4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems

4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems

These problems supply a large share of the word problems on any senior-high entrance test, and they all run on one idea: two quantities are locked together by a multiplier, and the item hides the multiplier somewhere in the sentence. Write the multiplier down explicitly and the arithmetic becomes short enough to finish without a calculator.

1. Ratios

A ratio compares two quantities of the same kind and is written $a : b$, $\frac{a}{b}$, or "$a$ to $b$". Ratios carry no units when the quantities match, and they simplify like fractions: $18 : 24 = 3 : 4$.

Partitioning a total. When a total $T$ is split in the ratio $a : b : c$, count the parts first: $N = a + b + c$. Each share is its own number of parts over $N$, times $T$:

$\text{share of } a = \frac{a}{a+b+c} \times T$

Three cousins divide ₱10,800 in the ratio $2 : 3 : 4$. There are $9$ parts, so one part is ₱1,200 and the shares are ₱2,400, ₱3,600 and ₱4,800 — which add back to ₱10,800, the check you should always run.

Ratios given with a difference. If two ropes are in the ratio $3 : 5$ and differ by 12 cm, the difference is $5 - 3 = 2$ parts, so one part is 6 cm and the ropes are 18 cm and 30 cm.

Chaining ratios. To merge $A : B = 3 : 2$ with $B : C = 5 : 4$, scale each ratio until the shared term matches: multiply the first by 5 and the second by 2, giving $A : B : C = 15 : 10 : 8$.

Adding to one side breaks the ratio. A class of 45 with boys : girls $= 3 : 2$ has 27 boys and 18 girls. Enrol 5 more boys and the ratio becomes $32 : 18 = 16 : 9$; you cannot just add 5 to the "3".

2. Proportion and variation

A proportion states that two ratios are equal and is solved by cross-multiplication:

$\frac{a}{b} = \frac{c}{d} \iff ad = bc$

Direct: if 3 kg of galunggong costs ₱480, then 7 kg costs $\frac{480}{3}(7) = 1{,}120$, or ₱1,120. Inverse: if 12 painters finish a job in 15 days, the constant is $12 \times 15 = 180$ worker-days, so 18 painters need $\frac{180}{18} = 10$ days — more workers, fewer days. Combined: suppose $y$ varies directly as $x$ and inversely as $z^2$, with $y = 12$ when $x = 6$ and $z = 2$. Then $12 = \frac{k(6)}{4}$ gives $k = 8$, and at $x = 9$, $z = 3$ we get $y = \frac{8(9)}{9} = 8$.

3. Percentage

Percent means "per hundred", so $r% = \frac{r}{100}$. Everything flows from one equation:

$\text{part} = \text{rate} \times \text{base}$

• What is 15% of ₱2,400? $0.15 \times 2{,}400 = 360$, so ₱360.
• ₱360 is what percent of ₱2,400? $\frac{360}{2{,}400} = 0.15 = 15%$.
• ₱360 is 15% of what amount? $\frac{360}{0.15} = 2{,}400$, so ₱2,400.

Percent change always divides by the original value:

$\text{percent change} = \frac{\text{new} - \text{old}}{\text{old}} \times 100\%$

Climbing from 250 to 320 is a $\frac{70}{250} = 28%$ increase, but falling back from 320 to 250 is only a $\frac{70}{320} = 21.875%$ decrease. The same peso movement gives two different percentages because the base changed. For the same reason a 25% rise followed by a 25% fall leaves $1.25 \times 0.75 = 0.9375$ of the original — a 6.25% net loss, not a wash.

Percent of a percent. If 60% of a class are girls and 25% of those girls joined the chorale, the chorale girls are $0.60 \times 0.25 = 0.15$, or 15% of the class.

Reverse percentage. Philippine retail tags are usually VAT (value-added tax) inclusive at 12%. A tagged price of ₱1,344 therefore hides a base of $\frac{1{,}344}{1.12} = 1{,}200$, or ₱1,200. Taking 12% off the tag would wrongly give ₱1,182.72.

Discounts, markup and margin

A discount of $d$ multiplies the price by $(1-d)$. Successive discounts multiply; they never add. A backpack listed at ₱1,800 with 25% off, then a further 10% off the reduced price, costs $1{,}800(0.75)(0.90) = 1{,}215$, so ₱1,215. The single equivalent discount is $1 - (0.75)(0.90) = 0.325 = 32.5%$ — not the 35% that adding would suggest.

Markup is measured against cost; margin is measured against selling price. They are never equal for the same peso profit.

Convert between them with $\text{margin} = \frac{\text{markup}}{1 + \text{markup}}$ and $\text{markup} = \frac{\text{margin}}{1 - \text{margin}}$.

Simple and compound interest

Simple interest is charged on the principal only: $I = Prt$ and $A = P(1 + rt)$, with $t$ in years. Compound interest is charged on principal plus accumulated interest:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

where $n$ is the number of compounding periods per year.

The widening gap is interest earned on interest. If the rate compounds quarterly, use $\frac{r}{4}$ per period and $4t$ periods — the annual rate is never used raw.

4. Work-rate problems

Turn every stated time into a rate: a worker who finishes a job in $t$ hours completes $\frac1t$ of it per hour. Rates add; times do not.

$\frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2} \quad \Longrightarrow \quad T = \frac{t_1 t_2}{t_1 + t_2}$

Ana repaints a room in 6 hours and Ben in 4 hours. Together they do $\frac16 + \frac14 = \frac{5}{12}$ of the room per hour, so $T = \frac{12}{5} = 2.4$ hours, i.e. 2 hours 24 minutes. Averaging the two times to "5 hours" is nonsense — two people must beat the faster one working alone.

For pipes, a drain is a negative rate. A pipe that fills a tank in 6 hours working against a drain that empties it in 9 hours nets $\frac16 - \frac19 = \frac{3}{18} - \frac{2}{18} = \frac1{18}$ of the tank per hour, so filling takes 18 hours. If the drain were faster than the fill pipe the net rate would be negative and the tank would never fill.

5. Mixture and alligation

Mixtures balance the solute, not the percentages:

$C_1V_1 + C_2V_2 = C_m\left(V_1 + V_2\right)$

Water counts as $0%$ and a pure substance as $100%$. Adding 10 g of pure silver to 40 g of an alloy that is 25% silver gives $0.25(40) + 1.00(10) = 20$ g of silver inside 50 g of metal, which is exactly 40%.

The alligation shortcut hands you the mixing ratio without algebra: take each stock's distance from the target, then cross them. The parts of one stock are proportional to the other stock's distance from the target.

   70% stock ────┐                    ┌──── |20 - 30| = 10 parts of the 70%
                  ├──  target 30%  ───┤
   20% stock ────┘                    └──── |70 - 30| = 40 parts of the 20%

   ratio 70% : 20%  =  10 : 40  =  1 : 4

6. Worked example 1 — a helper leaves partway

Problem. Mariel can encode a batch of enrolment forms in 10 hours; Kiko can encode the same batch in 15 hours. They start together, but after 4 hours Kiko is pulled into another office and Mariel finishes alone. How long does the whole batch take?

1. Rates: Mariel $\frac1{10}$ and Kiko $\frac1{15}$ of the batch per hour. Together, $\frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac16$ of the batch per hour.
2. In their 4 hours together they finish $4 \times \frac16 = \frac23$ of the batch.
3. Work remaining $= 1 - \frac23 = \frac13$.
4. Mariel alone needs $\frac{1/3}{1/10} = \frac{10}{3}$ hours $= 3$ hours 20 minutes.
5. Total elapsed time $= 4 + \frac{10}{3} = \frac{22}{3}$ hours, i.e. 7 hours 20 minutes. Note that the question asks for total time, not for Mariel's solo stretch — answering "3 hours 20 minutes" is the standard trap.

7. Worked example 2 — mixtures

Problem. (a) A laboratory aide needs 24 L of a 30% alcohol solution but has only 70% stock and 20% stock. How much of each? (b) A café in Bacolod blends beans costing ₱720/kg with beans costing ₱480/kg to produce 30 kg of blend costing ₱560/kg. How many kilograms of each?

(a) Let $x$ be the litres of 70% stock; the rest, $24 - x$, is 20% stock.

1. $0.70x + 0.20(24 - x) = 0.30(24)$
2. $0.70x + 4.8 - 0.20x = 7.2$
3. $0.50x = 2.4 \Rightarrow x = 4.8$
4. So 4.8 L of the 70% stock and 19.2 L of the 20% stock.
5. Alligation check: $\lvert 20 - 30 \rvert = 10$ and $\lvert 70 - 30 \rvert = 40$, so 70% : 20% $= 10 : 40 = 1 : 4$. One fifth of 24 L is 4.8 L. ✓

(b) Prices per kilogram behave exactly like concentrations.

1. Let $y$ be the kilograms of ₱720 beans, so $30 - y$ kg are ₱480 beans.
2. $720y + 480(30 - y) = 560(30) = 16{,}800$
3. $720y + 14{,}400 - 480y = 16{,}800 \Rightarrow 240y = 2{,}400 \Rightarrow y = 10$
4. The blend is 10 kg of the ₱720 beans and 20 kg of the ₱480 beans. Check: $10(720) + 20(480) = 7{,}200 + 9{,}600 = 16{,}800$, and $30 \times 560 = 16{,}800$. ✓

8. Common traps

• Parts are not values. "Shared in the ratio $4 : 7$" never means ₱4 and ₱7; find the value of one part first.
• Discounts do not add, and neither do successive increases. Multiply the factors instead.
• Markup is not margin. A 50% markup on cost is only a 33⅓% margin on the selling price.
• Inverse means invert. More workers means fewer days, so hold the product constant rather than cross-multiplying as if the relationship were direct.
• Never average times in work problems; add the rates and invert at the end.
• Every percentage needs its base. "10% more expensive" is meaningless until you know 10% of which price.
• In mixtures, pure substance is 100% and water is 0%. Dropping one of those terms from the solute balance is the usual wrong turn — and note that adding water leaves the amount of solute unchanged.