🔢 Quant

Rates, Work & Mixtures

Distance-rate-time, the average-speed trap, combined work via rates, and mixture conservation.

~6h

to master

One formula, three families

Distance, work, and mixtures all run on the same engine: $\text{amount} = \text{rate} \times \text{time}$ . The GMAT varies what accumulates — kilometres, completed jobs, or grams of salt.

Travel rates

$d = rt$ , used in any of its three solved forms. Keep units consistent before computing: 25 m/s for 5 minutes → $25 \times 300 = 7500$ m.
Unit conversions are rate multiplications: miles per hour ÷ gallons per hour = miles per gallon. Set up the fraction so unwanted units cancel visibly.

📐Core Rule

Average speed = total distance ÷ total time. Never the average of the speeds. Half a 600 km trip at 60 km/h (5 h) and half at 100 km/h (3 h) gives $\frac{600}{8} = 75$ km/h — not 80. The slow half eats more time, so it drags the average below the midpoint. Equal-distance legs always land below the simple average.

Test yourself on before reading on — the wrong answer "80" is option one for a reason.

Opposite directions / closing the gap: speeds add when approaching head-on, subtract when chasing. Time to meet = gap ÷ combined speed.

Work problems: think in rates per hour

If a machine finishes a job in $t$ hours, it completes $\frac{1}{t}$ of the job per hour. Rates add when workers run simultaneously.

📐Core Rule

Two workers with solo times $a$ and $b$ finish together in $T$ where $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{T}$ , i.e. $T = \dfrac{ab}{a+b}$ . Machines at 4 h and 5 h per batch together take $\frac{20}{9}$ h — always less than the faster solo time. If a "together" answer isn't smaller than every solo time, it's wrong.

The same identity runs backwards: given the together-time and one solo time, subtract rates to find the other worker — that's .

💡Exam Tip

Pick a concrete job size. Let the job be the LCM of the times (a 20-unit job for 4 h and 5 h workers → rates 5 and 4 units/h). Integer rates beat fraction juggling, especially when workers join or leave mid-job: track units completed, phase by phase.

Mixtures: track the pure stuff

Mixing problems are weighted averages. The reliable method: compute the amount of the pure ingredient (salt, ryegrass, alcohol, cost) contributed by each component, then divide by total quantity.

✏️Worked Example

How many litres of 15% salt solution must join 5 L of 8% solution to make 10%? Salt in = salt out: $0.15x + 0.08(5) = 0.10(x + 5)$ → $0.05x = 0.1$ → $x = 2$ litres. One equation, always the same shape: concentration × quantity, summed, equals final concentration × final quantity.

⚡Shortcut

The balance check: the final concentration must sit between the two inputs, closer to the bigger contributor. Mixing lots of 8% with a little 15% must land near 8% — use this to kill impossible options before solving, and to sanity-check after.

⚠️GMAT Trap

Cost mixtures are averages weighted by quantity, not by price. 6 kg at $1.20 plus 2 kg at $1.60 costs $\frac{6(1.20) + 2(1.60)}{8} = \$ 1.30 $/kg — the answer leans toward \$ 1.20 because more kilograms sit there, regardless of which price is larger.

Checklist

Units aligned before any arithmetic
Average speed via total/total — never midpoint of speeds
Work → convert times to rates; rates add
Together-time < every solo time (sanity check)
Mixtures → conserve the pure ingredient; final % between inputs

Sample Questions

22 practice questions

Medium

After cycling to a trailhead, Dana plans to walk east along a canal path, turn around, and return to the trailhead by the same path. After walking 8.5 kilometres east, she decides to walk for only 88 minutes more. If Dana walks at a constant rate of 8 minutes per kilometre, how many kilometres farther east can she walk and still be able to return to the trailhead in 88 minutes?

Medium

Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of $x$ units of Product R in 7 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of $3x$ units of Product R in 6 days?

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