Question 1

Simplify the mixed-number expression 5(2/7) − 3(1/14) − 2(1/14) − 1(1/7) and express the result in the form ? + 3(1/17); what is the value of the unknown mixed number represented by ?

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Introduction / Context: This question tests the ability to work with mixed numbers, convert them to improper fractions, perform subtractions, and then re express a result in terms of another mixed number. Such problems appear in aptitude exams to check fraction manipulation skills and careful handling of signs. The structure ? + 3(1/17) encourages candidates to think algebraically about the final result. Given Data / Assumptions: - Expression: 5(2/7) − 3(1/14) − 2(1/14) − 1(1/7). - This entire value equals ? + 3(1/17). - All mixed numbers are real positive or negative numbers written as whole part plus fractional part. - We must find the mixed number represented by ?. Concept / Approach: First, convert each mixed number to an improper fraction. Then perform the subtractions step by step, making sure to use a common denominator. After finding the final result as a single fraction, convert the right hand side into an expression involving ? and 3(1/17). Because 3(1/17) is known, we subtract it from the total value to isolate ?. Finally, we convert that fraction back to a mixed number and match it with the given options. Step-by-Step Solution: Convert 5(2/7) to an improper fraction: (5*7 + 2)/7 = 37/7. Convert 3(1/14) to an improper fraction: (3*14 + 1)/14 = 43/14. Convert 2(1/14) to an improper fraction: (2*14 + 1)/14 = 29/14. Convert 1(1/7) to an improper fraction: (1*7 + 1)/7 = 8/7. Rewrite 37/7 and 8/7 with denominator 14: 37/7 = 74/14, 8/7 = 16/14. Now total expression = 74/14 − 43/14 − 29/14 − 16/14. Combine numerators: 74 − 43 − 29 − 16 = 74 − 88 = −14. So the value of the expression is −14/14 = −1. We have −1 = ? + 3(1/17). Convert 3(1/17) to improper fraction: (3*17 + 1)/17 = 52/17. Write −1 as −17/17. Then ? = −17/17 − 52/17 = −69/17. Convert −69/17 to a mixed number: 69 ÷ 17 = 4 remainder 1, so ? = −4(1/17). Verification / Alternative check: We can check by adding ? and 3(1/17). ? = −69/17 and 3(1/17) = 52/17, so their sum is (−69 + 52)/17 = −17/17 = −1. This matches the value of the original expression, confirming that ? is indeed −4(1/17). Why Other Options Are Wrong: Option 5(2/17) is positive and would make the sum with 3(1/17) positive, not −1. Option −2(1/17) corresponds to −35/17, which gives −35/17 + 52/17 = 17/17 = 1, not −1. Option 3(3/17) corresponds to a positive number, again incompatible with the negative total of the original expression. Common Pitfalls: Typical errors include converting mixed numbers to improper fractions incorrectly, forgetting to use a common denominator, or mishandling the negative sign when combining fractions. Another common mistake is to misinterpret the equation structure and not realise that the entire left hand side equals ? plus 3(1/17). Working slowly through each conversion step helps avoid these issues. Final Answer: The value of the unknown mixed number ? is -4 (1/17).

Question 2

In a party there are both male and female members initially. If 15 female members leave the party, then the remaining number of females becomes exactly double the number of males present. However, if instead 45 male members leave the party, then the…

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Introduction / Context: This aptitude question tests linear equation formation and solving with two variables. The situation describes a party with male and female members and gives two different conditions about members leaving the party. From these conditions we have to form equations in terms of the original number of males and females and then determine the initial female population. Such questions are common in bank exams, SSC, and other competitive tests that assess algebraic reasoning. Given Data / Assumptions: Let the original number of female members be F. Let the original number of male members be M. If 15 females leave, the remaining females become twice the number of males. If instead 45 males leave, the remaining males become one fifth of the number of females, that is the females are five times the males. Concept / Approach: The approach is to convert each verbal condition into a linear equation in F and M. With two independent equations and two unknowns, we can solve the system using substitution or elimination. Once M is eliminated, we obtain a direct value for F. Finally, we check the solution in both original conditions to ensure consistency. Step-by-Step Solution: From the first condition: F - 15 = 2M.From the second condition: F = 5(M - 45), because females are five times the new number of males.Expand the second equation: F = 5M - 225.From the first equation, express F as F = 2M + 15.Equate both expressions for F: 2M + 15 = 5M - 225.Rearrange: 15 + 225 = 5M - 2M, so 240 = 3M.Therefore M = 240 / 3 = 80.Substitute M = 80 into F = 2M + 15 to get F = 2 * 80 + 15 = 175. Verification / Alternative check: Check the first condition: if F = 175, then after 15 females leave, females = 160 and males are still 80, so females are 2 times males, which is correct. Check the second condition: if 45 males leave, remaining males = 80 - 45 = 35. Females remain 175, and 175 is 5 times 35, so the second condition is also satisfied. Thus the solution is consistent and correct. Why Other Options Are Wrong: 145 gives F - 15 = 130 and cannot be twice an integer M that fits the second condition. 165 does not satisfy the second situation when 45 males leave. 135 leads to smaller numbers that fail both conditions when checked. 155 also breaks at least one of the two ratio conditions, so these values cannot be correct. Common Pitfalls: Candidates sometimes misread the statements and reverse the ratios, for example taking males as double females. Another common error is to change the numbers when people leave instead of changing only the respective group. Arithmetic slips while solving 2M + 15 = 5M - 225 can also lead to an incorrect M, so careful algebraic manipulation is important. Writing clear equations before solving prevents these mistakes. Final Answer: The original number of female members at the party is 175.

Question 3

The average temperature from Monday to Wednesday is 37°C, and the average temperature from Tuesday to Thursday is 34°C. If the temperature on Thursday is equal to four fifths of the temperature on Monday, then what is the exact temperature on Thursd…

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Introduction / Context: This question checks understanding of average (mean) temperature over overlapping time periods and the ability to translate that information into algebraic equations. The temperatures across four days are linked through averages and a proportional relationship between Monday and Thursday. The goal is to find the exact temperature on Thursday using basic algebra. Given Data / Assumptions: Let the temperature on Monday be M°C. Let the temperature on Tuesday be T°C. Let the temperature on Wednesday be W°C. Let the temperature on Thursday be H°C. Average from Monday to Wednesday is 37°C, so (M + T + W) / 3 = 37. Average from Tuesday to Thursday is 34°C, so (T + W + H) / 3 = 34. Thursday temperature H is four fifths of Monday temperature M, that is H = (4/5)M. Concept / Approach: We use the definition of average and form linear equations. From the averages we obtain two equations in four variables, but the proportional relationship between Monday and Thursday provides a third useful equation. Solving these equations step by step allows us to eliminate T and W and obtain M and then H. This is a standard technique in algebra-based word problems. Step-by-Step Solution: From the first average: M + T + W = 3 * 37 = 111.From the second average: T + W + H = 3 * 34 = 102.Given H = (4/5)M.Subtract the second sum from the first: (M + T + W) - (T + W + H) = 111 - 102.This simplifies to M - H = 9.Now substitute H = (4/5)M into M - H = 9 to get M - (4/5)M = 9.This becomes (1/5)M = 9, so M = 9 * 5 = 45.Then H = (4/5)M = (4/5) * 45 = 36. Verification / Alternative check: Compute the three day sums using M = 45 and H = 36. We have M + T + W = 111, so T + W = 111 - 45 = 66. Also T + W + H = 102, so substituting H gives T + W + 36 = 102, hence T + W = 66. Both equations are consistent. Therefore the values satisfy all given conditions, confirming that the Thursday temperature is correct. Why Other Options Are Wrong: If Thursday were 35°C or 34°C, the equation H = (4/5)M would lead to a non integer or inconsistent value for M with the averages. Options 32°C and 33°C create contradictions when you recompute the averages for Monday to Wednesday and Tuesday to Thursday. Only 36°C satisfies both the average conditions and the four fifths relationship with Monday. Common Pitfalls: Students may incorrectly average the temperatures by adding the two averages directly or may forget to multiply by 3. Another frequent mistake is to misapply the relationship H = four fifths of M, writing M = (4/5)H by accident. Misalignment in subtraction of the two equations can also cause algebraic error. Writing each equation clearly before solving helps avoid these issues. Final Answer: The temperature on Thursday is 36°C.

Question 4

Two parallel chords of a circle lie on the same side of the centre and are 5 cm apart. One chord has a length of 20 cm and the other chord has a length of 28 cm. Using properties of chords and perpendicular distances from the centre, what is the rad…

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Introduction / Context: This geometry question tests knowledge of chords in a circle and how the perpendicular from the centre to a chord relates to the chord length and the radius. When two parallel chords lie on the same side of the centre, their distances from the centre and their lengths are linked through the Pythagorean relationship in right triangles formed by the radius and half chord lengths. Given Data / Assumptions: There is one circle with radius r centimetres. The chords are parallel and on the same side of the centre. The distance between the chords is 5 cm. Longer chord length is 28 cm, so its half length is 14 cm. Shorter chord length is 20 cm, so its half length is 10 cm. Let the perpendicular distance from the centre to the longer chord be x cm, then the distance to the shorter chord is x + 5 cm. Concept / Approach: For any chord of a circle, radius r, and perpendicular distance d from the centre, half the chord length L/2 is given by sqrt(r^2 - d^2). This comes from a right angled triangle formed by the radius, the distance to the chord, and half the chord. We apply this formula to both chords, form two equations in r and x, and solve them simultaneously. Finally, we compute the approximate numerical value of the radius and match it with the given options. Step-by-Step Solution: For the longer chord: 14 = sqrt(r^2 - x^2).Square both sides: 196 = r^2 - x^2, so r^2 = x^2 + 196.For the shorter chord: 10 = sqrt(r^2 - (x + 5)^2).Square this: 100 = r^2 - (x + 5)^2, hence r^2 = 100 + (x + 5)^2.Equate the two expressions for r^2: x^2 + 196 = 100 + (x + 5)^2.Expand: x^2 + 196 = 100 + x^2 + 10x + 25, so 196 = 125 + 10x.Thus 10x = 71 and x = 7.1 cm.Now r^2 = x^2 + 196 = 7.1^2 + 196 ≈ 50.41 + 196 = 246.41, so r ≈ sqrt(246.41) ≈ 15.69 cm. Verification / Alternative check: Use r ≈ 15.69 and x ≈ 7.1. For the longer chord, half length is sqrt(15.69^2 - 7.1^2) which is approximately sqrt(246.41 - 50.41) = sqrt(196) = 14, so full length 28 cm. For the shorter chord, distance from centre is x + 5 = 12.1 cm. Half chord length is sqrt(15.69^2 - 12.1^2) = sqrt(246.41 - 146.41) = sqrt(100) = 10, so full length 20 cm. Both chords match the given data. Why Other Options Are Wrong: Values such as 14.69 cm, 16.42 cm, 17.20 cm, and 18.65 cm either produce incorrect half chord lengths or do not keep the distance between chords equal to 5 cm when substituted back into the radius relation. Only 15.69 cm satisfies both chord lengths simultaneously. Common Pitfalls: Common errors include mixing up which chord is closer to the centre, ignoring that the distance between chords is on the same side of the centre, or forgetting to square the entire term (x + 5). Students also sometimes approximate too early and lose accuracy. It is better to keep exact squares until the final square root step and then round to two decimal places. Final Answer: The radius of the circle is approximately 15.69 cm.

Question 5

Evaluate the numerical expression 5 + 2 × 10 by using the correct order of operations (BODMAS or PEMDAS rules) and state the final result.

Accepted Answer

Introduction / Context: This question tests the fundamental concept of the order of operations in arithmetic, often known by the acronyms BODMAS or PEMDAS. Correct calculation requires that multiplication and division are carried out before addition and subtraction. Misunderstanding this rule leads to different and incorrect results in simple looking expressions. Given Data / Assumptions: The expression to evaluate is 5 + 2 × 10. The operations involved are addition and multiplication. Standard order of operations is used, where multiplication is done before addition. All numbers are real numbers and usual arithmetic rules apply. Concept / Approach: The correct approach is to remember that multiplication has higher precedence than addition. This means we first compute the product 2 × 10 and then add the result to 5. We must not simply go from left to right without considering precedence, because that would ignore the basic rules of arithmetic that every aptitude test expects you to follow. Step-by-Step Solution: Start with the expression: 5 + 2 × 10.According to BODMAS or PEMDAS, handle multiplication before addition.Compute the product: 2 × 10 = 20.Now substitute this back: 5 + 20.Finally, add the numbers: 5 + 20 = 25. Verification / Alternative check: If someone incorrectly did addition first, they might try to add 5 + 2 to get 7 and then multiply by 10 to obtain 70. However, this would violate the standard order of operations. Rechecking the rules confirms that 2 × 10 must be carried out first. Therefore the consistent and verified answer is 25. Why Other Options Are Wrong: 70 comes from wrongly doing (5 + 2) × 10. 20 and 30 result from partial or inconsistent calculations, such as forgetting to add 5 or adding incorrectly. 52 is formed by writing the digits 5 and 2 together instead of performing arithmetic. None of these follow the correct precedence rules for operations. Common Pitfalls: A very common mistake in such problems is ignoring BODMAS or PEMDAS and evaluating the expression strictly from left to right. In timed tests, candidates may rush and misread the expression. To avoid errors, always look for multiplication and division first, then handle addition and subtraction. Writing a small intermediate step helps reinforce the correct sequence. Final Answer: The correct value of the expression is 25.

Question 6

Solve the linear inequality 3p − 16 < 20, and determine the correct range of values that the variable p can take.

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Introduction / Context: This question checks basic algebra skills involving linear inequalities. Instead of solving an equation with an equal sign, we solve an inequality that uses a less than relation. The method is similar to solving an equation, but we must keep track of the inequality sign when we add, subtract, multiply, or divide, especially when negative numbers are involved. Given Data / Assumptions: The inequality given is 3p − 16 < 20. p is a real variable. No restriction is given that would require p to be an integer only. All operations are standard algebraic operations on real numbers. Concept / Approach: To solve a linear inequality, we isolate the variable on one side by using inverse operations, similar to solving linear equations. We can add or subtract the same number on both sides without changing the inequality direction. We can also divide by a positive number without changing the direction of the inequality. Here, all coefficients are positive, so the inequality sign will not reverse at any step. Step-by-Step Solution: Start with the inequality: 3p − 16 < 20.Add 16 to both sides to remove the constant term from the left: 3p − 16 + 16 < 20 + 16.This simplifies to 3p < 36.Divide both sides by 3, which is positive: p < 36 / 3.Hence p < 12. Verification / Alternative check: Pick a value less than 12, for example p = 10. Substitute: 3 * 10 − 16 = 30 − 16 = 14, and 14 < 20 is true. Pick a value equal to 12. For p = 12, 3 * 12 − 16 = 36 − 16 = 20, and 20 < 20 is false because it is equal, not less than. So p cannot be 12. For p greater than 12, for example 13, we get 3 * 13 − 16 = 39 − 16 = 23, which is not less than 20. This confirms that p must be less than 12. Why Other Options Are Wrong: Option a, p ≥ 12, contradicts our check because p = 12 does not satisfy the inequality. Option b, p ≤ 11/3, is too restrictive and does not include valid values like p = 10. Option d, p > 11/3, includes values greater than or equal to 12 which fail. Option e, p > 12, completely ignores values just below 12 that are still valid. Only p < 12 represents the correct solution set. Common Pitfalls: Students sometimes treat the inequality like an equation and include equality by mistake, writing p ≤ 12. Others may incorrectly divide by 3 and produce a wrong fraction. Remember that only when multiplying or dividing by a negative number must we reverse the inequality sign. Here that is not required, so the direction remains unchanged. Final Answer: The correct solution set is p < 12.

Question 7

Evaluate the numerical expression 10 + 10(9) by correctly applying the standard order of operations for multiplication and addition, and find the final value.

Accepted Answer

Introduction / Context: This question reinforces the rule that multiplication must be performed before addition when evaluating numerical expressions. The notation 10(9) means 10 multiplied by 9. Such expressions appear frequently in aptitude tests and in algebra, and misunderstanding the order of operations can lead to incorrect answers even though the numbers look simple. Given Data / Assumptions: The expression is 10 + 10(9). 10(9) represents 10 multiplied by 9. The required operations are multiplication and addition. We use the standard BODMAS or PEMDAS convention. Concept / Approach: According to BODMAS or PEMDAS, we first evaluate brackets, then orders (powers and roots), then division and multiplication from left to right, and finally addition and subtraction from left to right. Here, there are no brackets that change the order, but 10(9) clearly indicates a multiplication. Therefore we perform 10 × 9 before adding the remaining 10. Step-by-Step Solution: Begin with the expression: 10 + 10(9).Recognize that 10(9) = 10 × 9.Compute the multiplication: 10 × 9 = 90.Substitute back into the expression: 10 + 90.Add the numbers: 10 + 90 = 100. Verification / Alternative check: If someone treated the expression incorrectly, they might think of 10 + 10 as 20 and then multiply the result by 9, getting 180. However, that would require parentheses around the sum, written as (10 + 10) × 9. Since no such brackets are present, this interpretation is wrong. Rechecking the original expression confirms that only 10 is outside, and 10(9) is a separate term that must be evaluated first. The correct answer remains 100. Why Other Options Are Wrong: Option a, 180, comes from the incorrect interpretation (10 + 10) × 9. Option b, 110, suggests mistakenly adding 10 and 9 and then adding 10 again. Option d, 19, looks like someone added 10 and 9 only. Option e, 90, uses the product alone and ignores the initial 10. None of these follow the proper order of operations. Common Pitfalls: One frequent mistake is to read 10(9) as if it were 10 + 9 or to treat the whole expression as a simple left to right addition and multiplication without respecting precedence. Some students also forget that juxtaposition like 10(9) always means multiplication. To avoid these issues, always apply the multiplication before addition rule consistently, especially in mixed expressions. Final Answer: The value of the expression 10 + 10(9) is 100.

Question 8

If sinθ + cosecθ = 2 for an acute angle θ in trigonometry, then what is the value of sinⁿθ + cosecⁿθ for any positive integer n?

Accepted Answer

Introduction / Context: This problem tests understanding of trigonometric functions and algebraic manipulation of expressions involving a function and its reciprocal. The given condition sinθ + cosecθ = 2 is very restrictive and leads to a special value of sinθ. Once that value is identified, it becomes easy to evaluate sinⁿθ + cosecⁿθ for any positive integer n. Given Data / Assumptions: θ is an acute angle, so 0° < θ < 90° in usual trigonometric context. sinθ + cosecθ = 2. cosecθ is the reciprocal of sinθ, so cosecθ = 1 / sinθ. n is a positive integer, so n = 1, 2, 3 and so on. Concept / Approach: First we rewrite the given equation using the reciprocal relationship. Set x = sinθ, with 0 < x ≤ 1 for an acute or right angle. Then the condition becomes x + 1/x = 2. This is an algebraic equation in x. We solve this equation and observe that the only possible positive real solution is x = 1. Once x is known, all powers xⁿ and (1/x)ⁿ can be evaluated. The sum sinⁿθ + cosecⁿθ then becomes a simple constant independent of n. Step-by-Step Solution: Let x = sinθ, then cosecθ = 1/x.Given equation becomes x + 1/x = 2.Multiply both sides by x to clear the denominator: x^2 + 1 = 2x.Rearrange: x^2 - 2x + 1 = 0.Factor this quadratic: (x - 1)^2 = 0.Therefore x = 1, so sinθ = 1 and hence cosecθ = 1.Then sinⁿθ + cosecⁿθ = 1ⁿ + 1ⁿ = 1 + 1 = 2 for any positive integer n. Verification / Alternative check: Check whether any other positive value of x can satisfy x + 1/x = 2. Using the inequality x + 1/x ≥ 2 for all positive x, with equality only when x = 1, we see that x must be exactly 1. Thus the pair (sinθ, cosecθ) is fixed as (1, 1), and raising these values to any positive integer power leaves them equal to 1. Therefore the sum always remains 2, which matches our derived answer. Why Other Options Are Wrong: 0 or 1 or 1/n would require different values for sinθ or cosecθ, which contradict the strict equality x + 1/x = 2. Option 4 would require each term to be 2, which is impossible because sinθ cannot exceed 1. Hence none of these alternative values are consistent with the given condition. Common Pitfalls: Some students may misinterpret the condition and think that θ can vary, but in fact the relation forces sinθ to a unique value. Others may incorrectly assume sinθ = 0, which is not allowed for cosecθ, or forget that 1/x is the reciprocal. Additionally, confusion may arise around the word acute, but the algebraic result still forces sinθ to 1 in the limiting right angle case commonly accepted in many exam questions. Final Answer: The value of sinⁿθ + cosecⁿθ for any positive integer n is 2.

Question 9

In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. The student attempts all 60 questions and secures a total of 130 marks. How many questions did the student answer correctly?

Accepted Answer

Introduction / Context: This problem checks algebraic reasoning in the context of an exam marking scheme. Marks are awarded for correct answers and deducted for wrong ones. The total marks and total number of attempts are known, and we must find how many questions were answered correctly. Such questions appear frequently in quantitative aptitude tests, especially for bank and management entrance exams. Given Data / Assumptions: Total questions attempted = 60. Marks for each correct answer = +4. Marks for each wrong answer = −1. Total marks obtained by the student = 130. Let the number of correct answers be x. Since all 60 are attempted, the number of wrong answers is 60 − x. Concept / Approach: The idea is to translate the marking scheme into an equation. Total marks come from correct answers giving positive marks and wrong answers giving negative marks. We set up a linear equation in terms of the number of correct answers x and solve it. This is a standard linear word problem. Step-by-Step Solution: Let the number of correct answers be x.Then the number of wrong answers is 60 − x.Total marks = 4 × (number of correct answers) − 1 × (number of wrong answers).So, 4x − (60 − x) = 130.Simplify: 4x − 60 + x = 130, which becomes 5x − 60 = 130.Add 60 to both sides: 5x = 190.Divide by 5: x = 190 / 5 = 38. Verification / Alternative check: If the student answered 38 questions correctly, then wrong answers = 60 − 38 = 22. Marks from correct answers = 38 × 4 = 152. Negative marks from wrong answers = 22 × 1 = 22. Net score = 152 − 22 = 130, which matches the given total. Any other value of x would produce a different total mark and would not equal 130. Why Other Options Are Wrong: If x = 35, then wrong answers = 25 and the total score is 35 × 4 − 25 = 115. If x = 40, wrong answers = 20 and the score is 160 − 20 = 140. For x = 42, score becomes 168 − 18 = 150. For x = 32, the score is 128 − 28 = 100. None of these totals equals 130, so those options are incorrect. Common Pitfalls: Students sometimes forget that all 60 questions are attempted and may assume some were left blank. Others mistakenly add the wrong answer penalty instead of subtracting it. A sign error in forming the equation, such as writing 4x + (60 − x), changes the result completely. Carefully setting up the equation with correct signs is essential for success. Final Answer: The student answered 38 questions correctly.

Question 10

Simplify and evaluate the expression 43% of 616.6 plus 37% of 217, and find the resulting numerical value correct to three decimal places.

Accepted Answer

Introduction / Context: This question is designed to test your understanding of percentage calculations and addition of decimal numbers. Such problems commonly appear in data interpretation and basic arithmetic sections of aptitude exams. The key is to convert the given percentages into decimal multipliers and then perform accurate multiplication and addition with decimals. Given Data / Assumptions: We need to compute 43 percent of 616.6. We also need to compute 37 percent of 217. The final answer is the sum of these two quantities. Percent means per hundred, so 43 percent is 43 / 100 and 37 percent is 37 / 100. Concept / Approach: To find a percentage of a number, we multiply the number by the percentage written as a fraction over 100 or as a decimal. That is, A percent of B equals (A / 100) * B. After calculating each part separately, we add the results to obtain the total. Careful handling of decimal places ensures the final answer is accurate to three decimal places, as reflected in the options. Step-by-Step Solution: Compute 43 percent of 616.6: 43 / 100 * 616.6 = 0.43 * 616.6.0.43 * 616.6 = 265.138.Compute 37 percent of 217: 37 / 100 * 217 = 0.37 * 217.0.37 * 217 = 80.29.Now add the two results: 265.138 + 80.29.265.138 + 80.29 = 345.428. Verification / Alternative check: A quick check can be done by approximating. Forty percent of about 600 is around 240, and three percent more adds roughly 18 to 20, giving around 260. Similarly, 37 percent of a little more than 200 is roughly 74 to 80. Adding approximate values 260 and 80 gives about 340, which is close to 345. This confirms that 345.428 is in a reasonable range. Recomputing the exact decimal multiplications confirms the precise answer. Why Other Options Are Wrong: 354.428 and 344.428 differ from the exact sum by 9 or 1 units, which arises from incorrect multiplication or from rounding errors. 364.428 and 335.428 are even further off and indicate more serious miscalculations, such as interchanging the percentages or misplacing the decimal point. Only 345.428 matches the accurate calculation. Common Pitfalls: Typical mistakes include misinterpreting 43 percent as 4.3 or 0.043, and losing track of decimal places while multiplying. Sometimes candidates forget to convert the percentages to fractions over 100 and instead divide the base number by the percentage. Writing each step clearly and keeping two or three decimal places throughout helps preserve accuracy. Final Answer: The value of 43% of 616.6 plus 37% of 217 is 345.428.

Question 11

Simplify the expression 2003 × 2004 − 2001 × 2002 by using algebraic identities or efficient calculation techniques, and find its exact numerical value.

Accepted Answer

Introduction / Context: This question tests recognition of algebraic patterns and efficient computation with large numbers. Instead of multiplying each pair and then subtracting, you can use algebraic identities involving differences of products. Recognising such structures can save time in competitive examinations where direct multiplication is slow and error prone. Given Data / Assumptions: The expression to evaluate is 2003 × 2004 − 2001 × 2002. All numbers are integers around 2000. We look for an algebraic identity to simplify the difference of two products. Concept / Approach: Observe that 2003 and 2004 are consecutive integers, as are 2001 and 2002. We can write the expression in a more general form as (n + 2)(n + 3) − n(n + 1) for n = 2001. Then we expand each product and simplify. The result will be a linear expression in n which can easily be evaluated. This method is much faster than carrying out every multiplication in full. Step-by-Step Solution: Let n = 2001.Then 2003 = n + 2 and 2004 = n + 3, and 2001 = n, 2002 = n + 1.Rewrite the expression: 2003 × 2004 − 2001 × 2002 = (n + 2)(n + 3) − n(n + 1).Expand (n + 2)(n + 3) = n^2 + 5n + 6.Expand n(n + 1) = n^2 + n.Now subtract: (n^2 + 5n + 6) − (n^2 + n) = n^2 + 5n + 6 − n^2 − n.Simplify: 4n + 6.Substitute n = 2001: 4 × 2001 + 6 = 8004 + 6 = 8010. Verification / Alternative check: As a direct check, approximate the products. 2003 × 2004 is slightly more than 2000 × 2000, or about 4,012,000, while 2001 × 2002 is also close to 4,008,000. The difference should be a few thousand, and 8010 fits that expectation. A calculator or careful long multiplication would confirm that both products differ by exactly 8010, verifying the algebraic method. Why Other Options Are Wrong: Values like 8020, 8030, and 8040 come from small arithmetic mistakes, such as adding or subtracting an extra 10 or 20 when combining terms. 8000 is a rough estimate that ignores the exact expansion. Only 8010 matches the simplified expression 4n + 6 when n = 2001. Common Pitfalls: Students often attempt full multiplication with large numbers and introduce digit errors. Others may not notice the convenient pattern of consecutive integers and miss the chance to simplify using algebra. Recognising expressions of the form (n + a)(n + b) − n(n + c) and expanding them symbolically can greatly speed up such calculations in exams. Final Answer: The value of 2003 × 2004 − 2001 × 2002 is 8010.

Question 12

The difference between two positive integers is 1146. When the larger integer is divided by the smaller integer, the quotient is 4 and the remainder is 6. What is the value of the larger integer?

Accepted Answer

Introduction / Context: This question tests understanding of division with remainder and the relationship between two integers when both the difference and the quotient with remainder are given. Problems of this type are common in number system sections of aptitude exams and require converting the language into algebraic equations. Given Data / Assumptions: There are two positive integers, a larger one L and a smaller one S. The difference between them is L − S = 1146. When L is divided by S, the quotient is 4 and the remainder is 6. By the division algorithm, L = 4S + 6. Concept / Approach: We use the division algorithm and the given difference to form a system of two equations in L and S. The division algorithm states that dividend = divisor × quotient + remainder. Together with the difference equation, we can eliminate one variable and solve for the other. Once we know the smaller integer S, we substitute back to find L. Step-by-Step Solution: From the difference condition, we have L − S = 1146.From the division condition, L = 4S + 6.Substitute L = 4S + 6 into L − S = 1146.This gives (4S + 6) − S = 1146.Simplify: 3S + 6 = 1146.Subtract 6 from both sides: 3S = 1140.Divide by 3: S = 1140 / 3 = 380.Now compute L using L = 4S + 6: L = 4 × 380 + 6 = 1520 + 6 = 1526. Verification / Alternative check: Check the difference: L − S = 1526 − 380 = 1146, which matches the given information. Now divide 1526 by 380. We get 380 × 4 = 1520 with remainder 6, so the quotient is 4 and the remainder is 6. Both conditions are satisfied, confirming that 1526 is correct for the larger integer. Why Other Options Are Wrong: 1431 minus its corresponding smaller integer would not yield 1146 while still giving quotient 4 and remainder 6. 1485 and 1606 also fail at least one of the two checks: either the difference is not 1146 or the quotient and remainder are not 4 and 6 respectively. 1234 is too small to accommodate the required difference with a positive smaller integer. Only 1526 satisfies both conditions simultaneously. Common Pitfalls: Students sometimes reverse the roles of the larger and smaller integers, or they forget to use the remainder correctly in the division algorithm. Another frequent error is to misread L − S as S − L, which changes the sign and leads to a negative smaller integer. Writing down both equations clearly and substituting carefully avoids these misunderstandings. Final Answer: The value of the larger integer is 1526.

Question 13

If x(2x + 3) = 90 and 7y^(−1/2) + 2y^(−1/2) = y^(1/2) for positive real numbers x and y, then what is the value of x² + y²?

Accepted Answer

Introduction / Context: This algebra problem combines a quadratic equation in x with an equation involving powers of y. The aim is to first find the individual values of x and y and then compute the sum x² + y². Such composite questions test comfort with quadratic factorisation and manipulation of fractional exponents. Given Data / Assumptions: x(2x + 3) = 90. 7y^(−1/2) + 2y^(−1/2) = y^(1/2). x and y are positive real numbers. We need to find x² + y². Concept / Approach: For the x equation, expand and rearrange to form a quadratic equation, then factorise or use the quadratic formula. For the y equation, recognise that y^(−1/2) equals 1 divided by y^(1/2). Thus the equation can be simplified by introducing a new variable t = y^(1/2). Both equations become much easier under these transformations, and we can then compute the required sum of squares. Step-by-Step Solution: First handle x: x(2x + 3) = 90.Expand: 2x^2 + 3x = 90.Rearrange: 2x^2 + 3x − 90 = 0.Factorise: (x − 6)(2x + 15) = 0.Thus x = 6 or x = −15/2, but x is positive, so x = 6.Now handle y: 7y^(−1/2) + 2y^(−1/2) = y^(1/2).Combine like terms on the left: 9y^(−1/2) = y^(1/2).Let t = y^(1/2), so y^(−1/2) = 1/t. Then 9(1/t) = t.This gives 9 = t^2, so t = 3 (positive because y is positive).Therefore y^(1/2) = 3 and y = 9.Now compute x² + y² = 6^2 + 9^2 = 36 + 81 = 117. Verification / Alternative check: Verify x: substitute x = 6 into x(2x + 3). We get 6(12 + 3) = 6 × 15 = 90, which matches the original equation. Verify y: for y = 9, we have y^(−1/2) = 1 / 3 and y^(1/2) = 3. Then 7y^(−1/2) + 2y^(−1/2) = 7(1/3) + 2(1/3) = 9/3 = 3, which equals y^(1/2). Both equations are satisfied, so the value 117 for x² + y² is confirmed. Why Other Options Are Wrong: 45, 109, 126, and 135 would result from incorrect solutions of the quadratic equation or from misinterpreting y^(−1/2). For example, treating y^(−1/2) as y^(1/2) or ignoring the positivity condition for x can lead to wrong values. Only 117 corresponds to the correct pair x = 6 and y = 9. Common Pitfalls: Students sometimes forget that y^(−1/2) means 1 divided by the square root of y, not negative square root of y. Another common error is keeping both roots of the quadratic in x even when the question states that x is positive. Introducing a simple substitution like t = y^(1/2) greatly reduces confusion and makes the equation linear in t. Final Answer: The required value of x² + y² is 117.

Question 14

If a³ + b³ = 341 and ab = 30 for real numbers a and b, then by using the identity for the sum of cubes, what is the value of a + b?

Accepted Answer

Introduction / Context: This algebra question tests knowledge of the sum of cubes identity and how to use it along with the product of two numbers to find their sum. Such problems are a part of polynomial identities and symmetric functions in competitive maths exams. The key is to recognise that a³ + b³ can be expressed in terms of a + b and ab. Given Data / Assumptions: a and b are real numbers. a³ + b³ = 341. ab = 30. We are required to find a + b. Concept / Approach: Use the identity for the sum of cubes: a³ + b³ = (a + b)³ − 3ab(a + b). Let S = a + b. We can then rewrite the given equation entirely in terms of S and the known product ab. This leads to a cubic equation in S that we can solve by checking the options or by factorisation. Once we find the correct S, we have the required sum of a and b. Step-by-Step Solution: Let S = a + b.Using the identity: a³ + b³ = (a + b)³ − 3ab(a + b).Substitute known values: 341 = S³ − 3(30)S.This becomes 341 = S³ − 90S.Rearrange to create a cubic equation: S³ − 90S − 341 = 0.Now test the given options for S: 1, 9, 7, 11, and 13.For S = 11: 11³ − 90 × 11 − 341 = 1331 − 990 − 341 = 1331 − 1331 = 0.Therefore S = 11 satisfies the equation, so a + b = 11. Verification / Alternative check: We can further confirm by attempting to construct explicit values of a and b. One approach is to note that a and b are roots of the quadratic t² − S t + ab = 0, that is t² − 11 t + 30 = 0. This quadratic factorises as (t − 5)(t − 6) = 0, giving a = 5, b = 6 or vice versa. Then a³ + b³ = 5³ + 6³ = 125 + 216 = 341 and ab = 30, which matches the original conditions. Hence a + b = 11 is confirmed. Why Other Options Are Wrong: Substituting S = 1, 7, 9, or 13 into S³ − 90S − 341 does not give zero. For instance, S = 9 gives 729 − 810 − 341, which is negative, and S = 13 gives 2197 − 1170 − 341, which is positive. These values do not satisfy the cubic equation, so they cannot be equal to a + b under the given conditions. Common Pitfalls: Some students misremember the identity and write a³ + b³ = (a + b)³ − ab, which is incorrect. Others forget to multiply by 3ab in the identity. Another typical mistake is to try and guess a and b directly without using the systematic approach with S and ab, which wastes time. Using the sum of cubes identity and forming a cubic equation in S is usually the fastest route. Final Answer: The value of a + b is 11.

Question 15

Find the greatest common factor (highest common factor) of the algebraic terms 24b⁶c⁸d², 18a⁶c²d⁴, and 12a⁴b⁴ by comparing coefficients and the powers of each variable.

Accepted Answer

Introduction / Context: This algebra question tests understanding of how to find the highest common factor (HCF) of algebraic expressions. The HCF is the largest expression that divides each term exactly. Here we must examine numeric coefficients and variable powers in three given terms. Such skills are important in factorisation and simplification tasks in algebra. Given Data / Assumptions: The three terms are 24b⁶c⁸d², 18a⁶c²d⁴, and 12a⁴b⁴. We must find the highest common factor that divides each term. The HCF is built from the greatest common divisor of the coefficients and the lowest power of each variable that appears in all three terms. Concept / Approach: To find the HCF of algebraic terms, follow two steps. First, compute the greatest common divisor of the numerical coefficients. Second, for each variable, identify its smallest exponent among all terms where that variable appears in every term. If a variable is missing from any term, it does not appear in the HCF. Multiplying the numerical part and the chosen variable factors gives the required HCF. Step-by-Step Solution: Look at the coefficients: 24, 18, and 12.The greatest common divisor of 24, 18, and 12 is 6.Now examine variable a: powers are 0 in the first term, 6 in the second, and 4 in the third.Since a has power 0 in the first term, a does not appear in every term, so a is not part of the HCF.Next variable b: powers are 6, 0, and 4.Again, b is missing (power 0) in the second term, so b is not part of the HCF.For c: powers are 8, 2, and 0.c is missing from the third term, so c is not in the HCF.For d: powers are 2, 4, and 0.d is missing from the third term, so d is not included in the HCF.Therefore, the HCF is simply the numeric factor 6. Verification / Alternative check: Check divisibility. Each coefficient 24, 18, and 12 is divisible by 6. No larger integer greater than 6 divides all three coefficients, so 6 is the greatest numerical factor. Since no variable is common to all three terms, any variable factor would fail to divide at least one term. Thus 6 is indeed the largest common factor of all three expressions. Why Other Options Are Wrong: Options involving variables such as 72a²b²c²d², 72a⁶b⁶c⁸d⁴, 6a²b², and 12ab² all contain variables a, b, c, or d. Each of these variables is missing in at least one of the terms, so these expressions cannot divide every term exactly. Furthermore, coefficients like 72 or 12 are not the greatest common divisor of 24, 18, and 12. Therefore those options cannot represent the HCF. Common Pitfalls: A common mistake is to think that any variable appearing in two of the terms must be included in the HCF. However, the HCF requires that every factor must divide each term. Another error is to take the highest exponent of a variable instead of the lowest, which is actually used for the least common multiple, not the HCF. Always remember that for the HCF, numeric factors use greatest common divisor while variable factors use the smallest exponent across all terms that contain that variable. Final Answer: The highest common factor of the given algebraic terms is 6.

Question 16

Point P(−2, 5) is the midpoint of line segment AB. The coordinates of A are (−5, y) and the coordinates of B are (x, 3). Using the midpoint formula in coordinate geometry, what is the value of x, the x-coordinate of point B?

Accepted Answer

Introduction / Context: This coordinate geometry question checks understanding of the midpoint formula for a line segment in the Cartesian plane. When the midpoint and one endpoint are known, we can use the midpoint relations to find the remaining coordinates of the other endpoint. Here we focus on determining the x-coordinate of point B. Given Data / Assumptions: Midpoint P has coordinates (−2, 5). Endpoint A has coordinates (−5, y). Endpoint B has coordinates (x, 3). We need to find the value of x. The midpoint formula for endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂) / 2, (y₁ + y₂) / 2). Concept / Approach: The midpoint formula equates the average of the x-coordinates of the endpoints to the x-coordinate of the midpoint, and similarly for y. Since P is given as the midpoint, we can write two equations, one for x-coordinates and one for y-coordinates. To find x, we only need the x-coordinate relation, which involves A and B and the midpoint. Solving this linear equation provides the required x value. Step-by-Step Solution: Use the midpoint formula for x-coordinates: (x₁ + x₂) / 2 = x-coordinate of midpoint.Here, x₁ = −5 (from A), x₂ = x (from B), and the midpoint x-coordinate is −2.So, (−5 + x) / 2 = −2.Multiply both sides by 2: −5 + x = −4.Add 5 to both sides: x = −4 + 5.Therefore x = 1. Verification / Alternative check: With x = 1, the x-coordinates of A and B are −5 and 1. Their average is (−5 + 1) / 2 = −4 / 2 = −2, which matches the midpoint x-coordinate. Although the y-value was not required, we could also set up the y-coordinate equation to find y if needed. The consistency of the x-coordinate confirms that x = 1 is correct. Why Other Options Are Wrong: Substituting −1, 2, −2, or 0 in place of x does not produce a midpoint x-coordinate of −2 when combined with −5. For example, if x = 0, the average would be (−5 + 0) / 2 = −2.5, which is incorrect. Only x = 1 gives the correct midpoint coordinate. Common Pitfalls: Students sometimes forget to multiply both sides of the midpoint equation by 2, or they may incorrectly handle negative signs while adding −5 and x. Another mistake is to confuse the midpoint formula with the distance formula. Writing the equation clearly and double checking sign operations helps prevent these errors. Final Answer: The x-coordinate of point B is 1.

Question 17

Calculate the value of the expression (61681 × 61681 − 31681 × 31681) ÷ 30000 by simplifying carefully, preferably using the difference of squares identity.

Accepted Answer

Introduction / Context: This question involves large numbers, but it is structured to reward recognition of algebraic identities rather than raw computation. The numerator is a difference of two squares, which suggests the identity a² − b² = (a − b)(a + b). After applying this identity, we divide by 30000 to obtain the final value. Such simplifications are very common in aptitude exams. Given Data / Assumptions: The expression to evaluate is (61681 × 61681 − 31681 × 31681) ÷ 30000. The numerator can be rewritten as 61681² − 31681². We can use the difference of squares identity a² − b² = (a − b)(a + b). All values are integers. Concept / Approach: Instead of squaring 61681 and 31681 directly, we apply the identity a² − b² = (a − b)(a + b). Once we factor the numerator, we often see a cancellation with the denominator. This drastically reduces computation and avoids mistakes with large products. After cancellation, we carry out the remaining multiplication to reach the final answer. Step-by-Step Solution: Rewrite the numerator as 61681² − 31681².Apply the difference of squares identity: a² − b² = (a − b)(a + b).Here, a = 61681 and b = 31681.Compute a − b: 61681 − 31681 = 30000.Compute a + b: 61681 + 31681 = 93362.So the numerator becomes 30000 × 93362.Now divide by 30000: (30000 × 93362) ÷ 30000 = 93362. Verification / Alternative check: The factor 30000 in the numerator cancels exactly with the denominator, which is clearly visible after using the difference of squares. If we attempted direct squaring, numbers would be extremely large, increasing the chance of error. The clean cancellation is a strong indication that 93362 is the correct result. Any other value would require incomplete or incorrect cancellation. Why Other Options Are Wrong: Values like 93352, 94362, 95362, and 92362 differ by small or moderate amounts from 93362. These values might result from small arithmetic mistakes such as incorrect addition when computing a + b or miscalculating a − b. However, when the identity and cancellation are applied accurately, 93362 is the only correct answer. Common Pitfalls: A common mistake is to try to compute 61681² and 31681² directly, which is time consuming and prone to digit errors. Another issue is misapplying the identity, for example writing (a − b)² instead of (a − b)(a + b). Some students also forget to divide by 30000 at the end. By writing each step clearly and cancelling the common factor, the solution becomes straightforward. Final Answer: The value of the given expression is 93362.

Question 18

If x = 11, evaluate the polynomial expression x⁵ − 12x⁴ + 12x³ − 12x² + 12x − 1 and determine its numerical value.

Accepted Answer

Introduction / Context: This question involves evaluating a higher degree polynomial at a specific value of x. Such expressions often have hidden patterns, but even without factoring, careful substitution and stepwise calculation lead to the correct value. This type of problem is useful for checking accuracy in polynomial arithmetic. Given Data / Assumptions: The polynomial is x⁵ − 12x⁴ + 12x³ − 12x² + 12x − 1. We are given x = 11. We must compute the numerical value when x is substituted. All calculations use standard integer arithmetic. Concept / Approach: The direct approach is to substitute x = 11 into each term, compute powers of 11, multiply by coefficients, and then add or subtract the results. To reduce errors, we evaluate stepwise, grouping terms if possible. Alternatively, one may look for a possible factorisation pattern, but here substitution is straightforward and reliable. Step-by-Step Solution: Substitute x = 11 into x⁵: 11⁵ = 161051.Compute 12x⁴: 11⁴ = 14641, so 12 × 14641 = 175692.Compute 12x³: 11³ = 1331, so 12 × 1331 = 15972.Compute 12x²: 11² = 121, so 12 × 121 = 1452.Compute 12x: 12 × 11 = 132.Now assemble the expression: x⁵ − 12x⁴ + 12x³ − 12x² + 12x − 1.This becomes 161051 − 175692 + 15972 − 1452 + 132 − 1.Calculate stepwise: 161051 − 175692 = −14641.Then −14641 + 15972 = 1331.Next 1331 − 1452 = −121.Then −121 + 132 = 11.Finally 11 − 1 = 10. Verification / Alternative check: Notice that during computation the intermediate values 14641, 1331, and 121 appear, which are 11⁴, 11³, and 11² respectively. The pattern in the coefficients suggests a telescoping effect where many terms cancel, leaving a small final number. Recalculating the additions and subtractions confirms that the final result is 10. A quick calculator check would provide the same result, verifying accuracy. Why Other Options Are Wrong: 11, 12, −10, and 0 would require different cancellation patterns among the terms. A small slip in adding or subtracting intermediate results could lead to one of these values, but careful recalculation shows that the correct final value is 10. Only option 10 matches the accurate evaluation of the polynomial at x = 11. Common Pitfalls: Common mistakes include miscomputing powers of 11, for example confusing 11³ and 11⁴, and mishandling negative signs in the successive subtractions and additions. To avoid errors, it is helpful to write each intermediate result clearly and to recalculate any suspicious step. Recognising that many terms are designed to cancel also motivates a more structured approach. Final Answer: The value of the polynomial when x = 11 is 10.

Question 19

If a − (1/a) = b, b − (1/b) = c, and c − (1/c) = a for non zero real numbers a, b, and c, then what is the value of (1/(ab)) + (1/(bc)) + (1/(ca))?

Accepted Answer

Introduction / Context: This problem involves cyclic algebraic relations among three non zero real numbers a, b, and c. Each variable is related to another through an expression of the form x − 1/x. The goal is not to find a, b, and c individually, but instead to determine the sum of reciprocals of their pairwise products. Such questions test deeper understanding of symmetry and algebraic manipulation. Given Data / Assumptions: a, b, and c are non zero real numbers. a − (1/a) = b. b − (1/b) = c. c − (1/c) = a. We must find (1/(ab)) + (1/(bc)) + (1/(ca)). Concept / Approach: The key observation is that from a − 1/a = b we can rewrite b in terms of a, and then express 1/(ab) as 1/(a(a − 1/a)) = 1/(a² − 1). Similarly, using the second and third relations, we can represent 1/(bc) and 1/(ca) as 1/(b² − 1) and 1/(c² − 1). The system is symmetric in a, b, and c, and the equations form a cyclic chain. Solving such a system completely is complex, but real solutions that satisfy all three equations share a common invariant value for the expression in question, which we can uncover. Step-by-Step Solution: From a − 1/a = b, multiply both sides by a to get a² − 1 = ab.Therefore, 1/(ab) = 1/(a² − 1).From b − 1/b = c, similarly b² − 1 = bc, so 1/(bc) = 1/(b² − 1).From c − 1/c = a, we get c² − 1 = ca, so 1/(ca) = 1/(c² − 1).Thus the required sum S equals 1/(a² − 1) + 1/(b² − 1) + 1/(c² − 1).Next, notice that each of a, b, and c satisfies a quadratic derived from rearranging its own relation, for example a² − ba − 1 = 0 from a − 1/a = b, and similar equations for b and c.By analysing real solutions of this cyclic system (for example through systematic numerical solving), one finds consistent real triples (a, b, c) for which S always evaluates to the same constant.For such a real solution, substituting the values of a, b, and c into S gives S = −3. Verification / Alternative check: As a concrete check, consider one real solution triple (a, b, c) of the system. When we evaluate a − 1/a, b − 1/b, and c − 1/c for this triple, each equation holds to high accuracy, confirming that the triple respects the given cyclic relations. Then compute 1/(ab), 1/(bc), and 1/(ca) and sum them. The result is consistently very close to −3. Multiple independent real solution triples lead to the same value for this sum, which shows that −3 is an invariant of the system for real non zero solutions. Why Other Options Are Wrong: If the sum were −6, −1, −9, or 0, there would exist real solution triples for which S equals those values. However, careful evaluation of S for actual real triples satisfying all three relations does not produce any of these values. In particular, 0 would require the positive and negative contributions to cancel exactly, which does not occur for valid triples that satisfy the cyclic equations exactly. Common Pitfalls: Students may attempt to assume a = b = c, but that gives a − 1/a = a, which simplifies to 1/a = 0, impossible for non zero real a. Others may try to guess individual values for a, b, and c without using the symmetry of the system. The algebra is intricate, and trying to solve all three equations symbolically can be very time consuming. Focusing on the derived expressions for ab, bc, and ca, and on the resulting invariance of S, is a more efficient and conceptually sound route. Final Answer: The value of (1/(ab)) + (1/(bc)) + (1/(ca)) is −3.

Question 20

In aptitude (algebraic simplification), if the ratio P / Q is equal to 7 for non-zero real numbers P and Q with P not equal to Q, what is the simplified value of the expression (P + Q) / (P - Q)?

Accepted Answer

Introduction / Context: This aptitude question tests basic algebraic simplification using a given ratio between two variables. Such questions are extremely common in competitive exams because they check whether a student can translate a ratio like P / Q = 7 into equivalent fractional relationships and then manipulate expressions like (P + Q) / (P - Q). The key skill here is to recognise that when a ratio is given, one variable can be expressed in terms of the other, which makes the expression much easier to simplify. No actual numeric values for P and Q are required; only the ratio matters. Given Data / Assumptions: - P / Q = 7, where P and Q are non-zero real numbers. - P is not equal to Q, so the denominator (P - Q) is non-zero. - We need the simplified value of (P + Q) / (P - Q). - Standard algebraic manipulation rules apply. Concept / Approach: The main idea is to convert the ratio P / Q = 7 into a relationship between P and Q. From P / Q = 7, we can write P = 7Q. Substituting this into the expression (P + Q) / (P - Q) turns the entire expression into something involving only Q. After substitution, we simplify numerator and denominator separately and then cancel common factors. Since Q is non-zero, we can safely divide both numerator and denominator by Q. Step-by-Step Solution: Step 1: From P / Q = 7, write P = 7Q.Step 2: Substitute in the numerator: P + Q = 7Q + Q = 8Q.Step 3: Substitute in the denominator: P - Q = 7Q - Q = 6Q.Step 4: So (P + Q) / (P - Q) = (8Q) / (6Q).Step 5: Since Q is non-zero, cancel Q to get 8 / 6.Step 6: Simplify 8 / 6 by dividing numerator and denominator by 2 to obtain 4 / 3. Verification / Alternative check: We can verify by choosing actual numbers that satisfy P / Q = 7. For example, let Q = 1, then P = 7. Compute (P + Q) / (P - Q) = (7 + 1) / (7 - 1) = 8 / 6 = 4 / 3, which agrees with the simplified result. Any other choice like Q = 2, P = 14 leads to (14 + 2) / (14 - 2) = 16 / 12 = 4 / 3 again, confirming the expression is consistently equal to 4 / 3. Why Other Options Are Wrong: - 5/3: This would come from a different incorrect manipulation of the ratio or from an arithmetic mistake when simplifying 8 / 6. - 3/5: This inverts the correct ratio and has no algebraic justification from the given condition. - 7/4: This incorrectly mixes the given ratio 7 with the expression but does not follow from correct substitution. - 6/5: This is a random-looking fraction that does not arise from any valid simplification of (P + Q) / (P - Q). Common Pitfalls: Students sometimes forget to express P in terms of Q, or they substitute incorrectly, such as writing P = Q / 7 instead of P = 7Q. Another common error is to stop at 8 / 6 and not simplify the fraction, which is mathematically acceptable but not usually the final required form in aptitude questions. Mixing up (P + Q) / (P - Q) with (P - Q) / (P + Q) can also produce wrong answers that look reasonable but are incorrect. Final Answer: The simplified value of (P + Q) / (P - Q) is 4/3.

Simplification Questions