Question 1

In Maa Yatri Temple each orphan receives bananas, oranges and grapes in the ratio 3 : 2 : 7 in terms of dozens. The weight of one grape is 24 g, the weights of a banana and an orange are in the ratio 4 : 5, and the weight of one orange is 150 g. Wha…

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Introduction / Context: This problem combines ratio, unit conversion and basic arithmetic in a context of fruit distribution to orphans. Although the initial distribution is given in dozens, the question finally asks for a ratio in terms of total weight of each fruit type. Converting counts to weights using the given mass of each fruit and then simplifying the resulting ratio is the central task. Given Data / Assumptions: The number of dozens of bananas, oranges and grapes per orphan is 3 : 2 : 7. One dozen equals 12 individual fruits. The weight of one grape is 24 g. The weight of one banana and one orange are in the ratio 4 : 5. The weight of one orange is 150 g. We assume all fruits are of uniform size and weight. Concept / Approach: The basic approach is to convert the ratios in dozens into actual counts of fruits, then multiply by the weight of each single fruit to obtain total mass. Banana weight is found using the given ratio with the orange. Once total weights for bananas, oranges and grapes are computed in grams, we form a three term ratio and simplify it by dividing through by a common factor to obtain the simplest integer ratio. Step-by-Step Solution: Bananas: 3 dozens means 3 * 12 = 36 bananas. Oranges: 2 dozens means 2 * 12 = 24 oranges. Grapes: 7 dozens means 7 * 12 = 84 grapes. Weight of one orange is 150 g. Banana to orange weight ratio is 4 : 5, so banana weight = (4 / 5) * 150 = 120 g. Weight of one grape is 24 g. Total banana weight = 36 * 120 = 4320 g. Total orange weight = 24 * 150 = 3600 g. Total grape weight = 84 * 24 = 2016 g. Weight ratio = 4320 : 3600 : 2016. Divide all three numbers by 48 to simplify: 4320 / 48 = 90, 3600 / 48 = 75, 2016 / 48 = 42. Therefore, the simple ratio is 90 : 75 : 42. Verification / Alternative check: We can verify by trying another common factor, such as 24. Dividing 4320, 3600 and 2016 by 24 gives 180, 150 and 84. Dividing these again by 2 returns 90, 75 and 42, confirming that the simplified ratio is consistent. There is no larger common factor that can reduce 90, 75 and 42 further while keeping all terms integral, so this is the simplest ratio. Why Other Options Are Wrong: The option 180 : 150 : 82 retains a wrong grape weight and does not correspond to correct multiplication. The ratio 75 : 42 : 90 mixes the order of fruits. Options 45 : 30 : 21 and 60 : 50 : 28 are other arbitrary reductions or misinterpretations of dozens or weights and do not match the given numeric data when calculated carefully. Common Pitfalls: Common mistakes include forgetting that each dozen corresponds to 12 fruits, misapplying the ratio of weights between bananas and oranges, or mixing up which numbers correspond to bananas, oranges and grapes when forming the final ratio. Another frequent slip is failing to fully simplify the ratio to its lowest terms. Final Answer: The ratio of the total weight of bananas, oranges and grapes that each orphan receives is 90 : 75 : 42.

Question 2

Out of 135 applicants for a post, 60 are graduates and 80 have work experience. What is the ratio of the minimum possible number of graduate applicants with experience to the maximum possible number of graduate applicants with experience?

Accepted Answer

Introduction / Context: This is a classic set theory and counting problem based on overlapping groups. We are given the number of graduates, the number of applicants with experience and the total number of applicants. The question asks for the possible extremes of overlap between the two groups and then requests the ratio of minimum to maximum overlap. This is a common style of aptitude question about unions and intersections of sets. Given Data / Assumptions: Total number of applicants = 135. Number of graduates (set G) = 60. Number of applicants with experience (set E) = 80. We assume all applicants are counted once in the total. We must find the minimum and maximum possible number of applicants who are both graduate and experienced, that is the size of G intersection E. Concept / Approach: The maximum intersection occurs when one smaller group is entirely contained within the other group. The minimum intersection is found by spreading the groups apart as much as possible without exceeding the total number of applicants. The standard formula using inclusion exclusion is helpful: size of union = size of G + size of E minus size of intersection. By using the fact that the union cannot exceed the total of 135, we can derive the minimum intersection. The maximum intersection is simply the size of the smaller group. Step-by-Step Solution: Let x be the number of applicants who are both graduates and experienced. The union G ∪ E contains G + E − x applicants. So G + E − x ≤ total applicants. 60 + 80 − x ≤ 135. 140 − x ≤ 135, so x ≥ 5. Therefore, the minimum possible intersection size is 5. The maximum intersection occurs when all graduates are experienced, so maximum x = 60 (the smaller of 60 and 80). Thus the ratio of minimum to maximum intersection is 5 : 60. Simplify 5 : 60 by dividing both terms by 5 to get 1 : 12. Verification / Alternative check: We can check the extreme cases. For the minimum overlap of 5, the union has 60 + 80 − 5 = 135 applicants, exactly matching the total. For the maximal overlap of 60, the union has 60 + 80 − 60 = 80 applicants, which is allowed because the union can be less than or equal to the total. This confirms that 5 and 60 are valid values and that the ratio 1 : 12 is consistent. Why Other Options Are Wrong: A ratio such as 5 : 1 or 1 : 5 reverses or miscalculates the comparison. The option 12 : 1 swaps the order of minimum and maximum. Ratios like 2 : 5 do not arise from any reasonable interpretation of the intersection values. Only 1 : 12 correctly reflects minimum 5 and maximum 60. Common Pitfalls: A common error is to think the minimum intersection is 0, ignoring the total limit on union size. Another mistake is to assume maximum intersection is the larger of the two group sizes rather than the smaller. Remembering the inclusion exclusion relationship and that the union cannot exceed the total avoids these mistakes. Final Answer: The required ratio of minimum to maximum number of graduates with experience is 1 : 12.

Question 3

The salaries of Karthik, Arun and Akhil are in the ratio 4 : 5 : 6 respectively. Their salaries are then increased by 50 percent, 60 percent and 50 percent respectively. What will be the new ratio of the salaries of Akhil, Arun and Karthik (in that …

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Introduction / Context: This question tests ratio manipulation and percentage increase applied to each term of a ratio. The initial ratio of salaries is given, then different percentage increases are applied to each person. Finally, the ratio is asked in a different order than originally stated, so careful reading is important. Such problems model real life scenarios where different employees receive different raise percentages. Given Data / Assumptions: Initial salary ratio: Karthik : Arun : Akhil = 4 : 5 : 6. Karthik gets a 50 percent increase. Arun gets a 60 percent increase. Akhil gets a 50 percent increase. We must find the new ratio in the order Akhil : Arun : Karthik. Concept / Approach: To handle this, we convert each original salary ratio term into an algebraic value, apply the relevant percentage increase, and obtain new terms. Since ratios can be scaled freely, we do not need actual currency amounts. After calculating the increased values, we arrange them according to the desired order and simplify the ratio if necessary. Percentage increase by p percent is implemented by multiplying the original quantity by (1 + p / 100). Step-by-Step Solution: Let the initial salaries be 4x, 5x and 6x for Karthik, Arun and Akhil respectively. Karthik gets 50 percent extra, so new salary of Karthik = 4x * 1.5 = 6x. Arun gets 60 percent extra, so new salary of Arun = 5x * 1.6 = 8x. Akhil gets 50 percent extra, so new salary of Akhil = 6x * 1.5 = 9x. So new salaries in the order Karthik : Arun : Akhil are 6x : 8x : 9x. We need the ratio Akhil : Arun : Karthik, which is 9x : 8x : 6x. Cancel x to obtain 9 : 8 : 6. Verification / Alternative check: We can pick a simple value like x = 1000. Then initial salaries are 4000, 5000 and 6000. After the increases: Karthik gets 6000, Arun gets 8000 and Akhil gets 9000. Writing them in the requested order Akhil : Arun : Karthik gives 9000 : 8000 : 6000. Dividing through by 1000 yields 9 : 8 : 6, confirming the ratio derived algebraically. Why Other Options Are Wrong: Option 6 : 8 : 9 is the ratio Karthik : Arun : Akhil, not the requested Akhil : Arun : Karthik. Options 8 : 6 : 9, 6 : 5 : 4 and 9 : 6 : 8 do not match any proper reordering of the computed new salaries. They either confuse the order or misapply the percentage increases. Common Pitfalls: Students often forget that the question changes the order of names in the final ratio. Another error is to add the same percentage to every term or to treat the ratio numbers directly as percentages. The correct method is to multiply each term by its respective multiplying factor and only then form the ratio. Final Answer: The new ratio of the salaries of Akhil, Arun and Karthik is 9 : 8 : 6.

Question 4

In an examination, a candidate secured 504 marks out of a maximum mark M. If the maximum mark M is scaled so that it becomes 800, the candidate would have secured 384 marks. What is the value of M?

Accepted Answer

Introduction / Context: This question examines proportional reasoning and scaling of marks. A candidate's performance is expressed as a fraction of the maximum marks. When the maximum possible marks change due to scaling, the candidate's marks change in the same proportion. By equating the performance fractions before and after scaling, we can solve for the original maximum mark M. The twist is that the correct numerical value does not appear directly among the numeric options, so careful interpretation of the option “None of the above” is required. Given Data / Assumptions: Original score of the candidate = 504 marks. Original maximum marks = M. After scaling, maximum marks become 800. Scaled score of the candidate would then be 384 marks. The candidate's performance fraction remains the same under scaling. Concept / Approach: The fraction of marks obtained is score divided by maximum marks. Initially, this fraction is 504 / M. After scaling, the fraction is 384 / 800. Since both fractions represent the same performance, we set them equal and solve for M. Once the value of M is found, we compare it with the listed numeric options. If it does not match, then the correct choice is “None of the above”. Step-by-Step Solution: Performance before scaling: 504 / M. Performance after scaling: 384 / 800. Equate the fractions: 504 / M = 384 / 800. Cross multiply: 504 * 800 = 384 * M. Compute left side: 504 * 800 = 403200. So 403200 = 384 * M. Therefore M = 403200 / 384. Simplify the division to obtain M = 1050. Verification / Alternative check: Check the fractions with M = 1050. The original performance is 504 / 1050, which simplifies to 0.48. The scaled performance is 384 / 800, which also equals 0.48. Since these are equal, M = 1050 is consistent. Now, comparing with the numeric options 910, 1000 and 970, none of them equals 1050, so the correct option is that none of the listed numeric values matches the true maximum mark. Why Other Options Are Wrong: Choosing 910, 1000 or 970 would produce a performance fraction different from 384 / 800. If M were 1000, for example, 504 / 1000 would be 0.504, which is not equal to 0.48. Similar discrepancies appear for 910 and 970. Only M = 1050 yields the correct equality between the two performance fractions. Common Pitfalls: A typical mistake is to assume M is 800 or to misinterpret scaling as simple addition instead of proportional change. Another error is to stop after computing M and pick the closest numeric option instead of recognizing that the correct value is absent and that the “None of the above” choice must be used. Always verify proportional relationships carefully when options include such a choice. Final Answer: The original maximum mark M is 1050, so the correct choice is None of the above.

Question 5

A mixture contains alcohol and water in the ratio 4 : 3. After 5 litres of water are added to the mixture, the ratio becomes 4 : 5. What is the quantity of alcohol in the original mixture?

Accepted Answer

Introduction / Context: This problem is about mixture and ratio. We are told the initial ratio of alcohol to water in a mixture and how the ratio changes after adding a known amount of water. Questions of this type are standard in aptitude tests and require forming algebraic equations from ratios, then solving for the unknown quantity, here the volume of alcohol in litres. Given Data / Assumptions: Initial ratio of alcohol : water = 4 : 3. A certain base quantity corresponds to these ratio parts. 5 litres of water are added to the mixture. After addition, the ratio becomes alcohol : water = 4 : 5. No alcohol is added or removed, so the quantity of alcohol stays the same. We need the volume of alcohol in litres in the original mixture. Concept / Approach: When a mixture is described by a ratio, we can express each component as a multiple of a common factor. Let alcohol be 4x litres and water be 3x litres initially. Adding 5 litres of water changes the water quantity but not the alcohol quantity. The new ratio equation then connects 4x and 3x + 5 to 4 : 5. Solving this equation gives x, and hence the amount of alcohol. This method is widely used for ratio based mixture problems. Step-by-Step Solution: Let initial alcohol = 4x litres. Let initial water = 3x litres. Total initial mixture = 7x litres. After adding 5 litres of water, water becomes 3x + 5 litres. Alcohol remains 4x litres. New ratio alcohol : water = 4 : 5. So 4x / (3x + 5) = 4 / 5. Cross multiply to get 5 * 4x = 4 * (3x + 5). This simplifies to 20x = 12x + 20. Then 20x − 12x = 20, so 8x = 20, giving x = 2.5. Alcohol quantity = 4x = 4 * 2.5 = 10 litres. Verification / Alternative check: Substitute back into the scenario. If x = 2.5, then initial alcohol is 10 litres and initial water is 7.5 litres. After adding 5 litres of water, water becomes 12.5 litres. The new ratio is 10 : 12.5. Dividing both terms by 2.5 gives 4 : 5, which matches the stated new ratio. This confirms that the calculation is correct and that 10 litres of alcohol is consistent with all the given information. Why Other Options Are Wrong: Values such as 8 or 12 litres come from incorrect handling of the ratio equation or arithmetic mistakes while simplifying. The larger numbers 18 and 22 litres do not satisfy the new ratio when 5 litres of water are added. Substituting any of these wrong values into the ratio will not produce 4 : 5. Common Pitfalls: A frequent mistake is to change both alcohol and water when the problem states that only water is added. Another is misinterpreting the ratio 4 : 5 as meaning a fixed difference rather than a proportional relationship. Carefully setting algebraic expressions for the initial and final situations helps avoid these errors. Final Answer: The quantity of alcohol in the original mixture is 10 litres.

Question 6

A, B and C have amounts of money in the ratio 3 : 4 : 5. First B gives one fourth of his money to A and one fourth to C. Then C gives one sixth of his resulting amount to A. What is the final ratio of the amounts of A, B and C respectively?

Accepted Answer

Introduction / Context: This question involves successive transfers of money between three people whose initial holdings are in a given ratio. It tests the ability to handle ratios and fractional transfers carefully. After two stages of giving and receiving, we need to determine the final ratio of their amounts. Such problems appear frequently in aptitude tests to examine systematic reasoning with ratios. Given Data / Assumptions: Initial ratio of A : B : C = 3 : 4 : 5. B gives one fourth of his amount to A. B then gives one fourth of his original amount to C as well (so he gives a total of half his starting amount). After this, C gives one sixth of his new amount to A. No money enters or leaves the group from outside. We must find the final ratio A : B : C. Concept / Approach: We can assume a common multiplier for their starting amounts, convert the ratio into algebraic values and then apply the described transfers step by step. Because only proportions matter, choosing a convenient multiplier simplifies arithmetic. After both transactions are applied, we express the final amounts as a ratio and simplify it to its lowest terms. This approach keeps the reasoning clear and avoids confusion caused by fractions. Step-by-Step Solution: Let the initial amounts be A = 3k, B = 4k and C = 5k. First, B gives one fourth of his amount to A: one fourth of 4k is k. So A becomes 3k + k = 4k. B loses k and becomes 4k − k = 3k at this moment. Next, B gives one fourth of his original amount (that is again k) to C. So B finally becomes 3k − k = 2k. C receives k and becomes 5k + k = 6k. Now C gives one sixth of his amount to A: one sixth of 6k is k. So C becomes 6k − k = 5k. A receives k and becomes 4k + k = 5k. Final amounts are A = 5k, B = 2k and C = 5k. Thus the final ratio A : B : C = 5 : 2 : 5. Verification / Alternative check: We can test with a concrete value, for example k = 100. Then initial amounts are 300, 400 and 500. B gives 100 to A and 100 to C, so intermediate amounts are A = 400, B = 200, C = 600. Now C gives one sixth of 600, which is 100, to A. After this, A = 500, B = 200 and C = 500. The ratio 500 : 200 : 500 simplifies exactly to 5 : 2 : 5, confirming the algebraic solution. Why Other Options Are Wrong: Ratios like 4 : 3 : 5 or 3 : 4 : 5 mistakenly reflect initial or intermediate states instead of the final distribution. The ratio 5 : 4 : 3 misorders the values, and 6 : 3 : 5 does not correspond to any step in the process. Only 5 : 2 : 5 matches the final amounts after all transfers are complete. Common Pitfalls: A frequent source of error is using B's reduced amount, rather than his original amount, to compute the second transfer. The problem states that B gives one fourth to A and one fourth to C, which both refer to the original 4k. Another common mistake is to forget that C's one sixth transfer is based on C's updated amount, not the initial 5k. Final Answer: The final ratio of the amounts of A, B and C is 5 : 2 : 5.

Question 7

At a casino in Mumbai, there are three tables A, B and C. The payoffs at table A, B and C are 10 : 1, 20 : 1 and 30 : 1 respectively. A man bets Rs 200 at each table and wins at exactly two of the three tables. What is the maximum possible differenc…

Accepted Answer

Introduction / Context: This question tests understanding of odds and payoffs in a gambling context, combined with basic profit comparison. A man places equal bets on three different tables with different payoff ratios and wins on any two of them. We must analyse the possible total earnings across the different winning combinations and compute the maximum difference between them. It is a nice applied ratio and multiplication problem. Given Data / Assumptions: Three tables A, B and C with payoffs 10 : 1, 20 : 1 and 30 : 1 respectively. The man bets Rs 200 on each table, so he bets Rs 600 in total. Winning at a table with odds p : 1 gives profit of p times the stake and also returns the stake to the player. He wins at exactly two tables and loses at the remaining one. We must find the maximum possible difference between his total earnings (net profit) among all such win combinations. Concept / Approach: First, compute profit from a single winning bet at each table. Then, for each possible pair of winning tables (A and B, A and C, B and C), sum the profits from the wins and subtract the loss of the stake at the losing table. This gives three possible total profits. The maximum difference in earnings is the difference between the highest and the lowest of these profits. The stake of Rs 200 is common, so the calculation remains straightforward. Step-by-Step Solution: Stake at each table = Rs 200. At table A, odds 10 : 1 mean profit = 10 * 200 = Rs 2000. At table B, odds 20 : 1 mean profit = 20 * 200 = Rs 4000. At table C, odds 30 : 1 mean profit = 30 * 200 = Rs 6000. If he wins at A and B and loses at C, total profit = 2000 + 4000 − 200 = Rs 5800. If he wins at A and C and loses at B, total profit = 2000 + 6000 − 200 = Rs 7800. If he wins at B and C and loses at A, total profit = 4000 + 6000 − 200 = Rs 9800. The smallest profit is Rs 5800 and the largest profit is Rs 9800. Maximum difference in earnings = 9800 − 5800 = Rs 4000. Verification / Alternative check: We can also focus only on the two winning tables since in each scenario exactly one Rs 200 loss occurs. For winnings from tables A and B, combined profit is 2000 + 4000 = 6000. For A and C, it is 2000 + 6000 = 8000. For B and C, it is 4000 + 6000 = 10000. Subtracting a fixed 200 loss from each gives 5800, 7800 and 9800, which differ pairwise by 2000 and 4000. Hence, the maximum gap between any two outcomes remains Rs 4000. Why Other Options Are Wrong: Values like Rs 2500 or Rs 2000 arise if one mixes up higher and lower profits or forgets to consider all three cases. Rs 1500 and Rs 3000 are other arbitrary differences that do not match the computed gap between 5800 and 9800. Only Rs 4000 is consistent with the full analysis of all possibilities. Common Pitfalls: Some learners confuse odds expressed as 10 : 1 with a simple multiplier of 11. Here we are interested in profit, so we take the 10 times payment, not adding the stake when computing profit. Others forget to subtract the lost stake at the losing table or consider only one winning combination. Systematically listing all win pairs and computing each outcome prevents such mistakes. Final Answer: The maximum possible difference between his earnings across the different winning combinations is Rs 4000.

Question 8

Three simple ratios are given: 5 : 6, 7 : 10 and 6 : 5. What is the equivalent compound ratio formed by combining these three ratios in order?

Accepted Answer

Introduction / Context: This question is about compound ratios. When several simple ratios are given, we can combine them to form an equivalent single ratio by multiplying corresponding terms. This concept is useful in advanced ratio and proportion problems where multiple relations need to be summarised into one collective ratio. Given Data / Assumptions: First ratio: 5 : 6. Second ratio: 7 : 10. Third ratio: 6 : 5. We are asked for the equivalent compound ratio of these three ratios. The compound ratio is formed by multiplying all antecedents together and all consequents together. Concept / Approach: For ratios a : b, c : d and e : f, the compound ratio is (a * c * e) : (b * d * f). This simply extends the idea that ratios compare multiplicative factors. After multiplying, we may simplify the resulting ratio by dividing both sides by their greatest common divisor to obtain a reduced form. Step-by-Step Solution: First ratio = 5 : 6. Second ratio = 7 : 10. Third ratio = 6 : 5. Multiply all antecedents: 5 * 7 * 6 = 210. Multiply all consequents: 6 * 10 * 5 = 300. So the compound ratio is 210 : 300. Find a common factor of 210 and 300, which is 30. Divide: 210 / 30 = 7 and 300 / 30 = 10. Therefore, the simplified compound ratio is 7 : 10. Verification / Alternative check: We can check the simplification by observing that 210 and 300 both end in zero, so dividing by 10 yields 21 : 30. Then noticing that both are multiples of 3, dividing by 3 gives 7 : 10 directly. This is the same result obtained by dividing by 30 in one step, confirming that 7 : 10 is fully simplified. Why Other Options Are Wrong: Ratios such as 7 : 9 and 9 : 7 involve incorrect products or incomplete simplification. The option 7 : 4 reverses and miscalculates the effect of multiplying the given ratios. Option 10 : 7 is simply the inverse of the correct ratio and does not represent the compound ratio in the given order. Common Pitfalls: One common mistake is to add the ratios term by term rather than multiplying them, which is not how compound ratios work. Others may incorrectly match antecedent with consequent when multiplying. Remember that we always multiply first terms together and second terms together to retain the correct structure of the ratio. Final Answer: The equivalent compound ratio of 5 : 6, 7 : 10 and 6 : 5 is 7 : 10.

Question 9

The marks obtained by Vijay and Amith are in the ratio 4 : 5, and the marks obtained by Amith and Abhishek are in the ratio 3 : 2. What is the ratio of the marks obtained by Vijay and Abhishek?

Accepted Answer

Introduction / Context: This question requires combining two ratios that share a common term in order to deduce the ratio between two other persons. We are given the ratio of marks of Vijay and Amith and the ratio of marks of Amith and Abhishek, and we must find the ratio of Vijay to Abhishek. This is a standard technique in ratio and proportion problems and is very useful in many competitive examinations. Given Data / Assumptions: Vijay : Amith = 4 : 5. Amith : Abhishek = 3 : 2. All marks are positive numbers. We are required to compute Vijay : Abhishek. Concept / Approach: The method is to express all marks in terms of a single variable by using the common person between the given ratios, which is Amith. From Vijay : Amith = 4 : 5, we can write Vijay and Amith in terms of one variable. From Amith : Abhishek = 3 : 2, we can write Amith and Abhishek in terms of another variable. Equating Amith's values from both expressions allows us to find a consistent set of values and then form the desired ratio Vijay : Abhishek. Step-by-Step Solution: Let Vijay's marks be 4x and Amith's marks be 5x. Also, from the second ratio, let Amith's marks be 3y and Abhishek's marks be 2y. Since Amith has the same marks in both descriptions, set 5x = 3y. This gives y = (5x) / 3. Now Abhishek's marks are 2y = 2 * (5x / 3) = 10x / 3. Vijay's marks are 4x. Therefore Vijay : Abhishek = 4x : (10x / 3). Multiply both terms by 3 to clear the denominator: 12x : 10x. Cancel x to get 12 : 10, which simplifies to 6 : 5. Verification / Alternative check: We may choose numerical values that satisfy both initial ratios. For example, let Amith have 15 marks. Then Vijay must have (4 / 5) * 15 = 12 marks. From Amith : Abhishek = 3 : 2 with Amith = 15, Abhishek has (2 / 3) * 15 = 10 marks. So Vijay : Abhishek = 12 : 10, which simplifies to 6 : 5. This matches the algebraic result. Why Other Options Are Wrong: The option 4 : 5 repeats the Vijay : Amith ratio, which is not requested. The ratio 3 : 2 is the Amith : Abhishek ratio, not Vijay : Abhishek. The ratio 1 : 3 is unrelated to the given proportions, and 5 : 6 is simply the inverse of the correct ratio. Only 6 : 5 correctly represents Vijay : Abhishek. Common Pitfalls: A common error is to multiply or divide the ratios incorrectly, for example directly combining 4 : 5 and 3 : 2 to get 12 : 10 without making sure which pair of names is being compared. Always use the common term (here Amith) to align the ratios properly before forming the new ratio. Final Answer: The ratio of the marks obtained by Vijay and Abhishek is 6 : 5.

Question 10

If k : l = 4 : 3 and l : m = 5 : 3, what is the combined ratio k : l : m?

Accepted Answer

Introduction / Context: This problem is a straightforward ratio chaining question. We are given relationships between k and l, and between l and m. The goal is to connect these pieces of information and determine a single three term ratio involving k, l and m. Problems like this highlight how ratios can be combined when one quantity is common in both ratios. Given Data / Assumptions: k : l = 4 : 3. l : m = 5 : 3. We must find the triple ratio k : l : m. All quantities are positive and proportional as stated. Concept / Approach: The key is to make the intermediate term l consistent in both ratios so that they can be combined. We express k and l in terms of one variable from the first ratio, and l and m in terms of another variable from the second ratio. Then we equate the two expressions for l, find the relationship between the variables, and compute k, l and m in comparable units. Finally, we write k : l : m and simplify the numbers if needed. Step-by-Step Solution: From k : l = 4 : 3, let k = 4x and l = 3x. From l : m = 5 : 3, let l = 5y and m = 3y. Equate l from both descriptions: 3x = 5y. So y = 3x / 5. Now m = 3y = 3 * (3x / 5) = 9x / 5. To eliminate denominators, multiply all terms by 5. k becomes 4x * 5 = 20x. l becomes 3x * 5 = 15x. m becomes (9x / 5) * 5 = 9x. Thus k : l : m = 20x : 15x : 9x = 20 : 15 : 9. Verification / Alternative check: We can check by reconstructing the original pairwise ratios from 20 : 15 : 9. For k : l, we obtain 20 : 15, which reduces to 4 : 3 as required. For l : m, we obtain 15 : 9, which reduces to 5 : 3. Since both original ratios are satisfied, 20 : 15 : 9 is consistent and correct. Why Other Options Are Wrong: Ratios such as 18 : 24 : 11 or 9 : 15 : 1 do not reduce to the given pairwise ratios when tested. The ratio 21 : 7 : 3 yields k : l = 3 : 1, which does not match 4 : 3. Similarly, 12 : 9 : 5 is inconsistent with l : m = 5 : 3. Only 20 : 15 : 9 satisfies both given conditions simultaneously. Common Pitfalls: A common mistake is to simply place the numbers side by side as 4 : 3 : 5 : 3 without properly aligning the common term. Others may try to average or mix numbers incorrectly. Always treat the shared variable carefully and ensure that the value of l is the same in both representations before combining ratios. Final Answer: The combined ratio of k : l : m is 20 : 15 : 9.

Question 11

An amount of Rs 782 is divided into three parts in the proportion 1/2 : 2/3 : 3/4. What is the value of the first part?

Accepted Answer

Introduction / Context: This question involves dividing a sum of money into three parts according to a ratio given in fractional form. Instead of simple integers, the ratio uses fractions such as 1/2, 2/3 and 3/4. The main idea is to convert these fractional parts into an equivalent integer ratio and then use standard proportional division to allocate the total amount. This is a useful skill for handling more complex ratio scenarios. Given Data / Assumptions: Total amount to be divided = Rs 782. The three parts are proportional to 1/2, 2/3 and 3/4. We must find the first part corresponding to 1/2. The entire amount is exactly divisible according to this ratio. Concept / Approach: To simplify calculations, we first convert 1/2, 2/3 and 3/4 into a common integer ratio by finding a common denominator. Multiplying each fraction by the least common multiple of the denominators yields equivalent whole numbers. Once we have an integer ratio, we sum the ratio parts to find the total number of shares. Each share corresponds to a fixed amount of money. Multiplying the value per share by the number of shares in the first part gives the required answer. Step-by-Step Solution: Fractions: 1/2, 2/3 and 3/4. Least common multiple of denominators 2, 3 and 4 is 12. Multiply each fraction by 12: 1/2 * 12 = 6, 2/3 * 12 = 8, 3/4 * 12 = 9. So the integer ratio is 6 : 8 : 9. Sum of ratio parts = 6 + 8 + 9 = 23 parts. Total amount Rs 782 corresponds to 23 parts. Value of one part = 782 / 23 = 34. The first part corresponds to 6 parts. First part = 6 * 34 = Rs 204. Verification / Alternative check: We can compute the other two parts to confirm consistency. The second part is 8 * 34 = Rs 272, and the third part is 9 * 34 = Rs 306. Adding them gives 204 + 272 + 306 = Rs 782, which matches the total amount. Since the distribution is exact and consistent, Rs 204 for the first part is correct. Why Other Options Are Wrong: Rs 408 is double the correct first part and corresponds to misreading the ratio or doubling 6 parts incorrectly. Rs 412 and Rs 220 do not respect the integer ratio when the remaining parts are computed. Rs 272 is the value of the second part, not the first. Only Rs 204 fits the correct proportional division. Common Pitfalls: Some learners attempt to use the fractions directly without converting to an integer ratio, which can lead to complicated arithmetic. Others may incorrectly use the numerators 1, 2 and 3 alone, ignoring denominators. Always convert fractional ratios using a common denominator so that the parts are truly proportional. Final Answer: The first part of the division is Rs 204.

Question 12

A sum of money is divided among K, L and M in such a way that for each rupee K receives, L receives 45 paise and M receives 30 paise. If the share of L is Rs 27, what is the total amount of money?

Accepted Answer

Introduction / Context: This question is about sharing a sum of money among three people according to a ratio expressed in rupees and paise. The distribution rule is that for every rupee K receives, L and M receive specified amounts in paise. We are also told the actual share of L and must deduce the total amount distributed. It is a practical ratio and proportion problem with unit conversion between rupees and paise. Given Data / Assumptions: For each rupee K gets, L gets 45 paise and M gets 30 paise. 1 rupee = 100 paise. L's actual share is Rs 27. We need to find the total of all three shares together. Concept / Approach: We first convert all amounts to a common unit, typically paise, to avoid confusion. If we treat K's unit share as 100 paise, then L's share is 45 paise and M's share is 30 paise for the same underlying unit. This gives a ratio between K, L and M. Once we express L's actual amount in terms of this ratio, we can find the scaling factor and therefore compute the total amount corresponding to the entire ratio. Step-by-Step Solution: For each rupee K gets, K's share in paise is 100. L's share for the same unit is 45 paise. M's share for the same unit is 30 paise. Thus K : L : M = 100 : 45 : 30. Divide by 5 to simplify: 20 : 9 : 6. So the ratio is K : L : M = 20 : 9 : 6. L's share corresponds to 9 parts. We are told L receives Rs 27. One part is therefore 27 / 9 = Rs 3. Total parts = 20 + 9 + 6 = 35. Total amount = 35 * 3 = Rs 105. Verification / Alternative check: We can compute each person's share explicitly. K gets 20 * 3 = Rs 60. L gets 9 * 3 = Rs 27, which matches the question. M gets 6 * 3 = Rs 18. Adding them gives 60 + 27 + 18 = Rs 105. This confirms that the total amount calculated is consistent with both the ratio and L's individual share. Why Other Options Are Wrong: Totals such as Rs 96, Rs 147 or Rs 159 do not yield L's share as exactly Rs 27 when the ratio 20 : 9 : 6 is applied. Rs 135 is another distracting value that does not produce integer shares respecting the given distribution rule. Only Rs 105 aligns with all the data and the established ratio. Common Pitfalls: One common mistake is to use the numbers 1, 0.45 and 0.30 directly as the ratio without converting to a uniform unit such as paise, which can lead to confusion. Another is to forget to simplify the ratio before scaling up. Working in paise and then converting back to rupees at the end helps maintain clarity. Final Answer: The total amount of money divided among K, L and M is Rs 105.

Question 13

Pooja, Shipra and Mounika are three sisters, and Pooja and Shipra are twins of equal age. The sum of the present ages of Pooja and Shipra is equal to the present age of Mounika. Three years earlier, the ratio of the age of Pooja to the age of Mounik…

Accepted Answer

Introduction / Context: This is a classic age problem involving ratios and simple linear equations. We are told that Pooja and Shipra are twins with equal ages, that their combined age equals Mounika's present age, and that three years earlier there was a given ratio between Pooja's and Mounika's ages. Using this information, we need to determine Shipra's age three years in the future. Given Data / Assumptions: Pooja and Shipra are twins, so their present ages are equal. The sum of the present ages of Pooja and Shipra equals Mounika's present age. Three years earlier, Pooja's age : Mounika's age = 2 : 7. We must find Shipra's age three years hence. All ages are measured in whole years. Concept / Approach: Let the present age of each twin be x years. Then Mounika's present age is 2x years, since it equals the sum of the twins' ages. We then use the ratio condition from three years earlier, when Pooja's age was x − 3 and Mounika's age was 2x − 3. Setting their ratio to 2 : 7 gives an equation in x. Solving for x gives the present age of each twin, from which we can add three years to find Shipra's age three years hence. Step-by-Step Solution: Let present age of Pooja = x years. Since Pooja and Shipra are twins, present age of Shipra = x years. Sum of ages of Pooja and Shipra = x + x = 2x years. Given that Mounika's present age = 2x years. Three years earlier, Pooja's age = x − 3 years. Three years earlier, Mounika's age = 2x − 3 years. The ratio (x − 3) : (2x − 3) = 2 : 7. Form the equation (x − 3) / (2x − 3) = 2 / 7. Cross multiply: 7(x − 3) = 2(2x − 3). This gives 7x − 21 = 4x − 6. So 7x − 4x = −6 + 21, which is 3x = 15. Hence x = 5. So present age of Shipra is 5 years. Shipra's age three years hence will be 5 + 3 = 8 years. Verification / Alternative check: Using x = 5, the present ages are: Pooja = 5, Shipra = 5 and Mounika = 10. Three years earlier, their ages were: Pooja = 2, Mounika = 7. The ratio 2 : 7 matches the given condition. Also, the sum of present ages of Pooja and Shipra is 5 + 5 = 10, which equals Mounika's present age. This confirms that the assumptions and calculations are consistent. Why Other Options Are Wrong: Values such as 12, 16, 18 or 21 years do not align with the derived present age of 5 years for Shipra. If Shipra were to be any of those ages three years hence, her current age would be inconsistent with both the ratio condition and the sum condition involving Mounika. Only 8 years fits both constraints. Common Pitfalls: A common error is to misinterpret the statement about the sum of the ages of Pooja and Shipra and to treat it as a ratio instead of an equality. Another mistake is to apply the ratio at the wrong time, using present ages instead of ages three years earlier. Always pay close attention to time shifts in age problems. Final Answer: Shipra's age three years hence will be 8 years.

Question 14

The total number of students in three classes of a school is 333. The numbers of students in class 1 and class 2 are in the ratio 3 : 5, and the numbers of students in class 2 and class 3 are in the ratio 7 : 11. What is the strength of the class th…

Accepted Answer

Introduction / Context: This question is about combining two ratios involving three related quantities and then using the total sum to find individual values. Here, the three quantities are the numbers of students in three different classes. We know the total number of students and two pairwise ratios. The goal is to determine which class has the highest strength and what that number is. Given Data / Assumptions: Total students in three classes = 333. Let the numbers in classes 1, 2 and 3 be C1, C2 and C3. C1 : C2 = 3 : 5. C2 : C3 = 7 : 11. We must find the largest of C1, C2 and C3. Concept / Approach: The idea is to express all three class strengths using a single variable. From the first ratio, express C1 and C2 in terms of one parameter. From the second ratio, express C2 and C3 in terms of another parameter. Equating the two expressions for C2 links the parameters and allows us to write C1, C2 and C3 in terms of a single variable. With these expressions, we use the total of 333 to find the variable, then compute the exact strengths and identify the maximum. Step-by-Step Solution: From C1 : C2 = 3 : 5, let C1 = 3k and C2 = 5k. From C2 : C3 = 7 : 11, let C2 = 7m and C3 = 11m. Equate C2 values: 5k = 7m. So m = (5k) / 7. Then C3 = 11m = 11 * (5k / 7) = 55k / 7. Total students C1 + C2 + C3 = 333. So 3k + 5k + 55k / 7 = 333. Combine terms: (3k + 5k) = 8k, so we have 8k + 55k / 7. Express with denominator 7: (56k + 55k) / 7 = 333. So 111k / 7 = 333. Multiply both sides by 7: 111k = 2331. Thus k = 2331 / 111 = 21. Now C1 = 3k = 63, C2 = 5k = 105, C3 = 55k / 7 = 55 * 3 = 165. The highest strength is C3 = 165. Verification / Alternative check: Add the class sizes 63, 105 and 165 to confirm the total: 63 + 105 = 168, and 168 + 165 = 333, which matches the given total. The ratios C1 : C2 = 63 : 105 simplify to 3 : 5, and C2 : C3 = 105 : 165 simplify to 7 : 11. Since both pairwise ratios and the total are satisfied, the values are correct. Why Other Options Are Wrong: Numbers such as 125, 135 or 155 do not match the computed strengths of any class. They arise if the algebra is done incorrectly or if a wrong class is chosen as the largest. The value 105 is the size of class 2, not the maximum strength. Only 165 correctly corresponds to the largest class size. Common Pitfalls: Students may attempt to combine ratios by guessing or by simply adding them without aligning the common term C2. Others may forget to express all classes in terms of a single variable before applying the total. Carefully equating C2 and then using the total prevents such errors. Final Answer: The strength of the class with the highest number of students is 165.

Question 15

At a game of billiards, A can give B 20 points in a game of 60, and A can give C 15 points in a game of 60. How many points can C give B in a game of 90?

Accepted Answer

Introduction / Context: This billiards problem is a standard relative speed and advantage question disguised in scoring terms. A can give B a certain number of points in a fixed game and also give C a different number of points in the same sized game. From these advantages, we infer the relative strengths of A, B and C. Then, for a game of a different size, we compute how many points C can give B. Given Data / Assumptions: A can give B 20 points in 60, meaning when A scores 60, B scores 40. A can give C 15 points in 60, meaning when A scores 60, C scores 45. Scores are proportional to players' skill levels or scoring rates. We must find how many points C can give B in a game up to 90 points. Concept / Approach: We treat each player's scoring ability as a rate and use ratios. From the statements, we deduce A : B and A : C. Then we derive B : C and use it to determine the relative scores when C reaches 90 points. The difference between C's and B's scores at that time is the number of points C can give B in a 90 point game. Step-by-Step Solution: A gives B 20 points in 60, so when A scores 60, B scores 40. Thus A : B = 60 : 40 = 3 : 2. A gives C 15 points in 60, so when A scores 60, C scores 45. Thus A : C = 60 : 45 = 4 : 3. From A : B = 3 : 2, take A = 3r and B = 2r. From A : C = 4 : 3, take A = 4s and C = 3s. Equate A: 3r = 4s, so r = 4s / 3. Then B = 2r = 2 * (4s / 3) = 8s / 3. C = 3s. Thus B : C = (8s / 3) : 3s. Multiply both sides by 3 to clear denominator: 8s : 9s = 8 : 9. So B : C = 8 : 9. In a game of 90 points, if C reaches 90, B's score is (8 / 9) * 90 = 80. Therefore, C can give B 90 − 80 = 10 points in a game of 90. Verification / Alternative check: We can also choose convenient scores. Suppose the skill units are set so that C scores 9 units while B scores 8 units in the same time. If C aims for 90 points, B will reach 80 points under the same conditions, confirming the 10 point difference. This is consistent with the derived ratio B : C = 8 : 9 and supports the conclusion. Why Other Options Are Wrong: Differences of 9, 11, 8 or 12 points do not match the 8 : 9 ratio when scaled to a 90 point target for C. For example, if the difference were 9 points, B would have to score 81 while C scores 90, which does not maintain the B : C ratio of 8 : 9. Only a 10 point difference preserves the correct ratio at that game length. Common Pitfalls: A common mistake is to average the advantages or directly subtract 20 and 15 without converting them into ratios. Another is to forget that the same time frame or scoring opportunity is implied when comparing players. Always convert each advantage statement into a ratio, find relative strengths and then apply them to the new game length. Final Answer: In a game of 90 points, C can give B 10 points.

Question 16

At Stewar farm, the ratio between the number of sheep and the number of horses is 4 : 15. Each horse is fed 110 ounces of horse food per day, and the farm needs a total of 13,200 ounces of horse food per day. What is the number of sheep on the farm?

Accepted Answer

Introduction / Context: This question combines a ratio of animals with a daily feed requirement for one type of animal. The feed is given only to horses, whose number we infer from the total feed needed. Once we know the number of horses, we can use the sheep to horses ratio to determine how many sheep are present. This is a standard type of ratio and unit multiplication question. Given Data / Assumptions: Ratio of sheep to horses = 4 : 15. Each horse eats 110 ounces of horse food per day. Total horse food required per day = 13,200 ounces. Sheep are not consuming this particular horse food. We need to find the number of sheep. Concept / Approach: First, compute how many horses are present by dividing the total daily feed by the feed per horse. This gives the count of horses. Then, using the ratio 4 : 15, we interpret 15 ratio units as the number of horses and 4 ratio units as the number of sheep. By finding the value of one ratio unit, we multiply it by 4 to get the number of sheep. This method keeps the reasoning clearly separated into counting horses and then applying the ratio. Step-by-Step Solution: Total daily horse food = 13,200 ounces. Food per horse per day = 110 ounces. Number of horses = 13,200 / 110. Compute: 13,200 / 110 = 120 horses. Sheep : horses = 4 : 15. So 15 ratio parts correspond to 120 horses. Value of one ratio part = 120 / 15 = 8. Sheep correspond to 4 ratio parts, so sheep = 4 * 8 = 32. Verification / Alternative check: We can cross check by reconstructing the ratio. If sheep are 32 and horses are 120, the ratio sheep : horses is 32 : 120. Dividing both by 8 gives 4 : 15, which matches the original ratio. The feed per horse times the number of horses is 110 * 120 = 13,200 ounces, which matches the total daily requirement. Both ratio and feed data are satisfied. Why Other Options Are Wrong: Values like 36, 44, 52 and 60 do not maintain the exact ratio 4 : 15 when the number of horses is 120. For example, with 36 sheep the ratio to 120 horses would be 3 : 10, not 4 : 15. Only 32 sheep give the correct proportional relationship and satisfy all the numerical conditions. Common Pitfalls: Some learners mistakenly use the total 13,200 as if it included food for sheep or mix up which animal count the ratio applies to. Another frequent error is to miscalculate 13,200 / 110 or misinterpret 4 : 15 as horses to sheep instead of sheep to horses. Keeping track of which quantity corresponds to which part of the ratio is essential. Final Answer: The number of sheep on the farm is 32.

Question 17

What is the average of the first five positive even numbers that are divisible by 9?

Accepted Answer

Introduction / Context: This question asks for the average of a specific finite sequence of numbers: the first five positive even numbers that are also divisible by 9. It tests understanding of multiples, even numbers and the formula for the mean of a list of numbers. The structure of the sequence makes the calculations simple once the correct numbers are identified. Given Data / Assumptions: We consider only positive even numbers. The numbers must be divisible by 9. We need the first five such numbers. We then compute their arithmetic mean. Concept / Approach: A number divisible by 9 is a multiple of 9. To also be even, it must be a multiple of 18, since 18 is the least common multiple of 2 and 9. Therefore, the required numbers form an arithmetic progression with first term 18 and common difference 18. Once we list the first five terms, we can either sum them and divide by 5 or use the property that the mean of an arithmetic progression equals the average of the first and last terms. Step-by-Step Solution: The smallest positive multiple of 18 is 18. The first five positive even numbers divisible by 9 are: 18, 36, 54, 72 and 90. Confirm each is divisible by 9: 18 / 9 = 2, 36 / 9 = 4, 54 / 9 = 6, 72 / 9 = 8, 90 / 9 = 10. Now compute the sum: 18 + 36 + 54 + 72 + 90. 18 + 36 = 54, so the sum becomes 54 + 54 + 72 + 90. 54 + 54 = 108, and 72 + 90 = 162. Total sum = 108 + 162 = 270. Average = sum / number of terms = 270 / 5 = 54. Verification / Alternative check: Since the numbers form an arithmetic progression with first term 18 and last term 90, the mean of the sequence can also be found as (first term + last term) / 2 = (18 + 90) / 2 = 108 / 2 = 54. This matches the result obtained by direct summation and division, confirming the calculation. Why Other Options Are Wrong: Options such as 50, 48, 60 or 62 do not correspond to the mean of the correctly identified sequence. These values might arise from mistakes like picking wrong multiples of 9, forgetting that the numbers must be even, or miscomputing the sum or the division. Only 54 is consistent with both the definition of the sequence and the arithmetic mean formula. Common Pitfalls: A common error is to list multiples of 9 that are not even, or to list the first five even numbers and then check divisibility by 9 incorrectly. Another mistake is to misapply the mean formula, for example dividing by an incorrect number of terms. Recognising that the sequence is made of multiples of 18 and using the arithmetic progression approach makes this problem quick and reliable to solve. Final Answer: The average of the first five positive even numbers divisible by 9 is 54.

Question 18

A grandmother wants to distribute all her assets worth Rs 52,500 among her three grandsons in the ratio 1/35 : 1/21 : 1/15. What is the value of the smallest share among these three portions?

Accepted Answer

Introduction / Context: This ratio and proportion question is about distributing a fixed amount of money among three people in a given fractional ratio. The twist is that the shares are expressed as fractions, and we have to convert that information into actual rupee amounts and then identify the smallest share. Such questions are common in aptitude exams because they test comfort with fractions, ratios and basic arithmetic in a practical context involving inheritance or division of money. Given Data / Assumptions: - Total assets of the grandmother are Rs 52,500. - The three grandsons share the money in the ratio 1/35 : 1/21 : 1/15. - The ratio refers to the amounts received by each grandson respectively. - We are asked to find the smallest of the three final rupee amounts. Concept / Approach: The key idea is that even if the terms of a ratio look complicated, they can be treated as proportional parts. We first convert the fractional terms 1/35, 1/21 and 1/15 into a common framework. Let the common proportional factor be k. Then the three shares are k*(1/35), k*(1/21) and k*(1/15). The sum of these three expressions must equal the total amount of Rs 52,500. Once we find k, we can compute each share and then select the smallest share among them. Step-by-Step Solution: Step 1: Let the three shares be k*(1/35), k*(1/21) and k*(1/15). Step 2: Add the three fractional parts: 1/35 + 1/21 + 1/15. Step 3: The sum of these fractions is 1/7, so k * (1/7) = 52,500. Step 4: Solve for k to get k = 52,500 * 7 = 3,67,500. Step 5: First grandson share = 3,67,500 * (1/35) = Rs 10,500. Step 6: Second grandson share = 3,67,500 * (1/21) = Rs 17,500. Step 7: Third grandson share = 3,67,500 * (1/15) = Rs 24,500. Step 8: Compare the three amounts: 10,500, 17,500 and 24,500. The smallest is Rs 10,500. Verification / Alternative check: A quick check is to confirm that the three shares add up to the original total. Add 10,500 + 17,500 + 24,500 which equals 52,500, so the distribution is consistent. A second check is to compute the ratio of the three rupee amounts and see whether it simplifies to 1/35 : 1/21 : 1/15 after multiplying by a common factor. Since the method used directly followed from the algebraic form, the result is reliable. Why Other Options Are Wrong: Rs 12,800 does not correspond to any of the three computed shares and would make the total distribution larger than Rs 52,500 if used as the smallest share. Rs 15,000 is larger than the true smallest share and also does not match any calculated portion. Rs 17,500 is actually the middle share, not the smallest one, so it cannot be the correct answer to the question that specifically asks for the smallest share. Common Pitfalls: A common mistake is to treat 1/35 : 1/21 : 1/15 as 1 : 35 : 21 : 15 or to invert the fractions accidentally. Another frequent error is to ignore the fractional nature of the ratio and attempt to convert them directly to percentages without a clear plan. Some learners also forget to ensure that the sum of the three calculated shares is equal to the total assets, which is an important consistency check. Final Answer: Thus, the smallest of the three shares received by the grandsons is Rs 10,500.

Question 19

Three product supporters handled 440, 360 and 300 calls respectively last month. This month the total job will consist of 1,500 calls, and each supporter will increase workload in the same proportion as before. By how many additional calls will the …

Accepted Answer

Introduction / Context: This question tests understanding of proportional increase when workloads scale up while preserving the same relative distribution between workers. The situation comes from a customer support or product support context, where three supporters handle calls. When the total number of calls changes, their individual loads change proportionally, and the goal is to focus specifically on how much extra work the second supporter will do in the new month compared to the previous month. Given Data / Assumptions: - Last month calls handled: first supporter 440, second supporter 360, third supporter 300. - Total calls last month = 440 + 360 + 300 = 1,100 calls. - Total calls this month are planned at 1,500 calls. - Work is divided in the same ratio as earlier among the three supporters. - We need the increase in the number of calls for the second supporter only. Concept / Approach: If workloads increase while keeping relative contributions constant, each person’s share is scaled by the same factor. The scaling factor is new total calls divided by old total calls. Once that factor is known, we multiply the second supporter’s old workload by this factor to get the new workload. The difference between the new workload and the original workload gives the additional number of calls for the second supporter. Step-by-Step Solution: Step 1: Compute last month total calls = 440 + 360 + 300 = 1,100. Step 2: Compute the scaling factor = new total / old total = 1,500 / 1,100 = 15 / 11. Step 3: Second supporter old workload = 360 calls. Step 4: Second supporter new workload = 360 * (15 / 11). Step 5: Calculate 360 * 15 / 11 = 5,400 / 11 which is approximately 490.9 calls. Step 6: Since calls are discrete, we interpret this as about 491 calls in practice. Step 7: Increase in calls for the second supporter = 491 minus 360 = 131 calls. Verification / Alternative check: We can check proportionality by computing how many calls the other supporters would get with the same factor 15 / 11 and see whether the total comes close to 1,500. The first supporter would get 440 * 15 / 11, and the third would get 300 * 15 / 11. Summing the approximate values together gives a total near 1,500, showing the scaling factor is used correctly. Slight rounding to the nearest whole call is acceptable in such word problems where call counts must be integers. Why Other Options Are Wrong: 160 calls and 600 calls represent much larger increases than justified by the scaling factor. They would give the second supporter a workload that is not consistent with proportional division. The value 491 calls is close to the new workload, but the question asks for the increase in calls, not the total. Therefore, 491 is not the correct final response; instead we need the difference which is 131. Common Pitfalls: Learners often confuse total new workload with the increment and mistakenly select a value near 491 instead of computing the increase above 360. Another pitfall is recalculating a completely new ratio instead of scaling the old distribution. Forgetting to compute the scaling factor from total calls can also lead to arbitrary or incorrect allocations among supporters. Final Answer: The second supporter will handle an additional 131 calls this month compared to last month.

Question 20

A sum of Rs 7,444 is divided among 5 ladies, 3 gents and 3 girls. The shares of one lady, one gent and one girl are in the ratio 7 : 4 : 3 respectively. What is the total share received by all the gents together?

Accepted Answer

Introduction / Context: This question combines ratios with total amount distribution. The amount is divided among three groups: ladies, gents and girls. Each individual from a group receives money according to a fixed ratio, and there are different numbers of individuals in each group. The problem is to compute the total share of all gents, which requires converting the ratio into actual rupee values based on the total Rs 7,444. Given Data / Assumptions: - Total amount to be divided is Rs 7,444. - Number of ladies = 5. - Number of gents = 3. - Number of girls = 3. - Ratio of the share of a single lady, a single gent and a single girl is 7 : 4 : 3. - We need the combined share of all gents together. Concept / Approach: We first convert the ratio into total parts contributed by each group. A single lady gets 7 parts, so 5 ladies get 5 * 7 parts. A single gent gets 4 parts, so 3 gents get 3 * 4 parts, and a single girl gets 3 parts, so 3 girls get 3 * 3 parts. The sum of all these parts corresponds to the whole Rs 7,444. Once we know the value of one part, we can calculate the total share received by all gents. Step-by-Step Solution: Step 1: For ladies, total parts = 5 * 7 = 35. Step 2: For gents, total parts = 3 * 4 = 12. Step 3: For girls, total parts = 3 * 3 = 9. Step 4: Total parts for all people = 35 + 12 + 9 = 56 parts. Step 5: Value of one part = 7,444 / 56. Step 6: Compute 7,444 / 56 = 1,861 / 14 which is approximately 132.93. Step 7: Total share of all gents = 12 parts (from Step 2) multiplied by value per part. Step 8: Gent share = 12 * (1,861 / 14) = (12 * 1,861) / 14 = 11,166 / 7 which is about Rs 1,595.14. Verification / Alternative check: To verify, compute the total share for each group and ensure the sum is Rs 7,444. Ladies share = 35 parts * value per part, gents share = 12 parts * value per part, and girls share = 9 parts * value per part. Adding all three should give exactly Rs 7,444, confirming that each part value and the gent share are correctly computed. This check helps prevent arithmetic mistakes when dividing or multiplying fractional values. Why Other Options Are Wrong: Rs 1,793.40 and Rs 1,551.50 are close but do not match the precise calculation from the part value and the number of parts allotted to gents. Rs 595.14 is far too small and corresponds roughly to the share of a much smaller number of parts, not the full 12 parts reserved for gents. Only Rs 1,595.14 matches the correct part based computation and thus is the accurate combined share of all gents. Common Pitfalls: A common error is to treat the ratio 7 : 4 : 3 as if it applies directly to the total number of people rather than to the share of a single person in each category. Another mistake is to forget to multiply the per person share by the number of people in that group. Some learners also miscalculate the value of one part by dividing 7,444 incorrectly or rounding too early, which leads to inaccurate final answers. Final Answer: The total share received by all the gents together is Rs 1,595.14 (approximate to two decimal places).

Ratio and Proportion Questions