Question 1

In a cricket match, Virat, Dhoni and Rohit together score 285 runs. The ratio of Virat’s runs to Rohit’s runs is 3 : 2 and the ratio of Rohit’s runs to Dhoni’s runs is also 3 : 2. How many runs did Virat score in that match?

Accepted Answer

Introduction / Context: This question tests understanding of chained ratios, where several people or quantities are compared in linked proportions. Here, the runs scored by Virat, Dhoni and Rohit are related through two given ratios, and we must convert these into a single combined relationship to find the exact runs scored by Virat. Given Data / Assumptions: • Total runs scored by Virat, Dhoni and Rohit together = 285. • Virat : Rohit = 3 : 2. • Rohit : Dhoni = 3 : 2. • All runs are whole numbers and the ratios are exact. Concept / Approach: When we have chained ratios, the key idea is to express all quantities in terms of a common variable using both ratios. We first write Virat and Rohit in one ratio and Rohit and Dhoni in another. By matching the common player Rohit between the two ratios, we can express Virat, Rohit and Dhoni all in terms of a single parameter. Then we use the total sum to solve for the parameter and finally obtain Virat’s individual score. Step-by-Step Solution: Step 1: From Virat : Rohit = 3 : 2, let Virat = 3x and Rohit = 2x.Step 2: From Rohit : Dhoni = 3 : 2, let Rohit = 3y and Dhoni = 2y.Step 3: Since Rohit is the same player in both expressions, set 2x = 3y. This gives x = 3y / 2.Step 4: Substitute x into Virat’s runs: Virat = 3x = 3 * (3y / 2) = 9y / 2.Step 5: Rohit = 3y and Dhoni = 2y, so total runs = (9y / 2) + 3y + 2y = (9y / 2) + 5y = (9y + 10y) / 2 = 19y / 2.Step 6: Set 19y / 2 = 285. So 19y = 570, giving y = 30.Step 7: Virat’s runs = 9y / 2 = 9 * 30 / 2 = 270 / 2 = 135. Verification / Alternative check: Using y = 30, Rohit’s runs are 3y = 90, and Dhoni’s runs are 2y = 60. Check the totals: Virat 135 + Rohit 90 + Dhoni 60 = 285, which matches the given total. Also verify ratios: Virat : Rohit = 135 : 90 = 3 : 2, and Rohit : Dhoni = 90 : 60 = 3 : 2. All conditions are satisfied, so the value for Virat’s runs is consistent and correct. Why Other Options Are Wrong: Option 154 would give remaining runs of 285 - 154 = 131 for Rohit and Dhoni combined, which cannot be split to maintain both given ratios simultaneously. Option 112 and option 106 similarly fail when you try to assign runs to Rohit and Dhoni while keeping both ratios and the total sum correct. Only 135 leads to integer values that exactly satisfy all constraints. Common Pitfalls: A frequent mistake is to add ratios directly instead of converting them into a combined relation. Another error is to assume Virat : Dhoni is directly 3 : 2, which is not given. It is essential to use Rohit as the common link and set the different expressions for Rohit equal to each other so that a single variable description is obtained. Final Answer: The number of runs scored by Virat in the match is 135.

Question 2

Maneela lends a total of Rs. 8000, part at 5% per annum and the rest at 6% per annum simple interest. If the total interest she earns in 2 years is Rs. 820, in what ratio should the Rs. 8000 be divided between the two rates?

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Introduction / Context: This question checks understanding of simple interest and allocation of a total investment between two different interest rates. The aim is to determine how much money is placed at each rate when the overall interest earned over a fixed time period is known. It is a typical mixture-of-rates question based on simple interest. Given Data / Assumptions: • Total amount lent by Maneela = Rs. 8000. • Part of the amount is lent at 5% per annum simple interest. • The remaining part is lent at 6% per annum simple interest. • Time period for both investments = 2 years. • Total interest received after 2 years = Rs. 820. Concept / Approach: Let the amount lent at 5% be x rupees, and the amount lent at 6% be 8000 − x rupees. For simple interest, interest = Principal * Rate * Time / 100. We form an equation using the total interest from both parts and set it equal to Rs. 820. Solving this equation gives the individual principal values and thus the ratio between them. This is a direct algebraic application of the simple interest formula. Step-by-Step Solution: Step 1: Let the amount lent at 5% be x and at 6% be (8000 − x).Step 2: Interest from x at 5% for 2 years = x * 5 * 2 / 100 = 0.10x.Step 3: Interest from (8000 − x) at 6% for 2 years = (8000 − x) * 6 * 2 / 100 = 0.12(8000 − x).Step 4: Total interest is given as 820, so 0.10x + 0.12(8000 − x) = 820.Step 5: Simplify: 0.10x + 960 − 0.12x = 820 ⇒ −0.02x + 960 = 820 ⇒ −0.02x = −140 ⇒ x = 7000.Step 6: Amount at 5% = 7000 and amount at 6% = 8000 − 7000 = 1000.Step 7: Required ratio (5% : 6%) = 7000 : 1000 = 7 : 1. Verification / Alternative check: Compute the actual interest using the found amounts. From Rs. 7000 at 5% for 2 years, interest = 7000 * 5 * 2 / 100 = Rs. 700. From Rs. 1000 at 6% for 2 years, interest = 1000 * 6 * 2 / 100 = Rs. 120. Total interest = 700 + 120 = Rs. 820, which matches the given value. This confirms that the ratio 7 : 1 is correct. Why Other Options Are Wrong: Ratio 13 : 5 would correspond to very different amounts and would not produce a total interest of Rs. 820 over 2 years at the given rates. Ratios 15 : 7 and 2 : 7 also lead to principal allocations that, when plugged into the simple interest formula, give total interest either more or less than Rs. 820. Thus they conflict with the problem data. Common Pitfalls: One common mistake is to treat 5% and 6% as if they can be averaged directly without accounting for the different amounts invested at each rate. Another error is to forget that the time period is the same for both parts and to incorrectly use different times. Setting up the algebraic equation correctly from the total interest is the most reliable method. Final Answer: The Rs. 8000 should be divided in the ratio 7 : 1 between 5% and 6% simple interest.

Question 3

In a class of 39 students, the ratio of boys to girls is 2 : 1. Radhika is one of the girls. She ranks 15th from the top among all students and 8th from the bottom among the girls. How many boys are there below Radhika in the overall class ranking?

Accepted Answer

Introduction / Context: This question blends ratio concepts with positional ranking and counting. It is common in aptitude tests where you must interpret ranks from both the top and bottom, and combine this with the total number of students and the ratio of boys to girls to determine how many of a particular group lie above or below a given person. Given Data / Assumptions: • Total students in the class = 39. • Ratio of boys : girls = 2 : 1. • Radhika is a girl. • Radhika is 15th from the top among all students. • Radhika is 8th from the bottom among the girls. Concept / Approach: First, convert the ratio of boys to girls into exact numbers. Then, use Radhika’s rank among girls from the bottom to find her rank among girls from the top. Once we know how many girls are above her, we subtract that from her overall rank to find how many boys are above her. Finally, subtract boys above from total boys to get the number of boys below her. Step-by-Step Solution: Step 1: Let boys : girls = 2 : 1. Total parts = 3. Since total students are 39, boys = 2/3 * 39 = 26 and girls = 1/3 * 39 = 13.Step 2: Radhika is a girl and is 8th from the bottom among 13 girls.Step 3: Her rank among girls from the top is 13 − 8 + 1 = 6. So there are 5 girls above her.Step 4: Radhika is 15th from the top in the entire class, so among the first 15 students: 5 are girls above her and the 6th is Radhika herself.Step 5: Therefore, number of boys above Radhika = 15 − 6 = 9.Step 6: Total boys = 26, so boys below Radhika = 26 − 9 = 17. Verification / Alternative check: We can check consistency: there are 13 girls; 5 above Radhika, 1 is Radhika, and 7 below her among girls, which matches the 8th-from-bottom statement. In the entire class, students below Radhika are 39 − 15 = 24; among these, 7 are girls and 17 are boys, which exactly matches the boys-below result we computed. All counts are consistent. Why Other Options Are Wrong: Option 9 is the number of boys above her, not below her. Options 11 and 15 do not satisfy the total count of boys and the distribution above and below Radhika when you work through the details. They contradict either the total number of boys or the given ranking information. Common Pitfalls: Students often confuse ranks from the top and bottom, forgetting to add 1 when converting between them. Another typical mistake is to ignore that the ratio of boys to girls must first be converted into actual numbers before counting people above and below a specific student. Careful step-by-step reasoning avoids these errors. Final Answer: The number of boys below Radhika in the overall ranking is 17.

Question 4

If 12 men can reap 120 acres of land in 36 days, how many acres of land can 54 men reap in 54 days, assuming the work rate per man remains constant?

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Introduction / Context: This is a standard work and manpower question where the total work done depends on the number of workers and the number of days they work, assuming a constant rate of work per person. Such problems commonly appear in aptitude tests to check understanding of direct proportionality between workers, time and work done. Given Data / Assumptions: • 12 men can reap 120 acres of land in 36 days. • We must find how many acres 54 men can reap in 54 days. • Each man works at a constant rate throughout the period. • Work done is directly proportional to (number of men) * (number of days). Concept / Approach: Total work can be expressed in man-days and then converted to actual acres using proportionality. First find how many acres are done per man-day. Then, compute the total man-days in the second scenario and multiply by the per man-day rate to get the number of acres reaped by 54 men in 54 days. Step-by-Step Solution: Step 1: In the first situation, total man-days = 12 men * 36 days = 432 man-days.Step 2: These 432 man-days correspond to 120 acres, so each man-day covers 120 / 432 acres.Step 3: Simplify 120 / 432 = 5 / 18. So one man-day reaps 5/18 acres.Step 4: In the second situation, total man-days = 54 men * 54 days = 2916 man-days.Step 5: Acres reaped = man-days * acres per man-day = 2916 * (5 / 18).Step 6: Compute 2916 / 18 = 162, then multiply 162 * 5 = 810 acres. Verification / Alternative check: We can also use proportionality directly: the number of men increases from 12 to 54 (factor 54 / 12 = 4.5), and days increase from 36 to 54 (factor 54 / 36 = 1.5). Total work is multiplied by 4.5 * 1.5 = 6.75. So new acres = 120 * 6.75 = 810 acres, confirming the earlier calculation. Why Other Options Are Wrong: 710 acres and 760 acres are both too small; they correspond to lower effective work rates than given by the original data. 860 acres is larger than 810 acres and does not match any integer multiple of the base scenario when proportional factors are correctly computed. Only 810 acres is consistent with the given numbers. Common Pitfalls: A common mistake is to scale only by men or only by days, instead of taking both changes into account. Another error is to miscalculate the per man-day rate by dividing incorrectly or not simplifying the fraction properly. Always compute total man-days first, then apply the rate logically. Final Answer: The 54 men can reap 810 acres of land in 54 days.

Question 5

Paint is required to be thinned so that the final mixture has paint and water in the ratio 2 : 1.5. Due to a mistake, a painter has 6 litres of mixture that is exactly half paint and half water. What should he add to obtain the correct proportions?

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Introduction / Context: This question is about mixtures and ratios. The painter wants a final mixture with a specified ratio of paint to water, but the current mixture has equal amounts of paint and water. We must determine what to add so that the corrected mixture satisfies the required ratio. Such mixture questions test understanding of ratio adjustment and proportional reasoning. Given Data / Assumptions: • Required ratio of paint : water is 2 : 1.5. • The painter mistakenly has 6 litres of mixture with half paint and half water. • Therefore, current paint = 3 litres and current water = 3 litres. • We are allowed to add only paint, only water, or a combination, as in the options. Concept / Approach: First, simplify the desired ratio. The ratio 2 : 1.5 can be scaled by 2 to remove the decimal, giving 4 : 3. This means that in the final mixture, paint : water should be 4 : 3. Let the extra paint added be p litres and extra water added be w litres. Then form an equation so that (3 + p) : (3 + w) = 4 : 3 and check which option satisfies this relationship. We will see that only adding 1 litre of paint and no extra water yields the desired ratio. Step-by-Step Solution: Step 1: Simplify the required ratio 2 : 1.5 to 4 : 3 by multiplying both terms by 2.Step 2: Currently, the mixture has 3 litres of paint and 3 litres of water.Step 3: Suppose we add only paint as in option A: new paint = 3 + 1 = 4 litres, new water = 3 litres.Step 4: The new ratio becomes paint : water = 4 : 3, which matches the required proportion.Step 5: Check that no other option is needed since we already meet the exact target ratio by this simple adjustment. Verification / Alternative check: If we consider the general condition (3 + p) : (3 + w) = 4 : 3, cross-multiplying gives 3(3 + p) = 4(3 + w). That is 9 + 3p = 12 + 4w. For option A, p = 1 and w = 0, so left side is 9 + 3 * 1 = 12 and right side is 12 + 0 = 12, which satisfies the equation. This confirms that adding only 1 litre of paint gives the desired ratio. Why Other Options Are Wrong: Option B (1 litre water) would give 3 litres paint and 4 litres water, resulting in a ratio 3 : 4, which is not 4 : 3. Option C (½ litre water and 1 litre paint) leads to 4 litres paint and 3.5 litres water, giving a ratio 4 : 3.5, which does not simplify to 4 : 3. Option D (½ litre paint and 1 litre water) produces 3.5 litres paint and 4 litres water, ratio 3.5 : 4, again not equal to 4 : 3. Common Pitfalls: A common error is to treat 2 : 1.5 as 2 : 1 directly without simplifying correctly, which changes the target mixture. Another frequent mistake is to think equal additions of paint and water will preserve or create the required ratio, which is not true except in special cases. Properly simplifying the required ratio and systematically testing each option prevents these mistakes. Final Answer: The painter should add 1 litre of paint to obtain the correct proportion of paint to water.

Question 6

Rs. 2600 is divided into three parts A, B and C in the ratio 1/2 : 1/3 : 1/4. By how much is share A greater than share C?

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Introduction / Context: This problem is about dividing a sum of money into three parts according to a ratio expressed with fractional terms. It tests the ability to convert fractional ratios into simple integer ratios and then compute exact amounts for each part before comparing them. Such questions are standard in aptitude and financial reasoning sections. Given Data / Assumptions: • Total amount to be divided = Rs. 2600. • The ratio of A : B : C is 1/2 : 1/3 : 1/4. • We must find how much more A receives compared to C. • All shares are non-negative and sum to the total amount. Concept / Approach: Fractions are converted to a common base so that ratio terms become integers. The denominators are 2, 3 and 4. The least common multiple of these denominators is 12. By multiplying each fraction by 12, we can convert 1/2, 1/3 and 1/4 into 6, 4 and 3 respectively. These integers give us the ratio A : B : C = 6 : 4 : 3. Then we distribute Rs. 2600 according to these 6 + 4 + 3 = 13 parts, and finally compute the difference between A and C. Step-by-Step Solution: Step 1: Original ratio is 1/2 : 1/3 : 1/4.Step 2: Convert each fraction using LCM of denominators 2, 3 and 4, which is 12.Step 3: Multiply each fraction by 12: (1/2)*12 = 6, (1/3)*12 = 4, (1/4)*12 = 3. So ratio becomes A : B : C = 6 : 4 : 3.Step 4: Total ratio parts = 6 + 4 + 3 = 13.Step 5: Value of one part = 2600 / 13 = 200.Step 6: Share of A = 6 * 200 = Rs. 1200; share of C = 3 * 200 = Rs. 600.Step 7: Difference A − C = 1200 − 600 = Rs. 600. Verification / Alternative check: We can also compute B to confirm the sum: B = 4 * 200 = Rs. 800. Then total = 1200 + 800 + 600 = Rs. 2600, which matches the original amount. This confirms that our conversion of the fractional ratio to integer parts and subsequent calculations are correct. Why Other Options Are Wrong: Option 300 would imply A is only Rs. 300 more than C, which would require different underlying shares than 1200 and 600. That contradicts the ratio 6 : 4 : 3 derived from the fractions. Options 150 and 75 similarly do not match the true difference obtained from correctly dividing the amount according to the given ratio. Common Pitfalls: One common error is to treat the fractional ratio values directly as the shares without converting to simple integers, which leads to incorrect partitioning. Another mistake is to divide Rs. 2600 by the sum of the denominators instead of the sum of the scaled integer parts. Always convert fractions to an equivalent integer ratio before distributing the total amount. Final Answer: Share A is greater than share C by Rs. 600.

Question 7

Two vessels P and Q contain mixtures of milk and water in the ratios 3 : 2 and 5 : 3 respectively. If 10 litres of mixture are removed from vessel P and poured into vessel Q, and then the ratio of milk to water in vessel Q becomes 8 : 5, what was th…

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Introduction / Context: This question combines ratio concepts with mixture transfer between two vessels. We are given initial ratios of milk and water in both vessels, a quantity of mixture transferred from one vessel to another, and the final ratio in one vessel. We must work backwards to find the original amount of water in that vessel. This tests algebraic modeling of mixtures and ratios. Given Data / Assumptions: • Vessel P has milk : water = 3 : 2. • Vessel Q has milk : water = 5 : 3. • From vessel P, 10 litres of its mixture is removed and poured into vessel Q. • After this transfer, the ratio milk : water in vessel Q becomes 8 : 5. • We must find the initial quantity of water in vessel Q (in litres). Concept / Approach: We use the fact that when we remove 10 litres from P with ratio 3 : 2, the removed liquid contains milk and water in the same ratio, i.e., 6 litres milk and 4 litres water. For vessel Q, we let its initial volume be V litres. Using its initial ratio, we can express initial milk and water in Q in terms of V. After adding 6 litres of milk and 4 litres of water from P, we use the final ratio 8 : 5 to set up and solve an equation for V. From V, we compute the initial water quantity in Q. Step-by-Step Solution: Step 1: In vessel P, ratio 3 : 2 means that in 10 litres of mixture, milk = (3/5) * 10 = 6 litres and water = (2/5) * 10 = 4 litres.Step 2: Let initial total mixture in vessel Q be V litres. With ratio 5 : 3, milk in Q is (5/8)V and water is (3/8)V.Step 3: After transferring from P to Q, milk in Q = (5/8)V + 6 and water in Q = (3/8)V + 4.Step 4: The final ratio in Q is milk : water = 8 : 5, so ((5/8)V + 6) / ((3/8)V + 4) = 8 / 5.Step 5: Cross-multiply: 5 * ((5/8)V + 6) = 8 * ((3/8)V + 4).Step 6: Simplify left side: 5 * (5V/8 + 6) = (25V/8) + 30. Simplify right side: 8 * (3V/8 + 4) = 3V + 32.Step 7: Set (25V/8) + 30 = 3V + 32. Multiply by 8: 25V + 240 = 24V + 256, so V = 16.Step 8: Initial water in Q = (3/8)V = (3/8)*16 = 6 litres. Verification / Alternative check: With V = 16, initial milk in Q = (5/8)*16 = 10 litres, water = 6 litres. After adding 6 litres milk and 4 litres water from P, we get milk = 16 litres and water = 10 litres. The ratio 16 : 10 simplifies to 8 : 5, which matches the final ratio given. Thus, the calculation is consistent and confirms that the initial water in Q is 6 litres. Why Other Options Are Wrong: Option 10 litres or 12 litres for initial water would lead to different initial totals V and would not result in a final milk : water ratio of 8 : 5 after adding 6 litres milk and 4 litres water. Option 16 litres is impossible because it would imply all of vessel Q initially is water, which contradicts the given ratio 5 : 3. Common Pitfalls: Common mistakes include assuming that only water or only milk is transferred, forgetting that the transferred 10 litres maintains the ratio from P. Another error is ignoring that V is the total mixture in Q, not directly the water content. Setting up the equation carefully using the correct expressions for milk and water before and after transfer is crucial. Final Answer: The initial quantity of water in vessel Q was 6 litres.

Question 8

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

Accepted Answer

Introduction / Context: This question examines the combined effect of percentage increases on a given ratio. It is a classic aptitude problem where initial ratios are scaled by different percentage factors, and then the final ratio must be simplified. Such questions are useful for learning how proportional changes affect multiple related quantities. Given Data / Assumptions: • Initial ratio of seats for Mathematics : Physics : Biology = 5 : 7 : 8. • Mathematics seats are increased by 40%. • Physics seats are increased by 50%. • Biology seats are increased by 75%. • We must find the ratio of the new numbers of seats after these increases. Concept / Approach: First, represent the initial numbers of seats as 5x, 7x and 8x for Mathematics, Physics and Biology. Then, apply the respective percentage increases by multiplying each by (1 + percentage increase). After calculating the new effective quantities, we find their ratio and simplify by dividing by the greatest common factor or scaling to small integers. Step-by-Step Solution: Step 1: Let original seats be M = 5x, P = 7x, B = 8x.Step 2: Increase Mathematics by 40%: new M = 5x * 1.40 = 7x.Step 3: Increase Physics by 50%: new P = 7x * 1.50 = 10.5x.Step 4: Increase Biology by 75%: new B = 8x * 1.75 = 14x.Step 5: The new ratio is 7x : 10.5x : 14x.Step 6: Cancel x to get 7 : 10.5 : 14. Multiply each term by 2 to clear the decimal: 14 : 21 : 28.Step 7: Divide by 7: 14 / 7 = 2, 21 / 7 = 3, 28 / 7 = 4. So the simplified ratio is 2 : 3 : 4. Verification / Alternative check: One can take actual numbers instead of 5x, 7x and 8x, for example, 5, 7 and 8. Applying the same percentage increases gives new values: 5 * 1.4 = 7, 7 * 1.5 = 10.5, 8 * 1.75 = 14. These three numbers again simplify to the ratio 2 : 3 : 4 when scaled appropriately, confirming the result is independent of the specific value of x. Why Other Options Are Wrong: Ratios 1 : 2 : 3, 3 : 4 : 5 and 4 : 5 : 6 do not arise from multiplying 5 : 7 : 8 by 1.4, 1.5 and 1.75 respectively. If you substitute these ratios back into the percentage changes, they do not match the proportional increases given in the problem. Common Pitfalls: Students sometimes incorrectly add the percentages directly to the ratio numbers (for example, adding 40 to 5 instead of increasing 5 by 40%), or apply the same percentage increase to all terms. Another mistake is to forget to clear decimals properly when simplifying ratios. Always multiply all terms by a suitable common factor to avoid decimal confusion. Final Answer: The new ratio of seats for Mathematics, Physics and Biology is 2 : 3 : 4.

Question 9

Maneela, Raghu and Aravind have some jems each. Five times the number of jems with Raghu equals seven times the number of jems with Maneela, and five times the number with Maneela equals seven times the number with Aravind. What is the minimum total…

Accepted Answer

Introduction / Context: This question involves ratios and least common multiples to find the smallest integer solution that satisfies two linked proportional relationships. Each person holds some number of jems, and there are conditions relating multiples of the counts. We must find the minimum total number of jems that satisfies both ratio conditions exactly. Given Data / Assumptions: • Let Maneela have M jems, Raghu have R jems and Aravind have A jems. • Condition 1: 5R = 7M. • Condition 2: 5M = 7A. • All numbers M, R and A must be positive integers. • We seek the minimum possible value of M + R + A under these constraints. Concept / Approach: From 5R = 7M, we get R = (7/5)M, so M must be a multiple of 5 for R to be an integer. From 5M = 7A, we get A = (5/7)M, so M must also be a multiple of 7 for A to be an integer. Therefore, M must be a multiple of the least common multiple of 5 and 7, which is 35. Taking M = 35 gives the smallest positive integer solution that satisfies both relations. Then we compute R and A and sum them to get the minimum total. Step-by-Step Solution: Step 1: From 5R = 7M, write R = 7M / 5.Step 2: From 5M = 7A, write A = 5M / 7.Step 3: For R to be an integer, M must be divisible by 5; for A to be an integer, M must be divisible by 7. So M is a multiple of LCM(5, 7) = 35.Step 4: Take the smallest possible M = 35.Step 5: Then R = 7M / 5 = 7 * 35 / 5 = 7 * 7 = 49.Step 6: Also, A = 5M / 7 = 5 * 35 / 7 = 5 * 5 = 25.Step 7: Total jems = M + R + A = 35 + 49 + 25 = 109. Verification / Alternative check: Check both conditions: 5R = 5 * 49 = 245 and 7M = 7 * 35 = 245, so the first condition is satisfied. Also, 5M = 5 * 35 = 175 and 7A = 7 * 25 = 175, so the second condition is satisfied. Since we chose the smallest possible M that is a multiple of both 5 and 7, the corresponding totals are minimal. Why Other Options Are Wrong: Options 108, 107 and 110 are close to 109 but cannot be expressed as M + R + A with positive integers satisfying both 5R = 7M and 5M = 7A. Any smaller M than 35 would make at least one of R or A fractional, which is not allowed. Therefore, 109 is the only feasible minimum total. Common Pitfalls: Students sometimes try random values and check conditions, which is inefficient and error prone. Others mistakenly think they can take M as 5 or 7 without checking integrality for both R and A. The correct approach is to use the least common multiple to ensure that all ratios yield integers. Carefully using LCM and verifying both conditions avoids these mistakes. Final Answer: The minimum total number of jems that all three can have together is 109.

Question 10

Two containers each initially hold 1 litre: the first has pure water and the second has pure milk. Five cups of water are taken from the first container and mixed thoroughly into the second. Then five cups of this new mixture are taken from the seco…

Accepted Answer

Introduction / Context: This is a classic puzzle about exchanging equal volumes of liquids between two containers. One container starts with pure water and the other with pure milk. After transferring equal volumes in both directions, we are asked to compare the final proportion of milk in one container with the final proportion of water in the other. The key concept is conservation: what leaves one container in terms of a substance must appear in the other container. Given Data / Assumptions: • First container initially: 1 litre of pure water, 0 milk. • Second container initially: 1 litre of pure milk, 0 water. • First step: take 5 cups of water from the first container and add to the second, then mix thoroughly. • Second step: take 5 cups of this mixture from the second container and add back to the first container. • Cup size is the same in both transfers and is smaller than 1 litre so operations are feasible. Concept / Approach: We do not need the exact cup volume; the answer depends only on symmetry and conservation. After the first transfer, only water has moved from the first to the second. After the second transfer, some of this water is moved back along with some milk. The net effect is that the amount of milk present in the first container at the end is equal to the amount of water that remains in the second container at the end. From this, we infer that the proportion of milk in one equals the proportion of water in the other. Step-by-Step Solution: Step 1: Initially, first container has 1 litre water and 0 milk; second container has 0 water and 1 litre milk.Step 2: After transferring 5 cups of water from the first to the second, the first container has less water and still no milk. The second now has 1 litre milk plus 5 cups water.Step 3: Mix the second container thoroughly so its composition is uniform.Step 4: When 5 cups of mixture are transferred back to the first container, this mixture contains both milk and water in the same proportions as in the second container at that moment.Step 5: Whatever volume of milk leaves the second container ends up in the first, and the same volume of water going back to the first container reduces the original water imbalance.Step 6: By conservation, the amount of milk that leaves the second and moves to the first equals the amount of water that remains in the second after the exchange, leading to equal fractions. Verification / Alternative check: A more formal way is to note that the total amount of water in the system stays constant, as does the total amount of milk. The first container starts with only water, and the second starts with only milk. After the two transfers, the amount of milk in the first container must be exactly equal to the amount of water that has ended up in the second container. Since both containers hold the same total volume (1 litre each) throughout, the fraction of milk in the first equals the fraction of water in the second. Hence A and B are equal. Why Other Options Are Wrong: Option A < B or A > B would imply that the net exchanges favored one liquid over the other, which contradicts the symmetry and conservation of volume in both directions. Option cannot be determined is incorrect because, despite not knowing the exact cup size, the qualitative relationship can be deduced from conservation principles. Common Pitfalls: Students often think that because water is transferred first, there will always be more water in the second container or more milk in the second container. Others assume we need the exact cup size to solve the problem. In fact, equal volume exchanges and conservation of total quantities make the comparison independent of cup volume, and symmetry gives the relationship directly. Final Answer: The proportion of milk in the first container equals the proportion of water in the second container, so A = B.

Question 11

Two solutions have milk and water in the ratios 7 : 5 and 6 : 11 respectively. In what proportion should these two solutions be mixed so that the resulting solution has 1 part milk and 2 parts water, or is this impossible with the given solutions?

Accepted Answer

Introduction / Context: This is a mixture and ratio question involving two different milk-water solutions. We are given their initial ratios and asked to find the mixing proportion that would produce a final mixture with a target milk-to-water ratio of 1 : 2. An important concept in such problems is that the resulting concentration of milk must lie between the concentrations of the individual solutions if we only mix them without adding anything else. Given Data / Assumptions: • Solution 1 has milk : water = 7 : 5. • Solution 2 has milk : water = 6 : 11. • The target final ratio is milk : water = 1 : 2. • We mix only these two solutions; no extra milk or water is added separately. Concept / Approach: First, convert each ratio into the fraction of milk in that solution. For 7 : 5, total parts = 12, so the milk fraction is 7/12. For 6 : 11, total parts = 17, so the milk fraction is 6/17. The target ratio 1 : 2 corresponds to a milk fraction of 1/(1 + 2) = 1/3. When you mix two solutions of concentrations C1 and C2, the resulting concentration must lie between C1 and C2 provided you use positive quantities of each. If the target concentration is outside this range, it is impossible to obtain by mixing only those two solutions. Step-by-Step Solution: Step 1: Compute the milk fraction in solution 1: C1 = 7/12 ≈ 0.5833.Step 2: Compute the milk fraction in solution 2: C2 = 6/17 ≈ 0.3529.Step 3: Compute the target milk fraction for ratio 1 : 2: Ct = 1 / (1 + 2) = 1/3 ≈ 0.3333.Step 4: Compare Ct with C1 and C2. We have C1 ≈ 0.5833 and C2 ≈ 0.3529, both of which are greater than 0.3333.Step 5: Since both available solutions are richer in milk than the target, any convex combination (mixture) of them will also have a milk fraction between 0.3529 and 0.5833, never as low as 0.3333.Step 6: Therefore, it is impossible to obtain a mixture with milk : water = 1 : 2 using only these two solutions. Verification / Alternative check: We can attempt to set up an equation. Suppose we mix x litres of solution 1 and y litres of solution 2. Milk in mixture = (7/12)x + (6/17)y. Total volume = x + y. The required condition is [(7/12)x + (6/17)y] / (x + y) = 1/3. When we simplify this equation, we end up with a condition that cannot be satisfied with positive x and y simultaneously, reinforcing the argument that the target concentration lies outside the achievable range. Why Other Options Are Wrong: Options 7 : 12, 8 : 13 and 9 : 4 all correspond to some assumed mixing ratios, but none of them can produce the required final milk fraction of 1/3 because any mixture of the two given solutions remains between 7/12 and 6/17 in milk concentration. Thus, these numeric mixing ratios do not satisfy the concentration requirement. Common Pitfalls: Students often try to solve for specific x : y without first checking whether the target concentration is feasible. Another common error is to average the ratios incorrectly or to think that any target between the two water ratios, rather than milk fractions, is achievable. Always compare the actual concentration fractions and ensure that the target lies between them before attempting detailed algebra. Final Answer: The desired mixture with milk : water = 1 : 2 cannot be obtained by mixing only the two given solutions, so the correct choice is “Not possible with these two solutions”.

Question 12

A group for social work has 64 boys and 40 girls. During a membership drive, the same number of boys and girls join the group. If after this the ratio of boys to girls becomes 4 : 3, how many members does the group have now in total?

Accepted Answer

Introduction / Context: This question deals with adjusting a ratio when equal numbers are added to each category. The group has boys and girls, and after adding the same number of each, the overall ratio changes to a new given value. We must use algebra to determine how many new members joined and then compute the new total membership. Given Data / Assumptions: • Initial number of boys = 64. • Initial number of girls = 40. • The same number of boys and girls join the group; let this number be x. • After this, the ratio of boys to girls becomes 4 : 3. • We need the new total number of members in the group. Concept / Approach: When equal numbers are added to both categories in a ratio, we introduce a variable x representing the additional boys and the same x additional girls. We then express the new ratio (64 + x) : (40 + x) and set it equal to 4 : 3. Solving this proportion equation for x gives the number of new boys and girls added, from which we can easily compute the total membership. Step-by-Step Solution: Step 1: Let x boys and x girls join. New boys = 64 + x, new girls = 40 + x.Step 2: The final ratio is (64 + x) : (40 + x) = 4 : 3.Step 3: Set up proportion: (64 + x)/(40 + x) = 4/3.Step 4: Cross-multiply: 3(64 + x) = 4(40 + x).Step 5: Expand: 192 + 3x = 160 + 4x.Step 6: Rearrange: 192 − 160 = 4x − 3x ⇒ 32 = x.Step 7: New number of boys = 64 + 32 = 96; new number of girls = 40 + 32 = 72.Step 8: New total members = 96 + 72 = 168. Verification / Alternative check: Check the final ratio: 96 : 72. Divide both terms by 24: 96 / 24 = 4, 72 / 24 = 3. The simplified ratio is indeed 4 : 3, matching the condition given. This confirms that adding 32 boys and 32 girls is consistent, making the new total membership 168. Why Other Options Are Wrong: Totals 201, 147 and 154 would correspond to different values of x, and if you plug those back into (64 + x)/(40 + x), the resulting ratio is not 4 : 3. Therefore, those totals contradict the given ratio condition. Common Pitfalls: Some students incorrectly add 4 and 3 to the original numbers or assume a fixed number based on guesswork, rather than setting up the proper algebraic equation. Another mistake is to forget that the same number is added to both boys and girls. Expressing the new numbers clearly as 64 + x and 40 + x and using ratio equality avoids such errors. Final Answer: The group now has a total of 168 members.

Question 13

In a coaching institute, the ratio of the number of students preparing for B.Tech to those preparing for MBA is 4 : 5. The ratio of the total fees collected from B.Tech students to that collected from MBA students is 25 : 16. If the total fees colle…

Accepted Answer

Introduction / Context: This question relates student counts and fee collections via ratios. We are given the ratio of numbers of students in two streams and a separate ratio of total fees collected from each stream. Using these ratios, we must determine the absolute amount of fees collected from MBA students given the overall total. It tests proportional reasoning and working with multiple ratios. Given Data / Assumptions: • Number of B.Tech students : number of MBA students = 4 : 5. • Fees collected from all B.Tech students : fees collected from all MBA students = 25 : 16. • Total fees collected from both streams together = Rs. 1.62 lakh = Rs. 162,000. • We must find the fee amount from MBA students only. Concept / Approach: Here, we do not need to know the exact number of students or the individual fee per student. The key is that the total fee from B.Tech and MBA students together is divided in the ratio 25 : 16. We can treat this as two parts of a whole, where 25 parts correspond to B.Tech total fees and 16 parts to MBA total fees. Adding these gives the total number of ratio parts, and then we scale these parts to match the real total of Rs. 162,000. Step-by-Step Solution: Step 1: Let total B.Tech fee = 25k and total MBA fee = 16k for some positive k.Step 2: Then total fee collected from all students = 25k + 16k = 41k.Step 3: We are told that 41k = Rs. 162,000.Step 4: Solve for k: k = 162000 / 41.Step 5: The total MBA fee is 16k = 16 * (162000 / 41) = (16 * 162000) / 41.Step 6: Compute 16 * 162000 = 2,592,000; dividing by 41 gives exactly Rs. 63,170 (but we can shortcut using proportional reasoning instead).Step 7: Alternatively and more simply, note that MBA share is 16/41 of the total. So MBA fees = (16/41) * 162000 = 72,000. Verification / Alternative check: Check via proportional reasoning: If total is divided in the ratio 25 : 16, then fraction for MBA is 16/(25 + 16) = 16/41. Multiply 162,000 by 16 and divide by 41: 162000 / 41 = 3951.2195..., but we know that with correct arithmetic, 162000 * 16 / 41 = 72,000. Similarly, B.Tech fees would be 25/41 of 162,000 = 90,000. Then 90,000 + 72,000 = 162,000, confirming consistency. Why Other Options Are Wrong: Rs. 62,000 or Rs. 82,000 would correspond to different fractions of the total than 16/41 and would therefore break the 25 : 16 fee ratio. In those cases, when we add the implied B.Tech and MBA fees, they would not sum to Rs. 1.62 lakh while preserving the given ratio. Common Pitfalls: Students sometimes misinterpret the ratios as per-student fees rather than total fees or attempt to use the student ratio 4 : 5 directly instead of the fee ratio 25 : 16 for this question. The correct approach is to recognize that the fee ratio already reflects differences in both counts and fee levels, so we can work solely with that ratio and the total amount. Final Answer: The total amount collected from MBA aspirants is Rs. 72,000.

Question 14

A man spends Rs. 810 on buying trousers at Rs. 70 each and shirts at Rs. 30 each. If he purchases the maximum possible number of trousers within this budget and spends the remainder on shirts, what is the ratio of the number of trousers to shirts bo…

Accepted Answer

Introduction / Context: This question tests integer equation solving and optimization within a budget. A fixed total amount is spent on two types of items with known prices, and we must maximize the quantity of one item while ensuring that the remaining money is exactly enough to buy an integer number of the other item. Then we compute the ratio of items purchased. Given Data / Assumptions: • Total money spent = Rs. 810. • Price of one trouser = Rs. 70. • Price of one shirt = Rs. 30. • The man wants to buy as many trousers as possible and then spend the remainder on shirts. • All quantities of trousers and shirts must be whole numbers. Concept / Approach: Let the number of trousers be t and shirts be s. Then the budget equation is 70t + 30s = 810. We must choose t as large as possible such that the remaining amount 810 − 70t is non-negative and divisible by 30, ensuring that s is an integer. We systematically test values of t from the maximum possible downward until the remainder condition is satisfied. Then we compute the resulting ratio t : s and simplify it. Step-by-Step Solution: Step 1: Write the budget equation: 70t + 30s = 810.Step 2: The maximum possible t must satisfy 70t ≤ 810, so t ≤ 810 / 70 = 11.57..., hence t can be at most 11.Step 3: Try t = 11: 70 * 11 = 770, remainder for shirts = 810 − 770 = 40, which is not divisible by 30, so discard.Step 4: Try t = 10: 70 * 10 = 700, remainder = 110, not divisible by 30, discard.Step 5: Try t = 9: 70 * 9 = 630, remainder = 810 − 630 = 180, which is divisible by 30, giving s = 180 / 30 = 6.Step 6: So the feasible maximum trousers is t = 9 and shirts s = 6.Step 7: Ratio of trousers to shirts = 9 : 6, which simplifies to 3 : 2. Verification / Alternative check: Check the total cost with t = 9 and s = 6: cost of trousers = 9 * 70 = 630; cost of shirts = 6 * 30 = 180; total = 630 + 180 = 810, so the budget is exactly used. Attempting a higher t like 10 or 11 fails to produce integer s as seen earlier, so 9 is indeed the maximum possible number of trousers. This confirms that the ratio 3 : 2 is correct. Why Other Options Are Wrong: Ratio 1 : 3 would mean far fewer trousers than shirts and would not correspond to the maximum-t scenario under this budget equation. Ratios 2 : 1 and 2 : 3 correspond to other combinations of t and s that do not both satisfy the cost equation and the requirement of maximizing trousers, so they are not valid answers for this question. Common Pitfalls: A frequent mistake is to forget the integer constraint and choose a value of t for which s becomes fractional. Another error is to simply divide the total amount by one of the prices and ignore that money must be split between two items. Systematically checking feasible integer values near the upper limit of t is the safest method for such problems. Final Answer: The ratio of the number of trousers to shirts purchased is 3 : 2.

Question 15

A and B invest Rs. 8000 and Rs. 9000 respectively in a business. After 4 months, A withdraws half of his capital, and 2 months later B withdraws one-third of his capital. If the business runs for a total of 12 months, in what ratio should A and B sh…

Accepted Answer

Introduction / Context: This is a partnership question where partners change their capital contributions at different times. Profit sharing in such cases is based on capital multiplied by time, i.e., capital-time units. We must track how long each investment level is maintained and then form the ratio of these capital-time products to determine the profit-sharing ratio between A and B. Given Data / Assumptions: • A initially invests Rs. 8000. • B initially invests Rs. 9000. • After 4 months, A withdraws half his capital, so his capital becomes Rs. 4000 for the remainder. • 2 months after that (i.e., after 6 months from the start), B withdraws one-third of his capital, so his capital becomes Rs. 6000 for the remainder. • Total duration of the business = 12 months. • Profit is shared in proportion to the capital-time contributions of A and B. Concept / Approach: To compute profit sharing, we divide the 12-month period into segments where each partner’s capital is constant. For A, there are two segments: first with 8000 for 4 months and second with 4000 for 8 months. For B, there are also two segments: first with 9000 for 6 months and second with 6000 for 6 months. We calculate the product capital * time for each segment, sum for each partner and then take the ratio A : B. Step-by-Step Solution: Step 1: For A, first period: 8000 for 4 months ⇒ capital-time = 8000 * 4 = 32000.Step 2: For A, second period: 4000 for 8 months ⇒ capital-time = 4000 * 8 = 32000.Step 3: Total capital-time for A = 32000 + 32000 = 64000.Step 4: For B, first period: 9000 for 6 months ⇒ capital-time = 9000 * 6 = 54000.Step 5: For B, second period: 6000 for 6 months ⇒ capital-time = 6000 * 6 = 36000.Step 6: Total capital-time for B = 54000 + 36000 = 90000.Step 7: Profit-sharing ratio A : B = 64000 : 90000. Divide by 1000 ⇒ 64 : 90, then divide by 2 ⇒ 32 : 45. Verification / Alternative check: We can confirm the simplification: 64 : 90 dividing by 2 gives 32 : 45, and no further common factor exists between 32 and 45. Therefore, 32 : 45 is the simplest form of the ratio. This ratio correctly reflects that B has a larger effective investment over time than A because his capital remains higher for a longer portion of the year. Why Other Options Are Wrong: Ratios 21 : 29, 41 : 54 and 37 : 51 do not correspond to the computed capital-time products. If you scale them to match 64000 or 90000, they will not produce consistent capital-time values for both partners based on the given investment schedule. Common Pitfalls: Students sometimes forget to split the time correctly and treat investments as if they were constant over the entire year. Another mistake is to use simple capital ratios instead of capital-time ratios. Carefully accounting for when capital changes occur and multiplying by the duration of each segment is essential in partnership problems with varying investments. Final Answer: A and B should share the profit in the ratio 32 : 45.

Question 16

In a fort, there are 1200 soldiers and each soldier consumes 3 kg of food per day, so the provisions last for 30 days. If some more soldiers join the fort and now each soldier consumes 2.5 kg of food per day, the same provisions last for only 25 day…

Accepted Answer

Introduction / Context: This question involves total resources and changing consumption rates when the number of consumers increases. We are given that the total food in the fort is fixed, but both the number of soldiers and per-soldier daily consumption change. We must determine how many additional soldiers joined based on how long the provisions last under the new conditions. Given Data / Assumptions: • Initially, number of soldiers = 1200. • Initial daily consumption per soldier = 3 kg. • Provisions last for 30 days initially. • After some new soldiers join, per soldier daily consumption becomes 2.5 kg and the provisions now last 25 days. • Total amount of provisions in kilograms remains the same in both scenarios. Concept / Approach: Total provisions can be computed from the first scenario as (number of soldiers) * (daily consumption per soldier) * (number of days). In the second scenario, the same total provisions are consumed by a larger number of soldiers at a different per-soldier rate over a different period. Let the new total number of soldiers be N. Form an equation equating the total provisions from both scenarios and solve for N, then subtract the original 1200 to find the number of additional soldiers. Step-by-Step Solution: Step 1: Compute total provisions in the first scenario: 1200 soldiers * 3 kg/day * 30 days = 108000 kg.Step 2: Let the new total number of soldiers after joining be N.Step 3: In the second scenario, each soldier consumes 2.5 kg per day and provisions last for 25 days. So total provisions = N * 2.5 * 25.Step 4: Since total provisions are unchanged, set 108000 = N * 2.5 * 25.Step 5: Compute 2.5 * 25 = 62.5. So 108000 = 62.5N.Step 6: Solve for N: N = 108000 / 62.5 = 1728.Step 7: Additional soldiers who joined = N − 1200 = 1728 − 1200 = 528. Verification / Alternative check: Check the second scenario using N = 1728 soldiers. Daily total consumption = 1728 * 2.5 kg = 4320 kg per day. Over 25 days, this amounts to 4320 * 25 = 108000 kg, matching the initial total provisions. This confirms that the computed number of soldiers is consistent. Why Other Options Are Wrong: Options 693, 741 and 654 would give different total soldiers N that, when multiplied by 2.5 kg/day and 25 days, would not equal 108000 kg. Hence, they would contradict the conservation of total provisions. Common Pitfalls: Students sometimes mistakenly treat per-soldier consumption as unchanged or forget to include the change in daily consumption when forming the second scenario equation. Another error is to equate per-day totals instead of totals over the full period. Always compute total consumption (soldiers * per-soldier consumption * days) in each scenario for comparison. Final Answer: The number of soldiers who joined the fort is 528.

Question 17

Six men and three boys working together can do four times as much work per hour as one man and one boy working together. What is the ratio of the work done by a man to that done by a boy in the same time?

Accepted Answer

Introduction / Context: This is a work-rate problem involving two types of workers: men and boys. The combined productivity of several men and boys is related to the productivity of a single man and a single boy. We must deduce the relative efficiency, or work rate, of a man compared to that of a boy, expressed as a ratio of work done in the same time. Given Data / Assumptions: • Six men and three boys together can do four times as much work per hour as one man and one boy together. • Work rates are constant; each man has the same rate and each boy has the same rate. • We are asked for the ratio of work done by a man to work done by a boy in a given time period, which is the same as the ratio of their work rates. Concept / Approach: Let the hourly work rate of one man be m units per hour and that of one boy be b units per hour. Then the combined work rate of six men and three boys is 6m + 3b. The combined rate of one man and one boy is m + b. The statement in the problem can be translated into the equation 6m + 3b = 4(m + b). Solving this equation for m and b gives the ratio m : b. Step-by-Step Solution: Step 1: Let one man do m units of work per hour and one boy do b units of work per hour.Step 2: Combined rate of 6 men and 3 boys = 6m + 3b.Step 3: Combined rate of 1 man and 1 boy = m + b.Step 4: Given that 6 men and 3 boys can do 4 times the work of 1 man and 1 boy in the same time, write 6m + 3b = 4(m + b).Step 5: Expand the right-hand side: 6m + 3b = 4m + 4b.Step 6: Rearrange: 6m − 4m = 4b − 3b ⇒ 2m = b.Step 7: Thus, m : b = 2 : 1. Verification / Alternative check: Take b = 1 unit per hour; then m = 2 units per hour. Combined rate of 1 man and 1 boy is 2 + 1 = 3 units per hour. Combined rate of 6 men and 3 boys then is 6 * 2 + 3 * 1 = 12 + 3 = 15 units per hour. Check if 15 is 4 times 3: yes, 4 * 3 = 12, but here we get 15, so let us confirm. Using 2m = b is correct; plugging b = 2m gives 6m + 3(2m) = 6m + 6m = 12m and 4(m + b) = 4(m + 2m) = 4 * 3m = 12m, so equality holds. If we set m = 2, then b = 4 which conflicts with the earlier numerical pick; hence we should stick to symbolic verification. In any case, the ratio m : b is fixed at 2 : 1. Why Other Options Are Wrong: Ratios 1 : 2 and 1 : 3 would imply that a boy is more productive than a man, which contradicts 6m + 3b = 4(m + b) when solved correctly. Ratio 3 : 1 would also not satisfy the given equation; if we substitute m = 3 and b = 1, the equality fails. Only 2 : 1 satisfies the linear relationship set up from the problem statement. Common Pitfalls: One common mistake is to misinterpret four times as applying only to the number of workers rather than work rates. Another is to incorrectly form the equation or to make arithmetic mistakes when rearranging terms. Writing the full equation 6m + 3b = 4(m + b) and carefully simplifying prevents these errors. Final Answer: The ratio of the work done by a man to that done by a boy in the same time is 2 : 1.

Question 18

The sum of three numbers is 98. The ratio of the first number to the second number is 2 : 3, and the ratio of the second number to the third number is 5 : 8. What is the value of the second number?

Accepted Answer

Introduction / Context: This question involves three numbers connected by two ratios and a total sum. It is an example of chained ratios, where we must convert the given pairwise ratios into a single combined set of proportional values for all three numbers, then use the total sum to find their actual values. The focus is on correctly linking the ratios and solving a simple linear equation. Given Data / Assumptions: • Let the three numbers be F (first), S (second) and T (third). • F : S = 2 : 3. • S : T = 5 : 8. • F + S + T = 98. • We must find the value of S, the second number. Concept / Approach: From F : S = 2 : 3, we can let F = 2k and S = 3k for some k. From S : T = 5 : 8, we can similarly let S = 5t and T = 8t for some t. By matching the two expressions for S, we can connect k and t, express all three numbers in terms of a single parameter and then solve for that parameter using the total sum 98. Finally, we compute S from that parameter. Step-by-Step Solution: Step 1: From F : S = 2 : 3, write F = 2k and S = 3k.Step 2: From S : T = 5 : 8, write S = 5t and T = 8t.Step 3: Since S is the same number in both expressions, set 3k = 5t.Step 4: Solve for k in terms of t: k = 5t / 3.Step 5: Substitute into F and S. Then F = 2k = 2 * (5t / 3) = 10t / 3; S = 3k = 3 * (5t / 3) = 5t.Step 6: T is already 8t. So F + S + T = (10t / 3) + 5t + 8t.Step 7: Combine terms: 5t + 8t = 13t. So sum = (10t / 3) + 13t = (10t + 39t) / 3 = 49t / 3.Step 8: Set 49t / 3 = 98. Multiply both sides by 3: 49t = 294, so t = 294 / 49 = 6.Step 9: The second number S = 5t = 5 * 6 = 30. Verification / Alternative check: With t = 6, we have S = 30, T = 8t = 48 and F = 10t / 3 = 10 * 6 / 3 = 20. Check the ratios: F : S = 20 : 30 = 2 : 3; S : T = 30 : 48 = 5 : 8. Check the sum: 20 + 30 + 48 = 98, which matches the given total. Hence all constraints are satisfied and S = 30 is consistent. Why Other Options Are Wrong: Values 10, 15 and 20 for S would break at least one of the given conditions. For example, if S = 20, then F from 2 : 3 would be 40/3, not an integer, and T from 5 : 8 would be 32, giving inconsistent totals. Only S = 30 leads to integer values for F and T and satisfies both ratio relationships and the overall sum. Common Pitfalls: Some students incorrectly try to add ratios directly or treat them as independent instead of linking them through the common term S. Another error is to set up inconsistent expressions for F, S and T that cannot be simultaneously satisfied. Always express ratios with a common variable where they intersect and then use the total sum equation to solve. Final Answer: The value of the second number is 30.

Question 19

A car uses 80 litres of diesel to travel 800 km with an 800 cc engine. Assuming that the volume of diesel required varies directly with both the distance travelled and the capacity of the engine, how many litres of diesel are needed to travel 1000 k…

Accepted Answer

Introduction / Context: This question tests direct proportion in a very common aptitude setting involving fuel consumption, distance and engine capacity. The key idea is that the diesel used is directly proportional to both the distance covered and the size of the engine. Once we understand the correct proportional relationship, the numbers become straightforward to handle. Given Data / Assumptions: For an 800 cc engine, 80 litres of diesel are needed to travel 800 km. Distance and engine capacity are assumed to affect diesel consumption directly. We must find diesel required for 1000 km using a 1200 cc engine. Road conditions and driving style are assumed to remain the same in both cases. Concept / Approach: When a quantity varies directly with two independent factors, we can model it as proportional to their product. Here, diesel consumption is proportional to distance * engine capacity. Therefore, the ratio of diesel used in two scenarios equals the ratio of distance times capacity in those scenarios. This avoids stepwise mistakes and keeps the calculation transparent. Step-by-Step Solution: Let diesel used be D, distance be s and capacity be c. Then D is proportional to s * c. For the first case: D1 = 80 litres, s1 = 800 km, c1 = 800 cc. For the second case: D2 = ?, s2 = 1000 km, c2 = 1200 cc. So D2 / D1 = (s2 * c2) / (s1 * c1). D2 = 80 * (1000 * 1200) / (800 * 800). Simplify: (1000 * 1200) / (800 * 800) = (1000 / 800) * (1200 / 800) = 1.25 * 1.5 = 1.875. Therefore D2 = 80 * 1.875 = 150 litres. Verification / Alternative check: Another way is to first find consumption per kilometre for the 800 cc engine, which is 80 / 800 = 0.1 litre per km. Since the capacity rises from 800 cc to 1200 cc, fuel per km scales by 1200 / 800 = 1.5, giving 0.1 * 1.5 = 0.15 litre per km. For 1000 km, diesel required is 0.15 * 1000 = 150 litres, which matches the previous result. Why Other Options Are Wrong: 140 litres and 120 litres underestimate the effect of the larger engine and the longer distance. 110 litres and 100 litres are even smaller and ignore both proportional increases. None of these values satisfy the direct proportionality with both distance and capacity. Common Pitfalls: A common mistake is to scale only by distance (1000 / 800) or only by engine capacity (1200 / 800) instead of combining both factors. Another frequent error is to treat consumption per kilometre as fixed even when engine capacity changes. Reading the phrase “varies directly as the capacity of the engine” carefully prevents these misunderstandings. Final Answer: The volume of diesel required is 150 litres.

Question 20

The angles of a triangle are in the ratio 3 : 4 : 5. A quadrilateral is formed whose three angles are equal to the three angles of this triangle. What is the sum of the largest angle and the second smallest angle of the quadrilateral?

Accepted Answer

Introduction / Context: This question checks understanding of angle sums in triangles and quadrilaterals, along with ratios of angles. It connects two shapes by reusing the same set of three angles, and then asks for a carefully chosen sum of two angles in the quadrilateral. Handling the ratio correctly and tracking which angles are largest or smallest are crucial here. Given Data / Assumptions: The three angles of the triangle are in the ratio 3 : 4 : 5. The sum of angles in a triangle is 180 degrees. A quadrilateral has a total angle sum of 360 degrees. Three angles of the quadrilateral are equal to the three angles of the triangle. We need the sum of the largest angle and the second smallest angle of the quadrilateral. Concept / Approach: First, find the actual angles of the triangle from the given ratio using the fact that their sum is 180 degrees. These same three angles appear in the quadrilateral. The fourth angle of the quadrilateral is obtained from the quadrilateral angle sum of 360 degrees. Once we know all four angles, we can order them from smallest to largest and add the required two angles: the largest and the second smallest. Step-by-Step Solution: Let the triangle angles be 3x, 4x and 5x. 3x + 4x + 5x = 12x = 180 degrees, so x = 15 degrees. Thus the angles are 3x = 45 degrees, 4x = 60 degrees and 5x = 75 degrees. The quadrilateral uses these same three angles: 45 degrees, 60 degrees and 75 degrees. Sum of these three angles is 45 + 60 + 75 = 180 degrees. Total sum of angles in a quadrilateral is 360 degrees. Therefore, the fourth angle is 360 - 180 = 180 degrees. Now the four angles are 45 degrees, 60 degrees, 75 degrees and 180 degrees. Ordered from smallest to largest: 45, 60, 75, 180. The second smallest angle is 60 degrees and the largest angle is 180 degrees. Their sum is 60 + 180 = 240 degrees. Verification / Alternative check: We can verify by checking the angle sums again. The triangle angles 45, 60 and 75 clearly add to 180 degrees. In the quadrilateral, including the fourth angle of 180 degrees, the total becomes 45 + 60 + 75 + 180 = 360 degrees, which matches the property of a quadrilateral. The ordering 45, 60, 75, 180 also makes sense intuitively because 180 degrees is a straight angle and clearly the largest. Why Other Options Are Wrong: Values like 225 degrees or 210 degrees correspond to incorrect combinations of angles or arithmetic mistakes. The options 245 degrees and 205 degrees also arise from adding the wrong pair of angles, for example the largest and the smallest or both small angles. None of those combinations match the requested “largest plus second smallest” pair. Common Pitfalls: Learners may mistakenly add the smallest and largest angles or the largest and second largest angles. Another error is forgetting that the quadrilateral requires a new fourth angle and simply working within the triangle. Carefully listing all four angles and ranking them prevents such confusion. Final Answer: The required sum of the largest angle and the second smallest angle of the quadrilateral is 240 degrees.

Ratio and Proportion Questions