Pair Of Linear Equations In Two Variables Test

Pair Of Linear Equations In Two Variables

This is Pair of Linear Equations in Two Variables Test-04 for CBSE class 10 Maths.. There are 15 questions in this test with each question having around four answer choices.

Questions & Answers

If 2x – 3y = 11 and (a + b)x – (a + b – 3)y = 4a + b has infinite number of solutions, then

A
a = 9 and b = 3
B
a = 9 and b = – 3
C
a = – 9 and b = – 3
D
a = – 9 and b = 3
Correct

The system of equations 6x + 3y = 6xy and 2x + 4y = 5xy has

A
many solutions
B
no solution
C
two solutions
Correct
D
one solution

The value of ‘k’ for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has a non zero solution is

A
0
B
2
C
8
D
6
Correct

Given that 2x + 3y = 11, 2x – 4y = – 24 and y = mx + 3, then the value of ‘m’ is

A
– 1
Correct
B
0
C
2
D
1

The solution of 3x – 5y = – 16 and 2x + 5y = 31

A
x = – 3, y = 5
B
x = 3, y = – 5
C
x = 3, y = 5
Correct
D
x = – 3, y = – 5

The system of linear equations \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) has a unique solution if

A
\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)
B
none of these
C
\(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\)
Correct
D
\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\)

If x = α and y = β is the solution of the equations x – y = 2 and x + y = 4, then

A
\(\alpha {\text{ }} = {\text{ }}3{\text{ }}and{\text{ }}\beta {\text{ }} = {\text{ - }}1\)
B
\(\alpha {\text{ }} = {\text{ - }}3{\text{ }}and{\text{ }}\beta {\text{ }} = {\text{ }}1\)
C
\(\alpha {\text{ }} = {\text{ }}3{\text{ }}and{\text{ }}\beta {\text{ }} = {\text{ }}1\)
Correct
D
\(\alpha {\text{ }} = {\text{ 1 }}and{\text{ }}\beta {\text{ }} = {\text{ 3}}\)

The solution of px + qy = p – q and qx – py = p + q is

A
x = 0 and y = 0
B
x = 1 and y = 1
C
x = 1 and y = – 1
Correct
D
x = – 1 and y = 1

Solution of \(\frac{{{a^2}}}{x} - \frac{{{b^2}}}{y} = 0\) and \(\frac{{{a^2}b}}{x} + \frac{{{b^2}a}}{y} = a + b\) where \(x,y \ne 0\) is

A
\(x = - {a^2}{\text{ }}and{\text{ }}y = - {b^2}\)
B
\(x = {a^2}{\text{ }}and{\text{ }}y = - {b^2}\)
C
\(x = {a^2}{\text{ }}and{\text{ }}y = {b^2}\)
Correct
D
\(x = - {a^2}{\text{ }}and{\text{ }}y = {b^2}\)

Solution of \(\frac{x}{a} + \frac{y}{b} = 2\) and \(ax - by = {a^2} - {b^2}\) is

A
\(x{\text{ }} = {\text{ }}{a^2}and{\text{ }}y{\text{ }} = {\text{ }}{b^2}\)
B
\(x{\text{ }} = {\text{ - }}{a^2}and{\text{ }}y{\text{ }} = {\text{ - }}{b^2}\)
C
x = – a and y = – b
D
x = a and y = b
Correct

The solution of 217x + 131y = 913 and 131x + 217y = 827 is

A
x = 2 and y = 3
B
x = 2 and y =2
C
x = 3 and y = 3
D
x = 3 and y = 2
Correct

The value of ‘a’ so that the point (3, a) lies on the line represented by 2x – 3y = 5 is

A
1
B
\(\frac{{ - 1}}{3}\)
C
– 1
D
\(\frac{1}{3}\)
Correct

If (x + 1) is a factor of \(2{x^3} + a{x^2} + 2bx + 1\), then the values of ‘a’ and ‘b’, given that 2a – 3b = 4 are

A
a = 5 and b = 2
Correct
B
a = – 5 and b = 2
C
a = 5 and b = – 2
D
a = – 5 and b = – 2

The angles of a triangle are\(x^\circ ,{\text{ }}y^\circ and{\text{ }}40^\circ \) . The difference between the two angles ‘x’ and ‘y’ is\(30^\circ \) , then

A
\(x^\circ = {\text{ }}65^\circ and{\text{ }}y^\circ = {\text{ }}95^\circ \)
B
\(x^\circ = {\text{ }}85^\circ and{\text{ }}y^\circ = {\text{ }}55^\circ \)
Correct
C
\(x^\circ = {\text{ }}75^\circ and{\text{ }}y^\circ = {\text{ }}45^\circ \)
D
none of these

5 pencils and 7 pens together cost Rs.50 whereas 7 pencils and 5 pens together cost Rs.46. The cost of 1 pen is

A
Rs.3
B
Rs.5
Correct
C
Rs.6
D
Rs.4