# Math Help - Bronze is an alloy of copper and zinic in the ratio of 8:3 by mass.

1. ## Bronze is an alloy of copper and zinic in the ratio of 8:3 by mass.

Bronze is an alloy of copper and zinc in the ratio of 8:3 by mass.

What mass of copper and zinc would be used to manufacture a bronze figure of 3.63 kilogrammes???

(i) $\frac {5}{8} + \frac{9}{16} - \frac{3}{4}$

(ii) $\frac {7}{10} - \frac{3}{8} / \frac{3}{4} +\frac {6}{5}$

Steven

2. Originally Posted by Steven777
Bronze is an alloy of copper and zinc in the ratio of 8:3 by mass.

What mass of copper and zinc would be used to manufacture a bronze figure of 3.63 kilogrammes???
If you add the ratio component $8+3 = 11$, you can see that the total part to be considered is $11$ of which $8$ part is of copper and $3$ part is zinc.

Thus $3.63 \div 11 = 0.33$kg is the ratio component of one part.

Therefore the part for copper is $8$ times the single part $0.33 \times 8 = ...$ and the part for zinc is $3$ times the single part $0.33 \times 3=...$

In the end be sure to add both part to see if you get the original composition, does it add up to $3.63$ kg?

Originally Posted by Steven777

(i) $\frac {5}{8} + \frac{9}{16} - \frac{3}{4}$

(ii) $\frac {7}{10} - \frac{3}{8} / \frac{3}{4} +\frac {6}{5}$

Steven
Make the denominator a common factor by multiplying numerator and denominator by a common values. Simplify the final answer.

(i) $\frac{(5)(16)(4)}{(8)(16)(4)} + \frac{(9)(8)(4)}{(16)(8)(4)} - \frac{(3)(16)(8)}{(4)(16)(8)}$

For the second one, make the division of fraction into multiplication of fraction. To multiply fraction, the numerators are multiplied and the denominators are multiplied. Then do as the other stages (Same steps).

(ii) $\frac {7}{10} - \frac{\frac{3}{8}}{\frac{3}{4}} +\frac {6}{5} = \frac {7}{10} - \left(\frac{3}{8}\right)\left(\frac{4}{3}\right) +\frac {6}{5} = \frac {7}{10} - \frac{12}{24} +\frac {6}{5} = ...$

If you find that the values are too high, then you can simply the fraction before proceeding with the arithmetics.

3. Hello, Steven!

Bronze is an alloy of copper and zinc in the ratio of 8:3 by mass.
What mass of copper and zinc would be used to manufacture a bronze figure of 3.63 kg?

We are given . . . . $\text{copper : zinc} \:=\:8:3 \quad\Rightarrow\quad C:Z \;=\;8a:3a$

We have: . $8a + 3a \:=\:3.63 \quad\Rightarrow\quad 11a \:=\:3.63 \quad\Rightarrow\quad a \:=\:0.33$

Therefore: . $\begin{array}{ccccccccc}C &=& 8(0.33) &=& 2.64\text{ kg} \\ Z &=& 3(0.33) &=& 0.99\text{ kg} \end{array}$

$(1)\;\;\frac{5}{8} + \frac{9}{16} - \frac{3}{4}$
. . ${\color{blue}\frac{2}{2}}\cdot\frac{5}{8} + \frac{9}{16} - {\color{blue}\frac{4}{4}}\cdot\frac{3}{4} \;=\;\frac{10}{16} + \frac{9}{16} -\frac{12}{16} \;=\;\frac{10 + 9 - 12}{16} \;=\;\frac{7}{16}$
$(2)\;\;\frac{\frac {7}{10} - \frac{3}{8}}{\frac{3}{4} +\frac {6}{5}}$
Multiply by $\tfrac{40}{40}\!:\;\;\frac{40\left(\frac{7}{10} - \frac{3}{8}\right)} {40\left(\frac{3}{4} + \frac{6}{5}\right)} \;=\;\frac{28-15}{30+48} \;=\;\frac{13}{78} \;=\;\frac{1}{6}$