Maths Olympiad Prep

Library / /6 of 30

, 2022

Algebra Difficulty 2.0 Junior Prove it Canada

Alice and Bello contributed to the cost of starting a new
business. The ratio of Alice’s contribution to Bello’s contribution
was 3:83:8.

If the cost of starting the new business
was $9240, what was Bello’s contribution to this starting cost?
Alice and Bello divided up all profits in
the first year of the business in the same ratio, 3:83:8. Alice’s share of the first year’s
total profit was $1881\$1881. What was
the total profit of the business for the first year?
In the second year, the business was
changed so the share of that year’s profits for Alice and Bello was in
the ratio of 3:(8+x)3:(8+x). If the profit
for the second year was $6400\$6400 and
Bello’s share of that profit was $5440\$5440, determine the value of xx.

Solution

The ratio of Alice’s contribution to Bello’s contribution was
3:83:8.

This means that for every $3+$8=$11\$3+\$8=\$11 contributed to starting the
business, Bello contributed 8,andsoBellocontributed8, and so Bello contributed 811$\dfrac{8}{11}\$ of the cost of starting the
new business.

If the cost of starting the new business was $9240, then Bello’s
contribution to this cost was 811×$9240=$6720\dfrac{8}{11}\times\$9240=\$6720.
The ratio of Alice’s share of the first year profits to Bello’s
share was 3:83:8.

This means that for every $3+$8=$11\$3+\$8=\$11 of profit in the first year,
Alice’s share was $3.

Let the first year profit be $P\$P.

Since Alice’s share of the first year profit was 1881,then1881, then 311×\dfrac{3}{11}\times P=1881or or P=1881×113P=\dfrac{1881\times11}{3},andso, and so P=206913=6897$.P=\dfrac{20\,691}{3}=6897\$.

The total profit of the business for the first year was $6897.
The ratio of Alice’s share of the second year profits to Bello’s
share was 3:(8+x)3:(8+x).

This means that for every $3+$(8+x)=$(11+x)\$3+\$(8+x)=\$(11+x) of profit in the
second year, Bello’s share was (8+x)$.

Since Bello’s share of the second year profit was $5440, and the profit
that year was 6400,then6400, then (8+x)(11+x)×\dfrac{(8+x)}{(11+x)}\times
6400=5440$.

Solving this equation, we get ENV0(8+x)(11+x)×6400=54406400(8+x)=5440(11+x)20(8+x)=17(11+x)(dividing both sides by 320)160+20x=187+17x3x=27\begin{align*} \dfrac{(8+x)}{(11+x)}\times 6400&=5440\\ 6400(8+x) &= 5440(11+x)\\ 20(8+x) &= 17(11+x) & \text{(dividing both sides by 320)}\\ 160+20x&=187+17x\\ 3x&=27\end{align*} and so x=9x=9.

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.