Question 1

Ages with present sum and past ratio: The present ages of A and B are in the ratio 3 : 5 and their sum is 56 years. What was the ratio of their ages 7 years ago?

Accepted Answer

Introduction / Context: This is a classic ages and ratios problem. We are given a current ratio and total, and then asked for the ratio at a past time. The technique is to first compute actual present ages and then shift both by the same amount for the past scenario. Given Data / Assumptions: Present ratio A : B = 3 : 5. A + B = 56 years. We need the ratio (A − 7) : (B − 7). Concept / Approach: Let A = 3k and B = 5k, so A + B = 8k = 56 ⇒ k = 7. Then find actual ages and subtract 7 years from both to get the past ages, and finally simplify that ratio if needed. Step-by-Step Solution: A = 3k = 21 years; B = 5k = 35 years.Seven years ago: A = 14, B = 28.Required ratio = 14 : 28 = 1 : 2. Verification / Alternative check: Check present sum: 21 + 35 = 56 (fits). The past ratio reduces perfectly to 1 : 2, which is consistent with equal subtraction from both ages. Why Other Options Are Wrong: 3 : 4 or 2 : 3 do not match the computed 14 : 28. 3 : 5 repeats the present ratio, not the past one. Common Pitfalls: Subtracting 7 only from the numerator term instead of both persons’ ages. Forgetting to compute exact ages before applying the 7-year shift. Final Answer: 1 : 2

Question 2

Percent-to-ratio conversion: Thirty-five percent of one number equals twice seventy-five percent of another number. Find the ratio of the first number to the second number.

Accepted Answer

Introduction / Context: Translating percentage statements into algebraic equations enables quick ratio determination. This question asks you to equate percent portions of two unknown numbers and express the result as a clean integer ratio. Given Data / Assumptions: Let the first number be x and the second be y. 35% of x equals twice 75% of y. Concept / Approach: Convert percentages to decimals and set up the equation. Then isolate x/y to get a ratio. Multiply numerator and denominator to clear decimals and express the ratio with integers in lowest terms. Step-by-Step Solution: 0.35x = 2 * 0.75y ⇒ 0.35x = 1.5y.x / y = 1.5 / 0.35.1.5 / 0.35 = 150 / 35 = 30 / 7.Therefore, the required ratio is 30 : 7. Verification / Alternative check: Pick y = 7 (for convenience). Then x = 30 satisfies 0.35 * 30 = 10.5 and 2 * 0.75 * 7 = 10.5, verifying the equality precisely. Why Other Options Are Wrong: 35 : 6, 23 : 7, and 32 : 9 do not satisfy 0.35x = 1.5y when tested. Common Pitfalls: Mishandling percent-to-decimal conversion (e.g., mixing 35 with 0.35). Forgetting to simplify the final ratio to lowest integer terms. Final Answer: 30 : 7

Question 3

Milk–water adjustment in a mixture: An 85 kg milk mixture contains milk and water in the ratio 27 : 7. How many kilograms of water should be added to make the ratio 3 : 1 (milk : water)?

Accepted Answer

Introduction / Context: This mixture problem involves keeping the amount of milk constant while adding water to reach a new target ratio. The technique is to compute current quantities from the initial ratio, then enforce the desired final ratio to solve for the added water. Given Data / Assumptions: Total mixture = 85 kg. Initial milk : water = 27 : 7. Final required milk : water = 3 : 1. Only water is added; milk quantity stays the same. Concept / Approach: Find initial milk and water using parts: total parts = 27 + 7 = 34. Then set up an equation where final milk remains 27 parts’ worth and final water equals 1 part relative to milk 3 parts (i.e., milk : water = 3 : 1). Step-by-Step Solution: Initial milk = 85 * 27/34 = 67.5 kg; initial water = 85 * 7/34 = 17.5 kg.Add x kg water ⇒ milk stays 67.5; water becomes 17.5 + x.Target ratio milk : water = 3 : 1 ⇒ 67.5 : (17.5 + x) = 3 : 1.67.5 = 3 * (17.5 + x) ⇒ 67.5 = 52.5 + 3x ⇒ 3x = 15 ⇒ x = 5 kg. Verification / Alternative check: Final totals: milk = 67.5, water = 22.5; 67.5 : 22.5 simplifies to 3 : 1, and new total becomes 90 kg, consistent with adding 5 kg water. Why Other Options Are Wrong: 6.5, 7.25, and 8 kg overshoot the water needed and do not yield exactly 3 : 1. Common Pitfalls: Accidentally changing the milk amount when only water is added. Forming milk : total ratio instead of milk : water ratio. Final Answer: 5 kg

Question 4

Three-part division with fractional conditions: ₹ 1870 is divided into three parts so that half of the first part, one-third of the second part, and one-sixth of the third part are all equal. Find the third part.

Accepted Answer

Introduction / Context: This partition problem gives fractional equalities on each part. Turning those into a single parameter makes the algebra straightforward and lets us recover each part from the total sum. Given Data / Assumptions: Total = ₹ 1870. a/2 = b/3 = c/6. We need c (the third part). Concept / Approach: Let a/2 = b/3 = c/6 = k. Then a = 2k, b = 3k, c = 6k. Use a + b + c = 1870 to find k and then compute c directly as 6k. Step-by-Step Solution: a = 2k, b = 3k, c = 6k.Sum = 2k + 3k + 6k = 11k = 1870.k = 1870 / 11 = 170.Third part c = 6k = 6 * 170 = ₹ 1020. Verification / Alternative check: Check the equalized fractions: a/2 = 340/2 = 170, b/3 = 510/3 = 170, c/6 = 1020/6 = 170, all consistent. Why Other Options Are Wrong: ₹ 510 and ₹ 680 are the other parts (b and a) or partial values, not the third part. ₹ 850 does not satisfy the fractional equalities. Common Pitfalls: Mistakenly using the reciprocals of fractions and assigning wrong multiples to parts. Final Answer: Rs. 1020

Question 5

Population composition with adult gender ratio: In a town, 80% of the population are adults. Among adults, men and women are in the ratio 9 : 7. If the number of adult women is 4.2 lakh, find the total population of the town.

Accepted Answer

Introduction / Context: Population structure problems commonly require moving between subgroup counts and overall totals using ratios and percentages. Here we move from adult women to adult total, and then to the overall population using the adult percentage. Given Data / Assumptions: Adults = 80% of total population. Adult men : adult women = 9 : 7. Adult women = 4.2 lakh. Concept / Approach: If adult women represent the “7 parts” in the 9 : 7 ratio, total adults correspond to 9 + 7 = 16 parts. Compute adult total from women, then scale up from 80% to get the town’s population. Step-by-Step Solution: Adult total = (16/7) * 4.2 lakh = 9.6 lakh.Adults are 80% of total ⇒ Total population = Adult total / 0.8.Total = 9.6 / 0.8 = 12 lakh. Verification / Alternative check: Adults = 12 × 0.8 = 9.6 lakh; women share = 7/16 of 9.6 = 4.2 lakh, consistent with the given data. Why Other Options Are Wrong: 9.6 lakh is adult total, not overall population. 9.8 and 11.6 lakh do not match the ratio-percentage chain. Common Pitfalls: Using 7/9 instead of 7/16 for the women fraction of adult total. Forgetting to scale adults to overall total using 80%. Final Answer: 12 lakh

Question 6

Coins inheritance and post-change ratio: Mr. Shrimant has 2505 gold coins divided among his sons Bharat, Parat, and Marat. Afterwards, Bharat sells 30 coins, Parat donates 30 coins, and Marat loses 25 coins. The numbers then stand in the ratio 46 : …

Accepted Answer

Introduction / Context: This problem mixes a total-sum constraint with a ratio that applies after specific adjustments to each person’s count. The key is to convert the post-change ratio into equations for the original numbers and then use the overall total to solve for the scaling factor. Given Data / Assumptions: Total coins initially = 2505. Post-change ratio (Bharat : Parat : Marat) = 46 : 41 : 34. Adjustments: Bharat −30, Parat −30, Marat −25 (respectively). Concept / Approach: Let the post-change numbers be 46t, 41t, and 34t. Then original numbers are Bharat = 46t + 30, Parat = 41t + 30, Marat = 34t + 25. The sum of originals equals 2505, which will determine t. Then compute Parat’s original figure. Step-by-Step Solution: (46t + 30) + (41t + 30) + (34t + 25) = 2505.121t + 85 = 2505 ⇒ 121t = 2420 ⇒ t = 20.Parat originally = 41t + 30 = 820 + 30 = 850. Verification / Alternative check: Originals: 46*20+30 = 950; 41*20+30 = 850; 34*20+25 = 705; total = 2505, consistent. Why Other Options Are Wrong: 705 and 950 are Marat’s and Bharat’s initial counts, not Parat’s. 800 does not satisfy the total when paired with the others. Common Pitfalls: Applying +30 instead of −30 (or vice versa) to the wrong son. Forgetting to add back lost/donated/sold coins to reach original counts. Final Answer: 850

Question 7

Equalized expressions from three shares: ₹ 600 is divided among A, B, and C such that the following three values are equal: 40 more than 2/5 of A’s share, 20 more than 2/7 of B’s share, and 10 more than 9/17 of C’s share. Find B’s share.

Accepted Answer

Introduction / Context: This algebraic division question equates three different linear expressions of A, B, and C to a common value. Expressing each share in terms of that common value, and then using the total, allows us to solve the system easily. Given Data / Assumptions: Total = ₹ 600. 2/5 A + 40 = 2/7 B + 20 = 9/17 C + 10 = k (say). Concept / Approach: From the equalities, write A, B, and C in terms of k: A = (5/2)(k − 40), B = (7/2)(k − 20), C = (17/9)(k − 10). Sum them to 600 to solve for k, then find B explicitly. Step-by-Step Solution: A = 5/2 (k − 40) = 2.5k − 100.B = 7/2 (k − 20) = 3.5k − 70.C = 17/9 (k − 10) = (17/9)k − 170/9.Sum: (5/2 + 7/2 + 17/9)k − (100 + 70 + 170/9) = 600.(71/9)k − 1700/9 = 600 ⇒ 71k − 1700 = 5400 ⇒ 71k = 7100 ⇒ k = 100.B = 7/2 (100 − 20) = 7/2 * 80 = ₹ 280. Verification / Alternative check: Compute A and C: A = 5/2 * 60 = 150; C = 17/9 * 90 ≈ 170; total 150 + 280 + 170 = 600. All three expressions equal k = 100, by construction. Why Other Options Are Wrong: ₹ 150 and ₹ 170 (not listed) align with A or C, not B. ₹ 185 and ₹ 285 come from arithmetic slips in solving for k or substituting into B. Common Pitfalls: Mistakes in fractional arithmetic when summing coefficients. Forgetting that each expression equals the same k. Final Answer: ₹ 280

Question 8

New ratio after percentage raises: Akash, Bablu, and Chintu have salaries in the ratio 2 : 3 : 5. Their salaries increase by 15%, 10%, and 20% respectively. What is the new ratio of their salaries?

Accepted Answer

Introduction / Context: Percentage changes on ratio terms can be handled by applying the change to each term and then simplifying to an integer ratio. This keeps relative comparisons correct after the increases. Given Data / Assumptions: Initial ratio A : B : C = 2 : 3 : 5. Raises: +15%, +10%, +20% respectively. Concept / Approach: Apply multipliers to each term: 1.15 for +15%, 1.10 for +10%, 1.20 for +20%. Then clear decimals by multiplying by a common factor (e.g., 20 or 10) and reduce if needed. Step-by-Step Solution: New A = 2 × 1.15 = 2.30.New B = 3 × 1.10 = 3.30.New C = 5 × 1.20 = 6.00.Scale by 10 ⇒ 23 : 33 : 60. Verification / Alternative check: Pick a base (e.g., ₹ 2000, ₹ 3000, ₹ 5000). After applying increases, their absolute values maintain the proportionality 23 : 33 : 60 when normalized, confirming the result. Why Other Options Are Wrong: 3 : 3 : 10 and 20 : 22 : 40 arise from mis-scaling or skipping percentage application. “None of these” is invalid since 23 : 33 : 60 is attainable exactly. Common Pitfalls: Adding percentages directly to the ratio numbers without using the correct multipliers. Forgetting to clear decimals to express the ratio in integers. Final Answer: 23 : 33 : 60

Question 9

Expenses vary with square of income: In January, Arun’s income and expenses were ₹ 15000 and ₹ 9000, respectively. His monthly expenses vary directly as the square of his monthly income. Find his income at which expenses exactly equal income.

Accepted Answer

Introduction / Context: This models a nonlinear cost pattern: expenses ∝ (income)^2. With one observed data point, we can find the proportionality constant and then compute the break-even income where expense equals income. Given Data / Assumptions: Income I₁ = 15000; Expense E₁ = 9000. E = k * I^2 for some constant k > 0. Find I such that E = I. Concept / Approach: Use the January data to solve for k. Then set I = k * I^2 and solve the quadratic I^2 − (1/k) I = 0 for the positive, nonzero root. Step-by-Step Solution: k = E₁ / I₁^2 = 9000 / (15000^2) = 9000 / 225000000 = 1 / 25000.At break-even: I = k * I^2 ⇒ I = (1/25000) * I^2.I^2 − 25000 I = 0 ⇒ I(I − 25000) = 0.Nonzero solution: I = ₹ 25000. Verification / Alternative check: At I = 25000, E = (1/25000) * 25000^2 = 25000, matching the break-even requirement exactly. Why Other Options Are Wrong: ₹ 15000 is the observed income, not the break-even point. ₹ 2000 and ₹ 35000 do not satisfy E = I under the derived k. Common Pitfalls: Treating expenses as proportional to income (linear) instead of income squared (quadratic). Arithmetic errors when computing k or solving the quadratic. Final Answer: ₹ 25000

Question 10

Leaps and stride length → speed ratio: A cat takes 5 leaps for every 4 leaps of a dog. Also, 3 dog leaps cover the same distance as 4 cat leaps. Find the ratio of the cat’s speed to the dog’s speed.

Accepted Answer

Introduction / Context: Speed equals strides per unit time multiplied by stride length. We are given a frequency ratio (leaps per equal time) and a stride-length relation between the animals. Combine both to obtain the speed ratio. Given Data / Assumptions: Cat leaps : Dog leaps (in same time) = 5 : 4. 3 dog leaps = 4 cat leaps (distance equality). All leaps occur over the same time window used for the frequency comparison. Concept / Approach: Let Lc and Ld be cat and dog leap lengths. From 3Ld = 4Lc ⇒ Ld = (4/3)Lc. Speeds over the same time T: Vc : Vd = (5Lc) : (4Ld). Substitute Ld to get a pure number ratio. Step-by-Step Solution: Ld = (4/3) Lc from 3Ld = 4Lc.Vc : Vd = 5Lc : 4Ld = 5Lc : 4*(4/3)Lc = 5 : 16/3 = 15 : 16. Verification / Alternative check: Pick Lc = 3 and one time unit; then Ld = 4. Cat covers 5*3 = 15 units, dog covers 4*4 = 16 units ⇒ ratio 15 : 16. Why Other Options Are Wrong: 11 : 15 and 15 : 11 ignore the stride-length relation. 16 : 15 inverts the correct order. Common Pitfalls: Multiplying or dividing the frequency and length ratios inconsistently. Confusing “3 dog leaps = 4 cat leaps” with the reverse. Final Answer: 15 : 16

Question 11

Monthly-to-annual income via ratio scaling: Sita, Riya, and Kunal earn monthly incomes in the ratio 84 : 76 : 89. If Riya’s annual income is ₹ 456000, find the sum of Sita’s and Kunal’s annual incomes.

Accepted Answer

Introduction / Context: Transforming a ratio into actual figures starts by finding the value of one “ratio part.” Since we know Riya’s annual total, we can deduce her monthly, compute the part value, and then obtain the others’ annual amounts. Given Data / Assumptions: Monthly ratio S : R : K = 84 : 76 : 89. Riya’s annual = ₹ 456000 ⇒ monthly = ₹ 38000. Find annual(Sita) + annual(Kunal). Concept / Approach: Each “part” equals Riya’s monthly divided by 76. Multiply the part by 84 for Sita’s monthly and by 89 for Kunal’s monthly; then annualize (×12) and add. Step-by-Step Solution: Value per part = 38000 / 76 = 500.Sita monthly = 84 * 500 = 42000; Kunal monthly = 89 * 500 = 44500.Annuals: Sita = 42000 * 12 = 504000; Kunal = 44500 * 12 = 534000.Sum = 504000 + 534000 = ₹ 1038000. Verification / Alternative check: Riya monthly = 76 * 500 = 38000 ⇒ annual 456000 matches the given. Ratios hold exactly when all are divided by 500. Why Other Options Are Wrong: ₹ 195000 is far too small (monthly-level figure). ₹ 983500 and ₹ 1130000 are off the computed sum from exact parts. Common Pitfalls: Using annual ratio parts directly without first converting Riya’s annual to monthly. Final Answer: ₹ 1038000

Question 12

Reconstruct original amounts from shifted ratio: ₹ 2186 is distributed among A, B, and C. If their amounts, after subtracting ₹ 26, ₹ 28, and ₹ 32 respectively, are in the ratio 9 : 13 : 8, find the amount originally given to A.

Accepted Answer

Introduction / Context: When a ratio applies after fixed adjustments, translate the ratio into variables for the post-change amounts, then add back the adjustments to obtain originals. Use the total to determine the scale factor. Given Data / Assumptions: Total = ₹ 2186. After changes: (A − 26) : (B − 28) : (C − 32) = 9 : 13 : 8. Concept / Approach: Let A − 26 = 9t, B − 28 = 13t, C − 32 = 8t. Then A = 9t + 26, B = 13t + 28, C = 8t + 32. Sum equals the total and yields t. Substitute back to find A. Step-by-Step Solution: A + B + C = (9t + 26) + (13t + 28) + (8t + 32) = 30t + 86.30t + 86 = 2186 ⇒ 30t = 2100 ⇒ t = 70.A = 9t + 26 = 630 + 26 = ₹ 656. Verification / Alternative check: Check post-change: A − 26 = 630, B − 28 = 910, C − 32 = 560 ⇒ ratio 630 : 910 : 560 = 9 : 13 : 8 (divide by 70). Sum of originals: 656 + 938 + 592 = 2186 confirms the total. Why Other Options Are Wrong: ₹ 575, ₹ 640, and ₹ 672 do not satisfy the total with the given post-change ratio. Common Pitfalls: Mistaking the 9 : 13 : 8 as the original ratio rather than the ratio after subtraction. Final Answer: ₹ 656

Question 13

Income allocation and trust savings: A person gives 2/5 of his income to his elder son and 30% to his younger son. The rest is saved in three trusts A, B, and C in the ratio 3 : 5 : 2. If the difference between the amounts received by the two sons i…

Accepted Answer

Introduction / Context: This problem combines fractional allocations with a remainder split in a specified ratio. The given difference between sons’ shares identifies total income, after which the savings portion can be divided among the trusts. Given Data / Assumptions: Elder son = 2/5 of income. Younger son = 30% = 3/10 of income. Difference between sons = ₹ 2000. Savings to A : B : C = 3 : 5 : 2. Concept / Approach: Difference between sons = (2/5 − 3/10) * Income = 1/10 * Income. Thus income = 10 * difference. The remainder after giving to sons is 1 − (2/5 + 3/10) = 3/10 of income, to be split 3 : 5 : 2 among A, B, C. Step-by-Step Solution: Income = 10 * 2000 = ₹ 20000.Savings portion = 3/10 of 20000 = ₹ 6000.Trust ratio total parts = 3 + 5 + 2 = 10.Trust C share = 2/10 of 6000 = ₹ 1200. Verification / Alternative check: Sons receive ₹ 8000 and ₹ 6000 (difference ₹ 2000). Savings ₹ 6000 split as A = ₹ 1800, B = ₹ 3000, C = ₹ 1200, matching the ratio 3 : 5 : 2. Why Other Options Are Wrong: ₹ 1000 and ₹ 1140 do not correspond to the 2/10 share of ₹ 6000. ₹ 1256 is not a clean part of the given ratio. Common Pitfalls: Mistaking the difference as 1/5 instead of 1/10 of income. Dividing the total income in the 3 : 5 : 2 ratio instead of only the savings. Final Answer: ₹ 1200

Question 14

Employee counts and wages ratio: In a factory, employees of types A, B, and C are in the ratio 9 : 13 : 18, with 54 employees of type C. Their per-employee wages are in the ratio 10 : 7 : 4. If each type-B employee earns ₹ 1400, find the total wages…

Accepted Answer

Introduction / Context: When counts and per-head wages follow separate ratios, the total payout to a category is the product of the number of employees in that category and the per-head wage. Known values allow solving the common wage factor and thus the required total. Given Data / Assumptions: Counts A : B : C = 9 : 13 : 18; C count = 54 ⇒ scaling factor = 3. Per-head wages A : B : C = 10 : 7 : 4. Per-head wage for B = ₹ 1400. Concept / Approach: From counts, A = 9*3 = 27 employees, B = 13*3 = 39 employees. From per-head wages, if B is 7k = 1400, then k = 200 and A per-head = 10k = ₹ 2000. Multiply by A’s count to get total A wages. Step-by-Step Solution: k = 1400 / 7 = 200.A per-head wage = 10k = ₹ 2000.A employees = 9*3 = 27.Total A wages = 27 * 2000 = ₹ 54000. Verification / Alternative check: C’s per-head wage would be 4k = ₹ 800; B has 39 employees at ₹ 1400, confirming k = 200 is consistent across categories. Why Other Options Are Wrong: ₹ 51000, ₹ 56000, ₹ 59000 arise from miscounting A employees or miscomputing per-head wage from the ratio. Common Pitfalls: Using the count ratio as the wage ratio or vice versa. Forgetting to scale counts using the given C count. Final Answer: ₹ 54000

Question 15

The average age of students in a class is 15.8 years. The average age of the boys is 16.4 years and the average age of the girls is 15.4 years. What is the ratio of the number of boys to the number of girls in the class?

Accepted Answer

Introduction / Context: This problem uses weighted averages to connect the overall average age of a class with the separate averages for boys and girls. You are given all three averages and asked to find the ratio of the number of boys to the number of girls. Such questions help learners understand how subgroup averages combine to produce a single overall average. Given Data / Assumptions: - Let the number of boys be B and the number of girls be G. - Overall average age of all students = 15.8 years. - Average age of boys = 16.4 years. - Average age of girls = 15.4 years. - We need to find the ratio B : G. Concept / Approach: The total age of the boys is 16.4 * B and the total age of the girls is 15.4 * G. The overall average times the total number of students gives another expression for the total age, namely 15.8 * (B + G). Equating these two expressions and simplifying gives a linear relation between B and G that can be converted into a ratio. Step-by-Step Solution: Step 1: Total age of boys = 16.4 * B.Step 2: Total age of girls = 15.4 * G.Step 3: Total age of all students = 16.4B + 15.4G.Step 4: Overall average age = 15.8, so total age can also be written as 15.8 * (B + G).Step 5: Equate the two expressions: 16.4B + 15.4G = 15.8(B + G).Step 6: Expand the right side: 16.4B + 15.4G = 15.8B + 15.8G.Step 7: Rearrange terms: 16.4B - 15.8B = 15.8G - 15.4G.Step 8: This gives 0.6B = 0.4G.Step 9: Divide both sides by 0.2 to simplify: 3B = 2G.Step 10: Therefore, B / G = 2 / 3 and the ratio B : G = 2 : 3. Verification / Alternative Check: Assume there are 2 boys and 3 girls to match the ratio 2 : 3. Total age of boys is 2 * 16.4 = 32.8, and total age of girls is 3 * 15.4 = 46.2. Combined total age = 79.0. Total students = 5. Average age = 79.0 / 5 = 15.8, which matches the given overall average. This numerical check confirms that the ratio 2 : 3 is correct. Why Other Options Are Wrong: Ratios such as 1 : 2, 3 : 4 or 3 : 5 do not satisfy the equation 16.4B + 15.4G = 15.8(B + G). Substituting these ratios will give an overall average different from 15.8, so they are inconsistent with the data in the problem. Common Pitfalls: Some students treat this as a simple average of 16.4 and 15.4 without weighting, which ignores the different numbers of boys and girls. Others might set up the equation incorrectly by mixing up the coefficients. Carefully forming the total age expressions and simplifying step by step avoids these errors. Final Answer: The ratio of the number of boys to the number of girls is 2 : 3.

Question 16

Two numbers are in the ratio 3 : 5. If 6 is added to each of the numbers, the new ratio becomes 2 : 3. What are the original two numbers?

Accepted Answer

Introduction: This is a classic ratio and proportion question where you are given an initial ratio and a transformation (adding the same number to both terms) that produces a new ratio. The task is to find the original numbers that satisfy both conditions. Given Data / Assumptions: The original ratio of the two numbers is 3 : 5. When 6 is added to each number, the ratio becomes 2 : 3. The numbers are positive real numbers, usually treated as integers in such problems. We need to determine the original two numbers. Concept / Approach: When numbers are in the ratio 3 : 5, we can represent them as 3k and 5k for some positive constant k. After adding 6 to each number, we are told that the new ratio is 2 : 3. We set up an equation based on this new ratio and solve for k, then compute the original numbers. Step-by-Step Solution: Step 1: Represent the numbers.Let the numbers be 3k and 5k.Step 2: Add 6 to each.New numbers: 3k + 6 and 5k + 6.Step 3: Use the new ratio 2 : 3.(3k + 6) / (5k + 6) = 2 / 3Step 4: Cross-multiply.3 * (3k + 6) = 2 * (5k + 6)9k + 18 = 10k + 12Step 5: Solve for k.9k + 18 = 10k + 12 ⇒ k = 6Step 6: Find the original numbers.First number = 3k = 3 * 6 = 18Second number = 5k = 5 * 6 = 30 Verification / Alternative check: Check with the transformed ratio: 18 + 6 = 24 and 30 + 6 = 36. Now 24 : 36 simplifies to 2 : 3 by dividing both terms by 12. This confirms that 18 and 30 satisfy the condition. Why Other Options Are Wrong: 21 and 35: Adding 6 gives 27 and 41; the ratio 27 : 41 is not 2 : 3. 30 and 50: Adding 6 gives 36 : 56, which simplifies to 9 : 14, not 2 : 3. 24 and 40: Adding 6 gives 30 : 46, not in the ratio 2 : 3. 12 and 20: Adding 6 gives 18 : 26, simplifies to 9 : 13, not 2 : 3. Common Pitfalls: Students sometimes incorrectly assume that the ratio difference directly gives the added number, or they try trial and error with options. A systematic algebraic approach using k makes the solution clean and error free. Final Answer: The original numbers are 18 and 30.

Question 17

A bag contains 50 P, 25 P and 10 P coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type respectively.

Accepted Answer

let ratio be x. Hence no. of coins be 5x ,9x , 4x respectively Now given total amount = Rs.206 => (.50)(5x) + (.25)(9x) + (.10)(4x) = 206 we get x = 40 => No. of 50p coins = 200 => No. of 25p coins = 360 => No. of 10p coins = 160

Question 18

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:

Accepted Answer

Let the third number be x. Then, first number = 120% of x =120x/100 = 6x/5 Second number =150% of x = 150x/100 = 3x/2 Ratio of first two numbers = 6x/5 : 3x/2 = 12x : 15x = 4 : 5

Question 19

A mixture contains alcohol and water in the ratio 4 : 3. If 5 liters of water is added to the mixture, the ratio becomes 4: 5. Find the quantity of alcohol in the given mixture.

Accepted Answer

Let the quantity of alcohol and water be 4x litres and 3x litres respectively 4x/(3x+5) = 4/5 20x = 4(3x+5) 8x = 20 x = 2.5 Quantity of alcohol = (4 x 2.5) litres = 10 litres.

Question 20

If Rs. 782 be divided into three parts, proportional to 12:23:34 , then the first part is?

Accepted Answer

R s . 782 * 6 23 Given ratio = 1 2 : 2 3 : 3 4 = 6:8:9 1st part = R s . 782 * 6 23 = Rs. 204

Ratio and Proportion Questions