Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 43

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Weekly Quiz Competition

Question 1
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Dhruv earned some money. He spent $$\cfrac { 1 }{ 3 } $$ of the money on magazines and $$\cfrac { 1 }{ 4 } $$ of the money on a snack. Which of the following fractions represents the part of money he did not spend?

A
$$\cfrac { 5 }{ 12 } $$

B
$$\cfrac { 1 }{ 2 } $$

C
$$\cfrac { 2 }{ 3 } $$

D
$$\cfrac { 5 }{ 7 } $$

Solution

Let the money earned by Dhruv be $$x$$.
Then, money on magazines is $$\dfrac{1}{3}x$$, and money spent on snack is $$\dfrac{1}{4}x$$.
So, money not spent is clearly, $$x-\dfrac{x}{3}-\dfrac{x}{4}=\dfrac{5x}{12}$$.
Hence, option A is correct.
Question 2
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Half of a number is added to $$18$$, then the sum is $$46$$. The number is ______ .

A
$$92$$

B
$$56$$

C
$$65$$

D
$$0$$

Solution

$$\Rightarrow$$ Let the required number be $$x$$.
$$\therefore$$ According to given condition,
$$\Rightarrow$$ $$\dfrac{x}{2}+18=46$$

$$\Rightarrow$$ $$\dfrac{x}{2}=46-18$$

$$\Rightarrow$$ $$\dfrac{x}{2}=28$$

$$\Rightarrow$$ $$x=28\times 2$$

$$\therefore$$ $$x=56.$$

$$\therefore$$ The Required number is $$56$$
Question 3
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The total cost of $$3$$ tables and $$2$$ chairs is Rs. $$745$$. If one table costs Rs. $$40$$ more than that of one chair, find the price of each table and chair respectively.

A
Rs. $$160$$, Rs.$$120$$

B
Rs. $$165$$, Rs.$$125$$

C
Rs.$$200$$, Rs.$$160$$

D
RS.$$120$$, RS.$$80$$

Solution

Let cost of a chair be $$x$$, then cost of table will be $$40+x$$.
It is given that,
$$3(40+x)+2x=745$$
$$\Rightarrow 120+5x=745$$
$$\Rightarrow x=\dfrac{745-120}{5}=\dfrac{625}{5}=125$$ $$=$$ cost of chair,
Hence, cost of table $$= 40+x=40+125=165$$.
Hence, B is the right option.
Question 4
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If $$6(2a-1)+8=14$$, the value of '$$a$$' is ______ .

A
$$-1$$

B
$$3\cfrac { 1 }{ 12 } $$

C
$$\cfrac{4}{10}$$

D
$$1$$

Solution

Give equation is $$6(2a-1)+8=14$$
To find out: The value of $$a$$

Solving the given equation, we get:
$$6(2a-1)+8=14$$
$$\Rightarrow$$ $$12a-6+8=14$$
$$\Rightarrow$$ $$12a+2=14$$
$$\Rightarrow$$ $$12a=14-2$$
$$\Rightarrow$$ $$12a=12$$
$$\Rightarrow$$ $$a=\dfrac{12}{12}$$

$$\therefore$$ $$a=1.$$

Hence, the value of $$a$$ is $$1$$.
Question 5
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Value of $$x$$ in $$\cfrac { x }{ 4 } +\cfrac { 1 }{ 2 } =4$$ is ______ .

A
$$28$$

B
$$-28$$

C
$$14$$

D
$$-14$$

Solution

$$\Rightarrow$$ $$\dfrac{x}{4}+\dfrac{1}{2}=4$$

$$\Rightarrow$$ $$\dfrac{x}{4}+\dfrac{1\times 2}{2\times 2}=4$$

$$\Rightarrow$$ $$\dfrac{x}{4}+\dfrac{2}{4}=4$$

$$\Rightarrow$$ $$\dfrac{x+2}{4}=4$$

$$\Rightarrow$$ $$x+2=16$$
$$\therefore$$ $$x=14$$
Question 6
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The perimeter of a triangle is $$45\ cm$$. Length of the second side is twice the length of first side. The third side is $$5$$ more than the first side. Find the length of each sides.

A
$$11,19,15$$

B
$$10,20,15$$

C
$$10,16,19$$

D
$$13,15,17$$

Solution

Let the length of first side $$=x$$
Length of second side $$=2x$$
Length of third side $$=x+5$$
Perimeter $$=45cm$$
$$x+2x+x+5=45\\ 4x=45-5\\ 4x=40\\ x=10$$
So the sides are
$$x=10$$
$$2x=2\times 10=20\\ x+5=10+5=15$$
Option $$B$$ is correct.
Question 7
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Form an equation of the form $$ax + b = c$$, where $$a$$, $$b$$ and $$c$$ are constants, such that the solution of the equation is $$x = 4$$.

A
2x + 5 = 15

B
7x + 2 = 10

C
5x + 4 = 16

D
3x + 4 = 16

Solution

$$(A)\,\,\, 2x + 5 = 15$$
$$\Rightarrow$$ $$2x = 10$$
$$\Rightarrow$$ $$ x = 5$$

$$(B)\,\,\, 7x + 2 = 10$$
$$\Rightarrow$$ $$ 7x = 8$$
$$\Rightarrow$$ $$ x = \dfrac{8}{7}$$

$$(C)\,\,\, 5x + 4 = 16$$
$$\Rightarrow$$ $$ 5x = 12$$
$$\Rightarrow$$ $$ x = \dfrac{12}{5}$$

$$(D)\,\,\, 3x + 4 = 16$$
$$\Rightarrow$$ $$ 3x = 12$$
$$\Rightarrow$$ $$ x = 4$$

$$\therefore$$ In option $$D$$ we get, $$x$$ as $$4$$ so,correct answer is option $$D$$
Question 8
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If $$\dfrac {9}{5}$$ of a number is 45, what is $$\dfrac {1}{5}$$ of the same number?

A
$$5$$

B
$$25$$

C
$$30$$

D
$$81$$

Solution

Let the number be $$x$$.
$$\Rightarrow$$ Then, according to the given condition,
$$\Rightarrow$$ $$\dfrac{9}{5}\,\,of\,\,x=45$$

$$\Rightarrow$$ $$\dfrac{9}{5}\times x=45$$
$$\Rightarrow$$ $$x=45\times \dfrac{5}{9}$$
$$\Rightarrow$$ $$x=25$$
$$\therefore$$ Number is $$25$$.
$$\therefore$$ $$\dfrac{1}{5}$$ of the number = $$\dfrac{1}{5}\times 25=5$$
Question 9
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If $$\dfrac {2}{5}(5x+1)+\dfrac {3}{5} = 1$$, then what is the value of 5 x?

A
$$\dfrac {-1}{5}$$

B
$$1$$

C
$$0$$

D
$$\dfrac {1}{5}$$

Solution

$$\dfrac{2}{5}(5x+1)+\dfrac{3}{5}=1$$

$$\Rightarrow$$ $$2x+\dfrac{2}{5}+\dfrac{3}{5}=1$$

$$\Rightarrow$$ $$2x+\dfrac{5}{5}=1$$

$$\Rightarrow$$ $$2x+1=1$$

$$\Rightarrow$$ $$2x=0$$

$$\therefore$$ $$x=0$$

$$\therefore$$ $$x=0$$ then, $$5x=0$$
Question 10
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The denominator of a rational number is greater than its numerator by 3. If 3 is subtracted from the numerator and 2 is added to the denominator, the new number becomes $$\displaystyle \frac{1}{5}$$. Then the original fraction was _________.

A
$$\displaystyle \frac{7}{11}$$

B
$$\displaystyle \frac{3}{5}$$

C
$$\displaystyle \frac{5}{8}$$

D
$$\displaystyle \frac{4}{7}$$

Solution

Let, the numerator of rational fraction be $$x$$
$$\therefore$$ the denomiantor $$=x+3$$
A.T.Q,
$$\dfrac{x-3}{x+3+2}=\dfrac{1}{5}$$
$$\implies \dfrac{x-3}{x+5}=\dfrac{1}{5}$$
$$\implies 5x-15=x+5$$
$$\implies 5x-x=5+15$$
$$\implies 4x=20$$
$$\implies x=\dfrac{20}{4}=5$$
$$\therefore$$ The original rational fraction is $$\dfrac{5}{5+3}=\dfrac{5}{8}$$

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Linear Equations in One Variable Test - 43

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Linear Equations in One Variable Test - 43

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Question 1

$$\cfrac { 5 }{ 12 } $$

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { 2 }{ 3 } $$

$$\cfrac { 5 }{ 7 } $$

Solution

Question 2

$$92$$

$$56$$

$$65$$

$$0$$

Solution

Question 3

Rs. $$160$$, Rs.$$120$$

Rs. $$165$$, Rs.$$125$$

Rs.$$200$$, Rs.$$160$$

RS.$$120$$, RS.$$80$$

Solution

Question 4

$$-1$$

$$3\cfrac { 1 }{ 12 } $$

$$\cfrac{4}{10}$$

$$1$$

Solution

Question 5

$$28$$

$$-28$$

$$14$$

$$-14$$

Solution

Question 6

$$11,19,15$$

$$10,20,15$$

$$10,16,19$$

$$13,15,17$$

Solution

Question 7

2x + 5 = 15

7x + 2 = 10

5x + 4 = 16

3x + 4 = 16

Solution

Question 8

$$5$$

$$25$$

$$30$$

$$81$$

Solution

Question 9

$$\dfrac {-1}{5}$$

$$1$$

$$0$$

$$\dfrac {1}{5}$$

Solution

Question 10

$$\displaystyle \frac{7}{11}$$

$$\displaystyle \frac{3}{5}$$

$$\displaystyle \frac{5}{8}$$

$$\displaystyle \frac{4}{7}$$

Solution

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