Ratio & Proportion practice questions — GCSE Maths

Test the product $xy$ first: $$2\cdot 18=36,3\cdot 8=24,6\cdot 2=12.$$ These are not equal, so $y$ is not inversely proportional to $x$ itself.

Now test $x^{2}y$: $$2^2\cdot 18=72,3^2\cdot 8=72,6^2\cdot 2=72.$$ This is constant across all three pairs, so $x^{2}y=72$ for every measurement, which means $y$ is inversely proportional to $x^2$ with constant of proportionality $72$: $$y=\frac{72}{x^2}\mathrm{.}$$

The seniors term is $4$ in the first ratio and $6$ in the second, and $\mathrm{lcm}(4,6)=12$.

Scale each ratio so the seniors term becomes $12$: $$3:4=9:12,6:5=12:10.$$

So juniors : seniors : veterans $=9:12:10$.

The $9$ parts of juniors are $90$ students, so one part is $10$ students. The veterans occupy $10$ parts: $$10\cdot 10=100.$$

Write $y={\displaystyle \frac{k}{x^2}}$, so that $k=y{x}^2$. The given pair $x=2$, $y=12$ fixes the constant: $$k=12\cdot 2^2=48.$$ The relationship is therefore $y={\displaystyle \frac{48}{x^2}}$. Setting $y=3$: $$3=\frac{48}{x^2}⟹x^2=\frac{48}{3}=16.$$ Since $x>0$, only the positive root is admissible: $$x=4.$$

Write the starting amounts as $7k$ and $2k$, so the total is $9k$.

After the transfer the total is still $9k$, and $4:5$ splits it into $4+5=9$ parts of the same size $k$, so Ali ends with $4k$ and Ben with $5k$.

Ali's loss is exactly the $15$ stickers he handed over: $$7k-4k=15\Rightarrow 3k=15\Rightarrow k=5.$$

Ben started with $2k$: $$2\cdot 5=10.$$

Let one part have mass $k$ grams, so copper $=7k$, zinc $=5k$ and tin $=3k$.

The copper is $120$ g heavier than the tin: $$7k-3k=120\Rightarrow 4k=120\Rightarrow k=30.$$

The whole bar is $7+5+3=15$ parts, so its mass is $15k$: $$15\cdot 30=450.$$

The two statements combine into a single relationship with one constant: $$T=\frac{k{v}^2}{w}\mathrm{.}$$ After the changes, $v$ becomes $1.2v$ and $w$ becomes $0.9w$. Substituting these into the same formula: $$T_{\text{new}}=\frac{k(1.2v{)}^2}{0.9w}=\frac{1.44}{0.9}\cdot \frac{k{v}^2}{w}\mathrm{.}$$ The multiplier is $$\frac{1.44}{0.9}=1.6,$$ so $T_{\text{new}}=1.6T$. A multiplier of $1.6$ means the new value is $160\mathrm{\%}$ of the old one, and the increase is the excess over the original: $$\text{increase}=60\mathrm{\%}\mathrm{.}$$