# Question 557ac

Sep 22, 2016

$17 \frac{1}{7} \text{ Kg of the P50 rice should be added to 15 Kg of P35 rice}$

$\textcolor{red}{\text{The explanation makes this a lot longer than just doing the calculation}}$

#### Explanation:

There are several ways of approaching this problem. I tend to use a method related to a straight line graph equation but just using the gradient. This turns out to be ratios.

$\textcolor{b l u e}{\text{Preamble}}$

Let rice type 1 be ${R}_{1} \to \text{ selling at P50.00 per Kg}$

Let rice type 2 be ${R}_{2} \to \text{ selling at P15.00 per Kg}$

Let the final weight be $w$

The target blend will cost P35.00

If the mix were to be all ${R}_{1}$ the cost would be P50
If the mix were to be all ${R}_{2}$ the cost would be P15

If we have all ${R}_{1}$ then there is no ${R}_{2}$
If we have no ${R}_{1}$ then it is all ${R}_{2}$

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$\textcolor{b l u e}{\text{Solving the question}}$

Using the above information and plot the content of only ${R}_{1}$ against cost of the mix and we have our model.

Using ratio $\to$ the gradient of part is the gradient of all

$\left(\text{change in along")/("change in up}\right) \to \frac{100}{50 - 35} \equiv \frac{x}{43 - 35}$

$\implies \frac{100}{15} = \frac{x}{8}$

$x = \frac{8 \times 100}{15} = 53 \frac{1}{3}$ but this is percent

=> R_1 =53 1/3%" of the mix "w

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If R_1->53 1/3% then R_2->(100-53 1/3)%

=> R_2=46 2/3%#

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But we are told that the quantity of ${R}_{2}$ is fixed at 15 Kg implying that the whole weight $\left(w\right)$ is such that:

${R}_{2} = \frac{46 \frac{2}{3}}{100} \times w = 15$

$\implies w = 15 \times \frac{100}{46 \frac{2}{3}} = 32 \frac{1}{7} \text{ Kg}$
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${R}_{1} \to P 50.00 \text{rice}$

Thus the weight of ${R}_{1} = w - 15 \text{ "=" "32 1/7 -15" " =" } 17 \frac{1}{7} K g$

Sep 24, 2016

The two types of rice should be mixed in the ratio of $8 : 7$

#### Explanation:

We are only concerned with the price per Kilo, not an actual number of Kilograms.

Let's consider a whole kilo, split into two parts:

Whatever part is the first rice, ${R}_{1}$ the rest will be the second rice ${R}_{2}$

Let the amount of ${R}_{1}$ be x.

Then the amount of ${R}_{2}$ is $1 - x$

The cost of ${R}_{1}$ is $50 x$

The cost of ${R}_{2}$ is $35 \left(1 - x\right)$

Together the value of the mixture is $43$

$50 x + 35 \left(1 - x\right) = 43$

$50 x + 35 - 35 x = 43$

$15 x = 43 - 35$

$15 x = 8$

$x = \frac{8}{15}$

$\therefore 1 - x = \frac{7}{15}$

The two types of rice should be mixed in the ratio of:

Expensive rice : cheaper rice
$8 : 7$

As a check - average the two prices:

$\frac{50 + 35}{2} = 42.50$

As the required price of P43 is slightly more than P42.50, a ratio of $8 : 7$ seems reasonable,