Question 1

What is $$ \frac{\frac{7}{8} 	imes \frac{7}{8} + \frac{5}{6} 	imes \frac{5}{6} + \frac{7}{8} 	imes \frac{5}{3}}{\frac{7}{8} 	imes \frac{7}{8} - \frac{5}{6} 	imes \frac{5}{6}} $$ equal to?

Accepted Answer

Concept & Formula: This problem masks fundamental algebraic identities inside complex-looking fraction multiplication. The two governing formulas to recognize here are: (a + b)^2 = a^2 + b^2 + 2ab a^2 - b^2 = (a - b)(a + b) Step-by-Step Solution: Let's substitute the repeating fractions with variables to spot the pattern more easily: • Let a = (7/8) • Let b = (5/6) Look closely at the numerator: a × a + b × b + a × (5/3) The trick lies in the fraction (5/3) . Notice that (5/3) is exactly double (5/6) . Therefore, (5/3) = 2 × b . Substitute this back into the numerator: Numerator = a^2 + b^2 + a(2b) = a^2 + b^2 + 2ab Which simplifies to (a + b)^2 . Now look at the denominator: Denominator = a × a - b × b = a^2 - b^2 Which expands to (a - b)(a + b) . Combine the simplified numerator and denominator: Expression = ((a + b)^2/(a - b)(a + b)) Cancel out one (a + b) term from top and bottom: Expression = (a + b/a - b) Now, evaluate (a + b) and (a - b) by substituting the fractions back. Find a common denominator for 8 and 6, which is 24. • a = (7/8) = (21/24) • b = (5/6) = (20/24) Calculate the sum and difference: • a + b = (21/24) + (20/24) = (41/24) • a - b = (21/24) - (20/24) = (1/24) Finally, divide the sum by the difference: (41/24/1/24) = 41 × (24/1) = 41 Exam Strategy & Shortcut: Whenever you see a a^2 + b^2 + \dots structure, actively look for how the remaining term represents 2ab. Here, seeing 5/3 alongside 5/6 immediately signals 2b. You can then instantly simplify the whole block of math down to (a+b)/(a-b) , requiring only one basic fraction addition and subtraction. Common Pitfall: Failing to recognize 5/3 as 2 × 5/6 causes students to abandon the algebraic approach entirely and attempt brute-force arithmetic, which under exam conditions almost always results in a miscalculation given the large mixed denominators. Final Answer: Therefore, the correct answer is 41.

Question 2

If $ x = 5 $, $ y = 3 $, the value of $ \frac{x^3 - y^3}{x^2 - y^2} - \frac{3xy}{x + y} $ will be

Accepted Answer

Concept & Formula: This problem tests your ability to simplify rational algebraic expressions before inserting numerical values. The goal is to combine the two fractions into a much simpler form. The necessary factorization formulas are: x^3 - y^3 = (x - y)(x^2 + xy + y^2) x^2 - y^2 = (x - y)(x + y) Step-by-Step Solution: Let's start with the first fraction in the expression: (x^3 - y^3/x^2 - y^2) Factor both the numerator and the denominator using the standard formulas: = ((x - y)(x^2 + xy + y^2)/(x - y)(x + y)) Cancel out the common (x - y) term from the top and bottom: = (x^2 + xy + y^2/x + y) Now, bring back the second part of the original expression and subtract it: (x^2 + xy + y^2/x + y) - (3xy/x + y) Because both fractions now share a common denominator ( x + y ), we can combine their numerators: = (x^2 + xy + y^2 - 3xy/x + y) Combine the like terms ( +xy - 3xy = -2xy ): = (x^2 - 2xy + y^2/x + y) Notice that the numerator is now a perfect square trinomial. Factor it: = ((x - y)^2/x + y) The expression is fully simplified. Now substitute the given values: x = 5 and y = 3 . = ((5 - 3)^2/5 + 3) = ((2)^2/8) = (4/8) = (1/2) Exam Strategy & Shortcut: Instead of algebraic manipulation, you can use direct substitution ("Value Putting") immediately, as the numbers are very small. Fraction 1: (5^3 - 3^3/5^2 - 3^2) = (125 - 27/25 - 9) = (98/16) = (49/8) Fraction 2: (3(5)(3)/5 + 3) = (45/8) Difference: (49/8) - (45/8) = (4/8) = (1/2) . For small integers, basic arithmetic is often faster than performing multi-step polynomial factorization. Common Pitfall: When simplifying algebraically, students sometimes mistakenly cancel (x^2 - y^2) out of the denominator without properly factoring the numerator, or miscalculate the combination +xy - 3xy as +2xy , which leads to an entirely incorrect final value. Final Answer: Therefore, the correct answer is (1/2) .

Question 3

The value of  $$\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}$$ is

Accepted Answer

Concept & Formula: This fraction simplifies using the conditional identity for the sum of cubes: If A + B + C = 0, then A^3 + B^3 + C^3 = 3ABC. Step-by-Step Solution: • Let the numerator components be A = x - y, B = y - z, and C = z - x. • Notice that A + B + C = (x - y) + (y - z) + (z - x) = 0. • Therefore, the numerator becomes 3(x - y)(y - z)(z - x). • Now, let the denominator components be P = x^2 - y^2, Q = y^2 - z^2, and R = z^2 - x^2. • Notice that P + Q + R = (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = 0. • Therefore, the denominator becomes 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2). • Expand the denominator using the difference of squares a^2 - b^2 = (a-b)(a+b): 3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x) • Divide the simplified numerator by the simplified denominator: (3(x - y)(y - z)(z - x)/3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)) • Cancel the common terms 3(x - y)(y - z)(z - x) from top and bottom: = (1/(x + y)(y + z)(z + x)) • Rewrite using a negative exponent: = [(x + y)(y + z)(z + x)]^{-1} Exam Strategy & Shortcut: Whenever you see a sum of three cubes in a pattern that looks cyclical (like x-y, y-z, z-x), immediately check if they sum to zero. If they do, instantly replace them with 3 times their product. This visual recognition avoids lengthy expansions. Common Pitfall: A common error is to try expanding (x - y)^3 individually. This creates a massive, unmanageable polynomial that guarantees a loss of time and inevitable calculation errors. Final Answer: Therefore, the correct answer is [(x + y) (y + z) (z + x)]^{-1}.

Question 4

If $a = 29$, $b = 24$, $c = 27$, the value of $a^3 + b^3 + c^3 - 3abc$ is

Accepted Answer

Concept & Formula: When the values of a, b, and c are large but relatively close to each other, computing their cubes directly is inefficient. Instead, use the specialized form of the cubic identity involving squares of differences: a^3 + b^3 + c^3 - 3abc = (1/2)(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2] Step-by-Step Solution: • Given: a = 29, b = 24, c = 27. • First, calculate the sum of the variables: a + b + c = 29 + 24 + 27 = 80 • Next, calculate the differences between each pair of variables: a - b = 29 - 24 = 5 b - c = 24 - 27 = -3 c - a = 27 - 29 = -2 • Square these differences: (a - b)^2 = 5^2 = 25 (b - c)^2 = (-3)^2 = 9 (c - a)^2 = (-2)^2 = 4 • Sum the squares of the differences: 25 + 9 + 4 = 38 • Plug everything into the specialized formula: (1/2) × (80) × (38) • Simplify the calculation: = 40 × 38 = 1520 Exam Strategy & Shortcut: Using the difference-of-squares formula (1/2)(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] is the ultimate shortcut here. It reduces large multi-digit cube calculations to basic addition and single-digit squaring. Common Pitfall: The most common trap is attempting brute-force calculation (29^3 + 24^3 + 27^3 - \dots). This will consume several minutes and likely result in an arithmetic mistake. Final Answer: Therefore, the correct answer is 1520.

Question 5

$(32)^3 + (79)^3 - (111)^3 + 3 	imes 32 	imes 79 	imes 111$ is equal to

Accepted Answer

Concept & Formula: This expression evaluates to zero based on a clever disguised application of the conditional identity: If a + b + c = 0, then a^3 + b^3 + c^3 - 3abc = 0. Step-by-Step Solution: • Look closely at the structure of the expression: (32)^3 + (79)^3 + (-111)^3 - 3(32)(79)(-111). • Notice that the negative signs have been distributed. The expression exactly matches the form a^3 + b^3 + c^3 - 3abc. • Let a = 32, b = 79, and c = -111. • Note that in the given expression, subtracting (111)^3 is mathematically identical to adding (-111)^3. • Similarly, adding + 3 × 32 × 79 × 111 is identical to subtracting -3 × (32) × (79) × (-111). • Check the sum of the variables: a + b + c = 32 + 79 + (-111) = 111 - 111 = 0 • Because a + b + c = 0, the entire expression a^3 + b^3 + c^3 - 3abc must equal 0. Exam Strategy & Shortcut: Any time an aptitude question asks you to add and subtract massive cubed numbers alongside a product of those same numbers, it is always testing the a+b+c=0 rule. Don't calculate anything; just verify that the base numbers sum to zero. Common Pitfall: Failing to recognize that c is negative (-111) and incorrectly assuming the formula doesn't apply because the signs look slightly different than the standard textbook formula. Final Answer: Therefore, the correct answer is 0.

Question 6

If $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$ and $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ where $a, b, c, p, q, r$ are non-zero real numbers, then $\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2}$ is equal to

Accepted Answer

Concept & Formula: This problem is a disguised application of the algebraic identity for the square of a trinomial: (X + Y + Z)^2 = X^2 + Y^2 + Z^2 + 2(XY + YZ + ZX) Step-by-Step Solution: • To make the problem visually simpler, substitute the fractions with capital variables: Let X = (p/a), Y = (q/b), and Z = (r/c). • The first given equation becomes: X + Y + Z = 1 • The second given equation represents their reciprocals: (1/X) + (1/Y) + (1/Z) = 0 • Combine the reciprocal equation into a single fraction by finding a common denominator (XYZ): (YZ + XZ + XY/XYZ) = 0 • Since the fraction equals zero, its numerator must be zero: XY + YZ + ZX = 0 • Now we need to find the value of (p^2/a^2) + (q^2/b^2) + (r^2/c^2), which translates to finding X^2 + Y^2 + Z^2. • Use the trinomial square identity: (X + Y + Z)^2 = X^2 + Y^2 + Z^2 + 2(XY + YZ + ZX) • Substitute the known values (X+Y+Z=1 and XY+YZ+ZX=0): 1^2 = X^2 + Y^2 + Z^2 + 2(0) 1 = X^2 + Y^2 + Z^2 Exam Strategy & Shortcut: By using the substitution variable method (changing complex fractions into X, Y, Z), you reduce visual clutter, which drastically decreases the cognitive load and prevents transcription errors under exam pressure. Common Pitfall: A standard error is misinterpreting (a/p) + (b/q) + (c/r) = 0 and wrongly assuming it implies the original fractions square to 0. Always combine reciprocal sums properly to find the pairwise product sum. Final Answer: Therefore, the correct answer is 1.

Question 7

A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the …

Accepted Answer

Concept & Strategy: This problem sets up a linear equivalence based on the fencepost principle: Number of posts = (Total Length / Interval) + 1. By equating the number of posts bought under two different spacing scenarios, we can solve for the total length. Step-by-Step Solution: • Given: Scenario A (Planned): 6m intervals. Requires P posts. Scenario B (Actual): Posts bought were P - 5. Scenario C (Adjusted): Spacing them 8m apart perfectly uses the bought posts. • Let L be the total length of the side of the property. Let N be the number of posts actually bought. • Create an equation for the *required* posts based on the 6m interval: Required Posts = (L/6) + 1 • Create an equation for the *bought* posts (N). We know he bought 5 less than required: N = ((L/6) + 1) - 5 N = (L/6) - 4 • Create an equation for the *bought* posts based on the 8m interval working perfectly: N = (L/8) + 1 • Set the two expressions for N equal to each other to solve for L: (L/6) - 4 = (L/8) + 1 (L/6) - (L/8) = 5 • Find a common denominator (24) to solve: (4L - 3L/24) = 5 (L/24) = 5 L = 120 metres • Now substitute L = 120 back into either equation to find N (posts bought): N = (120/8) + 1 N = 15 + 1 = 16 Exam Strategy & Shortcut: Use Option Elimination based on divisibility. The length must be perfectly divisible by 8 (since spacing them 8m apart worked perfectly with no remainder). Both 100 and 120 are in the options. 100 / 8 = 12.5 (invalid). 120 / 8 = 15. This immediately isolates the 120m options. 15 gaps require 16 posts. Thus, 120m and 16 posts is the only logical answer. Common Pitfall: Forgetting the "+1" rule for posts. If you use Posts = (L/Interval) without adding 1 for the end post, the algebraic setup will fail, leading to an incorrect length or a negative number of posts. Final Answer: Therefore, the correct answer is 120 m, 16.

Question 8

Sweets were distributed equally among 64 children. After giving 7 sweets to each child 15 sweets were left out. Total how many sweets were there?

Accepted Answer

Concept & Logic: This is a basic arithmetic problem involving multiplication and addition to find an original total quantity from distributed amounts and a remainder. Step-by-Step Solution: • Given: Number of children = 64. Sweets given to each = 7. Sweets leftover = 15. • First, calculate the total number of sweets successfully distributed to the children: Distributed Sweets = Children × Sweets per child Distributed Sweets = 64 × 7 • Perform the multiplication: 64 × 7 = 448. • Next, add the leftover sweets to the distributed amount to find the initial total: Total Sweets = Distributed + Leftover Total Sweets = 448 + 15 = 463 Exam Strategy & Shortcut: Use the unit digit concept to save time. 64 × 7 \rightarrow The unit digit is 4 × 7 = 28 \rightarrow ends in 8. Add the remainder 15 \rightarrow The unit digit is 8 + 5 = 13 \rightarrow ends in 3. Look at the options: only option (d) 463 ends in a 3. You can select it without doing the full calculation. Common Pitfall: A simple calculation error during 64 × 7. Students sometimes forget to carry over correctly. Unit digit verification acts as a great safeguard against this. Final Answer: Therefore, the correct answer is 463.

Question 9

A man had two sons. To the elder, he left 5/11 of his property, to the younger 5/11 of the remainder, the rest to the widow. Find the share of the sons if the widow gets ₹ 3600.

Accepted Answer

Concept & Logic: This is a successive distribution problem. When a portion is given from the "remainder", the new fraction must be multiplied by the remaining fraction of the total, not simply subtracted from the total. Remaining\ Fraction = 1 - Fraction\ Given Step-by-Step Solution: • Given: Elder son gets (5/11) of the total property. Younger gets (5/11) of the remainder. Widow gets ₹ 3600. Let total property be P. • Calculation: Elder son's share = (5/11)P. • Deduction: The remainder of the property after the elder son's share is P - (5/11)P = (6/11)P. • Calculation: Younger son's share is (5/11) of this remainder: Younger = (5/11) × (6/11)P = (30/121)P. • Deduction: The widow gets whatever is left from the remainder after the younger son is paid: Widow's fraction = (6/11)P - (30/121)P = (66P - 30P/121) = (36/121)P. • Calculation: Equate the widow's fractional share to her actual amount: (36/121)P = 3600 P = (3600 × 121/36) = 12100. • Calculation: Find the exact share of both sons: Elder son = (5/11) × 12100 = 5500. Younger son = (30/121) × 12100 = 3000. • Conclusion: The shares are ₹ 5500 and ₹ 3000, which do not match options (a), (b), or (c). Exam Strategy & Shortcut: Work with ratios. The fractions involve 11, so let the total be a multiple of 11 × 11 = 121 units. Total = 121 units. Elder = (5/11) of 121 = 55 units. Remainder = 66 units. Younger = (5/11) of 66 = 30 units. Remainder (Widow) = 36 units. We know 36 units = ₹ 3600, so 1 unit = ₹ 100. Thus, Elder = 5500, Younger = 3000. These are not in the options, confirming "None of these". Common Pitfall: Treating "(5/11) of the remainder" as simply (5/11) of the total. Students who do this incorrectly calculate the widow's share as 1 - (5/11) - (5/11) = (1/11), leading to completely wrong valuations for the property. Final Answer: Therefore, the correct answer is None of these.

Question 10

Anand, Bijoy, Chetan and Dharma together have ₹ 47 with them. Anand and Bijoy together have ₹ 27; Chetan and Anand have ₹ 25 and Dharma and Anand have ₹ 23. How much money does Bijoy have?

Accepted Answer

Concept & Logic: This is a system of linear equations where one variable (Anand) acts as a common link across multiple subset sums. Summing the overlapping equations allows us to extract the value of the common variable efficiently. Sum\ of\ Subsets = Multiplier × Common\ Variable + Total\ Sum Step-by-Step Solution: • Given: Let their money be A, B, C, D. Total: A + B + C + D = 47 Equation 1: A + B = 27 Equation 2: C + A = 25 Equation 3: D + A = 23 • Calculation: Add Equations 1, 2, and 3 together: (A + B) + (C + A) + (D + A) = 27 + 25 + 23 3A + B + C + D = 75 • Substitution: We know that A + B + C + D = 47. We can rewrite our summed equation as: 2A + (A + B + C + D) = 75 2A + 47 = 75 • Calculation: Solve for A (Anand): 2A = 75 - 47 2A = 28 A = 14 • Calculation: The question asks for Bijoy's money (B). Substitute A into Equation 1: 14 + B = 27 \implies B = 13 Exam Strategy & Shortcut: Add the three partial sums: 27 + 25 + 23 = 75. Notice that this sum counts Anand (A) three times, while counting everyone else (B, C, D) exactly once. So, Sum = 2A + Total. 75 = 2A + 47 \implies 2A = 28 \implies A = 14. Since A + B = 27, Bijoy (B) is simply 27 - 14 = 13. This can be done mentally in seconds. Common Pitfall: Attempting to solve for each individual person (C and D) before answering the question. This wastes valuable exam time. Always focus strictly on finding the specific variable requested (in this case, B, which only requires finding A). Final Answer: Therefore, the correct answer is ₹ 13.

Question 11

When a sum of money was distributed to 12 persons instead of 16 persons, each person got ₹ 400 more. What was the sum?

Accepted Answer

Concept & Logic: This problem involves inverse variation. The total sum is constant. The difference in the per-person share between the two scenarios forms a simple algebraic equation based on fractions of the total sum. (Total/Smaller\ Group) - (Total/Larger\ Group) = Difference\ in\ Share Step-by-Step Solution: • Given: Let the total sum of money be x. Original scenario: Divided among 16 persons. Share = (x/16). New scenario: Divided among 12 persons. Share = (x/12). Difference in share = ₹ 400. • Calculation: Set up the equation representing the difference: (x/12) - (x/16) = 400 • Calculation: Find the Lowest Common Multiple (LCM) of 12 and 16, which is 48. Rewrite the fractions: (4x/48) - (3x/48) = 400 (x/48) = 400 • Calculation: Solve for x: x = 400 × 48 = 19200. Exam Strategy & Shortcut: Use ratio mapping to avoid algebra. The ratio of people is 12 : 16, which simplifies to 3 : 4. Because the sum is constant, the ratio of their individual shares is inversely proportional: 4 : 3. The difference in shares is 1 unit (4 - 3 = 1 unit). We are told this difference is ₹ 400. So, 1 unit = ₹ 400. The share when divided among 12 people (the 4-unit share) is 4 × 400 = 1600. Total sum = Share × Number of people = 1600 × 12 = 19200. Common Pitfall: Writing the equation backwards as (x/16) - (x/12) = 400, which results in a negative value for the total sum. Always remember that dividing by a smaller number yields a larger share, so (x/12) must come first. Final Answer: Therefore, the correct answer is ₹ 19200.

Question 12

After distributing the sweets equally among 25 children, 8 sweets remain. Had the number of children been 28, 22 sweets would have been left after equally distributing. What was the total number of sweets?

Accepted Answer

Concept & Logic: This problem tests the application of the Remainder Theorem and modular arithmetic. We are looking for a number N that leaves specific remainders when divided by two different divisors. N = D_1 × Q_1 + R_1 N = D_2 × Q_2 + R_2 Step-by-Step Solution: • Given: Let the total number of sweets be N. Condition 1: N \div 25 leaves a remainder of 8. So, N = 25a + 8. Condition 2: N \div 28 leaves a remainder of 22. So, N = 28b + 22. • Deduction: We can test the given options to see if they satisfy both conditions. • Calculation: Let's test Option (c) 358: 358 \div 25 = 14 with a remainder of 8. (Matches Condition 1) 358 \div 28 = 12 with a remainder of 358 - (28 × 12) = 358 - 336 = 22. (Matches Condition 2) • Deduction: 358 works perfectly. However, is it the *only* answer? According to the Chinese Remainder Theorem, the general solution for such a system is N = N_0 + k × LCM(D_1, D_2), where N_0 is the smallest positive solution. LCM(25, 28) = 700. So, any number in the form N = 358 + 700k (where k is an integer \ge 0) will satisfy both conditions. Valid numbers are 358, 1058, 1758, 2458, etc. • Conclusion: Because no upper limit or exact constraint is given for the total number of sweets, there are infinitely many valid answers. We cannot determine the exact total. Exam Strategy & Shortcut: When a question asks "What was the *total* number..." (implying a single unique absolute value) but only provides remainder conditions without an upper bound or a quotient relationship, immediately suspect "Data inadequate". Check the options. While 358 works, the structural lack of constraints means the answer cannot be uniquely determined. Common Pitfall: The most common trap is testing the options, finding that 358 works perfectly for both conditions, and immediately marking (c) as the answer. Examiners include 358 specifically to bait students who don't realize that multiple solutions exist (like 1058) and thus the data is technically inadequate to find *the* unique total. Final Answer: Therefore, the correct answer is Data inadequate.

Question 13

Vidushi and Sanya distributed ₹ 100 each in charity. Vidushi distributes money to 5 more people than Sanya and Sanya gives each ₹ 1 more than Vidushi. How many people are recipients of the charity?

Accepted Answer

Concept & Logic: This problem links the number of recipients to the amount given per recipient via an inverse relationship, forming a quadratic equation. The total amount given by each person is fixed at ₹ 100. (Total\ Amount/Number\ of\ People) = Amount\ per\ Person Step-by-Step Solution: • Given: Both distribute ₹ 100. Let the number of people Sanya distributes to be S. Then, the number of people Vidushi distributes to is S + 5. • Deduction: Amount Sanya gives to each person = (100/S). Amount Vidushi gives to each person = (100/S+5). • Calculation: The problem states Sanya gives ₹ 1 more per person than Vidushi. (100/S) - (100/S+5) = 1 • Calculation: Factor out 100 and find a common denominator: 100 ( ((S+5) - S/S(S+5)) ) = 1 100 ( (5/S^2 + 5S) ) = 1 500 = S^2 + 5S • Calculation: Solve the quadratic equation: S^2 + 5S - 500 = 0 Find factors of -500 that add to +5. They are +25 and -20. (S + 25)(S - 20) = 0. Since S must be positive, S = 20. • Conclusion: Sanya distributes to 20 people. Vidushi distributes to 20 + 5 = 25 people. Total recipients = 20 + 25 = 45. Exam Strategy & Shortcut: Bypass the quadratic equation by looking for two integer factors of 100 whose difference is related to the problem. We need two numbers of people that divide 100 perfectly (since usually amounts are integers in these contexts), and their difference must be 5. Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100. Look at 20 and 25. Their difference is 5. Check the amounts: (100/20) = 5 and (100/25) = 4. The difference in amounts is 5 - 4 = 1. This perfectly matches all conditions in the question! Total people = 20 + 25 = 45. Common Pitfall: Solving for S (20) and selecting an option if it were available, or solving for Vidushi (25). Read the final question carefully: it asks "How many people are recipients of the charity?", meaning the *total combined* number of recipients (20 + 25 = 45). Final Answer: Therefore, the correct answer is 45.

Question 14

In a classroom the number of boys is three times the number of girls. Which of the following numbers does not represent the total number of students in the classroom?

Accepted Answer

Concept & Logic: This problem relies on basic ratios and divisibility rules. Since the number of human beings (students) must be an integer, the total number of students must be a multiple of the sum of the ratio units. Total\ Units = Ratio_1 + Ratio_2 Total\ Population = Integer × Total\ Units Step-by-Step Solution: • Given: Number of boys = 3 × Number of girls. • Deduction: Let the number of girls be x. Then, the number of boys is 3x. • Calculation: The total number of students in the classroom is the sum of boys and girls. Total = x + 3x = 4x. • Conclusion: Since x (number of girls) must be a whole number, the total number of students (4x) must be a perfect multiple of 4. • Evaluation: Check the given options against divisibility by 4: (a) 40 = 4 × 10 (Valid) (b) 42 = 4 × 10.5 (Invalid, results in a fractional person) (c) 44 = 4 × 11 (Valid) (d) 48 = 4 × 12 (Valid) • Conclusion: 42 cannot represent the total number of students. Exam Strategy & Shortcut: Simply identify the ratio: Boys : Girls = 3 : 1. Sum of ratio parts = 3 + 1 = 4. The correct answer must be the option that is not divisible by 4. Looking at the options: 40, 44, and 48 are clearly multiples of 4. 42 is not. Done. Common Pitfall: Overthinking the problem or trying to set up complex algebraic systems. Recognizing that populations must be discrete integers immediately unlocks the "divisibility by ratio sum" trick. Final Answer: Therefore, the correct answer is 42.

Question 15

The expression $\frac{1}{x-1} - \frac{1}{x+1} - \frac{2}{x^2+1} - \frac{4}{x^4+1}$ is equal to

Accepted Answer

Concept & Formula: This involves stepwise algebraic simplification of rational expressions. Like the previous problem, it relies heavily on creating common denominators using the difference of squares. Formula: (a - b)(a + b) = a^2 - b^2 Step-by-Step Solution: • Solve the expression in pairs from left to right. Take the first two terms: (1/x-1) - (1/x+1) • Find the common denominator (x-1)(x+1) = x^2 - 1: ((x+1) - (x-1)/x^2 - 1) = (2/x^2 - 1) • Substitute this result back into the expression, taking the next term: (2/x^2 - 1) - (2/x^2 + 1) • Find the common denominator (x^2 - 1)(x^2 + 1) = x^4 - 1: (2(x^2 + 1) - 2(x^2 - 1)/x^4 - 1) = (2x^2 + 2 - 2x^2 + 2/x^4 - 1) = (4/x^4 - 1) • Substitute this result back, taking the final term: (4/x^4 - 1) - (4/x^4 + 1) • Find the common denominator (x^4 - 1)(x^4 + 1) = x^8 - 1: (4(x^4 + 1) - 4(x^4 - 1)/x^8 - 1) = (4x^4 + 4 - 4x^4 + 4/x^8 - 1) = (8/x^8 - 1) Exam Strategy & Shortcut: Notice the predictable doubling pattern. The numerator doubles at every step (1 \rightarrow 2 \rightarrow 4 \rightarrow 8), and the denominator powers double with a minus sign (x^2-1 \rightarrow x^4-1 \rightarrow x^8-1). Knowing this structural behavior, you can directly leap to (8/x^8 - 1) without performing manual fraction subtractions. Common Pitfall: A frequent mistake is mishandling the negative signs during numerator expansion (e.g., getting zero in the numerator). Always use parentheses when subtracting binomials in the numerator to ensure the negative distributes correctly across both terms. Final Answer: Therefore, the correct answer is (8/x^8-1).

Question 16

Krishan has some hens and some cows. If the total number of animal heads is 59 and the total number of feet is 190, how many cows does Krishan have?

Accepted Answer

Concept & Strategy: This is a standard system of linear equations word problem. We need to define variables for the two distinct items (hens and cows) and create two separate equations based on the given totals for heads and feet. Step-by-Step Solution: • Set up variables based on known attributes: Let H be the number of hens. (1 head, 2 feet) Let C be the number of cows. (1 head, 4 feet) • Create Equation 1 (Heads): The total number of animal heads is 59. H + C = 59 • Create Equation 2 (Feet): The total number of feet is 190. 2H + 4C = 190 • Simplify Equation 2: Divide the entire feet equation by 2 to make it easier to work with: H + 2C = 95 • Solve the system of equations: We now have: Eq 1: H + C = 59 Eq 2: H + 2C = 95 Subtract Eq 1 from Eq 2 to eliminate H: (H + 2C) - (H + C) = 95 - 59 C = 36 • Krishan has 36 cows. (If asked, hens would be 59 - 36 = 23). Exam Strategy & Shortcut: The "Assume All" Shortcut: Assume all 59 animals are the one with *fewer* legs (Hens). If all were hens, total feet = 59 × 2 = 118 feet. Given total feet = 190. Difference = 190 - 118 = 72 extra feet. These extra feet must belong to the cows, which have 2 *extra* feet each compared to a hen. Number of cows = 72 / 2 = 36. This method avoids algebra entirely and can be done quickly on paper. Common Pitfall: A common error is confusing which variable represents the 4-legged animal when setting up the feet equation, inadvertently writing 4H + 2C = 190, leading to completely inverted answers. Always double-check your biological facts before writing the coefficients. Final Answer: Therefore, the correct answer is 36.

Question 17

In a party 15 people shake their hands with each other. How many times did the hand-shakes take place?

Accepted Answer

Concept & Formula: When every person in a group shakes hands with every other person exactly once, the total number of handshakes is a basic combination problem. The number of unique pairs that can be formed from n items is given by: Total Handshakes = (n(n - 1)/2) Step-by-Step Solution: • Given: The total number of people, n = 15. • Calculation: Substitute n = 15 into the combination formula. • Total Handshakes = (15 × (15 - 1)/2) • Total Handshakes = (15 × 14/2) • Total Handshakes = 15 × 7 = 105 Exam Strategy & Shortcut: For handshake problems, do not waste time writing the formula. Immediately multiply the given number by the number just below it, and halve it. Mentally calculating 15 × 7 = 105 takes less than 3 seconds. Common Pitfall: A very common mistake is using n(n - 1) which gives 210. This happens when students confuse handshakes (where A shaking B's hand is the same as B shaking A's hand) with gift exchanges (which are directional). Always divide by 2 for handshakes. Final Answer: Therefore, the correct answer is 105.

Question 18

A man has 1044 candles. After burning, he can make a new candle from 9 stubs left behind. Find the maximum number of candles that can be made.

Accepted Answer

Concept & Logic: This problem requires an iterative calculation. Creating a new candle from stubs allows you to burn that newly created candle, which then leaves behind another stub. You must repeat the division process until the remaining stubs are fewer than the required amount (9) to make a new candle. Step-by-Step Solution: • Initial Step: The man burns 1044 candles. This leaves 1044 stubs. • First Remake: Number of new candles made = (1044/9) = 116. • He burns these 116 candles. This leaves 116 stubs. • Second Remake: Number of new candles made = (116/9) = 12 new candles, with 8 stubs left over. • He burns these 12 candles, leaving 12 stubs. • Total stubs available now = 12 (new) + 8 (left over) = 20 stubs. • Third Remake: Number of new candles made = (20/9) = 2 new candles, with 2 stubs left over. • He burns these 2 candles, leaving 2 stubs. Total stubs = 2 + 2 = 4. • Since 4 < 9, no more candles can be made. • Total new candles: 116 + 12 + 2 = 130. Exam Strategy & Shortcut: Follow the chain of division carefully and maintain a running tally of remainders. Don't discard remainders until the very end, as combining them with new stubs often yields enough material for additional rounds. Common Pitfall: The most common mistake is stopping after the first division. Students calculate (1044/9) = 116 and select option (a). You must remember that newly created candles also leave stubs when burned! Final Answer: Therefore, the correct answer is 130.

Question 19

In a test, a candidate secured 336 marks out of maximum marks $x$. If the maximum marks $x$ had been converted into 400 marks, he would have secured 192 marks. What was the maximum marks of the test?

Accepted Answer

Concept & Formula: This problem requires setting up a direct proportion. The ratio of the marks obtained to the maximum marks remains constant, even if the maximum marks scale is converted. \frac{Obtained Marks_1}{Maximum Marks_1} = \frac{Obtained Marks_2}{Maximum Marks_2} Step-by-Step Solution: • Given: • Initial state: Obtained = 336, Maximum = x. • Converted state: Obtained = 192, Maximum = 400. • Equation: Set up the proportion: • (336/x) = (192/400) • Calculation: • Simplify the right side fraction first: (192/400) divides cleanly by 8 to give (24/50), and dividing by 2 gives (12/25). • Now, (336/x) = (12/25) • Cross-multiply to solve for x: 12x = 336 × 25 • x = (336 × 25/12) • Divide 336 by 12: 336 / 12 = 28. • x = 28 × 25 • x = 700 Exam Strategy & Shortcut: Recognize the multiplier scale. The new base is 400. The ratio is (192/400). Observe that 192 is slightly less than half of 400. Thus, 336 must be slightly less than half of x. This means x is slightly more than double 336 (which is 672). The only logical option nearby is 700. Alternatively, multiplying 28 × 25 is quickly done as 28 × (100/4) = 7 × 100 = 700. Common Pitfall: Getting bogged down in massive cross-multiplication without simplifying the fractions first. Doing 336 × 400 = 134400 and then trying to divide by 192 by hand wastes precious exam time and increases the likelihood of arithmetic errors. Always simplify fractions first. Final Answer: Therefore, the correct answer is 700.

Question 20

The marks scored in an examination are converted from 50 to 10 for the purpose of internal assessment. The highest marks were 47 and the lowest were 14. The difference between the maximum and the minimum internal assessment scores is

Accepted Answer

Concept & Formula: When a set of values is scaled by a constant factor, the difference between any two values in the set is also scaled by that exact same factor. Scaled Difference = Original Difference × Scale Factor Step-by-Step Solution: • Scale Factor: The base marks are converting from 50 to 10. • Scale Factor = (10/50) = (1/5). This means every score is divided by 5. • Original Difference: Calculate the difference between the highest and lowest original marks. • Original Maximum = 47, Original Minimum = 14. • Original Difference = 47 - 14 = 33. • Calculation: Apply the scale factor to the original difference. • Scaled Difference = 33 × (1/5) = (33/5) • Scaled Difference = 6.6 Exam Strategy & Shortcut: Instead of converting both the maximum and the minimum scores individually ((47/5) = 9.4 and (14/5) = 2.8) and then subtracting them (9.4 - 2.8 = 6.6), it is vastly faster to subtract the raw scores first (33) and divide by the scale factor once ((33/5) = 6.6). Common Pitfall: Converting the individual numbers first. While mathematically correct, dealing with decimals like 9.4 and 2.8 under time pressure invites subtraction errors. Stick to integers as long as possible before dividing. Final Answer: Therefore, the correct answer is 6.6.

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