Question 1

Equal amounts of water were poured into two empty jars of different capacities, which made one jar $\frac{1}{4}$ full and the other jar $\frac{1}{3}$ full. If the water in the jar with lesser capacity is, then poured into the jar with the greater ca…

Accepted Answer

Concept & Logic: Equal quantities of volume fill different ratios of two distinct capacities. The jar that becomes (1/3) full with the same water must have a smaller capacity than the jar that becomes (1/4) full. Thus, the jar that is (1/4) full is the larger jar. Step-by-Step Solution: • Given: • Volume of water added to both jars is identical (W). • Jar 1 (Larger Capacity C_1): (1/4) full \implies W = (1/4)C_1 \implies C_1 = 4W • Jar 2 (Smaller Capacity C_2): (1/3) full \implies W = (1/3)C_2 \implies C_2 = 3W • Calculation: • The question states that the water from the smaller capacity jar (C_2) is poured into the larger capacity jar (C_1). • Note that the smaller jar already contained exactly W amount of water. • The larger jar also already contained exactly W amount of water. • When combined into the larger jar, the total volume of water becomes W + W = 2W. • The fraction of the larger jar filled is: Fraction = \frac{Total Volume of Water}{Capacity of Larger Jar} = (2W/C_1) = (2W/4W) = (1/2) Exam Strategy & Shortcut: Direct Value Trick: Assume the equal water amount poured into both jars is 1 unit. • Larger jar capacity = 4 units (since 1 unit fills (1/4) of it). • Smaller jar capacity = 3 units (since 1 unit fills (1/3) of it). • Combining both water contents means we have 1 + 1 = 2 units of water inside the larger jar. • Fraction of larger jar filled = (2/4) = (1/2). Common Pitfall: A common point of confusion is parsing "poured into the jar with the greater capacity". Students sometimes assume the smaller jar's total volume capacity (3W) is being poured over, or they forget that the larger jar already contains its initial portion of water. Final Answer: Therefore, the correct answer is 1/2.

Question 2

Mr. X spends $\frac{1}{4}$ of his income on food and $\frac{1}{3}$ less than what he spent on food, on the education of his children. What fraction of his income did he spend on food and education?

Accepted Answer

Concept & Formula: The problem requires computing the total fraction spent by summing up the relative component allocations. The key structural nuance is parsing "(1/3) less than what he spent on food". Total Fraction = Fraction_{food} + Fraction_{education} Step-by-Step Solution: • Given: • Fraction spent on food = (1/4) • Fraction spent on education = (1/4) - ((1/3) × (1/4)) • Calculation: • Find the education expenditure fraction: Education Expenditure = (1/4) × (1 - (1/3)) = (1/4) × (2/3) = (1/6) • Sum the expenditure fractions for food and education: Total Spent = (1/4) + (1/6) • The lowest common multiple of 4 and 6 is 12: Total Spent = (3/12) + (2/12) = (5/12) Exam Strategy & Shortcut: Variable Multiplier Option: Let Mr. X's total income be 12 units (LCM of 4 and 3). • Amount spent on food = (1/4) × 12 = 3 units. • Amount spent on education is (1/3) less than food = 3 - ((1/3) × 3) = 3 - 1 = 2 units. • Combined spending = 3 + 2 = 5 units. • Required fraction = (5/12). Common Pitfall: Misinterpreting "(1/3) less than what he spent on food" as a absolute arithmetic deduction from overall income (i.e., doing (1/4) - (1/3)), which produces negative results, or subtracting it directly from the whole. It means a reduction relative to the value of the food allocation itself. Final Answer: Therefore, the correct answer is 5/12.

Question 3

A body of 7300 troops is formed of 4 battalions so that $\frac{1}{2}$ of the first, $\frac{2}{3}$ of the second, $\frac{3}{4}$ of the third and $\frac{4}{5}$ of the fourth are all composed of the same number of men. How many men are there in the sec…

Accepted Answer

Concept & Formula: When fractional parts of multiple independent variables are equivalent, we equate them to a constant scaling metric factor k to express each variable relative to a single shared parameter. If (1/2)A = (2/3)B = (3/4)C = (4/5)D = k, then: A = 2k, \quad B = (3/2)k, \quad C = (4/3)k, \quad D = (5/4)k Step-by-Step Solution: • Given: • Total troops = 7300 • Let the sizes of the four battalions be B_1, B_2, B_3, B_4. • (1/2)B_1 = (2/3)B_2 = (3/4)B_3 = (4/5)B_4 = k • Calculation: • Express each battalion in terms of k: B_1 = 2k B_2 = (3/2)k B_3 = (4/3)k B_4 = (5/4)k • Sum the battalions up to equate to the total population: 2k + (3/2)k + (4/3)k + (5/4)k = 7300 • Find the common denominator for the fractions (LCM of 2, 3, 4 is 12): ( (24 + 18 + 16 + 15/12) ) k = 7300 (73/12)k = 7300 \implies k = 1200 • Compute the size of the second battalion (B_2): B_2 = (3/2) × 1200 = 1800 Exam Strategy & Shortcut: Write down the ratios of the battalions directly by reciprocating the coefficients: B_1 : B_2 : B_3 : B_4 = 2 : (3/2) : (4/3) : (5/4) Multiply through by the LCM (12) to clear out fractional remainders: B_1 : B_2 : B_3 : B_4 = 24 : 18 : 16 : 15 Sum of ratio parts = 24 + 18 + 16 + 15 = 73. Since total troops = 7300, 1 unit value = (7300/73) = 100. Second battalion size = 18 × 100 = 1800. Common Pitfall: A standard error is failing to invert the fractional coefficients properly when shifting expressions into proportional value components, leading to an incorrect aggregate sum equation. Final Answer: Therefore, the correct answer is 1800.

Question 4

After reading $\frac{3}{5}$ of the biology homework on Monday night, Sanjay read $\frac{1}{3}$ of the remaining homework on Tuesday night. What fraction of the original homework would Sanjay have to read on Wednesday night to complete the assignment?

Accepted Answer

Concept & Strategy: This is a sequential remainder tracking problem. We establish the left-over work iteratively after each night's individual contribution until reaching Wednesday night. Step-by-Step Solution: • Given: • Fraction read on Monday = (3/5) • Calculation: • Remaining homework after Monday: 1 - (3/5) = (2/5) • Fraction read on Tuesday = (1/3) of the remaining homework: Tuesday Read = (1/3) × (2/5) = (2/15) • Fraction left for Wednesday night = Remaining after Monday - Tuesday Read Wednesday Remaining = (2/5) - (2/15) • Convert to a shared denominator: Wednesday Remaining = (6/15) - (2/15) = (4/15) Exam Strategy & Shortcut: Remainder Factor Multiplication: Each step leaves a specific fractional balance behind: • If Sanjay reads (3/5) on Monday, the remainder factor is (1 - (3/5)) = (2/5). • If he reads (1/3) of that balance on Tuesday, the fraction left for Wednesday is (1 - (1/3)) = (2/3) of the previous remainder. • Total fraction remaining for Wednesday = (2/3) × (2/5) = (4/15). Common Pitfall: Mistakenly treating Tuesday's consumption fraction as an absolute subtraction from the absolute start point (e.g., calculating 1 - (3/5) - (1/3) = (1/15)). Ensure you evaluate expressions dependent on the phrase "of the remaining". Final Answer: Therefore, the correct answer is 4/15.

Question 5

The highest score in an inning was $\frac{3}{11}$ of the total and the next highest was $\frac{3}{11}$ of the remainder. If the scores differed by 9, the total score was

Accepted Answer

Concept & Strategy: This problem requires formulating an equation for successive remainders. By defining the total score as an algebraic variable or choosing a perfect denominator square value, we can model the relationship between the highest and second-highest scores. Step-by-Step Solution: • Given: • Highest score = (3/11) of Total Score • Next highest score = (3/11) of Remainder Score • Difference between these two scores = 9 • Calculation: • Let the total score be S. • Highest Score = (3/11)S • Remainder = S - (3/11)S = (8/11)S • Next Highest Score = (3/11) × ((8/11)S) = (24/121)S • Setup the difference equation: (3/11)S - (24/121)S = 9 • Equate the denominators by scaling up (3/11) to (33/121): (33/121)S - (24/121)S = 9 (9/121)S = 9 \implies S = 121 Exam Strategy & Shortcut: Intelligent Choice of Variable Multiplier: Since the fraction (3/11) is applied twice sequentially, let the total score be 11 × 11 = 121 units. • Highest score = (3/11) × 121 = 33 units. • Remainder = 121 - 33 = 88 units. • Next highest score = (3/11) × 88 = 24 units. • Difference between the two scores = 33 - 24 = 9 units. • Since the actual difference given is exactly 9, our assumed base scale value matches perfectly. Therefore, total score = 121. Common Pitfall: A standard error is taking the second highest score as (3/11) of the total score instead of the remainder, resulting in an incorrect difference step. Always isolate the phrase "of the remainder". Final Answer: Therefore, the correct answer is 121.

Question 6

Ganeshi's monthly income is twice Jassi's monthly income. Two-third of Jassi's monthly income is equal to Sukhvinder's monthly income. If Sukhvinder's annual income is ₹ 2.34 lakhs, what is Ganeshi's monthly income?

Accepted Answer

Concept & Strategy: The problem links three distinct entities through consecutive multipliers. The critical operational task is maintaining consistency between annual units and monthly units before performing the substitutions. Step-by-Step Solution: • Given: • Ganeshi's monthly income (G) = 2 × Jassi's monthly income (J) • Sukhvinder's monthly income (S) = (2/3) × Jassi's monthly income (J) • Sukhvinder's annual income = ₹ 2,34,000 • Calculation: • Step 1: Find Sukhvinder's monthly income (S): S = (2,34,000/12) = ₹ 19,500 • Step 2: Use the relation between S and J to find J: 19,500 = (2/3) × J \implies J = 19,500 × (3/2) = 9,750 × 3 = ₹ 29,250 • Step 3: Compute Ganeshi's monthly income (G): G = 2 × J = 2 × 29,250 = ₹ 58,500 Exam Strategy & Shortcut: Combine the variable equations down into a direct structural ratio layout: (G/S) = (2J/\frac{2){3}J} = 2 × (3/2) = 3 This directly tells us that Ganeshi's monthly income is exactly 3 times Sukhvinder's monthly income. • Sukhvinder's monthly income = (2,34,000/12) = 19,500. • Ganeshi's monthly income = 19,500 × 3 = 58,500. Common Pitfall: A common mistake is using the annual figure (2.34 lakhs) directly without converting it to a monthly scale first, leading to massive scale mismatches in the options. Final Answer: Therefore, the correct answer is ₹ 58500.

Question 7

The cost of 36 kg of rice last year was ₹ 1044 and the cost of 24 kg of rice this year is ₹ 768. What is the difference between the cost per kg of rice last year and the cost per kg of rice this year?

Accepted Answer

Concept & Formula: This problem requires computing individual unit rates (cost per kg) for two distinct periods and finding their arithmetic variance. Unit Cost = \frac{Total Cost}{Total Quantity} Step-by-Step Solution: • Given: • Last year: Cost of 36 kg = ₹ 1044 • This year: Cost of 24 kg = ₹ 768 • Calculation: • Find the cost per kg last year: Cost per kg_{last year} = (1044/36) = ₹ 29 • Find the cost per kg this year: Cost per kg_{this year} = (768/24) = ₹ 32 • Calculate the absolute difference between the two unit prices: Difference = 32 - 29 = ₹ 3 Exam Strategy & Shortcut: Simplify through rapid division: • (1044/36): Divide both parts by 9 first \rightarrow (116/4) = 29. • (768/24): Divide both parts by 8 first \rightarrow (96/3) = 32. • Subtract directly: 32 - 29 = 3. Common Pitfall: Avoid mixed combinations, such as subtracting quantities or total values before dividing. The unit rates must be evaluated separately before applying the subtraction. Final Answer: Therefore, the correct answer is ₹ 3.

Question 8

The cost of 5 pendants and 8 chains is ₹ 1,45,785. What would be the cost of 15 pendants and 24 chains?

Accepted Answer

Concept & Strategy: Similar to other simultaneous expression variants, we look for a linear transformation scalar. If the desired item mix is a direct multiple of the baseline set, we multiply the total cost by that factor. Step-by-Step Solution: • Given: • Let a pendant be p and a chain be c. • 5p + 8c = 1,45,785 • Calculation: • We need to determine the cost of 15p + 24c. • Factor out the shared scale factor from the target equation: 15p + 24c = 3 × (5p + 8c) • Substitute the known value into the expression: Cost = 3 × 1,45,785 Cost = 4,37,355 Exam Strategy & Shortcut: • Detect that 5 × 3 = 15 and 8 × 3 = 24. • The price must be multiplied by exactly 3. • Estimate through unit digit multiplication or rounding: 3 × 5 = 15 \implies ending digit must be 5. • 3 × 1,45,000 \approx 4,35,000. This instantly points towards option (b). Common Pitfall: Do not fall for the distractor trap "Cannot be determined". While you cannot find the individual cost of a single pendant or a single chain, the collective group price is completely determinable. Final Answer: Therefore, the correct answer is ₹ 4,37,355.

Question 9

The cost of 13 kg of sugar is ₹ 195, the cost of 17 kg of rice is ₹ 544 and the cost of 21 kg of wheat is ₹ 336. What is the total cost of 21 kg of sugar, 26 kg of rice and 19 kg of wheat?

Accepted Answer

Concept & Formula\n\nThis problem uses the unitary method, where we first find the price per unit (kilogram) for each commodity and then multiply it by the required quantities.\n\Unit Price = \frac{\Total Cost}{\Total Quantity}\n\n### Step-by-Step Solution\n\n* Given:\n * 13 kg of sugar = ₹ 195\n * 17 kg of rice = ₹ 544\n * 21 kg of wheat = ₹ 336\n\n* Calculation of Unit Prices:\n * Cost of 1 kg of sugar = \(195/13) = ₹ 15\n * Cost of 1 kg of rice = \(544/17) = ₹ 32\n * Cost of 1 kg of wheat = \(336/21) = ₹ 16\n\n* Calculation of New Total Cost:\n * Cost of 21 kg of sugar = 21 \× 15 = 315\n * Cost of 26 kg of rice = 26 \× 32 = 832\n * Cost of 19 kg of wheat = 19 \× 16 = 304\n \n \Total Cost = 315 + 832 + 304 = 1451\n\n### Exam Strategy & Shortcut\n\nUse unit digit analysis after finding the unit rates to save calculation time under pressure:\n* Sugar: 1 \× 5 \rightarrow 5\n* Rice: 6 \× 2 \rightarrow 2\n* Wheat: 9 \× 6 \rightarrow 4\n\nSum of unit digits = 5 + 2 + 4 = 11, so the last digit must be 1. Looking at the options, only ₹ 1451 ends in 1.\n\n### Common Pitfall\n\nErrors usually arise from simple division mistakes when calculating the individual per-kg rates (e.g., 544 \div 17). Practice tables up to 20 to ensure fast and error-free mental division.\n\n### Final Answer\n\nTherefore, the correct answer is ₹ 1451.:

Question 10

There are 200 questions on a 3 hour examination. Among these questions are 50 Mathematics problems. It is suggested that twice as much time be spent on each Maths problem as for each other question. How many minutes should be spent on Mathematics pr…

Accepted Answer

Concept & Formula\n\nThis problem requires setting up a linear equation based on time distribution. Total time available must equal the combined time spent on math questions and non-math questions.\n\Total Time = (N_{\math} \× T_{\math}) + (N_{\other} \× T_{\other})\n\n### Step-by-Step Solution\n\n* Given:\n * Total questions = 200\n * Mathematics questions = 50\n * Other questions = 200 - 50 = 150\n * Total exam time = 3 hours = 3 \× 60 = 180 minutes\n\n* Deduction:\n Let the time spent on each non-math question be x minutes.\n Therefore, the time spent on each mathematics problem is 2x minutes.\n\n* Equation Setup:\n (50 \× 2x) + (150 \× x) = 180\n 100x + 150x = 180\n 250x = 180\n x = \(180/250) = \(18/25) minutes\n\n* Target Calculation:\n We need the total time spent on Mathematics problems, which is 50 \× 2x = 100x:\n \Total Maths Time = 100 \× \(18/25) = 4 \× 18 = 72 \ minutes\n\n### Exam Strategy & Shortcut\n\nUse the ratio method for time distribution:\n* Total relative weight for Maths = 50 \× 2 = 100\n* Total relative weight for Others = 150 \× 1 = 150\n* Total weight ratio = 100 : 150 = 2 : 3\n\nMaths takes \(2/2+3) = \(2/5) of the total exam time:\n\Maths Time = \(2/5) \× 180 \ minutes = 2 \× 36 = 72 \ minutes\n\n### Common Pitfall\n\nStudents often solve for x (time per non-math question) or 2x (time per math question) and select that as the final answer, forgetting that the question asks for the total time spent on all 50 Mathematics problems.\n\n### Final Answer\n\nTherefore, the correct answer is 72.:

Question 11

A class starts at 10 a.m. and lasts till 1.27 p.m. Four periods are held during this interval. After every period, 5 minutes are given free to the students. The exact duration of each period is

Accepted Answer

Concept & Logic\n\nTo find the exact duration of a single period, determine the total duration of the school session, subtract the total interval time allocated for breaks, and divide the remaining time equally by the number of periods.\n\Total Period Time = \Total Duration - \Total Break Time\n\n### Step-by-Step Solution\n\n* Calculate Total Duration:\n From 10:00 a.m. to 1:27 p.m. is 3 hours and 27 minutes.\n \Total minutes = (3 \× 60) + 27 = 180 + 27 = 207 \ minutes\n\n* Calculate Total Break Time:\n There are 4 periods. Breaks occur *between* periods. Therefore, there are exactly 3 breaks (after the 1st, 2nd, and 3rd periods).\n \Total break time = 3 \× 5 = 15 \ minutes\n\n* Calculate Duration of Each Period:\n \Time for periods = 207 - 15 = 192 \ minutes\n \Duration of one period = \(192/4) = 48 \ minutes\n\n### Exam Strategy & Shortcut\n\nBe absolutely precise on the number of breaks: N periods arranged linearly will always have exactly N - 1 breaks between them. Once you arrive at 192 minutes, a quick glance at the options shows that 4 \× 50 = 200, so the answer must be slightly less than 50, pointing directly to 48.\n\n### Common Pitfall\n\nMany students assume there are 4 breaks for 4 periods, calculating 4 \× 5 = 20 minutes of free time. This leads to 207 - 20 = 187, which is not cleanly divisible by 4, causing confusion and wasted time.\n\n### Final Answer\n\nTherefore, the correct answer is 48 minutes.:

Question 12

A woman works continuously for 7 days in the kitchen and refuses to work on the 8th day when her husband takes over. If she starts her work on a Sunday, then the 10th time she rests will be on which day of the week?

Accepted Answer

Concept & Logic\n\nThe schedule follows a strict recurring cycle. The woman works for 7 days and rests on the 8th day. Thus, each cycle consists of exactly 8 days, and the rest day always occurs on the last day of that cycle.\n\n### Step-by-Step Solution\n\n* Identify the cycle:\n * 1 cycle = 7 working days + 1 rest day = 8 days.\n * The rest day of any cycle n occurs at the end of n complete cycles.\n\n* Find total days elapsed for 10 rest periods:\n To reach her 10th rest day, she needs to complete exactly 10 full cycles.\n \Total days = 10 \× 8 = 80 \ days\n\n* Determine the day of the week:\n * She starts her first cycle on a Sunday (Day 1).\n * The 80th day from the start of the sequence can be found using modular arithmetic (odd days).\n \Number of weeks = \(80/7) = 11 \ weeks with a remainder of 3 \ days\n \n Since Day 1 is Sunday, we count forward:\n * Remainder 1 = Sunday\n * Remainder 2 = Monday\n * Remainder 3 = Tuesday\n\n### Exam Strategy & Shortcut\n\nEach cycle is 8 days long. Since 8 \equiv 1 \pmod 7, each cycle shifts the rest day by exactly 1 day of the week forward relative to the cycle's start day cycle positioning.\n* Cycle 1 starts on Sunday \rightarrow Rest day is Day 8 (8 \equiv 1 \rightarrow Sunday).\n* Since each 8-day cycle advances the week day calendar position by 1 day (8 - 7 = 1):\n* 10 cycles will advance the rest day by 10 \× 1 = 10 days from the base pattern baseline.\n* 10 days \equiv 3 days forward from Sunday \rightarrow Monday, Tuesday.\n\n### Common Pitfall\n\nStudents often mistakenly calculate the total days as 10 \× 7 + 10 = 80, but then add 80 days *to* Sunday (making Sunday = 0), which results in a off-by-one calendar day error. Remember that Day 1 itself is Sunday.\n\n### Final Answer\n\nTherefore, the correct answer is Tuesday.:

Question 13

If $a * b = 2a - 3b + ab$, then $3 * 5 + 5 * 3$ is equal to

Accepted Answer

Concept & Logic: This problem requires evaluating a non-commutative binary operation for two different pairs of inputs and adding their results. Because the operation is not commutative (a * b 
eq b * a), you must calculate each term separately. Step-by-Step Solution: • Given: a * b = 2a - 3b + ab • Step 1: Calculate 3 * 5 Substitute a = 3 and b = 5: 3 * 5 = 2(3) - 3(5) + (3)(5) 3 * 5 = 6 - 15 + 15 3 * 5 = 6 • Step 2: Calculate 5 * 3 Substitute a = 5 and b = 3: 5 * 3 = 2(5) - 3(3) + (5)(3) 5 * 3 = 10 - 9 + 15 5 * 3 = 16 • Step 3: Add the results (3 * 5) + (5 * 3) = 6 + 16 = 22 Exam Strategy & Shortcut: Notice that the +ab term is commutative and will be the same for both 3 * 5 and 5 * 3. You can group the expression algebraically before computing fully: (2a_1 - 3b_1 + a_1b_1) + (2a_2 - 3b_2 + a_2b_2) where the pairs are reversed. Sum = 2(3+5) - 3(5+3) + 2(15) = 2(8) - 3(8) + 30 = 16 - 24 + 30 = 22. This algebraic grouping can sometimes be faster than computing two separate arithmetic chains. Common Pitfall: Assuming the operation is commutative (3 * 5 = 5 * 3) and just calculating one side and doubling it. Always check if the formula is symmetrical. Here, 2a - 3b is asymmetrical, so the terms must be calculated independently. Final Answer: Therefore, the correct answer is 22.

Question 14

Directions: Each question below consists of a number. Your task is to follow the following steps in order to find the result from the given alternatives.  Step I: Multiply the number by 3 and add the square of the number.  Step II: Divide the result…

Accepted Answer

Concept & Logic: This is a sequential algorithm problem. You must take an initial input value and pass it through a series of mathematical transformations, using the output of each step as the strict input for the next. Step-by-Step Solution: • Given Input: x = 6 • Step I: Multiply the number by 3 and add the square of the number. Result_1 = 3(6) + 6^2 Result_1 = 18 + 36 = 54 • Step II: Divide the result obtained after Step I by 5 and add 6. Result_2 = (54/5) + 6 Result_2 = 10.8 + 6 = 16.8 • Step III: Divide the result obtained after Step II by 3 and then square the resulting number. Result_3 = ( (16.8/3) )^2 Result_3 = (5.6)^2 Calculate 5.6 × 5.6: 5.6^2 = 31.36 Exam Strategy & Shortcut: Instead of fully calculating 5.6^2, use unit digit and magnitude estimation. 5^2 = 25 and 6^2 = 36, so the answer must be around 31. The unit digit of 5.6 × 5.6 will end in a 6 (since 6 × 6 = 36). Looking at the options, 31.36 is the only one in the correct range and ending with the correct decimal digit. Common Pitfall: Applying mathematical operations in the wrong order within a step. For example, in Step II, students might add 6 first and then divide by 5, rather than dividing the previous result strictly by 5 first before adding the constant 6. Read the sequence phrasing literally. Final Answer: Therefore, the correct answer is 31.36.

Question 15

Directions: Each question below consists of a number. Your task is to follow the following steps in order to find the result from the given alternatives.  Step I: Multiply the number by 3 and add the square of the number.  Step II: Divide the result…

Accepted Answer

Concept & Logic: This problem requires following a predefined mathematical algorithm strictly sequentially. When dealing with decimals, converting to fractions often helps maintain precision through intermediate steps. Step-by-Step Solution: • Given Input: x = 2.5. It is often easier to use the fraction form x = (5/2). • Step I: Multiply the number by 3 and add the square of the number. Result_1 = 3(2.5) + (2.5)^2 Result_1 = 7.5 + 6.25 = 13.75 • Step II: Divide the result obtained after Step I by 5 and add 6. Result_2 = (13.75/5) + 6 Result_2 = 2.75 + 6 = 8.75 • Step III: Divide the result obtained after Step II by 3 and then square the resulting number. Result_3 = ( (8.75/3) )^2 Convert 8.75 to a fraction to make squaring easier: 8.75 = (35/4). Result_3 = ( (\frac{35/4)}{3} )^2 = ( (35/12) )^2 Result_3 = (1225/144) Perform the final division: 1225 \div 144 \approx 8.5069 Rounding to two decimal places gives 8.51. Exam Strategy & Shortcut: Avoid heavy fraction squaring by using smart decimal estimation at Step 3. (8.75/3) is slightly less than 3 (since (9/3) = 3). Let's approximate it: 8.75 \div 3 \approx 2.916. Since 3^2 = 9, the answer must be slightly less than 9. Looking at the options, 8.10 is too low (that would require 2.84^2), making 8.51 the only logical approximation. Common Pitfall: Struggling with the decimal division in Step II (13.75 \div 5). A quick trick for dividing any number by 5 is to double the number and divide by 10. 13.75 × 2 = 27.5. 27.5 \div 10 = 2.75. Final Answer: Therefore, the correct answer is 8.51 (approx.).

Question 16

If $\oplus$ is an operation such that $a \oplus b = \begin{cases} 2a, & 	ext{when } a > b \ a + b, & 	ext{when } a < b \ a^2, & 	ext{when } a = b \end{cases}$  then $\left[ \frac{(5 \oplus 7) + (4 \oplus 4)}{3(5 \oplus 5) - (15 \oplus 11) - 3} …

Accepted Answer

Concept & Logic: This involves evaluating a piecewise mathematical function. For each operation in parentheses, you must first compare the two numbers to determine which of the three conditional rules applies, and then execute that specific rule. Step-by-Step Solution: • Given Rules: • If a > b, use 2a • If a 11 (a > b). So use 2a = 2(15) = 30. • Denominator Total = 3(25) - 30 - 3 = 75 - 30 - 3 = 42. • Step 3: Calculate Final Fraction Result = (28/42) Divide numerator and denominator by their greatest common divisor, 14: Result = (2/3) Exam Strategy & Shortcut: Process the conditionals systematically to avoid cognitive overload. Write down the intermediate values directly above the expression on your scratch paper: [ (12) + (16) ] / [ 3(25) - (30) - 3 ] Doing this explicitly prevents you from accidentally combining the multiplier 3 into the conditional logic of (5 \oplus 5). Common Pitfall: Misreading the greater-than/less-than signs or applying the rule to the wrong variable. For example, in (15 \oplus 11) where a > b, the rule is 2a. Some students mistakenly calculate 2b = 22. Always trace back to the exact definition carefully. Final Answer: Therefore, the correct answer is (2/3).

Question 17

At Srinagar, starting at 9 a.m. on a certain day, snow began to fall at a rate of $1\frac{1}{4}$ inches every two hours until 3 p.m. If there was already $2\frac{1}{4}$ inches of snow on the ground at 9 a.m., how many inches of snow was on the groun…

Accepted Answer

Concept & Logic: This problem involves calculating a total accumulation by multiplying a rate over a specific time interval and adding it to an initial baseline quantity. Step-by-Step Solution: • Given: • Initial snow on the ground = 2(1/4) = 2.25 inches. • Rate of snowfall = 1(1/4) = 1.25 inches per 2 hours. • Time duration (from 9 a.m. to 3 p.m.) = 6 hours. • Calculation: First, determine the snowfall per hour: Hourly rate = (1.25/2) = 0.625 inches/hour Next, calculate the total snow fallen during the 6-hour window: Total fall = 6 × 0.625 = 3.75 inches Finally, add this new accumulation to the initial amount on the ground: Total snow = 2.25 + 3.75 = 6 inches Exam Strategy & Shortcut: Since the rate is given per 2 hours, and 6 hours is exactly three 2-hour intervals, skip the per-hour calculation entirely. Simply multiply the interval rate by the number of intervals: 3 × 1(1/4) = 3(3/4) inches. Add this directly to the initial baseline: 2(1/4) + 3(3/4) = 6 inches. This mental math pathway avoids decimal conversions completely. Common Pitfall: Students often forget to add the initial snow amount (2(1/4)) to the accumulated snowfall, incorrectly choosing option (a) 3(3/4). Always reread the final question line to ensure all starting conditions are factored into the final answer. Final Answer: Therefore, the correct answer is 6.

Question 18

A squirrel starts climbing up a tree at the speed of 6 metres a minute but after every 6 metres it slips down 4 metres in the next minute. It will be able to reach the top 120 metres high in

Accepted Answer

Concept & Strategy: This is a classic "cyclic progress" problem (often framed as a monkey climbing a pole). The crucial insight is that during the *last* cycle, the target is reached during the upward phase, ending the journey before any backward slip occurs. Step-by-Step Solution: • Given: • Target height = 120 metres. • Climb phase = 6 metres in 1 minute. • Slip phase = 4 metres in 1 minute. • Deduction: Calculate the net progress of one complete cycle: Cycle duration = 1 + 1 = 2 minutes Net gain per cycle = 6 - 4 = 2 metres • Calculation: Subtract the final unabated climb (6m) from the total height to find the distance that must be covered via the slow, slipping cycles: Cyclic distance = 120 - 6 = 114 metres Calculate the number of complete cycles required to cover 114 metres: Cycles = (114/2) = 57 cycles Calculate the time taken for these 57 cycles: 57 × 2 = 114 minutes Finally, add the 1 minute for the last 6-metre climb to reach the 120m top: Total Time = 114 + 1 = 115 minutes Exam Strategy & Shortcut: Use the standardized formula for alternating positive/negative work problems to find the number of pair cycles: Cycles = \frac{Total Distance - Final Jump}{Net Gain} Cycles = (120 - 6/2) = 57. Total time is 57 × 2 = 114 mins, plus 1 minute for the final jump = 115 mins. Common Pitfall: Dividing the total height (120) by the net cycle gain (2) directly to get 60 cycles, and thus concluding 120 minutes. This ignores the physical reality that once the top is reached at 115 minutes, the squirrel grabs the edge and won't slip down again. Final Answer: Therefore, the correct answer is 115 minutes.

Question 19

Ram had ₹ 1000 in his Savings Bank Account. Every month in the first week he needs money, so he withdraws ₹ 500, but by the end of the month, he deposits ₹ 750. After how many months the original amount will grow three times?

Accepted Answer

Concept & Logic: This problem requires tracking a monthly balance with sequential deductions and additions. You must find the net change over a full month and use it to determine when a fixed final threshold is reached at the end of a cycle. Step-by-Step Solution: • Given: • Initial balance = ₹ 1000 • Target balance = 3 × 1000 = ₹ 3000 • Monthly withdrawal = ₹ 500 • Monthly deposit = ₹ 750 • Deduction: Find the net monthly growth after a full cycle (withdrawal + deposit). Net Monthly Increase = 750 - 500 = ₹ 250 We need to find the number of months m it takes to reach ₹ 3000 at the end of a month. • Calculation: After m months, the balance at the end of the month is expressed as: Balance = 1000 + 250m Set the equation to reach the target balance: 1000 + 250m = 3000 250m = 2000 m = 8 Let's verify the 8th month: Start of month 8 balance = ₹ 2750. First week withdrawal = 2750 - 500 = ₹ 2250. End of month deposit = 2250 + 750 = ₹ 3000. Target reached perfectly. Exam Strategy & Shortcut: Simply use the net monthly gain as a linear rate. Required Growth = 3000 - 1000 = 2000. Since each month ends with a net +250, divide the required growth by the net gain: 2000 \div 250 = 8. Common Pitfall: Applying the "monkey climbing" logic (where the target might be hit mid-cycle). Here, the negative action (withdrawal) happens *first*, causing a mid-month dip, and the positive peak happens *last* with the deposit. Thus, the target is hit at the end of the month, and standard linear rate division works without needing to adjust for an early threshold cross. Final Answer: Therefore, the correct answer is 8 months.

Question 20

The growth of a medicinal plant of height 1 metre, is 1 cm/day and it is growing straight vertically upward. A man cuts 6 cm from the top, at a regular interval of every 11 days. Its height will be 2 metres after

Accepted Answer

Concept & Strategy: This is a sequential growth and reduction problem. Care must be taken to distinguish between the continuous daily growth and the discrete periodic cuts. We must track the daily height to see exactly when it hits the threshold. Step-by-Step Solution: • Given: • Initial height = 1 m = 100 cm • Target height = 2 m = 200 cm • Growth rate = 1 cm/day • Cut = 6 cm every 11 days. • Deduction: In one 11-day period, the plant grows 11 cm and is then cut by 6 cm. Net growth per 11 days = 11 - 6 = 5 cm • Calculation: Find how many 11-day cycles it takes to get near 200 cm. After 19 cycles (19 × 11 = 209 days): Height = 100 + 19(5) = 195 cm Now, track the daily growth from day 210 onwards (before the 20th cut happens on day 220): • Day 210: 196 cm • Day 211: 197 cm • Day 212: 198 cm • Day 213: 199 cm • Day 214: 200 cm The plant reaches exactly 2 metres on the 214th day. Since 214 is not listed among the options, the correct choice is "None of these". Exam Strategy & Shortcut: Evaluate the widely accepted "flawed" logic vs. the strict mathematical reality. Flawed logic: Target growth = 100 cm. Net rate = 5 cm / 11 days. Time = (100 \div 5) × 11 = 220 days. Strict math: At day 209, height is 195cm. It grows 1cm/day for 5 days to hit 200cm exactly at day 214. Since 214 is missing, confidently choose "None of these". Common Pitfall: Blindly assuming the target height is only reached exactly at the end of a full cycle and calculating (100 \div 5) × 11 = 220 days (Option a). This ignores the intra-cycle peak where the plant hits 200 cm organically on day 214 before the next cut occurs. Final Answer: Therefore, the correct answer is None of these.

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