Question #557ac

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Question #557ac

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Answer 1 · 2016-09-22T21:59:20

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

$The explanation makes this a lot longer than just doing the calculation$ $color(red)("The explanation makes this a lot longer than just doing the calculation")$

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.

$Preamble$ $color(blue)("Preamble")$

Let rice type 1 be $R_{1} \to selling at P50.00 per Kg$ $R_1->" selling at P50.00 per Kg"$

Let rice type 2 be $R_{2} \to selling at P15.00 per Kg$ $R_2->" selling at P15.00 per Kg"$

Let the final weight be $w$ $w$

The target blend will cost P35.00

If the mix were to be all $R_{1}$ $R_1$ the cost would be P50
If the mix were to be all $R_{2}$ $R_2$ the cost would be P15

If we have all $R_{1}$ $R_1$ then there is no $R_{2}$ $R_2$
If we have no $R_{1}$ $R_1$ then it is all $R_{2}$ $R_2$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Solving the question$ $color(blue)("Solving the question")$

Using the above information and plot the content of only $R_{1}$ $R_1$ against cost of the mix and we have our model.

Tony B

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

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

$\Rightarrow \frac{100}{15} = \frac{x}{8}$ $=>100/15=x/8$

$x = \frac{8 \times 100}{15} = 53 \frac{1}{3}$ $x=(8xx100)/15 = 53 1/3$ but this is percent

$\Rightarrow R_{1} = 53 \frac{1}{3} % of the mix w$ $=> R_1 =53 1/3%" of the mix "w$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If $R_{1} \to 53 \frac{1}{3} %$ $R_1->53 1/3%$ then $R_{2} \to (100 - 53 \frac{1}{3}) %$ $R_2->(100-53 1/3)%$

$\Rightarrow R_{2} = 46 \frac{2}{3} %$ $=> R_2=46 2/3%$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

But we are told that the quantity of $R_{2}$ $R_2$ is fixed at 15 Kg implying that the whole weight $(w)$ $(w)$ is such that:

$R_{2} = \frac{46 \frac{2}{3}}{100} \times w = 15$ $R_2=(46 2/3)/100 xx w =15$

$\Rightarrow w = 15 \times \frac{100}{46 \frac{2}{3}} = 32 \frac{1}{7} Kg$ $=> w=15xx 100/(46 2/3) = 32 1/7" Kg"$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$R_{1} \to P 50.00 rice$ $R_1-> P50.00 "rice"$

Thus the weight of $R_{1} = w - 15 = 32 \frac{1}{7} - 15 = 17 \frac{1}{7} K g$ $R_1 = w-15" "=" "32 1/7 -15" " =" " 17 1/7 Kg$

Answer 2 · 2016-09-24T09:45:09

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

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, $R_{1}$ $R_1$ the rest will be the second rice $R_{2}$ $R_2$

Let the amount of $R_{1}$ $R_1$ be x.

Then the amount of $R_{2}$ $R_2$ is $1 - x$ $1-x$

The cost of $R_{1}$ $R_1$ is $50 x$ $50x$

The cost of $R_{2}$ $R_2$ is $35 (1 - x)$ $35(1-x)$

Together the value of the mixture is $43$ $43$

$50 x + 35 (1 - x) = 43$ $50x+ 35(1-x) = 43$

$50 x + 35 - 35 x = 43$ $50x+35-35x = 43$

$15 x = 43 - 35$ $15x = 43-35$

$15 x = 8$ $15x = 8$

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

$:. 1-x = 7/15$

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:

$(50+35)/2 = 42.50$

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

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