Free Pair of Linear Equations in Two Variables 01 Practice Test - 10th Grade

Question 1

If (2,2) lies on $4 x + 5 y = k$ , the value of $k$ is

___

SOLUTION

Solution :
If the point (2,2) lies on the given linear equation, then it must satisfy the equation.

Putting the value of x and y in the equation we get:

$\Rightarrow 4 (2) + 5 (2) = k$

$\Rightarrow k = 18$

Thus, the value of k is 18.

Question 2

Which of the following equations have x = 7, y = 3 as solution?

x + y = 4

x + 3y = 16

x - y + 4 = 0

x + y = 7

SOLUTION

Solution : B

The solution of the linear equation should satisfy the equation. The given values of x and y are 7 and 3 respectively.

Among the given equations, only the equation x + 3y = 16 is able to satisfy the equation.

$\Rightarrow 7 + 3 (3) = 16$

Hence $x = 7 a n d y = 3$ is the solution of x + 3y = 16.

Question 3

Half the perimeter of a rectangular room is 46 m, and its length is 6 m more than its breadth. What is the length and breadth of the room?

2 m, 20 m

26 m, 20 m

56 m, 40 m

2 m, 3 m

SOLUTION

Solution : B

Let $l$ and $b$ be the length and breadth of the room respectively.
Then, the perimeter of the room is $2 (l + b)$ metres.

From the question,

$l = 6 + b$ ...(1)

$\frac{1}{2} \times 2 (l + b) = 46$

$\Rightarrow l + b = 46$ ...(2)

Let's solve these two equations using substitution method.

On substituting the value of $l$ from (1) in (2), we get

$6 + b + b = 46$

$\Rightarrow$ $6 + 2 b = 46$

$\Rightarrow 2 b = 40$

$\Rightarrow b = 20 m$

$\Rightarrow l = 20 + 6 = 26 m$

Therefore, length and breadth are 26 m and 20 m long respectively.

Question 4

Which of the following is a solution to $3 x + 4 y = 38 ?$

(3, 4)

(6, 5)

(2, 19)

(3, 12)

SOLUTION

Solution : B

The solution of a linear equation satisfies the equation.

$3 \times 3 + 4 \times 4 = 9 + 16 = 25 \neq 38 \to$ (3, 4) does not satisfy the equation.

$3 \times 6 + 4 \times 5 = 18 + 20 = 38 \to$ (6, 5) satisfies the equation.

$3 \times 2 + 4 \times 19 = 6 + 76 = 82 \neq 38 \to$ (2, 19) does not satisfy the equation.

$3 \times 3 + 4 \times 12 = 9 + 48 = 57 \neq 38 \to$ (3, 12) does not satisfy the equation.

Question 5

______ satisfies the equation $3 x + 4 y = 2$ .

(2, 1)

(-2, 1)

(2, -1)

(2, 2)

SOLUTION

Solution : C

The solutions of a linear equation must be able to satisfy the equation. Among the given options, point (2,-1) satisfies the given equation. Substituing the value of x and y in the equation we get:
3(2) + 4(-1) = 2.

Question 6

On the basis of the graph shown below which shows the graphical representation of a pair of linear equations, the pair of linear equations has _______________solution(s).

a unique

infinite

four

zero

SOLUTION

Solution : A

If the graph of linear equations represented by the lines intersect at only one point, then the point is its only solution.

Here, the lines meet at only one point (1,-1). Therefore, the given pair of equations has a unique solution.

Question 7

Solve the following system of equations:
$8 v - 3 u = 5 u v$
$6 v - 5 u = - 2 u v$

u = 0, v = 0

u = \frac{22}{31}, v = \frac{11}{23}

C. Both A and B

u = \frac{31}{22}, v = \frac{23}{11}

SOLUTION

Solution : B

Divide the given equations by uv,
$8 v - 3 u = 5 u v \Rightarrow \frac{8}{u} - \frac{3}{v} = 5 . . . (1)$
$6 v - 5 u = - 2 u v \Rightarrow \frac{6}{u} - \frac{5}{v} = - 2 . . . (2)$

Assume $\frac{1}{u} = x$ and $\frac{1}{v} = y$
Put the values of $\frac{1}{u}$ and $\frac{1}{v}$ in (1) and (2)
$8 x - 3 y = 5 . . . (3)$
$6 x - 5 y = - 2 . . . (4)$

Solving equations (3) and (4) we get x and y
Multiply (3) with 5 and (4) with 3 to equate the coefficients of y.
$40 x - 15 y = 25$
$18 x - 15 y = - 6$

$\Rightarrow x = \frac{31}{22} \Rightarrow u = \frac{22}{31}$

Substituting x in (3)
$8 \times \frac{31}{22} - 3 y = 5 \to 3 y = \frac{69}{11}$

$\to y = \frac{23}{11} \Rightarrow v = \frac{11}{23}$

So, $u = \frac{22}{31}$ and $v = \frac{11}{23}$

Question 8

If $y = \frac{1}{2} (3 x + 7)$ is written in the form $a x + b y + c = 0,$ which of the following are possible values of a, b and c?

$a = + 3, b = - 2, c = + 7$

$a = + 3, b = + 2, c = + 7$

$a = - 3, b = + 2, c = - 7$

$a = - 3, b = - 2, c = - 7$

SOLUTION

Solution : A and C

The given equation is $y = \frac{1}{2} (3 x + 7)$ .

Simplifying the equation we get
$2 y - 3 x - 7 = 0$

Rewriting the equation in standard form ax + by + c = 0 we get

$- 3 x + 2 y - 7 = 0$

$∴$ On comparing the above equation with standard form ax + by +c = 0, we get the values of a, b and c is -3, 2 and -7 respectively.

But the equation can also be written as,

$3 x - 2 y + 7 = 0$
(By multiplying both the sides by (-1))

$∴$ On comparing the above equation with standard form ax + by +c = 0, we get the value of a, b and c is +3, -2 and +7 respectively.

Question 9

The solution of the pair of linear equations
$x + 2 y = 5$ and $7 x + 3 y = 13$ is (2, 1).

True

False

SOLUTION

Solution : B

$x + 2 y = 5$ ...(1)

$7 x + 3 y = 13$ ...(2)

Let's solve these equations using elimination method.
On multiplying the first equation by 7, we get

$7 x + 14 y = 35$ ...(3)

On subtracting (2) from (3), we have

$7 x + 14 y = 35$

$- 7 x - 3 y = - 13$
_______________

$11 y = 22$

$\Rightarrow y = 2$

On subsituting value of y in $x + 2 y = 5$ , we get

$x + 2 (2) = 5$

$\Rightarrow x = 1$

$\Rightarrow x = 1$ and $y = 2$

Hence, the solution of the given equations is (1, 2).

Question 10

Solve the equations for $x$ and $y$ .

$2 x - 3 y = 7$

$5 x + y = 9$

3,4

2, -1

2,2

3, -1

SOLUTION

Solution : B

Given

$2 x - 3 y = 7$ ...(i)

$5 x + y = 9$ ...(ii)

Rearranging (ii), we get $y = 9 - 5 x$ ...(iii)

Substituting (iii) in (i), we get

$2 x - 3 (9 - 5 x) = 7$

$\Rightarrow$ $17 x = 34$

$\Rightarrow x = 2$ .

Substituting the value of $x$ in (i), we get

$2 (2) - 3 y = 7$

$\Rightarrow y = - 1$ .

Free Pair of Linear Equations in Two Variables 01 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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