Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 44

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Question 1
1 / -0

Arpan cut a cake in two equal parts and one of these two pieces is again cut into equal parts. Each small part is of $$20$$ grams. If there are seven parts of the whole cake, what was the weight of the original cake?

A
$$140$$ grams

B
$$280$$ grams

C
$$240$$ grams

D
$$300$$ grams

Solution

Let the weight of the cake is $$x$$ grams.
If he cuts it into two equal parts, then the weight of one part will be $$\dfrac{x}{2}$$.
Again he cuts one part into equal parts.
Since total seven parts are there, so he has cut one piece into $$6$$ equal parts.
The weight of the small part is $$\dfrac{x}{2\times 6}$$ and it is given that the weight of small part is $$20$$ grams.
$$\Rightarrow 20= \dfrac{x}{12}$$
$$\Rightarrow x=240$$
Weight of cake is $$240$$ grams.
Question 2
1 / -0

Divide Rs. $$3010$$ among A, B, C respectively in such a way that A gets double of what B gets and B gets double of what C gets.

A
Rs. $$1720$$, Rs. $$860$$, Rs. $$430$$

B
Rs. $$1640$$, Rs. $$920$$, Rs. $$450$$

C
Rs. $$1690$$, Rs. $$825$$, Rs. $$495$$

D
Rs. $$1800$$, Rs. $$400$$, Rs. $$810$$

Solution

Let the amount of money C gets be Rs. x
$$\therefore$$ Amount of money B gets $$=$$Rs. $$2x$$
Amount of money A gets $$=Rs. 4x$$
According to question,
$$4x+2x+x=3010$$
$$\Rightarrow 7x=3010\Rightarrow x=\displaystyle\frac{3010}{7}=430$$
$$\therefore$$ Amount of money A gets $$=Rs. (4\times 430)$$
$$=Rs. 1720$$
Amount of money B gets $$=Rs. (2\times 430)$$
$$=Rs. 860$$
And amount of money C get $$=Rs. 430$$.
Question 3
1 / -0

Which of the following statement do not hold in solving the equation $$15 + 3x = 3$$?

A
$$3x = 3 -15$$

B
$$15 - 3 = -3x$$

C
$$15+\dfrac {3x}{3}=3$$

D
$$\dfrac {15}{3}$$ + $$\dfrac {3x}{3}$$ = $$\dfrac {3}{3}$$

Solution

The given equation is,
$$\Rightarrow$$ $$3x=-12$$, so $$x=-4$$
$$\Rightarrow$$ $$(A)$$ $$3x=3-15$$, here we get $$x=-4$$
$$\Rightarrow$$ $$(B)$$ $$15-3=-3x$$, here we get $$x=-4$$
$$\Rightarrow$$ $$(C)$$ $$15+\dfrac{3x}{3}=3$$, here we get $$x=-12$$

$$\Rightarrow$$ $$(D)$$ $$\dfrac{15}{3}+\dfrac{3x}{3}=\dfrac{3}{3}$$, here we get $$x=-4$$.
$$\therefore$$ Option $$(C)$$ do not hold in solving given equation.
Question 4
1 / -0

Solve for x : $$\dfrac {6x-2}{9}+\dfrac {3x+5}{18} = \dfrac {1}{3}.$$

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

B
$$\dfrac {2}{3}$$

C
$$\dfrac {3}{5}$$

D
$$\dfrac {8}{3}$$

Solution

$$\dfrac{6x-2}{9}+\dfrac{3x+5}{18}=\dfrac{1}{3}$$

$$\Rightarrow$$ $$\dfrac{12x-4+3x+5}{18}=\dfrac{1}{3}$$

$$\Rightarrow$$ $$\dfrac{15x+1}{18}=\dfrac{1}{3}$$

$$\Rightarrow$$ $$15x+1=\dfrac{1}{3}\times 18$$

$$\Rightarrow$$ $$15x+1=6$$

$$\Rightarrow$$ $$15x=5$$

$$\Rightarrow$$ $$x=\dfrac{5}{15}$$

$$\therefore$$ $$x=\dfrac{1}{3}$$
Question 5
1 / -0

A boy was asked to multiply a given number by $$\dfrac{8}{17}.$$ Instead, he divided the given number by $$\dfrac{8}{17}$$ and got the result $$225$$ more than what he should have got if he had multiplied the number by $$\dfrac{8}{17}.$$ The given number was

A
$$8$$

B
$$17$$

C
$$64$$

D
$$136$$

Solution

Let the number be $$x$$

Result when calculated wrongly $$=\dfrac{17x}{8} \quad \quad ...(I)$$

Result when calculated correctly $$=\dfrac{8x}{17}\quad \quad ...(II)$$

$$(I)-(II)=225$$

$$\Rightarrow \dfrac{17x}{8}-\dfrac{8x}{17}=225$$

$$\Rightarrow \dfrac{289-64}{136}x=225$$

$$\Rightarrow \dfrac{225}{136}x=225$$

$$\Rightarrow x=136$$
Question 6
1 / -0

A drum of kerosene oil is $$\cfrac{3}{4}$$ full. When $$15$$ liters of oil is drawn from it, it is $$\cfrac{7}{12}$$ full. The capacity of the drum is _____.

A
$$45$$ litres

B
$$90$$ litres

C
$$60$$ litres

D
Can't be determined

Solution

Let the capacity of drum $$= x$$ liters.
Initially, the drum was $$\dfrac{3x}{4}$$ part filled. When $$15$$ liters oil is drawn then it becomes $$\dfrac{7x}{12}$$ part.
Now, according to the question,
$$\dfrac{3x}{4}$$$$-\dfrac{7x}{12}=15$$
$$\dfrac{9x - 7x}{12} = 15$$ { $$LCM(4,12)=12$$ }
$$2x = 180$$
so, $$x = 90$$ litres. Hence, B is right option.
Question 7
1 / -0

$$A$$ has certain amount in his account. He gives half of this to his eldest son and one third of the remaining to his youngest son. The amount left with him now is

A
$$\cfrac { 1 }{ 3 } $$ of the original

B
$$\cfrac { 2 }{ 5 } $$ of the original

C
$$\cfrac { 3 }{ 4 } $$ of the original

D
$$\cfrac { 1 }{ 6 } $$ of the original

Solution

Let $$A$$ have $$x$$ amount.

Hence, according to the problem,
$$x-\cfrac { x }{ 2 } -\left(\cfrac { 1 }{ 3 }\right)\cfrac { 1 }{ 2 } x$$

$$\Rightarrow \cfrac { 1 }{ 2 } x-\cfrac16 x$$

$$\Rightarrow \cfrac { 2 }{ 6 } x$$

$$\Rightarrow \cfrac { 1 }{ 3 } x$$

It is the one-third of the original amount $$x$$.

Hence, the amount left with him now is $$\cfrac { 1 }{ 3 }$$ of the original.
Question 8
1 / -0

A number when added to its half gives $$36$$. Find the number.

A
$$24$$

B
$$28$$

C
$$20$$

D
$$18$$

Solution

Let the number be $$x$$
$$\Rightarrow x+\cfrac { x }{ 2 } =36\\ \Rightarrow \cfrac { 3x }{ 2 } =36\\ \Rightarrow x=24$$
Question 9
1 / -0

The two digit number having the unit digit $$x + 5$$ and ten's digit $$x - 5$$ is _______.

A
$$11x + 55$$

B
$$11x - 45$$

C
$$9x - 55$$

D
$$2x + 10$$

Solution

In a two digit number,
The unit digit $$= x + 5$$
Ten's digit $$= x - 5$$
$$\therefore$$ The number
$$= 1(x + 5) + 10 (x - 5)$$
$$= x + 5 + 10x - 50$$
$$= 11x - 45$$.
Question 10
1 / -0

The sum of 4 consecutive numbers is 94. Then the last number is

A
22

B
27

C
36

D
25

Solution

solution:
let the first number be $$ x$$ then all 4 consecutive numbers will be
$$x , x+1,x+2,x+3$$
sum will be $$x + x+1 + x+2 + x+3 = 4x +6$$
$$4x +6 = 94$$
$$4x = 88$$
$$x = 22$$
then , the last number is $$x+3 = 25$$
hence the correct opt: D

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Linear Equations in One Variable Test - 44

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Linear Equations in One Variable Test - 44

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SHARING IS CARING

Question 1

$$140$$ grams

$$280$$ grams

$$240$$ grams

$$300$$ grams

Solution

Question 2

Rs. $$1720$$, Rs. $$860$$, Rs. $$430$$

Rs. $$1640$$, Rs. $$920$$, Rs. $$450$$

Rs. $$1690$$, Rs. $$825$$, Rs. $$495$$

Rs. $$1800$$, Rs. $$400$$, Rs. $$810$$

Solution

Question 3

$$3x = 3 -15$$

$$15 - 3 = -3x$$

$$15+\dfrac {3x}{3}=3$$

$$\dfrac {15}{3}$$ + $$\dfrac {3x}{3}$$ = $$\dfrac {3}{3}$$

Solution

Question 4

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

$$\dfrac {2}{3}$$

$$\dfrac {3}{5}$$

$$\dfrac {8}{3}$$

Solution

Question 5

$$8$$

$$17$$

$$64$$

$$136$$

Solution

Question 6

$$45$$ litres

$$90$$ litres

$$60$$ litres

Can't be determined

Solution

Question 7

$$\cfrac { 1 }{ 3 } $$ of the original

$$\cfrac { 2 }{ 5 } $$ of the original

$$\cfrac { 3 }{ 4 } $$ of the original

$$\cfrac { 1 }{ 6 } $$ of the original

Solution

Question 8

$$24$$

$$28$$

$$20$$

$$18$$

Solution

Question 9

$$11x + 55$$

$$11x - 45$$

$$9x - 55$$

$$2x + 10$$

Solution

Question 10

22

27

36

25

Solution

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