Algebra replaces an unknown number with a symbol while preserving arithmetic rules. The objective is to translate a condition faithfully, solve the resulting equation and test the result against the original wording. Many “number problems” are short linear equations in disguise.
1. Linear equations
Perform the same valid operation on both sides of an equation. If 3x+7=25, subtract 7 and divide by 3 to obtain x=6. When fractions appear, multiply through by a common non-zero denominator to clear them. Be careful not to multiply one side only.
Worked example 1. Solve (x−2)/3+(x+1)/2=7. Multiply both sides by 6: 2(x−2)+3(x+1)=42. Therefore 5x−1=42 and x=43/5. Substituting 8.6 into the original gives 2.2+4.8=7.
In word problems, choose a variable with a unit. “A number is five more than twice another” can be written A=2B+5. If the second number is an age, keep years attached to the interpretation, even though the symbol itself is unitless in the algebraic line.
2. Two equations in two unknowns
Elimination or substitution solves simultaneous linear equations. If x+y=30 and x−y=8, add to get 2x=38, so x=19 and y=11. A ratio condition can be converted to a linear equation: x:y=3:4 means 4x=3y, assuming y≠0.
Worked example 2. Two tickets cost ₹230. One adult ticket is ₹30 more than one child ticket. Let adult=A and child=C. Then A+C=230 and A−C=30. Adding yields A=130 and C=100. Both satisfy the price and difference conditions.
3. Identities and factorisation
Useful identities include (a+b)²=a²+2ab+b², (a−b)²=a²−2ab+b² and a²−b²=(a−b)(a+b). An identity holds for all allowed values; an equation holds only for values solving it. Factorisation can simplify arithmetic or reveal roots.
Worked example 3. Calculate 103² without long multiplication. Use (100+3)²=10,000+600+9=10,609. The cross term 2×100×3 is essential.
If xy and x+y are known, x²+y²=(x+y)²−2xy. If x−y and xy are known, x²+y²=(x−y)²+2xy. These are identities, so no need to solve for x and y individually unless the question asks for them.
4. Quadratics and roots
A quadratic ax²+bx+c=0 with a≠0 can sometimes be factored. For x²−5x+6=0, (x−2)(x−3)=0, so x=2 or x=3. Both candidates must be checked in the original word problem: a negative or zero length may be mathematically valid for the equation but invalid for the physical context.
The discriminant b²−4ac indicates the number of real roots: positive gives two distinct real roots, zero one repeated real root, negative no real roots. The quadratic formula x=[−b±√(b²−4ac)]/(2a) works when factorisation is not obvious. For elementary exam questions, try simple factor pairs before invoking the formula.
5. Inequalities and restrictions
Multiplying or dividing an inequality by a negative number reverses its direction. Denominators cannot be zero. Squaring both sides can introduce extra candidate solutions, so substitute them into the original equation. For example, x=−2 cannot solve √x=2 in real arithmetic even if a careless square operation seems to permit it.
Worked example 4. Solve −2x<8. Divide by −2 and reverse the sign: x>−4. Test x=0: the original gives 0<8, confirming the direction.
6. Model check
After solving, ask whether the variable means a number, price, count or distance. A non-integer answer to “number of people” usually signals a model or arithmetic error. A negative solution to a pure number equation may be acceptable; a negative time normally is not. The original wording, not merely the transformed equation, decides.
Recall before practice
1. Explain why clearing denominators must affect every term on both sides. 2. Expand (a−b)² correctly. 3. State how a negative multiplier changes an inequality. 4. Explain why a quadratic's roots require a context check.
Chapter 19 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. If a + b + c = 0 and abc = −384, then the value of a³ + b³ + c³ is:
A. 1152 B. −384 C. −1152 D. −1149
2. The sum of two numbers is 96 and their difference is 76. Find the larger number.
A. 48 B. 86 C. 20 D. 10
3. 1 notebook and 2 erasers cost ₹51, while 4 notebooks and 5 erasers cost ₹168. What is the cost of one notebook?
A. ₹39 B. ₹17 C. ₹27 D. ₹12
4. If x/y = 13/4, then the value of (x + y)/(x − y) is:
A. 13/9 B. 17/9 C. 9/17 D. 17/13
5. If 5x − 5/x = 20, then the value of x² + 1/x² is:
A. 402 B. 18 C. 14 D. 16
6. Simplify: (9m + 8n)(9m − 8n)
A. 81m² + 64n² B. 18m² − 16n² C. 81m² − 144mn − 64n² D. 81m² − 64n²
7. If x + y = 11 and xy = 30, then the value of x² + y² is:
A. 91 B. 121 C. 181 D. 61
8. If x² − 9x + 1 = 0, then the value of x² + 1/x² is:
A. 79 B. 81 C. 77 D. 83
9. The roots of the equation x² + 4x + 3 = 0 are:
A. Not real (imaginary) B. Real, rational and unequal C. Real and equal D. Real, irrational and unequal
10. The roots of the equation x² + 2x + 1 = 0 are:
A. Real and equal B. Real, rational and unequal C. Real, irrational and unequal D. Not real (imaginary)
11. The roots of the equation x² − 5x + 4 = 0 are:
A. Real, rational and unequal B. Real, irrational and unequal C. Real and equal D. Not real (imaginary)
12. The sum of two numbers is 176 and their difference is 56. Find the larger number.
A. 120 B. 88 C. 60 D. 116
13. If 2x − 5y = −20 and x − 2y = −7, then the value of x + y is:
A. −1 B. −11 C. 12 D. 11
14. 1 bag and 6 caps cost ₹87, while 5 bags and 1 cap cost ₹145. What is the cost of one bag?
A. ₹10 B. ₹27 C. ₹37 D. ₹32
15. When 7 times a number is increased by 7, the result is 287. Find the number.
A. 47 B. 280 C. 40 D. 42
16. 5 apples and 6 oranges cost ₹294, while 6 apples and 4 oranges cost ₹276. What is the cost of 5 apples and 2 oranges?
A. ₹208 B. ₹180 C. ₹198 D. ₹210
17. If 4x + 3y = 17 and x − 3y = 23, then the value of x − y is:
A. −13 B. 13 C. 14 D. 3
18. If x + y = 5 and xy = 6, then the value of x³ + y³ is:
A. 107 B. 35 C. 125 D. 215
19. The sum of the digits of a two-digit number is 13. The number obtained by reversing the digits is 45 more than the original number. Find the original number.
A. 58 B. 49 C. 40 D. 94
20. If 4x − 6y = −60 and 2x − 7y = −46, then the value of x − y is:
A. −12 B. −13 C. 13 D. −5
21. If x² − 8x + 1 = 0, then the value of x² + 1/x² + 6 is:
A. 56 B. 72 C. 68 D. 70
22. If 4 is added to the numerator of a fraction, it becomes 13/11; if 4 is added to the denominator instead, it becomes 3/5. Find the fraction.
A. 11/9 B. 10/12 C. 9/11 D. 9/12
23. If 2 is added to the numerator of a fraction, it becomes 11/16; if 1 is added to the denominator instead, it becomes 9/17. Find the fraction.
A. 9/17 B. 10/17 C. 16/9 D. 9/16
24. If m + n = 4 and mn = 3, then the value of m³ + n³ is:
A. 28 B. 100 C. 64 D. 55
25. 5 apples and 4 oranges cost ₹98, while 4 apples and 1 orange cost ₹52. What is the cost of 5 apples and 2 oranges?
A. ₹80 B. ₹70 C. ₹74 D. ₹84
26. If x + 3y = 16 and 4x + 6y = 22, then the value of x + y is:
A. −12 B. 3 C. 2 D. −2
27. The difference of two numbers is 18. 3 times the larger number added to 2 times the smaller number gives 304. Find the larger number.
A. 68 B. 50 C. 100 D. 59
28. If x = −6 is a root of the equation x² + kx − 12 = 0, then the other root is:
A. 6 B. 2 C. −2 D. 4
29. If a + b = 31 and a − b = 9, then the value of ab is:
A. 220 B. 440 C. 260.5 D. 880
30. If 3 is added to the numerator of a fraction, it becomes 12/13; if 3 is added to the denominator instead, it becomes 9/16. Find the fraction.
A. 13/9 B. 10/14 C. 9/14 D. 9/13
31. The difference of two numbers is 42. 2 times the larger number added to 5 times the smaller number gives 238. Find the smaller number.
A. 42 B. 64 C. 22 D. 43
32. In a test of 40 questions, 4 marks are awarded for every correct answer and 1 mark is deducted for every wrong answer. Pooja attempted all the questions and scored 130 marks. How many questions did Pooja answer correctly?
A. 6 B. 32 C. 36 D. 34
33. If a/b + b/a = 6, then the value of (a³ + b³)/(ab(a + b)) is:
A. 4 B. 7 C. 6 D. 5
34. If a + b + c = 0 and a² + b² + c² = 56, then the value of ab + bc + ca is:
A. −56 B. 28 C. −28 D. 56
35. If x + y = 5, then the value of x³ + y³ + 15xy is:
A. 125 B. 110 C. 25 D. 15
36. If m + n = 10 and mn = 16, then the value of m/n + n/m is:
A. 17/4 B. 25/4 C. 5/8 D. 33/4
37. If x² + y² + z² = 62 and xy + yz + zx = 41, where x + y + z > 0, then the value of x + y + z is:
A. 12 B. 14 C. 144 D. 103
38. In a test of 40 questions, 2 marks are awarded for every correct answer and 1 mark is deducted for every wrong answer. Asha attempted all the questions and scored 71 marks. How many questions did Asha answer correctly?
A. 37 B. 35 C. 39 D. 3
39. 1 cup and 5 plates cost ₹520, while 3 cups and 2 plates cost ₹520. What is the cost of 4 cups and 5 plates?
A. ₹880 B. ₹890 C. ₹1,080 D. ₹920
40. If 3x + 3/x = 9, then the value of x² + 1/x² is:
A. 11 B. 7 C. 79 D. 9
41. If 2x + 2/x = 4, then the value of x³ + 1/x³ is:
A. 5 B. 14 C. 2 D. 8
42. The sum of a positive number greater than 1 and its reciprocal is 37/6. Find the number.
A. 7 B. 5 C. 6 D. 37
43. For what value of k does the pair of equations x + 4y = 3 and (k − 6)x + (k + 3)y = k have infinitely many solutions?
A. 9 B. −9 C. 8 D. 10
44. If a = (√7 − 2)^(1/3), then the value of (a − 1/a)³ + 3(a − 1/a) is:
A. (4√7 − 4)/3 B. (2√7 − 8)/3 C. −√7 D. 2√7 − 4
45. The sum of a positive number greater than 1 and its reciprocal is 145/12. Find the number.
A. 12 B. 11 C. 145 D. 13
46. How many common solutions (x, y) do the inequalities 3x + 2y < 2 and 6x + 4y ≥ 10 have?
A. 0 B. 1 C. 2 D. Infinitely many
47. If x = 272, y = 274 and z = 276, then the value of x³ + y³ + z³ − 3xyz is:
A. 19728 B. 822 C. 3288 D. 9864
48. Sanjay is 4 years older than Rohit, and the product of their ages is 357. What is Rohit's age?
A. 16 B. 18 C. 17 D. 21
49. If 4x² − 11x + 4 = 0, then the value of x² + 1/x² + 3 is:
A. 201/16 B. 41/16 C. 169/16 D. 137/16
50. Meena is 6 years older than Geeta, and the product of their ages is 720. What is Geeta's age?
A. 25 B. 24 C. 30 D. 23
Chapter 19 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. C. a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 0, so a³ + b³ + c³ = 3abc = 3 × (−384) = −1152.
APAR26-19-36 | a + b + c = 0 forms | Easy
2. B. x + y = 96, x − y = 76. Adding: 2x = 172 ⇒ x = 86; subtracting: 2y = 20 ⇒ y = 10. The larger number is 86.
APAR26-19-01 | Two numbers, linear pair | Easy
3. C. Let one notebook cost ₹x and one eraser ₹y: x + 2y = 51 and 4x + 5y = 168. Eliminate one variable: multiply the first equation by 4 and the second by 1 and subtract ⇒ (8 − 5)y = 204 − 168 ⇒ y = 12; then x = (51 − 2 × 12)/1 = 27. Required cost = ₹27.
APAR26-19-03 | Two-variable cost system | Easy
4. B. By componendo and dividendo, (x + y)/(x − y) = (13 + 4)/(13 − 4) = 17/9 = 17/9. (Or put x = 13k, y = 4k and simplify.)
APAR26-19-32 | Componendo and dividendo | Easy
5. B. Divide by 5: x − 1/x = 20/5 = 4. Square: (x − 1/x)² = x² + 1/x² − 2 ⇒ x² + 1/x² = 4² + 2 = 18. (Sign trap: the minus form gives +2 when expanded.)
APAR26-19-31 | x + 1/x family | Easy
6. D. Use (A + B)(A − B) = A² − B² with A = 9m, B = 8n: (9m)² − (8n)² = 81m² − 64n². The cross terms ±144mn cancel, so no middle term appears.
APAR26-19-35 | (a + b)(a − b) | Easy
7. D. x² + y² = (x + y)² − 2xy = 11² − 2 × 30 = 121 − 60 = 61. (Sign trap: the + case carries −3xy(x + y).)
APAR26-19-34 | Cubes from sum and product | Easy
8. A. Since x ≠ 0, divide by x: x + 1/x = 9. x² + 1/x² = (x + 1/x)² − 2 = 9² − 2 = 79.
APAR26-19-33 | x ± 1/x from a quadratic | Easy
9. B. Discriminant D = b² − 4ac = (4)² − 4 × 1 × (3) = 16 − 12 = 4. D > 0 and a perfect square (2²), so the roots are real, rational and unequal.
APAR26-19-04 | Nature of roots | Easy
10. A. Discriminant D = b² − 4ac = (2)² − 4 × 1 × (1) = 4 − 4 = 0. D = 0, so the roots are real and equal.
APAR26-19-09 | Nature of roots | Easy
11. A. Discriminant D = b² − 4ac = (−5)² − 4 × 1 × (4) = 25 − 16 = 9. D > 0 and a perfect square (3²), so the roots are real, rational and unequal.
APAR26-19-08 | Nature of roots | Easy
12. D. x + y = 176, x − y = 56. Adding: 2x = 232 ⇒ x = 116; subtracting: 2y = 120 ⇒ y = 60. The larger number is 116.
APAR26-19-02 | Two numbers, linear pair | Easy
13. D. Multiply to match the y-coefficients and eliminate: (−4 − −5)x = 40 − 35 ⇒ x = 5; substitute back: y = 6. Hence x + y = 11. (Check by substituting in both equations.)
APAR26-19-07 | Solving a linear pair | Easy
14. B. Let one bag cost ₹x and one cap ₹y: x + 6y = 87 and 5x + y = 145. Eliminate one variable: multiply the first equation by 5 and the second by 1 and subtract ⇒ (30 − 1)y = 435 − 145 ⇒ y = 10; then x = (87 − 6 × 10)/1 = 27. Required cost = ₹27.
APAR26-19-05 | Two-variable cost system | Easy
15. C. Let the number be x: 7x + 7 = 287 ⇒ 7x = 287 − 7 = 280 ⇒ x = 280/7 = 40. (Adding 7 instead of subtracting gives 42.)
APAR26-19-06 | Framing a linear equation | Easy
16. C. Let one apple cost ₹x and one orange ₹y: 5x + 6y = 294 and 6x + 4y = 276. Eliminate one variable: multiply the first equation by 6 and the second by 5 and subtract ⇒ (36 − 20)y = 1764 − 1380 ⇒ y = 24; then x = (294 − 6 × 24)/5 = 30. Required cost = 5 × 30 + 2 × 24 = ₹198.
APAR26-19-16 | Two-variable cost system | Medium
17. B. Multiply to match the y-coefficients and eliminate: (−12 − 3)x = −51 − 69 ⇒ x = 8; substitute back: y = −5. Hence x − y = 13. (Check by substituting in both equations.)
APAR26-19-19 | Solving a linear pair | Medium
18. B. x³ + y³ = (x + y)³ − 3xy(x + y) = 5³ − 3 × 6 × 5 = 125 − 90 = 35. (Sign trap: the + case carries −3xy(x + y).)
APAR26-19-38 | Cubes from sum and product | Medium
19. B. Let the tens digit be t and the units digit be u: t + u = 13, and (10t + u) − (10u + t) = 9(t − u) = −45 ⇒ t − u = −5. Adding: 2t = 8 ⇒ t = 4, u = 9. Number = 49 (the reversed number 94 is the classic wrong pick).
APAR26-19-24 | Two-digit number from digits | Medium
20. B. Multiply to match the y-coefficients and eliminate: (−28 − −12)x = 420 − 276 ⇒ x = −9; substitute back: y = 4. Hence x − y = −13. (Check by substituting in both equations.)
APAR26-19-14 | Solving a linear pair | Medium
21. C. Since x ≠ 0, divide by x: x + 1/x = 8. x² + 1/x² = (x + 1/x)² − 2 = 8² − 2 = 62; adding 6 gives 68.
APAR26-19-43 | x ± 1/x from a quadratic | Medium
22. C. Let the fraction be x/y. (x + 4)/y = 13/11 ⇒ 11x − 13y = -44; x/(y + 4) = 3/5 ⇒ 5x − 3y = 12. Solving the pair gives x = 9, y = 11, so the fraction is 9/11. (Check: 13/11 = 13/11 and 9/15 = 3/5.)
APAR26-19-12 | Fraction from two conditions | Medium
23. D. Let the fraction be x/y. (x + 2)/y = 11/16 ⇒ 16x − 11y = -32; x/(y + 1) = 9/17 ⇒ 17x − 9y = 9. Solving the pair gives x = 9, y = 16, so the fraction is 9/16. (Check: 11/16 = 11/16 and 9/17 = 9/17.)
APAR26-19-21 | Fraction from two conditions | Medium
24. A. m³ + n³ = (m + n)³ − 3mn(m + n) = 4³ − 3 × 3 × 4 = 64 − 36 = 28. (Sign trap: the + case carries −3mn(m + n).)
APAR26-19-46 | Cubes from sum and product | Medium
25. C. Let one apple cost ₹x and one orange ₹y: 5x + 4y = 98 and 4x + y = 52. Eliminate one variable: multiply the first equation by 4 and the second by 5 and subtract ⇒ (16 − 5)y = 392 − 260 ⇒ y = 12; then x = (98 − 4 × 12)/5 = 10. Required cost = 5 × 10 + 2 × 12 = ₹74.
APAR26-19-17 | Two-variable cost system | Medium
26. C. Multiply to match the y-coefficients and eliminate: (6 − 12)x = 96 − 66 ⇒ x = −5; substitute back: y = 7. Hence x + y = 2. (Check by substituting in both equations.)
APAR26-19-22 | Solving a linear pair | Medium
27. A. Let the numbers be x (larger) and y: x − y = 18 and 3x + 2y = 304. Substitute x = y + 18: 3(y + 18) + 2y = 304 ⇒ 5y = 250 ⇒ y = 50, x = 68. Required: 68.
APAR26-19-18 | Two numbers, linear pair | Medium
28. B. Substitute x = −6: (−6)² + k(−6) − 12 = 0 ⇒ 24 − 6k = 0 ⇒ k = 4. Product of roots = −12, so the other root = −12/(−6) = 2 (check: sum of roots = −k = −4).
APAR26-19-23 | Root given, find parameter | Medium
29. A. (a + b)² − (a − b)² = 4ab ⇒ 4ab = 31² − 9² = 961 − 81 = 880 ⇒ ab = 880/4 = 220. (Forgetting to divide by 4 gives 880.)
APAR26-19-45 | 4ab = (a + b)² − (a − b)² | Medium
30. D. Let the fraction be x/y. (x + 3)/y = 12/13 ⇒ 13x − 12y = -39; x/(y + 3) = 9/16 ⇒ 16x − 9y = 27. Solving the pair gives x = 9, y = 13, so the fraction is 9/13. (Check: 12/13 = 12/13 and 9/16 = 9/16.)
APAR26-19-10 | Fraction from two conditions | Medium
31. C. Let the numbers be x (larger) and y: x − y = 42 and 2x + 5y = 238. Substitute x = y + 42: 2(y + 42) + 5y = 238 ⇒ 7y = 154 ⇒ y = 22, x = 64. Required: 22.
APAR26-19-20 | Two numbers, linear pair | Medium
32. D. Let correct = c, wrong = 40 − c. Then 4c − (1)(40 − c) = 130 ⇒ 4c − (40 − c) = 130 ⇒ 5c = 170 ⇒ c = 34. Wrong = 40 − 34 = 6. Answer: 34. (Ignoring the negative marking gives 130/4.)
APAR26-19-13 | Marking scheme equation | Medium
33. D. a/b + b/a = 6 ⇒ a² + b² = 6ab. Then a³ + b³ = (a + b)(a² − ab + b²) = (a + b)(6ab − ab) = 5ab(a + b), so the ratio = 5.
APAR26-19-37 | a/b + b/a relations | Medium
34. C. (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) ⇒ 0 = 56 + 2(ab + bc + ca) ⇒ ab + bc + ca = −56/2 = −28. (Dropping the minus sign gives 28.)
APAR26-19-42 | a + b + c = 0 forms | Medium
35. A. (x + y)³ = x³ + y³ + 3xy(x + y) = x³ + y³ + 3xy × 5 = x³ + y³ + 15xy. So the expression is (x + y)³ = 5³ = 125, independent of the individual values.
APAR26-19-40 | Cube of a binomial | Medium
36. A. m/n + n/m = (m² + n²)/mn = [(m + n)² − 2mn]/mn = (100 − 32)/16 = 68/16 = 17/4. Express the target through m + n and mn instead of solving for m and n separately.
APAR26-19-44 | Symmetric expressions (two variables) | Medium
37. A. (x + y + z)² = 62 + 2 × 41 = 144 ⇒ x + y + z = √144 = 12 (positive root). The identity (x + y + z)² = Σx² + 2Σxy links the three quantities; forgetting the factor 2 is the common error.
APAR26-19-39 | Square of a trinomial | Medium
38. A. Let correct = c, wrong = 40 − c. Then 2c − (1)(40 − c) = 71 ⇒ 2c − (40 − c) = 71 ⇒ 3c = 111 ⇒ c = 37. Wrong = 40 − 37 = 3. Answer: 37. (Ignoring the negative marking gives 71/2.)
APAR26-19-15 | Marking scheme equation | Medium
39. A. Let one cup cost ₹x and one plate ₹y: x + 5y = 520 and 3x + 2y = 520. Eliminate one variable: multiply the first equation by 3 and the second by 1 and subtract ⇒ (15 − 2)y = 1560 − 520 ⇒ y = 80; then x = (520 − 5 × 80)/1 = 120. Required cost = 4 × 120 + 5 × 80 = ₹880.
APAR26-19-11 | Two-variable cost system | Medium
40. B. Divide by 3: x + 1/x = 9/3 = 3. Square: (x + 1/x)² = x² + 1/x² + 2 ⇒ x² + 1/x² = 3² − 2 = 7. (Sign trap: the plus form gives −2 when expanded.)
APAR26-19-41 | x + 1/x family | Medium
41. C. Divide by 2: x + 1/x = 4/2 = 2. Cube: (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x) ⇒ x³ + 1/x³ = 2³ − 3 × 2 = 8 − 6 = 2. (Sign trap: the plus form gives −3(x + 1/x) when expanded.)
APAR26-19-48 | x + 1/x family | Difficult
42. C. x + 1/x = 37/6 ⇒ 6x² − 37x + 6 = 0 ⇒ (6x − 1)(x − 6) = 0 ⇒ x = 6 or 1/6. Since the number exceeds 1, x = 6.
APAR26-19-30 | Number plus reciprocal | Difficult
43. A. Condition for infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂. From a₁/a₂ = b₁/b₂: 1/(k − 6) = 4/(k + 3) ⇒ 1(k + 3) = 4(k − 6) ⇒ k = 9. At k = 9 the second equation is 3x + 12y = 9, i.e. exactly 3 × (first equation), so c₁/c₂ also equals 1/3: infinitely many solutions.
APAR26-19-28 | Consistency of a linear pair | Difficult
44. B. (a − 1/a)³ + 3(a − 1/a) = a³ − 1/a³. Here a³ = √7 − 2 and 1/a³ = 1/(√7 − 2) = (2 + √7)/3 after rationalising (norm -2² − 7 × 1² = −3). So a³ − 1/a³ = (√7 − 2) − ((2 + √7)/3) = (2√7 − 8)/3.
APAR26-19-50 | a³ ± 1/a³ with surds | Difficult
45. A. x + 1/x = 145/12 ⇒ 12x² − 145x + 12 = 0 ⇒ (12x − 1)(x − 12) = 0 ⇒ x = 12 or 1/12. Since the number exceeds 1, x = 12.
APAR26-19-26 | Number plus reciprocal | Difficult
46. A. Divide the second inequality by 2: 3x + 2y ≥ 5. We would need 3x + 2y < 2 and 3x + 2y ≥ 5 simultaneously, but 5 ≥ 2, so no point satisfies both. Answer: 0.
APAR26-19-29 | Linear inequalities | Difficult
47. D. x³ + y³ + z³ − 3xyz = (x + y + z) × ½[(x − y)² + (y − z)² + (z − x)²] = 822 × ½[4 + 4 + 16] = 822 × 12 = 9864. (Never cube the numbers; only the differences matter.)
APAR26-19-47 | x³ + y³ + z³ − 3xyz | Difficult
48. C. Let Rohit's age be x, so Sanjay's age is x + 4: x(x + 4) = 357 ⇒ x² + 4x − 357 = 0 ⇒ (x − 17)(x + 21) = 0 ⇒ x = 17. Rohit = 17 years, Sanjay = 21 years; required 17 years.
APAR26-19-25 | Quadratic from ages | Difficult
49. D. Since x ≠ 0, divide by 4x: x + 1/x = 11/4. x² + 1/x² = (x + 1/x)² − 2 = 11/4² − 2 = 89/16; adding 3 gives 137/16.
APAR26-19-49 | x ± 1/x from a quadratic | Difficult
50. B. Let Geeta's age be x, so Meena's age is x + 6: x(x + 6) = 720 ⇒ x² + 6x − 720 = 0 ⇒ (x − 24)(x + 30) = 0 ⇒ x = 24. Geeta = 24 years, Meena = 30 years; required 24 years.
APAR26-19-27 | Quadratic from ages | Difficult