Percent Problems: Free Response

Convert the percent to a decimal rate, $24\% = 0.24$, and substitute into $p = rw$ with $w = 175$:

$p = 0.24 \times 175 = 42.$

So $24\%$ of $175$ is $42$.

The rate and the part are known, and the whole is unknown. Substitute into $p = rw$:

$51 = 0.12\,w.$

Divide both sides by the rate:

$w = \frac{51}{0.12} = 425.$

Check it forward: $12\%$ of $425$ is $0.12 \times 425 = 51$, the part given.

In $p = rw$, whichever letter is missing is isolated by ordinary algebra, and that choice is what decides the operation.

$p = rw \;\Longrightarrow\; p = rw \text{ (direct)}, \qquad w = \frac{p}{r} \text{ (divide)}.$

In part A, $w$ and $r$ were both known and $p$ was missing, so the equation was already solved for $p$: just multiply. In part B, $p$ and $r$ were known and $w$ was missing, so $w$ sat multiplied by $r$ inside $rw = p$, and isolating $w$ meant dividing both sides by $r$. The rule is not about the story in the problem, it is about which letter the algebra has left alone: whenever the missing letter is multiplied by a known one, undo that multiplication with a division.

A $40\%$ markup multiplies the wholesale cost by the factor $1 + 0.40 = 1.40$ to reach the shelf price, so

$1.40\,c = 294.$

Divide both sides by the factor:

$c = \frac{294}{1.40} = 210.$

The wholesale cost was $210$ dollars.

The markup factor is again $1.40$. With shelf price $161$ and wholesale cost $c$:

$1.40\,c = 161 \;\Longrightarrow\; c = \frac{161}{1.40} = 115.$

The lamp's wholesale cost was $115$ dollars.

The shelf price was produced by one multiplication,

$\text{shelf} = 1.40\,c.$

To recover $c$, undo that single multiplication with the matching division,

$c = \frac{\text{shelf}}{1.40},$

exactly as parts A and B did. Subtracting $40\%$ of the shelf price instead treats the shelf price, rather than the wholesale cost, as the $100\%$ base. That is not what a markup means: the $40\%$ was always a percent of the wholesale cost, so reversing it has to divide the shelf price by the factor built from that same original base, not carve a percent off the result.

A percent statement always names its base, the number the percent is a fraction of, and "$30\%$ off" means $30\%$ of the original price is removed, so

$\text{sale price} = (1 - 0.30) \times \text{original},$

not the other way around. Priya took $30\%$ of $56$, the amount she paid after the discount, so her $16.80$ is $30\%$ of the wrong number entirely. The original price is the base the $30\%$ was always measured against, and it is exactly the quantity she does not yet know, so it cannot also be the number she takes $30\%$ of.

A $30\%$ discount multiplies the original price by the factor $1 - 0.30 = 0.70$, so with sale price $56$ and original $P$:

$0.70\,P = 56.$

Divide by the factor:

$P = \frac{56}{0.70} = 80.$

Check it forward: $30\%$ of $80$ is $24$, and $80 - 24 = 56$, the price Priya actually paid.

Compare your original price from part B to Priya's number: the gap is

$80 - 72.80 = 7.20$

dollars, so Priya's answer is too low. This is not a coincidence of these particular numbers. A percent is always a fraction of its base, so a bigger base gives a bigger dollar amount for the same rate. Since the $56$ dollar sale price is smaller than the true $80$ dollar original, $30\%$ of the sale price is necessarily smaller than $30\%$ of the original would have been. Adding back a too-small amount to $56$ can only land short of $80$, never past it, so the direction of Priya's error is forced by which of the two bases is larger, not by any arithmetic slip.

Check the increase first:

$120 \times (1 + 0.25) = 120 \times 1.25 = 150.$

Now find the decrease factor that reverses it. Let $r$ be the needed rate:

$(1 - r) \times 150 = 120.$

Divide by $150$:

$1 - r = \frac{120}{150} = 0.8,$

so $r = 0.20$. A $20\%$ decrease on $150$ returns it to $120$.

The increase: $80 \times 1.25 = 100$. To reverse it, solve

$(1 - r) \times 100 = 80 \;\Longrightarrow\; 1 - r = \frac{80}{100} = 0.8 \;\Longrightarrow\; r = 0.20.$

The needed decrease is again $20\%$, the same rate as part A even though the starting quantity is different.

A $25\%$ increase always multiplies the starting quantity, call it $x$, by $1.25$. Reversing it means multiplying that increased amount by some factor $1 - r$ to land back on $x$:

$(1 - r)(1.25\,x) = x.$

For any positive $x$, divide both sides by $x$. That division is the whole move, and it is legal exactly because $x \neq 0$. What is left mentions the two factors and nothing else:

$(1 - r)(1.25) = 1 \;\Longrightarrow\; 1 - r = \frac{1}{1.25} = 0.8 \;\Longrightarrow\; r = 0.20.$

The starting quantity has vanished from the equation, so the answer cannot depend on it: $1.25$ and $1 - r$ describe the relationship between the two amounts, not either amount itself. Scaling the start up or down scales the increased amount and the target by the identical factor, so their ratio, which is exactly what $1 - r$ computes, stays fixed at $0.8$. That is why parts A and B, with starting quantities of $120$ and $80$, both needed the same $20\%$. The one starting quantity this argument cannot reach is $0$: there is nothing to divide out, $0$ rises to $0$, and every decrease brings it straight back, so no single rate is forced.

Use $I = Prt$ for each.

Investor A, with $P = 500$, $r = 0.06$, $t = 4$:

$I = 500 \times 0.06 \times 4 = 120.$

Investor B, with $P = 800$, $r = 0.04$, $t = 3$:

$I = 800 \times 0.04 \times 3 = 96.$

Compare the three ingredients across the two accounts.

$\frac{800}{500} = 1.6, \qquad \frac{0.06}{0.04} = 1.5, \qquad \frac{4}{3} \approx 1.33.$

Principal favors Investor B by a factor of $1.6$. Rate favors Investor A by a factor of $1.5$. Time also favors Investor A, $4$ years against $3$, a factor of $\frac{4}{3}$. So B holds the single biggest edge, and neither of A's edges beats it on its own. Since $I = Prt$ multiplies all three together, though, A's two edges multiply as well:

$1.5 \times \frac{4}{3} = 2.0,$

and $2.0$ beats $1.6$. That is the whole story of the gap: $\frac{2.0}{1.6} = 1.25$, which is exactly $\frac{120}{96}$, the ratio of the two interest totals. A ends up ahead despite lending less money because two multiplied edges outrun one larger edge.

Solve $I = Prt$ for $r$, using your interest total from part A for Investor A, with $P = 800$ and $t = 3$:

$800 \times r \times 3 = 120 \;\Longrightarrow\; 2400\,r = 120 \;\Longrightarrow\; r = \frac{120}{2400} = 0.05.$

So Investor B would need a $5\%$ rate, higher than the $4\%$ actually used. That direction makes sense: with the same larger principal and the same shorter time, only a bigger rate can make up the gap that part B's comparison identified.