Question 1

Two sums are divided as follows: • First sum: shared among P, Q, and R in the ratio 5 : 6 : 7. • Second sum: shared equally between S and T. If S receives ₹ 2100 less than P, determine how much Q receives from the first sum.

Accepted Answer

Introduction / Context: The question mixes two different distributions: one among P, Q, R in a given ratio and another between S and T equally. It links them by stating S gets ₹ 2100 less than P. We must decide whether this information is enough to compute how much Q receives. Given Data / Assumptions: P : Q : R = 5 : 6 : 7 in the first sum (unknown total). S = T in the second sum (unknown total). S = P − ₹ 2100 (difference only). No relation between the two totals is provided. Concept / Approach: Amounts in proportional division require the total sum. A single difference constraint between P and S cannot determine the absolute total for the first sum unless a second independent equation links the totals. Step-by-Step Analysis: Let first total be X. Then P = 5X/18, Q = 6X/18 = X/3, R = 7X/18. Let second total be Y. Then S = Y/2 and T = Y/2. Given: S = P − 2100 ⇒ Y/2 = 5X/18 − 2100. Unknowns X and Y appear in a single equation; this underdetermines the system. Q = X/3 depends on X, which cannot be uniquely found without another independent condition. Verification / Alternative check: Choose example pairs (X, Y) satisfying Y/2 = 5X/18 − 2100; Q = X/3 will vary with X. Hence Q is not uniquely determined. Why Other Options Are Wrong: Specific rupee amounts imply a unique X, which we cannot establish from the data. Therefore, any numeric choice is unjustified. Common Pitfalls: Assuming the two totals are equal or that one is derived from the other. The question never states such a link, so we must not impose it. Final Answer: Couldn't be determined

Question 2

Nandita scores 80% across five subjects (Hindi, Science, Mathematics, English, Sanskrit), each with a maximum of 105 marks. Her scores are: Hindi = 89, Sanskrit = 92, Mathematics = 98, English = 81. How many marks did she score in Science?

Accepted Answer

Introduction / Context: This is a straightforward total-score completion problem with percentages. We use the overall percentage across equal-maximum subjects to find the missing subject's marks. Given Data / Assumptions: Five subjects; each maximum = 105. Overall percentage = 80%. Known scores: Hindi 89, Sanskrit 92, Mathematics 98, English 81. Science score = ? Concept / Approach: Compute total maximum and total obtained from the percentage, subtract known subject scores, and the remainder is Science. Step-by-Step Solution: Total maximum = 5 * 105 = 525. Total obtained = 80% of 525 = 0.8 * 525 = 420. Sum of known scores = 89 + 92 + 98 + 81 = 360. Science = 420 − 360 = 60. Verification / Alternative check: Add Science back: 360 + 60 = 420; divide by 525 → 420/525 = 0.8 = 80%. Why Other Options Are Wrong: 65, 70, 75, and 68 do not yield the 80% overall when summed with the other four marks. Common Pitfalls: Misadding the four given scores or miscomputing 80% of 525. Double-check arithmetic. Final Answer: 60

Question 3

In a factory, the number of employees decreases in the ratio 8 : 7, while the per-employee salary increases in the ratio 5 : 6. Determine whether the total wage bill increases or decreases, and by what ratio.

Accepted Answer

Introduction / Context: When one factor (headcount) decreases and another factor (salary per person) increases, the overall wage bill changes by the product of the individual change factors. We must compute that combined effect. Given Data / Assumptions: Employees change 8 → 7 (multiplier = 7/8). Salary per employee changes 5 → 6 (multiplier = 6/5). Total wage bill ∝ employees * salary per employee. Concept / Approach: Multiply the two multipliers to get the overall change: (7/8) * (6/5). If this product is greater than 1, the total increases; if less than 1, it decreases. Step-by-Step Solution: Overall multiplier = (7/8) * (6/5) = 42/40 = 21/20. Since 21/20 > 1, the total wage bill increases. Expressed as a ratio, the increase is 21 : 20 (increased). Verification / Alternative check: Try a baseline: 8 workers at ₹5 each → total 40. After change: 7 workers at ₹6 each → total 42. 42 : 40 = 21 : 20 (increased). Why Other Options Are Wrong: 19 : 20 increased and 20 : 19 decreased reflect incorrect multiplications; 21 : 20 decreased has wrong direction; “None of above” is incorrect because a precise ratio exists. Common Pitfalls: Adding ratios instead of multiplying multipliers, or inverting one of the ratios. Final Answer: 21 : 20 increased

Question 4

Out of 120 applicants: 70 are males and 80 have a driver’s license. What is the ratio of the minimum to the maximum possible number of males having a driver’s license?

Accepted Answer

Introduction / Context: This is an inclusion–exclusion style bounds question. We must find the minimum and maximum possible number of males with licenses, given totals for males and total licensed applicants. Given Data / Assumptions: Total applicants = 120. Males = 70 ⇒ Females = 50. Licensed applicants = 80. Concept / Approach: Maximum males with licenses occurs when as many licenses as possible are assigned to males: min(males, licensed) = min(70, 80) = 70. Minimum males with licenses occurs when as many licenses as possible go to females: max(0, licensed − females) = 80 − 50 = 30. Step-by-Step Solution: Minimum males with license = 30. Maximum males with license = 70. Required ratio (min : max) = 30 : 70 = 3 : 7. Verification / Alternative check: Construct extreme cases: Case min: all 50 females licensed ⇒ remaining 30 licenses to males ⇒ 30 males licensed. Case max: all licenses to males until exhausted ⇒ 70 males licensed. Why Other Options Are Wrong: 1 : 3, 2 : 3, 5 : 7, and 3 : 5 do not match the computed bounds. Common Pitfalls: Forgetting to cap by the number of females (for license allocation) or by the number of males when maximizing. Final Answer: 3 : 7

Question 5

A 60 L mixture has acid : water = 2 : 1 (i.e., 40 L acid and 20 L water). How many liters of water must be added so that the final ratio becomes acid : water = 1 : 2?

Accepted Answer

Introduction / Context: We are changing the composition of a mixture by adding only water until the new ratio is reached. This is a direct proportion setup. Given Data / Assumptions: Total = 60 L with acid : water = 2 : 1 ⇒ acid = 40 L, water = 20 L. Add x liters of water. Target ratio: acid : water = 1 : 2. Concept / Approach: Acid remains 40 L (no acid added). Final water becomes 20 + x. Enforce the target ratio 40 : (20 + x) = 1 : 2 and solve for x. Step-by-Step Solution: 40 / (20 + x) = 1 / 2 ⇒ 2 * 40 = 20 + x. 80 = 20 + x ⇒ x = 60 L. Verification / Alternative check: New mixture: 40 L acid, 80 L water ⇒ ratio 1 : 2 as desired. Why Other Options Are Wrong: 55, 50, 45, 40 do not produce 1 : 2 when added to the 20 L of water. Common Pitfalls: Changing acid accidentally or using total volumes instead of component-wise ratio. Final Answer: 60

Question 6

Teas priced at ₹ 126/kg and ₹ 135/kg are mixed with a third variety in the ratio 1 : 1 : 2. If the resulting mixture is worth ₹ 153/kg, what is the price per kg of the third variety?

Accepted Answer

Introduction / Context: This is a weighted-average problem across three components. The average price equals the total cost divided by total quantity, weighted by the mixing ratio. Given Data / Assumptions: Prices: ₹ 126/kg, ₹ 135/kg, and ₹ x/kg. Mixing ratio: 1 : 1 : 2 (total 4 parts). Mixture price = ₹ 153/kg. Concept / Approach: Use weighted average: (126*1 + 135*1 + x*2) / 4 = 153. Solve for x to find the third variety's price. Step-by-Step Solution: (126 + 135 + 2x) / 4 = 153. 126 + 135 + 2x = 612. 2x = 612 − 261 = 351. x = 351 / 2 = ₹ 175.5 per kg. Verification / Alternative check: Substitute back: total cost = 126 + 135 + 2*175.5 = 612; divide by 4 = 153. Correct. Why Other Options Are Wrong: Values like ₹ 170, ₹ 169.5, ₹ 180, ₹ 165 do not satisfy the weighted-average equation with ratio 1 : 1 : 2. Common Pitfalls: Averaging the prices arithmetically without weighting or miscounting the two parts of the third variety. Final Answer: ₹ 175.5

Question 7

Find the compound ratio (product of ratios) of 2 : 7, 5 : 3, and 4 : 7. Express your answer as a : b in simplest terms.

Accepted Answer

Introduction / Context: The compound ratio of several ratios is obtained by multiplying them termwise: (a1 : b1) × (a2 : b2) × (a3 : b3) = (a1*a2*a3) : (b1*b2*b3). Given Data / Assumptions: Ratios: 2 : 7, 5 : 3, and 4 : 7. Concept / Approach: Multiply numerators together and denominators together, then simplify if needed. Step-by-Step Solution: Numerators: 2 * 5 * 4 = 40. Denominators: 7 * 3 * 7 = 147. Compound ratio = 40 : 147. Verification / Alternative check: Consider fractions: (2/7) * (5/3) * (4/7) = 40/147, matching the ratio. Why Other Options Are Wrong: 147 : 40 swaps order; 147 : 30 and 30 : 147 arise from incorrect multiplication. Common Pitfalls: Forgetting to multiply all numerators together and denominators together, or simplifying prematurely. Final Answer: 40 : 147

Question 8

Given A : B = 3 : 4 and B : C = 8 : 9, determine A : B : C in simplest integer form.

Accepted Answer

Introduction / Context: To merge two ratios with a common term, equalize the common term and then combine to get a three-term ratio. Given Data / Assumptions: A : B = 3 : 4; B : C = 8 : 9. Concept / Approach: Make the B terms equal by scaling the first ratio so B = 8. Then read off A and C accordingly. Step-by-Step Solution: From A : B = 3 : 4, scale by 2 ⇒ A : B = 6 : 8. From B : C = 8 : 9. Combine to get A : B : C = 6 : 8 : 9. Verification / Alternative check: Check A : B = 6 : 8 = 3 : 4 and B : C = 8 : 9; both hold. Why Other Options Are Wrong: The other triples do not preserve both source ratios simultaneously. Common Pitfalls: Adding ratios or averaging instead of equalizing the common term. Final Answer: 6 : 8 : 9

Question 9

If (a + b) : (a − b) = 5 : 3, find (a^2 + b^2) : (a^2 − b^2) in simplest form.

Accepted Answer

Introduction / Context: Using the given ratio of (a + b) to (a − b), we can extract values for a and b up to a common factor, then compute the requested ratio for the squares. Given Data / Assumptions: (a + b) : (a − b) = 5 : 3 with real numbers allowing a > b for positivity. Concept / Approach: Let a + b = 5k and a − b = 3k. Solve for a and b, then compute a^2 + b^2 and a^2 − b^2 = (a − b)(a + b). Step-by-Step Solution: Add: (a + b) + (a − b) = 5k + 3k ⇒ 2a = 8k ⇒ a = 4k. Then b = (a + b) − a = 5k − 4k = k. a^2 + b^2 = (4k)^2 + k^2 = 16k^2 + k^2 = 17k^2. a^2 − b^2 = (a − b)(a + b) = 3k * 5k = 15k^2. Ratio = 17 : 15. Verification / Alternative check: Choose k = 1 ⇒ a = 4, b = 1 ⇒ (a + b)/(a − b) = 5/3 and (a^2 + b^2)/(a^2 − b^2) = 17/15. Why Other Options Are Wrong: 25 : 9, 4 : 1, and 16 : 1 derive from squaring the given ratio or misapplying identities. Common Pitfalls: Squaring 5 : 3 directly; remember we are not finding (a + b)^2 : (a − b)^2 but a^2 + b^2 : a^2 − b^2. Final Answer: 17 : 15

Question 10

Given x / (2y) = 6 / 7, evaluate (x − y) / (x + y) + 14/19.

Accepted Answer

Introduction / Context: We first transform the given ratio into a simple parametric form for x and y, then compute the target expression. This tests algebraic manipulation with ratios. Given Data / Assumptions: x/(2y) = 6/7 ⇒ x/y = 12/7. Concept / Approach: Parameterize x and y as a common multiple of the ratio, compute (x − y)/(x + y), and then add 14/19. Step-by-Step Solution: Let x = 12t and y = 7t. (x − y)/(x + y) = (12t − 7t)/(12t + 7t) = 5t/19t = 5/19. Add 14/19 ⇒ 5/19 + 14/19 = 19/19 = 1. Verification / Alternative check: Substitute t = 1: x = 12, y = 7 ⇒ (12 − 7)/(12 + 7) + 14/19 = 5/19 + 14/19 = 1. Why Other Options Are Wrong: 5, 4, 3, 2 are not equal to the simplified sum; they ignore the fractional addition step. Common Pitfalls: Forgetting to reduce the fraction (x − y)/(x + y) properly or misadding numerator terms. Final Answer: 1

Question 11

Sanjay and Suresh together weigh 120 kg. Sanjay weighs 30 kg more than Suresh. What is the ratio of Suresh’s weight to Sanjay’s weight (expressed as a decimal)?

Accepted Answer

Introduction / Context: This is a linear equation problem. We convert the statements into algebra, find individual weights, and then compute the requested ratio as a decimal. Given Data / Assumptions: Total weight = 120 kg; Sanjay = Suresh + 30 kg. Concept / Approach: Solve for Suresh and Sanjay directly, then compute Suresh/Sanjay as a decimal. Step-by-Step Solution: Let Suresh = s; Sanjay = s + 30. Sum: s + (s + 30) = 120 ⇒ 2s = 90 ⇒ s = 45. Sanjay = 45 + 30 = 75. Ratio (Suresh/Sanjay) = 45 / 75 = 3/5 = 0.6. Verification / Alternative check: Check totals: 45 + 75 = 120 and difference is 30, both conditions satisfied. Why Other Options Are Wrong: 0.4, 0.3, 0.25, 0.5 are not equal to 45/75. Common Pitfalls: Using Sanjay/Suresh instead of Suresh/Sanjay, or miscomputing the decimal form. Final Answer: 0.6

Question 12

In a school, the ratio of boys to girls is 4 : 5. After 100 girls leave, the ratio becomes 6 : 7. How many boys are there in the school initially?

Accepted Answer

Introduction / Context: Two consecutive ratio conditions enable solving for actual counts. We set up variables for initial numbers and apply the change to girls to get the second ratio. Given Data / Assumptions: Boys : Girls initially = 4 : 5. After 100 girls leave, Boys : Girls = 6 : 7. Concept / Approach: Parameterize initial counts as 4x and 5x, update the girls by subtracting 100, and equate to the new ratio 6 : 7. Step-by-Step Solution: Initial: Boys = 4x, Girls = 5x. After departure: Boys = 4x, Girls = 5x − 100. Ratio condition: 4x / (5x − 100) = 6 / 7. Cross-multiply: 28x = 30x − 600 ⇒ 2x = 600 ⇒ x = 300. Boys = 4x = 1200. Verification / Alternative check: Then girls initially 1500; after 100 leave → 1400; 1200 : 1400 simplifies to 6 : 7. Correct. Why Other Options Are Wrong: 1800, 1000, 1500, 1400 do not satisfy both ratio conditions when tested. Common Pitfalls: Changing the number of boys by mistake or failing to simplify the final ratio for checking. Final Answer: 1200

Question 13

In a class, boys : girls = 4 : 5 initially. If 10 more boys join, the new ratio becomes 6 : 5. How many girls are there in the class?

Accepted Answer

Introduction / Context: This problem uses initial ratio and a post-change ratio after a fixed increment to one group to find the original scale and the number of girls. Given Data / Assumptions: Boys : Girls = 4 : 5 initially; after 10 boys join → 6 : 5. Concept / Approach: Express boys and girls as 4x and 5x; add 10 to boys and set the ratio to 6 : 5; solve for x, then compute girls. Step-by-Step Solution: Initial: B = 4x, G = 5x. After joining: (4x + 10) / 5x = 6 / 5. Cross-multiply: 5(4x + 10) = 6(5x) ⇒ 20x + 50 = 30x ⇒ 10x = 50 ⇒ x = 5. Girls = 5x = 25. Verification / Alternative check: Boys initially 20; after +10 → 30; ratio 30 : 25 = 6 : 5; consistent. Why Other Options Are Wrong: 20, 30, 35 do not satisfy both the initial and final ratio conditions. Common Pitfalls: Adding 10 to both groups or misapplying the ratio after the change. Final Answer: 25

Question 14

A school has 8,670 students. If boys are 4,545, find the ratio of boys to girls in simplest terms.

Accepted Answer

Introduction / Context: We compute girls from the total and then reduce the ratio boys : girls to lowest terms. Given Data / Assumptions: Total = 8,670; Boys = 4,545; Girls = Total − Boys. Concept / Approach: Find girls, then simplify the ratio using a common divisor. Step-by-Step Solution: Girls = 8,670 − 4,545 = 4,125. Boys : Girls = 4,545 : 4,125. Divide both by 15: 4,545/15 = 303; 4,125/15 = 275. Simplified ratio = 303 : 275. Verification / Alternative check: Compute 303*15 = 4,545 and 275*15 = 4,125, confirming the reduction. Why Other Options Are Wrong: 275 : 303 inverts; 11 : 12 and 12 : 11 do not match the exact counts. Common Pitfalls: Arithmetic mistakes when subtracting or simplifying by the wrong common divisor. Final Answer: 303 : 275

Question 15

A truck covers 640 km in 10 hours. A car covers the same distance in 8 hours. What is the ratio of the speed of the truck to the speed of the car?

Accepted Answer

Introduction / Context: For equal distances, the speed is inversely proportional to time taken. Alternatively, compute speeds directly and take their ratio. Given Data / Assumptions: Distance = 640 km for both; Truck time = 10 h; Car time = 8 h. Concept / Approach: Compute speeds: v = distance/time. Then form the ratio truck : car. Step-by-Step Solution: Truck speed = 640 / 10 = 64 km/h. Car speed = 640 / 8 = 80 km/h. Ratio (truck : car) = 64 : 80 = 4 : 5. Verification / Alternative check: Using inverse proportionality: times 10 : 8 reduce to 5 : 4, so speeds are 4 : 5. Why Other Options Are Wrong: 3 : 4, 1 : 2, 5 : 6 do not match the computed speeds. Common Pitfalls: Forgetting to simplify the ratio or mixing the order (car : truck instead of truck : car). Final Answer: 4 : 5

Question 16

A certain distance is covered at speed v1. If half of this distance is covered in double the original time, what is the ratio of the two speeds (v1 : v2)?

Accepted Answer

Introduction / Context: Speed is distance divided by time. Comparing two scenarios for the same benchmark distance allows a direct speed-ratio calculation. Given Data / Assumptions: Let original distance be D in time T at speed v1. New scenario covers D/2 in time 2T at speed v2. Concept / Approach: Compute v1 and v2 symbolically and then take the ratio v1 : v2. Step-by-Step Solution: v1 = D / T. v2 = (D/2) / (2T) = D / (4T) = v1 / 4. Thus v1 : v2 = (v1) : (v1/4) = 4 : 1. Verification / Alternative check: Choose D = 100 and T = 10 ⇒ v1 = 10. New: 50 in 20 ⇒ v2 = 2.5 ⇒ v1 : v2 = 10 : 2.5 = 4 : 1. Why Other Options Are Wrong: 1 : 4, 2 : 1, 1 : 2, 3 : 1 contradict v2 = v1/4. Common Pitfalls: Misreading “half of the distance in double the time” and inverting the ratio. Final Answer: 4 : 1

Question 17

A person distributes pens among A, B, C, and D in the ratio 1/3 : 1/4 : 1/5 : 1/6. What is the minimum total number of pens so that each gets a whole number of pens?

Accepted Answer

Introduction / Context: When shares are proportional to fractions, choose a common parameter so each share becomes an integer. The least such total comes from using the least common multiple of denominators. Given Data / Assumptions: Shares ∝ 1/3, 1/4, 1/5, 1/6. Concept / Approach: Let each share be k times the respective fraction. To make shares integers, pick k as LCM of the denominators. Then sum the integer shares for the minimum total. Step-by-Step Solution: LCM(3, 4, 5, 6) = 60. Shares: 60*(1/3) = 20; 60*(1/4) = 15; 60*(1/5) = 12; 60*(1/6) = 10. Minimum total pens = 20 + 15 + 12 + 10 = 57. Verification / Alternative check: Any smaller scaling will yield a non-integer for at least one recipient because 60 is the least common multiple. Why Other Options Are Wrong: 45, 65, 75, 60 are not the minimal achievable integer-sum with integral shares under the given proportions. Common Pitfalls: Adding fractions directly or failing to use LCM to clear denominators. Final Answer: 57

Question 18

Divide ₹ 990 into three parts such that half of the first part, one-third of the second part, and one-fifth of the third part are all equal. Find the three parts.

Accepted Answer

Introduction / Context: The condition equates particular fractions of the three parts. Converting those equalities to a common parameter allows us to express each part in terms of that parameter and then use the total to find exact values. Given Data / Assumptions: Let parts be x, y, z such that (1/2)x = (1/3)y = (1/5)z. Concept / Approach: Set the equal value to k. Then solve x, y, z in terms of k and sum to 990 to find k. Finally compute each part. Step-by-Step Solution: (1/2)x = k ⇒ x = 2k. (1/3)y = k ⇒ y = 3k. (1/5)z = k ⇒ z = 5k. Sum: x + y + z = 2k + 3k + 5k = 10k = 990 ⇒ k = 99. Thus x = 198, y = 297, z = 495. Verification / Alternative check: Check the condition: half of 198 is 99, one-third of 297 is 99, one-fifth of 495 is 99; all equal. Why Other Options Are Wrong: Other triplets do not satisfy both the total 990 and the fractional equalities simultaneously. Common Pitfalls: Equating x : y : z directly to 1/2 : 1/3 : 1/5 rather than converting via a common k. Final Answer: 198, 297, 495

Question 19

Find the quantity x that must be added to both terms of the ratio a : b so that the new ratio becomes c : d. Express x in terms of a, b, c, and d.

Accepted Answer

Introduction / Context: We want to adjust a ratio by adding the same amount x to both terms so that it changes to a specified ratio c : d. This requires algebraic manipulation of proportions. Given Data / Assumptions: Desired: (a + x) : (b + x) = c : d with real values and c ≠ d to avoid division by zero in the final expression. Concept / Approach: Translate the ratio to a fraction equality and cross multiply. Solve the linear equation for x. Step-by-Step Solution: (a + x)/(b + x) = c/d. Cross multiply: d(a + x) = c(b + x). da + dx = cb + cx. Group x terms: dx − cx = cb − da. x(d − c) = cb − da = −(ad − bc). x = (cb − da)/(d − c) = (ad − bc)/(c − d). Verification / Alternative check: Substitute x = (ad − bc)/(c − d) back into (a + x)/(b + x) and simplify; the result reduces to c/d. Why Other Options Are Wrong: The other formulas do not satisfy the derived linear equation for x; they result from misgrouping or mis-cross-multiplication. Common Pitfalls: Sign errors when moving terms or inverting (c − d) to (d − c). Both forms are equivalent up to a sign flip; ensure consistency with the numerator. Final Answer: (ad − bc) / (c − d)

Question 20

Proportional division with linked ratios: A total of ₹ 7000 is divided among A, B, and C. The shares of A and B are in the ratio 2 : 3, and the shares of B and C are in the ratio 4 : 5. Find the exact amount received by B.

Accepted Answer

Introduction / Context:This problem tests chained-ratio division of a fixed sum of money. We are given two compatible ratios that share a common term (B), and we must use them together to determine each person’s share, especially B’s amount. Given Data / Assumptions: Total amount = ₹ 7000. A : B = 2 : 3. B : C = 4 : 5. All shares are positive and add up to the total. Concept / Approach:Let A = 2x and B = 3x from the first ratio. Also let B = 4y and C = 5y from the second ratio. Since B is common, equate 3x = 4y to connect the two parameterizations. Use the total sum to solve for x (or y), then compute B. Step-by-Step Solution:From 3x = 4y ⇒ y = 3x/4.A + B + C = 2x + 3x + 5y = 5x + 5y.Substitute y: total = 5x + 5*(3x/4) = 5x + 15x/4 = 35x/4.35x/4 = 7000 ⇒ x = 800.B = 3x = 2400. Verification / Alternative check:Compute all shares: A = 1600, B = 2400, C = 3000 (since y = 600 and C = 5y = 3000). The ratios 2 : 3 and 4 : 5 are satisfied and the sum is 7000. Why Other Options Are Wrong: ₹ 1600 and ₹ 2000 are smaller than B’s required 3-part share. ₹ 3000 corresponds to C’s share when constraints are satisfied. Common Pitfalls: Treating the two ratios independently without equating the common term B. Forgetting to use the total to fix the scale of the ratios. Final Answer:₹ 2400

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