyle k=5\)

And so we conclude 5 pounds of the inferior tea must be mixed with each pound of the superior.

 

[NotaMathPerson]

Thanks very much for your help!