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CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 8 - MCQExams.com
CBSE
Class 8 Maths
Linear Equations In One Variable
Quiz 8
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 75 questions and secures 125 marks, the number of questions he attempted correctly, is
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0%
35
0%
40
0%
42
0%
46
Explanation
If the number of correct answer be x, then
the number of incorrect answer is $$\displaystyle (75- x)$$.
$$\displaystyle \therefore \quad 4x-\left( 75-x \right) =125\quad or\quad x=40$$
$$\displaystyle \frac { 4x }{ 9 } -8=8-\frac { 5x }{ 9 } $$, find $$x$$.
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0%
$$-16$$
0%
$$16$$
0%
$$9$$
0%
$$\dfrac{3}{16}$$
Explanation
Given, $$\displaystyle \frac { 4x }{ 9 } -8=8-\frac { 5x }{ 9 } $$.
Transposing $$x$$ terms to one side, we get,
$$\Longrightarrow\dfrac { 4x }{ 9 } +\dfrac { 5x }{ 9 } =8+8$$
$$\Longrightarrow\displaystyle \frac { 9x }{ 9 } =16$$
$$\Longrightarrow x=16$$.
Hence, option $$B$$ is correct.
If $$\displaystyle 5x-35=55+18x$$, then $$x$$.
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0%
is a fraction
0%
is an integer
0%
is a rational number
0%
cannot be solved
Explanation
Given, $$5x-35=55+18x$$.
Transposing $$x$$ terms to one side, we get,
$$\Longrightarrow 5x-18x=55+35$$
$$\Longrightarrow -13x=90$$
$$\Longrightarrow x=-\dfrac{90}{13}$$, which is a fraction of the form $$\dfrac{p}{q}$$ ; $$q \ne 0$$ and $$p$$ & $$q$$ are co-primes.
Hence, options $$A$$ and $$C$$ are correct.
A man has Rs. $$x$$ with him. He gave $$\dfrac{1}{4}$$th to his wife, $$\dfrac{1}{3}$$rd to his son and Rs. $$1000$$ to his daughter. Find $$x$$.
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0%
$$2400$$
0%
$$2200$$
0%
$$2000$$
0%
None of the above.
Explanation
The total amount of money with the man = Rs. $$x$$.
$$\displaystyle \frac { 1 }{ 4 } x+\frac { 1 }{ 3 } x+1000=x$$
$$\Longrightarrow\displaystyle x-\frac { 1 }{ 4 } x-\frac { 1 }{ 3 } x=1000$$
$$\Longrightarrow\displaystyle \frac { 12x-3x-4x }{ 12 } =1000$$
$$\Longrightarrow\displaystyle 5x=1000\times 12$$
$$x$$ = Rs. $$2400$$
The sum of $$5$$, $$9$$ and a number is $$50$$. Find the number.
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0%
$$35$$
0%
$$36$$
0%
$$37$$
0%
$$38$$
Explanation
Let the number be $$ x$$
According to the question,
$$\displaystyle 5+9+x=50$$
$$\therefore \displaystyle x=50-14$$
$$\therefore \displaystyle x=36$$
The sum of two numbers is $$30$$. One of the number exceeds the other by $$10$$.
Find the number.
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0%
$$10, -20$$
0%
$$-10, 20$$
0%
$$10, 20$$
0%
$$-10, -20$$
Explanation
Let the required number be $$x$$.
Then, the other number will be $$x+10$$
According to the question, sum of the two numbers is $$30$$.
Therefore, $$\displaystyle x+x+10=30$$
$$\therefore \displaystyle 2x=20$$
$$\therefore \displaystyle x=10$$
The numbers are $$10$$ and $$10+10=20$$.
Solve
$$8(3+2x)=13x$$
and find the value of $$ x$$
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0%
$$-8$$
0%
$$8$$
0%
$$\dfrac{1}{8}$$
0%
$$\dfrac{-1}{8}$$
Explanation
Given,
$$ \displaystyle 8(3+2x)=13x$$
$$\Rightarrow 24+16x=13x$$ ...[By Distribution Law]
Transposing $$x$$ terms to one side, we get,
$$\Rightarrow\displaystyle 16x-13x = -24$$
$$\Rightarrow\displaystyle 3x=-24$$
$$\Rightarrow x=\dfrac{-24}{3}$$
$$\Rightarrow x=-8$$
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?
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0%
$$3$$
0%
$$4$$
0%
$$6$$
0%
$$8$$
Explanation
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.
The four consecutive numbers add up to $$74$$. What are these integers?
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0%
$$18, 19, 20, 21$$
0%
$$17, 18, 19, 20$$
0%
$$20, 18, 18, 19$$
0%
$$16, 17, 18, 19$$
Explanation
Let the first integer$$ = p$$
$$\displaystyle \therefore $$ the second integer $$= p+1$$
and the third consecutive integer $$= p+2$$ and fourth $$= p+3$$
$$\displaystyle \because $$ Sum of four consecutive numbers = $$74$$
$$\displaystyle p+(p+1)+(p+2)+(p+3)=74$$
$$\Longrightarrow \displaystyle 4p+6=74$$
$$\Longrightarrow 4p=68$$
$$\Longrightarrow \displaystyle p=\frac { 68 }{ 4 } $$
$$\therefore p=17$$
1 st integer $$= 17$$
2nd integer $$= 18$$
3rd integer $$= 19$$
4 th integer $$= 20$$
So, option B is correct.
The sum of three consecutive even integers is $$72$$. Find the smallest number
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0%
$$24$$
0%
$$22$$
0%
$$26$$
0%
None
Explanation
Let the three consecutive numbers be $$ x, x+2$$ and $$x+4$$.
Sum = $$\displaystyle x+x+2+x+4$$
$$\therefore 3x + 6 = 72$$
$$\Longrightarrow 3x=66$$
$$\Longrightarrow x=22$$
The sum of $$6$$ consecutive numbers is $$105$$. Find the smallest number.
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0%
$$13$$
0%
$$15$$
0%
$$14$$
0%
$$12$$
Explanation
Let the 6 consecutive numbers be $$x, x+1, x+2, x+3, x+4$$ and $$x+5$$
$$\Longrightarrow \displaystyle x+x+1+x+2+x+3+x+4+x+5=105$$
$$\Longrightarrow \displaystyle 6x+15=105$$
$$\therefore 6x=90$$
$$\therefore x=15$$
The sum of three consecutive multiples of $$5$$ is $$45$$. Which is the smallest of the three multiples?
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0%
$$10$$
0%
$$5$$
0%
$$15$$
0%
$$20$$
Explanation
Let $$5x$$ be the smallest multiple of $$5$$.
Then, the three consecutive multiples shall be $$5x$$, $$5x+5$$ and $$5x+10$$
According to the question,
$$\displaystyle 5x+5x+5+5x+10=45$$
$$\therefore \displaystyle 5x+5x+5x=45-15$$
$$\therefore \displaystyle 15x=30$$
$$\therefore \displaystyle x=2$$
$$\displaystyle \therefore $$ Smallest multiple $$= 5x = 5\times 2 = 10$$
The length of a rectangle is $$ 1\frac { 3 }{ 4 } $$ of its breadth. If perimeter is $$66$$. find the length of rectangle.
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0%
$$12$$
0%
$$21$$
0%
$$20$$
0%
$$11$$
Explanation
Let breadth of rectangle be $$x$$
Length = $$\displaystyle 1\frac { 3 }{ 4 } \times x=\frac { 7 }{ 4 } x$$ ....given
According to the question,
$$\displaystyle 2\left( \frac { 7 }{ 4 } x+x \right) =66$$
$$\Longrightarrow 2\left( \dfrac { 7x+4x }{ 4 } \right) =66$$
$$\Longrightarrow\displaystyle 11x=66\times 2$$
$$\Longrightarrow\displaystyle x=12$$
Length =$$\dfrac { 7 }{ 4 } \times 12=21$$
Reduce the following linear equation: $$6t - 1 = t - 11$$
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0%
$$t=-1$$
0%
$$t=-2$$
0%
$$t=-3$$
0%
$$t=-4$$
Explanation
$$6t - 1 = t - 11$$
$$6t - t = -11 + 1$$
$$5t = -10$$
$$t = -2$$
Solve the linear equation: $$5x - 12 = 2x + 18$$
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0%
$$x=8$$
0%
$$x=9$$
0%
$$x=10$$
0%
$$x=11$$
Explanation
$$5x - 12 = 2x + 18$$
$$5x - 2x = 18 + 12$$
$$3x = 30$$
$$x = 10$$
Reduce the linear equation: $$x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$$
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0%
$$x=12$$
0%
$$x=10$$
0%
$$x=8$$
0%
$$x=-6$$
Explanation
Given, $$x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$$
L.C.M of the denominator $$3$$ and $$6$$ is $$6$$.
Multiplying both the sides by $$6$$, we get
$$6x + 18 - 4x + x = 0$$
$$7x - 4x + 18 = 0$$
$$3x = -18$$
$$x = -6$$
Reduce the linear equation: $$\dfrac{x}{2}+\dfrac{2x}{4}= 10$$
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0%
$$x=6$$
0%
$$x=7$$
0%
$$x=8$$
0%
$$x=10$$
Explanation
Given, $$\dfrac{x}{2}+\dfrac{2x}{4}= 10$$
L.C.M of the denominator $$2$$ and $$4$$ is $$4$$.
Multiplying both the sides by $$4$$, we get
$$2x + 2x = 40$$
$$4x = 40$$
$$x = 10$$
The sum of one fifth, one third and one ninth of a number is $$29$$. Find the number.
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0%
$$45$$
0%
$$29$$
0%
$$54$$
0%
None of the above.
Explanation
Let the required number be $$ x$$
According to the question, $$\displaystyle \frac { x }{ 5 } +\frac { x }{ 3 } +\frac { x}{ 9 } =29$$
L.C.M of $$5,3,9$$ is $$45.$$
Then the equation becomes :
$$\displaystyle \frac { x \times 9 }{ 5 \times 9 } +\frac { x \times 15 }{ 3 \times 15 } +\frac { x \times 5}{ 9 \times 5 } =29$$
$$\Longrightarrow\displaystyle \frac { 9x+15x+5x }{ 45 } =29$$
$$\Longrightarrow\displaystyle 29x=29\times 45$$
$$\Longrightarrow\displaystyle x=\frac { 29\times 45 }{ 29 } $$
$$\therefore x=45$$
Solve linear equation:
$$m - \dfrac {m - 1}{2} = 1 - \dfrac {m - 2}{3}$$.
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0%
$$m = \dfrac {4}{5}$$
0%
$$m = \dfrac {3}{5}$$
0%
$$m = \dfrac {8}{5}$$
0%
$$m = \dfrac {7}{5}$$
Explanation
Solve for $$x$$:
$$x+2\left (5x-\dfrac { 5 }{ 2 }\right )=4(x+1)-2$$.
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0%
$$-2$$
0%
$$ 2$$
0%
$$1$$
0%
$$0$$
Explanation
Given, $$x+2\left (5x-\dfrac{5}{2}\right)=4(x+1)-2$$
$$\Rightarrow x+10x-5=4x+4-2$$
...[By Distribution Law]
$$\Rightarrow 11x-5=4x+2$$
.
Transposing $$x$$ terms to one side, we get,
$$\Rightarrow 11x-4x=2+5$$
$$\Rightarrow 7x=7$$
$$\Rightarrow x=\dfrac{7}{7}=1$$.
Hence, option $$C$$ is correct.
Ram is $$5$$ times as old as Shyam. If their difference of age is $$8$$ years, how old is Ram?
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0%
$$8$$ years
0%
$$10$$ years
0%
$$12$$ years
0%
$$5$$ years
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None of these
Explanation
Let the age of Shyam be $$ x$$ so,
age of Ram is $$ 5x$$.
As per the problem, the difference of their ages is $$8$$ years.
$$5x-x = 8$$
$$\Rightarrow 4x = 8$$
$$\Rightarrow x = 2$$
So, Rams age $$= 5x$$
$$ = 5\times 2 $$
$$= 10$$ years.
There are $$42$$ integers in a group. If there are $$5$$ times as many odd integers as there are even integers, how many numbers are even integers?
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0%
$$7$$
0%
6
0%
5
0%
9
Explanation
Let numbers of even integers be $$x$$
Then numbers of odd integers are $$5x$$.
$$\therefore x+5x=42$$
$$ 6x=42$$
$$\Rightarrow x=\dfrac{42}{6}$$
$$=7$$
Solve for $$t$$:
$$\displaystyle\frac{t}{2}+4=\displaystyle\frac{3}{4}t-5$$.
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0%
$$4$$
0%
$$9$$
0%
$$18$$
0%
$$36$$
Explanation
Given, $$ \dfrac{1}{2}t+4=\dfrac{3}{4}t-5$$.
Transposing $$t$$ terms to one side, we get,
$$\Rightarrow \dfrac{1}{2}t-\dfrac{3}{4}t=-5-4$$
$$\Rightarrow \dfrac{2t-3t}{4}=-9$$
…[Cross-multiplying the denominators on the LHS]
$$\Rightarrow \dfrac{-t}{4}=-9$$
$$\Rightarrow -t=-36$$
...[Multiplying $$4$$ on both sides]
$$\Rightarrow t=36$$.
Hence, option $$D$$ is correct.
The sum of four consecutive odd positive integers is $$160$$. Find the largest among them.
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0%
37
0%
39
0%
41
0%
43
0%
45
Explanation
Let the four consecutive odd positive integers are $$x,x+2,x+4,x+6$$
$$\therefore x+x+2+x+4+x+6=160$$
$$ 4x+12=160$$
$$ 4x=160-12$$
$$ 4x=148$$
$$x=\dfrac{148}{4}$$
$$=37$$
$$\therefore $$ Largest odd integer is $$37+6=43$$.
If the number of boys in a class is thrice the number of girls and the total number of boys is 84, calculate the number of girls in the class.
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0%
$$28$$
0%
$$20$$
0%
$$30$$
0%
$$252$$
Explanation
Let the number of girls are $$x$$
Then number of boys $$=3x$$
$$\Rightarrow 3x=84$$
$$\Rightarrow x=\dfrac{84}{3}=28$$
Then number of girls $$=28$$
The sum of five positive, consecutive integers is $$115$$. Find the value of the smallest integer among these.
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0%
$$21$$
0%
$$22$$
0%
$$23$$
0%
$$24$$
0%
$$25$$
Explanation
Let the five consecutive integers are $$x,x+1,x+2,x+3,x+4$$
$$\therefore x+x+1+x+2+x+3+x+4=115$$
$$\Rightarrow 5x+10=115$$
$$\Rightarrow 5x=115-10$$
$$\Rightarrow 5x=105$$
$$\Rightarrow x=\dfrac{105}{5}=21$$
The smallest integer $$=21$$
If $$\dfrac{2}{3}(5x+7)=8x$$,then $$x$$ is equal to:
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0%
$$1$$
0%
$$0$$
0%
$$2$$
0%
$$-2$$
Explanation
Given, $$ \dfrac{2}{3}(5x+7)=8x$$.
$$\Rightarrow \dfrac{10x}{3}+\dfrac{14}{3}=8x$$
...[By Distribution Law]
$$\Rightarrow \dfrac{10x+14}{3}=8x$$
…[Cross-multiplying the denominators on the LHS]
$$\Rightarrow 10x+14=24x$$ ...[Multiplying $$3$$ on both sides].
Transposing $$x$$ terms to one side, we get,
$$\Rightarrow 10x-24x=-14$$
$$\Rightarrow -14x=-14$$
$$\Rightarrow x=\dfrac{-14}{-14}=1$$.
Hence, option $$A$$ is correct.
A club has $$27$$ employees. If there are seven more women than men in the club, calculate the number of men in the club.
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0%
$$10$$
0%
$$8$$
0%
$$6$$
0%
$$9$$
Explanation
Let the number of men in the club $$=x$$
Then number of women $$=$$ $$x+7$$
Then according to the question,
$$\Rightarrow x+x+7=27$$
$$\Rightarrow 2x=27-7$$
$$\Rightarrow 2x=20$$
$$\Rightarrow x=\dfrac{20}{2}=10$$
Then number of men $$=10$$
$$280$$ meals are to be prepared by three chefs. Every chef has its own speed but the combined output of all three is modelled by the equation $$8x+4x+2x=280$$. If $$x$$ is a positive integer, which of the following could $$8x$$ represent in the equation?
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0%
The total meal output by the slowest chef, who made $$40$$ meals
0%
The total meal output by the fastest chef, who made $$160$$ meals
0%
The total meal output by the fastest chef, who made $$80$$ meals
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The different between the output between the slowest and fastest chef, which would be $$120$$ meals
Explanation
In given equation $$8x+4x+2x=280$$
Then $$14x=280$$
$$\therefore x=20$$
Then $$x$$ is $$20$$ meals
Then mean of $$8x$$ in equation is
$$8x=8\times 20=160$$ meals
Then the meal output by the fastest chef, who made $$160$$ meals.
The value of $$d$$ which satisfies the expression $$\cfrac{4(d+3)-9}{8}=\cfrac{10-(2-d)}{6}$$ is:
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0%
$$\cfrac{23}{16}$$
0%
$$\cfrac{23}{8}$$
0%
$$\cfrac{25}{8}$$
0%
$$\cfrac{25}{4}$$
Explanation
Given, $$\dfrac {4(d+3)-9}{8}=\dfrac {10-
(2-d)}{6}$$.
Cross multiplying, we get,
$$24(d+3)-54=80-8(2-d)$$
$$\Rightarrow 24d+72-54=80-16+8d$$
...[By Distribution Law]
$$\Rightarrow 24d+18=64+8d$$.
Transposing $$d$$ terms to one side, we get,
$$\Rightarrow 24d-8d=64-18$$
$$\Rightarrow 16d=46$$
$$\Rightarrow d=\dfrac{23}{8}$$.
Hence, option $$B$$ is correct.
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