Question 1

The sum of three numbers is 98. If first:second = 2:3 and second:third = 5:8, find the second number.

Accepted Answer

Problem restatement Three numbers have chained ratios; determine the second number given the total is 98. Given data First:Second = 2:3 Second:Third = 5:8 Sum = 98 Concept/Approach Convert chained ratios to a single triplet by matching the common term (the second number) and then scale to the total. Step-by-step calculation Let First:Second = 2k:3k; Second:Third = 5m:8mMatch seconds: 3k = 5m ⇒ k = 5t, m = 3tTriplet = First:Second:Third = 10t : 15t : 24tSum = 49t = 98 ⇒ t = 2Second = 15t = 30 Verification/Alternative Numbers are 20, 30, 48; sums to 98 and ratios check out. Common pitfalls Adding the two ratios directly instead of linking via the common second term. Final Answer 30

Question 2

Boys and girls are in the ratio 7:8. After a 20% increase in boys and 10% increase in girls, what is the new ratio?

Accepted Answer

Problem restatement Apply percentage increases to both parts of a ratio and simplify. Given data Initial ratio Boys:Girls = 7:8 Boys increase by 20%; Girls by 10% Concept/Approach Multiply each term by its respective growth factor, then reduce the resulting ratio. Step-by-step calculation New boys = 7 × 1.2 = 8.4New girls = 8 × 1.1 = 8.8New ratio = 8.4 : 8.8 = divide by 0.4 ⇒ 21 : 22 Common pitfalls Adding percentages to the ratio numbers instead of applying multiplicative factors. Final Answer 21 : 22

Question 3

If 40% of one number equals two-thirds of another, what is the ratio of the first number to the second?

Accepted Answer

Problem restatement Relate two numbers via a percentage equality and express their ratio. Given data 0.4A = (2/3)B Concept/Approach Solve for A/B by dividing both sides by 0.4B. Step-by-step calculation A / B = [(2/3)] / 0.4 = (2/3) × (1/0.4) = (2/3) × (5/2) = 5/3 Common pitfalls Treating percentages additively or misplacing the decimal (0.4 vs 4%). Final Answer 5:3

Question 4

Ravi and Sumit's salaries are in the ratio 2:3. After increasing each by Rs. 4000, the ratio becomes 40:57. Find Sumit's original salary.

Accepted Answer

Problem restatement Two salaries in ratio 2:3 are each increased by the same amount (Rs. 4000), producing a new ratio 40:57. Determine Sumit's original salary. Given data Let Ravi = 2x, Sumit = 3x After increase: (2x + 4000) : (3x + 4000) = 40 : 57 Concept/Approach Translate the ratio into an equation via cross-multiplication and solve for x. Step-by-step calculation 57(2x + 4000) = 40(3x + 4000)114x + 228000 = 120x + 1600006x = 68000 ⇒ x = 68000 / 6 = 11333.333…Sumit = 3x = 3 × 11333.333… = Rs. 34,000 Verification/Alternative New salaries: Ravi = 2x + 4000 = 22666.666… + 4000 = 26666.666…; Sumit = 34000 + 4000 = 38000; ratio = 26666.666… : 38000 = 40 : 57. Common pitfalls Attempting to keep x integral; here x is fractional but the salaries after multiplication are valid currency amounts. Final Answer Rs. 34,000

Question 5

Rs. 782 is divided in the ratio 1/2 : 2/3 : 3/4. Find the first part.

Accepted Answer

Problem restatement Divide Rs. 782 in the proportion 1/2 : 2/3 : 3/4 and compute the first share. Concept/Approach Clear the fractional ratio by multiplying each term by the LCM of denominators (12) to get an integer ratio, then apply the standard part formula. Step-by-step calculation (1/2 : 2/3 : 3/4) → multiply by 12 → 6 : 8 : 9Total parts = 6 + 8 + 9 = 23Value per part = 782 / 23 = 34First part = 6 × 34 = Rs. 204 Verification/Alternative Other parts: 8 × 34 = Rs. 272; 9 × 34 = Rs. 306; sum = 204 + 272 + 306 = 782. Common pitfalls Adding the fractions directly or forgetting to scale them to a common denominator. Final Answer Rs. 204

Question 6

A and B together have Rs. 1210. If (4/15) of A equals (2/5) of B, how much does B have?

Accepted Answer

Problem restatement Two amounts sum to Rs. 1210 and satisfy (4/15)A = (2/5)B. Find B. Concept/Approach Use the proportionality to express A in terms of B, then apply the total to solve for B. Step-by-step calculation (4/15)A = (2/5)B ⇒ multiply by 15: 4A = 6B ⇒ 2A = 3B ⇒ A = (3/2)BA + B = 1210 ⇒ (3/2)B + B = (5/2)B = 1210B = 1210 × (2/5) = Rs. 484 Verification/Alternative A = (3/2) × 484 = 726; A + B = 726 + 484 = 1210; condition check: (4/15)A = 193.6, (2/5)B = 193.6. Common pitfalls Inverting the proportion (taking A = (2/3)B) or mishandling fraction arithmetic. Final Answer Rs. 484

Question 7

In a bag, coins of 25 paise, 10 paise, and 5 paise are in the ratio 1 : 2 : 3. If the total amount is ₹30, how many 5 paise coins are there?

Accepted Answer

Problem restatement The numbers of 25p, 10p, and 5p coins are in the ratio 1 : 2 : 3. The total value is ₹30 (= 3000 paise). Find the count of 5p coins. Given data Counts (25p : 10p : 5p) = 1 : 2 : 3 → let them be x, 2x, 3x. Total value = 3000 paise. Coin values: 25p, 10p, 5p. Concept/Approach Total value (in paise) equals the sum of (number × denomination) across coin types. Solve for x, then compute 3x for 5p coins. Step-by-step calculation Total value = 25x + 10(2x) + 5(3x) = 25x + 20x + 15x = 60x 60x = 3000 ⇒ x = 3000 ÷ 60 = 50 5p coins = 3x = 3 × 50 = 150 Verification/Alternative Check value: 25×50 + 10×100 + 5×150 = 1250 + 1000 + 750 = 3000 paise = ₹30. Common pitfalls Using ₹30 directly without converting to paise; the ratio is on counts, but values must be in paise. Final Answer 150

Question 8

Seats for Mathematics, Physics, and Biology are in the ratio 5 : 7 : 8. If they are increased by 40%, 50%, and 75% respectively, what is the ratio of the increased seats?

Accepted Answer

Problem restatement Apply percentage increases to a ratio and express the new counts in simplest ratio form. Given data Original ratio (Math : Physics : Biology) = 5 : 7 : 8. Increases: 40% (M), 50% (P), 75% (B). Concept/Approach Multiply each term by its growth factor (1 + rate) and then simplify the resulting ratio. Step-by-step calculation New M = 5 × 1.40 = 7.00 New P = 7 × 1.50 = 10.50 New B = 8 × 1.75 = 14.00 Ratio = 7 : 10.5 : 14 = multiply by 2 → 14 : 21 : 28 Simplify by 7 → 2 : 3 : 4 Verification/Alternative Any common scalar multiple (e.g., 14 : 21 : 28) is equivalent; simplest form is 2 : 3 : 4. Common pitfalls Adding percentages directly to the ratio numbers (e.g., 5+40) instead of scaling by the correct factor. Final Answer 2 : 3 : 4

Question 9

A sum of money is to be divided among A, B, C, and D in the ratio 5 : 2 : 4 : 3. If C receives ₹1000 more than D, what is B's share?

Accepted Answer

Problem restatement Money is split in the ratio 5 : 2 : 4 : 3 (A : B : C : D). Given C − D = ₹1000, find B's share. Given data Shares ∝ 5k, 2k, 4k, 3k. 4k − 3k = ₹1000 ⇒ k = ₹1000. Concept/Approach Translate the ratio into parts with a common multiplier k and use the given difference to find k. Step-by-step calculation k = 1000 B's share = 2k = 2 × ₹1000 = ₹2000 Verification/Alternative C = 4k = ₹4000; D = 3k = ₹3000 → difference ₹1000 (checks out). Final Answer ₹2000

Question 10

Find the fourth proportional to 5, 8, and 15.

Accepted Answer

Problem restatement For numbers a : b = c : x, find x with a = 5, b = 8, c = 15. Concept/Approach Fourth proportional x satisfies 5 : 8 = 15 : x ⇒ x = (8 × 15) ÷ 5. Step-by-step calculation x = (8 × 15) ÷ 5 = 120 ÷ 5 = 24 Verification/Alternative 5 : 8 = 15 : 24 → both ratios equal 5÷8 = 0.625 and 15÷24 = 0.625. Final Answer 24

Question 11

Two numbers are 20% and 50% more than a third number. What is the ratio of these two numbers?

Accepted Answer

Problem restatement Let the third number be t. Then the two numbers are 1.2t and 1.5t. Find their ratio. Concept/Approach Ratios scale out common factors: (1.2t : 1.5t) = (1.2 : 1.5). Step-by-step calculation 1.2 : 1.5 = 12 : 15 = 4 : 5 Verification/Alternative Pick t = 10 → numbers 12 and 15 → ratio 4 : 5. Final Answer 4 : 5

Question 12

The salaries of A, B, and C are in the ratio 2 : 3 : 5. After increments of 15%, 10%, and 20% respectively, what is the new ratio of their salaries?

Accepted Answer

Problem restatement Apply distinct percentage increases to each term of the ratio 2 : 3 : 5 and express the new ratio in simplest form. Given data A's increase = 15% ⇒ factor 1.15. B's increase = 10% ⇒ factor 1.10. C's increase = 20% ⇒ factor 1.20. Concept/Approach Scale each base ratio term by its factor and clear decimals to get integer ratio. Step-by-step calculation New A = 2 × 1.15 = 2.30 New B = 3 × 1.10 = 3.30 New C = 5 × 1.20 = 6.00 Ratio = 2.30 : 3.30 : 6.00 = multiply by 10 → 23 : 33 : 60 Verification/Alternative No further common divisor across 23, 33, and 60; thus already in simplest form. Common pitfalls Adding percentages directly to the ratio numbers (e.g., 2+15) instead of multiplying by growth factors. Final Answer 23 : 33 : 60

Question 13

A 60-litre mixture has milk and water in the ratio 2 : 1. How much water must be added to make the ratio 1 : 2?

Accepted Answer

Problem restatement Start with 60 L mixture where milk : water = 2 : 1. Add water to make milk : water = 1 : 2. Find the added water. Given data Total initially = 60 L; milk = 40 L; water = 20 L. Add x litres of water. Concept/Approach Milk quantity stays 40 L; water becomes 20 + x. Set the target ratio 40 : (20 + x) = 1 : 2 and solve for x. Step-by-step calculation 40 : (20 + x) = 1 : 2 ⇒ (20 + x) = 80 x = 80 − 20 = 60 L Verification/Alternative New amounts: milk 40 L, water 80 L → ratio 1 : 2. Total becomes 120 L. Common pitfalls Changing the milk amount; only water is added. Final Answer 60 litres

Question 14

If 0.75 : x :: 5 : 8, find x.

Accepted Answer

Problem restatement Solve the proportion 0.75 : x = 5 : 8 for x. Concept/Approach Use equality of ratios: 0.75/x = 5/8 ⇒ x = 0.75 × 8 ÷ 5. Step-by-step calculation x = (0.75 × 8) ÷ 5 = 6 ÷ 5 = 1.2 Verification/Alternative Check: 0.75 : 1.2 = 0.625 and 5 : 8 = 0.625 → consistent. Final Answer 1.20

Question 15

Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. What is the smaller number?

Accepted Answer

Problem restatement Original numbers are 3k and 5k. After subtracting 9 from each, the ratio becomes 12 : 23. Find the smaller original number. Given data Original: 3k, 5k. After subtraction: (3k − 9) : (5k − 9) = 12 : 23. Concept/Approach Cross-multiply to solve for k, then compute 3k. Step-by-step calculation 23(3k − 9) = 12(5k − 9) 69k − 207 = 60k − 108 9k = 99 ⇒ k = 11 Smaller number = 3k = 33 Verification/Alternative Check new ratio: (33 − 9) : (55 − 9) = 24 : 46 = 12 : 23. Common pitfalls Using (3k − 9)/(5k − 9) = 23/12 inadvertently (reversing the ratio). Final Answer 33

Question 16

Boys-to-girls ratio from class counts In a class of 95 students, there are 45 girls. What is the ratio of the number of boys to the number of girls?

Accepted Answer

Introduction / Context: This is a direct ratio problem: compute counts of boys and girls and express them in simplest integer ratio form. Such questions check arithmetic accuracy and reduction to simplest terms. Given Data / Assumptions: Total students = 95. Girls = 45. Boys = Total − Girls = 95 − 45 = 50. Concept / Approach: Required ratio = Boys : Girls = 50 : 45. Divide both terms by their greatest common divisor (5) to simplify the ratio to lowest terms. Step-by-Step Solution: Boys = 95 − 45 = 50.Raw ratio = 50 : 45.Divide by 5 ⇒ 10 : 9. Verification / Alternative check: 10 + 9 = 19 ratio parts; if each part represented 5 students, 19 * 5 = 95 total, consistent with the class size. Why Other Options Are Wrong: Ratios 9:10 or 7:8 invert or misstate the counts; 9:11 is not equivalent to 50:45; 5:4 is 1.25, not matching 50/45 = 1.111… Common Pitfalls: Accidentally giving Girls:Boys instead of Boys:Girls, or forgetting to reduce to simplest form. Final Answer: 10:9

Question 17

Chaining ratios across three persons The ratio of money with Ram and Gopal is 7 : 17, and the ratio with Gopal and Krishan is 7 : 17. If Ram has ₹ 490, how much money does Krishan have?

Accepted Answer

Introduction / Context: When two pairwise ratios share a common person, we can “chain” them by equalizing the common term to derive a three-person ratio. Then actual amounts follow by scaling to a given person’s value. Given Data / Assumptions: Ram : Gopal = 7 : 17. Gopal : Krishan = 7 : 17. Ram has ₹ 490. Concept / Approach: Make Gopal’s parts in both ratios equal. LCM(17, 7) = 119, so scale the first ratio by 7 and the second by 17 to set Gopal = 119 parts in both. Then read off Ram and Krishan’s parts and scale to rupees using Ram’s given amount. Step-by-Step Solution: Scaled Ram:Gopal = 7*7 : 17*7 = 49 : 119.Scaled Gopal:Krishan = 7*17 : 17*17 = 119 : 289.Combined three-way ratio: Ram : Gopal : Krishan = 49 : 119 : 289.Given Ram = ₹ 490 ⇒ 49 parts = 490 ⇒ 1 part = ₹ 10.Krishan = 289 parts = 289 * 10 = ₹ 2,890. Verification / Alternative check: Gopal = 119 * 10 = ₹ 1,190; then Ram:Gopal = 490:1,190 simplifies to 7:17, and Gopal:Krishan = 1,190:2,890 simplifies to 7:17, confirming consistency. Why Other Options Are Wrong: Other rupee amounts do not maintain both pairwise ratios when combined with Ram = ₹ 490. Common Pitfalls: Adding ratios or averaging them. The correct technique is to equalize the common person’s parts before forming the combined ratio. Final Answer: Rs. 2890

Question 18

Price difference resolves individual prices from a 3:2 ratio The prices of a scooter and a television set are in the ratio 3 : 2. If the scooter costs ₹ 6,000 more than the television, what is the television’s price?

Accepted Answer

Introduction / Context: Price-ratio with a known difference lets us solve for individual prices. Let the common ratio multiplier be x and translate the difference into an equation in x, then compute each price. Given Data / Assumptions: Scooter : Television = 3 : 2. Scooter costs ₹ 6,000 more. Concept / Approach: Let scooter = 3x and television = 2x. Then difference 3x − 2x = x = ₹ 6,000. Hence television = 2x = 2 * 6,000. Step-by-Step Solution: x = ₹ 6,000.Television price = 2x = 2 * 6,000 = ₹ 12,000. Verification / Alternative check: Then scooter = 3x = ₹ 18,000. The ratio 18,000 : 12,000 simplifies to 3 : 2 and the difference is ₹ 6,000, confirming correctness. Why Other Options Are Wrong: ₹ 6,000 is the difference, not the TV price; ₹ 10,000 and ₹ 9,000 break the ratio-difference relationship; ₹ 18,000 is the scooter price, not the TV price. Common Pitfalls: Swapping which item is larger or treating 6,000 as one of the ratio parts rather than the difference of 1 part. Final Answer: Rs. 12000

Question 19

Mean proportional in a continued proportion If 8 : y = y : 18 (i.e., 8/y = y/18), find the value of y.

Accepted Answer

Introduction / Context: In a continued proportion a : b = b : c, the middle term b is the mean proportional, satisfying b^2 = a * c. This question directly applies that identity to solve for y. Given Data / Assumptions: 8 : y = y : 18. Concept / Approach: Set up the equality 8 / y = y / 18, cross-multiply, and solve the resulting quadratic y^2 = 8 * 18. Take the positive root since ratios of magnitudes are nonnegative here. Step-by-Step Solution: 8 / y = y / 18 ⇒ y^2 = 8 * 18 = 144.y = √144 = 12 (taking the positive value). Verification / Alternative check: Check: 8 : 12 = 2 : 3 and 12 : 18 = 2 : 3; ratios match, confirming y = 12. Why Other Options Are Wrong: 144 and 72 are products or doubles, not the mean proportional; 26 is arbitrary; 36 would imply 8:36 ≠ 36:18. Common Pitfalls: Taking the product 8*18 as the answer or forgetting to take the square root to find the mean proportional. Final Answer: 12

Question 20

Equal areas: square side to circle radius A circle and a square have the same area. What is the ratio of the side of the square to the radius of the circle?

Accepted Answer

Introduction / Context: Geometric area equivalence allows deriving ratios among characteristic lengths. Here, a square and a circle share the same area; we are asked for the ratio of the square’s side to the circle’s radius. Given Data / Assumptions: Square side = s. Circle radius = r. Areas are equal. Concept / Approach: Area(square) = s^2. Area(circle) = π * r^2. Equality s^2 = π r^2 implies s = r * √π. Therefore the required ratio s : r = √π : 1. Step-by-Step Solution: Set s^2 = π r^2.Taking square roots (positive lengths): s = r * √π.Hence s : r = √π : 1. Verification / Alternative check: If r = 1, then s = √π gives areas π and π, equal as required. Any consistent scaling preserves the ratio √π : 1. Why Other Options Are Wrong: 1 : √π inverts the ratio; 1 : π and π : 1 mix area and length scales wrongly; 2 : √π has no basis in the area relationship. Common Pitfalls: Confusing circumference or diameter with area formula, or inverting the requested ratio accidentally. Final Answer: √π : 1

Ratio and Proportion Questions