Percentages — Practice Set
Bank code: ARI-PCT · Section: Arithmetic · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
Every percent question is really: per cent of what? Turn each percent into a multiplier, then multiply.
Every question below is live in the question bank under the ID shown — the sheet and the portal are the same questions. Attempt a level with the clock running, then check Part C.
Part A — Questions
Quantitative Comparison — the four choices are always the same, so they are not reprinted: (A) Column A is greater · (B) Column B is greater · (C) The two quantities are equal · (D) The relationship cannot be determined from the information given
Level 1 · Easy — 5 questions · ~4 min
warm-up — these must be automatic
Q1 · ARI-PCT-002 · MCQ · 45s
A shirt is priced at $80. During a sale it is marked 25% off. What is the sale price?
(A) $20
(B) $55
(C) $60
(D) $64
(E) $100
Q2 · ARI-PCT-003 · Numeric Entry · 45s
45 is what percent of 150? Enter your answer as a number.
Numeric entry — write the number.
Q3 · ARI-PCT-008 · QC · 45s
A price increases from $50 to $60.
Column A: The percent increase Column B: 20%
Q4 · ARI-PCT-011 · Numeric Entry · 55s
After a 20% discount, a product sells for $240. What was the price before the discount, in dollars? Enter your answer as a number.
Numeric entry — write the number.
Q5 · ARI-PCT-014 · MCQ · 60s
A number is increased by 10%, and the result is then decreased by 10%. What is the net percent change?
(A) 0%
(B) -1%
(C) +1%
(D) -2%
(E) -10%
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · ARI-PCT-021 · MCQ · 60s
A laptop's price is reduced from $1,200 to $900. By what percent was the price reduced?
(A) 20%
(B) 25%
(C) 30%
(D) 33 1/3%
(E) 75%
Q7 · ARI-PCT-023 · QC · 75s
A jacket costs $150. Store A offers a 30% discount followed by an additional 10% off the discounted price. Store B offers a flat 40% discount.
Column A: Final price at Store A Column B: Final price at Store B
Q8 · ARI-PCT-025 · Select all that apply · 85s
The price of an item increased from $P to $1.3P, where P > 0. Which of the following statements must be true? Select all that apply.
(A) The price increased by 30%
(B) The price increased by $0.30
(C) The new price is 130% of the original
(D) The new price is greater than the original
(E) The percent increase depends on the value of P
Q9 · ARI-PCT-026 · Numeric Entry · 75s
After a 20% increase followed by a 20% decrease, a number becomes 576. What was the original number? Enter your answer as a number.
Numeric entry — write the number.
Q10 · ARI-PCT-027 · QC · 60s
Column A: The percent increase when a quantity goes from 80 to 100 Column B: The percent decrease when a quantity goes from 100 to 80
Q11 · ARI-PCT-033 · MCQ · 85s
A shopkeeper sells two items for $240 each. On one he makes a 20% profit; on the other he takes a 20% loss. Overall, what is his profit or loss percent?
(A) No profit, no loss
(B) 4% loss
(C) 4% profit
(D) 2% loss
(E) 8% loss
Q12 · ARI-PCT-043 · MCQ · 70s
In a town, 55% of residents own a car and 40% own a bike. If 15% own both, what percent own neither?
(A) 5%
(B) 15%
(C) 20%
(D) 25%
(E) 30%
Q13 · ARI-PCT-050 · Numeric Entry · 85s
In a class, the boys averaged 75% on a test and the girls averaged 80%. The class as a whole averaged 78%. What percent of the class are boys? Enter your answer as a number.
Numeric entry — write the number.
Level 3 · Hard — 12 questions · ~18 min
where 162+ is won or lost
Q14 · ARI-PCT-051 · MCQ · 95s
A retailer sells an item at a 25% profit on cost. The cost price then rises by 20% while the selling price is held unchanged. What is the new profit or loss percent?
(A) 4.17% profit
(B) 4% profit
(C) 4.17% loss
(D) 5% profit
(E) 5% loss
Q15 · ARI-PCT-052 · QC · 80s
N and p are positive. A number N is increased by p%, and the result is then decreased by p%.
Column A: The final number Column B: N
Q16 · ARI-PCT-053 · Numeric Entry · 85s
The price of a commodity falls by 20%. By what percent must consumption increase so that total expenditure is unchanged? Enter your answer as a number.
Numeric entry — write the number.
Q17 · ARI-PCT-054 · MCQ · 80s
In a school, 60% of students passed Math, 50% passed Science, and 25% failed both subjects. What percent of students passed BOTH subjects?
(A) 10%
(B) 15%
(C) 25%
(D) 35%
(E) 45%
Q18 · ARI-PCT-055 · Select all that apply · 105s
A store offers successive discounts of 20%, 10%, and 5% on an item. Which of the following are true? Select all that apply.
(A) The effective discount is 35%
(B) The effective discount is less than 35%
(C) The final price is 68.4% of the original
(D) The order in which the three discounts are applied does not affect the final price
(E) A single discount of 35% would give a lower final price than these three discounts
Q19 · ARI-PCT-060 · Select all that apply · 95s
A price P increases by x% to become Q, where 0 < P < Q. Which of the following expressions correctly represent x? Select all that apply.
(A) (Q - P)/P x 100
(B) (Q/P - 1) x 100
(C) (Q/P) x 100 - 100
(D) (Q - P)/Q x 100
(E) 100(Q - P)/P
Q20 · ARI-PCT-061 · Numeric Entry · 80s
A's income is 30% more than B's. B's income is 20% less than C's. A's income is what percent of C's income? Enter your answer as a number.
Numeric entry — write the number.
Q21 · ARI-PCT-062 · MCQ · 85s
In 2020 the ratio of A's salary to B's salary was 3:4. In 2021 A's salary rose by 25% and B's salary fell by 25%. What is the new ratio of A's salary to B's salary?
(A) 3:4
(B) 4:3
(C) 5:4
(D) 5:3
(E) 15:16
Q22 · ARI-PCT-063 · QC · 75s
A product's price increases by 10% each year for 7 years.
Column A: The total percent increase over the 7 years Column B: 70%
Q23 · ARI-PCT-066 · MCQ · 95s
A dishonest dealer claims to sell goods at cost price, but his 'kilogram' weight is actually 900 grams. What is his real profit percent?
(A) 9.09%
(B) 10%
(C) 11.11%
(D) 12.5%
(E) 20%
Q24 · ARI-PCT-069 · MCQ · 90s
An employee's salary increases by 10% every year. After how many complete years will the salary first exceed 150% of the current salary?
(A) 4
(B) 5
(C) 6
(D) 8
(E) 10
Q25 · ARI-PCT-070 · Select all that apply · 100s
In a class, 70% of students passed English and 60% passed Math. Which of the following could be the percent who passed both? Select all that apply.
(A) 25%
(B) 30%
(C) 40%
(D) 55%
(E) 65%
Level 4 · Extreme — 5 questions · ~9 min
165+ — expect to need the insight, not the grind
Q26 · ARI-PCT-071 · MCQ · 115s
A merchant mixes tea costing $40 per kg with tea costing $60 per kg and sells the mixture at $60 per kg, making a 25% profit on his average cost. In what ratio, by weight, did he mix the two varieties?
(A) 1:1
(B) 2:3
(C) 3:2
(D) 3:1
(E) 4:1
Q27 · ARI-PCT-073 · Numeric Entry · 90s
A sum of money invested at compound interest amounts to $5,000 in 2 years and $5,500 in 3 years. What is the annual rate of interest, as a percent? Enter your answer as a number.
Numeric entry — write the number.
Q28 · ARI-PCT-074 · Select all that apply · 110s
A company's profit changed by x% in year 1 and by y% in year 2 (x and y may be negative), and the overall change over the two years was exactly +44%. Which of the following (x, y) pairs are possible? Select all that apply.
(A) x = 20, y = 20
(B) x = 10, y = 30
(C) x = 44, y = 0
(D) x = 20, y = 21
(E) x = 100, y = -28
Q29 · ARI-PCT-075 · MCQ · 100s
Numbers A, B, C and D satisfy: A is 20% more than B, B is 25% less than C, and C is 60% of D. A is what percent of D?
(A) 54%
(B) 55%
(C) 60%
(D) 66%
(E) 72%
Q30 · ARI-PCT-077 · QC · 115s
An item's price increases by r% each year for n years.
Column A: The total percent increase when r = 10 and n = 10 Column B: The total percent increase when r = 5 and n = 20
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | ARI-PCT-002 |
easy | MCQ | C |
| 2 | ARI-PCT-003 |
easy | Numeric Entry | 30 |
| 3 | ARI-PCT-008 |
easy | QC | C |
| 4 | ARI-PCT-011 |
easy | Numeric Entry | 300 |
| 5 | ARI-PCT-014 |
easy | MCQ | B |
| 6 | ARI-PCT-021 |
medium | MCQ | B |
| 7 | ARI-PCT-023 |
medium | QC | A |
| 8 | ARI-PCT-025 |
medium | Select all that apply | A, C, D |
| 9 | ARI-PCT-026 |
medium | Numeric Entry | 600 |
| 10 | ARI-PCT-027 |
medium | QC | A |
| 11 | ARI-PCT-033 |
medium | MCQ | B |
| 12 | ARI-PCT-043 |
medium | MCQ | C |
| 13 | ARI-PCT-050 |
medium | Numeric Entry | 40 |
| 14 | ARI-PCT-051 |
hard | MCQ | A |
| 15 | ARI-PCT-052 |
hard | QC | B |
| 16 | ARI-PCT-053 |
hard | Numeric Entry | 25 |
| 17 | ARI-PCT-054 |
hard | MCQ | D |
| 18 | ARI-PCT-055 |
hard | Select all that apply | B, C, D, E |
| 19 | ARI-PCT-060 |
hard | Select all that apply | A, B, C, E |
| 20 | ARI-PCT-061 |
hard | Numeric Entry | 104 |
| 21 | ARI-PCT-062 |
hard | MCQ | C |
| 22 | ARI-PCT-063 |
hard | QC | A |
| 23 | ARI-PCT-066 |
hard | MCQ | C |
| 24 | ARI-PCT-069 |
hard | MCQ | B |
| 25 | ARI-PCT-070 |
hard | Select all that apply | B, C, D |
| 26 | ARI-PCT-071 |
extreme hard | MCQ | C |
| 27 | ARI-PCT-073 |
extreme hard | Numeric Entry | 10 |
| 28 | ARI-PCT-074 |
extreme hard | Select all that apply | A, C, E |
| 29 | ARI-PCT-075 |
extreme hard | MCQ | A |
| 30 | ARI-PCT-077 |
extreme hard | QC | B |
Part C — Worked Solutions
Q1 · ARI-PCT-002 — Answer: C
Step 1 — turn the discount into a multiplier. Paying after 25% off means paying 100% - 25% = 75% of the price, so the multiplier is 0.75. Step 2 — multiply. 0.75 x 80 = 60. Check: the discount is 0.25 x 80 = $20, and 80 - 20 = $60. Correct. Sale price = $60.
Trap. Choosing $20, which is the discount rather than the price paid. The second trap is reading '25% off' as '$25 off', which gives $55.
Q2 · ARI-PCT-003 — Answer: 30
Step 1 — classify. This is 'find the percent': part is given, whole is given. Step 2 — the base is the number after 'of', so 150 goes on the bottom. 45/150 = 3/10 = 0.3. Step 3 — convert to percent: 0.3 x 100 = 30. Answer: 30.
Trap. Dividing the other way, 150/45 = 3.33, and answering 333. Whatever follows the word 'of' is always the denominator.
Q3 · ARI-PCT-008 — Answer: C
Percent change = (New - Old)/Old x 100. Step 1 — the change: 60 - 50 = 10. Step 2 — the base is the ORIGINAL value, 50. Step 3 — 10/50 x 100 = 0.2 x 100 = 20%. Column A = 20% = Column B, so the two quantities are equal.
Trap. Dividing the $10 change by the new price: 10/60 = 16.7%, which makes Column B look greater. Percent change divides by the value you started from, never the value you ended at.
Q4 · ARI-PCT-011 — Answer: 300
Step 1 — build the multiplier. A 20% discount means the buyer pays 80%, so 0.80. Step 2 — write the equation forwards: 0.80 x P = 240. Step 3 — reverse it by dividing: P = 240/0.80 = 300. Check: 20% of 300 = 60, and 300 - 60 = 240. Correct. Answer: 300.
Trap. Adding 20% back to the sale price: 240 x 1.20 = 288. The 20% was taken off $300, not off $240, so you must divide by the multiplier rather than add the percent back on.
Q5 · ARI-PCT-014 — Answer: B
Step 1 — pick 100 as the starting value (legal, since the question is all percents). Step 2 — up 10%: 100 x 1.10 = 110. Step 3 — down 10%, applied to 110: 110 x 0.90 = 99. Step 4 — net change = (99 - 100)/100 x 100 = -1%. The one-line version: 1.10 x 0.90 = 0.99, a 1% decrease.
Trap. Answering 0% because +10% and -10% look like they cancel. They cannot: the rise is 10% of 100, but the fall is 10% of the larger number 110, so more comes off than went on. Up x% then down x% always loses exactly x^2/100 percent - here 100/100 = 1%.
Q6 · ARI-PCT-021 — Answer: B
Step 1 — the change: 1200 - 900 = 300. Step 2 — the base is the original price, 1200. Step 3 — 300/1200 = 1/4 = 25%. Answer: 25%.
Trap. Dividing by the new price: 300/900 = 33 1/3%, which is choice D. A second trap is 900/1200 = 75%, which is the fraction of the price that REMAINS, not the reduction.
Q7 · ARI-PCT-023 — Answer: A
Store A - build one chain: 150 x 0.70 x 0.90. 0.70 x 0.90 = 0.63, so 150 x 0.63 = $94.50. Store B: 150 x 0.60 = $90.00. $94.50 > $90.00, so Column A is greater.
Trap. Reading 30% then 10% as a combined 40% and choosing C. Successive discounts never add: 0.70 x 0.90 = 0.63, an effective 37% off, because the 10% is taken from an already reduced price. Two successive discounts are always weaker than their sum.
Q8 · ARI-PCT-025 — Answer: A, C, D
A - percent increase = (1.3P - P)/P x 100 = 0.3P/P x 100 = 30%. The P cancels, so this is true for every P. TRUE. B - the increase is 0.3P dollars, not $0.30. That would only be right if P = $1. FALSE. C - 1.3P = 130% of P by definition of the multiplier. TRUE. D - since P > 0, 1.3P > P. TRUE. E - the P cancelled in A, so the percent increase is 30% for every P. FALSE. Answer: A, C, D.
Trap. Picking B by reading the 0.3 as a dollar amount instead of a coefficient, and picking E because P is unknown. When the unknown cancels out of the ratio, the percent is fixed even though the dollar change is not.
Q9 · ARI-PCT-026 — Answer: 600
Step 1 — build the whole chain first: 1.20 x 0.80 = 0.96. Step 2 — 0.96 x N = 576. Step 3 — reverse by dividing: N = 576/0.96 = 600. Check: 600 -> 720 -> 720 x 0.80 = 576. Correct. Answer: 600.
Trap. Assuming +20% and -20% cancel, so the original 'must' be 576. The chain is 0.96, a 4% net fall, which is exactly 20^2/100 percent.
Q10 · ARI-PCT-027 — Answer: A
Column A: change = 20, base = 80 (where you started), so 20/80 x 100 = 25%. Column B: change = 20, base = 100 (where you started), so 20/100 x 100 = 20%. 25% > 20%, so Column A is greater.
Trap. Both moves cover the same 20 units, so students pick C. The base flips: going up you divide by 80, coming down you divide by 100. A rise and the fall that undoes it are never the same percent.
Q11 · ARI-PCT-033 — Answer: B
Selling price is given, so work backwards to each cost. Step 1 — profit item: SP = 1.20 x CP, so CP = 240/1.20 = $200. Step 2 — loss item: SP = 0.80 x CP, so CP = 240/0.80 = $300. Step 3 — totals: cost = 200 + 300 = $500; revenue = 2 x 240 = $480. Step 4 — loss = $20 on a cost of $500 = 20/500 x 100 = 4% loss.
Trap. Assuming equal selling prices with equal +20%/-20% cancel out. They cannot: the loss item cost more ($300 vs $200), so the loss is taken on a bigger base. Whenever two items sell for the SAME price at +x% and -x%, the result is always a loss of x^2/100 percent - here 4%.
Q12 · ARI-PCT-043 — Answer: C
Step 1 — union by inclusion-exclusion: car OR bike = 55 + 40 - 15 = 80%. Step 2 — neither is the complement of the union: 100 - 80 = 20%. Answer: 20%.
Trap. Adding 55 + 40 = 95 and answering 5%. That double-counts the 15% who own both, so the union is overstated and 'neither' comes out too small. Subtract the overlap once before taking the complement.
Q13 · ARI-PCT-050 — Answer: 40
Method 1 - weighted average. Let b be the fraction of boys. 75b + 80(1 - b) = 78 75b + 80 - 80b = 78 -5b = -2, so b = 0.4 = 40%. Method 2 - alligation (faster). Distance from boys' mean to the class mean = 78 - 75 = 3. Distance from the class mean to the girls' mean = 80 - 78 = 2. The group sizes are in the INVERSE ratio of these distances, so boys : girls = 2 : 3, and boys = 2/5 = 40%. Answer: 40.
Trap. Answering 60. The class average of 78 sits closer to the girls' 80, which means girls are the LARGER group - so boys must be the smaller share. In alligation the bigger group is the one nearer the overall mean, which flips the ratio you first write down.
Q14 · ARI-PCT-051 — Answer: A
Step 1 — pick CP = 100. Then SP = 1.25 x 100 = 125. Step 2 — the new cost: 100 x 1.20 = 120. The selling price stays at 125. Step 3 — profit = 125 - 120 = 5, and profit percent is always taken on COST. 5/120 x 100 = 500/120 = 4.166... = 4.17% profit.
Trap. Two errors compete here. First, assuming a 20% cost rise must wipe out the margin and produce a loss - it does not, because the original markup was large enough to absorb it. Second, dividing by the selling price: 5/125 = 4%, which is choice B. Profit percent always divides by cost, and here that means the NEW cost of 120.
Q15 · ARI-PCT-052 — Answer: B
Final = N x (1 + p/100) x (1 - p/100). This is a difference of squares: (1 + p/100)(1 - p/100) = 1 - p^2/10000. Since p > 0, the term p^2/10000 is strictly positive, so the multiplier is strictly less than 1. With N > 0, the final number is always less than N, whatever p is. Column B is greater. The loss is exactly p^2/100 percent - for p = 10 that is 1%, for p = 20 it is 4%, for p = 50 it is 25%.
Trap. Choosing C because the same p goes up and comes back down. The rise is p% of N, but the fall is p% of the LARGER number N(1 + p/100), so more is removed than was added. Note the answer does not depend on the size of p, so D is also wrong.
Q16 · ARI-PCT-053 — Answer: 25
Expenditure = price x consumption, so for expenditure to be unchanged the two multipliers must multiply to 1. Step 1 — price multiplier = 0.80. Step 2 — consumption multiplier = 1/0.80 = 1.25. Step 3 — 1.25 means a 25% increase. Check with numbers: price 100 x quantity 10 = 1000. New price 80, so quantity must be 1000/80 = 12.5, up from 10 - a 25% rise. Correct. Answer: 25.
Trap. Answering 20 to mirror the 20% price fall. Test it: 0.80 x 1.20 = 0.96, so expenditure would still be 4% lower. A fall of 1/5 needs a rise of 1/4 - a drop of 1/(n+1) always needs a rise of 1/n.
Q17 · ARI-PCT-054 — Answer: D
Step 1 — 'failed both' is the complement of 'passed at least one', so the union is 100 - 25 = 75%. Step 2 — inclusion-exclusion: (passed Math) + (passed Science) - (passed both) = (passed at least one). 60 + 50 - both = 75 Step 3 — both = 110 - 75 = 35%. Check the four regions: both 35, Math only 25, Science only 15, neither 25. They sum to 100. Correct. Answer: 35%.
Trap. Treating the 25% who failed both AS the overlap and answering 25% (choice C). Those 25% sit OUTSIDE both circles; the overlap has to be computed. The other slip is stopping at 75% - that is the union, the answer to a different question.
Q18 · ARI-PCT-055 — Answer: B, C, D, E
Build the chain: 0.80 x 0.90 x 0.95. 0.80 x 0.90 = 0.72, and 0.72 x 0.95 = 0.684. So the final price is 68.4% of the original and the effective discount is 100 - 68.4 = 31.6%. A - 31.6% is not 35%. FALSE. B - 31.6% < 35%. TRUE. C - final price = 68.4% of the original. TRUE. D - multiplication is commutative, so any order gives 0.684. TRUE. E - a single 35% discount leaves 65% of the price, and 65% < 68.4%, so it is the cheaper deal. TRUE. Answer: B, C, D, E.
Trap. A is the obvious bait - adding 20 + 10 + 5. The subtler one is E: because successive discounts are weaker than their sum, the single 35% discount actually beats the three, so a student who reasons 'successive discounts must be better for the customer' gets E backwards.
Q19 · ARI-PCT-060 — Answer: A, B, C, E
The definition: x = (change / original) x 100 = (Q - P)/P x 100. A - that is the definition itself. TRUE. B - Q/P - 1 = (Q - P)/P, so multiplying by 100 gives the same thing. TRUE. C - (Q/P) x 100 - 100 = 100(Q/P - 1) = 100(Q - P)/P. Same expression rearranged. TRUE. D - divides by Q, the NEW price. That is the percent decrease going from Q back down to P, a different (smaller) number. FALSE. E - identical to A, just written with the 100 in front. TRUE. Answer: A, B, C, E.
Trap. D is the classic wrong formula and it looks symmetric with A, so it gets picked. The percent rise from P to Q and the percent fall from Q back to P are never equal, because they use different denominators.
Q20 · ARI-PCT-061 — Answer: 104
Step 1 — pick C = 100, since C is the base at the end of the chain. Step 2 — B is 20% less than C: B = 0.80 x 100 = 80. Step 3 — A is 30% more than B: A = 1.30 x 80 = 104. Step 4 — A as a percent of C: 104/100 x 100 = 104%. One-line version: 1.30 x 0.80 = 1.04. Answer: 104.
Trap. Netting the percents to 30 - 20 = 10 and answering 110. The 30% is measured on B while the 20% is measured on C - different bases, so they cannot be added. Chain the multipliers instead.
Q21 · ARI-PCT-062 — Answer: C
Step 1 — write the salaries as 3k and 4k so the ratio is preserved. Step 2 — apply each multiplier to its own term: new A = 3k x 1.25 = 3.75k new B = 4k x 0.75 = 3k Step 3 — the ratio: 3.75k : 3k = 3.75 : 3 = 375 : 300 = 5 : 4. Answer: 5:4.
Trap. Expecting the +25% and -25% to cancel and leave the ratio at 3:4 (choice A) - they act on different salaries, so nothing cancels except k. Choice E, 15:16, is the other live error: raising A to 3.75 but forgetting to shrink B, which leaves 3.75:4 = 15:16.
Q22 · ARI-PCT-063 — Answer: A
The price multiplier is 1.10^7. Build it in pieces: 1.10^2 = 1.21, so 1.10^4 = 1.21^2 = 1.4641, and 1.10^3 = 1.331. 1.10^7 = 1.10^4 x 1.10^3 = 1.4641 x 1.331 = 1.9487. That is a 94.87% increase, comfortably above 70%. Column A is greater. You do not even need the exact value: 1.10^7 = 1.10^4 x 1.10^3 > 1.4 x 1.3 = 1.82, already a 82% rise.
Trap. Multiplying 10% x 7 = 70% and choosing C. That is simple-interest thinking. Once there is more than one period, each year's rise is computed on a larger amount, so compound growth always exceeds the simple sum.
Q23 · ARI-PCT-066 — Answer: C
Set the true cost at $1 per gram. Step 1 — the customer is charged for 1000 g, so revenue = $1000. Step 2 — the dealer actually hands over 900 g, so his cost = $900. Step 3 — profit = 1000 - 900 = $100, taken on a cost of $900. 100/900 x 100 = 11.11%. General rule for false weights: profit % = (true weight - false weight)/false weight x 100.
Trap. Dividing by 1000 to get 10% (choice B). The 1000 is what he charged for - that is revenue, not cost. Profit percent is always taken on what he gave up, the 900 g actually delivered. Choice A, 9.09%, is 100/1100, from adding the shortfall to the kilo instead of subtracting it; choice D, 12.5%, is 100/800, from mistaking the false weight for 800 g.
Q24 · ARI-PCT-069 — Answer: B
We need the smallest whole n with 1.10^n > 1.50. 1.10^2 = 1.21 1.10^3 = 1.331 1.10^4 = 1.4641 - still below 1.50 1.10^5 = 1.61051 - above 1.50 So the salary first exceeds 150% of its current value after 5 complete years. Answer: 5.
Trap. Reading '150% of the current salary' as 'a 150% increase', which means 2.5 times the salary and takes 10 years (1.10^10 = 2.594) - choice E. 150% OF a number is 1.5 times it; a 150% INCREASE makes it 2.5 times. A second error is stopping at year 4 because 1.4641 looks close enough to 1.5.
Q25 · ARI-PCT-070 — Answer: B, C, D
The overlap of two sets is squeezed between two bounds. Maximum: the overlap can never exceed the smaller set, so both <= min(70, 60) = 60%. Minimum: 70 + 60 = 130, but only 100% of the class exists, so at least 130 - 100 = 30% must be counted twice. So both >= 30%. The overlap therefore satisfies 30% <= both <= 60%. A (25%) is below the minimum. B (30%) is exactly at the minimum, achievable when everyone passed at least one. C (40%) and D (55%) are inside the range. E (65%) exceeds the smaller set. Answer: B, C, D.
Trap. Assuming the overlap could be anything from 0 to 60 and including A. When the two percents sum to more than 100 there is not enough room to keep the groups apart, so a forced overlap of (sum - 100) exists. E catches the opposite error - forgetting that the overlap cannot be bigger than the smaller group.
Q26 · ARI-PCT-071 — Answer: C
Step 1 — recover the cost from the selling price. A 25% profit means SP = 1.25 x cost, so cost = 60/1.25 = $48 per kg. Step 2 — alligation on the average cost of 48: distance from the $40 tea: 48 - 40 = 8 distance from the $60 tea: 60 - 48 = 12 Step 3 — the quantities are in the INVERSE ratio of these distances: cheap : dear = 12 : 8 = 3 : 2. Check: (3 x 40 + 2 x 60)/5 = (120 + 120)/5 = 240/5 = $48. Correct. Answer: 3:2.
Trap. Two traps stack. First, taking 25% OFF the selling price (60 x 0.75 = 45) instead of dividing by 1.25 - profit is measured on cost, so the cost is SP/1.25 = 48, not 45. Second, writing the ratio as 8:12 = 2:3 (choice B): in alligation the quantities are the inverse of the distances, so the tea whose price is CLOSER to the average is the one you need MORE of.
Q27 · ARI-PCT-073 — Answer: 10
The insight: under compound interest every year multiplies the balance by the same factor (1 + r). The balance goes from $5,000 at the end of year 2 to $5,500 at the end of year 3, so that factor is: 5500/5000 = 1.10 Therefore 1 + r = 1.10, and r = 0.10 = 10%. Why the shortcut works algebraically: P(1+r)^3 / P(1+r)^2 = (1+r), and both P and the exponents cancel. Answer: 10.
Trap. Setting up P(1+r)^2 = 5000 and P(1+r)^3 = 5500 as a system in two unknowns and grinding through it. Dividing one equation by the other eliminates P in one step. The other error is computing 500/5500 = 9.09%, using the END balance as the base instead of the balance the interest was earned on.
Q28 · ARI-PCT-074 — Answer: A, C, E
The condition is (1 + x/100)(1 + y/100) = 1.44. A - 1.20 x 1.20 = 1.44. TRUE. B - 1.10 x 1.30 = 1.43, not 1.44. FALSE. C - 1.44 x 1.00 = 1.44. TRUE. D - 1.20 x 1.21 = 1.452, not 1.44. FALSE. E - 2.00 x 0.72 = 1.44. TRUE. Answer: A, C, E.
Trap. Every option except E has x + y within a whisker of 40, so additive reasoning cannot tell A (40) from B (40) or D (41) - and adding to the target 44 points only at C, which happens to be correct for the wrong reason. The test is the PRODUCT of the multipliers, never the sum of the percents: 1.10 x 1.30 = 1.43 and 1.20 x 1.21 = 1.452 both miss 1.44. E checks whether you will accept a negative second year, which the stem allows.
Q29 · ARI-PCT-075 — Answer: A
Work from D, the base at the far end of the chain, and attach each percent to its own base. C = 0.60D B = C reduced by 25% = 0.75C = 0.75 x 0.60D = 0.45D A = B increased by 20% = 1.20B = 1.20 x 0.45D = 0.54D So A = 54% of D. One line: 1.20 x 0.75 x 0.60 = 0.54. Answer: 54%.
Trap. Adding and subtracting the percents - 60 - 25 + 20 = 55 - which is planted as choice B. Each percent is measured against a different quantity, so they cannot be combined additively. Choice E, 72%, is the second live error: chaining 1.20 x 0.60 and dropping the 25% cut entirely.
Q30 · ARI-PCT-077 — Answer: B
Column A: 1.10^10. Build it up - 1.10^2 = 1.21, 1.10^4 = 1.4641, 1.10^5 = 1.61051, and 1.10^10 = 1.61051^2 = 2.5937. That is a 159.4% increase. Column B: 1.05^20. Here 1.05^2 = 1.1025, 1.05^4 = 1.2155, 1.05^5 = 1.2763, 1.05^10 = 1.2763^2 = 1.6289, and 1.05^20 = 1.6289^2 = 2.6533. That is a 165.3% increase. 165.3% > 159.4%, so Column B is greater. The principle: for a fixed product r x n, spreading the growth over more compounding periods always ends higher, because interest gets more chances to earn interest.
Trap. Both columns have r x n = 100, so students read them as equivalent and choose C. That would only hold under simple growth. Under compounding the number of periods matters on its own, and more periods at a smaller rate wins.