Algebra (SSC) (Part-4)

Total Questions: 50

21. If (x - 2) is a factor of (2x² + 12kx - 25k), then what is the value of k? [SSC MTS 01/09/2023 (2nd Shift)]

Correct Answer: (b) 8
Solution:

Put x = 2 in equation

2x² + 12kx − 25k = 0

2 × (2)² + 12k × 2 − 25k = 0 ⇒ k = 8

22. Solve the following equation. [SSC CHSL 02/08/2023 (1st Shift)]

Correct Answer: (d) 81/8
Solution:

2x + 2/x = 5 ⇒ x + 1/x = 5/2

x³ + 1/x³ = (5/2)³ − 3 × (5/2) = 65/8

x³ + 1/x³ + 2 = 65/8 + 2 = 81/8

23. If (a - b) = 9 and (a³ - b³) = 4401, find the value of ab. [SSC CHSL 02/08/2023 (2nd Shift)]

Correct Answer: (d) 136
Solution:

(a³ − b³) = (a − b){(a − b)² + 3ab}

4401 = 9(81 + 3ab)

4401 = 9(81 + 3ab)

489 − 81 = 3ab ⇒ ab = 408/3 = 136

24. Solve the following equation: [SSC CHSL 02/08/2023 (2nd Shift)]

Correct Answer: (c) 25.5
Solution:

According to question,

a + 1/a = 6 ⇒ a² + 1/a² = (6)² − 2 = 34

Now,
3/4 (a² + 1/a²) = 3/4 × 34 = 51/2 = 25.5

25. Solve the following equation: [SSC CHSL 02/08/2023 (3rd Shift)]

Correct Answer: (b) √10
Solution:

a = 1/(a − √6) ⇒ a − 1/a = −√6

a + 1/a = √((√6)² + 4) = √10

26. Solve the following equation: [SSC CHSL 03/08/2023 (1st Shift)]

Correct Answer: (c) 81
Solution:

(a + b) = 30 ⇒ a² + b² = 900 − 2ab

√(a/b) = 8/3 + √(b/a)

(a − b)/√(ab) = 8/3

⇒ a² + b² − 2ab / ab = 64/9

900 − 4ab / ab = 64/9

⇒ ab = 8100 / 100 = 81

27. If 2𝓍 + 3y = 9, and 𝓍y = 3, what is 8𝓍³ + 27y³ ? [SSC CHSL 03/08/2023 (1st Shift)]

Correct Answer: (c) 243
Solution:

2x + 3y = 9 and xy = 3

Then,
8x³ + 27y³ = (2x + 3y)[(2x + 3y)² − 3(2x × 3y)]

= (9)[(9)² − 18xy] = (9)[81 − 18 × 3]

= 243

28. Solve the following equation: [SSC CHSL 03/08/2023 (3rd Shift)]

Correct Answer: (b) 10/3√3
Solution:

a² + 1/a² = 7/3

a − 1/a = √(7/3 − 2) = 1/√3

Now,
a³ − 1/a³ = (1/√3)³ + 3 × 1/√3

= 1/(3√3) + √3 = 10/(3√3)

29. If a + b = 10 and a² + b² = 58, find the value of ab. [SSC CHSL 03/08/2023 (3rd Shift)]

Correct Answer: (d) 21
Solution:

a + b = 10 and a² + b² = 58

(a + b)² = a² + b² + 2ab

100 = 58 + 2ab ⇒ ab = 21

30. If (a + b + c) ≠ 0, then (a + b + c) (a² + b² + c²- ab - bc - ca) is equal to: [SSC CHSL 03/08/2023 (4th Shift)]

Correct Answer: (c) a³ + b³ + c³ - 3abc
Solution:

a³ + b³ + c³ − 3abc

= (a + b + c)(a² + b² + c² − ab − bc − ca)