Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

If $$13$$ is added to one-half of a certain number, the result is $$37$$. The original number is equal to

A
$$24$$

B
$$40$$

C
$$48$$

D
$$61$$

Solution

Let the original number is $$x$$
Then according to the question, we have
$$\dfrac{x}{2}+13=37$$
$$\Rightarrow \dfrac{x+26}{2}=37$$
$$\Rightarrow x+26=74$$
$$\Rightarrow x=74-26$$
$$\Rightarrow x=48$$
Therefore, the original number is $$48$$.
Question 2
1 / -0

If $$16+4x$$ is $$10$$ more than $$14$$, what is the value of $$8x$$?

A
$$2$$

B
$$6$$

C
$$16$$

D
$$80$$

Solution

According to question, we have
$$16+4x=14+10$$
$$\Rightarrow 4x=8$$
$$\Rightarrow x=2$$
So,
$$8x=8\times2=16$$
Therefore, option C is correct.
Question 3
1 / -0

The monthly membership fee for an online television and movie service is $ $$9.80$$. The cost of viewing television shows online is included in the membership fee, but there is an additional fee of $ $$1.50$$ to rent each movie online. For one month, Jill's membership and movie rental fees were $ $$12.80$$ . How many movies did Jill rent online that month?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Equation for monthly bill when $$ m $$ number of movies are rented per month $$ = 9.80 + 1.5m $$

Given, $$ 9.80 + 1.5m = 12.80 $$
$$ => 1.5m = 3 $$
$$ => m = 2 $$

Thus, Jill rented two movies that month
Question 4
1 / -0

Last week Raul worked $$11$$ more hours than Angelica. If they worked a combined total of $$59$$ hours, how many hours did Angelica work last week?

A
$$24$$

B
$$35$$

C
$$40$$

D
$$48$$

Solution

If Angelica worked $$ a $$ hours, then Raul worked $$ a + 11 $$ hours.
Total hours worked $$ = a + a + 11 = 59 $$
$$ \Rightarrow 2a = 48 $$
$$ \Rightarrow a = 24 $$ hours
Question 5
1 / -0

The ratio of adults to girls to boys on a class field trip was $$1:4:5$$. If the trip included $$6$$ more boys than girls, how many adults were with the group?

A
$$3$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

Given, the ratio of adults to girls to boys on a class field trip is $$1:4:5$$
Let $$x$$ be the common variable
Let number of adults be $$x$$, number of girls be $$4x$$ and number of boys be $$5x$$.
Since the trip included $$6$$ more boys than girls,
$$5x-4x=6$$
$$x=6$$
Therefore $$6$$ adults were with the group.
Question 6
1 / -0

If $$\dfrac{5}{x+3} = \dfrac{1}{x}+\dfrac{1}{2x}$$, calculate the value of $$x$$.

A
$$\dfrac{3}{14}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{6}{13}$$

D
$$\dfrac{3}{4}$$

E
$$\dfrac{9}{7}$$

Solution

Given, $$\dfrac { 5 }{ x+3 } =\dfrac { 1 }{ x } +\dfrac { 1 }{ 2x } $$
Taking RHS:
$$\dfrac { 1 }{ x } +\dfrac { 1 }{ 2x } $$
LCM of these is $$2x$$
$$\Rightarrow \dfrac { 2 }{ 2x } +\dfrac { 1 }{ 2x } \\ \Rightarrow \dfrac { 3 }{ 2x } $$
Now taking LHS:
$$\dfrac { 5 }{ x+3 } $$
It is given, LHS $$=$$ RHS
$$\dfrac { 5 }{ x+3 } =\dfrac { 3 }{ 2x } $$
$$\Rightarrow 5\times 2x=3\times (x+3)\\ \Rightarrow 10x=3x+9\\ \Rightarrow x=\dfrac {9}{7}$$
Question 7
1 / -0

A food truck sells salads for $ $$6.50$$ each and drinks for $ $$2.00$$ each. The food truck's revenue from selling a total of $$209$$ salads and drinks in one day was $ $$836.50$$. How many salads were sold that day?

A
$$77$$

B
$$93$$

C
$$99$$

D
$$105$$

Solution

Let $$ s $$ be the number of salads sold.
Then number of drinks sold $$ = 209 - s $$

Since the earning was $$ 836.50 $$, we have
$$ 6.5s + 2(209-s) = 836.5 $$
$$ \Rightarrow 6.5s + 418 -2s = 836.5 $$
$$ \Rightarrow 4.5s = 418.5 $$
$$ \Rightarrow s = 93 $$

Thus, $$ 93 $$ salads were sold.
Question 8
1 / -0

Solve the equation: $$\dfrac{12x+3}{2} = \dfrac{x+1}{4}$$.

A
$$-10$$

B
$$15$$

C
$$\dfrac {-5}{23}$$

D
$$\dfrac {37}{12}$$

Solution

Given, $$\dfrac{12x+3}{2} = \dfrac{x+1}{4}$$

$$\Rightarrow 4(12x+3)=2(x+1)$$...[Multiply both sides by $$4$$]

$$\Rightarrow 48x+12=2x+2$$ ...[By Distribution Law]

Transposing $$x$$ terms to one side, we get,
$$\Rightarrow 48x-2x=2-12$$

$$\Rightarrow 46x=-10$$

$$\Rightarrow 23x=-5$$

$$\therefore x=-\dfrac{5}{23}$$.

Hence, option $$C$$ is correct.
Question 9
1 / -0

Solve the equation:
$$-(x-5)+5(x-9)=2(x+8)-(2x+5)$$.

A
$$x=5$$

B
no solutions

C
$$x=\dfrac { 51 }{ 4 } $$

D
$$x=-8$$

Solution

Given, $$-\left( x-5 \right) +5\left( x-9 \right) =2\left( x+8 \right) -\left( 2x+5 \right) $$

$$\implies$$ $$-x+5+5x-45=2x+16-2x-5$$ ...[By Distribution Law]

$$\implies$$ $$4x-40=11$$ ....[Simplifying]

$$\implies$$ $$4x=11+40$$ ...[Adding $$40$$ on both sides]

$$\implies$$ $$4x=51$$

$$\implies$$ $$x=\cfrac { 51 }{ 4 } $$.

Hence, option $$C$$ is correct.
Question 10
1 / -0

If $$\dfrac{19}{5x+17} = \dfrac{19}{31}$$, then find $$x $$.

A
$$0.4$$

B
$$1.4$$

C
$$2.8$$

D
$$3.4$$

Solution

Given $$\dfrac{19}{5x+17} = \dfrac{19}{31}$$
$$\Rightarrow 19\times 31=19(5x+17)\\ \Rightarrow 31=5x+17\\ \Rightarrow 5x=14\\ \Rightarrow x=\dfrac {14}{5}=2.8$$

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Linear Equations in One Variable Test - 39

Result

Linear Equations in One Variable Test - 39

Score

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Rank

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SHARING IS CARING

Question 1

$$24$$

$$40$$

$$48$$

$$61$$

Solution

Question 2

$$2$$

$$6$$

$$16$$

$$80$$

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

$$24$$

$$35$$

$$40$$

$$48$$

Solution

Question 5

$$3$$

$$4$$

$$6$$

$$8$$

Solution

Question 6

$$\dfrac{3}{14}$$

$$\dfrac{1}{3}$$

$$\dfrac{6}{13}$$

$$\dfrac{3}{4}$$

$$\dfrac{9}{7}$$

Solution

Question 7

$$77$$

$$93$$

$$99$$

$$105$$

Solution

Question 8

$$-10$$

$$15$$

$$\dfrac {-5}{23}$$

$$\dfrac {37}{12}$$

Solution

Question 9

$$x=5$$

no solutions

$$x=\dfrac { 51 }{ 4 } $$

$$x=-8$$

Solution

Question 10

$$0.4$$

$$1.4$$

$$2.8$$

$$3.4$$

Solution

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