# simultaneous problem solving? year 10 extension work

• Jun 29th 2008, 11:23 PM
white
simultaneous problem solving? year 10 extension work
eeek (Doh) these questions are confusing . year 10 extension work the teacher gave me.. need help. thank you

q.1an investor received $1400 interest per annum from a sum of money , with part of it invested at 10% and the remainder at 7% simple interest. This investor found that if she interchanged the amounts she had invested she could increase her return by$90 per annum. calculate the total amount invested.

q.2a tea wholesaler blends together three types of tea that normally sell for $10,$11,$12 per kilogram so as to obtain 100 kilograms f tea worth$11.20 per kilogram. If the same amounts of the two higher priced teas are used, calculate how much of each type must be used in the blend.
• Jun 30th 2008, 05:10 AM
mr fantastic
Quote:

Originally Posted by white
[snip]
q.2a tea wholesaler blends together three types of tea that normally sell for $10,$11,$12 per kilogram so as to obtain 100 kilograms f tea worth$11.20 per kilogram. If the same amounts of the two higher priced teas are used, calculate how much of each type must be used in the blend.

I have a hunch Soroban is doing q1.

I'll let you contemplate where the following two equations come from that you must solve simultaneously:

100 = x + 2y .... (1)

1120 = 10x + 23y .... (2)

I get 40 kg each of the two higher priced teas and 20 kg of the lowest priced tea in the 100 kg of tea. Therefore .....
• Jun 30th 2008, 05:35 AM
Soroban
Hello, white!

I'll set these up for you . . .

Quote:

1. An investor received $1400 interest per annum from a sum of money with part of it invested at 10% and the remainder at 7% simple interest. This investor found that if she interchanged the amounts she had invested, she could increase her return by$90 per annum.
Calculate the total amount invested.

Let: . $\begin{array}{ccc}x &=& \text{amount at 10\%} \\ y &=& \text{amount at 7\%} \end{array}$

$x$ dollars at 10% produces $0.10x$ dollars in interest.
$y$ dollars at 7% produces $0.07y$ dollars in interest.

The total interest is $1400: . $0.10x + 0.07y \:=\:1400\;\;{\color{blue}[1]}$ If the amounts are interchanged, the total interest is$1490.

. . $0.07x + 0.10y \:=\:1490\;\;{\color{blue}[2]}$

Multiply [1] and [2] by 100: . $\begin{array}{ccc}10x + 7y &=& 140,000 \\ 7x + 10y &=& 149,000 \end{array}$

Quote:

2. A tea wholesaler blends together three types of tea that normally sell
for $10,$11, $12 per kg, as to obtain 100 kilograms of tea worth$11.20 per kg.
If the same amounts of the two higher priced teas are used,
calculate how much of each type must be used in the blend.

Let: . $\begin{array}{ccc}x &=& \text{amount of \10 tea} \\ y &=& \text{amount of \11 tea} \\ y &=& \text{amount of \12 tea}\end{array}$

The total amount of tea is 100 kg: . $x + 2y \:=\:100\;\;{\color{blue}[1]}$

$x$ kg of $10 tea has a value of: $10x$ dollars. $y$ kg of$11 tea has a value of: $11y$ dollars.
$y$ kg of $12 tea has a value of: $12y$ dollrs. . . The total value is: . $10x + 23y$ dollars. But the mixture is 100 kg worth$11.20 per kg.
. . Its total value is: . $100 \times 11.20 \:=\:1120$

There is our second equation: . $10x + 23y \:=\:1120\;\;{\color{blue}[2]}$