# algebra word problem

• Dec 12th 2008, 07:07 AM
777
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.
• Dec 13th 2008, 07:07 AM
Prove It
Quote:

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.

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

The area of the circles are as follows...

$A_1 = \pi r_1^2$

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

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

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

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

So the area between the two outer circles is therefore

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

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

$1:\frac{5}{3}$ or $3:5$.
• Dec 13th 2008, 08:43 AM
Soroban
Hello, 777!

We can talk our way through this one . . .

Quote:

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

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

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

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

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

• Dec 13th 2008, 09:48 AM
Soroban
Hello, 777!

#5 is a tricky one . . .

Quote:

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

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

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

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

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

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

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

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