Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Reduce the following linear equation: $$6t - 1 = t - 11$$

A
$$t=-1$$

B
$$t=-2$$

C
$$t=-3$$

D
$$t=-4$$

Solution

$$6t - 1 = t - 11$$

$$6t - t = -11 + 1$$
$$5t = -10$$
$$t = -2$$
Question 2
1 / -0

Solve the linear equation: $$5x - 12 = 2x + 18$$

A
$$x=8$$

B
$$x=9$$

C
$$x=10$$

D
$$x=11$$

Solution

$$5x - 12 = 2x + 18$$

$$5x - 2x = 18 + 12$$
$$3x = 30$$
$$x = 10$$
Question 3
1 / -0

Reduce the linear equation: $$x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$$

A
$$x=12$$

B
$$x=10$$

C
$$x=8$$

D
$$x=-6$$

Solution

Given, $$x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$$
L.C.M of the denominator $$3$$ and $$6$$ is $$6$$.
Multiplying both the sides by $$6$$, we get
$$6x + 18 - 4x + x = 0$$
$$7x - 4x + 18 = 0$$
$$3x = -18$$
$$x = -6$$
Question 4
1 / -0

Reduce the linear equation: $$\dfrac{x}{2}+\dfrac{2x}{4}= 10$$

A
$$x=6$$

B
$$x=7$$

C
$$x=8$$

D
$$x=10$$

Solution

Given, $$\dfrac{x}{2}+\dfrac{2x}{4}= 10$$
L.C.M of the denominator $$2$$ and $$4$$ is $$4$$.
Multiplying both the sides by $$4$$, we get
$$2x + 2x = 40$$
$$4x = 40$$
$$x = 10$$
Question 5
1 / -0

The sum of one fifth, one third and one ninth of a number is $$29$$. Find the number.

A
$$45$$

B
$$29$$

C
$$54$$

D
None of the above.

Solution

Let the required number be $$ x$$

According to the question, $$\displaystyle \frac { x }{ 5 } +\frac { x }{ 3 } +\frac { x}{ 9 } =29$$

L.C.M of $$5,3,9$$ is $$45.$$

Then the equation becomes :
$$\displaystyle \frac { x \times 9 }{ 5 \times 9 } +\frac { x \times 15 }{ 3 \times 15 } +\frac { x \times 5}{ 9 \times 5 } =29$$

$$\Longrightarrow\displaystyle \frac { 9x+15x+5x }{ 45 } =29$$

$$\Longrightarrow\displaystyle 29x=29\times 45$$

$$\Longrightarrow\displaystyle x=\frac { 29\times 45 }{ 29 } $$

$$\therefore x=45$$
Question 6
1 / -0

Solve linear equation:
$$m - \dfrac {m - 1}{2} = 1 - \dfrac {m - 2}{3}$$.

A
$$m = \dfrac {4}{5}$$

B
$$m = \dfrac {3}{5}$$

C
$$m = \dfrac {8}{5}$$

D
$$m = \dfrac {7}{5}$$

Solution
Question 7
1 / -0

Solve for $$x$$:
$$x+2\left (5x-\dfrac { 5 }{ 2 }\right )=4(x+1)-2$$.

A
$$-2$$

B
$$ 2$$

C
$$1$$

D
$$0$$

Solution

Given, $$x+2\left (5x-\dfrac{5}{2}\right)=4(x+1)-2$$

$$\Rightarrow x+10x-5=4x+4-2$$ ...[By Distribution Law]

$$\Rightarrow 11x-5=4x+2$$.

Transposing $$x$$ terms to one side, we get,
$$\Rightarrow 11x-4x=2+5$$

$$\Rightarrow 7x=7$$

$$\Rightarrow x=\dfrac{7}{7}=1$$.

Hence, option $$C$$ is correct.
Question 8
1 / -0

Ram is $$5$$ times as old as Shyam. If their difference of age is $$8$$ years, how old is Ram?

A
$$8$$ years

B
$$10$$ years

C
$$12$$ years

D
$$5$$ years

E
None of these

Solution

Let the age of Shyam be $$ x$$ so, age of Ram is $$ 5x$$.

As per the problem, the difference of their ages is $$8$$ years.
$$5x-x = 8$$
$$\Rightarrow 4x = 8$$
$$\Rightarrow x = 2$$

So, Rams age $$= 5x$$
$$ = 5\times 2 $$
$$= 10$$ years.
Question 9
1 / -0

There are $$42$$ integers in a group. If there are $$5$$ times as many odd integers as there are even integers, how many numbers are even integers?

A
$$7$$

B
6

C
5

D
9

Solution

Let numbers of even integers be $$x$$
Then numbers of odd integers are $$5x$$.
$$\therefore x+5x=42$$
$$ 6x=42$$
$$\Rightarrow x=\dfrac{42}{6}$$
$$=7$$
Question 10
1 / -0

Solve for $$t$$:
$$\displaystyle\frac{t}{2}+4=\displaystyle\frac{3}{4}t-5$$.

A
$$4$$

B
$$9$$

C
$$18$$

D
$$36$$

Solution

Given, $$ \dfrac{1}{2}t+4=\dfrac{3}{4}t-5$$.

Transposing $$t$$ terms to one side, we get,
$$\Rightarrow \dfrac{1}{2}t-\dfrac{3}{4}t=-5-4$$

$$\Rightarrow \dfrac{2t-3t}{4}=-9$$ …[Cross-multiplying the denominators on the LHS]

$$\Rightarrow \dfrac{-t}{4}=-9$$

$$\Rightarrow -t=-36$$ ...[Multiplying $$4$$ on both sides]

$$\Rightarrow t=36$$.

Hence, option $$D$$ is correct.

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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 37

Result

Linear Equations in One Variable Test - 37

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SHARING IS CARING

Question 1

$$t=-1$$

$$t=-2$$

$$t=-3$$

$$t=-4$$

Solution

Question 2

$$x=8$$

$$x=9$$

$$x=10$$

$$x=11$$

Solution

Question 3

$$x=12$$

$$x=10$$

$$x=8$$

$$x=-6$$

Solution

Question 4

$$x=6$$

$$x=7$$

$$x=8$$

$$x=10$$

Solution

Question 5

$$45$$

$$29$$

$$54$$

None of the above.

Solution

Question 6

$$m = \dfrac {4}{5}$$

$$m = \dfrac {3}{5}$$

$$m = \dfrac {8}{5}$$

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

Solution

Question 7

$$-2$$

$$ 2$$

$$1$$

$$0$$

Solution

Question 8

$$8$$ years

$$10$$ years

$$12$$ years

$$5$$ years

None of these

Solution

Question 9

$$7$$

6

5

9

Solution

Question 10

$$4$$

$$9$$

$$18$$

$$36$$

Solution

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