Ratios & Proportions — Practice Set
Bank code: ARI-RAT · Section: Arithmetic · 30 questions — Easy 5 · Medium 8 · Hard 12 · Extreme 5
A ratio is not an amount. Write ak and bk first — until you have named the multiplier you have no numbers at all.
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-RAT-003 · Numeric Entry · 45s
The ratio of boys to girls in a class is 4:3. If there are 28 boys, how many students are in the class? Enter your answer as a number.
Numeric entry — write the number.
Q2 · ARI-RAT-008 · MCQ · 45s
$1,200 is divided among three people in the ratio 2:3:5. How much does the person with the largest share receive?
(A) $120
(B) $240
(C) $360
(D) $600
(E) $720
Q3 · ARI-RAT-010 · MCQ · 45s
If p:q = 4:7, what is p:(p+q)?
(A) 4:7
(B) 4:11
(C) 4:3
(D) 7:11
(E) 11:7
Q4 · ARI-RAT-016 · QC · 45s
m and n are positive with m:n = 5:9.
Column A: m + n Column B: 2m
Q5 · ARI-RAT-020 · Select all that apply · 60s
The ratio of A to B is 3:4, and A and B are positive. Which of the following must be true? Select all that apply.
(A) A < B
(B) A = 0.75B
(C) B/A = 4/3
(D) A + B = 7
(E) A:B = 6:8
Level 2 · Medium — 8 questions · ~10 min
two or three steps, one planted trap each
Q6 · ARI-RAT-021 · MCQ · 80s
The ratio of men to women on a committee is 5:3. If 4 more women join and no men leave, the ratio becomes 5:4. How many men are on the committee?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30
Q7 · ARI-RAT-022 · Numeric Entry · 85s
The ratio of Priya's age to Raj's age is 4:5. Six years ago the ratio was 3:4. What is Priya's current age? Enter your answer as a number.
Numeric entry — write the number.
Q8 · ARI-RAT-023 · MCQ · 75s
If a:b = 3:4 and b:c = 5:6, what is a:b:c?
(A) 3:4:5
(B) 15:20:24
(C) 3:5:6
(D) 18:24:20
(E) 15:24:20
Q9 · ARI-RAT-026 · Select all that apply · 90s
A solution of salt and water is in the ratio 1:4 by mass. Which of the following must be true? Select all that apply.
(A) Salt is 20% of the solution by mass
(B) Water is 80% of the solution by mass
(C) If 100 g of water is added, the new salt:water ratio is 1:5
(D) The mass of salt is one quarter the mass of water
(E) The mass of water is one quarter the mass of salt
Q10 · ARI-RAT-032 · QC · 65s
x varies inversely with y. When x = 6, y = 8.
Column A: y when x = 4 Column B: 12
Q11 · ARI-RAT-034 · Numeric Entry · 70s
$7,200 is divided among A, B and C in the ratio 2:3:4. How many dollars more does C receive than A? Enter your answer as a number.
Numeric entry — write the number.
Q12 · ARI-RAT-035 · MCQ · 80s
The ratio of petrol to kerosene in a mixture is 7:3. When 10 litres of kerosene is added, the ratio becomes 7:4. How many litres of petrol are in the original mixture?
(A) 30
(B) 40
(C) 49
(D) 63
(E) 70
Q13 · ARI-RAT-036 · QC · 60s
a:b = 2:5 and c:d = 3:7, and all four quantities are positive.
Column A: a/b Column B: c/d
Level 3 · Hard — 12 questions · ~19 min
where 162+ is won or lost
Q14 · ARI-RAT-051 · MCQ · 110s
The ratio of A's income to B's income is 3:2, and the ratio of their expenditures is 5:3. If each of them saves $500 per month, what is A's monthly income?
(A) $1,500
(B) $2,000
(C) $2,500
(D) $3,000
(E) $3,500
Q15 · ARI-RAT-052 · QC · 80s
a, b, c and d are positive, with a:b = 3:4 and c:d = 5:6.
Column A: (ac):(bd) Column B: 15:24
Q16 · ARI-RAT-053 · Numeric Entry · 95s
Three quantities satisfy A:B = 2:3 and B:C = 4:5. If A + B + C = 140, what is B? Enter your answer as a number.
Numeric entry — write the number.
Q17 · ARI-RAT-054 · MCQ · 105s
x, y, a and b are positive with x:y = a:b. Which of the following is NOT necessarily true?
(A) (x + y):(x - y) = (a + b):(a - b), given x is not equal to y
(B) (2x + y):(2x - y) = (2a + b):(2a - b), given 2x is not equal to y
(C) x^2:y^2 = a^2:b^2
(D) (x + y):y = (a + b):b
(E) (x + 2):(y + 2) = (a + 2):(b + 2)
Q18 · ARI-RAT-055 · Select all that apply · 100s
The ratio of boys to girls in a class is 4:5. After n boys join and no one leaves, the ratio becomes 1:1. If the class originally had fewer than 50 students, which of the following could be the value of n? Select all that apply.
(A) 1
(B) 4
(C) 5
(D) 8
(E) 10
Q19 · ARI-RAT-057 · Numeric Entry · 90s
If x:y:z = 2:3:5 and x^2 + y^2 + z^2 = 152, what is z? Enter your answer as a number.
Numeric entry — write the number.
Q20 · ARI-RAT-058 · MCQ · 95s
In 70 litres of juice, the ratio of water to concentrate is 5:2. Water is boiled off until the ratio is 3:2. How many litres of water were removed?
(A) 10
(B) 14
(C) 20
(D) 28
(E) 35
Q21 · ARI-RAT-061 · MCQ · 100s
If a/b = b/c = c/d = 2/3, what is (a + b + c)/(b + c + d)?
(A) 2/3
(B) 4/9
(C) 8/27
(D) 1
(E) 3/2
Q22 · ARI-RAT-064 · QC · 85s
x and y are positive with x/y = 4/5.
Column A: (x + 4)/(y + 4) Column B: x/y
Q23 · ARI-RAT-065 · Select all that apply · 95s
a:b = 3:5, with a and b positive. Which of the following are equal to b/(a + b)? Select all that apply.
(A) 5/8
(B) (b/a)/(1 + b/a)
(C) 1/(1 + a/b)
(D) 5/(3 + 5)
(E) [(b - a)/b] x [b/(b - a)]
Q24 · ARI-RAT-068 · QC · 75s
p, q and r are positive, with p:q = 2:3 and q:r = 3:5.
Column A: p + r Column B: 2q
Q25 · ARI-RAT-069 · MCQ · 90s
If a:b:c = 2:3:4, what is the value of (a^2 + b^2 + c^2)/(ab + bc + ca)?
(A) 29/26
(B) 29/24
(C) 29/23
(D) 29/28
(E) 29/22
Level 4 · Extreme — 5 questions · ~9 min
165+ — expect to need the insight, not the grind
Q26 · ARI-RAT-072 · QC · 105s
x, y and z are positive, with x:y = 2:3 and y:z = 4:5.
Column A: x(y + z) Column B: y(x + z)
Q27 · ARI-RAT-074 · MCQ · 100s
If x:y = 3:4, y:z = 2:3 and z:w = 5:6, what is x:w?
(A) 5:12
(B) 5:9
(C) 3:8
(D) 5:6
(E) 1:4
Q28 · ARI-RAT-077 · Numeric Entry · 110s
p:q = 3:5 with p + q = 40, and r:s = 7:3 with r - s = 16. What is the value of (p + r)/(q + s)? Enter your answer as a decimal rounded to two places.
Numeric entry — write the number.
Q29 · ARI-RAT-078 · MCQ · 120s
A cash box holds 50-cent, 25-cent and 10-cent coins in the ratio 5:9:4 by number. If the total value of the coins is $206, how many 25-cent coins are there?
(A) 72
(B) 90
(C) 144
(D) 180
(E) 360
Q30 · ARI-RAT-080 · Select all that apply · 125s
a, b, c and d are positive with a/b = c/d = k. Which of the following expressions equal k for all such values? Select all that apply.
(A) (a + c)/(b + d)
(B) (a - c)/(b - d), given b is not equal to d
(C) (ma + nc)/(mb + nd), for any positive m and n
(D) sqrt(ac)/sqrt(bd)
(E) (a + c)/(b - d), given b is not equal to d
Part B — Answer Key
| Q | ID | Level | Type | Answer |
|---|---|---|---|---|
| 1 | ARI-RAT-003 |
easy | Numeric Entry | 49 |
| 2 | ARI-RAT-008 |
easy | MCQ | D |
| 3 | ARI-RAT-010 |
easy | MCQ | B |
| 4 | ARI-RAT-016 |
easy | QC | A |
| 5 | ARI-RAT-020 |
easy | Select all that apply | A, B, C, E |
| 6 | ARI-RAT-021 |
medium | MCQ | C |
| 7 | ARI-RAT-022 |
medium | Numeric Entry | 24 |
| 8 | ARI-RAT-023 |
medium | MCQ | B |
| 9 | ARI-RAT-026 |
medium | Select all that apply | A, B, D |
| 10 | ARI-RAT-032 |
medium | QC | C |
| 11 | ARI-RAT-034 |
medium | Numeric Entry | 1600 |
| 12 | ARI-RAT-035 |
medium | MCQ | E |
| 13 | ARI-RAT-036 |
medium | QC | B |
| 14 | ARI-RAT-051 |
hard | MCQ | D |
| 15 | ARI-RAT-052 |
hard | QC | C |
| 16 | ARI-RAT-053 |
hard | Numeric Entry | 48 |
| 17 | ARI-RAT-054 |
hard | MCQ | E |
| 18 | ARI-RAT-055 |
hard | Select all that apply | A, B, C |
| 19 | ARI-RAT-057 |
hard | Numeric Entry | 10 |
| 20 | ARI-RAT-058 |
hard | MCQ | C |
| 21 | ARI-RAT-061 |
hard | MCQ | A |
| 22 | ARI-RAT-064 |
hard | QC | A |
| 23 | ARI-RAT-065 |
hard | Select all that apply | A, B, C, D |
| 24 | ARI-RAT-068 |
hard | QC | A |
| 25 | ARI-RAT-069 |
hard | MCQ | A |
| 26 | ARI-RAT-072 |
extreme hard | QC | B |
| 27 | ARI-RAT-074 |
extreme hard | MCQ | A |
| 28 | ARI-RAT-077 |
extreme hard | Numeric Entry | 1.16 |
| 29 | ARI-RAT-078 |
extreme hard | MCQ | E |
| 30 | ARI-RAT-080 |
extreme hard | Select all that apply | A, B, C, D |
Part C — Worked Solutions
Q1 · ARI-RAT-003 — Answer: 49
Step 1 — name the multiplier: boys = 4k, girls = 3k. Step 2 — find k: 4k = 28, so k = 7. Step 3 — the question asks for the TOTAL, which is 7 parts: 7k = 7 x 7 = 49. Check: boys 28, girls 3 x 7 = 21, and 28 + 21 = 49. Correct. Answer: 49.
Trap. Answering 21 - that is the number of girls, not the class. The ratio 4:3 is part-to-part, so the whole is 4 + 3 = 7 parts, not 4 or 3.
Q2 · ARI-RAT-008 — Answer: D
Step 1 — total parts = 2 + 3 + 5 = 10. Step 2 — one part = 1200/10 = $120. Step 3 — the largest share is 5 parts: 5 x 120 = $600. Check: 240 + 360 + 600 = $1,200. Correct. Answer: $600.
Trap. Answering $120 (the value of one part) or $240 (the smallest share). Choice E, $720, comes from dropping the third term and splitting $1,200 on 2:3 alone, which gives 480 and 720. Always compute the value of ONE part first - here 1200/10 = 120 - then multiply by the number of parts the question asked for.
Q3 · ARI-RAT-010 — Answer: B
Step 1 — name the multiplier: p = 4k, q = 7k. Step 2 — p + q = 4k + 7k = 11k. Step 3 — p : (p + q) = 4k : 11k = 4 : 11, since k cancels. Answer: 4:11.
Trap. Choosing 7:11, which is q as a share of the whole, not p. This is the part-to-part versus part-to-whole switch: 4:7 compares p with q, while 4:11 compares p with the total.
Q4 · ARI-RAT-016 — Answer: A
Step 1 — name the multiplier: m = 5k, n = 9k, with k > 0. Step 2 — Column A = m + n = 14k. Step 3 — Column B = 2m = 10k. Step 4 — 14k > 10k for every positive k, so the answer does not depend on k at all. Column A is greater.
Trap. Choosing D because no actual values are given. Once both columns are written in terms of the same k, the k cancels out of the comparison - a ratio question is 'not determinable' only if the two columns scale differently.
Q5 · ARI-RAT-020 — Answer: A, B, C, E
Write A = 3k and B = 4k with k > 0. A - 3k < 4k for every positive k. TRUE. B - A/B = 3/4 = 0.75, so A = 0.75B. TRUE. C - B/A = 4k/3k = 4/3. TRUE. D - A + B = 7k, which equals 7 only if k = 1. The ratio does not fix the size. FALSE. E - 6:8 divides down to 3:4, the same ratio. TRUE. Answer: A, B, C, E.
Trap. Choice D is the whole point of the question: a ratio fixes the proportion, never the amounts. A and B could be 30 and 40, or 300 and 400. Any statement that pins down an actual value from a ratio alone is false.
Q6 · ARI-RAT-021 — Answer: C
Step 1 — name the multiplier: men = 5k, women = 3k. Step 2 — the invariant is the men: their count never changes. Only the women's count moves. 5k/(3k + 4) = 5/4 Step 3 — cross-multiply: 4 x 5k = 5 x (3k + 4), so 20k = 15k + 20. Step 4 — 5k = 20, so k = 4. Step 5 — men = 5k = 20. Check: 20 men, 12 women; after 4 join, 20:16 = 5:4. Correct. Answer: 20.
Trap. Answering 25 or 30 by solving for the wrong quantity - 5k + 4 or the new total - instead of the men. The structural error behind both is giving the second ratio a NEW multiplier (men = 5m, women = 4m), which severs the link to the first ratio; the single k must carry through both statements.
Q7 · ARI-RAT-022 — Answer: 24
Step 1 — name the multiplier for the PRESENT: Priya = 4k, Raj = 5k. Step 2 — six years ago BOTH were six years younger: (4k - 6)/(5k - 6) = 3/4 Step 3 — cross-multiply: 4(4k - 6) = 3(5k - 6), so 16k - 24 = 15k - 18. Step 4 — k = 6. Step 5 — Priya = 4k = 24. Check: now 24 and 30; six years ago 18 and 24, and 18:24 = 3:4. Correct. Answer: 24.
Trap. Subtracting 6 from only one age, or trying to work with the ratios directly (4 - 6 and 5 - 6 give negatives). Time moves for everyone: whatever you add or subtract must be applied to every person in the ratio.
Q8 · ARI-RAT-023 — Answer: B
Step 1 — b appears in both ratios but with different values (4 and 5), so scale each ratio until the b terms match. The LCM of 4 and 5 is 20. Step 2 — a:b = 3:4, multiply both terms by 5 to get 15:20. Step 3 — b:c = 5:6, multiply both terms by 4 to get 20:24. Step 4 — now b is 20 in both, so a:b:c = 15:20:24. Check: 15:20 divides to 3:4, and 20:24 divides to 5:6. Correct. Answer: 15:20:24.
Trap. Writing 3:4:6 or 3:5:6 by stapling the two ratios together without matching the shared term. The b in '3:4' and the b in '5:6' are the same quantity, so they must be represented by the same number before the chain means anything.
Q9 · ARI-RAT-026 — Answer: A, B, D
Write salt = k and water = 4k, so the whole solution is 5k. A - salt/solution = k/5k = 1/5 = 20%. TRUE. B - water/solution = 4k/5k = 4/5 = 80%. TRUE. C - adding 100 g of water gives k : (4k + 100), which is 1:5 only if k = 100. Nothing in the question fixes k. FALSE. D - salt/water = k/4k = 1/4. TRUE. E - water/salt = 4k/k = 4, so the water is four TIMES the salt, not one quarter of it. FALSE. Answer: A, B, D.
Trap. C is the main bait: an absolute amount (100 g) cannot be combined with a ratio until the ratio has been anchored to a real quantity. E is the direction trap - 1:4 read left to right means salt is the quarter, and reversing the two names reverses the fraction.
Q10 · ARI-RAT-032 — Answer: C
Step 1 — inverse variation means the PRODUCT is constant: xy = c. Step 2 — c = 6 x 8 = 48. Step 3 — when x = 4: y = 48/4 = 12. Column A = 12 = Column B, so the two quantities are equal.
Trap. Scaling y the same direction as x. Here x falls by a factor of 6/4 = 1.5, so students shrink y as well and get around 5.3. Under inverse variation the two move in opposite directions: x divided by 1.5 means y multiplied by 1.5, and 8 x 1.5 = 12.
Q11 · ARI-RAT-034 — Answer: 1600
Step 1 — total parts = 2 + 3 + 4 = 9. Step 2 — one part = 7200/9 = $800. Step 3 — the difference between C and A is 4 - 2 = 2 parts, so 2 x 800 = $1,600. Check the long way: A = 1600, B = 2400, C = 3200; 3200 - 1600 = 1600, and 1600 + 2400 + 3200 = 7200. Correct. Answer: 1600.
Trap. Dividing 7200 by the largest part, or by 3 people, instead of by the 9 parts. The other error is answering $3,200 (C's whole share) rather than the difference the question asked for. Once one part is known, any question about the split is a single multiplication - so find one part first, always.
Q12 · ARI-RAT-035 — Answer: E
Step 1 — name the multiplier: petrol = 7k, kerosene = 3k. Step 2 — spot the invariant: no petrol is added or removed, so the petrol stays at 7k throughout. 7k/(3k + 10) = 7/4 Step 3 — cross-multiply: 4 x 7k = 7 x (3k + 10), so 28k = 21k + 70. Step 4 — 7k = 70, so k = 10, and petrol = 7k = 70 litres. Check: 70 petrol, 30 kerosene; add 10 kerosene to get 70:40 = 7:4. Correct. Shortcut worth seeing: the petrol figure 7 is unchanged in both ratios, so the ratios are already on a common scale - the kerosene went from 3 parts to 4 parts, meaning 10 litres is exactly one part, so petrol = 7 x 10 = 70. Answer: 70.
Trap. Adding the 10 litres to both components, which gives k = 30/7 and a petrol figure near choice A. Only kerosene was poured in. Choice B, 40, is the new kerosene volume rather than the petrol. Notice the '7' is identical in both ratios - that is a gift, telling you the two ratios are already on the same scale, so one part is exactly the 10 litres added.
Q13 · ARI-RAT-036 — Answer: B
Column A = 2/5 and Column B = 3/7. Cross-multiply to compare without decimals: 2 x 7 = 14 against 5 x 3 = 15. The product on the Column B side is larger, so 3/7 > 2/5. Column B is greater. Decimal check: 2/5 = 0.400 and 3/7 = 0.4286. Correct.
Trap. Comparing numerators and denominators separately - 2 < 3 and 5 < 7, so 'it could go either way', which leads to D. Two positive fractions can always be ordered by cross-multiplication, so D is never right here.
Q14 · ARI-RAT-051 — Answer: D
Two independent ratios need two independent multipliers. Step 1 — incomes: A = 3k, B = 2k. Expenditures: A = 5m, B = 3m. Step 2 — savings = income - expenditure, and both save 500: 3k - 5m = 500 ... (i) 2k - 3m = 500 ... (ii) Step 3 — eliminate m. Multiply (i) by 3 and (ii) by 5: 9k - 15m = 1500 10k - 15m = 2500 Step 4 — subtract the first from the second: k = 1000. Step 5 — A's income = 3k = $3,000. Check: from (ii), 2000 - 3m = 500 so m = 500. Incomes 3000 and 2000; expenditures 2500 and 1500; savings 500 and 500. Correct. Answer: $3,000.
Trap. Every wrong option here is a right calculation answering the wrong question: $2,000 is B's income, $2,500 is A's expenditure, $1,500 is B's expenditure. The setup error that causes the rest is using the SAME letter for both ratios - income and expenditure are unrelated quantities and each needs its own multiplier.
Q15 · ARI-RAT-052 — Answer: C
Step 1 — name separate multipliers: a = 3k, b = 4k, c = 5m, d = 6m. Step 2 — ac = 3k x 5m = 15km, and bd = 4k x 6m = 24km. Step 3 — (ac):(bd) = 15km : 24km = 15:24, since km cancels. Column B is also 15:24, so the two are equal. Note that 15:24 both reduce to 5:8, but a ratio and its reduced form are the same ratio - the comparison is unaffected.
Trap. Choosing D because a, b, c, d are unknown, or B because Column A 'ought to be simplified to 5:8' and 5:8 looks smaller than 15:24. Ratios multiply term by term - (a/b) x (c/d) = ac/bd - and 15:24 and 5:8 are the same number.
Q16 · ARI-RAT-053 — Answer: 48
Step 1 — B appears in both ratios as 3 and as 4, so scale until the two match. The LCM of 3 and 4 is 12. A:B = 2:3, multiply by 4 to get 8:12. B:C = 4:5, multiply by 3 to get 12:15. Step 2 — the chain: A:B:C = 8:12:15. Step 3 — total parts = 8 + 12 + 15 = 35, so one part = 140/35 = 4. Step 4 — B = 12 parts = 12 x 4 = 48. Check: A = 32, B = 48, C = 60; sum 140; 32:48 = 2:3 and 48:60 = 4:5. Correct. Answer: 48.
Trap. Dividing 140 by the parts of a chain that was never made consistent, for example using 2:3:5 (total 10) or 2:4:5 (total 11). Both give a whole number, so the arithmetic feels fine while the answer is wrong - the only defence is checking that both original ratios still hold at the end.
Q17 · ARI-RAT-054 — Answer: E
x:y = a:b means x = ta and y = tb for some positive t. Everything hinges on whether t survives the operation. A - (x + y)/(x - y) = (ta + tb)/(ta - tb) = t(a + b)/[t(a - b)] = (a + b)/(a - b). The t cancels. TRUE. B - (2ta + tb)/(2ta - tb) = t(2a + b)/[t(2a - b)] = (2a + b)/(2a - b). The t cancels. TRUE. C - (ta)^2/(tb)^2 = t^2 a^2/(t^2 b^2) = a^2/b^2. TRUE. D - (x + y)/y = (ta + tb)/(tb) = (a + b)/b. The t cancels. TRUE. E - (ta + 2)/(tb + 2). The 2 is not multiplied by t, so it does not cancel. Take a:b = 1:2 with t = 3, so x = 3 and y = 6: (3 + 2)/(6 + 2) = 5/8, while (1 + 2)/(2 + 2) = 3/4. Different. FALSE. Answer: E.
Trap. Assuming every algebraic move preserves a ratio. The test is simple: multiplying, dividing, squaring, and adding or subtracting MULTIPLES of the terms all keep the common factor t in play so it cancels. Adding a plain CONSTANT does not - it sits outside the scaling and destroys the proportion.
Q18 · ARI-RAT-055 — Answer: A, B, C
Step 1 — name the multiplier: boys = 4k, girls = 5k, k a positive integer. Step 2 — after n boys join the counts are equal: 4k + n = 5k, so n = k. Step 3 — apply the size limit. The original class was 4k + 5k = 9k students, and 9k < 50 gives k <= 5. Step 4 — so n = k can be 1, 2, 3, 4 or 5. A (1) yes. B (4) yes. C (5) yes - the class had 45 students, still under 50. D (8) would need 72 students. E (10) would need 90. Answer: A, B, C.
Trap. Missing that n is forced to equal k, and instead testing values at random. The other error is ignoring the 'fewer than 50' clause and accepting every option, since without it any positive integer n works. On the GRE a size constraint tacked onto a ratio question is never decoration - it is what makes the answer finite.
Q19 · ARI-RAT-057 — Answer: 10
Step 1 — name the multiplier: x = 2k, y = 3k, z = 5k. Step 2 — substitute into the given equation: (2k)^2 + (3k)^2 + (5k)^2 = 4k^2 + 9k^2 + 25k^2 = 38k^2 Step 3 — 38k^2 = 152, so k^2 = 4 and k = 2 (positive, since the quantities are). Step 4 — z = 5k = 10. Check: x = 4, y = 6, z = 10, and 16 + 36 + 100 = 152. Correct. Answer: 10.
Trap. Squaring only the ratio numbers and writing 2^2 + 3^2 + 5^2 = 38 = 152, which gives no k at all, or squaring k but not the coefficients (2k^2 instead of (2k)^2 = 4k^2). Every term of the ratio carries the k, so the k gets squared along with its coefficient.
Q20 · ARI-RAT-058 — Answer: C
Step 1 — split the original 70 litres. Total parts = 5 + 2 = 7, so one part = 10 litres. Water = 5 x 10 = 50 L, concentrate = 2 x 10 = 20 L. Step 2 — spot the invariant: only water leaves, so the concentrate stays at 20 L throughout. Step 3 — in the new ratio 3:2, the concentrate is 2 parts = 20 L, so one new part = 10 L and the water is 3 x 10 = 30 L. Step 4 — water removed = 50 - 30 = 20 litres. Check: 30:20 = 3:2. Correct. Answer: 20 litres.
Trap. Subtracting the removed amount from BOTH components, i.e. solving (50 - x):(20 - x) = 3:2, which gives a negative x and no usable answer. Nothing left the concentrate. Choice A, 10, is one original part; choice D, 28, comes from applying 3:2 to the whole 70 litres and forgetting the concentrate is fixed. Anchor the new ratio to the component that did not change.
Q21 · ARI-RAT-061 — Answer: A
Step 1 — pick clean numbers by starting at the far end. Let d = 27. c/d = 2/3, so c = 18. b/c = 2/3, so b = 12. a/b = 2/3, so a = 8. Step 2 — substitute: (a + b + c) = 8 + 12 + 18 = 38 (b + c + d) = 12 + 18 + 27 = 57 Step 3 — 38/57 = 2/3. Why it must come out this way: a = (2/3)b, b = (2/3)c, c = (2/3)d, so the whole numerator is (2/3) times the whole denominator term by term. Adding matched terms keeps the ratio. Answer: 2/3.
Trap. Assuming three links in the chain compound, giving (2/3)^3 = 8/27, or two links giving 4/9. The chain does compound between a and d - a/d is indeed 8/27 - but the sums here are matched term for term, each pair in the ratio 2:3, so the combined ratio is still 2:3.
Q22 · ARI-RAT-064 — Answer: A
Step 1 — x/y = 4/5 < 1, so x < y. Step 2 — adding the same positive amount to the top and bottom of a fraction pulls it towards 1. Below 1, that means it goes UP. Step 3 — test it with the smallest case, x = 4 and y = 5: (4 + 4)/(5 + 4) = 8/9 = 0.889, which is greater than 4/5 = 0.8. Step 4 — test a scaled case, x = 40 and y = 50: 44/54 = 0.815, still greater than 0.8. The direction never changes, so Column A is greater. Proof if you want it: (x + 4)/(y + 4) - x/y = 4(y - x)/[y(y + 4)], which is positive whenever y > x.
Trap. Believing that adding the same number to both parts leaves a ratio alone - that is true for MULTIPLYING both parts, not adding. Note the rule has a direction: a fraction below 1 rises, a fraction above 1 falls, and both are moving towards 1.
Q23 · ARI-RAT-065 — Answer: A, B, C, D
The target: b/(a + b) = 5k/(3k + 5k) = 5/8. A - 5/8 exactly. TRUE. B - b/a = 5/3, so this is (5/3)/(1 + 5/3) = (5/3)/(8/3) = 5/8. TRUE. C - a/b = 3/5, so this is 1/(1 + 3/5) = 1/(8/5) = 5/8. TRUE. D - 5/(3 + 5) = 5/8, the part-over-whole written straight from the ratio numbers. TRUE. E - the two factors are reciprocals of each other, so the product is 1, not 5/8. FALSE. Answer: A, B, C, D.
Trap. Choice E is designed to look elaborate enough to be right. Simplify before evaluating: (b - a)/b times b/(b - a) cancels completely to 1. Choices B and C are the same identity written from the two ends - dividing top and bottom of b/(a + b) by a gives B, and dividing by b gives C.
Q24 · ARI-RAT-068 — Answer: A
Step 1 — the q term is already 3 in both ratios, so the chain needs no rescaling: p:q:r = 2:3:5. Step 2 — write them on one multiplier: p = 2k, q = 3k, r = 5k. Step 3 — Column A = p + r = 2k + 5k = 7k. Step 4 — Column B = 2q = 6k. Step 5 — 7k > 6k for every positive k, so the k cancels out of the comparison. Column A is greater.
Trap. Choosing D because no actual values are given. The whole comparison reduces to 7 against 6 once both columns are on the same multiplier. Watch also for the case where the shared term does NOT already match - here q is 3 in both ratios, which is why no rescaling was needed; if it had been 3 and 4 you would have to build 8:12:20 first.
Q25 · ARI-RAT-069 — Answer: A
Step 1 — every term on top and bottom is a product of two of the quantities, so the k^2 cancels and you can simply use a = 2, b = 3, c = 4. Step 2 — numerator: 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29. Step 3 — denominator: ab + bc + ca = (2)(3) + (3)(4) + (4)(2) = 6 + 12 + 8 = 26. Step 4 — the value is 29/26. Formal check with the multiplier: numerator = 29k^2, denominator = 26k^2, and the k^2 divides out. Answer: 29/26.
Trap. Miscounting the third product in the denominator. Writing ab twice - 6 + 12 + 6 = 24 - gives 29/24, choice B; the correct third term is ca = 4 x 2 = 8. The structural point is that top and bottom are both degree 2, so the multiplier cancels and you may substitute the bare ratio numbers 2, 3, 4.
Q26 · ARI-RAT-072 — Answer: B
Step 1 — build the chain. y is 3 in the first ratio and 4 in the second, so scale to the LCM 12: x:y = 2:3 = 8:12 y:z = 4:5 = 12:15 So x:y:z = 8:12:15, meaning x = 8k, y = 12k, z = 15k. Step 2 — Column A = 8k(12k + 15k) = 8k x 27k = 216k^2. Step 3 — Column B = 12k(8k + 15k) = 12k x 23k = 276k^2. Step 4 — 276k^2 > 216k^2 for every positive k, so Column B is greater. A cleaner way to see it: expand both columns. Column A = xy + xz and Column B = xy + yz. The xy term is common to both, so the whole comparison is xz against yz, i.e. x against y - and x < y. Column B wins without any arithmetic.
Trap. Choosing D because no actual values are given. Every term in both columns is degree 2 in k, so k^2 cancels and the comparison is completely determined. The other slip is failing to rescale y to 12 before chaining, which gives the wrong numbers for z even though the final letter happens to survive.
Q27 · ARI-RAT-074 — Answer: A
When only the two ENDS of a chain are wanted, multiply the fractions rather than building the full chain. Step 1 — x/w = (x/y) x (y/z) x (z/w), because y and z cancel telescopically. Step 2 — x/w = (3/4) x (2/3) x (5/6). Step 3 — cancel before multiplying: the 3s cancel, leaving (1/4) x 2 x (5/6) = 10/24 = 5/12. Step 4 — so x:w = 5:12. Check by building the full chain: scale to x:y:z:w = 30:40:60:72, and 30:72 = 5:12. Correct. Answer: 5:12.
Trap. Writing one link upside down - flipping y:z to 3:2 gives (3/4) x (3/2) x (5/6) = 15/16, and a student who then reaches for the nearest listed answer picks D, 5:6. Each fraction must be written 'earlier term over later term' so the middle terms cancel. Building the full four-term chain also works but takes three rescalings.
Q28 · ARI-RAT-077 — Answer: 1.16
Step 1 — resolve the first ratio using the SUM. Parts = 3 + 5 = 8, and the sum is 40, so one part = 5. Then p = 15 and q = 25. Step 2 — resolve the second ratio using the DIFFERENCE. The gap is 7 - 3 = 4 parts, and that gap equals 16, so one part = 4. Then r = 28 and s = 12. Step 3 — p + r = 15 + 28 = 43, and q + s = 25 + 12 = 37. Step 4 — 43/37 = 1.1621..., which rounds to 1.16. Check: 15:25 reduces to 3:5, and 28:12 reduces to 7:3, with 28 - 12 = 16. Correct. Answer: 1.16.
Trap. Using the same handle for both ratios - dividing 16 by the total parts (10) instead of by the difference in parts (4), which makes r and s wrong. Read what the given number describes: a SUM is divided by the sum of the parts, a DIFFERENCE by the difference of the parts. The other error is combining the two ratios into one, which is invalid since they describe unrelated quantities and carry different multipliers.
Q29 · ARI-RAT-078 — Answer: E
Step 1 — the ratio counts COINS, but the total given is VALUE, so each part must be weighted by what that coin is worth. Numbers: 5k fifty-cent coins, 9k twenty-five-cent coins, 4k ten-cent coins. Step 2 — value in dollars: 5k x 0.50 = 2.50k 9k x 0.25 = 2.25k 4k x 0.10 = 0.40k Total = 2.50k + 2.25k + 0.40k = 5.15k Step 3 — 5.15k = 206, so k = 40. Step 4 — the number of 25-cent coins = 9k = 9 x 40 = 360. Check: 200 half-dollars ($100) + 360 quarters ($90) + 160 dimes ($16) = $206. Correct. Answer: 360.
Trap. Treating 5:9:4 as a ratio of VALUES rather than of coin counts and splitting $206 in that ratio. Choice D, 180, is the other live error: solving 5.15k = 206 correctly but halving, or dividing the total by 2 somewhere in the cents-to-dollars conversion. The ratio counts coins; the money is a weighted total, so each part must be multiplied by what that coin is worth.
Q30 · ARI-RAT-080 — Answer: A, B, C, D
The key substitution: a = kb and c = kd. A - (kb + kd)/(b + d) = k(b + d)/(b + d) = k. TRUE. B - (kb - kd)/(b - d) = k(b - d)/(b - d) = k, and b is not equal to d so the denominator is non-zero. TRUE. C - (mkb + nkd)/(mb + nd) = k(mb + nd)/(mb + nd) = k. TRUE. Note A is the case m = n = 1; B is the same identity with a minus sign, which works for exactly the same reason. D - sqrt(ac) = sqrt(kb x kd) = k sqrt(bd), so the expression is k sqrt(bd)/sqrt(bd) = k. TRUE (all quantities positive, so the roots are real). E - (kb + kd)/(b - d) = k(b + d)/(b - d), and (b + d)/(b - d) is not 1 in general. Test a:b = c:d = 1:2 with b = 2, d = 4: (1 + 2)/(2 - 4) = -1.5, not 0.5. FALSE. Answer: A, B, C, D.
Trap. Rejecting B on the assumption that subtracting must break the property - it does not, because the same k factors out of the numerator exactly as it does in A. E is the one that fails, and it fails for a structural reason: the numerator and denominator no longer pair the same terms, so nothing cancels. Every one of A to D is really the same identity - k factors out whenever top and bottom use the MATCHING combination of terms.