Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If $$p$$ is added to $$-5$$, the result is $$7$$. Hence $$3p=$$________

A
$$6$$

B
$$18$$

C
$$36$$

D
$$4$$

Solution

Given number is $$p$$
Adding $$-5,$$ we get number
$$p-5$$
But given that this is equal to $$7$$
$$\therefore p-5=7$$
Add $$5$$ on both sides, we get
$$p-5+5=7+5$$
$$\Rightarrow p=12$$
$$\therefore 3p=3\times 12=36$$
Question 2
1 / -0

Kishore has Rs.$$p$$. Hari has Rs.$$3$$ more than Kishore and Ashok has Rs.$$7$$ less than Hari. Altogether the three boys have Rs.$$20$$. How much money does Kishore have?

A
Rs.$$30$$

B
Rs.$$10$$

C
Rs.$$7$$

D
Rs.$$3$$

Solution

Given : Kishore have Rs $$p$$
And Hari have Rs $$3$$ more than Kishore
Then, Hari have $$p+3$$ Rs
Ashok have Rs $$7$$ less than Hari
Then, Ashok have $$p+3-7$$ Rs
But three boys all together have Rs $$20$$
$$\therefore p+(p+3)+(p+3-7)=20$$
$$\Rightarrow p+p+3+p+3-7=20$$
$$\Rightarrow p+p+p=20-3-3+7$$
$$\Rightarrow 3p=21$$
$$\Rightarrow p=7$$
Question 3
1 / -0

If one-third of a two digit number exceeds its one-fourth by $$8$$, then what is the sum of the digits of the number?

A
$$6$$

B
$$13$$

C
$$15$$

D
$$17$$

Solution

Let the two digit number be $$xy$$ = $$10x+y$$.
So, accoding to the problem.
$$(\dfrac {1}{3}(10x + y)$$$$-(10x + y)\dfrac {1}{4}) = 8$$
$$(10x + y)\left (\dfrac {1}{3} - \dfrac {1}{4}\right ) = 8$$
$$\Rightarrow 10x + y = 96$$
$$\Rightarrow x = 9, y = 6$$
$$\therefore x + y = 15$$
Question 4
1 / -0

If $$1$$ is subtracted from the numerator of a fraction it becomes $$\dfrac 13$$, and if $$5$$ added to the denominator, the fraction becomes $$\dfrac 14$$. Which fraction shall result if $$1$$ is subtracted from the numerator and $$5$$ is added to the denominator?

A
$$\dfrac 5{12}$$

B
$$\dfrac 7{12}$$

C
$$\dfrac 1{8}$$

D
$$\dfrac 23$$

Solution

Let the fraction be $$\dfrac {x}{y}$$
$$\dfrac {x - 1}{y} = \dfrac {1}{3}$$ and $$\dfrac {x}{y + 5} = \dfrac {1}{4}$$
$$\Rightarrow 3(x-1)=y\; and\; 4x=y+5$$......eqn(1)
$$\Rightarrow 3x-3=y\; and\; 4x=y+5$$.
$$\Rightarrow 3x-3=4x-5$$.
$$\Rightarrow x=-3+5$$.
$$\Rightarrow x = 2$$ and susbtitute in eqn (1) we get $$y = 3$$. So, the fraction is $$\dfrac {2}{3}$$.
New fraction $$= \dfrac {2 - 1}{3 + 5} = \dfrac {1}{8}$$.
Question 5
1 / -0

If $$14$$ is taken away from one - fifth of a number, the result is $$20$$. The equation expressing this statement is:

A
$$\dfrac{x}{5}-14=20$$

B
$$x-\left (\dfrac{14}{5}\right)=\left (\dfrac{20}{5}\right)$$

C
$$x-14=\left (\dfrac{20}{5}\right)$$

D
$$x+4\left (\dfrac{14}{5}\right)=20$$

Solution

Let the number be $$x$$.
As per the question, we can write,
One- fifth of a number, i.e. $$\dfrac {1}{5}x$$.
Now $$14$$ is taken away from this means $$14$$ is subtracted i.e., $$\dfrac {x}{5}-14$$.
The result is equal to $$20$$.
Thus all this can be written as ,
$$\dfrac {x}{5}-14=20$$
Question 6
1 / -0

If the sum of two numbers is $$84$$ and their difference is $$30$$, the numbers are:

A
$$-57$$ and $$27$$

B
$$57$$ and $$27$$

C
$$57$$ and $$-27$$

D
$$-57$$ and $$-27$$

Solution

Given, the sum of two numbers is $$84$$.
Let the smaller number be $$x$$ then, the larger number will be $$84-x$$.

Given, their difference is $$30$$.
$$84-x-x=30$$
$$\Rightarrow 84-2x=30$$
$$\Rightarrow 84-30=2x$$
$$\Rightarrow 54=2x$$
$$\Rightarrow x=27$$

Smaller number is $$27$$.
Now, larger number is,
$$84-x=84-27$$
$$=57$$
Question 7
1 / -0

If five times a number increased by 8 is 83, then the number is:

A
$$12$$

B
$$15$$

C
$$11$$

D
$$2$$

Solution

Given that, five times a number increased by $$8$$ is $$83$$.
To find out: The number.

Let the number be $$x$$.
As per the question, we have
$$5x+8=83$$
$$\Rightarrow 5x=83-8$$
$$\Rightarrow 5x=75$$
$$\Rightarrow x=15$$

Hence, the number is $$15$$.
Question 8
1 / -0

If $$\cfrac{(x + 0.7x)}{2} = 0.85$$, then the value of $$x$$ is:

A
$$2$$

B
$$1$$

C
$$-1$$

D
$$0$$

Solution

Given, $$\dfrac {x+0.7x}{2}=0.85$$

Multiply $$2$$ on both the sides,

$$\Rightarrow x+0.7x=1.7$$

$$\Rightarrow 1.7x=1.7$$

Divide both sides by $$1.7$$, we get

$$\Rightarrow x=1$$
Question 9
1 / -0

The numerator of a fraction is $$6$$ less than the denominator. If $$3$$ is added to the numerator, the fraction is equal to $$\cfrac{2}{3}$$, find the original fraction

A
$$\dfrac 39$$

B
$$\dfrac 69$$

C
$$\dfrac 37$$

D
None of these

Solution

Let denominator $$=x$$

thus numerator $$=x-6$$

By condition $$\dfrac{x-6+3}{x}=\dfrac{2}{3}$$

$$3(x-3)=2x$$

$$3x-9=2x$$

$$x=9$$

So denominator $$=x=9$$

thus numerator $$\Rightarrow x-6=9-6=3$$

Original fraction $$=\dfrac{3}{9}$$
Question 10
1 / -0

Tank $$A$$ contains $$5$$ times as much water as Tank $$B$$. How much water must be transfer from Tank $$A$$ to Tank $$B$$ so that each tank contains $$45$$ liters of water?

A
$$30\ litres$$

B
$$60\ litres$$

C
$$75\ litres$$

D
$$45\ litres$$

Solution

Let quantity of water in tank $$B$$ be $$x$$ litres.
Then quantity of water in the tank $$A = 5x$$ litres
Total water in both the tanks $$= x + 5x$$
$$= 90\ liters$$
$$\Rightarrow 6x = 90 \Rightarrow x = 15$$
$$\therefore$$ Water in tank $$B = 15\ litres$$
$$\Rightarrow$$ Water in tank $$A = 15\times 5 = 75\ litres$$
So, quantity of water to be transferred from tank $$A$$ to tank $$B$$, so that each tank contains $$45\ litres = 75 - 30 = 45\ litres$$
Therefore, $$30\ liters$$ of water should be transferred.

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Linear Equations in One Variable Test - 42

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Linear Equations in One Variable Test - 42

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SHARING IS CARING

Question 1

$$6$$

$$18$$

$$36$$

$$4$$

Solution

Question 2

Rs.$$30$$

Rs.$$10$$

Rs.$$7$$

Rs.$$3$$

Solution

Question 3

$$6$$

$$13$$

$$15$$

$$17$$

Solution

Question 4

$$\dfrac 5{12}$$

$$\dfrac 7{12}$$

$$\dfrac 1{8}$$

$$\dfrac 23$$

Solution

Question 5

$$\dfrac{x}{5}-14=20$$

$$x-\left (\dfrac{14}{5}\right)=\left (\dfrac{20}{5}\right)$$

$$x-14=\left (\dfrac{20}{5}\right)$$

$$x+4\left (\dfrac{14}{5}\right)=20$$

Solution

Question 6

$$-57$$ and $$27$$

$$57$$ and $$27$$

$$57$$ and $$-27$$

$$-57$$ and $$-27$$

Solution

Question 7

$$12$$

$$15$$

$$11$$

$$2$$

Solution

Question 8

$$2$$

$$1$$

$$-1$$

$$0$$

Solution

Question 9

$$\dfrac 39$$

$$\dfrac 69$$

$$\dfrac 37$$

None of these

Solution

Question 10

$$30\ litres$$

$$60\ litres$$

$$75\ litres$$

$$45\ litres$$

Solution

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