Linear Equation with two variables
Q11. If 3 chairs and 1 table cost Rs. 900 and 5 chairs and 3 tables cost Rs. 2400 then the cost of 2 chairs and 2 tables is
(a) Rs. 105
(b) Rs. 115
(c) Rs. 125
(d) Rs. 150
Answer: (d) Rs. 150
Let cost of 1 chair = Rs. x
cost of 1 table = Rs. y
A/Q 3x + y = 900
⇒ 9x + 3y = 2700 ------- (1)
5x + 3y = 2400 ------ (2)
(1)-(2) ⇒ 4x = 300
⇒ x = 75
(1)⇒ 9X75 + 3y = 2700
⇒ 675 + 3y = 2700
⇒ y = 2025/3 = 675
∴ Cost of two chair = Rs. 2 X 75 = Rs. 150
Q12. For which value of K, the system of equations kx - 10y - 3 = 0 and 3x - 5y - 7 = 0 has no solution ?
(a) 5
(b) 6
(c) 8
(d) 10
Answer: (b) 6
For no solution we have
| a1 | = | b1 | ≠ | c1 | |
| a2 | b2 | c2 |
| ∴ | k | = | -10 | ≠ | -3 |
| 3 | -5 | -7 |
| ⇒ | k | = 2 |
| 3 |
⇒ k = 6
Q13. Solve x - 4 = 5/x
(a) 5, -1
(b) 6, -2
(c) 7, -3
(d) 10, -7
Answer: (a) 5, -1
x - 4 = 5/x
⇒ x2 - 4x - 5 = 0
⇒ x2 - 5x + x - 5 = 0
⇒ x(x - 5) + 1(x - 5) = 0
⇒ (x - 5) (x + 1) = 0
∴ x = 5 or x = -1
∴ x = 5, -1
Q14. The roots of the equation 2x2 - 11x + 15 = 0 are
(a) 2, 1/3
(b) 2, 1/5
(c) 3, 3/7
(d) 3, 5/2
Answer: (d) 3, 5/2
2x2 - 11x + 15 = 0
⇒ 2x2 - 6x - 5x + 15 = 0
⇒ 2x(x - 3) - 5(x - 3) = 0
⇒ (x - 3)(2x - 5) = 0
∴ x - 3 = 0 or 2x - 5 = 0
⇒ x = 3 ⇒ x = 5/2
∴ x = 3, 5/2
Q15. If x2 - 3x - 1 = 0 then the value of x - 1/x is
(a) 2
(b) 3
(c) 5
(d) 9
Answer: (b) 3
x2 - 3x - 1 = 0
⇒ x2 - 1 = 3x
⇒ (x2 - 1)/x = 3x/x
⇒ x - 1/x = 3
Q16. If 2x2 + 12x + 18 = 0 then the value of x is
(a) -3
(b) -1
(c) 4
(d) 7
Answer: (a) -3
2x2 + 12x + 18 = 0
⇒ x2 + 16x + 9 = 0
⇒ x2 + 2.x.3 + 32 = 0
⇒ (x + 3)2 = 0
⇒ x + 3 = 0
⇒ x = -3
Q17. If the roots of ax2 + bx + c = 0 be equal then the value of c is
(a) b2/3a
(b) a2/4b
(c) b2/4a
(d) a3/4b
Answer: (c) b2/4a
We have b2 - 4ac = 0
⇒ c = b2/4a
Q18. If one root of x2 - 5kx + 6 = 0 is 4, then the value of k is
(a) 5/7
(b) 9/8
(c) 11/7
(d) 11/10
Answer: (d) 11/10
The equation is x2 - 5kx + 6 = 0
⇒ 42 - 5.k.4 + 6 = 0
⇒ 16 - 20k + 6 = 0
⇒ 20k = 22
⇒ k = 22/20 = 11/10
Q19. For what value of k, the equation x2 + 2(k - 4)x + 2k = 0 has equal roots ?
(a) 5, 7
(b) 7, 3
(c) 8, 2
(d) 9, 5
Answer: (c) 8, 2
For equal roots
b2 - 4ac = 0
⇒ {2(k-4)}2 - 4.1.2k = 0
⇒ 4(k-4)2 - 8k = 0
⇒ 4k2 - 32k + 64 - 8k = 0
⇒ 4k2 - 40k + 64 = 0
⇒ k2 - 10k + 16 = 0
⇒ (k - 8)(k - 2) = 0
∴ k = 8 or k = 2
k = 8, 2
Q20. If α, β be roots of the equation ax2 + bx + c = 0, then the value of (α2/β + β2/α) is
(a) (3abc-b3)/a2c
(b) (3abc-a3)/a3c
(c) (2abc-b2)/a2c
(d) (2abc-b3)/a3c
Answer: (a) (3abc-b3)/a2c
α + β = -b/a αβ = c/a
| Now | α2 | + | β2 | = | α3 + β3 | = | (α + β)3 - 3αβ (α + β) |
| β | α | αβ | αβ |
| = | (-b/a)3 - 3c/a(-b/a) | = | -b3/a3 + 3bc/a2 | = | 3abc - b3 |
| c/a | c/a | a2c |
Practice Test Exam