Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Solve the following equation: $$\cfrac{3y+4}{2-6y}=\cfrac{-2}{5}$$

A
$$2$$

B
$$-4$$

C
$$4$$

D
$$-8$$

Solution

Given, $$\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}$$
On cross multiplying, we get
$$5(3y+4)=-2(2-6y)$$
$$\Rightarrow 15y+20=-4+12y$$
$$\Rightarrow 3y=-24$$
$$\Rightarrow y=-8$$
Question 2
1 / -0

Solve the following linear equation. If $$\cfrac{3t-2}{4}-\cfrac{2t+3}{3} = \cfrac{2}{3}-t$$, then $$t $$ is equal to

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given $$\cfrac { 3t-2 }{ 4 } -\cfrac { (2t+3) }{ 3 } = \cfrac { 2 }{ 3 } - t$$
L.C.M. of $$4,3,3$$ is $$ 12$$
Multiplying with 12 on both sides of the equation, we get
$$\Rightarrow 3(3t-2) - 4(2t+3) = 8 - 12t$$
$$\Rightarrow 9t - 6 -(8t +12) = 8 -12t$$
$$\Rightarrow t -18 = 8 -12t$$
$$\Rightarrow 13 t = 26$$
$$\Rightarrow t = 2$$
Question 3
1 / -0

Solve the following equation: $$\cfrac{9x}{7-6x}=15$$

A
$$x = \cfrac{25}{72}$$

B
$$x = \cfrac{35}{33}$$

C
$$x = \cfrac{45}{39}$$

D
$$x = \cfrac{22}{45}$$

Solution

Given, $$\dfrac{9x}{7-6x}=15$$
$$9x=105-90x$$
Add $$90x$$ on both the sides, we get
$$9x+90x=105-90x+90x $$
$$99x=105$$
$$\therefore x=\dfrac{105}{99}=\dfrac{35}{33}$$
Question 4
1 / -0

Linear equation in one variable is :

A
$$2x = y$$

B
$$y^2 = 3y + 5$$

C
$$4x - y = 5$$

D
$$3t + 5 = 9t- 7$$

Solution

A linear equation with one variable is an equation with only one variable and the highest power of the variable is $$1$$.

In first option, there are $$2$$ variables, $$x$$ and $$y$$.

In the second option, there is only a $$1$$ variable but the highest power is $$2$$ hence, it is a quadratic equation.

In the third option, there are $$2$$ variables, $$x$$ and $$y$$.

In the fourth option, there is only $$1$$ variable and the highest power is $$1$$ as well.

Hence, option $$D$$ is correct,
Question 5
1 / -0

Three consecutive integers add up to $$51$$. What are these integers?

A
$$15, 16, 17$$

B
$$16, 17, 18$$

C
$$10, 11, 12$$

D
$$18, 19, 20$$

Solution

Let the three consecutive numbers be $$x, x+1, x+2$$
Sum of the numbers is $$51$$.
$$\therefore x + x+ 1+ x+2 = 51$$
$$3x + 3 = 51$$
$$3x = 48$$
$$x = 16$$
The numbers would be $$x=16$$,
$$x+1=16+1=17$$,
$$x+2=16+2=18$$
So, the numbers are $$16, 17, 18$$.
Question 6
1 / -0

Which of the following is the solution of the equation $$\displaystyle \frac{7y+4}{y+2}=\frac{-4}{3}$$ ?

A
$$\displaystyle y = -\frac{4}{5}$$

B
$$\displaystyle y = \frac{4}{5}$$

C
$$\displaystyle y = -\frac{5}{4}$$

D
$$\displaystyle y = \frac{5}{4}$$

Solution

$$\dfrac{7y+4}{y+2}=\dfrac{-4}{3}$$
$$21y+12=-4y-8$$
$$25y=-20$$
$$y=\dfrac{-20}{25} = \dfrac{-4}{5}$$
Question 7
1 / -0

Solve the following equation: $$\cfrac{8x-3}{3x}=2$$

A
$$x = \cfrac{5}{2}$$

B
$$x = \cfrac{5}{6}$$

C
$$x = \cfrac{9}{8}$$

D
$$x = \cfrac{3}{2}$$

Solution

Given, $$\dfrac{8x-3}{3x}=2$$
Multiply $$3x$$ on both the sides, we get
$$8x-3=6x$$
$$8x-6x=3$$
$$2x=3$$
$$\therefore x=\dfrac{3}{2}$$
Question 8
1 / -0

Solve the following linear equation: $$m-\cfrac{m-1}{2} = 1-\cfrac{m-2}{3}$$

A
$$m = \cfrac{7}{5}$$

B
$$m = \cfrac{2}{3}$$

C
$$m = \cfrac{10}{3}$$

D
$$m = \cfrac{3}{8}$$

Solution

Given, $$m - \cfrac { (m-1) }{ 2 } = 1-\cfrac { (m-2) }{ 3 }$$
L.C.M. of $$2$$ and $$3$$ is $$6$$
Multiplying with 6 on both sides of the equation, we get
$$6m -3(m-1) = 6 - 2(m-2)$$
$$ 6m - 3m +3 = 6 - 2m +4$$
$$5m = 7$$
$$ m = \cfrac { 7 }{ 5 }$$
Question 9
1 / -0

Solve the following equation: $$\cfrac{z}{z+15}=\cfrac{4}{9}$$

A
$$2$$

B
$$9$$

C
$$12$$

D
$$16$$

Solution

Given, $$\dfrac{z}{z+15}=\dfrac{4}{9}$$
$$9z=4(z+15)$$
$$9z=4z+60$$
Subtract $$4z$$ from both the sides, we get
$$9z-4z=4z+60-4z$$
$$5z=60$$
$$\therefore z=12$$
Question 10
1 / -0

Two numbers are in the ratio of $$6:3$$. If they differ by $$18$$, what are the numbers?

A
$$17, 26$$

B
$$42, 76$$

C
$$15, 13$$

D
$$36, 18$$

Solution

Let the two numbers be $$6x$$ and $$3x$$.

They are in the ratio of $$6:3$$.

By the given condition, we have

$$6x -3x =18$$
$$\therefore 3x = 18$$

$$\therefore x =6$$

Therefore, the numbers would be $$6x=6\times 6=36$$ and $$3x=3\times 6=18$$.

So, the numbers are $$36$$ and $$18$$.

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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 15

Result

Linear Equations in One Variable Test - 15

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Question 1

$$2$$

$$-4$$

$$4$$

$$-8$$

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

$$x = \cfrac{25}{72}$$

$$x = \cfrac{35}{33}$$

$$x = \cfrac{45}{39}$$

$$x = \cfrac{22}{45}$$

Solution

Question 4

$$2x = y$$

$$y^2 = 3y + 5$$

$$4x - y = 5$$

$$3t + 5 = 9t- 7$$

Solution

Question 5

$$15, 16, 17$$

$$16, 17, 18$$

$$10, 11, 12$$

$$18, 19, 20$$

Solution

Question 6

$$\displaystyle y = -\frac{4}{5}$$

$$\displaystyle y = \frac{4}{5}$$

$$\displaystyle y = -\frac{5}{4}$$

$$\displaystyle y = \frac{5}{4}$$

Solution

Question 7

$$x = \cfrac{5}{2}$$

$$x = \cfrac{5}{6}$$

$$x = \cfrac{9}{8}$$

$$x = \cfrac{3}{2}$$

Solution

Question 8

$$m = \cfrac{7}{5}$$

$$m = \cfrac{2}{3}$$

$$m = \cfrac{10}{3}$$

$$m = \cfrac{3}{8}$$

Solution

Question 9

$$2$$

$$9$$

$$12$$

$$16$$

Solution

Question 10

$$17, 26$$

$$42, 76$$

$$15, 13$$

$$36, 18$$

Solution

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