Question 1

Rs. 6300 is divided among A, B and C in the ratio 1/2 : 1 : 3/5. What is the share of B in rupees?

Accepted Answer

Introduction / Context:This question deals with the division of a total amount of money among three people in a ratio that involves fractions. Such ratio problems are widely asked in aptitude exams to test a candidate's understanding of fractional ratios and basic arithmetic with proportions. The key is to convert fractional ratios into simple integer ratios and then allocate the total amount according to these parts. Given Data / Assumptions: Total amount = Rs. 6300. Ratio of shares of A, B and C is 1/2 : 1 : 3/5. We need to find the share of B. All amounts are non negative and the entire sum is distributed. Concept / Approach:When a total amount is divided in a given ratio, each share is equal to (individual ratio term / sum of ratio terms) * total amount. Here, the ratio terms are fractions: 1/2, 1 and 3/5. It is easier to convert them to equivalent integers. We do this by multiplying all terms by the least common multiple of the denominators. Once the equivalent integer ratio is obtained, we can compute B's share using the standard formula. Step-by-Step Solution:Given ratio: A : B : C = 1/2 : 1 : 3/5.Find a common denominator for 2 and 5, which is 10.Multiply each term by 10: (1/2) * 10 = 5, 1 * 10 = 10, (3/5) * 10 = 6.So the simplified integer ratio is A : B : C = 5 : 10 : 6.Total parts = 5 + 10 + 6 = 21.Each part is worth 6300 / 21 rupees.6300 / 21 = 300, so each part is Rs. 300.B has 10 parts, so B's share = 10 * 300 = Rs. 3000. Verification / Alternative check:Check A's share: 5 parts * 300 = Rs. 1500.Check B's share: 10 parts * 300 = Rs. 3000.Check C's share: 6 parts * 300 = Rs. 1800.Total = 1500 + 3000 + 1800 = Rs. 6300, which matches the original amount, so the division is correct. Why Other Options Are Wrong:Rs. 3300, 2700, 2400 and 2100 do not satisfy the fractional ratio when the corresponding shares of A and C are computed. For example, if B received Rs. 2700, the equivalent part value would be 2700 / 10 = 270, which would give A = 5 * 270 = 1350 and C = 6 * 270 = 1620. The total would then be 1350 + 2700 + 1620 = 5670, not 6300. Hence these alternatives are incorrect. Common Pitfalls:The main errors occur when candidates forget to convert fractional ratios to integers or mishandle the denominators. Some students also mistakenly divide 6300 by 3 instead of by the sum of ratio parts. Always remember that the sum used in the denominator is the sum of all ratio terms after simplification, not the number of persons. Final Answer:The share of B is Rs. 3000.

Question 2

If 6A = 7B = 5C, what is the ratio A : B : C implied by this relationship?

Accepted Answer

Introduction / Context:This problem is similar to other chained proportional equations such as 3A = 7B = 13C. It examines whether the student can convert the expression 6A = 7B = 5C into a meaningful ratio for A, B and C. Questions of this type are foundational for ratio and proportion and often appear in aptitude tests and entrance examinations. Given Data / Assumptions: We are given 6A = 7B = 5C. A, B and C are positive quantities. The task is to find the ratio A : B : C. Concept / Approach:Whenever we have 6A = 7B = 5C, we introduce a constant k such that 6A = 7B = 5C = k. From this, we express A, B and C in terms of k, giving A = k / 6, B = k / 7 and C = k / 5. This leads to a ratio in terms of fractions. We then remove denominators by multiplying through by the least common multiple of 6, 7 and 5. The resulting integer ratio is our final answer. Step-by-Step Solution:Assume 6A = 7B = 5C = k.Then A = k / 6, B = k / 7 and C = k / 5.So A : B : C = (k / 6) : (k / 7) : (k / 5).Cancel the common factor k to obtain 1 / 6 : 1 / 7 : 1 / 5.Find the least common multiple of 6, 7 and 5, which is 210.Multiply each term by 210: (1 / 6) * 210 = 35, (1 / 7) * 210 = 30, (1 / 5) * 210 = 42.Therefore, A : B : C = 35 : 30 : 42. Verification / Alternative check:Take A = 35, B = 30 and C = 42.Compute 6A = 6 * 35 = 210, 7B = 7 * 30 = 210 and 5C = 5 * 42 = 210.All three expressions are equal, which confirms that 6A = 7B = 5C is satisfied by A : B : C = 35 : 30 : 42. Why Other Options Are Wrong:Options such as 42 : 30 : 35 or 30 : 35 : 42 represent permutations of the correct ratio but do not preserve the condition 6A = 7B = 5C when checked. The option 5 : 7 : 6 confuses coefficients with the actual ratio and ignores the reciprocal transformation. Therefore, only 35 : 30 : 42 is consistent with the original equality. Common Pitfalls:Students sometimes mistakenly take the direct ratio of A : B : C as 6 : 7 : 5, which is incorrect because those are the multipliers of A, B and C, not the variables themselves. Another mistake is to forget cancelling the common constant k before clearing denominators. Always convert the chained equality into fractional expressions, then clear denominators systematically. Final Answer:The required ratio is 35 : 30 : 42.

Question 3

The salary of Gaganjyot is 7/6 times the salary of Hafiz and the salary of Sayed is 8/7 times the salary of Hafiz. What is the ratio of Gaganjyot's salary to Sayed's salary?

Accepted Answer

Introduction / Context:This question examines how to combine two comparative statements that are both expressed in terms of a third person's salary. It is a classic application of ratios and fractions. The goal is to find the ratio between Gaganjyot and Sayed even though both salaries are initially given in relation to Hafiz. Converting all information to a common base (Hafiz) and then simplifying is the main technique here. Given Data / Assumptions: Let Hafiz's salary be H. Gaganjyot's salary = (7 / 6) * H. Sayed's salary = (8 / 7) * H. We need to find the ratio of Gaganjyot's salary to Sayed's salary. Concept / Approach:When multiple quantities are expressed as fractions of a common quantity, we can express each explicitly and then divide one by the other to find the required ratio. Here, the ratio Gaganjyot : Sayed is equal to [(7 / 6) * H] : [(8 / 7) * H]. After cancelling H, the ratio reduces to a fraction involving only numbers. Simplifying that fraction yields the final ratio. Step-by-Step Solution:Assume Hafiz's salary = H.Then Gaganjyot's salary = (7 / 6) * H.Sayed's salary = (8 / 7) * H.Required ratio = Gaganjyot : Sayed = (7 / 6) * H : (8 / 7) * H.Cancel the common factor H to get (7 / 6) : (8 / 7).Write this as a single fraction: (7 / 6) / (8 / 7) = (7 / 6) * (7 / 8).Compute: (7 / 6) * (7 / 8) = 49 / 48.Therefore, Gaganjyot : Sayed = 49 : 48. Verification / Alternative check:Take a convenient value for H, for example H = 48.Then Gaganjyot's salary = (7 / 6) * 48 = 56.Sayed's salary = (8 / 7) * 48 ≈ 54.857, but for ratio work it is more convenient to work algebraically.Direct algebraic simplification already showed Gaganjyot : Sayed = 49 : 48, which is the simplest integer form of the ratio. Why Other Options Are Wrong:Options like 3 : 4 or 4 : 3 treat the multipliers 7/6 and 8/7 very roughly and ignore the actual division (7 / 6) / (8 / 7). The options 48 : 49 and 7 : 8 invert or simplify the fraction incorrectly. Only 49 : 48 exactly matches the fraction 49 / 48 obtained after proper simplification. Common Pitfalls:A frequent mistake is to attempt to take the ratio 7/6 : 8/7 as 7 : 8 or 6 : 7 directly without doing the correct fractional division. Some students also multiply the fractions incorrectly by adding or subtracting numerators and denominators. The safe method is to always convert a : b where a and b are fractions into a / b and then multiply by the reciprocal of b to simplify. Final Answer:The ratio of Gaganjyot's salary to Sayed's salary is 49 : 48.

Question 4

Rs. 782 is divided into three parts in the ratio 12 : 23 : 34. What is the amount of the first part?

Accepted Answer

Introduction / Context:This problem is a straightforward application of ratio and proportion where a fixed sum of money is divided into three parts according to a given ratio. The question asks specifically for the first part. These questions are very common in bank exams, SSC, and other competitive tests, and they reinforce the idea that ratios represent relative shares of a total quantity. Given Data / Assumptions: Total amount = Rs. 782. Ratio of the three parts = 12 : 23 : 34. The first part corresponds to the term 12 in the ratio. All three parts are non negative and add up exactly to Rs. 782. Concept / Approach:If a sum S is divided into parts in the ratio a : b : c, then the first part is (a / (a + b + c)) * S, the second part is (b / (a + b + c)) * S and the third part is (c / (a + b + c)) * S. Here, a = 12, b = 23 and c = 34. Once we compute the sum of the ratio terms, we can easily compute the value of one part by simple multiplication and division. Step-by-Step Solution:Given ratio of parts = 12 : 23 : 34.Compute the total of ratio terms: 12 + 23 + 34 = 69.Total amount S = Rs. 782.Value of one ratio unit = 782 / 69.Compute 782 / 69 = 11.34 approximately, but we want the exact first part.First part = 12 * (782 / 69) = (12 * 782) / 69.Calculate the numerator: 12 * 782 = 9384.Now divide 9384 by 69 to get 136.Therefore, the first part is Rs. 136. Verification / Alternative check:Compute all three parts to verify the total.First part: 12 * (782 / 69) = 136.Second part: 23 * (782 / 69) = 23 * 136 / 12 ≈ 260.67, but more precisely 260 is near when rounded; in exam contexts the first part is usually the focus.Third part: 34 * (782 / 69) gives the remaining share so that the total is 782.Adding the three exact parts computed using the ratio method gives back the original total 782, so the division is consistent with the ratio 12 : 23 : 34. Why Other Options Are Wrong:Values like Rs. 182, Rs. 190, Rs. 196 or Rs. 204 do not fit the ratio correctly. For instance, if the first part were Rs. 182, then the value per ratio unit would be 182 / 12, and the implied total amount would not equal Rs. 782 when the three parts are recomputed. Only Rs. 136 leads to a consistent division of the total in the ratio 12 : 23 : 34. Common Pitfalls:Many students mistakenly divide 782 directly by 3 or confuse the ratio itself with the actual amounts. Another error is to miscalculate the sum of the ratio terms or to perform division incorrectly. Always remember the formula: first part = (first ratio term / sum of all ratio terms) * total amount. Careful arithmetic avoids unnecessary mistakes. Final Answer:The amount of the first part is Rs. 136.

Question 5

The average age of all students in a class is 15.8 years. The average age of boys alone is 16.4 years and the average age of girls alone 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 question involves weighted averages. The overall average age of a group is given, along with the averages of two subgroups (boys and girls). The task is to find the ratio of the number of boys to girls. Such problems are common in data interpretation and arithmetic sections, and they check whether the student can connect the combined average with the weighted contributions from subgroups. Given Data / Assumptions: Average age of the entire class = 15.8 years. Average age of boys = 16.4 years. Average age of girls = 15.4 years. Let the number of boys be B and the number of girls be G. We assume all averages are correctly computed from actual ages. Concept / Approach:The overall average is a weighted average of subgroup averages. Using total age of boys as 16.4 * B and total age of girls as 15.4 * G, the overall mean is (16.4 * B + 15.4 * G) / (B + G) = 15.8. This gives an equation connecting B and G. Solving for the ratio B : G gives the required answer. This is a standard technique whenever an overall average and subgroup averages are known. Step-by-Step Solution:Let B = number of boys and G = number of girls.Total age of boys = 16.4 * B.Total age of girls = 15.4 * G.Total number of students = B + G, and overall average = 15.8.So (16.4 * B + 15.4 * G) / (B + G) = 15.8.Multiply both sides by (B + G): 16.4B + 15.4G = 15.8B + 15.8G.Rearrange terms: 16.4B - 15.8B = 15.8G - 15.4G.0.6B = 0.4G.Divide both sides by 0.2: 3B = 2G.So B / G = 2 / 3.Therefore, the ratio of boys to girls is 2 : 3. Verification / Alternative check:Assume B = 2k and G = 3k for some positive k.Total students = 5k.Total age = 16.4 * 2k + 15.4 * 3k = 32.8k + 46.2k = 79k.Overall average = 79k / 5k = 15.8 years, which matches the given overall average.So the ratio 2 : 3 is consistent with all given data. Why Other Options Are Wrong:Ratios like 1 : 2, 3 : 4 or 3 : 5 do not satisfy the weighted average equation. If any of these ratios are substituted into the equation, the resulting overall average is different from 15.8. Only B : G = 2 : 3 keeps the overall average equal to 15.8 years. Common Pitfalls:Students sometimes try to solve these questions by guessing rather than setting up the weighted average formula. Another mistake is to subtract averages directly without considering how the numbers of students weigh each average. Always represent total ages using average * number of people, add them, and then divide by total people to get the overall average equation. Final Answer:The required ratio of boys to girls is 2 : 3.

Question 6

Divide Rs. 80 among three people in the proportion 3 : 6 : 7. What are the three resulting shares?

Accepted Answer

Introduction / Context:This is a basic ratio division question. The total amount of Rs. 80 has to be split among three persons in the ratio 3 : 6 : 7. Such problems are often asked to check whether candidates can correctly convert a ratio into actual numerical parts and ensure that the entire total is exactly distributed. Given Data / Assumptions: Total amount = Rs. 80. Ratio of shares = 3 : 6 : 7. We need to find each individual share that matches both the ratio and the sum. Concept / Approach:If a sum S is divided into parts in the ratio a : b : c, then the sum of the ratio terms is a + b + c. Each part is obtained by multiplying S by the respective ratio term and dividing by the total of the ratio terms. That is, first share = (a / (a + b + c)) * S, second share = (b / (a + b + c)) * S and so on. We apply this directly here with a = 3, b = 6 and c = 7. Step-by-Step Solution:Given ratio = 3 : 6 : 7.Total of ratio terms = 3 + 6 + 7 = 16.Total amount S = Rs. 80.Value of one ratio unit = 80 / 16 = 5.First share = 3 * 5 = Rs. 15.Second share = 6 * 5 = Rs. 30.Third share = 7 * 5 = Rs. 35.Therefore, the three shares are Rs. 15, Rs. 30 and Rs. 35. Verification / Alternative check:Add the three shares: 15 + 30 + 35 = 80, which matches the original total.Check ratio: 15 : 30 : 35 simplifies to 3 : 6 : 7 when divided by 5, so the division respects the given ratio. Why Other Options Are Wrong:Options where the three amounts do not add to Rs. 80 or do not simplify to the ratio 3 : 6 : 7 are incorrect. For example, Rs. 10, Rs. 35 and Rs. 40 sum to 85, not 80. Similarly, Rs. 10, Rs. 40 and Rs. 35 give a total of 85. Only Rs. 15, Rs. 30 and Rs. 35 both sum to 80 and maintain the 3 : 6 : 7 ratio. Common Pitfalls:A common error is to treat ratio terms as direct rupee values without scaling them to match the total sum, leading to a mismatch in the final total. Another pitfall is arithmetic mistakes when computing the sum of ratio terms or dividing the total amount. Always compute the unit value as total divided by the sum of ratio terms and then multiply each term by that unit value. Final Answer:The three shares are Rs. 15, Rs. 30 and Rs. 35 respectively.

Question 7

The speeds of three racers are in the ratio 3 : 4 : 6. What is the ratio of the times taken by them to cover the same distance?

Accepted Answer

Introduction / Context:This question explores the inverse relationship between speed and time for a fixed distance. When distance is constant, time is inversely proportional to speed: higher speed means less time, and lower speed means more time. The problem gives the ratio of speeds of three racers and asks for the ratio of times taken to cover the same distance. Such inverse proportion questions are very common in time, speed and distance topics. Given Data / Assumptions: Ratio of speeds of three racers = 3 : 4 : 6. They all cover the same distance. We need the ratio of time taken by each racer. All speeds are positive and non zero. Concept / Approach:For a fixed distance, Time = Distance / Speed. This means time is inversely proportional to speed. If speeds are in the ratio v1 : v2 : v3, then times are in the ratio 1 / v1 : 1 / v2 : 1 / v3. Therefore, if speeds are 3 : 4 : 6, then times are 1 / 3 : 1 / 4 : 1 / 6. We then find an equivalent integer ratio by taking a common multiple of the denominators. Step-by-Step Solution:Speeds are in ratio 3 : 4 : 6.Let the common distance be D.Time taken by each racer: T1 = D / (3k), T2 = D / (4k), T3 = D / (6k) for some constant k.So T1 : T2 : T3 = 1 / 3 : 1 / 4 : 1 / 6.To remove denominators, multiply all terms by 12 (the least common multiple of 3, 4 and 6).(1 / 3) * 12 = 4, (1 / 4) * 12 = 3, (1 / 6) * 12 = 2.Therefore, T1 : T2 : T3 = 4 : 3 : 2. Verification / Alternative check:Assume the distance D = 12 units and speeds are proportional to actual speeds 3, 4 and 6 units per hour.Time for first racer = 12 / 3 = 4 hours.Time for second racer = 12 / 4 = 3 hours.Time for third racer = 12 / 6 = 2 hours.This confirms the time ratio 4 : 3 : 2. Why Other Options Are Wrong:The ratio 3 : 4 : 6 is the speed ratio, not the time ratio. Ratios 6 : 4 : 3 or 1 : 2 : 3 do not correspond to the inverse of 3 : 4 : 6 when calculated properly. Option 2 : 3 : 5 is also unrelated. Only 4 : 3 : 2 matches the correct inverse proportional relation. Common Pitfalls:A typical mistake is to think time and speed are directly proportional and to answer 3 : 4 : 6. Another error is to invert the ratio partially or incorrectly. Always remember that for a fixed distance, time is inversely proportional to speed, so you must take reciprocals and then simplify. Final Answer:The ratio of times taken by the three racers is 4 : 3 : 2.

Question 8

The ratio of the weights of two solid spheres is 8 : 27, and the ratio of their weights per cubic centimetre (densities) is 8 : 1. What is the ratio of the radii of the two spheres?

Accepted Answer

Introduction / Context:This question combines ideas from ratio and basic geometry of spheres. The weight of a solid sphere is proportional to its volume multiplied by the density of its material. Volume of a sphere is proportional to the cube of its radius. Given the ratio of weights and the ratio of densities, we are asked to find the ratio of the radii. This type of problem appears in aptitude tests where understanding proportional relationships in geometry is important. Given Data / Assumptions: Weight ratio of the two spheres W1 : W2 = 8 : 27. Density ratio (weight per cubic centimetre) rho1 : rho2 = 8 : 1. Let radii of the spheres be r1 and r2. Weights are given by W = density * volume. Concept / Approach:For a sphere, volume V is proportional to r^3. Hence weight W is proportional to rho * r^3. Using proportionality, W1 / W2 = (rho1 * r1^3) / (rho2 * r2^3). We are given the ratio of W1 : W2 and rho1 : rho2, so we can substitute those and solve for r1^3 : r2^3, and then for r1 : r2 by taking cube roots. This is a standard method whenever radius and weight are linked through volume and density. Step-by-Step Solution:Let W1 and W2 be weights and rho1 and rho2 be densities.Given W1 : W2 = 8 : 27 and rho1 : rho2 = 8 : 1.For spheres, W is proportional to rho * r^3, so W1 / W2 = (rho1 * r1^3) / (rho2 * r2^3).Substitute ratios: 8 / 27 = (8 * r1^3) / (1 * r2^3).So 8 / 27 = 8 * r1^3 / r2^3.Divide both sides by 8: (8 / 27) / 8 = r1^3 / r2^3.This gives r1^3 / r2^3 = 1 / 27.Take cube roots on both sides: r1 / r2 = cube root of (1 / 27) = 1 / 3.Therefore, the ratio of radii r1 : r2 = 1 : 3. Verification / Alternative check:Assume r1 = 1 unit and r2 = 3 units.Then r1^3 = 1 and r2^3 = 27.Using rho1 : rho2 = 8 : 1, let rho1 = 8k and rho2 = k for some k.Then W1 = 8k * 1 = 8k and W2 = k * 27 = 27k.So W1 : W2 = 8k : 27k = 8 : 27, which matches the given weight ratio. Why Other Options Are Wrong:Ratios such as 2 : 3, 3 : 1 or 3 : 2 would give r1^3 : r2^3 as 8 : 27, 27 : 1 or 27 : 8. When these are combined with rho1 : rho2 = 8 : 1, they do not produce W1 : W2 = 8 : 27. Only r1 : r2 = 1 : 3 leads to the correct result when substituted back into the proportional relationship. Common Pitfalls:Some students mistakenly assume that the ratio of radii is directly 8 : 27 or confuse radius with volume. Others forget that volume of a sphere depends on the cube of radius and not on the square. Always set up the proportional relationship W proportional to rho * r^3 and carefully isolate the ratio of r1 to r2 from the given ratios. Final Answer:The ratio of the radii of the two spheres is 1 : 3.

Question 9

The monthly salaries of A and B are in the ratio 15 : 16 and the monthly salaries of B and C are in the ratio 17 : 18. If the total of their monthly salaries together is Rs. 1,87,450, what is the monthly income of C?

Accepted Answer

Introduction / Context:This question combines two ratio relationships and a total sum to find the individual salary of one person. We are given A : B and B : C and the total of A + B + C. From these, we need to derive A : B : C and then compute C's share. This type of chained ratio problem appears very frequently in aptitude exams and helps to test consistency in working with multiple related ratios. Given Data / Assumptions: Ratio of salaries of A and B is 15 : 16. Ratio of salaries of B and C is 17 : 18. Total of salaries A + B + C = Rs. 1,87,450. We assume all salaries are positive and fixed for the calculation. Concept / Approach:We first convert the two given ratios into a single combined ratio A : B : C. Let A = 15x and B = 16x from the first ratio. From the second ratio, let B = 17y and C = 18y. Since B is common, we set 16x = 17y and express everything in one variable. This gives us A, B and C in a consistent ratio. After that, the total salary is used to find the actual monetary value of one ratio unit, and then C's salary is obtained by multiplying that unit value with C's ratio term. Step-by-Step Solution:From A : B = 15 : 16, let A = 15x and B = 16x.From B : C = 17 : 18, let B = 17y and C = 18y.Since both expressions represent B, set 16x = 17y.So y = 16x / 17.Now express C in terms of x: C = 18y = 18 * (16x / 17) = 288x / 17.Thus A : B : C = 15x : 16x : 288x / 17.Multiply each term by 17 to clear the denominator: 15x * 17 : 16x * 17 : 288x.This gives A : B : C = 255 : 272 : 288.Sum of ratio terms = 255 + 272 + 288 = 815.Total salary A + B + C = Rs. 1,87,450, so each unit = 187450 / 815 = 230.C's salary = 288 * 230 = Rs. 66,240. Verification / Alternative check:Compute A and B using the same unit value.A = 255 * 230 = Rs. 58,650.B = 272 * 230 = Rs. 62,560.Check total: 58,650 + 62,560 + 66,240 = 1,87,450, which matches the given total.Check ratios: A : B = 58,650 : 62,560 simplifies to 15 : 16, and B : C = 62,560 : 66,240 simplifies to 17 : 18.Thus all conditions are satisfied. Why Other Options Are Wrong:Other figures such as Rs. 72,100 or Rs. 62,200 do not conform to the derived ratio 255 : 272 : 288. If any of these were taken as C's salary, the implied values of A and B would either fail to add up to Rs. 1,87,450 or fail to maintain the given pairwise ratios. Only Rs. 66,240 fits perfectly with both the total and the ratio conditions. Common Pitfalls:Candidates sometimes combine A : B and B : C incorrectly and directly multiply all numbers, leading to a wrong three way ratio. Another error is to forget to align the common term B by using a common multiple. Always express the common term (here B) consistently, find A : B : C as a single ratio, then use the total to scale up to actual values. Final Answer:The monthly income of C is Rs. 66,240.

Question 10

A company, during a period of inflation, reduced its staff in the ratio 5 : 3 and increased the average salary per employee in the ratio 7 : 8. By doing this, the company saved Rs. 55,000 in total monthly expenditure. What was the initial total expe…

Accepted Answer

Introduction / Context:This question mixes ratios with percentage style changes in a company's total expenditure on salaries. The staff strength is reduced, while the average salary per employee is increased. The result is a net reduction in total salary expenditure, which is given. We are asked to find the initial total expenditure. This is a typical application of multiplying ratios and comparing original and new totals in quantitative aptitude. Given Data / Assumptions: Staff is reduced in the ratio 5 : 3, meaning new staff = 3/5 of initial staff. Average salary per employee is increased in the ratio 7 : 8, meaning new average salary = 8/7 of the initial average salary. Due to these changes, the company saves Rs. 55,000 in total expenditure. We assume salary changes apply uniformly to all remaining employees. Concept / Approach:Let the original number of employees be N and the original average salary be S. Then the original total salary expenditure E0 = N * S. After the reduction, the number of employees becomes (3/5) * N, and the new average salary becomes (8/7) * S. Hence, new expenditure E1 = (3/5) * N * (8/7) * S. The difference E0 - E1 equals 55,000. Solving this equation in terms of E0 gives the initial total expenditure. Step-by-Step Solution:Let original number of employees = N and original average salary = S.Original expenditure E0 = N * S.New number of employees after reduction: N_new = (3 / 5) * N.New average salary after increase: S_new = (8 / 7) * S.New expenditure E1 = N_new * S_new = (3 / 5) * N * (8 / 7) * S.Simplify E1: (3 / 5) * (8 / 7) = 24 / 35, so E1 = (24 / 35) * N * S = (24 / 35) * E0.Given that savings = E0 - E1 = Rs. 55,000.So E0 - (24 / 35) * E0 = 55,000.E0 * (1 - 24 / 35) = 55,000.1 - 24 / 35 = 11 / 35.So (11 / 35) * E0 = 55,000.Therefore, E0 = 55,000 * (35 / 11) = 1,75,000. Verification / Alternative check:If original expenditure E0 = Rs. 1,75,000, then new expenditure E1 = (24 / 35) * 1,75,000.Compute E1: 1,75,000 * 24 / 35 = 1,20,000.Savings = E0 - E1 = 1,75,000 - 1,20,000 = 55,000, which matches the given savings.Thus the initial total expenditure is consistent with the information. Why Other Options Are Wrong:Other values, such as Rs. 1,55,000 or Rs. 1,60,000, do not produce a saving of Rs. 55,000 when multiplied by the factor 24 / 35. For example, if E0 = 1,60,000, then E1 = (24 / 35) * 1,60,000 = 1,09,714.29 approximately, and the saving would not be exactly Rs. 55,000. Only E0 = 1,75,000 gives the correct saving. Common Pitfalls:Some candidates confuse the direction of the ratios and incorrectly take new staff as 5/3 of old staff or new average as 7/8 of old average. Others subtract the ratios directly instead of using them as multiplicative factors. Always write original and new quantities explicitly, convert percentage like changes into multiplicative ratios, and then compare the products for expenditure. Final Answer:The initial total expenditure of the company was Rs. 1,75,000.

Question 11

The bus fare between two cities is increased in the ratio 5 : 11. If the original fare was Rs. 275, what is the increase in the fare in rupees?

Accepted Answer

Introduction / Context:This is a simple ratio application question involving an increase in bus fare. The phrase “increased in the ratio 5 : 11” means that the ratio of the old fare to the new fare is 5 : 11. Given the original fare, we need to find the absolute increase. Such questions test whether the student correctly interprets how ratios describe change in prices or fares. Given Data / Assumptions: Old bus fare (original fare) = Rs. 275. Old fare : new fare = 5 : 11. We need to find the increase = new fare - old fare. Concept / Approach:If the ratio of old fare to new fare is 5 : 11, we can let the old fare be 5k and the new fare be 11k for some positive constant k. Since we know the actual old fare (Rs. 275), we can solve for k and then compute the new fare as 11k. The required increase is the difference between new fare and old fare. This is a direct and standard way of using ratios to move from relative values to absolute values. Step-by-Step Solution:Let the old fare be 5k and the new fare be 11k.Given that the old fare is Rs. 275, so 5k = 275.Solve for k: k = 275 / 5 = 55.New fare = 11k = 11 * 55 = Rs. 605.Increase in fare = new fare - old fare = 605 - 275 = Rs. 330. Verification / Alternative check:Check the ratio of old to new: 275 : 605.Divide both numbers by 55: (275 / 55) : (605 / 55) = 5 : 11.This confirms that the new fare is correctly related to the old fare in the ratio 5 : 11. Why Other Options Are Wrong:An increase of Rs. 121 or Rs. 242 would correspond to new fares of Rs. 396 or Rs. 517 respectively. The ratios 275 : 396 or 275 : 517 do not simplify to 5 : 11. Rs. 605 is the new fare, not the increase. Therefore the only correct increase that matches the given ratio and original fare is Rs. 330. Common Pitfalls:Some students misinterpret 5 : 11 as meaning the fare becomes 5 / 11 of the original or perhaps increases by (11 - 5) parts only. Others subtract 5 from 11 and treat 6 parts as the increase without linking back to the original 5 parts. The correct approach is to treat both old and new fares as proportional to a common factor k and then use the known old fare to determine k. Final Answer:The increase in bus fare is Rs. 330.

Question 12

In an MBA selection process, the ratio of selected to unselected candidates was 11 : 2. If 40 fewer candidates had applied and 20 fewer had been selected, the ratio of selected to unselected would have become 10 : 1. How many candidates had original…

Accepted Answer

Introduction / Context:This problem involves working with ratios and a hypothetical change in the number of candidates. Initially, we know the ratio of selected to unselected candidates. After a hypothetical reduction in both total applicants and selected candidates, the ratio changes. The task is to find the original total number of applicants. These questions are common in exam settings and test skill with forming and solving algebraic equations from verbal ratio statements. Given Data / Assumptions: Initial ratio of selected to unselected candidates = 11 : 2. In a hypothetical case, 40 fewer candidates applied and 20 fewer were selected. Under this hypothetical scenario, the new ratio of selected to unselected is 10 : 1. We need the original number of candidates who applied. Concept / Approach:Let the number of selected candidates initially be 11k and unselected candidates be 2k, so total applicants N = 13k. After the hypothetical change, the total applicants become N - 40 and the selected become 11k - 20. The unselected become (N - 40) - (11k - 20). The new ratio (selected : unselected) is 10 : 1, which gives an equation in k. Solving this equation gives k, and hence N = 13k. Step-by-Step Solution:Let initial selected candidates = 11k and unselected candidates = 2k.Total initial applicants N = 11k + 2k = 13k.In the hypothetical situation, total applicants become N - 40 = 13k - 40.Selected candidates become 11k - 20.Unselected candidates in the hypothetical case = (13k - 40) - (11k - 20) = 13k - 40 - 11k + 20 = 2k - 20.We are told that the new ratio of selected to unselected is 10 : 1.So (11k - 20) / (2k - 20) = 10 / 1.Cross multiply: 11k - 20 = 10 * (2k - 20).11k - 20 = 20k - 200.Bring like terms together: 20k - 11k = 200 - 20.9k = 180, so k = 20.Therefore, original total applicants N = 13k = 13 * 20 = 260. Verification / Alternative check:Original: selected = 11 * 20 = 220, unselected = 2 * 20 = 40, total = 260.Initial ratio selected : unselected = 220 : 40 = 11 : 2, which matches the given ratio.Hypothetical: total = 260 - 40 = 220, selected = 220 - 20 = 200.Unselected in the hypothetical case = 220 - 200 = 20.New ratio selected : unselected = 200 : 20 = 10 : 1, which matches the condition. Why Other Options Are Wrong:If we assume other totals like 220, 300 or 340, we do not get integer values of selected and unselected candidates that satisfy both ratio conditions simultaneously. Only N = 260 allows consistent integer values that satisfy both the original and the hypothetical ratio statements. Common Pitfalls:Common errors include mixing up the direction of the hypothetical changes or forgetting that unselected candidates in the second case must be computed as total minus selected. Some students also misinterpret the 40 fewer applicants as referring only to selected candidates. Carefully defining selected, unselected and total in both scenarios and forming an equation using the given ratios avoids such mistakes. Final Answer:The original number of candidates who applied for the process was 260.

Question 13

Find two numbers such that their mean proportional is 16 and the third proportional to them is 1024. What are the two numbers?

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Introduction / Context:This question is similar to the earlier one involving mean proportional and third proportional, but uses different numerical values. The mean proportional (geometric mean) between two numbers and the third proportional to them are provided. We must compute the original pair of numbers. This tests deeper understanding of classical proportion terminology used in aptitude and algebra questions. Given Data / Assumptions: Let the two numbers be x and y. Their mean proportional is 16. The third proportional to x and y is 1024. We assume x and y are positive real numbers. Concept / Approach:The mean proportional m between x and y satisfies x : m = m : y, which implies m^2 = x * y. So x * y = 16^2 = 256. The third proportional T to x and y is defined by x : y = y : T, which implies T = y^2 / x. Given T = 1024, we obtain y^2 / x = 1024. Using x * y = 256 and y^2 / x = 1024, we can solve for x and y systematically. Step-by-Step Solution:Let the two numbers be x and y.Mean proportional m = 16 implies x * y = 16^2 = 256.Third proportional T = 1024 implies y^2 / x = 1024.From y^2 / x = 1024, we get y^2 = 1024 * x.From x * y = 256, we can write y = 256 / x.Substitute y = 256 / x into y^2 = 1024 * x.(256 / x)^2 = 1024 * x.65536 / x^2 = 1024 * x.Multiply both sides by x^2: 65536 = 1024 * x^3.x^3 = 65536 / 1024 = 64.So x = cube root of 64 = 4.Now use x * y = 256: 4 * y = 256, so y = 64. Verification / Alternative check:Check mean proportional: x * y = 4 * 64 = 256, geometric mean = sqrt(256) = 16, which matches the given mean proportional.Check third proportional: x : y = 4 : 64 = 1 : 16 and y : T must also be 1 : 16.With T = 1024, y : T = 64 : 1024 = 1 : 16, so 1024 is indeed the third proportional to 4 and 64. Why Other Options Are Wrong:Pairs like (4, 32), (8, 64) or (8, 32) do not satisfy both conditions simultaneously. For instance, (4, 32) gives product 128 and geometric mean sqrt(128), which is not 16. Similarly, other options fail either the mean proportional condition or the third proportional condition. Only 4 and 64 satisfy both. Common Pitfalls:Students sometimes confuse arithmetic mean with geometric mean and use (x + y) / 2 = 16 instead of x * y = 256. Another mistake is to misinterpret the definition of third proportional and write x^2 / y instead of y^2 / x. Consistently using the correct definitions and double checking each condition after solving helps to avoid these errors. Final Answer:The two numbers are 4 and 64.

Question 14

What is the third proportional to the two numbers 9 and 15?

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Introduction / Context:This question focuses on the classical definition of the third proportional of two numbers. It is a direct application of ratio and proportion, used frequently in elementary algebra and aptitude tests. Understanding these terms is important because they appear in many old style aptitude questions and also help in dealing with geometric mean and related topics. Given Data / Assumptions: The two given numbers are 9 and 15. We are asked to find the third proportional to these two numbers. We assume the standard definition of third proportional. Concept / Approach:If a, b and c are three numbers such that a : b = b : c, then c is called the third proportional to a and b. This definition gives the relation b^2 = a * c, or c = b^2 / a. In this problem, a = 9 and b = 15, so c will be equal to (15^2) / 9. We just need to compute this value and then match it with one of the options provided. Step-by-Step Solution:Let the third proportional to 9 and 15 be c.By definition, 9 : 15 = 15 : c.This implies 9 * c = 15 * 15.So 9 * c = 225.Therefore, c = 225 / 9.Compute 225 / 9 = 25.Hence, the third proportional is 25. Verification / Alternative check:Check the proportion: 9 : 15 should equal 15 : 25.Compute 9 / 15 = 3 / 5 = 0.6.Compute 15 / 25 = 3 / 5 = 0.6.Since both ratios are equal, the proportion holds and 25 is correct as the third proportional. Why Other Options Are Wrong:Values such as 21, 30, 45 or 36 do not satisfy the proportion 9 : 15 = 15 : c. If we take c = 21, we would get 15 / 21 = 5 / 7, which is not equal to 3 / 5. Similar checks show that 30, 45 and 36 also fail. Only c = 25 gives the correct equality of ratios. Common Pitfalls:A frequent mistake is to confuse third proportional with arithmetic progression or with geometric mean. Some students compute c as 15 * 9 or as (9 + 15) / 2 without using the proportion definition. The safest method is to always start from the equality a : b = b : c and cross multiply to obtain a * c = b^2. Final Answer:The third proportional to 9 and 15 is 25.

Question 15

What is the fourth proportional to the numbers 72, 168 and 150?

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Introduction / Context:This question uses the definition of fourth proportional, another classical term from ratio and proportion. It is a direct skill check: given three numbers, we are asked to find the fourth proportional. Such questions are straightforward once the definition and formula are remembered, and they form a base for more complicated proportional reasoning in aptitude exams. Given Data / Assumptions: First three numbers are 72, 168 and 150. We need to find the fourth proportional. We use the standard proportional definition. Concept / Approach:If a, b, c and d are four numbers such that a : b = c : d, then d is called the fourth proportional to a, b and c. The proportion a : b = c : d leads to the equation a * d = b * c, hence d = (b * c) / a. In this problem, a = 72, b = 168 and c = 150, so d = (168 * 150) / 72. Once we compute d, we select the matching option. Step-by-Step Solution:Let the fourth proportional to 72, 168 and 150 be d.By definition, 72 : 168 = 150 : d.So 72 * d = 168 * 150.Compute 168 * 150 = 168 * (100 + 50) = 16800 + 8400 = 25200.Thus 72 * d = 25200.So d = 25200 / 72.Simplify: divide numerator and denominator by 8 to get 3150 / 9 = 350.Therefore, d = 350. Verification / Alternative check:Check the proportion 72 : 168 = 150 : 350.Simplify 72 : 168 by dividing by 24 to get 3 : 7.Simplify 150 : 350 by dividing by 50 to get 3 : 7.Since both ratios are equal, 350 is the correct fourth proportional. Why Other Options Are Wrong:Values such as 450, 300, 400 or 250 do not maintain the equality 72 : 168 = 150 : d. For example, if d = 300, then 150 : 300 = 1 : 2 which is not equal to 3 : 7. Only 350 gives the same simplified ratio as 72 : 168, so the other options are incorrect. Common Pitfalls:Students often confuse fourth proportional with third proportional or use the wrong formula. Some may attempt to average numbers or simply guess close values. The proper approach is to write a : b = c : d, cross multiply, and solve for d. Keeping this template in mind ensures reliable solutions to all similar problems. Final Answer:The fourth proportional to 72, 168 and 150 is 350.

Question 16

A certain amount of money is divided between A and B in the ratio 5 : 6. If B receives Rs. 360, what is the total amount of money?

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Introduction / Context:This question is a standard application of ratio based distribution of money. We are given the ratio of amounts received by A and B, along with the actual amount received by B. From this, we have to compute the total amount shared between them. This is one of the most basic kinds of ratio questions used in competitive exams. Given Data / Assumptions: Money is divided between A and B in the ratio A : B = 5 : 6. B receives Rs. 360. We need to find the total amount = A + B. Concept / Approach:Let the common ratio factor be k. Then A = 5k and B = 6k. Since B's share is given as Rs. 360, we equate 6k = 360 and solve for k. Once k is known, A's share is 5k and the total amount is A + B = 11k. This is a direct and simple use of ratios to determine actual amounts. Step-by-Step Solution:Given A : B = 5 : 6.Let A = 5k and B = 6k.Given B = Rs. 360, so 6k = 360.Therefore, k = 360 / 6 = 60.Then A = 5k = 5 * 60 = Rs. 300.Total amount = A + B = 300 + 360 = Rs. 660. Verification / Alternative check:Check the ratio A : B using the found values.A : B = 300 : 360 = 5 : 6 after dividing both numbers by 60.This matches the given ratio, so the values are consistent. Why Other Options Are Wrong:If the total were Rs. 560, 680 or 580, the corresponding value of A obtained by subtracting 360 would not give the ratio 5 : 6. For example, if total = 720, then A would be 360 and the ratio A : B would be 360 : 360 = 1 : 1, which is not 5 : 6. Therefore only Rs. 660 satisfies both the given ratio and B's known share. Common Pitfalls:Some students mistakenly treat the ratio 5 : 6 as percentages or incorrectly assume that B's share is 6% of the total. Ratios express relative parts, not percentages directly. The reliable approach is always to assign a common factor k to each term, convert the known share into an equation, solve for k, and then derive the other amounts from it. Final Answer:The total amount of money is Rs. 660.

Question 17

Three numbers are in the ratio 3 : 2 : 5. If the sum of the squares of these three numbers is 1862, what is the middle number?

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Introduction / Context:This problem combines ratio with the sum of squares of numbers. We know the ratio of three numbers and the sum of their squares. The task is to identify the middle number. Such questions test algebraic manipulation skills and the ability to represent ratios algebraically using a common factor, then apply the given condition to solve for that factor. Given Data / Assumptions: Three numbers are in the ratio 3 : 2 : 5. Let the numbers be 3k, 2k and 5k for some positive k. The sum of the squares of these numbers is 1862. We need to find the value of the middle number, which is 2k. Concept / Approach:From the ratio, we represent the three numbers as 3k, 2k and 5k. The sum of their squares is then (3k)^2 + (2k)^2 + (5k)^2. We set this equal to 1862, and solve for k. Once k is known, the individual numbers can be found by multiplying the ratio terms with k. The middle number is simply 2k. Step-by-Step Solution:Let the three numbers be 3k, 2k and 5k.Compute their squares: (3k)^2 = 9k^2, (2k)^2 = 4k^2 and (5k)^2 = 25k^2.Sum of squares = 9k^2 + 4k^2 + 25k^2 = 38k^2.We are given that 38k^2 = 1862.So k^2 = 1862 / 38.Compute 1862 / 38 = 49.Thus k^2 = 49, so k = 7 (taking the positive value as we are dealing with magnitudes).The three numbers are then 3k = 21, 2k = 14 and 5k = 35.The middle number is 14. Verification / Alternative check:Check the sum of squares using the found numbers.21^2 = 441, 14^2 = 196 and 35^2 = 1225.Sum = 441 + 196 + 1225 = 1862, which matches the given sum.Thus the values 21, 14 and 35 satisfy both the ratio and the sum of squares condition. Why Other Options Are Wrong:If the middle number were 16, 13, 15 or 18, then the implied k and the corresponding numbers 3k, 2k and 5k would not give a sum of squares equal to 1862. Only 14 produces the exact required sum. Therefore, those options are incorrect even if they seem close in magnitude. Common Pitfalls:Some students mistakenly square the ratio terms and equate the sum of squared ratios directly to 1862, without introducing the variable k. Others may make arithmetic errors when computing the sum of squares. The correct process is to express the numbers as multiples of k, compute the squared sum, equate it to the given value, solve for k, and then determine the actual numbers. Final Answer:The middle number is 14.

Question 18

Before a battle, the ratio of captains to soldiers in an army was 2 : 7. During the war, 25 captains and 100 soldiers were martyred, and after these losses the new ratio of captains to soldiers became 3 : 10. How many soldiers were there in the army…

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Introduction / Context:Questions on captains and soldiers are classic ratio and proportion word problems used in aptitude exams. Here, we are told the initial ratio of captains to soldiers in an army, then given how many of each are lost in battle, and finally given the new ratio after the losses. Our task is to apply basic algebra on ratios to find the number of soldiers remaining after the war. This problem tests understanding of how to translate ratio information into equations and how to handle increases or decreases in the quantities while preserving the ratio relationships. Given Data / Assumptions: The initial ratio of captains to soldiers is 2 : 7. During the war, 25 captains are martyred. During the war, 100 soldiers are martyred. After the war, the new ratio of captains to soldiers is 3 : 10. We assume all quantities are whole numbers and ratios represent actual counts in the same proportion. Concept / Approach:The key idea is to represent the initial numbers of captains and soldiers in terms of a common multiplier using the given ratio 2 : 7. Then we subtract the numbers lost in the war from each group. The resulting numbers must satisfy the new ratio 3 : 10. This produces a simple linear equation in the unknown multiplier. Solving this equation gives us the actual original counts and therefore the final number of soldiers after the war. Step-by-Step Solution: Let the initial number of captains be 2k and the initial number of soldiers be 7k. After the war, captains remaining = 2k - 25. After the war, soldiers remaining = 7k - 100. We are told that the new ratio of captains to soldiers is 3 : 10. So, (2k - 25) / (7k - 100) = 3 / 10. Cross-multiply: 10(2k - 25) = 3(7k - 100). This gives 20k - 250 = 21k - 300. Rearrange: 21k - 20k = 300 - 250. So, k = 50. Initial soldiers = 7k = 7 * 50 = 350. Soldiers remaining after war = 350 - 100 = 250. Verification / Alternative check:Using k = 50, initial captains = 2 * 50 = 100. After war, captains = 100 - 25 = 75. Soldiers after war = 350 - 100 = 250. Check ratio: 75 : 250 = 3 : 10 (since 75/250 = 3/10). Therefore, the value is consistent and confirms that 250 soldiers remained after the war. Why Other Options Are Wrong: 200: This would give the ratio of captains to soldiers as 75 : 200 = 3 : 8, not 3 : 10. 150: This would result in 75 : 150 = 1 : 2, which is not equal to 3 : 10. 100: This gives 75 : 100 = 3 : 4, also not matching 3 : 10. Common Pitfalls:Students may incorrectly set up the equation by mixing initial and final numbers or by directly using 2 : 7 and 3 : 10 without including the losses. Another common mistake is to forget to subtract the martyred captains and soldiers before applying the second ratio. Some learners also attempt to guess values instead of forming a proper equation, which can lead to errors for less obvious numbers. Final Answer:The number of soldiers remaining in the army after the war is 250.

Question 19

A sum of Rs. 210 is to be divided among four people in the ratio 2 : 3 : 4 : 5. What are the respective amounts received by each share of this division?

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Introduction / Context:This is a straightforward ratio and proportion problem where a fixed sum of money must be divided among four parts in a given ratio. Such questions are common in aptitude tests and banking exams because they check whether a candidate can convert a verbal description of a ratio into actual numerical shares. Given Data / Assumptions: Total amount to be divided = Rs. 210. The ratio of the four parts is 2 : 3 : 4 : 5. All parts are directly proportional to these ratio values. We assume that the entire Rs. 210 is completely divided without any remainder. Concept / Approach:When an amount is divided in a given ratio, each share is calculated by multiplying the total amount by the fraction corresponding to that part. The fraction for each part is its ratio number divided by the sum of all ratio numbers. So, we first find the total of the ratio terms, then compute each share of Rs. 210 using simple multiplication and division. This method generalizes to any ratio-splitting question. Step-by-Step Solution: Step 1: Add the ratio terms: 2 + 3 + 4 + 5 = 14. Step 2: Each unit of the ratio is equal to 210 / 14. Compute this: 210 / 14 = 15. Step 3: The first part = 2 units = 2 * 15 = 30. Step 4: The second part = 3 units = 3 * 15 = 45. Step 5: The third part = 4 units = 4 * 15 = 60. Step 6: The fourth part = 5 units = 5 * 15 = 75. Verification / Alternative check:Add up the four shares: 30 + 45 + 60 + 75 = 210. This matches the original amount given in the question, so the division is correct and consistent with the required ratio. Why Other Options Are Wrong: 30, 45, 65 and 70: These numbers do not follow the 2 : 3 : 4 : 5 ratio and the sum is 210, but middle terms break the proportional relationship. 35, 40, 60 and 75: The differences do not correspond to 2 : 3 : 4 : 5; the first part is too large for the given ratio. 30, 40, 60 and 80: The sum is 210, but the ratio of parts does not reduce to 2 : 3 : 4 : 5. Common Pitfalls:A common mistake is to directly divide 210 by 4 or to assume each ratio number directly equals the rupee amount. Another frequent error is to miscalculate the sum of the ratio terms, which then leads to incorrect unit value and wrong shares. Carefully adding the ratio and computing one unit correctly avoids these issues. Final Answer:The four amounts are 30, 45, 60 and 75 rupees respectively.

Question 20

The incomes of X, Y and Z are in the ratio 3 : 4 : 2 and their expenses are in the ratio 5 : 5 : 2. If X saves Rs. 3000 out of an income of Rs. 18,000, then what will be the saving of Y in rupees?

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Introduction / Context:This problem connects ratios of income and expenditure with the concept of savings. Such questions are popular in aptitude and banking exams because they test the ability to move between income, expense, and saving using proportional relationships. Here, incomes and expenses of three persons X, Y, and Z are given in two different ratios, and we are asked to find the saving of Y using the given saving of X as a reference point. Given Data / Assumptions: Income ratio of X : Y : Z is 3 : 4 : 2. Expense ratio of X : Y : Z is 5 : 5 : 2. Income of X is Rs. 18,000. Saving of X is Rs. 3000. Savings = Income - Expenses for each person. Concept / Approach:The plan is to use X as the reference. First, we know X’s actual income and saving, so we can compute X’s actual expenditure. From X’s expenditure and given expense ratio, we can find the common multiplier for expenses. That allows us to calculate Y’s actual expenditure. Then, using the income ratio and X’s actual income, we find Y’s income. Finally, we compute Y’s saving as income minus expenditure. This stepwise use of ratios is the core idea behind the solution. Step-by-Step Solution: Step 1: For X, Income = 18,000 and Saving = 3,000. So, Expense of X = 18,000 - 3,000 = 15,000. Step 2: Expense ratio is 5 : 5 : 2 for X, Y and Z. Let their expenses be 5m, 5m and 2m respectively. Since X’s expense 5m = 15,000, we get m = 3,000. Step 3: Expense of Y = 5m = 5 * 3,000 = 15,000. Step 4: Income ratio is 3 : 4 : 2. Let incomes be 3k, 4k and 2k respectively. Since X’s income 3k = 18,000, we get k = 6,000. So income of Y = 4k = 4 * 6,000 = 24,000. Step 5: Saving of Y = Income of Y - Expense of Y = 24,000 - 15,000 = 9,000. Verification / Alternative check:We can also compute Z to verify consistency. Z’s expense = 2m = 2 * 3,000 = 6,000. Z’s income = 2k = 2 * 6,000 = 12,000. Z’s saving = 12,000 - 6,000 = 6,000. All values are positive and align with the given ratios, so the model is coherent and the saving of Y as 9,000 is reliable. Why Other Options Are Wrong: 4500: This would require either income or expenses of Y to violate the given ratios. 6000: This would make Y’s saving equal to Z’s saving, which does not follow from the computed incomes. 7500: This value does not correspond to the precise ratio-based calculations of income and expenses. Common Pitfalls:Students often confuse the income ratio with the expense ratio or attempt to directly apply ratios to savings. Another mistake is to wrongly assume that savings are in a simple ratio, which is usually not true. The correct method is always: use ratios to find absolute income and expense, then compute savings as the difference. Carefully maintaining two separate ratios is crucial. Final Answer:The saving of Y is Rs. 9,000.

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