Simple Equations Test

Result

Simple Equations Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

The average of 11, 12, 13, 14, and x is 13. The value of x is

A
$$17$$

B
$$21$$

C
$$15$$

D
None

Solution

$$\displaystyle \frac{11+12+13+14+x}{5}=13$$
$$\displaystyle \frac{50+x}{5}=13$$
$$50+x = 65$$
$$x = 65-50$$
$$x = 15$$
Hence, the correct option is $$C$$
Question 2
1 / -0

The average of four consecutive even numbers is $$15$$. The $$2$$nd highest number is

A
$$12$$

B
$$18$$

C
$$14$$

D
$$16$$

Solution

Let the numbers be $$x - 2,x,x + 2,x + 4$$

According to the question,

$$\dfrac{x-2 + x + x + 2 + x + 4}{4} = 15$$

$$4x + 4 = 60$$

$$4x = 56$$ $$\Rightarrow x = 14$$

Hence, the second highest number i.e. $$x + 2 = 14 + 2 = 16$$
Question 3
1 / -0

If the mean of $$x, x+2, x+4, x+6, x+8$$ is $$20$$ then $$x$$ is

A
$$32$$

B
$$16$$

C
$$8$$

D
$$4$$

Solution

mean= $$\dfrac{x+x+2+x+4+x+6+x+8}{5}=20$$
$$5x=80$$
$$x=16$$
Question 4
1 / -0

If the mean of $$2, 4, 6, 8, x, y$$ is $$5$$, then find the value of $$x + y $$?

A
$$6$$

B
$$8$$

C
$$10$$

D
$$12$$

Solution

Mean $$=\cfrac{{Sum \; of\; the\; terms}}{{Total\; number\; of\; terms}}$$

$$\Rightarrow \cfrac{2+4+6+8+x+y}{6}=5$$

$$\cfrac{20+x+y}{6}=5$$

$$20+x+y=5(6)$$

$$x+y=30-20=10$$
Question 5
1 / -0

The weight of Rahul is $$5\ kg$$ more than Sachin .The weight of Samir is $$12\ kg$$ less than double the weight of Sachin .If the total weight of all the three is $$93$$ form the equation

A
$$4x-7=93$$

B
x+5=0

C

$$x+ (x+5)+ (2x-12)=93$$

D
2x-12=21

Solution

Let weight of sachin $$=x \ kg$$

Acc. to question.

Weight of Rahul $$=(x+5)\ kg$$

Weight of Samir $$=(2x-12)\ kg$$

Total weight of all $$=93\ kg$$

That is ,

$$x+ (x+5)+ (2x-12)=93$$

$$4x-7=93$$
Question 6
1 / -0

Convert this statement in mathematical form.
The number is $$8$$ greater than $$80.$$

A
$$x+8=80$$

B
$$x-8=80$$

C
$$8x=80$$

D
$$x=80\times8$$

Solution

Let the number be $$x$$. Here, the given statement is the number is $$8$$ greater than $$80$$.

The word 'greater than' suggests addition. So, the verbal expression the number is $$8$$ greater than $$80$$ can be represented by the algebraic expression:

$$x=80+8\\ \Rightarrow x-8=80$$

Hence, the statement the number is $$8$$ greater than $$80$$ can be rewritten as $$x-8=80$$.
Question 7
1 / -0

$$6$$ more than twice the number is $$ 22$$. Express the statement in mathematical form.

A
$$2x+6=22$$

B
$$2x-6=22$$

C
$$2x+22=6$$

D
$$2x-22=6$$

Solution

Let the number be $$x.$$

Twice the number $$=2x$$
Six more than $$2x$$ will be $$6+2x.$$
This is equal to $$22.$$

Hence, the answer is $$6+2x=22.$$
Question 8
1 / -0

If $$6$$, $$p$$, $$12$$, $$8$$ and $$9$$ mean of the data is $$9$$ then $$p=$$ ?

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

Solution

Arithmetic mean,$$A=\dfrac { S }{ N } =\dfrac { 6+p+12+8+9 }{ 5 } =9$$

$$\dfrac { 35+p }{ 5 } =9$$

$$35+p=45$$

$$p=45-35=10$$
Question 9
1 / -0

If $$16t-4=0$$ then find $$3t$$

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

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

C
$$\dfrac{9}{4}$$

D
$$1$$

Solution

$$16t-4=0$$
Add $$4$$ both the sides

$$\Rightarrow$$$$16t=4$$
Multiply both the sides by $$16$$
$$\Rightarrow$$$$t=\dfrac 4{16}$$

$$\Rightarrow$$$$t=\dfrac 14$$

$$\Rightarrow$$$$3t=\dfrac 34$$
Question 10
1 / -0

The value of $$x$$ if $$23x+69=0$$

A
$$-3$$

B
$$0$$

C
$$3$$

D
$$4$$

Solution

$$23x+69=0$$
Subract both side by 69
$$\Rightarrow$$$$23x=-69$$
Divide both side by 23

$$\Rightarrow$$$$x=\dfrac{-69}{23}$$

$$\Rightarrow$$$$x=-3$$

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Re-Attempt Weekly Quiz Competition

Simple Equations Test - 18

Result

Simple Equations Test - 18

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SHARING IS CARING

Question 1

$$17$$

$$21$$

$$15$$

None

Solution

Question 2

$$12$$

$$18$$

$$14$$

$$16$$

Solution

Question 3

$$32$$

$$16$$

$$8$$

$$4$$

Solution

Question 4

$$6$$

$$8$$

$$10$$

$$12$$

Solution

Question 5

$$4x-7=93$$

x+5=0

$$x+ (x+5)+ (2x-12)=93$$

2x-12=21

Solution

Question 6

$$x+8=80$$

$$x-8=80$$

$$8x=80$$

$$x=80\times8$$

Solution

Question 7

$$2x+6=22$$

$$2x-6=22$$

$$2x+22=6$$

$$2x-22=6$$

Solution

Question 8

$$7$$

$$8$$

$$9$$

$$10$$

Solution

Question 9

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

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

$$\dfrac{9}{4}$$

$$1$$

Solution

Question 10

$$-3$$

$$0$$

$$3$$

$$4$$

Solution

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