Ratio Problems: Free Response

Write each count with the shared multiplier: cashews are $2k$, almonds are $3k$, and pretzels are $4k$. The extra fact is the total, so add the three expressions and set the sum equal to $63$:

$2k + 3k + 4k = 63.$

Combine the like terms and solve:

$9k = 63, \qquad k = 7.$

Each part is worth $7$ pieces. Multiply back through each term:

$2k = 14, \qquad 3k = 21, \qquad 4k = 28.$

So the batch has $14$ cashews, $21$ almonds, and $28$ pretzels.

Add the three counts to check the total:

$14 + 21 + 28 = 63,$

which matches. Divide every count by $7$ to check the ratio:

$14 : 21 : 28 = 2 : 3 : 4.$

Both checks pass. The reason one shared $k$ is required, rather than a separate multiplier for each term, is that the ratio is a statement about how the three quantities compare to one another. If cashews were scaled by one number and almonds by a different one, the new counts would no longer stand in the ratio $2 : 3 : 4$ at all, so the very thing the problem asserted about them would stop being true.

Write the two numbers as $3k$ and $5k$, and set their sum equal to the total:

$3k + 5k = 40.$

Combine and solve:

$8k = 40, \qquad k = 5.$

Multiply back through each term:

$3k = 15, \qquad 5k = 25.$

So the two actual amounts are $15$ and $25$, and $15 + 25 = 40$ checks.

One breaking case is enough to refute a claim about EVERY total. Part A supplies it: with ratio $3 : 5$ and total $40$,

$3k = 15, \qquad 5k = 25,$

so the actual amounts are $15$ and $25$, while the claim asserted they would be $3$ and $5$. Since $15 \neq 3$ and $25 \neq 5$, the claim fails on this case, and one failing case is all a universal claim needs to be refuted.

Be precise about the scope. The counterexample kills the claim that the ratio numbers ARE the amounts; it does not touch the ratio itself, since $15 : 25$ still reduces to $3 : 5$ by dividing both by $5$. The only total for which the claim would hold is $a + b$ itself, the one case where $k = 1$.

Write both shares with the one multiplier: the senior member gets $5k$ dollars and the junior member gets $3k$ dollars. The extra fact is the difference of $48$ dollars, and the senior share is the larger one, so

$5k - 3k = 48.$

Combine the like terms and solve:

$5k - 3k = 48, \qquad 2k = 48, \qquad k = 24.$

Multiply back through each term:

$5k = 120, \qquad 3k = 72.$

So the senior member gets $120$ dollars and the junior member gets $72$ dollars, and $120 - 72 = 48$ checks.

Add the two shares:

$120 + 72 = 192.$

Now split $192$ dollars fresh, in the ratio $5 : 3$: the parts number $5 + 3 = 8$, so

$8k = 192, \qquad k = 24,$

giving $5k = 120$ and $3k = 72$. These are the same two shares found from the difference alone, which is expected: a ratio together with either its total or the difference between two of its terms pins down the same single multiplier $k$, so either extra fact leads to the same amounts.

The shared quantity is rock, appearing as $4$ in the first ratio and $6$ in the second, so scale each ratio until rock reaches their least common multiple, $12$.

Scale the first ratio, pop to rock, by $3$:

$\text{pop} : \text{rock} = 3 : 4 = 9 : 12.$

Scale the second ratio, rock to jazz, by $2$:

$\text{rock} : \text{jazz} = 6 : 5 = 12 : 10.$

Both now agree that rock is $12$, so line them up:

$\text{pop} : \text{rock} : \text{jazz} = 9 : 12 : 10.$

The three scaled terms make $9 + 12 + 10 = 31$ parts, and the total is $217$ songs:

$31k = 217, \qquad k = 7.$

Multiply back through the jazz term:

$10k = 70.$

So there are $70$ jazz songs, alongside $9k = 63$ pop songs and $12k = 84$ rock songs, and $63 + 84 + 70 = 217$ checks.

To make rock read $24$, scale the first ratio (pop to rock $= 3 : 4$) by $6$ and the second (rock to jazz $= 6 : 5$) by $4$:

$\text{pop} : \text{rock} = 3 : 4 = 18 : 24, \qquad \text{rock} : \text{jazz} = 6 : 5 = 24 : 20.$

Lined up, this gives pop : rock : jazz $= 18 : 24 : 20$. Divide every term by $2$:

$18 : 24 : 20 = 9 : 12 : 10,$

the same ratio part A found using $12$. The two merges do not disagree, they agree once reduced, so the least common multiple was a convenient choice for keeping the numbers small, not a requirement for the merge to work.

Motorcycles never change, so they stay $2k$ throughout. Cars start at $5k$ and drop by $12$, becoming $5k - 12$. The new ratio of cars to motorcycles is stated as $3 : 2$, which as a proportion says

$\frac{5k - 12}{2k} = \frac{3}{2}.$

Cross-multiply to clear the fraction:

$2(5k - 12) = 3(2k).$

Expand and solve:

$10k - 24 = 6k, \qquad 4k = 24, \qquad k = 6.$

So originally there were $5k = 30$ cars and $2k = 12$ motorcycles. After $12$ cars leave, the lot holds $30 - 12 = 18$ cars and still $12$ motorcycles, and $18 : 12$ reduces to $3 : 2$ as stated.

The ratio right now is cars divided by the fixed $12$ motorcycles: $\frac{18}{12} = 1.5$. If, say, $6$ more cars left, the ratio would become $\frac{12}{12} = 1$; if $12$ more left, $\frac{6}{12} = 0.5$. Every additional departure lowers the numerator while the denominator, $12$, never changes, so the value of the ratio only ever falls from here.

$\frac{18}{12} = 1.5, \qquad \frac{12}{12} = 1, \qquad \frac{6}{12} = 0.5.$

Climbing to $7 : 2 = 3.5$ would need the ratio to more than double from its current $1.5$, and nothing in this situation can raise it: cars never increase and motorcycles never move. So the friend's suggestion is impossible; more departures push the ratio further down, not up.