Question 1

School enrollment ratio problem: The total number of students in a school is 2140. If 1200 are girls, how many are boys, and what is the simplified ratio of boys to girls?

Accepted Answer

Introduction / Context: This is a foundational ratio-and-proportion question that asks you to convert counts into a simplified ratio. You must first find the number of boys, then reduce the ratio of boys to girls to lowest terms. Given Data / Assumptions: Total students = 2140. Girls = 1200. Boys = Total − Girls. Ratio required = Boys : Girls in simplest form. Concept / Approach: Ratios compare two quantities by division. After computing the absolute number of boys, form the ratio Boys : Girls and divide both terms by their greatest common divisor (GCD) to simplify. Step-by-Step Solution: Boys = 2140 − 1200 = 940.Raw ratio = 940 : 1200.Divide both by 20 (a common factor): 940/20 : 1200/20 = 47 : 60. Verification / Alternative check: Check divisibility again: 47 is prime and does not divide 60; hence 47 : 60 is already in lowest terms. Why Other Options Are Wrong: 26 : 25 corresponds to unequal totals and does not match 940 : 1200. 18 : 13 is not proportional to 940 : 1200. 31 : 79 inverts the relative sizes; girls exceed boys here. Common Pitfalls: Forgetting to compute boys as the remainder after removing girls from the total. Not simplifying the ratio fully to lowest terms. Final Answer: 47 : 60

Question 2

Inverse-time concept for equal distances: The speeds of three cars are in the ratio 2 : 3 : 4. Find the ratio of the times taken by these cars to cover the same distance.

Accepted Answer

Introduction / Context: For a fixed distance, time is inversely proportional to speed. This problem tests whether you can correctly invert a speed ratio to obtain a time ratio and express it in integers. Given Data / Assumptions: Speed ratio = 2 : 3 : 4. Distance is the same for all cars. Time ∝ 1 / Speed. Concept / Approach: If v1 : v2 : v3 = 2 : 3 : 4, then t1 : t2 : t3 = 1/2 : 1/3 : 1/4. To avoid fractions, multiply by the least common multiple of denominators (LCM = 12). Step-by-Step Solution: Time ratio (fractions) = 1/2 : 1/3 : 1/4.Multiply each term by 12 ⇒ 6 : 4 : 3. Verification / Alternative check: Pick a sample distance, say 12 units. Times would be 12/2 = 6, 12/3 = 4, 12/4 = 3, confirming 6 : 4 : 3. Why Other Options Are Wrong: 2 : 3 : 4 repeats the speed ratio instead of inverting. 4 : 3 : 2 and 4 : 3 : 6 do not match 1/2 : 1/3 : 1/4 scaled. Common Pitfalls: Confusing direct and inverse proportion for time vs speed. Forgetting to scale fractional ratios to whole numbers. Final Answer: 6 : 4 : 3

Question 3

Combine chained ratios: If P : Q = 8 : 15 and Q : R = 3 : 2, find the compound ratio P : Q : R in simplest integers.

Accepted Answer

Introduction / Context: Chained ratios require aligning the common term. Here Q appears in both given ratios, so we scale them to the same Q before combining into a three-term ratio P : Q : R. Given Data / Assumptions: P : Q = 8 : 15. Q : R = 3 : 2. All terms represent positive quantities. Concept / Approach: Make Q equal in both ratios. Since Q is 15 in the first and 3 in the second, multiply the second ratio by 5 to get Q = 15 in both. Then read off P, Q, and R directly. Step-by-Step Solution: Scale Q : R = 3 : 2 by 5 ⇒ Q : R = 15 : 10.Now P : Q = 8 : 15 and Q : R = 15 : 10.Therefore, P : Q : R = 8 : 15 : 10. Verification / Alternative check: Let Q = 15 units. Then P = 8 units (from the first ratio) and R = 10 units (from the scaled second), confirming the compound ratio. Why Other Options Are Wrong: 8 : 15 : 7 and 7 : 15 : 8 invent an R that does not follow Q : R = 3 : 2. 10 : 15 : 8 swaps positions incorrectly. Common Pitfalls: Adding ratios instead of aligning the common term. Forgetting to scale the second ratio properly. Final Answer: 8 : 15 : 10

Question 4

Link three ratios to get P : S: Given P : Q = 8 : 15, Q : R = 5 : 8, and R : S = 4 : 5, find the ratio P : S.

Accepted Answer

Introduction / Context: When multiple linked ratios are given, proceed stepwise, keeping the middle terms consistent. This problem requires moving from P to S via Q and R. Given Data / Assumptions: P : Q = 8 : 15. Q : R = 5 : 8. R : S = 4 : 5. Concept / Approach: Assign variables to maintain consistency. Start with P = 8k and Q = 15k. Use Q : R to derive R, then apply R : S to get S. Finally simplify P : S. Step-by-Step Solution: Let P = 8k, Q = 15k.From Q : R = 5 : 8 ⇒ if Q = 15k = 5 × 3k, then R = 8 × 3k = 24k.From R : S = 4 : 5 ⇒ S = (5/4) × R = (5/4) × 24k = 30k.Thus P : S = 8k : 30k = 4 : 15. Verification / Alternative check: Convert all to a single base k and inspect: P = 8k, S = 30k. The ratio reduces cleanly by 2 to 4 : 15. Why Other Options Are Wrong: 2 : 15 halves P again unnecessarily. 3 : 19 and 7 : 15 do not match the chained calculations. Common Pitfalls: Mismatching the intermediary term Q or R when scaling ratios. Not simplifying the final ratio to lowest terms. Final Answer: 4 : 15

Question 5

Chaining three ratios to find A : D: If A : B = 2 : 3, B : C = 5 : 7, and C : D = 3 : 10, determine A : D in simplest form.

Accepted Answer

Introduction / Context: This problem requires careful linking of three ratios through the repeated variables B and C to arrive at the ratio between A and D. It highlights consistent scaling across chained proportional relations. Given Data / Assumptions: A : B = 2 : 3. B : C = 5 : 7. C : D = 3 : 10. Concept / Approach: Equate the middle terms successively. Express each variable in terms of a single base quantity (e.g., x) and simplify the final ratio A : D. Step-by-Step Solution: Let A = 2x, B = 3x.From B : C = 5 : 7 ⇒ if B = 3x = 5y, then y = 3x/5 and C = 7y = 21x/5.From C : D = 3 : 10 ⇒ if C = 21x/5 = 3z, then z = 7x/5 and D = 10z = 14x.Therefore A : D = 2x : 14x = 1 : 7. Verification / Alternative check: Normalize to x = 5 for easy integers: A = 10, D = 70 ⇒ A : D = 1 : 7 as obtained. Why Other Options Are Wrong: 2 : 7, 1 : 5, 5 : 1 are not consistent with the chained scaling. Common Pitfalls: Losing track of scaling constants when switching between the three ratios. Failing to simplify at the end. Final Answer: 1 : 7

Question 6

Third proportional computation: Calculate the third proportional to 15 and 30 (i.e., find x such that 15 : 30 = 30 : x).

Accepted Answer

Introduction / Context: The third proportional to a and b is the number x satisfying a : b = b : x. This is a direct application of cross-multiplication and proportional reasoning. Given Data / Assumptions: a = 15. b = 30. Find x such that 15 : 30 = 30 : x. Concept / Approach: For a : b = b : x, we have a * x = b * b ⇒ x = b^2 / a. Substitute the values and simplify to obtain an integer result. Step-by-Step Solution: x = b^2 / a = 30^2 / 15.30^2 = 900; 900 / 15 = 60. Verification / Alternative check: Check ratio equality: 15 : 30 = 1 : 2 and 30 : 60 = 1 : 2, so the proportion holds. Why Other Options Are Wrong: 55 and 65 do not satisfy 30 : x = 1 : 2. 15 repeats the first term and breaks the proportion. Common Pitfalls: Mistaking “third proportional” for “mean proportional.” Arithmetic slip in squaring b or dividing by a. Final Answer: 60

Question 7

Mean proportional (geometric mean): Find the mean proportional between 9 and 64 (i.e., find x such that 9 : x = x : 64).

Accepted Answer

Introduction / Context: The mean proportional (geometric mean) x between a and b satisfies a : x = x : b, or x^2 = a * b. This is a standard concept in ratio and proportion and geometric mean computations. Given Data / Assumptions: a = 9. b = 64. Find x such that x^2 = a * b. Concept / Approach: Compute the product a * b and then take the positive square root, since a and b are positive. Step-by-Step Solution: x^2 = 9 * 64 = 576.x = sqrt(576) = 24. Verification / Alternative check: Check proportionality: 9 : 24 = 3 : 8 and 24 : 64 = 3 : 8, confirming correctness. Why Other Options Are Wrong: 25, 27, 35 are not the square root of 576 and do not keep equal ratios with 9 and 64. Common Pitfalls: Confusing third proportional with mean proportional. Arithmetic mistakes in multiplying or in the square root. Final Answer: 24

Question 8

Solve a quadratic ratio identity: If P^2 + 4Q^2 = 4PQ, determine the ratio P : Q in lowest terms.

Accepted Answer

Introduction / Context: Algebraic identities often hide a perfect square. Recognizing and factoring them lets you deduce a fixed ratio between variables. Here, the expression rearranges into a perfect square in P and Q. Given Data / Assumptions: P^2 + 4Q^2 = 4PQ. P and Q are real numbers (nontrivial case). Concept / Approach: Bring all terms to one side and try to factor. If the left-hand side becomes a perfect square, set the factor to zero to find a linear relation between P and Q, which yields the ratio. Step-by-Step Solution: P^2 − 4PQ + 4Q^2 = 0.This is (P − 2Q)^2 = 0.Therefore P − 2Q = 0 ⇒ P = 2Q.Hence P : Q = 2 : 1. Verification / Alternative check: Substitute P = 2Q into the original: LHS = (2Q)^2 + 4Q^2 = 4Q^2 + 4Q^2 = 8Q^2; RHS = 4 * (2Q) * Q = 8Q^2, identity holds. Why Other Options Are Wrong: 1 : 3, 3 : 1, 1 : 2 do not satisfy the given identity when substituted. Common Pitfalls: Missing the perfect-square pattern and attempting unnecessary quadratic solutions. Final Answer: 2 : 1

Question 9

Algebra with a given ratio: If P : Q = 7 : 5, find the value of (5P − 2Q) : (3P + 2Q).

Accepted Answer

Introduction / Context: When the ratio of two variables is known, you can represent them with a common multiplier and evaluate composite linear combinations cleanly. This is a standard technique for simplifying expressions involving ratios. Given Data / Assumptions: P : Q = 7 : 5. Compute (5P − 2Q) : (3P + 2Q). Concept / Approach: Let P = 7k and Q = 5k for some k > 0. Substitute and simplify each linear combination to obtain a simple fractional ratio, which is independent of k. Step-by-Step Solution: P = 7k, Q = 5k.5P − 2Q = 5(7k) − 2(5k) = 35k − 10k = 25k.3P + 2Q = 3(7k) + 2(5k) = 21k + 10k = 31k.Ratio = 25k : 31k = 25/31. Verification / Alternative check: Pick k = 1 to verify numerically: P = 7, Q = 5 ⇒ (35 − 10) : (21 + 10) = 25 : 31, confirming the result. Why Other Options Are Wrong: 5/4, 6/5, 31/42 emerge from incorrect arithmetic on coefficients. Common Pitfalls: Losing the common factor k when simplifying, which can lead to wrong numerators/denominators. Final Answer: 25/31

Question 10

Numbers with shifted ratio: Two numbers are in the ratio 4 : 7. If 5 is subtracted from each number, the ratio becomes 1 : 2. Find the greater number.

Accepted Answer

Introduction / Context: Ratio problems with additive shifts are solved by expressing numbers via a common multiplier and then applying the shifted condition. Solving a simple linear equation yields the multiplier and thus the original numbers. Given Data / Assumptions: Original ratio = 4 : 7. Subtract 5 from each number ⇒ new ratio = 1 : 2. Find the larger original number. Concept / Approach: Let the numbers be 4x and 7x. After subtracting 5 from each, the ratio becomes (4x − 5) : (7x − 5) = 1 : 2. Cross-multiply and solve for x. Step-by-Step Solution: 2(4x − 5) = 1(7x − 5).8x − 10 = 7x − 5 ⇒ x = 5.Numbers are 4x = 20 and 7x = 35; greater number = 35. Verification / Alternative check: Check the shifted ratio: (20 − 5) : (35 − 5) = 15 : 30 = 1 : 2, correct. Why Other Options Are Wrong: 15 and 20 are the smaller number or an intermediate value. 40 does not satisfy the shifted condition when paired with its counterpart. Common Pitfalls: Applying the shift to only one term or forgetting to subtract from both numbers. Final Answer: 35

Question 11

Divide a sum by parts given one share: A sum is divided among three persons in the ratio 3 : 4 : 7. If the second person receives ₹ 12, what is the total sum?

Accepted Answer

Introduction / Context: When a total is split in a given ratio and one share is known, convert that share into its “number of parts” to find the value per part, and then multiply by the sum of parts to get the total amount. Given Data / Assumptions: Ratio = 3 : 4 : 7. Second share (4 parts) = ₹ 12. Total parts = 3 + 4 + 7 = 14. Concept / Approach: Value per part = Given share / Number of parts for that person. Then Total = (Value per part) * (Total parts). Step-by-Step Solution: Value per part = 12 / 4 = ₹ 3.Total = 14 * 3 = ₹ 42. Verification / Alternative check: Compute all shares: 3 parts = ₹ 9, 4 parts = ₹ 12, 7 parts = ₹ 21; sum = ₹ 42, consistent. Why Other Options Are Wrong: ₹ 21 and ₹ 9 are partial shares, not the total. ₹ 32 does not match the ratio arithmetic. Common Pitfalls: Mistaking the given ₹ 12 as the total instead of just the second person’s share. Final Answer: ₹ 42

Question 12

Linked price ratio with difference: The prices of a scooter and a television are in the ratio 8 : 7. A scooter costs ₹ 2000 more than a television. Find the price of the television.

Accepted Answer

Introduction / Context: When two prices are in a ratio and their difference is given, you can treat each price as a multiple of a common unit value. The difference equals the difference of ratio parts multiplied by that unit value. Given Data / Assumptions: Scooter : Television = 8 : 7. Scooter − Television = ₹ 2000. Concept / Approach: Let Scooter = 8k and Television = 7k. Then (8k − 7k) = k = 2000. Hence Television = 7k = 7 × 2000. Step-by-Step Solution: k = ₹ 2000.Television price = 7k = ₹ 14,000. Verification / Alternative check: Compute scooter price: 8k = ₹ 16,000; difference is ₹ 2000 and ratio 16000 : 14000 reduces to 8 : 7, confirming the setup. Why Other Options Are Wrong: ₹ 10000 and ₹ 18000 don’t maintain an 8 : 7 ratio with a ₹ 2000 gap. ₹ 28000 doubles the correct television price. Common Pitfalls: Forgetting that the difference corresponds to exactly one ratio part (8 − 7 = 1). Final Answer: ₹ 14000

Question 13

Walking rate vs time for same distance: The walking rates of A and B are in the ratio 2 : 3. If B takes 36 minutes to cover a fixed distance, how long does A take for the same distance?

Accepted Answer

Introduction / Context: For equal distances, time is inversely proportional to speed (or rate). If two people have a known rate ratio, their required times will be in the inverse ratio. Given Data / Assumptions: Rate(A) : Rate(B) = 2 : 3. Time is inversely proportional to rate. Time(B) = 36 minutes. Concept / Approach: Time(A) : Time(B) = 3 : 2. With Time(B) = 36 representing 2 parts, compute the value of one part and then scale up for A’s time (3 parts). Step-by-Step Solution: Let the time ratio be 3 : 2 (A : B).2 parts = 36 ⇒ 1 part = 18.A’s time = 3 parts = 54 minutes. Verification / Alternative check: Pick a distance, say 6 km. If A’s rate = 2 units and B’s rate = 3 units, then T(A) = 6/2 = 3 units of time and T(B) = 6/3 = 2 units. Scaling to B = 36 gives A = 54. Why Other Options Are Wrong: 24 and 21.6 are smaller than B’s time despite A being slower. 48 assumes a 4 : 3 inverse instead of 3 : 2. Common Pitfalls: Using direct instead of inverse proportion. Final Answer: 54min

Question 14

Sum of three numbers with chained ratios: The sum of three numbers is 68. The ratio of the first to the second is 2 : 3, and the ratio of the second to the third is 5 : 3. Find the second number.

Accepted Answer

Introduction / Context: This problem links three numbers using two ratios. Converting both to a single parameter lets you express all three numbers consistently and then use the total to solve for that parameter. Given Data / Assumptions: First : Second = 2 : 3. Second : Third = 5 : 3. Sum = 68. Concept / Approach: Let First = 2a, Second = 3a. Also Second = 5b, Third = 3b. Equate Second values to connect a and b, then write all three in terms of b and use the sum to solve for b. Step-by-Step Solution: 3a = 5b ⇒ a = 5b/3.First = 2a = 10b/3; Second = 5b; Third = 3b.Sum = 10b/3 + 5b + 3b = (10/3 + 8)b = 34b/3 = 68.b = 68 * 3 / 34 = 6; Second = 5b = 30. Verification / Alternative check: Numbers are 10b/3 = 20, 5b = 30, 3b = 18; total = 68 and ratios 2 : 3 and 5 : 3 hold. Why Other Options Are Wrong: 20 and 48 are the first and twice the third respectively, not the second. 58 exceeds the total logic. Common Pitfalls: Mismatching the “second” when equating across both ratios. Final Answer: 30

Question 15

Three numbers in ratio with a sum condition: Three numbers are in the ratio 3 : 4 : 5. The sum of the largest and the smallest equals the sum of the remaining number and 52. Find the smallest number.

Accepted Answer

Introduction / Context: This is a direct application of ratio representation with an extra linear relationship. Identifying which term is largest and smallest from the ratio lets you set up a simple equation in one variable. Given Data / Assumptions: Numbers ∝ 3 : 4 : 5 ⇒ let them be 3k, 4k, 5k. Largest + smallest = middle + 52. Concept / Approach: From 3k, 4k, 5k, the largest is 5k and the smallest is 3k; the “remaining number” is 4k. Use 5k + 3k = 4k + 52 to solve for k and then compute the smallest, 3k. Step-by-Step Solution: 5k + 3k = 4k + 52 ⇒ 8k = 4k + 52.4k = 52 ⇒ k = 13.Smallest = 3k = 39. Verification / Alternative check: Numbers are 39, 52, 65. Check: 65 + 39 = 104 and 52 + 52 = 104, condition satisfied. Why Other Options Are Wrong: 20 and 27 are smaller than the correct smallest derived from k = 13. 52 is the middle number, not the smallest. Common Pitfalls: Misidentifying the “remaining number” (it is the middle one, 4k). Final Answer: 39

Question 16

Shares from conditions on combined amounts: ₹ 5625 is divided among A, B, and C so that A receives 1/2 as much as B and C together receive, and B receives 1/4 of what A and C together receive. By how much does A’s share exceed B’s?

Accepted Answer

Introduction / Context: Here, shares are tied to sums of the other shares. Forming equations from the narrative constraints and combining them with the total allows you to solve for each person’s amount precisely. Given Data / Assumptions: Total = 5625. A = 1/2 (B + C) ⇒ 2A = B + C. B = 1/4 (A + C) ⇒ 4B = A + C. Concept / Approach: Use the two linear relations to express A and C in terms of B (or vice versa), then use A + B + C = 5625 to find numerical values. Finally compute A − B. Step-by-Step Solution: From 2A = B + C and 4B = A + C ⇒ eliminate C.From 2A = B + C ⇒ C = 2A − B.Sub into 4B = A + C ⇒ 4B = A + (2A − B) = 3A − B ⇒ 5B = 3A ⇒ A = (5/3)B.Then C = 2A − B = 2*(5/3 B) − B = 7/3 B.Total = A + B + C = (5/3 + 1 + 7/3)B = 5B ⇒ 5B = 5625 ⇒ B = 1125, A = 1875.Difference A − B = 1875 − 1125 = 750. Verification / Alternative check: Check conditions: B + C = 1125 + 2625 = 3750 ⇒ A = 1875 = 1/2 of 3750; A + C = 1875 + 2625 = 4500 ⇒ B = 1125 = 1/4 of 4500. Why Other Options Are Wrong: Other amounts do not satisfy both constraints simultaneously. Common Pitfalls: Losing track of variables when substituting for C. Final Answer: Rs. 750

Question 17

Find total from a two-person split: An amount was divided between Kavita and Reena in the ratio 4 : 3. If Reena’s share is Rs. 2400, what is the total amount?

Accepted Answer

Introduction / Context: This is a classic reverse-ratio question. Knowing one person’s share and the ratio, you can compute the total by finding the value of one ratio part and multiplying by the sum of parts. Given Data / Assumptions: Kavita : Reena = 4 : 3. Reena’s share (3 parts) = Rs. 2400. Concept / Approach: Value per part = 2400 / 3 = 800. Total parts = 4 + 3 = 7. Total amount = 7 * 800. Step-by-Step Solution: One part = Rs. 800.Total = 7 * 800 = Rs. 5600. Verification / Alternative check: Shares: Kavita = 4 * 800 = 3200; Reena = 2400; total = 5600; ratio 3200 : 2400 reduces to 4 : 3, correct. Why Other Options Are Wrong: Rs. 3200 is only Kavita’s share, not the total. Rs. 9600 contradicts the given share for Reena at 3 parts. Common Pitfalls: Swapping the 4 and 3 or adding incorrectly. Final Answer: Rs. 5600

Question 18

Share work payment by efficiency (rates): A, B, and C can complete a job in 20, 25, and 30 days respectively. They finish the work together for Rs. 2220. By how much does A’s share exceed B’s share?

Accepted Answer

Introduction / Context: When workers are paid for a completed job, the fair division is proportional to the work each performs. For constant-rate workers, work done is proportional to their rates, which are inversely proportional to their individual times to complete the task alone. Given Data / Assumptions: A = 20 days, B = 25 days, C = 30 days. Total payment = Rs. 2220. Share ∝ 1/day count. Concept / Approach: Compute rate proportions using a convenient LCM of days (e.g., 300). Then divide the total in the ratio of rates, and compare A and B’s shares. Step-by-Step Solution: LCM = 300 ⇒ A’s rate = 300/20 = 15, B’s = 300/25 = 12, C’s = 300/30 = 10.Ratio = 15 : 12 : 10; sum = 37.Each “ratio unit” = 2220 / 37 = 60.A’s share = 15 * 60 = 900; B’s share = 12 * 60 = 720.Difference = 900 − 720 = Rs. 180. Verification / Alternative check: C’s share = 10 * 60 = 600; total 900 + 720 + 600 = 2220, consistent. Why Other Options Are Wrong: 120 and 300 are arithmetic misreads of the 60-unit scaling. 600 is the entire share of C, not the difference A − B. Common Pitfalls: Dividing in the ratio of days (20 : 25 : 30) instead of rates. Final Answer: Rs. 180

Question 19

Monthly-to-annual salary via ratio with difference: A, B, and C earn monthly salaries in the ratio 2 : 3 : 5. If C’s monthly salary exceeds A’s by Rs. 1200, what is B’s annual salary?

Accepted Answer

Introduction / Context: Salary ratio problems with a known difference can be solved by representing each salary as a multiple of a common part. The difference maps to the difference in ratio parts times the value per part. Given Data / Assumptions: Monthly ratio A : B : C = 2 : 3 : 5. C − A = 1200 per month. Need B’s annual (12 months) salary. Concept / Approach: Let A = 2k, B = 3k, C = 5k. Then C − A = 3k = 1200 ⇒ k = 400. Hence B’s monthly salary is 3k and annual is 12 times this amount. Step-by-Step Solution: 3k = 1200 ⇒ k = 400.B (monthly) = 3k = 1200.B (annual) = 1200 * 12 = Rs. 14400. Verification / Alternative check: A = 2k = 800; C = 5k = 2000; indeed C − A = 1200, and the ratio 800 : 1200 : 2000 reduces to 2 : 3 : 5. Why Other Options Are Wrong: Rs. 24000 does not follow from k = 400. Rs. 1200 is B’s monthly, not annual salary; Rs. 2000 is unrelated. Common Pitfalls: Using annual difference 1200 * 12 incorrectly; the difference given is monthly. Final Answer: Rs. 14400

Question 20

Coin mixture value with ratio counts: A bag contains 25p, 10p, and 5p coins in the ratio 1 : 2 : 3. If the total value is Rs. 30, how many 5p coins are there?

Accepted Answer

Introduction / Context: Here the ratio governs counts, not values. You must convert each coin’s count into its rupee value, sum them, equate to the given total, and solve for the common count factor. Finally extract the number of 5p coins from the ratio. Given Data / Assumptions: Counts ratio (25p : 10p : 5p) = 1 : 2 : 3. Total value = Rs. 30. 1 rupee = 100 paise. Concept / Approach: Let counts be x, 2x, 3x. Convert paise to rupees: 25p = 0.25, 10p = 0.10, 5p = 0.05. The total value equation is 0.25x + 0.10(2x) + 0.05(3x) = 30. Solve for x, then compute 3x (the number of 5p coins). Step-by-Step Solution: Total value = 0.25x + 0.20x + 0.15x = 0.60x.0.60x = 30 ⇒ x = 50.Number of 5p coins = 3x = 150. Verification / Alternative check: Values: 25p coins ⇒ 50 * 0.25 = Rs. 12.50; 10p coins ⇒ 100 * 0.10 = Rs. 10; 5p coins ⇒ 150 * 0.05 = Rs. 7.50; total = Rs. 30. Why Other Options Are Wrong: 50 and 100 correspond to x or 2x, not 3x. 200 exceeds the required sum for the given total value. Common Pitfalls: Treating the ratio as values instead of counts. Forgetting to convert paise to rupees consistently. Final Answer: 150

Ratio and Proportion Questions