Question 1

Two numbers are in the ratio 3 : 5. If each number is increased by 10, the new ratio becomes 5 : 7. What are the original two numbers?

Accepted Answer

Introduction / Context: This ratio-and-proportion problem asks for two original numbers given their initial ratio and how that ratio changes after adding a constant to both numbers. It tests setting up a ratio equation and solving for the common multiplier. Given Data / Assumptions: Initial ratio of the two numbers = 3 : 5. 10 is added to each number. New ratio becomes 5 : 7. Concept / Approach: If numbers are in the ratio a : b, we can write them as a*x and b*x . When the same number is added to both, form a new ratio and solve for x. Then compute actual numbers. Step-by-Step Solution: Let numbers be 3x and 5x. After adding 10 to each: (3x + 10) : (5x + 10) = 5 : 7. Cross-multiply: 7*(3x + 10) = 5*(5x + 10). 21x + 70 = 25x + 50. Rearrange: 70 - 50 = 25x - 21x ⇒ 20 = 4x ⇒ x = 5. Numbers = 3x = 15 and 5x = 25. Verification / Alternative check: Add 10: 15 + 10 = 25 and 25 + 10 = 35. Ratio 25 : 35 simplifies to 5 : 7, which matches the condition. Why Other Options Are Wrong: 13, 22 does not give 5 : 7 after adding 10. 7, 9 and 3, 5 do not satisfy the new ratio when 10 is added. Common Pitfalls: Mixing up the order of terms, forgetting to cross-multiply correctly, or adding 10 to only one number. Always represent numbers as 3x and 5x before adding the constant to both. Final Answer: 15, 25

Question 2

The ratio of Sumit’s and Prakash’s present ages is 2 : 3. Sumit is 6 years younger than Prakash. What will be the ratio of their ages after 6 years?

Accepted Answer

Introduction / Context: This age problem uses ratios and a known difference between two people. We determine current ages from ratio plus difference, then project forward to find the future ratio. Given Data / Assumptions: Sumit : Prakash = 2 : 3 at present. Sumit is 6 years younger than Prakash. Find ratio after 6 years. Concept / Approach: Express ages as 2k and 3k. Use the given difference (3k − 2k = 6) to find k. Then add 6 years to each and form the new ratio. Step-by-Step Solution: Let Sumit = 2k, Prakash = 3k. Given 3k − 2k = 6 ⇒ k = 6. Present ages: Sumit = 12, Prakash = 18. After 6 years: Sumit = 18, Prakash = 24. Required ratio = 18 : 24 = 3 : 4. Verification / Alternative check: The difference remains 6 years before and after. The ratio after 6 years simplifies cleanly to 3 : 4, confirming the calculation. Why Other Options Are Wrong: 1 : 2 and 2 : 3 do not reflect the correct future ages. 3 : 8 is far too skewed considering the fixed 6-year gap. Common Pitfalls: Confusing which person is older, or forming the future ratio using wrong present values. Always compute k from the difference first, then update ages. Final Answer: 3 : 4

Question 3

The present ages of Vivek and Sumit are in the ratio 2 : 3. After 12 years, their ages will be in the ratio 11 : 15. What is Sumit’s current age?

Accepted Answer

Introduction / Context: This is a classic age-ratio problem involving present and future ratios. We must convert ratio statements into equations and solve for the scaling factor of the current ages. Given Data / Assumptions: Present ratio (Vivek : Sumit) = 2 : 3. After 12 years, ratio becomes 11 : 15. Find Sumit’s current age. Concept / Approach: If present ages are 2x and 3x, then after 12 years they are (2x + 12) and (3x + 12). Set their ratio to 11 : 15 and solve for x. Then compute Sumit’s age = 3x. Step-by-Step Solution: Let Vivek = 2x, Sumit = 3x. (2x + 12) : (3x + 12) = 11 : 15. 15(2x + 12) = 11(3x + 12). 30x + 180 = 33x + 132. 48 = 3x ⇒ x = 16. Sumit’s present age = 3x = 48 years. Verification / Alternative check: After 12 years: Vivek = 32 + 12 = 44; Sumit = 48 + 12 = 60; Ratio 44 : 60 reduces to 11 : 15, confirming the result. Why Other Options Are Wrong: 32 or 42 years do not satisfy the future ratio when checked. 56 years would make the future ratio exceed 11 : 15. Common Pitfalls: Forgetting to add 12 to both ages, or simplifying ratios incorrectly. Always check by substituting back into the future ratio condition. Final Answer: 48 years

Question 4

The ratio of a father’s age to his son’s age is 4 : 1, and the product of their present ages is 196. What will be the ratio of their ages after 5 years?

Accepted Answer

Introduction / Context: We combine ratio information with a product of ages to find actual present ages, and then compute a future ratio after a fixed number of years. This is a straightforward algebraic setup using ratios. Given Data / Assumptions: Father : Son = 4 : 1. Product of present ages = 196. Find ratio after 5 years. Concept / Approach: Represent the ages as 4x and x. Use the product 4x^2 = 196 to find x, compute present ages, add 5 to each, and take the ratio. Step-by-Step Solution: Let Father = 4x, Son = x. Product: 4x^2 = 196 ⇒ x^2 = 49 ⇒ x = 7. Present ages: Father = 28, Son = 7. After 5 years: Father = 33, Son = 12. Required ratio = 33 : 12 = 11 : 4. Verification / Alternative check: Recomputing the product with present ages: 28 * 7 = 196 (matches given). Future ratio calculation is consistent. Why Other Options Are Wrong: 3 : 1 and 10 : 3 do not match 33 : 12. 14 : 5 incorrectly scales the ages after adding 5 years. Common Pitfalls: Using 196 as a sum rather than a product, or forgetting to add 5 to both ages before forming the final ratio. Final Answer: 11 : 4

Question 5

Kamla and Savitri are aged in the ratio 6 : 5, and the sum of their present ages is 44 years. What will be the ratio of their ages after 8 years?

Accepted Answer

Introduction / Context: This problem uses an age ratio with a known sum to find present ages, and then asks for the ratio after a fixed time increment. It reinforces basic ratio partitions and linear updates to ages. Given Data / Assumptions: Kamla : Savitri = 6 : 5. Sum of present ages = 44 years. Find the ratio after 8 years. Concept / Approach: If 6x + 5x = 44, then x is the basic unit of age. Compute each age, then add 8 years to both and simplify the resulting ratio. Step-by-Step Solution: 6x + 5x = 44 ⇒ 11x = 44 ⇒ x = 4. Kamla = 6x = 24 years; Savitri = 5x = 20 years. After 8 years: Kamla = 32, Savitri = 28. Required ratio = 32 : 28 = 8 : 7. Verification / Alternative check: The sum remains consistent when adding the same number to both. The simplified ratio 8 : 7 directly results from dividing 32 and 28 by 4. Why Other Options Are Wrong: 5 : 6 and 7 : 8 invert or misstate the direction. 14 : 13 is not equivalent to 8 : 7 and does not come from the computed ages. Common Pitfalls: Forgetting to simplify the final ratio or miscomputing x from the sum 44. Always check by multiplication back to 44. Final Answer: 8 : 7

Question 6

Five years ago, a father’s age was 5 times his son’s age; two years from now, it will be 3 times the son’s age. What is the ratio of their present ages?

Accepted Answer

Introduction / Context: This age problem provides two separate time-based conditions. We convert each into an equation and solve the system to find present ages and their ratio. Given Data / Assumptions: 5 years ago: Father was 5 times the son. In 2 years: Father will be 3 times the son. Find present age ratio Father : Son. Concept / Approach: Translate conditions into equations for present ages F and S. Solve the two linear equations to find F and S, then form the ratio F : S. Step-by-Step Solution: From 5 years ago: F − 5 = 5(S − 5) ⇒ F = 5S − 20. From 2 years ahead: F + 2 = 3(S + 2) ⇒ F = 3S + 4. Equate: 5S − 20 = 3S + 4. 2S = 24 ⇒ S = 12; F = 3*12 + 4 = 40. Present ratio = 40 : 12 = 10 : 3. Verification / Alternative check: 5 years ago: 35 and 7 (35 = 5*7). In 2 years: 42 and 14 (42 = 3*14). Both conditions hold. Why Other Options Are Wrong: 5 : 2, 5 : 3, 11 : 5 do not match the derived 40 : 12 ratio. Common Pitfalls: Sign errors (adding instead of subtracting years) or equating the wrong expressions. Always write both equations clearly before solving. Final Answer: 10 : 3

Question 7

The average age of three girls is 20 years, and their ages are in the proportion 3 : 5 : 7. What is the age of the youngest girl?

Accepted Answer

Introduction / Context: We are given an average age and a ratio of ages. The task is to convert the ratio into actual ages using the average to find total and then allocate by ratio parts. Given Data / Assumptions: Average of 3 girls = 20 years ⇒ total = 60 years. Proportion of ages = 3 : 5 : 7. Find the youngest girl’s age. Concept / Approach: Sum of ratio parts is 3 + 5 + 7 = 15. Each part equals total/15. Multiply each part by the corresponding ratio number to get individual ages. Step-by-Step Solution: Total age = 3 * 20 = 60. Sum of parts = 15 ⇒ one part = 60 / 15 = 4. Ages: 3*4 = 12; 5*4 = 20; 7*4 = 28. Youngest = 12 years. Verification / Alternative check: Average of 12, 20, 28 is (60/3) = 20, which matches the given average. Why Other Options Are Wrong: 4 years, 6 years 8 months, and 8 years 3 months are inconsistent with both the total sum and the given ratio. Common Pitfalls: Forgetting to multiply the unit part properly or miscomputing the total (average * count). Ratio problems require careful handling of parts. Final Answer: 12 years

Question 8

A farmer has hens and cows. The total number of heads is 48 and the total number of feet is 140. How many hens are there?

Accepted Answer

Introduction / Context: This is a classic heads-and-legs (feet) problem using simultaneous linear equations. Each animal contributes 1 head; hens have 2 feet and cows have 4 feet. Given Data / Assumptions: Total heads = 48 (hens + cows). Total feet = 140. Hens have 2 feet; cows have 4 feet. Concept / Approach: Let h = number of hens, c = number of cows. Form two equations: h + c = 48 and 2h + 4c = 140. Solve the system to find h. Step-by-Step Solution: h + c = 48. 2h + 4c = 140 ⇒ divide by 2 ⇒ h + 2c = 70. Subtract: (h + 2c) − (h + c) = 70 − 48 ⇒ c = 22. Then h = 48 − 22 = 26. Verification / Alternative check: Check feet: 26 hens ⇒ 52 feet; 22 cows ⇒ 88 feet; total 52 + 88 = 140. Correct. Why Other Options Are Wrong: 22 or 23 or 24 hens do not simultaneously satisfy both equations. Common Pitfalls: Mixing up the number of feet per animal or making arithmetic slips while subtracting equations. Use systematic elimination to avoid errors. Final Answer: 26

Question 9

Six men, eight women, and six children complete a job for Rs. 950. If individual daily wages are in the ratio 4 : 3 : 2 respectively, what is the total amount earned by the children?

Accepted Answer

Introduction / Context: This problem distributes a total payment according to weighted shares. Each group member earns in fixed ratio units, so the total amount is split by total ratio shares. Given Data / Assumptions: Total payout = Rs. 950. Per person wage ratios: man : woman : child = 4 : 3 : 2. Counts: men = 6, women = 8, children = 6. Concept / Approach: Compute total ratio shares = (men * 4) + (women * 3) + (children * 2). Then value of one share = total amount / total shares. Finally, multiply by the children’s total shares. Step-by-Step Solution: Men shares = 6 * 4 = 24. Women shares = 8 * 3 = 24. Children shares = 6 * 2 = 12. Total shares = 24 + 24 + 12 = 60. Share value = 950 / 60 = 15.8333... Children’s amount = 12 * 15.8333... = Rs. 190. Verification / Alternative check: Sums for all groups equal 950 when each share is valued at 950/60, confirming the distribution is correct. Why Other Options Are Wrong: Rs. 195, Rs. 215, Rs. 230 correspond to different share allocations and do not match the exact proportional split. Common Pitfalls: Forgetting to multiply by the number of people in each group, or adding ratios incorrectly. Always compute aggregate shares before valuing each share. Final Answer: Rs. 190

Question 10

Two numbers are in the ratio 3 : 4 and their least common multiple (LCM) is 180. What is the first number?

Accepted Answer

Introduction / Context: The problem connects ratio with least common multiple (LCM). Represent numbers by a common multiplier and use known LCM behavior to find that multiplier, then recover the numbers. Given Data / Assumptions: Numbers are 3k and 4k. LCM = 180. Find the first number (3k). Concept / Approach: For relatively prime 3 and 4, LCM(3k, 4k) = 12k. Equate 12k with 180 to find k, then compute 3k. Step-by-Step Solution: LCM(3, 4) = 12. LCM(3k, 4k) = 12k. 12k = 180 ⇒ k = 15. First number = 3k = 45. Verification / Alternative check: Second number = 4k = 60. LCM(45, 60) is 180, confirming the calculation. Why Other Options Are Wrong: 15 or 20 are too small and do not give LCM 180 with their partners. 60 is the second number, not the first. Common Pitfalls: Confusing LCM with HCF, or forgetting that 3 and 4 are coprime which keeps LCM = product times k. Carefully apply LCM logic. Final Answer: 45

Question 11

A 60-litre mixture contains milk and water in the ratio 2 : 1 (i.e., 40 L milk and 20 L water). How much water must be added so that the new ratio of milk to water becomes 1 : 2?

Accepted Answer

Introduction / Context: This mixture problem requires adjusting the water quantity to achieve a target ratio. Milk quantity remains unchanged when only water is added; the ratio shift is achieved by increasing the water part. Given Data / Assumptions: Total initial mixture = 60 L with milk : water = 2 : 1. Milk = 40 L; water = 20 L. Add x litres of water so that milk : water = 1 : 2. Concept / Approach: Set up the ratio equation with milk fixed at 40 and water as 20 + x. Enforce 40 : (20 + x) = 1 : 2, then solve for x. Step-by-Step Solution: 40 / (20 + x) = 1 / 2. Cross-multiply: 40 * 2 = 20 + x. 80 = 20 + x ⇒ x = 60. Therefore, add 60 litres of water. Verification / Alternative check: New water = 20 + 60 = 80 L. New total = 120 L. Milk : water = 40 : 80 = 1 : 2, as desired. Why Other Options Are Wrong: 42, 56, or 77 L do not yield a 1 : 2 ratio when substituted into 40 : (20 + x). Common Pitfalls: Misreading the initial ratio or thinking milk changes. Only water is added, so milk remains 40 L. Set the ratio correctly to avoid algebra errors. Final Answer: 60 litres

Question 12

Rs. 1300 is divided among A, B, C, and D such that A/B = B/C = C/D = 2/3 (i.e., in continued proportion). What is A’s share?

Accepted Answer

Introduction / Context: The shares are in continued proportion with common ratio 3/2 from each to the next. We use a geometric progression approach to express all shares in terms of A and then match the total to Rs. 1300. Given Data / Assumptions: A/B = B/C = C/D = 2/3. Total sum = Rs. 1300. Find A’s share. Concept / Approach: From A/B = 2/3, we get B = (3/2)A; then C = (3/2)B = (3/2)^2 A; D = (3/2)^3 A. Sum these four terms and equate to 1300 to solve for A. Step-by-Step Solution: S = A + (3/2)A + (3/2)^2 A + (3/2)^3 A. Compute factors: 1 + 1.5 + 2.25 + 3.375 = 8.125. Thus, 8.125A = 1300 ⇒ A = 1300 / 8.125. 8.125 = 65/8 ⇒ A = 1300 * 8 / 65 = 160. Hence A’s share = Rs. 160. Verification / Alternative check: Compute other shares and sum to 1300; the arithmetic will check out with A = 160. Why Other Options Are Wrong: Rs. 140, Rs. 240, Rs. 320 do not satisfy the continued-proportion sum equaling 1300. Common Pitfalls: Treating the ratio as simple proportion instead of continued proportion or summing only two or three terms. All four must be included. Final Answer: Rs. 160

Question 13

Two equal glasses are respectively 1/3 and 1/4 full of milk. Each is then filled to the top with water, and the contents are mixed into a tumbler. What is the ratio of milk to water in the tumbler?

Accepted Answer

Introduction / Context: This is a mixture problem with equal-capacity vessels. Parts of each glass contain milk; the rest is water after topping up. We add both glasses to a tumbler and compute the overall ratio of milk to water. Given Data / Assumptions: Two equal glasses of capacity V. Milk volumes: V/3 and V/4 respectively. Each is filled to full with water; then the two are mixed. Concept / Approach: Total milk is the sum of milk from both glasses. Total mixture volume is 2V. Ratio milk:water is milk : (2V − milk). Step-by-Step Solution: Milk total = V/3 + V/4 = 7V/12. Total mixture = 2V ⇒ water = 2V − 7V/12 = 17V/12. Milk : Water = (7V/12) : (17V/12) = 7 : 17. Verification / Alternative check: Using any numerical V (e.g., V = 12 units) reproduces the same ratio 7 : 17, confirming independence from absolute capacity. Why Other Options Are Wrong: 7 : 5, 3 : 7, 11 : 23 are not consistent with the computed volumes 7V/12 milk and 17V/12 water. Common Pitfalls: Forgetting that the glasses are first topped with water, or assuming unequal capacities. Also, do not average ratios directly; compute actual quantities. Final Answer: 7 : 17

Question 14

Given a : b = c : d, evaluate (m a + n c) / (m b + n d). The expression simplifies to which ratio?

Accepted Answer

Introduction / Context: This property of proportional quantities states that if two ratios are equal, then any linear combination with the same coefficients in numerator and denominator preserves the common ratio. It is a fundamental result in ratio manipulation. Given Data / Assumptions: a : b = c : d. m and n are constants. We evaluate (m a + n c) / (m b + n d). Concept / Approach: If a/b = c/d = k, then a = kb and c = kd. Substitute into the expression to check simplification toward k (i.e., a : b). Step-by-Step Solution: Let a = kb and c = kd. Then (m a + n c) / (m b + n d) = (m kb + n kd) / (m b + n d). Factor k in numerator: = k (m b + n d) / (m b + n d) = k. Since a/b = k, the ratio equals a : b. Verification / Alternative check: Choose sample numbers satisfying a/b = c/d (e.g., 2/3 and 4/6). Any positive m, n will yield the same ratio value, confirming the identity. Why Other Options Are Wrong: m : n, na : mb, md : nc do not generally hold for arbitrary m, n; only a : b is invariant. Common Pitfalls: Treating m and n as multiplicative scalars of the whole ratio rather than components of a linear combination. The invariance depends on a/b = c/d. Final Answer: a : b

Question 15

A mixture initially has milk and water in the ratio 5 : 1. After adding 5 litres of water, the ratio becomes 5 : 2. What is the quantity of milk in the original mixture?

Accepted Answer

Introduction / Context: We are given a mixture ratio that changes after adding a known quantity of water. Since milk remains constant, we solve for the initial amounts consistent with both ratios. Given Data / Assumptions: Initial ratio milk : water = 5 : 1 ⇒ milk = 5x, water = x. After adding 5 L of water: milk : water = 5 : 2. Find milk quantity 5x. Concept / Approach: Use the relationship 5x : (x + 5) = 5 : 2. Solve for x, then compute milk = 5x. Step-by-Step Solution: (5x) / (x + 5) = 5 / 2. Cross-multiply: 10x = 5x + 25. 5x = 25 ⇒ x = 5. Milk = 5x = 25 litres. Verification / Alternative check: Initial water = x = 5 L; after adding 5 L, water becomes 10 L. Ratio milk : water = 25 : 10 = 5 : 2, as required. Why Other Options Are Wrong: 16, 32.5, or 22.75 litres do not satisfy the changed ratio when 5 L water is added. Common Pitfalls: Adding 5 L to both milk and water (incorrect) or misreading which component changes. Only water increases; milk stays the same. Final Answer: 25 litres

Question 16

A’s money is to B’s money as 4 : 5, and B’s money is to C’s money as 2 : 3. If A has Rs. 800, how much money does C have?

Accepted Answer

Introduction / Context: This problem composes two ratios to derive a three-term ratio A : B : C. Using A’s actual amount, we scale the ratio to find C’s amount. Given Data / Assumptions: A : B = 4 : 5. B : C = 2 : 3. A has Rs. 800; find C. Concept / Approach: Equalize the middle term (B) across both ratios and combine to get A : B : C. Then use A’s actual value to determine the scale factor and compute C. Step-by-Step Solution: Make B common: A : B = 4 : 5 ⇒ multiply by 2 ⇒ 8 : 10. B : C = 2 : 3 ⇒ multiply by 5 ⇒ 10 : 15. Thus A : B : C = 8 : 10 : 15. If A = 8 parts = Rs. 800 ⇒ 1 part = Rs. 100. C = 15 parts = 15 * 100 = Rs. 1500. Verification / Alternative check: B = 10 parts = Rs. 1000 satisfies both given ratios with A = 800 and C = 1500. Why Other Options Are Wrong: Rs. 1000 and Rs. 1200 mis-scale the ratio; Rs. 2000 overstates C’s share. Common Pitfalls: Adding ratios instead of forming a compound ratio, or not equalizing the B terms before combining. Always align the shared term correctly. Final Answer: Rs. 1500

Question 17

A 15-litre mixture contains 20% alcohol, the remainder being water. If 3 litres of water are added, what is the percentage of alcohol in the new mixture?

Accepted Answer

Introduction / Context: We are adjusting a mixture by adding water only. Alcohol amount remains unchanged; the total volume increases, reducing alcohol percentage. This tests percentage and ratio understanding in mixtures. Given Data / Assumptions: Initial volume = 15 L. Alcohol fraction = 20% ⇒ alcohol = 3 L, water = 12 L. Added water = 3 L. Concept / Approach: New alcohol quantity stays 3 L; new total volume = 18 L. Percentage alcohol = (alcohol / total) * 100%. Step-by-Step Solution: Initial alcohol = 0.20 * 15 = 3 L. Water increases from 12 L to 15 L. New total volume = 3 + 15 = 18 L. Alcohol percentage = (3 / 18) * 100% = 16 2/3 %. Verification / Alternative check: Decimal form: 3/18 = 0.1666..., which is 16.67% approximately, consistent with 16 2/3 %. Why Other Options Are Wrong: 17% and 18 1/2 % overestimate the alcohol content; 15% underestimates it. Only 16 2/3 % matches the exact calculation. Common Pitfalls: Changing alcohol amount mistakenly or forgetting to update the total volume after adding water. Always hold the solute constant when diluting with pure solvent. Final Answer: 16 2/3 %

Question 18

Vinay scored thrice as many marks in Mathematics as in English. The proportion of his marks in Mathematics and History is 4 : 3. If his total marks in Mathematics, English, and History is 250, what are his marks in English?

Accepted Answer

Introduction / Context: This marks-allocation problem uses two relationships: a direct multiple between Mathematics and English, and a ratio between Mathematics and History. We then apply the total-sum constraint to solve for each subject’s marks. Given Data / Assumptions: Maths = 3 * English. Maths : History = 4 : 3. Total of Maths + English + History = 250. Concept / Approach: Let English = x ⇒ Maths = 3x. From Maths : History = 4 : 3, History = (3/4)*Maths = (3/4)*3x = 9x/4. Sum the three expressions and set equal to 250, then solve for x. Step-by-Step Solution: Let E = x, M = 3x, H = 9x/4. Total: x + 3x + 9x/4 = 250. Combine: (4x + 12x + 9x)/4 = 25x/4 = 250. 25x = 1000 ⇒ x = 40. English = 40. Verification / Alternative check: Maths = 120; History = 90. Sum = 40 + 120 + 90 = 250, consistent. Why Other Options Are Wrong: 120 and 90 correspond to Maths and History, not English. 80 is not consistent with the given relationships. Common Pitfalls: Mixing up which subject is a multiple of which, or misapplying the 4 : 3 ratio. Always define variables clearly and substitute carefully. Final Answer: 40

Question 19

Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should gold and copper be mixed so that the mixture is 15 times as heavy as water?

Accepted Answer

Introduction / Context: This is a standard mean-density (alligation) problem. We are given relative densities with respect to water and asked for the mixing ratio to achieve a target mean density for the alloy. Given Data / Assumptions: Gold density ratio = 19. Copper density ratio = 9. Mixture target ratio = 15. Concept / Approach: Use weighted average: (19a + 9b) / (a + b) = 15, where a and b are amounts of gold and copper. Solve for a : b. Alternatively, apply alligation rule: parts are 15 − 9 and 19 − 15. Step-by-Step Solution: 19a + 9b = 15a + 15b. (19 − 15)a = (15 − 9)b ⇒ 4a = 6b. a / b = 6 / 4 = 3 / 2. Therefore, gold : copper = 3 : 2. Verification / Alternative check: Alligation method: Difference from mean—gold: 19 − 15 = 4; copper: 15 − 9 = 6. Ratio is 6 : 4 = 3 : 2 for gold : copper (reverse placement), confirming the same result. Why Other Options Are Wrong: 1 : 2 or 2 : 3 give a mean below or above 15. 19 : 135 is unrelated to the balanced mean. Common Pitfalls: Reversing the differences in alligation or taking arithmetic mean instead of weighted mean. Always use weighted average for mixture problems. Final Answer: 3 : 2

Question 20

Divide Rs. 600 among A, B, and C so that Rs. 40 more than 2/5 of A’s share, Rs. 20 more than 2/7 of B’s share, and Rs. 10 more than 9/17 of C’s share are all equal. What is A’s share?

Accepted Answer

Introduction / Context: This distribution problem sets three expressions (each a fraction of a share plus a constant) to be equal. We introduce a common value, express each share in terms of it, and satisfy the total sum constraint to find each person’s share—specifically A’s. Given Data / Assumptions: Let the common equal value be k. 2/5 of A + 40 = k ⇒ A = (5/2)(k − 40). 2/7 of B + 20 = k ⇒ B = (7/2)(k − 20). 9/17 of C + 10 = k ⇒ C = (17/9)(k − 10). A + B + C = 600. Concept / Approach: Express all three shares in terms of k, then add and set equal to 600. Solve for k, then compute A’s share directly using A = (5/2)(k − 40). Step-by-Step Solution: A = (5/2)(k − 40). B = (7/2)(k − 20). C = (17/9)(k − 10). Sum: (5/2)(k − 40) + (7/2)(k − 20) + (17/9)(k − 10) = 600. Solving gives k = 100. A = (5/2)(100 − 40) = (5/2)*60 = 150. Verification / Alternative check: Compute B = (7/2)*80 = 280; C = (17/9)*90 = 170; total 150 + 280 + 170 = 600. Also, checks: 2/5 of 150 + 40 = 60 + 40 = 100; 2/7 of 280 + 20 = 80 + 20 = 100; 9/17 of 170 + 10 = 90 + 10 = 100—consistent. Why Other Options Are Wrong: Rs. 280 and Rs. 170 correspond to B and C shares, not A. Rs. 200 does not satisfy all three equality conditions. Common Pitfalls: Forgetting to set all three expressions equal to a single k, or mishandling fractional coefficients. Work symbolically and then verify by back-substitution. Final Answer: Rs. 150

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