Pair of Linear Equations in Two Variables Test

Result

Pair of Linear Equations in Two Variables Test - 32

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A man gets Rs. 100 per day if he works, but he is fined by Rs. 10 per day if he is absent. In the whole month of April he received Rs. 1900 only. How many days did he work ?

A
20 days

B
22 days

C
29 days

D
23 days

Solution

Let the man works for $$x$$ days in April and he is absent for $$y$$ days.
Since, April has total 30 days.
$$\therefore x+y=30$$ .........(1)
According to question:
$$100x-10y=1900 \Rightarrow 10x-y=190$$ ....(2)
Adding (1) and (2). we get,
$$11x=220$$
$$\therefore x=20$$
Hence the man worked for $$20$$ days in April.
Question 2
1 / -0

If $$12x\, +\, 13y\, =\, 29\, and\, 13x\, +\, 12y\, =\, 21,$$ find $$x\, +\, y$$.

A
$$2$$

B
$$7$$

C
$$4$$

D
$$11$$

Solution

$$12x\, +\, 13y\, =\, 29\, and\, 13x\, +\, 12y\, =\, 21,$$
Add the two equations:
$$25x + 25y = 50
\\x + y = 2$$
Question 3
1 / -0

Solve the following pair of equations. $$25x-24y=197; \, \, 24x-25y=195$$

A
$$x=1,\, y=-3$$

B
$$x=9,\, y=-3$$

C
$$x=5,\, y=-3$$

D
$$x=3,\, y=-3$$

Solution

$$25x-24y=197$$                              ...(i)
$$24x-25y=195$$                           ...(ii)
On adding (i) and (ii) , we get
$$49x-49y=392$$
$$\Rightarrow x-y=8$$             ....(iii)
On subtracting (ii) from (i) , we get
$$x+y=2$$                         ....(iv)
On adding (iii) and (iv) , we get
$$2x=10$$
$$\Rightarrow x=5$$
By putting $$x=5$$ in (iv), we get
$$5+y=2$$
$$\Rightarrow y=-3$$
Question 4
1 / -0

Solve graphically:
$$\displaystyle 2x-3y=7 $$ and $$\displaystyle 5x+y=9$$

A
$$\displaystyle x=2,y=-6$$

B
$$\displaystyle x=8,y=-9$$

C
$$\displaystyle x=0,y=-3$$

D
$$\displaystyle x=2,y=-1$$

Solution

In equation $$2x-3y=7$$
Let $$y$$ =$$1$$
then $$x$$ will be $$5$$
The one point is $$(5,1)$$
Taking $$x=-1$$ we get $$y=-3$$
The other point is $$(-1,-3)$$
Join both the points by drawing line.
In equation $$5x+y=9$$
Let $$y$$ =$$-1$$
then $$x$$ will be $$2$$
The one point is $$(2,-1)$$
Taking $$x=-1$$ we get $$y=14$$
The other point is $$(-1,14)$$
Join both the points by drawing line.
Now the point of intersection of both lines is $$(2,-1)$$
Question 5
1 / -0

Solve the following simultaneous equations:
$$8x+13y-29= 0$$, $$12x-7y-17= 0$$

A
$$x=3;\, y=2$$

B
$$x=0;\, y=5$$

C
$$x=2;\, y=1$$

D
$$x=-1;\, y=3$$

Solution

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $$8x+13y=29$$ and write it as

$$x=\frac { 29-13y }{ 8 }.........(1)$$

Substitute the value of $$x$$ in Equation $$12x-7y=17$$. We get

$$12\left( \frac { 29-13y }{ 8 } \right) -7y=17\\ \Rightarrow 87-39y-14y=34\\ \Rightarrow 87-53y=34\\ \Rightarrow -53y=-53\\ \Rightarrow y=1$$

Therefore, $$y = 1$$

Substituting this value of $$y$$ in Equation (1), we get

$$x=\frac { 29-13 }{ 8 } =\frac { 16 }{ 8 } =2$$

Hence, the solution is $$x = 2, y = 1$$.
Question 6
1 / -0

Solve the following simultaneous equation :
$$ax\, +\, by\, =\, 5$$ and $$\, bx\, +\, ay\, =\, 3,$$ where $$a$$ and $$b$$ are constants.

A
$$\displaystyle x\, =\, \frac{5a\, -\, 3b}{a^2\, +\, b^2}, \quad\, y\, =\, \frac{3a\, -\, 5b}{a^2\, +\, b^2}$$

B
$$\displaystyle x\, =\, \frac{5a\, -\, 3b}{a^2\, -\, b^2}, \quad\, y\, =\, \frac{3a\, -\, 5b}{a^2\, -\, b^2}$$

C
$$\displaystyle x\, =\, \frac{3a\, -\, 5b}{a^2\, -\, b^2}, \quad\, y\, =\, \frac{5a\, -\, 5b}{a^2\, -\, b^2}$$

D
$$\displaystyle x\, =\, \frac{3a\, -\, 5b}{a^2\, +\, b^2}, \quad\, y\, =\, \frac{5a\, -\, 5b}{a^2\, +\, b^2}$$

Solution

$$ax + by = 5$$ ...(i)
$$bx + ay = 3$$ ...(ii)
Multiply (i) by $$a$$ and (ii) by $$b$$ and subtract
$$a^2x + aby - b^2x - aby = 5a - 3b$$
$$(a^2 - b^2) x = 5a - 3b$$

$$x = \dfrac{5a - 3b}{a^2 - b^2}$$
Similarly,
Multiply (i) by $$b$$ and (ii) by $$a$$ and subtract
$$abx + b^2y - abx - a^2y = 5b - 3a$$
$$(b^2 - a^2)y = 5b - 3a$$

$$y = \dfrac{5b - 3a}{b^2 - a^2}$$
Question 7
1 / -0

A triangle is formed by the straight lines: $$\displaystyle x + 2y - 3 = 0, 3x - 2y + 7 = 0$$ and $$\displaystyle y + 1 = 0$$.
Find graphically, the co-ordinates of the vertices of the triangle.

A
$$(-1,2), (5,-1), (-3,-1)$$

B
$$(-1,2), (7,4), (-3,-1)$$

C
$$(7,-4), (5,-1), (-3,-1)$$

D
$$(-12,5), (5,-1), (-3,-1)$$

Solution

Plotting the equation $$x+2y=0$$,$$3x-2y+7=0$$ and$$ y+1=0.$$ in the graph we get,the vertices of the triangle are $$(-1,2),(5,-1)$$ and $$ (-3,-1)$$
Question 8
1 / -0

Solve the following simultaneous equations :
$$3x-5y+1= 0$$, $$2x-y+3= 0$$

A
$$x= 2;\, y= -1$$

B
$$x= 7;\, y= -9$$

C
$$x= -1;\, y= -4$$

D
$$x= -2;\, y= -1$$

Solution

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $$2x-y+3=0$$ and write it as

$$y=3+2x...........(1)$$

Substitute the value of $$y$$ in Equation $$3x-5y+1=0$$ or $$3x-5y=-1$$. We get

$$3x-5(3+2x)=-1$$

i.e. $$3x-15-10x=-1$$

i.e. $$-7x=14$$

i.e. $$x=-2$$

Therefore, $$x=-2$$

Substituting this value of $$x$$ in the equation $$2x-y+3=0$$, we get

$$-4-y+3=0$$

i.e. $$-y-1=0$$

i.e. $$y=-1$$

Hence, the solution is $$x = -2, y = -1$$.
Question 9
1 / -0

Solve the following simultaneous equations:
$$3x+2y= 14$$, $$-x+4y= 7$$

A
$$x= 0;\, y= -4$$

B
$$x= 3;\, y= 2.5$$

C
$$x= 3.5;\, y= -2$$

D
$$x= 4.5;\, y= 3$$

Solution

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $$-x+4y=7$$ and write it as

$$x = 4y-7...........(1)$$

Substitute the value of $$x$$ in Equation $$3x+2y=14$$. We get

$$3(4y-7)+2y=14$$

i.e. $$12y-21+2y=14$$

i.e. $$14y=35$$

i.e. $$2y = 5$$

Therefore, $$y=\frac { 5 }{ 2 }=2.5$$

Substituting this value of $$y$$ in Equation (1), we get

$$x = 10-7=3$$

Hence, the solution is $$x = 3, y = 2.5$$.
Question 10
1 / -0

Solvw the following pair of linear equations:
$$2x\, +\, 5y\, =\, 13$$ and $$4x\, -\, 9y\, =7$$

A
$$(1,4)$$

B
$$(2,-3)$$

C
$$(-2,3)$$

D
$$(4,1)$$

Solution

$$2x+5y=13$$ ......$$(1)$$
$$4x-9y=7$$ ....$$(2)$$
Multiplying equation $$(1)$$ by $$4$$. we get,
$$8x+20y=52$$ ....(3)
Now, Multiplying equation $$(2)$$ by $$2$$. we get,
$$8x-18y=14$$ ......$$(4)$$
On subtracting $$(4)$$ from $$(3)$$. we get,
$$38y=38\\ \Rightarrow y=1$$
On substitute the value of $$y=1$$ in $$(1)$$. we get,
$$2x+5(1)=13 \\\Rightarrow 2x=8 \\\Rightarrow x=4$$
Thus solution of given equations is $$(4,1)$$
Option D is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 32

Result

Pair of Linear Equations in Two Variables Test - 32

Score

-

Rank

-

SHARING IS CARING

Question 1

20 days

22 days

29 days

23 days

Solution

Question 2

$$2$$

$$7$$

$$4$$

$$11$$

Solution

Question 3

$$x=1,\, y=-3$$

$$x=9,\, y=-3$$

$$x=5,\, y=-3$$

$$x=3,\, y=-3$$

Solution

Question 4

$$\displaystyle x=2,y=-6$$

$$\displaystyle x=8,y=-9$$

$$\displaystyle x=0,y=-3$$

$$\displaystyle x=2,y=-1$$

Solution

Question 5

$$x=3;\, y=2$$

$$x=0;\, y=5$$

$$x=2;\, y=1$$

$$x=-1;\, y=3$$

Solution

Question 6

$$\displaystyle x\, =\, \frac{5a\, -\, 3b}{a^2\, +\, b^2}, \quad\, y\, =\, \frac{3a\, -\, 5b}{a^2\, +\, b^2}$$

$$\displaystyle x\, =\, \frac{5a\, -\, 3b}{a^2\, -\, b^2}, \quad\, y\, =\, \frac{3a\, -\, 5b}{a^2\, -\, b^2}$$

$$\displaystyle x\, =\, \frac{3a\, -\, 5b}{a^2\, -\, b^2}, \quad\, y\, =\, \frac{5a\, -\, 5b}{a^2\, -\, b^2}$$

$$\displaystyle x\, =\, \frac{3a\, -\, 5b}{a^2\, +\, b^2}, \quad\, y\, =\, \frac{5a\, -\, 5b}{a^2\, +\, b^2}$$

Solution

Question 7

$$(-1,2), (5,-1), (-3,-1)$$

$$(-1,2), (7,4), (-3,-1)$$

$$(7,-4), (5,-1), (-3,-1)$$

$$(-12,5), (5,-1), (-3,-1)$$

Solution

Question 8

$$x= 2;\, y= -1$$

$$x= 7;\, y= -9$$

$$x= -1;\, y= -4$$

$$x= -2;\, y= -1$$

Solution

Question 9

$$x= 0;\, y= -4$$

$$x= 3;\, y= 2.5$$

$$x= 3.5;\, y= -2$$

$$x= 4.5;\, y= 3$$

Solution

Question 10

$$(1,4)$$

$$(2,-3)$$

$$(-2,3)$$

$$(4,1)$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.