1. ## algebra word problem

Please can anyone solve the following questions and show working pls.?

4. The radii of three concentric circles are in the ratio 1: 2: 3. Find the ratio of the area between the two inner circles to that between the two outer circles.

5. A dealer allows a discount of 12% on the marked price. How much per cent above the cost price, must he mark his goods so as to make a profit of 10%?

6. From a 5 litres of 20% solution of alcohol in water, 2 litres of solution is taken out and 2 litres of water is added to it. Find the strength of alcohol in the new solution.

2. Originally Posted by 777
Please can anyone solve the following questions and show working pls.?

4. The radii of three concentric circles are in the ratio 1: 2: 3. Find the ratio of the area between the two inner circles to that between the two outer circles.

5. A dealer allows a discount of 12% on the marked price. How much per cent above the cost price, must he mark his goods so as to make a profit of 10%?

6. From a 5 litres of 20% solution of alcohol in water, 2 litres of solution is taken out and 2 litres of water is added to it. Find the strength of alcohol in the new solution.

$\displaystyle r_2 = 2r_1$ and $\displaystyle r_3 = 3r_1$

The area of the circles are as follows...

$\displaystyle A_1 = \pi r_1^2$

$\displaystyle A_2 = \pi r_2^2$
$\displaystyle = \pi(2r_1)^2$
$\displaystyle = 4\pi r_1^2$
$\displaystyle = 4A_1$

The area between the two inner circles (which is known as a toros) is therefore

$\displaystyle T_1 = A_2 - A_1$
$\displaystyle = 4A_1 - A_1$
$\displaystyle = 3A_1$.

Similarly $\displaystyle A_3 = \pi r_3^2$
$\displaystyle = \pi(3r_1)^2$
$\displaystyle = 9\pi r_1^2$
$\displaystyle = 9A_1$.

So the area between the two outer circles is therefore

$\displaystyle T_2 = A_3 - A_2$
$\displaystyle = 9A_1 - 4A_1$
$\displaystyle = 5A_1$
$\displaystyle = \frac{5}{3}T_1$.

So the ratio of the inner toros to the outer toros is therefore

$\displaystyle 1:\frac{5}{3}$ or $\displaystyle 3:5$.

3. Hello, 777!

We can talk our way through this one . . .

6. From a 5 litres of 20% solution of alcohol in water,
2 litres of solution is taken out and 2 litres of water is added to it.
Find the strength of alcohol in the new solution.
Since water is being removed and replaced,
. . we will consider the amount of water at each stage.

. . It contains: .$\displaystyle 0.80 \times 5 \:=\:4.4$ liters of water.

Two liters of solution is removed.
. . This contains: .$\displaystyle 0.80 \times 2 \:=\:1.76$ liters of water.
So there are: .$\displaystyle 4.4 - 1.76 \:=\:2.64$ liters of water left.

Then two liters of pure water is added.
. . So there are: .$\displaystyle 2.64 + 2 \:=\:4.64$ liters of water.

We have 5 liters of new solution of which 4.64 liters is water.
. . It is: .$\displaystyle \frac{4.64}{5} \:=\:0.928 \:=\:92.8\%$ water.

Therefore, the new solution is: .$\displaystyle 100\% - 92.8\% \:=\:7.2\%$ alcohol.

4. Hello, 777!

#5 is a tricky one . . .

5. A dealer allows a discount of 12% on the selling price.
What percent above the cost must he mark his goods so as to make a profit of 10%?

We have: .$\displaystyle \text{(Cost)} + \text{(Markup)} \:=\:\text{(Selling price)} \quad\Rightarrow\quad C + M \:=\:S$ .[1]

Suppose Markup is $\displaystyle p$ percent of Cost: .$\displaystyle M \:=\:pC$

Then [1] becomes: .$\displaystyle S \:=\:C + pC \:=\:(1+p)C$

He offers a 12% discount, so he sells it for: .$\displaystyle 0.88(1+p)C$ .[2]

This amount is to give him a profit which is 10% of cost.
. . That is, the sale price should be: .$\displaystyle C + 0.10C \:=\:1.1C$ .[3]

Equate [2] and [3]: .$\displaystyle 0.88(1+p)C \;=\;1.1C$

Divide by $\displaystyle 0.88C\!:\;\;1+p \:=\:1.25 \quad\Rightarrow\quad p \:=\:0.25$

He should mark it up 25% of cost
. . for the sale price to bring a 10% profit.