GMAT · Applied problem solving · TOPIC LESSON
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
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01 / UNDERSTAND & APPLY
Convention for every solution-mixture example and practice in this lesson: concentration is ingredient volume divided by total solution volume, all volumes are additive, and pure water contributes zero ingredient. In removal-and-replacement problems, the mixture is thoroughly mixed before removal. These are stated mathematical assumptions for the exercises.
A concentration is ingredient amount divided by total mixture amount, using consistent units and a clearly stated percentage basis. When combining mixtures, add ingredient amounts and total amounts separately. The new concentration is their ratio, so it is a weighted average of the original concentrations.
Assume volumes are additive only when the mathematical model says so or supplies that convention. A mass percentage and a volume percentage cannot be combined without conversion information. With positive amounts of two distinct concentrations, the final concentration lies strictly between them.
ORIGINAL WORKED EXAMPLE
Mix 10 litres of 20% solution with 15 litres of 40% solution, with additive volumes. What is the resulting concentration?
Answer 32%.
ORIGINAL PRACTICE · 1
6 litres.
Let x be the added volume. Ingredient balance gives 2.4+0.5x=0.3(12+x), hence 0.2x=1.2 and x=6. Averaging 20% and 50% to get 35% would assume equal volumes, which are not given.
02 / UNDERSTAND & APPLY
Adding pure solvent changes total volume but leaves ingredient amount unchanged. Removing a well-mixed sample removes both ingredient and solvent in the current concentration. After replacing that sample with pure solvent, the ingredient fraction is multiplied by the fraction of the original volume retained.
For repeated equal-volume replacements in a well-mixed tank, multiply retention factors; do not subtract the same original ingredient amount each time. The tank must be thoroughly mixed before each removal for the uniform-concentration model to apply.
ORIGINAL WORKED EXAMPLE
A tank contains 20 litres of 30% solution. Five litres are removed after mixing and replaced with pure water. What is the new concentration?
Answer 22.5%.
ORIGINAL PRACTICE · 1
6 litres.
Ingredient stays at 4 litres. If water added is w, then 4/(10+w)=1/4, so total volume is 16 and w=6. Reducing the concentration by 15 percentage points does not mean adding 15% of the initial volume.
03 / UNDERSTAND & APPLY
Revenue is money received from sales. Cost is money spent under the stated model. Profit=revenue−cost and can be negative, in which case it is a loss. If price per item is p and q items are sold, revenue is pq only when the price is constant for all those items.
Separate items produced from items sold if the scenario distinguishes them. Unsold production may still incur cost without generating revenue. A requested profit per item may use sold units or produced units as its denominator; read the definition.
ORIGINAL WORKED EXAMPLE
A seller buys 40 items at $6 each and sells all of them at $9 each. What are revenue and profit?
Answer Revenue $360; profit $120.
ORIGINAL PRACTICE · 1
$80.
Revenue is 40×7=280; cost includes all 50 purchased items, 50×4=200. Profit is 80. Using only the cost of sold items would ignore the $40 spent on unsold stock.
04 / UNDERSTAND & APPLY
A fixed cost does not change with output over the range specified. A variable cost per unit contributes vq for q units. Total cost is F+vq and profit at constant selling price p is (p−v)q−F. The contribution p−v pays toward the fixed cost before producing profit.
A fixed cost per batch is not necessarily a fixed cost for unlimited production: another batch may trigger another charge. Likewise, a per-unit cost may change at a threshold. Use the stated range and avoid extending a simple formula beyond it.
ORIGINAL WORKED EXAMPLE
A workshop has a $180 setup cost and $7 material cost per item. It sells 50 items for $13 each. Find profit.
Answer $120.
ORIGINAL PRACTICE · 1
$7.
Total cost is 90+4×30=210; average is 210/30=7. The variable cost $4 omits the fixed cost spread across the 30 units.
05 / UNDERSTAND & APPLY
Break-even occurs when profit is zero. With positive contribution p−v, solve (p−v)q=F. If q must be an integer and the question asks for no loss or at least a specified profit, round the threshold upward and verify the neighboring integer.
If p≤v and F>0, no nonnegative output breaks even under the stated model. More sales cannot repair a negative contribution. “At least zero profit” includes exact break-even, while “positive profit” requires exceeding it.
ORIGINAL WORKED EXAMPLE
A firm has fixed cost $250, variable cost $9 per unit and selling price $14. What is the smallest integer sales quantity yielding positive profit?
Answer 51 units.
ORIGINAL PRACTICE · 1
15 units.
Require 7q−100≥0, so q≥100/7=14 2/7. Round up to 15; 14 units produce −$2 and 15 produce $5. Rounding to the nearest integer would fail the constraint.
06 / UNDERSTAND & APPLY
A piecewise rule uses different formulas in different ranges. Overtime commonly applies only to hours above a threshold; a quantity discount may apply only to additional units or to all units once a threshold is reached. These are different rules and must be stated explicitly.
For a proposed answer, check which range it belongs to before applying a formula. At boundaries, confirm whether equality belongs to the lower or upper range and whether there is a jump. Do not charge every hour at the overtime rate unless the rule says so.
ORIGINAL WORKED EXAMPLE
A worker earns $18 per hour for the first 40 hours of a week and $27 per hour for each additional hour. What is pay for 46 hours?
Answer $882.
ORIGINAL PRACTICE · 1
$26.
The first 100 cost $20; the next 50 cost $6. Total $26. Applying $0.12 to every page would incorrectly turn an incremental discount into an all-units discount.
07 / UNDERSTAND & APPLY
An optimization problem asks for the largest or smallest achievable value subject to all constraints. Write the resource inequality, nonnegativity and integer restrictions. A continuous optimum may not be attainable with whole items. Test nearby feasible counts, not only a decimal solution.
A value-per-resource shortcut can suggest which item to favor, but leftovers and minimum quantities can change the optimum. For two small integer choices, enumerate the constrained variable or reason from the remainder. Prove that no untested option can do better.
ORIGINAL WORKED EXAMPLE
A supplier buys small packs of 4 units for $7 or large packs of 7 units for $11. With a $40 budget, what is the maximum number of units obtainable?
Answer 25 units: 3 large packs and 1 small pack, costing $40.
ORIGINAL PRACTICE · 1
3.
With a adults, cost=9a+5(6−a)=30+4a. The budget requires a≤3, and a≥2. Thus 3 adults are feasible at $42; 4 would cost $46. Ignoring the fixed total of six tickets changes the problem.
08 / UNDERSTAND & APPLY
Mixture and business problems often share linear equations but conserve different things. A mixture conserves ingredient amount; a cost model totals specified expenses; a resource constraint caps what can be used. State the conserved or limited quantity before setting up algebra.
After solving, ask whether the answer lies within a sensible range and satisfies the actual question. A concentration between two positive source concentrations should lie between them; a claimed profit must account for all stated costs; an integer optimum needs a feasible construction.
ORIGINAL WORKED EXAMPLE
Can mixing only positive amounts of 10% and 25% solutions produce a 30% solution under an additive-volume model?
Answer No.
ORIGINAL PRACTICE · 1
No.
Profit is (8−9)q−50=−q−50, which is negative for every q≥0. Solving for a negative break-even quantity is not a feasible sales plan; the sign of contribution already establishes impossibility.
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