# Thread: simultaneous problem solving? year 10 extension work

1. ## simultaneous problem solving? year 10 extension work

eeek 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.

2. 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 .....

3. Hello, white!

I'll set these up for you . . .

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: . $\begin{array}{ccc}10x + 7y &=& 140,000 \\ 7x + 10y &=& 149,000 \end{array}$

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: . $10x + 23y \:=\:1120\;\;{\color{blue}[2]}$