Linear Equation with two variables
Home⇒Quantitative Aptitude⇒ Linear Equation with two variables
Q21. The system of equations 3x + 7y = 9 and 10x + ky = 30 has infinite number of solutions when
(a) k = 50/3
(b) k = 55/3
(c) k = 70/3
(d) k = 77/3
Answer: (c) k = 70/3
For infinite number of solution we have
a_{1} | = | b_{1} | = | c_{1} | |
a_{2} | b_{2} | c_{2} |
⇒ | 3 | = | 7 | = | -9 |
10 | k | -30 |
⇒ | 3 | = | 7 | = | 3 |
10 | k | 10 |
∴ 7/k = 3/10
⇒ 3k = 70
⇒ k = 70/3
Q22. If 3x - 5y = 5 and x/(x+y) = 5/7 then x + y = ?
(a) 5
(b) 7
(c) 11
(d) 13
Answer: (b) 7
3x - 5y = 5 ------- (1)
x/(x+y) = 5/7
⇒ 7x - 5y - 5x = 0
⇒ 2x - 5y = 0 ------ (2)
(1)-(2) ⇒ x = 5
(1)⇒ 3X5 - 5y = 5
⇒ - 5y = 5 - 15
⇒ y = 2
∴ x + y = 5 + 2 = 7
Q23. On solving 3x + y = 7 and 3y = 2 + 2x we have
(a) x = 23/11, y =8/11
(b) x = 21/11, y =9/11
(c) x = 17/11, y =7/11
(d) x = 13/11, y =5/11
Answer: (a) x = 23/11, y =8/11
3x + y = 7
⇒ 6x + 2y = 14 ------- (1)
3y - 2x - 2 = 0
⇒ 2x - 3y = 2
⇒ 6x - 9y = 6 --------- (2)
(1)-(2)⇒ 11y = 8
⇒ y = 8/11
(1)⇒ 6x + 2X(8/11) = 14
⇒ 6x = 14 - (16/11)
⇒ 6x = 138/11
⇒ x = 138/(11 X 6) = 23/11
∴ x = 23/11, y = 8/11
Q24. The system of equations 3x + 2y = 3 and 6x + 4y = 3 has
(a) exactly two solutions
(b) no solutions
(c) a unique solution
(d) infinitely many solutions
Answer: (b) no solutions
Here | a_{1} | = | 3 | = | 1 |
a_{2} | 6 | 2 |
b_{1} | = | 2 | = | 1 | |
b_{2} | 4 | 2 |
c_{1} | = | 3 | = 1 | |
c_{2} | 3 |
∴ | a_{1} | = | b_{1} | ≠ | c_{1} |
a_{2} | b_{2} | c_{2} |
∴ Given system has no solution
Q25. On solving 3x + y = 6 and 5y + x = 10, we get
(a) a = 5/7, y = 9/7
(b) a = 9/7, y = 11/7
(c) a = 10/7, y = 12/7
(d) a = 12/7, y = 13/7
Answer: (c) a = 10/7, y = 12/7
3x + y = 6 ------- (1)
5y + x = 10
3x + 15y = 30 ------- (2)
(2)-(1) ⇒ 14y = 24
⇒ y = 12/7
(1)⇒ 3x + 12/7 = 6
⇒ 3x = 6 - (12/7) = 30/7
⇒ x = 30/(3X7) = 10/7
∴ x = 10/7 y = 12/7
Q26. If 4x + 6y = 32 and 4x - 2y = 4 then 6y = ?
(a) 13
(b) 16
(c) 17
(d) 21
Answer: (d) 21
4x + 6y = 32 ------ (1) 4x - 2y = 4 ------ (2)
(1)-(2)⇒ 8y = 28
⇒ y = 28/8
⇒ 6y = 6 X (28/8) = 21
Q27. The solution of 2x + 3y = 2 and 3x + 2y = 2 can be represented by a point in the co-ordinate plane in
(a) 1st quadrant
(b) 2nd quadrant
(c) 3rd quadrant
(d) 4th quadrant
Answer: (a) 1st quadrant
2x + 3y = 2
⇒ 6x + 9y = 6 ------ (1)
3x + 2y = 2
⇒ 6x + 4y = 4 ------- (2)
(1)-(2) ⇒ 5y = 2
⇒ y = 2/5
(1)⇒ 6x + 9 X (2/5) = 6
⇒ 6x + 18/5 = 6
⇒ 6x = 6 - (18/5) = 12/5
⇒ x = 12/(5X6) = 2/5
The solution can be represented by the point (2/5, 2/5) which lies in 1st quadrant
Q28. If α, β are the roots of the equations x^{2} - px + q = 0 then the value of α^{2} + β^{2} is
(a) q - 2p
(b) p - 2q
(c) q^{2} - 2p
(d) p^{2} - 2q
Answer: (d) p^{2} - 2q
α + β = -(-p)/1 = p
αβ = q/1 = q
Now α^{2} + β^{2} = (α + β)^{2} - 2αβ = p^{2} - 2q
Q29. A and B solved a quadratic equation. In solving it, A made a mistake in the constant term and obtained the roots as 6 and 2, while B made a mistake in the co-efficient of x only and obtained the roots -7 and -1. Find the correct roots of the equation.
(a) 1, 5
(b) 2, 5
(c) 1, 7
(d) 2, 9
Answer: (c) 1, 7
As obtained by A, we have
α + β = 8 and αβ = 12
∴ The equation is x^{2} + 8x + 7 = 0
As obtained by B, we have α + β = -8, αβ = 7
∴ The equation is x^{2} + 8x + 7 = 0
Hence the correct equation is x^{2} - 8x + 7 = 0
Now x^{2} - 8x + 7 = 0 ⇒ x^{2} - 7x - x + 7 = 0
⇒ x(x - 7) - 1(x - 7) = 0
⇒ (x - 1) (x - 7) = 0
⇒ x = 7 or x = 1
Q30. The sum of the roots of the equation 3x^{2} + (p + q + r)x + pqr = 0 is equal to zero. What is the value of (p^{3} + q^{3} + r^{3}) = ?
(a) 2pqr
(b) 3pqr
(c) 2pq/r
(d) 3qr/p
Answer: (b) 3pqr
We have
-(p+q+r)/3 = 0
⇒ p + q + r = 0
⇒ p^{3} + q^{3} + r^{3} = 3pqr