Q: If 9x^2 + 16y^2 = 60 and 3x + 4y = 6, what is the value of the product xy?

Introduction / Context: This question combines simultaneous equations and algebraic identities to find the product xy without necessarily solving for x and y individually. It is a neat example of how squaring a linear equation and comparing with a quadratic relationship can reveal cross terms like xy. Given Data / Assumptions: 9x^2 + 16y^2 = 60. 3x + 4y = 6. We need to find the value of xy. Concept / Approach: Main ideas: Square the linear equation (3x + 4y)^2 to produce terms in x^2, y^2, and xy. Compare the result with the given quadratic relation 9x^2 + 16y^2 = 60. Subtract to isolate the xy term. Step-by-Step Solution: Start from 3x + 4y = 6. Square both sides: (3x + 4y)^2 = 6^2. Expand the left side: 9x^2 + 24xy + 16y^2 = 36. We are also given 9x^2 + 16y^2 = 60. Subtract the quadratic equation from the squared equation: (9x^2 + 24xy + 16y^2) − (9x^2 + 16y^2) = 36 − 60. This simplifies to 24xy = −24. Therefore xy = −24 / 24 = −1. Verification / Alternative check: We can solve the system explicitly to verify. From 3x + 4y = 6, express x in terms of y and substitute into 9x^2 + 16y^2 = 60. Solving this pair will give numerical values of x and y whose product equals −1, confirming the derived result without needing to show every intermediate step here. Why Other Options Are Wrong: Option b: 1 would require 24xy = 24, contradicting the calculated equation 24xy = −24. Option c and option d: −2 or 2 would require different constants on the right side of the equation obtained after subtraction. Option e: 0 would mean 3x + 4y squared equals 9x^2 + 16y^2, which happens only if xy = 0, inconsistent with 24xy = −24. Common Pitfalls: Students may incorrectly expand (3x + 4y)^2 or mis-handle the subtraction of the two quadratic equations, leading to sign mistakes. Another pitfall is trying to solve directly for x and y when a more elegant approach via comparison is faster and less error prone. Final Answer: The value of the product xy is −1.

Q: Find the expanded square of the binomial expression (7 − 4x) and write it as a standard quadratic expression in x.

Introduction / Context: This question is about expanding the square of a binomial, a very common algebra skill. Recognising and applying the identity for (a − b)^2 or (a + b)^2 allows you to quickly convert a compact expression into a standard polynomial form, which is heavily used in solving quadratic equations and simplifying algebraic expressions. Given Data / Assumptions: The binomial is (7 − 4x). We need to compute (7 − 4x)^2. The answer should be written as a quadratic expression in x. Concept / Approach: Main identity: (a − b)^2 = a^2 − 2ab + b^2. Here, a = 7 and b = 4x. Compute each term carefully: a^2, 2ab, and b^2. Step-by-Step Solution: Write the binomial square: (7 − 4x)^2. Identify a = 7 and b = 4x. Compute a^2 = 7^2 = 49. Compute 2ab = 2 * 7 * 4x = 56x. Compute b^2 = (4x)^2 = 16x^2. Apply the identity: (7 − 4x)^2 = a^2 − 2ab + b^2. So (7 − 4x)^2 = 49 − 56x + 16x^2. Rearrange into standard quadratic order: 16x^2 − 56x + 49. Verification / Alternative check: You can expand directly by multiplication: (7 − 4x)(7 − 4x). Multiply term by term: 7 * 7 = 49, 7 * (−4x) = −28x, (−4x) * 7 = −28x, and (−4x) * (−4x) = 16x^2. Adding these gives 49 − 56x + 16x^2, which matches the result from the identity. Why Other Options Are Wrong: Option a: 16x^2 − 28x + 49 uses 28x instead of 56x, forgetting that the middle term is twice the product ab. Option b and option c: These forms either have incorrect coefficients or wrong signs for the quadratic term and do not match the proper expansion. Option e: 49 + 56x + 16x^2 corresponds to (7 + 4x)^2, not (7 − 4x)^2, so the middle term has the wrong sign. Common Pitfalls: A frequent error is to forget the factor of 2 in the middle term or to mis-handle the sign when the binomial includes a minus. Carefully applying the identity and double checking the sign of 2ab prevents such mistakes. Final Answer: The expanded quadratic expression is 16x^2 − 56x + 49.

Q: Find the roots of the quadratic equation x^2 − 25x + 156 = 0 and identify the correct ordered pair of solutions.

Introduction / Context: This problem is a standard quadratic equation solving exercise. It checks whether you can factorise a quadratic expression of the form x^2 − 25x + 156 and correctly identify its two real roots. Such equations appear frequently in algebra and aptitude tests. Given Data / Assumptions: The equation is x^2 − 25x + 156 = 0. We look for real roots. We can attempt factorisation into (x − p)(x − q) = 0. Concept / Approach: Main ideas: For x^2 − 25x + 156, find two numbers whose product is 156 and whose sum is 25. Use these numbers to factor the quadratic expression. Set each factor equal to zero to find the roots. Step-by-Step Solution: We seek integers p and q such that p + q = 25 and pq = 156. Try factor pairs of 156: 12 × 13 = 156 and 12 + 13 = 25. So p = 12 and q = 13 satisfy the conditions. Write the quadratic as x^2 − 25x + 156 = (x − 12)(x − 13). Set each factor equal to zero: x − 12 = 0 or x − 13 = 0. Thus x = 12 or x = 13. Verification / Alternative check: Substitute x = 12 into the original equation: 12^2 − 25 * 12 + 156 = 144 − 300 + 156 = 0. For x = 13: 169 − 325 + 156 = 0. Both values satisfy the equation, confirming that they are indeed roots. Why Other Options Are Wrong: Option b: 25 and 1 have product 25, not 156, so they cannot be the roots of this quadratic. Option c: 9 and 16 have product 144 and sum 25, but the constant term in the equation is 156, not 144. Option d: 31 and 6 have product 186 and do not even sum to 25. Option e: 8 and 17 have product 136, which again is not equal to 156. Common Pitfalls: A frequent mistake is to miscalculate factor pairs or to rely on guessing without checking both the sum and product conditions. Another error is to factor incorrectly and then not verify the roots. Systematically listing factor pairs and checking them, or using the quadratic formula as a backup, helps avoid issues. Final Answer: The roots of the quadratic equation are 12 and 13.

Question 1

The point P(5, -2) divides the segment joining A(x, 0) and B(0, y) internally in the ratio 2 : 5. Using the section formula for internal division, find the exact values of x and y.

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Introduction / Context: This coordinate geometry question tests the section formula (also called the internal division formula). When a point divides a line segment joining two points in a given ratio m:n internally, its coordinates are a weighted average of the endpoints. Here, the endpoints are on the axes: A is on the X-axis at (x, 0) and B is on the Y-axis at (0, y). The dividing point P is given as (5, -2). Using the internal section formula separately for x-coordinate and y-coordinate gives two equations, allowing us to solve for x and y exactly. Given Data / Assumptions: • A = (x, 0) • B = (0, y) • P = (5, -2) • P divides AB internally in ratio 2 : 5 (AP : PB = 2 : 5) • Required: values of x and y Concept / Approach: Internal division (section) formula: If P divides A(x1, y1) and B(x2, y2) in ratio m:n internally, then: Px = (m*x2 + n*x1) / (m + n) Py = (m*y2 + n*y1) / (m + n) Here m = 2, n = 5, A = (x, 0) and B = (0, y). Substitute and solve the two equations: 5 = (2*0 + 5*x)/7 and -2 = (2*y + 5*0)/7. Step-by-Step Solution: 1) Identify values: A(x1, y1) = (x, 0), B(x2, y2) = (0, y), m = 2, n = 5 2) Apply section formula for x-coordinate: Px = (m*x2 + n*x1)/(m + n) = (2*0 + 5*x)/7 = 5 3) Solve for x: (5x)/7 = 5 5x = 35 x = 7 4) Apply section formula for y-coordinate: Py = (m*y2 + n*y1)/(m + n) = (2*y + 5*0)/7 = -2 5) Solve for y: (2y)/7 = -2 2y = -14 y = -7 6) Therefore: x = 7 and y = -7 Verification / Alternative check: With x = 7 and y = -7: A = (7, 0), B = (0, -7). Compute P using formula: Px = (2*0 + 5*7)/7 = 35/7 = 5, Py = (2*(-7) + 5*0)/7 = -14/7 = -2. It matches P(5, -2), confirming correctness. Why Other Options Are Wrong: • Any option with x = -7 would give Px = (5*(-7))/7 = -5, not +5. • Any option with y = +7 would give Py = (2*(7))/7 = 2, not -2. • x = ±3 values fail the x-coordinate equation immediately. Common Pitfalls: • Reversing the weights m and n in the section formula. • Using external division formula by mistake (signs change there). • Forgetting that A is (x, 0) and B is (0, y), so one coordinate is already zero at each endpoint. Final Answer: x = 7; y = -7

Question 2

A TV was sold at a profit of 5%. If it had been sold at a profit of 10%, the profit would have been ₹1,000 more. Find the cost price (CP) of the TV in ₹ by using the difference in profit percentages.

Accepted Answer

Introduction / Context: This profit and loss question tests understanding of profit percentage on cost price. The key idea is that the difference between two profit scenarios (10% profit versus 5% profit) is simply the difference in profit percentages applied to the same cost price. Because the cost price does not change, the extra profit amount directly corresponds to an extra 5% of the cost price. Once you set up that relationship, the cost price can be found in one clean division. These problems are common in aptitude because they reward seeing the shortcut rather than calculating two separate selling prices first. Given Data / Assumptions: • Profit at first sale = 5% of CP • Profit at second (hypothetical) sale = 10% of CP • Difference in profit amounts = ₹1,000 • Required: cost price (CP) Concept / Approach: Profit difference percentage = 10% - 5% = 5%. So the extra ₹1,000 profit equals 5% of CP. Write: 5% of CP = 1000. Convert 5% to fraction: 5% = 5/100 = 1/20. So (1/20)*CP = 1000, hence CP = 1000 * 20 = 20000. Step-by-Step Solution: 1) Compute the difference in profit percentages: 10% - 5% = 5% 2) This difference corresponds to the extra profit amount: 5% of CP = ₹1,000 3) Convert 5% to a fraction: 5% = 5/100 = 1/20 4) Form the equation: (1/20) * CP = 1000 5) Multiply both sides by 20: CP = 1000 * 20 = 20000 6) Therefore the cost price is ₹20,000. Verification / Alternative check: If CP = 20,000: 5% profit = 0.05 * 20,000 = 1,000. 10% profit = 0.10 * 20,000 = 2,000. Difference = 2,000 - 1,000 = 1,000, which matches the condition. So CP = 20,000 is correct. Why Other Options Are Wrong: • ₹5,000: 5% of 5,000 is only 250, not 1,000. • ₹10,000: 5% is 500, not 1,000. • ₹15,000: 5% is 750, not 1,000. • ₹25,000: 5% is 1,250, which would make the difference ₹1,250, not ₹1,000. Common Pitfalls: • Applying profit percent on selling price instead of cost price. • Calculating selling prices unnecessarily instead of using the percentage difference shortcut. • Treating the ₹1,000 as 5% of selling price rather than 5% of CP. Final Answer: ₹20,000

Question 3

In basic algebraic simplification, when the product expression 24a^2b^2 is divided by 6b^2, what is the resulting simplified expression written only in terms of the variable a?

Accepted Answer

Introduction / Context: This question tests basic algebraic simplification involving powers of variables and cancellation of common factors. Such problems are common in school algebra and aptitude exams and help you practice how to handle products of powers when one expression is divided by another. The goal is to recognise which factors cancel out completely and which powers remain in the final simplified result. Given Data / Assumptions: The original algebraic expression is 24a^2b^2. This expression is divided by 6b^2. a and b are real variables. Concept / Approach: The key ideas are: Division of coefficients is done as ordinary number division. For variables with powers, when dividing like bases, subtract the exponents: a^m / a^n = a^(m − n). If the power becomes zero, the variable factor becomes 1. Any common factor in numerator and denominator can be cancelled. Step-by-Step Solution: Start with the expression: 24a^2b^2 ÷ 6b^2. Rewrite it as a single fraction: (24a^2b^2) / (6b^2). Simplify the numerical coefficients: 24 / 6 = 4. Simplify the b terms using exponent rules: b^2 / b^2 = b^(2−2) = b^0 = 1. The a term is only in the numerator: a^2 remains unchanged. Therefore the simplified expression is 4a^2. Verification / Alternative check: We can verify by substituting simple numeric values for a and b. For example, let a = 2 and b = 3. Then 24a^2b^2 = 24 * 4 * 9 = 864. The denominator 6b^2 = 6 * 9 = 54. The fraction 864 / 54 = 16. Now evaluate 4a^2: 4 * 4 = 16. The values match, confirming that 4a^2 is correct. Why Other Options Are Wrong: Option b: 4ab keeps a factor of b, but b^2 cancels completely, so this is incorrect. Option c: 4b also ignores the remaining a^2, so it does not match the simplification. Option d: 4 drops both a^2 and b^2, losing the variable dependence. Option e: a^2b^2 / 6 is an unsimplified rearrangement and does not reflect the required division by 6b^2. Common Pitfalls: A common mistake is to divide coefficients correctly but ignore powers of variables, or to cancel only one factor of b instead of both b^2 terms. Another typical error is to think that everything cancels and the answer becomes just the coefficient. Careful application of exponent rules prevents these errors. Final Answer: The simplified result of dividing 24a^2b^2 by 6b^2 is 4a^2.

Question 4

If 1 / √(cosec^2 A − 1) = x for an acute angle A in trigonometry, then x is equal to which standard trigonometric ratio of A?

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Introduction / Context: This problem checks your understanding of basic trigonometric identities, especially the relation between cosec A and cot A. Questions of this type often appear in aptitude and competitive exams where you must quickly convert an expression involving one trigonometric function into another simpler function of the same angle. Given Data / Assumptions: 1 / √(cosec^2 A − 1) = x. A is an acute angle, so 0° < A < 90°. We are asked to express x in terms of a basic trigonometric ratio of A. Concept / Approach: Key trigonometric identities used: cosec^2 A − 1 = cot^2 A. For an acute angle, square roots of squared trigonometric ratios are positive: √(cot^2 A) = cot A. If x = 1 / cot A, then x = tan A, because tan A is the reciprocal of cot A. Step-by-Step Solution: Start with the given expression: x = 1 / √(cosec^2 A − 1). Use the identity cosec^2 A − 1 = cot^2 A. Substitute this into the square root: √(cosec^2 A − 1) = √(cot^2 A). For an acute angle, √(cot^2 A) = cot A (positive value). Therefore x = 1 / cot A. Since tan A is the reciprocal of cot A, we have x = tan A. Verification / Alternative check: Take a simple acute angle, for example A = 30°. Then cosec 30° = 2, so cosec^2 30° = 4. Now cosec^2 30° − 1 = 3. Hence x = 1 / √3. But tan 30° is also 1 / √3, confirming that x = tan A is correct for this sample value. Because trigonometric identities are general, this holds for all acute angles. Why Other Options Are Wrong: Option b: cot A is the denominator inside the reciprocal and therefore cannot equal x. Option c: sin A is unrelated to the identity cosec^2 A − 1 = cot^2 A in this context. Option d: cos A is also not directly obtained from this identity. Option e: sec A is the reciprocal of cos A, not of cot A, so it does not match the simplified value. Common Pitfalls: Students sometimes confuse the identity cosec^2 A − 1 = cot^2 A with sec^2 A − 1 = tan^2 A. Another common error is ignoring the fact that A is acute, which ensures that the square root remains positive. Correctly recognising and applying the identity is crucial to arrive at the right trigonometric ratio. Final Answer: The value of x is tan A.

Question 5

In trigonometry using radian measure, what is the exact value of cosec(3π/4), the cosecant of three pi by four radians?

Accepted Answer

Introduction / Context: This question tests your understanding of standard trigonometric values for special angles, here expressed in radians. Being able to convert angles between degrees and radians and recall exact values for sine, cosine, and related functions such as cosecant is very important in both school mathematics and many competitive exams. Given Data / Assumptions: The angle is 3π/4 radians. We are asked to find cosec(3π/4). cosec x is defined as 1 / sin x. We work with standard values on the unit circle. Concept / Approach: Key ideas: 3π/4 radians corresponds to 135 degrees. The sine of 135 degrees (or 3π/4) is sin(180° − 45°) which equals sin 45°. sin 45° = √2 / 2. cosec x is the reciprocal of sin x, so cosec x = 1 / sin x. Step-by-Step Solution: Convert 3π/4 radians to degrees: (3π/4) * (180° / π) = 3 * 45° = 135°. Recognise that 135° = 180° − 45° and lies in the second quadrant. In the second quadrant, sine is positive, so sin 135° = sin 45°. Use the standard value sin 45° = √2 / 2. Therefore, sin(3π/4) = √2 / 2. Compute cosec(3π/4) = 1 / sin(3π/4) = 1 / (√2 / 2) = 2 / √2 = √2. Verification / Alternative check: Using the unit circle, the point corresponding to 135° has coordinates (−√2 / 2, √2 / 2). The y coordinate is the sine value, which is √2 / 2. Taking its reciprocal again gives √2. This agrees with the analytic computation and confirms the answer. Why Other Options Are Wrong: Option a: −√2 would be the cosecant if sine were negative, but in the second quadrant sine is positive. Option c: 2 / √3 is associated with angles like 60°, not 135°. Option d: −2 / √3 is incorrect for the same reason and also has the wrong sign. Option e: −1 / √2 is simply the negative of sin 45°, not its reciprocal, and has an incorrect sign. Common Pitfalls: Students often forget the sign of trigonometric functions in different quadrants or mix up reciprocal values. Confusing 135° with 45° or 225° can also lead to errors. Always check the quadrant and remember that cosecant is the reciprocal of sine, not cosine or tangent. Final Answer: The exact value of cosec(3π/4) is √2.

Question 6

Simplify the trigonometric expression [sin A / (1 + cos A)] + [(1 + cos A) / sin A] and determine its final value in terms of a basic trigonometric function of A.

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Introduction / Context: This problem focuses on simplifying a compound trigonometric expression that mixes sine and cosine in fractional form. Such questions are designed to test your fluency with algebraic manipulation of trigonometric functions and your comfort with combining fractions and applying fundamental identities. Given Data / Assumptions: The expression is E = sin A / (1 + cos A) + (1 + cos A) / sin A. Angle A is such that sin A and 1 + cos A are non zero, so the expression is defined. We must express the final simplified result in terms of a single standard trigonometric function. Concept / Approach: Main ideas: Use a common denominator to add the two fractions. Apply the Pythagorean identity sin^2 A + cos^2 A = 1. Look for factors that cancel after simplification. Recognise that cosec A is 1 / sin A. Step-by-Step Solution: Let E = sin A / (1 + cos A) + (1 + cos A) / sin A. The common denominator is sin A (1 + cos A). Write the numerator as sin A * sin A + (1 + cos A) * (1 + cos A). This gives sin^2 A + (1 + 2cos A + cos^2 A). Group terms: sin^2 A + cos^2 A + 1 + 2cos A. Use sin^2 A + cos^2 A = 1, so numerator becomes 1 + 1 + 2cos A = 2(1 + cos A). Thus, E = 2(1 + cos A) / [sin A (1 + cos A)]. Cancel the factor (1 + cos A) from numerator and denominator to get E = 2 / sin A = 2cosec A. Verification / Alternative check: Choose A = 30°. Then sin 30° = 1 / 2 and cos 30° = √3 / 2. Compute the original expression numerically and compare with 2cosec 30°. We get: sin A / (1 + cos A) = (1 / 2) / (1 + √3 / 2) and (1 + cos A) / sin A = (1 + √3 / 2) / (1 / 2). When evaluated carefully the total equals 4, and 2cosec 30° = 2 * 2 = 4, which matches. Why Other Options Are Wrong: Option a: 2sec A would equal 2 / cos A, which does not match the simplified form 2 / sin A. Option c: 2tan A equals 2sin A / cos A and has a completely different dependence on A. Option d: 2cot A is 2cos A / sin A, again not equal to 2 / sin A for general A. Option e: 2 would require sin A = 1, which is not generally true for all A. Common Pitfalls: A frequent mistake is to incorrectly square (1 + cos A), or to forget to use the identity sin^2 A + cos^2 A = 1. Some students also cancel sin A or (1 + cos A) prematurely without having them as common factors in numerator and denominator. Keeping the algebra organized avoids these errors. Final Answer: The simplified value of the expression is 2cosec A.

Question 7

In parallelogram ABCD with vertices A(5, 0), B(−2, 3) and C(−1, 4), determine the equation of side AD in coordinate geometry.

Accepted Answer

Introduction / Context: This problem tests your understanding of coordinate geometry, specifically properties of a parallelogram and how to find the equation of a line when you know two points that lie on it. Such questions combine vector or midpoint ideas with straight line equations and are very popular in analytical geometry sections of exams. Given Data / Assumptions: Parallelogram ABCD has vertices A(5, 0), B(−2, 3) and C(−1, 4). We must find the equation of side AD. Vertices follow the order A → B → C → D around the parallelogram. Concept / Approach: Key concepts: In a parallelogram, the diagonals bisect each other, and opposite sides are parallel. For a parallelogram, position vectors satisfy A + C = B + D, so D = A + C − B. Once two points on line AD are known, its equation can be found using the two point or point slope form. Step-by-Step Solution: Use the vector relation: D = A + C − B. Compute A + C = (5, 0) + (−1, 4) = (4, 4). Subtract B: D = (4, 4) − (−2, 3) = (4 + 2, 4 − 3) = (6, 1). So vertex D is at (6, 1). Line AD passes through A(5, 0) and D(6, 1). Slope m = (1 − 0) / (6 − 5) = 1. Using point slope form with point A: y − 0 = 1(x − 5). Therefore y = x − 5 is the equation of side AD. Verification / Alternative check: Check that C also lies on a line parallel to AD. Slope of BC is (4 − 3) / (−1 − (−2)) = 1 / 1 = 1, the same as slope of AD. Since opposite sides of a parallelogram are parallel, this supports the correctness of side AD having slope 1 and equation y = x − 5 through point A. Why Other Options Are Wrong: Option a: y = 2x − 5 has slope 2, not matching the slope between A and D. Option b: y = x + 5 passes through (0, 5), not through point A(5, 0). Option c: y = 2x + 5 has both wrong slope and wrong intercept. Option e: y = −x + 5 has negative slope and does not pass through A(5, 0) or D(6, 1). Common Pitfalls: A common mistake is to misidentify the coordinates of D by adding instead of subtracting correctly, or to assume that C is diagonally opposite A without verifying the order of vertices. Another frequent error is mixing up x and y differences when computing slope. Carefully applying the vector relation and slope formula avoids these issues. Final Answer: The equation of side AD of the parallelogram is y = x − 5.

Question 8

A bag contains ₹1, 50 paise, and 10 paise coins in the ratio 6 : 3 : 2 and the total amount is ₹30.80. How many 50 paise coins are there in the bag?

Accepted Answer

Introduction / Context: This question is a standard coins and ratio problem from quantitative aptitude. It tests your ability to translate a ratio into actual numbers, handle mixed denominations, and set up an equation using the total monetary value. Such questions appear frequently in banking, clerical, and management entrance tests. Given Data / Assumptions: The bag contains coins of ₹1, 50 paise, and 10 paise. The ratio of their counts is 6 : 3 : 2. The total value of all coins together is ₹30.80. We need to determine the number of 50 paise coins. Concept / Approach: Main ideas: Let the common multiplying factor of the ratio be k, so numbers of coins are 6k, 3k, and 2k. Convert all coin values to rupees: 50 paise = ₹0.50 and 10 paise = ₹0.10. Form an equation for total value and solve for k. Use k to find the number of 50 paise coins, which is 3k. Step-by-Step Solution: Let the numbers of coins be: ₹1 coins = 6k, 50 paise coins = 3k, 10 paise coins = 2k. Values in rupees: ₹1 coin is 1, 50 paise is 0.5, and 10 paise is 0.1. Total value V = 6k * 1 + 3k * 0.5 + 2k * 0.1. Compute: V = 6k + 1.5k + 0.2k = 7.7k. Given total value is ₹30.80, so 7.7k = 30.80. Solve for k: k = 30.80 / 7.7 = 4. Number of 50 paise coins = 3k = 3 * 4 = 12. Verification / Alternative check: Check the total value using k = 4. Numbers of coins: ₹1 coins = 24, 50 paise = 12, 10 paise = 8. Total value = 24 * 1 + 12 * 0.5 + 8 * 0.1 = 24 + 6 + 0.8 = 30.8, that is ₹30.80. This matches the given total, confirming our solution. Why Other Options Are Wrong: Option a: 8 would make the total value smaller, and the equation 7.7k = 30.80 would not hold. Option b: 24 would correspond to k = 8, giving a total much larger than ₹30.80. Option d: 4 makes k = 4/3, which does not satisfy the original total value condition. Option e: 16 would give k = 16/3, again not consistent with the total sum. Common Pitfalls: Common mistakes include forgetting to convert paise to rupees, treating the ratio numbers as actual counts without using a common factor, and miscalculating the total value. Always maintain consistent units (rupees here) and carefully multiply each coin type by its value and its count before forming the equation. Final Answer: The bag contains 12 coins of 50 paise.

Question 9

In coordinate geometry, what is the image of the point (−3, 6) when it is reflected in the x-axis?

Accepted Answer

Introduction / Context: This problem checks your understanding of reflections in coordinate geometry, specifically reflection in the x axis. Such transformations are fundamental when working with graphs, symmetry, and geometric interpretations on the Cartesian plane. Given Data / Assumptions: Original point P has coordinates (−3, 6). The point is reflected in the x axis. We must find the coordinates of the reflected point. Concept / Approach: Key ideas: Reflection in the x axis keeps the x coordinate unchanged. The y coordinate changes sign: (x, y) becomes (x, −y). This comes from the fact that points symmetric about the x axis are equidistant above and below that axis. Step-by-Step Solution: Start with the original point P(−3, 6). Under reflection in the x axis, x remains the same, so x = −3. The y coordinate changes sign: y becomes −6. Therefore the image point P′ is (−3, −6). Verification / Alternative check: You can visualise the point on the coordinate plane. P(−3, 6) lies in the second quadrant. Reflecting in the x axis moves it straight down vertically to the point with the same x coordinate and the opposite y coordinate, which is (−3, −6) in the third quadrant. The vertical distance from the x axis is 6 units in both cases, confirming the reflection. Why Other Options Are Wrong: Option a: (3, 6) represents reflection in the y axis, not in the x axis. Option b: (6, −3) changes both coordinates and does not preserve the correct symmetric distance. Option d: (−6, 3) is reflection about the line y = x combined with sign changes and is not appropriate here. Option e: (3, −6) corresponds to reflection in the origin, changing both signs, rather than only flipping across the x axis. Common Pitfalls: A common mistake is to change the wrong coordinate or both coordinates. Always remember: reflection in the x axis affects only the y coordinate, and reflection in the y axis affects only the x coordinate. Keeping a mental picture of the coordinate plane helps avoid confusion. Final Answer: The image of the point after reflection in the x axis is (−3, −6).

Question 10

Simplify the algebraic expression (b^3 x^2 a^4 z^3) · (b^4 x^3 a^3 z^2) ÷ (a^2 b^4 z^3) and express the result in terms of a, b, x, and z.

Accepted Answer

Introduction / Context: This question assesses your skill in manipulating algebraic expressions with multiple variables and exponents. Simplifying products and quotients of powers is a fundamental algebra technique used in higher mathematics, physics, and engineering formulas. Given Data / Assumptions: Expression: (b^3 x^2 a^4 z^3) · (b^4 x^3 a^3 z^2) ÷ (a^2 b^4 z^3). Variables a, b, x, and z are non zero. We must simplify and express the final answer as a power of each variable. Concept / Approach: Key exponent rules: When multiplying like bases: a^m · a^n = a^(m + n). When dividing like bases: a^m ÷ a^n = a^(m − n). We can rearrange factors because multiplication is commutative and associative. Step-by-Step Solution: Consider the numerator: (b^3 x^2 a^4 z^3) · (b^4 x^3 a^3 z^2). Combine exponents for each variable: b^(3 + 4) x^(2 + 3) a^(4 + 3) z^(3 + 2). This gives b^7 x^5 a^7 z^5. Now divide by the denominator a^2 b^4 z^3. Apply exponent subtraction: a^(7 − 2) = a^5, b^(7 − 4) = b^3, z^(5 − 3) = z^2, and x^5 remains unchanged. Therefore the simplified expression is a^5 b^3 x^5 z^2, which can be written as b^3 x^5 a^5 z^2. Verification / Alternative check: As a numerical check, assign values a = 2, b = 3, x = 4, and z = 5. Compute the original expression and the simplified form with a calculator or carefully by hand. Both evaluations will yield the same numerical result, confirming that the algebraic simplification is correct. Why Other Options Are Wrong: Option a: b^2 x^4 a^6 z has incorrect exponents for b, x, a, and z compared to the correct subtraction and addition of exponents. Option b: b^3 x^2 a^4 z^3 reproduces part of the original numerator and fails to include the effect of the second factor and the denominator. Option d: b a z x severely underestimates the powers and ignores the exponent arithmetic. Option e: a^7 b^3 x^5 z^2 has the wrong exponent for a since 7 − 2 should give 5, not 7. Common Pitfalls: Students sometimes add instead of subtracting exponents during division, or forget to combine like bases across both numerators before dividing. Another mistake is misreading variables and mixing up exponents for x and z. Carefully grouping all like terms before applying exponent rules helps avoid these issues. Final Answer: The simplified expression is b^3 x^5 a^5 z^2.

Question 11

If x = a cos θ + b sin θ and y = b cos θ − a sin θ, what is the simplified value of x^2 + y^2 expressed purely in terms of a and b?

Accepted Answer

Introduction / Context: This question checks your understanding of trigonometric expressions combined with algebra and the Pythagorean identity. It is a classic example where trigonometric terms vanish after simplification, leaving a result that depends only on algebraic parameters a and b. Such forms appear in rotations and transformations in coordinate geometry and physics. Given Data / Assumptions: x = a cos θ + b sin θ. y = b cos θ − a sin θ. We must compute x^2 + y^2. Angle θ is real and sin θ and cos θ are defined. Concept / Approach: Key ideas: Expand x^2 and y^2 using algebraic identities. Combine like terms in cos^2 θ, sin^2 θ, and sin θ cos θ. Use the identity sin^2 θ + cos^2 θ = 1. Observe that cross terms cancel out, leaving an expression in a and b only. Step-by-Step Solution: Compute x^2: x^2 = (a cos θ + b sin θ)^2 = a^2 cos^2 θ + 2ab sin θ cos θ + b^2 sin^2 θ. Compute y^2: y^2 = (b cos θ − a sin θ)^2 = b^2 cos^2 θ − 2ab sin θ cos θ + a^2 sin^2 θ. Add x^2 and y^2. x^2 + y^2 = a^2 cos^2 θ + 2ab sin θ cos θ + b^2 sin^2 θ + b^2 cos^2 θ − 2ab sin θ cos θ + a^2 sin^2 θ. Notice that the +2ab sin θ cos θ and −2ab sin θ cos θ terms cancel. Group a terms: a^2 cos^2 θ + a^2 sin^2 θ = a^2 (cos^2 θ + sin^2 θ) = a^2. Group b terms: b^2 sin^2 θ + b^2 cos^2 θ = b^2 (sin^2 θ + cos^2 θ) = b^2. Therefore x^2 + y^2 = a^2 + b^2. Verification / Alternative check: Pick simple values, for example a = 1, b = 2, and θ = 30°. Compute x and y numerically and then find x^2 + y^2. Separately compute a^2 + b^2 = 1^2 + 2^2 = 5. Both calculations will give the same numerical result, confirming the formula. Why Other Options Are Wrong: Option a: ab would imply all squared terms vanish, which is not supported by the algebra. Option c: a^2 − b^2 would require different cancellation and sign patterns, which do not occur here. Option d: 1 would suggest the expression is independent of a and b, which is impossible because x and y scale with a and b. Option e: (a + b)^2 = a^2 + 2ab + b^2, but the cross term 2ab does not remain in the simplified expression. Common Pitfalls: A common mistake is mishandling the cross terms 2ab sin θ cos θ and failing to notice that they cancel when x^2 and y^2 are added. Another frequent error is not applying sin^2 θ + cos^2 θ = 1 correctly, which leaves unnecessary trigonometric expressions in the final result. Final Answer: The value of x^2 + y^2 is a^2 + b^2.

Question 12

If x = √2 + 1, then find the exact simplified value of x^4 − 1 / x^4 expressed in terms of √2.

Accepted Answer

Introduction / Context: This question combines algebraic manipulation with surds. It checks whether you can work with expressions involving √2, powers, and reciprocals, and then simplify to a neat exact form. Such techniques are very useful in algebra, number theory, and competitive exam problems that demand exact radical answers instead of decimals. Given Data / Assumptions: x = √2 + 1. We need to compute x^4 − 1 / x^4. All operations are exact, without decimal approximation. Concept / Approach: Main ideas: Simplify step by step using powers: first find x^2, then x^4. Use the idea that 1 / x can be simplified by rationalising the denominator. After finding 1 / x, compute 1 / x^2 and 1 / x^4 similarly. Subtract the two exact expressions to obtain a multiple of √2. Step-by-Step Solution: Start with x = √2 + 1. Compute x^2 = (√2 + 1)^2 = 2 + 2√2 + 1 = 3 + 2√2. Now compute x^4 = (x^2)^2 = (3 + 2√2)^2. (3 + 2√2)^2 = 9 + 12√2 + 8 = 17 + 12√2. Find 1 / x by rationalising: 1 / (√2 + 1) = (√2 − 1) / [(√2 + 1)(√2 − 1)] = √2 − 1. Then 1 / x^2 = (√2 − 1)^2 = 2 − 2√2 + 1 = 3 − 2√2. Next 1 / x^4 = (1 / x^2)^2 = (3 − 2√2)^2 = 9 − 12√2 + 8 = 17 − 12√2. Finally compute x^4 − 1 / x^4 = (17 + 12√2) − (17 − 12√2) = 24√2. Verification / Alternative check: You can approximate √2 as 1.414 and calculate x ≈ 2.414. Then x^4 ≈ 2.414^4 and 1 / x^4 ≈ 1 / 2.414^4. Numerically, their difference is very close to 24 * 1.414, validating the exact symbolic result. Why Other Options Are Wrong: Option a: 8√2 is too small and would correspond to only one third of the correct difference. Option b: 18√2 underestimates the true value; the algebraic computation clearly gives 24√2. Option c and option e: 6√2 or 12√2 are both significantly smaller than the correct multiple obtained from the exact expansion. Common Pitfalls: Typical mistakes include squaring incorrectly, especially when handling terms like 2√2, or failing to rationalise the reciprocal properly. Another error is to attempt to approximate early with decimals, which can hide the exact pattern. Working symbolically and systematically with surds keeps the steps clear. Final Answer: The exact value of x^4 − 1 / x^4 is 24√2.

Question 13

If 9x^2 + 16y^2 = 60 and 3x + 4y = 6, what is the value of the product xy?

Accepted Answer

Introduction / Context: This question combines simultaneous equations and algebraic identities to find the product xy without necessarily solving for x and y individually. It is a neat example of how squaring a linear equation and comparing with a quadratic relationship can reveal cross terms like xy. Given Data / Assumptions: 9x^2 + 16y^2 = 60. 3x + 4y = 6. We need to find the value of xy. Concept / Approach: Main ideas: Square the linear equation (3x + 4y)^2 to produce terms in x^2, y^2, and xy. Compare the result with the given quadratic relation 9x^2 + 16y^2 = 60. Subtract to isolate the xy term. Step-by-Step Solution: Start from 3x + 4y = 6. Square both sides: (3x + 4y)^2 = 6^2. Expand the left side: 9x^2 + 24xy + 16y^2 = 36. We are also given 9x^2 + 16y^2 = 60. Subtract the quadratic equation from the squared equation: (9x^2 + 24xy + 16y^2) − (9x^2 + 16y^2) = 36 − 60. This simplifies to 24xy = −24. Therefore xy = −24 / 24 = −1. Verification / Alternative check: We can solve the system explicitly to verify. From 3x + 4y = 6, express x in terms of y and substitute into 9x^2 + 16y^2 = 60. Solving this pair will give numerical values of x and y whose product equals −1, confirming the derived result without needing to show every intermediate step here. Why Other Options Are Wrong: Option b: 1 would require 24xy = 24, contradicting the calculated equation 24xy = −24. Option c and option d: −2 or 2 would require different constants on the right side of the equation obtained after subtraction. Option e: 0 would mean 3x + 4y squared equals 9x^2 + 16y^2, which happens only if xy = 0, inconsistent with 24xy = −24. Common Pitfalls: Students may incorrectly expand (3x + 4y)^2 or mis-handle the subtraction of the two quadratic equations, leading to sign mistakes. Another pitfall is trying to solve directly for x and y when a more elegant approach via comparison is faster and less error prone. Final Answer: The value of the product xy is −1.

Question 14

If (cos θ / (1 − sin θ)) + (cos θ / (1 + sin θ)) = 4 for an acute angle θ, then θ is equal to how many degrees?

Accepted Answer

Introduction / Context: This trigonometric equation problem asks you to simplify a sum of two fractions involving sine and cosine and then solve for an acute angle θ. It tests your ability to manipulate trigonometric expressions and apply identities to reach a standard value that can be compared with known special angles. Given Data / Assumptions: (cos θ / (1 − sin θ)) + (cos θ / (1 + sin θ)) = 4. 0° < θ < 90°, so θ is acute. Denominators 1 − sin θ and 1 + sin θ are non zero. Concept / Approach: Key concepts: Use a common denominator (1 − sin θ)(1 + sin θ). Apply the identity 1 − sin^2 θ = cos^2 θ. Simplify the resulting expression to involve only sec θ or cos θ. Equate the simplified form to 4 and solve for θ. Step-by-Step Solution: Let E = cos θ / (1 − sin θ) + cos θ / (1 + sin θ). Common denominator is (1 − sin θ)(1 + sin θ) = 1 − sin^2 θ = cos^2 θ. Numerator becomes cos θ(1 + sin θ) + cos θ(1 − sin θ). Factor cos θ: cos θ[(1 + sin θ) + (1 − sin θ)] = cos θ(2) = 2cos θ. Therefore E = (2cos θ) / cos^2 θ = 2 / cos θ = 2sec θ. Given that E = 4, we have 2sec θ = 4. So sec θ = 2, which implies cos θ = 1 / 2. For an acute angle, cos θ = 1 / 2 gives θ = 60°. Verification / Alternative check: Substitute θ = 60° back into the original expression. cos 60° = 1 / 2 and sin 60° = √3 / 2. Evaluate cos θ / (1 − sin θ) and cos θ / (1 + sin θ), then add them. After simplification, the result equals 4, confirming the correctness of θ = 60°. Why Other Options Are Wrong: Option b: 45° gives cos 45° = √2 / 2 and leads to a value that is not 4 when substituted. Option c: 30° gives cos 30° = √3 / 2 and creates a different sum. Option d and option e: 35° or 15° are not standard special angles that give sec θ = 2, and the equation would not hold for these values. Common Pitfalls: Errors often occur when finding the common denominator or expanding the numerators. Some students forget that (1 − sin θ)(1 + sin θ) equals cos^2 θ, or they try to solve the equation numerically instead of simplifying symbolically. Correct use of identities makes the problem straightforward. Final Answer: The acute angle θ satisfying the equation is 60°.

Question 15

Solve the decimal equation 3.352 − (9.759 − x) − 19.64 = 7.052 and find the value of x.

Accepted Answer

Introduction / Context: This question is a straightforward but slightly lengthy decimal arithmetic problem. It tests your ability to handle brackets, negative signs, and decimal subtraction and addition accurately, which is important in many real life financial and data calculations. Given Data / Assumptions: The equation is 3.352 − (9.759 − x) − 19.64 = 7.052. We must solve for x. All numbers are in decimal form and standard arithmetic rules apply. Concept / Approach: Main steps: Remove brackets carefully, noting that −(9.759 − x) changes signs. Combine like decimal terms on the left side. Isolate x on one side to obtain its value. Check the solution by substitution. Step-by-Step Solution: Start with 3.352 − (9.759 − x) − 19.64 = 7.052. Distribute the minus sign: 3.352 − 9.759 + x − 19.64 = 7.052. First combine 3.352 − 9.759 = −6.407. Then combine −6.407 − 19.64 = −26.047. So the equation becomes x − 26.047 = 7.052. Add 26.047 to both sides: x = 7.052 + 26.047. Compute the sum: 7.052 + 26.047 = 33.099. Therefore x = 33.099. Verification / Alternative check: Substitute x = 33.099 into the original equation: 3.352 − (9.759 − 33.099) − 19.64. Compute inside the bracket: 9.759 − 33.099 = −23.34. Then 3.352 − (−23.34) − 19.64 = 3.352 + 23.34 − 19.64 = 26.692 − 19.64 = 7.052. Since the left side equals the right side, the solution is correct. Why Other Options Are Wrong: Option a: −6.181 would give a very different value when substituted back, not equal to 7.052. Option b: 13.581 is too small and does not balance the equation. Option d: 39.803 is too large and leads to a left hand side much bigger than 7.052. Option e: 3.099 is far from the correct sum and fails the substitution check. Common Pitfalls: Common mistakes include mishandling the minus sign before the bracket, incorrectly subtracting decimals, or rounding too early. To avoid these errors, perform operations systematically and line up decimal points carefully if doing manual arithmetic. Final Answer: The solution to the equation is x = 33.099.

Question 16

Find the expanded square of the binomial expression (7 − 4x) and write it as a standard quadratic expression in x.

Accepted Answer

Introduction / Context: This question is about expanding the square of a binomial, a very common algebra skill. Recognising and applying the identity for (a − b)^2 or (a + b)^2 allows you to quickly convert a compact expression into a standard polynomial form, which is heavily used in solving quadratic equations and simplifying algebraic expressions. Given Data / Assumptions: The binomial is (7 − 4x). We need to compute (7 − 4x)^2. The answer should be written as a quadratic expression in x. Concept / Approach: Main identity: (a − b)^2 = a^2 − 2ab + b^2. Here, a = 7 and b = 4x. Compute each term carefully: a^2, 2ab, and b^2. Step-by-Step Solution: Write the binomial square: (7 − 4x)^2. Identify a = 7 and b = 4x. Compute a^2 = 7^2 = 49. Compute 2ab = 2 * 7 * 4x = 56x. Compute b^2 = (4x)^2 = 16x^2. Apply the identity: (7 − 4x)^2 = a^2 − 2ab + b^2. So (7 − 4x)^2 = 49 − 56x + 16x^2. Rearrange into standard quadratic order: 16x^2 − 56x + 49. Verification / Alternative check: You can expand directly by multiplication: (7 − 4x)(7 − 4x). Multiply term by term: 7 * 7 = 49, 7 * (−4x) = −28x, (−4x) * 7 = −28x, and (−4x) * (−4x) = 16x^2. Adding these gives 49 − 56x + 16x^2, which matches the result from the identity. Why Other Options Are Wrong: Option a: 16x^2 − 28x + 49 uses 28x instead of 56x, forgetting that the middle term is twice the product ab. Option b and option c: These forms either have incorrect coefficients or wrong signs for the quadratic term and do not match the proper expansion. Option e: 49 + 56x + 16x^2 corresponds to (7 + 4x)^2, not (7 − 4x)^2, so the middle term has the wrong sign. Common Pitfalls: A frequent error is to forget the factor of 2 in the middle term or to mis-handle the sign when the binomial includes a minus. Carefully applying the identity and double checking the sign of 2ab prevents such mistakes. Final Answer: The expanded quadratic expression is 16x^2 − 56x + 49.

Question 17

Solve the system of inequalities 7 + 3x ≥ 5 − x/2 and 2x + 3 ≤ 5 − 2x; which of the following values of x satisfies both conditions simultaneously?

Accepted Answer

Introduction / Context: This problem involves solving two linear inequalities in one variable and then finding which option satisfies both at the same time. Such questions are common in aptitude tests and help you understand how to work with inequality symbols, manipulate expressions, and interpret ranges of possible solutions. Given Data / Assumptions: First inequality: 7 + 3x ≥ 5 − x/2. Second inequality: 2x + 3 ≤ 5 − 2x. We need a value of x that satisfies both inequalities. Concept / Approach: Key ideas: Solve each inequality separately, obtaining a range of x values. Combine the two solution sets using intersection to see which x values work for both. Check each option against the final allowed interval. Step-by-Step Solution: First inequality: 7 + 3x ≥ 5 − x/2. Bring all x terms to one side: 3x + x/2 ≥ 5 − 7 = −2. 3x = 6x/2, so 6x/2 + x/2 = 7x/2. Thus 7x/2 ≥ −2, so 7x ≥ −4, giving x ≥ −4/7. Second inequality: 2x + 3 ≤ 5 − 2x. Bring x terms together: 2x + 2x ≤ 5 − 3 = 2. This gives 4x ≤ 2, so x ≤ 1/2. Combined solution: −4/7 ≤ x ≤ 1/2. Verification / Alternative check: Now test each option. Among the given values, 0 lies between −4/7 and 1/2, whereas 1 and 2 are greater than 1/2 and −1 and −2 are less than −4/7. Therefore x = 0 is the only value that satisfies both inequalities. Substituting x = 0 explicitly into both inequalities confirms that each side of both conditions is satisfied. Why Other Options Are Wrong: Option b: 1 violates x ≤ 1/2 because 1 is greater than 0.5. Option c: 2 also violates the second inequality and is far outside the allowed range. Option d: −1 is less than −4/7 and fails the first inequality. Option e: −2 is even smaller and does not satisfy x ≥ −4/7. Common Pitfalls: Mistakes often occur when handling fractions, especially with the term −x/2, or when flipping the inequality sign incorrectly after multiplying or dividing by negative numbers. In this problem, all multipliers were positive, so the direction of the inequality does not change. Carefully isolating x in each inequality prevents confusion. Final Answer: The value of x that satisfies both inequalities is 0.

Question 18

Find the roots of the quadratic equation x^2 − 25x + 156 = 0 and identify the correct ordered pair of solutions.

Accepted Answer

Introduction / Context: This problem is a standard quadratic equation solving exercise. It checks whether you can factorise a quadratic expression of the form x^2 − 25x + 156 and correctly identify its two real roots. Such equations appear frequently in algebra and aptitude tests. Given Data / Assumptions: The equation is x^2 − 25x + 156 = 0. We look for real roots. We can attempt factorisation into (x − p)(x − q) = 0. Concept / Approach: Main ideas: For x^2 − 25x + 156, find two numbers whose product is 156 and whose sum is 25. Use these numbers to factor the quadratic expression. Set each factor equal to zero to find the roots. Step-by-Step Solution: We seek integers p and q such that p + q = 25 and pq = 156. Try factor pairs of 156: 12 × 13 = 156 and 12 + 13 = 25. So p = 12 and q = 13 satisfy the conditions. Write the quadratic as x^2 − 25x + 156 = (x − 12)(x − 13). Set each factor equal to zero: x − 12 = 0 or x − 13 = 0. Thus x = 12 or x = 13. Verification / Alternative check: Substitute x = 12 into the original equation: 12^2 − 25 * 12 + 156 = 144 − 300 + 156 = 0. For x = 13: 169 − 325 + 156 = 0. Both values satisfy the equation, confirming that they are indeed roots. Why Other Options Are Wrong: Option b: 25 and 1 have product 25, not 156, so they cannot be the roots of this quadratic. Option c: 9 and 16 have product 144 and sum 25, but the constant term in the equation is 156, not 144. Option d: 31 and 6 have product 186 and do not even sum to 25. Option e: 8 and 17 have product 136, which again is not equal to 156. Common Pitfalls: A frequent mistake is to miscalculate factor pairs or to rely on guessing without checking both the sum and product conditions. Another error is to factor incorrectly and then not verify the roots. Systematically listing factor pairs and checking them, or using the quadratic formula as a backup, helps avoid issues. Final Answer: The roots of the quadratic equation are 12 and 13.

Question 19

Which of the following equations in x is not a quadratic equation after simplification of both sides?

Accepted Answer

Introduction / Context: This question asks you to identify which given equation is not quadratic once both sides are simplified. It tests your understanding of what defines a quadratic equation and your ability to expand and rearrange algebraic expressions correctly. Given Data / Assumptions: Several equations in x are listed. We must simplify each and check whether the final form contains an x^2 term. A quadratic equation in one variable has the general form ax^2 + bx + c = 0 with a not equal to zero. Concept / Approach: Main ideas: Expand both sides of each equation. Bring all terms to one side to see the degree of the resulting polynomial. If the x^2 term cancels out completely, the equation is no longer quadratic. Step-by-Step Solution: Option a: 3x(x + 5) + 11 = 2x(x − 2) + 6. LHS: 3x^2 + 15x + 11. RHS: 2x^2 − 4x + 6. Subtract RHS: x^2 + 19x + 5 = 0, which is quadratic. Option b: 4x(x + 3) + 7 = 4x(x − 11) + 9. LHS: 4x^2 + 12x + 7. RHS: 4x^2 − 44x + 9. Subtract RHS: (4x^2 + 12x + 7) − (4x^2 − 44x + 9) = 56x − 2 = 0. This simplifies to 56x − 2 = 0, which is linear, not quadratic. Option c: x(x + 2) + 15 = x(2x − 5) + 11 gives −x^2 + 7x + 4 = 0, which is quadratic. Option d: 4x^2 − 6x + 9 = 0 is already quadratic. Option e: x^2 + 4x + 4 = 0 is also clearly quadratic. Verification / Alternative check: For option b, solving 56x − 2 = 0 gives a single solution x = 1/28. A genuine quadratic equation would have at most two roots and must involve an x^2 term. Since the x^2 terms cancel completely, the equation reduces to degree one, confirming that it is not quadratic. Why Other Options Are Wrong: Options a, c, d, and e all result in expressions containing x^2 after simplification. They match the general form ax^2 + bx + c = 0 with a not equal to zero. Therefore, each of them is a quadratic equation in x. Common Pitfalls: The main pitfall is performing expansions or subtractions incorrectly, leading to wrong cancellations of x^2 terms. Another mistake is to judge the equation by its initial appearance without actually simplifying both sides. Always expand, collect like terms, and then check the highest power of x. Final Answer: The equation that is not quadratic after simplification is 4x(x + 3) + 7 = 4x(x − 11) + 9.

Question 20

In triangle PQR, S lies on PQ and T lies on PR such that ST is parallel to QR. If S divides PQ in the ratio 4 : 1 and QR = 15 cm, what is the length of segment ST?

Accepted Answer

Introduction / Context: This geometry question involves similar triangles formed when a line segment is drawn parallel to one side of a triangle. It tests your understanding of the basic proportionality theorem (also known as Thales theorem) and how lengths scale in similar triangles. Given Data / Assumptions: Triangle PQR has S on PQ and T on PR. ST is parallel to QR. S divides PQ in the ratio 4 : 1, meaning PS : SQ = 4 : 1. QR, the base of the larger triangle, is 15 cm. We need to find the length ST, the corresponding side in the smaller similar triangle. Concept / Approach: Main ideas: If a line segment is drawn parallel to one side of a triangle and intersects the other two sides, it creates a smaller triangle similar to the original. The ratio of corresponding sides in similar triangles equals the ratio of any pair of corresponding segments on the intersected sides. Here, triangle PST is similar to triangle PQR. The scale factor is PS / PQ. Step-by-Step Solution: Given PS : SQ = 4 : 1. Then PQ = PS + SQ, so PQ is in parts 4 + 1 = 5. Thus PS / PQ = 4 / 5. Triangles PST and PQR are similar since ST ∥ QR. Therefore ST / QR = PS / PQ = 4 / 5. Given QR = 15 cm, we get ST = (4 / 5) * 15. Compute ST: (4 / 5) * 15 = 12. So ST = 12 cm. Verification / Alternative check: You can also think in terms of area scaling or other corresponding sides. Any side of the smaller triangle is 4/5 of the corresponding side of the larger triangle. If the side ratio is 4/5, then the base ST must be 4/5 of 15 cm, which is 12 cm, matching our calculation. Why Other Options Are Wrong: Option a: 3 cm is much smaller and would correspond to a scale factor of 1/5, not 4/5. Option b: 5 cm does not relate to the 4 : 1 ratio and gives an incorrect scale factor. Option d: 10 cm would correspond to a factor 2/3, which is not derived from the given ratio 4 : 1. Option e: 7 cm has no simple proportional relationship with 15 cm and does not follow from the 4 : 1 division of PQ. Common Pitfalls: A common mistake is misinterpreting the ratio 4 : 1 as PS : PQ instead of PS : SQ, or forgetting to add the parts to find the whole PQ. Another error is to invert the ratio when relating ST to QR. Always keep track of which segment corresponds to which side in the similar triangles. Final Answer: The length of ST is 12 cm.

Simplification Questions