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
Age differences stay constant as people age together; age ratios generally change. If someone is x years old now, the age t years later is x+t and t years earlier is x−t. Historical ages must remain consistent with the scenario.
For a two-digit positive integer with tens digit a and units digit b, the value is 10a+b, with a from 1 to 9 and b from 0 to 9. Reversing digits gives 10b+a, but it is a two-digit number only if b≠0. The digit sum a+b is not the original number.
ORIGINAL WORKED EXAMPLE
A parent is three times a child’s current age. In 12 years, the parent will be twice the child’s age. Find both current ages.
Answer Child 12 years; parent 36 years.
ORIGINAL PRACTICE · 1
74.
Let digits be a,b. Then a+b=11 and (10a+b)−(10b+a)=9(a−b)=27, so a−b=3. Solving gives a=7,b=4. The answer 47 reverses which number is larger.
02 / UNDERSTAND & APPLY
For two groups A and B, the count in at least one is |A|+|B|−|A∩B|. The intersection was counted once in each group total, so subtract one copy. “Exactly one” removes the intersection twice from the sum: |A|+|B|−2|A∩B|.
If a total population N is given, neither=N−|A∪B|. Check that every derived cell count is nonnegative and that the intersection is no larger than either group. These constraints can expose an impossible data set.
ORIGINAL WORKED EXAMPLE
Among 80 people, 45 use service A, 30 use B and 12 use both. How many use neither?
Answer 17.
ORIGINAL PRACTICE · 1
37.
French only=22 and Spanish only=15, so exactly one=37. The union 47 includes the 10 bilingual attendees and answers “at least one,” not “exactly one.”
03 / UNDERSTAND & APPLY
A two-way table classifies each person by two yes/no attributes. The four interior cells are mutually exclusive and sum to the population. Row and column totals are margins, not extra people. Build missing cells by subtraction from a known margin.
The displayed original table represents 100 people: 18 attended a workshop and subscribe to a newsletter; 12 attended only; 22 subscribe only; 48 did neither. Workshop total is 30 and newsletter total is 40. Every required value is supplied both in text and in the figure.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
Using the displayed survey, what fraction of workshop attendees also subscribe to the newsletter?
Answer 3/5, or 60%.
ORIGINAL PRACTICE · 1
34.
The two exclusive cells contain 12 and 22 people, so 12+22=34. Adding margins 30+40 double-counts the 18 in both and includes them when they should be excluded.
04 / UNDERSTAND & APPLY
“At least k” means ≥k and “at most k” means ≤k. “More than k” and “fewer than k” are strict. For integer counts, more than 7 means at least 8, but for real measurements it can mean 7.1.
A capacity constraint may apply to a total rather than to each category separately. Translate the full sentence and attach units. After solving a strict bound, verify whether an endpoint is permitted before rounding an integer answer.
ORIGINAL WORKED EXAMPLE
A room holds at most 75 people. There are 18 staff and g guests. What is the largest allowed integer g?
Answer 57 guests.
ORIGINAL PRACTICE · 1
5.
The values are 5,6,7,8,9. Including 4 ignores “more than”; excluding 9 ignores “at most.”
05 / UNDERSTAND & APPLY
Define variables for quantities directly related by the problem. If two counts sum to a known total, one variable may suffice: x and total−x. If a ratio is known, use a common scale. State units so that coefficients have a clear meaning.
Fewer variables are not always better. Two transparent equations can be safer than one overloaded expression. Choose the representation that preserves constraints clearly and makes the requested quantity easy to recover.
ORIGINAL WORKED EXAMPLE
A box contains 24 coins, each worth either 5 or 10 cents, totaling 175 cents. How many 10-cent coins are there?
Answer 11.
ORIGINAL PRACTICE · 1
4k and 7k; the quantities are 20 and 35.
Their sum gives 11k=55, so k=5. Treating the quantities as k and k+3 mistakes a ratio difference for an absolute difference.
06 / UNDERSTAND & APPLY
When the requested result depends only on a ratio or percentage and no absolute size is specified, choose a convenient base such as 100 or a common multiple. All chosen values must satisfy every condition. A single example can reveal a scale-invariant value, but explain why scale cancels.
Testing numbers is not a general proof when the answer depends on an unconstrained variable. Avoid special values such as 0 or 1 if they accidentally make distinct expressions coincide. To disprove a universal claim, however, one valid counterexample is enough.
ORIGINAL WORKED EXAMPLE
A quantity rises by 20% and then falls by 25%. What is the net percent change?
Answer 10% decrease.
ORIGINAL PRACTICE · 1
7/5.
Choose a=2,b=5, or write a=2k,b=5k with k≠0. Then (2k+5k)/(5k)=7/5. The cancellation proves that the result does not depend on scale.
07 / UNDERSTAND & APPLY
When choices give possible values of the requested quantity, substitute them into a short original relationship. For a monotonic expression, starting near the middle can eliminate a whole direction. First confirm monotonicity over the allowed domain; quadratics and piecewise rules may not be monotonic.
Backsolving is least useful when each substitution is expensive or when choices represent transformed quantities rather than the variable in your equation. Keep track of what each option claims, and verify every original condition before accepting it.
ORIGINAL WORKED EXAMPLE
A positive integer n satisfies n(n+3)=70. The choices are 4,5,6,7,8. Find n by testing efficiently.
Answer 7.
ORIGINAL PRACTICE · 1
8.
Test 8: 12+4×8=44. Or solve 4v=32. The option 11 would arise from ignoring the fixed fee, but it is not consistent with the original cost rule.
08 / UNDERSTAND & APPLY
Bounds replace an unwieldy computation with a guaranteed interval. For positive quantities, increasing a numerator increases a quotient and increasing a denominator decreases it. Rounding both in the same direction does not automatically give a bound.
Estimation can choose among widely separated options, but close options demand tighter bounds or exact arithmetic. Decide how much precision the question needs before calculating. Keep strict and non-strict boundaries when the answer is an integer count.
ORIGINAL WORKED EXAMPLE
Which is closest to 398×0.251: 10,50,100,150 or 200?
Answer 100.
ORIGINAL PRACTICE · 1
30.
24×30=720≤725<744=24×31. Thus 30≤725/24<31 and its integer part is 30. Rounding to the nearest integer is a different operation, though it happens to match here.
09 / UNDERSTAND & APPLY
A no-calculator check should be cheaper than repeating the entire calculation. Use parity, last digits, approximate size, units or substitution into the original equation. A check can disprove a result quickly; passing one weak check does not prove it.
Reserve exact verification for the decisive relationship. If a count solution gives a noninteger, revisit the setup or determine whether the problem intentionally has no feasible solution. If a percent answer exceeds 100%, ask whether the comparison permits it rather than rejecting it automatically.
ORIGINAL WORKED EXAMPLE
A calculation claims 48×25=1,020. Find the correct value and a quick check.
Answer 1,200.
ORIGINAL PRACTICE · 1
45.
18%=0.18, so 250×0.18=(250/100)×18=2.5×18=45. Because 18% is below 100%, the part must be below 250; the claimed 450 immediately fails that bound.
KEEP THE BIG IDEAS
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