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CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 6 - MCQExams.com
CBSE
Class 8 Maths
Linear Equations In One Variable
Quiz 6
Two consecutive odd numbers are such that when the smaller number is subtracted from three times the bigger
number, the result is $$56$$. The bigger number is:
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0%
$$29$$
0%
$$31$$
0%
$$25$$
0%
$$27$$
Explanation
Let the two consecutive odd numbers are $$x$$ and $$x+2$$.
Two consecutive odd numbers are such that when the smaller number is subtracted from three times the bigger
number, the result is $$56$$.
Hence, equation is,
$$3(x+2)-x=56$$
$$\therefore$$ $$2x=50$$
$$\therefore$$ $$x=25$$
$$\therefore$$ Smaller number is $$25$$ and bigger number is $$27$$.
One number is 3 less than the two times of the other; if their sum is increased by 7 the result isFind the numbers.
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$$9 , 11$$
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$$11 , 13$$
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$$11 , 19$$
0%
$$9 , 13$$
Explanation
Let one number = x
other number = 2x-3
$$\displaystyle (x+2x-3)+7=37$$
3x = 33
x = 11
$$\displaystyle \therefore $$ The two number are 11 and 19
The total cost of three prizes is Rs.If the value of second prize is $$\displaystyle \frac{3}{4}$$th of the first and the value of 3rd prize is $$\displaystyle \frac{1}{2}$$ of the second prize;
find the value of first prize.
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Rs. $$900$$
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Rs. $$1500$$
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Rs. $$1200$$
0%
Rs. $$450$$
Explanation
Let First Prize be $$ x$$
Then Second Prize $$\displaystyle =\frac{3}{4}x$$
Third Prize $$\displaystyle =\frac{1}{2}\times \frac{3x}{4}=\frac{3x}{8}$$
$$\displaystyle x + \frac{3x}{4}+\frac{3x}{8}=2550$$ (given)
L.C.M of $$1,4,8$$ is $$8$$.
$$\displaystyle \frac{x \times 8}{1 \times 8}+\frac{3x \times 2}{4 \times 2}+\frac{3x}{8} =2550$$
$$\displaystyle \frac{8x+6x+x}{8} =2550$$
$$\displaystyle \frac{17x}{8}=2550$$
$$\displaystyle x = \frac{2550 \times 8}{17}$$
$$x=Rs.1200$$
$$\therefore$$ The value of first first prize is $$Rs.1200$$
If $$\displaystyle \frac{7x+2}{6}-\frac{x-4}{3}=\frac{x+3}{3}$$, then $$x=$$
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$$\displaystyle -\frac{4}{3}$$
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$$\displaystyle \frac{1}{3}$$
0%
$$\displaystyle \frac{-1}{3}$$
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$$\displaystyle 1\frac{1}{3}$$
Explanation
Given. $$\displaystyle \frac{7x+2}{6}-\frac{x-4}{3}=\frac{x+3}{3}$$.
Multiply throughout by $$3$$, we get
$$\therefore$$ $$\displaystyle \frac{7x+2}{2}-(x-4)=x+3$$
$$\therefore$$ $$\displaystyle \frac{7x+2}{2}=2x-1$$
On cross multiplying, we get
$$3x=-4$$
$$\therefore$$ $$x=\dfrac {-4}{3}$$
The sum of 4 consecutive integers isThen the greatest among them is:
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0%
19
0%
23
0%
17
0%
16
Explanation
Let the $$4$$ consecutive numbers be $$x$$, $$x+1$$, $$x+2$$, $$x+3$$
Given: The sum of $$4$$ consecutive numbers is $$70$$.
Therefore, $$(x)+(x+1)+(x+2)+(x+3)=70$$
$$4x+6=70$$
$$4x=64$$
$$x=16$$
Therefore,
$$x+1=17, x+2=18, x+3=19$$
Thus the greatest is $$19$$
Solve for $$x$$: $$\displaystyle \frac{1}{4}(5x+4)=\frac{1}{3}(2x-1)$$
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$$\displaystyle \frac{-16}{7}$$
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$$\displaystyle \frac{10}{7}$$
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$$\displaystyle \frac{8}{7}$$
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$$\displaystyle \frac{-15}{7}$$
Explanation
Given, $$\cfrac14(5x+4)=\cfrac13(2x-1)$$
$$\Rightarrow 3(5x+4)=4(2x-1)$$ ($$\text{cross multiply}$$)
$$\Rightarrow 15x+12=8x-4$$
$$\Rightarrow 15x-8x=-4-12$$
$$\Rightarrow 7x=-16$$
$$\Rightarrow x=-\cfrac{16}7$$
$$\displaystyle \frac{1}{5}$$th of a flagpole is black and $$\displaystyle \frac{1}{4}$$th is white and the remaining three metres is painted yellow then ffind the length of the flagpole
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$$\displaystyle 5\frac{5}{11}m$$
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$$\displaystyle \frac{60}{11}cm$$
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$$5 km$$
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None
Explanation
Let the length of flagpole be $$x$$ m
Length of flagpole which is black $$=\dfrac{1}{5}x$$ m
Length of flagpole which is white $$=\dfrac{1}{4}x$$ m
Length of flagpole which is yellow $$=3$$ m
A.T.Q,
$$\dfrac{1}{5}x+\dfrac{1}{4}x+3=x$$
$$\implies x-\dfrac{x}{5}-\dfrac{x}{4}=3$$
$$\implies \dfrac{20x-4x-5x}{20}=3$$
$$\implies 11x=3\times 20$$
$$\implies x=\dfrac{60}{11}$$
$$\implies x=5\dfrac{5}{11} m$$
$$\therefore $$ The length of flagpole $$=5\dfrac{5}{11}m$$
$$\displaystyle \frac{1}{2}$$ is subtracted from a number and the difference is multiplied by 4 , then 25 is added to the product and the sum is divided by 3 the result is equal toFind the number
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$$\displaystyle \frac{3}{5}$$
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$$\displaystyle \frac{7}{4}$$
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$$\displaystyle \frac{6}{7}$$
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$$\displaystyle \frac{2}{3}$$
Explanation
Let the number be $$x$$
Given $$\displaystyle \left \{ \left ( x-\frac{1}{2} \right )\times4+25\right \}\div 3=10$$
$$\displaystyle \left \{ 4x-2+25 \right \}\times\frac{1}{3}=10$$
$$\displaystyle \left ( \frac{4x+23}{3} \right )=10$$
$$ 4x+23=30$$
$$\displaystyle 4x=7$$
$$\Rightarrow x=\frac{7}{4}$$
In a piggy bank, the number of $$25$$ paise coins are five times the number of $$50$$ paise coins. If there are $$120$$ coins find the amount in the bank ?
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Rs. $$25$$
0%
Rs. $$10$$
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Rs. $$35$$
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Rs. $$40$$
Explanation
Let number of $$50$$ paise coins $$= x$$
then number of $$25$$ paise coins $$= 5x$$
Given total number of coins $$= 120$$
$$\Rightarrow x + 5x = 120$$
$$\Rightarrow 6x = 120$$
$$\displaystyle\Rightarrow x=\frac{120}{6}=20$$
$$\therefore$$ Number of $$50$$ paise coins $$= 20$$
Thus Amount $$20 \times50 =$$ Rs. $$10$$
And number of $$25$$ paise coins $$= 5 \times 20=100$$
Thus amount $$= 25 \times 100 =$$ Rs. $$25$$
Hence, total amount $$25 + 10 = $$ Rs. $$35$$.
Rina is $$x$$ years old. Her sister Bina is $$5$$ years older than her. If their ages add up to $$15$$, then Bina's age is
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0%
a multiple of $$5$$
0%
a multiple of $$6$$
0%
a multiple of $$7$$
0%
divisible by $$8$$
Explanation
Age of Rina $$=x $$
Age of Bina $$= (x + 5) $$
Sum of their ages $$= x +x + 5$$
According to the statement:
$$2x+ 5= 15$$
$$2x= 10$$
$$x=5$$
Hence, option A is correct.
Divide 224 into three parts so that the second will be twice the first and third will be twice the second
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0%
$$26, 52, 104$$
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$$24, 48, 96$$
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$$18, 36, 72$$
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$$32, 64, 128$$
Explanation
Let the first part be $$x$$
Then, second part $$= 2x$$
Third Part $$= 2 \times 2x = 4x$$
Given, $$x +2x + 4x = 224$$
$$\displaystyle \Rightarrow x=\frac{224}{7}=32$$
$$\displaystyle \therefore $$ The three parts are $$32, 2 \times 32, 4 \times 32$$ i.e. $$32,64,128$$.
Ages of two friends are in the ratio $$2 : 1$$. If sum of their ages is $$51$$ then their ages are
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0%
$$34$$ yrs, $$20$$ yrs
0%
$$34$$ yrs, $$17$$ yrs
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$$20$$ yrs, $$10$$ yrs
0%
$$30$$ yrs, $$15$$ yrs
Explanation
Given ages in the ratio $$2 : 1$$
Thus let their ages be $$2x$$ and $$x$$
Given, the sum of their ages is $$51$$
Therefore, $$2x + x = 51$$
$$\Rightarrow 3x = 51$$
$$\Rightarrow x=17$$
Hence, required age's are $$34$$ yrs and $$17$$ yrs.
Solve: $$\displaystyle 6x-1=2x+9$$
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0%
$$\displaystyle 1\frac{1}{2}$$
0%
$$\displaystyle -2\frac{1}{2}$$
0%
$$\displaystyle \frac{4}{10}$$
0%
$$\displaystyle 2\frac{1}{2}$$
Explanation
Given, $$6x - 1 = 2x + 9$$
Bringing all $$x$$ terms to one side and constants to another side.
$$ 6x - 2x = 9 + 1$$
$$\Rightarrow 4x = 10$$
$$\displaystyle\Rightarrow x =\frac{10}{4}=\frac{5}{2}=2\frac{1}{2}$$
Hence, solution is $$x=2\cfrac{1}{2}$$.
In a two digit number the digit at the unit's place is four times the digit in the ten's place and the sum of the digits is equal to 10 What is the number?
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0%
14
0%
44
0%
82
0%
28
Explanation
Let the digit at ten's place $$= x$$
Then digit at unit's place $$= 4x$$
Sum of digit $$= 10$$
$$\displaystyle \Rightarrow x+4x=10\, \Rightarrow 5x=10\Rightarrow x=2$$
$$\displaystyle \therefore $$ The number $$= 10 \times x + 4x$$
$$= 10 \times 2 + 4 \times 2 = 20 + 8=28$$
If the sum of one-half and one-fifth of a number exceeds one-third of that number by $$\displaystyle 7\frac{1}{3}$$ then the number is
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0%
15
0%
18
0%
20
0%
30
Explanation
Let the number be $$x$$
Given $$\displaystyle \left ( \frac{x}{2}+\frac{x}{5} \right )-\frac{x}{3}=7\frac{1}{3}=\frac{22}{3}$$
$$\displaystyle \Rightarrow \frac{15x+6x-10x}{30}=\frac{22}{3}$$
$$\displaystyle \Rightarrow \frac{11x}{30}=\frac{22}{3}\Rightarrow x=\frac{220}{11}=20$$
A train starts with full of passengers. At the first station, the train drops one-third of the passengers and takes in 96 more. At the next station one-half of the passengers on board get down while 12 new passengers get on board If the passengers on board now are 240 the number of passengers in the beginning was
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0%
540
0%
600
0%
444
0%
430
Explanation
Let the full number of passengers be $$x$$
Number of passengers at the first station , after drops one-third of the passenger
$$\displaystyle =\left ( x-\frac{1}{3} x\right )+96=\frac{2x}{3}+96$$
Number of passengers at the second station , after one-half get down and 12 added
$$\displaystyle =\frac{1}{2}\left ( \frac{2x}{3}+96 \right )+12=\frac{x}{3}+48+12=\frac{x}{3}+60$$
Given $$\displaystyle \frac{x}{3}+60=240\Rightarrow \frac{x}{3}=240-60=180$$
$$\displaystyle \Rightarrow x=180\times3= 540$$
Rs $$770$$ have to be divided among A, B and C such that A receives 2/9th of what B and C together receive. Then A's share is
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0%
Rs $$140$$
0%
Rs $$154$$
0%
Rs $$165$$
0%
Rs $$170$$
Explanation
Let, B and C together receives $$ Rs. x$$
$$\therefore $$ A's share $$=\dfrac{2}{9}x$$
A.T.Q,
$$x+\dfrac{2}{9}x=770$$
$$\implies \dfrac{9x+2x}{9}=770$$
$$\implies 11x=770 \times 9$$
$$\implies x=\dfrac{770 \times 9}{11}=630$$
A's Share $$ =\dfrac{2}{9}\times 630=140$$
A sum of Rs 36.90 is made up of 180 coins which are either 10 paise coins or 25 paise coins Determine the number of each type of coins
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0%
$$126$$ of $$10$$ p coins and $$54$$ of $$25$$ p coins
0%
$$54$$ of $$10$$ p coins and $$126$$ of $$25$$ p coins
0%
$$90$$ of $$10$$ p coins and $$90$$ of $$25$$ p coins
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$$54$$ of $$10$$ p coins and $$90$$ of $$25$$ p coins
Explanation
Given : Total no of coins $$=180$$
Let, the no of $$10$$ paise coins $$x$$
No of $$25$$ paise coins $$=180-x$$
Value of $$10$$ paise coins $$=10x $$
Value of $$25$$ paise coins $$=25(180-x)$$
A.T.Q,
$$10x+25(180-x)=3690$$
$$\implies 10x+4500-25x=3690$$
$$\implies 4500-3690=25x-10x$$
$$\implies 15x=810$$
$$\implies x=\dfrac{810}{15}=54$$
$$\therefore$$ No of $$10 $$ paise coins $$=54$$
No of $$25$$ paise coins $$=180-54=126$$
If the sum of four consecutive odd numbers is $$40$$, then the smallest number is
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0%
$$7$$
0%
$$9$$
0%
$$11$$
0%
$$13$$
Explanation
Let the first odd number be $$x$$.
Thus, the other consecutive numbers are $$x+2,x+4,x+6$$
Now, given that sum of all four numbers is 40,
$$x+ x+2+x+4+x+6=40$$
$$\Rightarrow 4x+12=40$$
$$\Rightarrow 4x=28$$
$$\Rightarrow x=7$$
Thus, the smallest number $$ 7$$.
If $$50$$ is subtracted from two-third of a number, the result is equal to sum of $$40$$ and one-fourth of that number. What is the number?
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$$256$$
0%
$$216$$
0%
$$186$$
0%
$$144$$
Explanation
Let the number be $$x$$.
According to the statement,
$$\displaystyle \frac{2x}{3}-50=\frac{x}{4}+40$$
$$\dfrac{2x}{3}-\dfrac{x}{4}=50+40$$
$$\dfrac{8x-3x}{12}=90$$
$$\dfrac{5x}{12}=90$$
$$x=\dfrac{90\times 12}{5}$$
$$=216$$
Out of three numbers, the first is twice the second and the second is thrice the third. If the average of the three numbers is $$10,$$ find the largest number?
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$$18$$
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$$14$$
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$$22$$
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$$24$$
Explanation
Let the third number be $$x$$.
Then, second number $$= 3x$$ and first number $$= 6x$$
Given, $$\displaystyle \frac{x + 3x + 6x}{3} = 10 \Rightarrow 10x = 30 \Rightarrow x = 3$$
$$\therefore$$ Largest number $$= 6x = 6 \times 3 = 18$$
In a zoo there are rabbits and pigeons. If their heads are counted these are 90 while their legs are 224 The number of pigeons in the zoo are
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0%
70
0%
68
0%
72
0%
22
Explanation
Let the number of rabbits be $$x$$
No of heads $$\displaystyle = 90\Rightarrow $$ No of pigeons $$= (90 - x)$$
Since a pigeon has $$2$$ legs and a rabbit has 4 legs
$$x \times 4 + (90 - x) \times 2 = 224$$
$$\displaystyle \Rightarrow $$ $$4x + 180 -2x = 224$$ $$\displaystyle \Rightarrow $$ $$2x = 224 - 180 = 44$$
$$\displaystyle \Rightarrow x = 22$$
$$\displaystyle \therefore $$ No of pigeons $$= 90 - 22 = 68$$
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?
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0%
$$\cfrac { 5 }{ 12 } $$
0%
$$\cfrac { 1 }{ 2 } $$
0%
$$\cfrac { 2 }{ 3 } $$
0%
$$\cfrac { 5 }{ 7 } $$
Explanation
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.
The sum of the digits of a three-digit number is $$16$$. If the ten's digit of the number is $$3$$ times the unit's digit and the unit's digit is one-fourth of the hundredth digit then what is the number?
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0%
$$446$$
0%
$$561$$
0%
$$682$$
0%
$$862$$
Explanation
Let the hundredth's digit of the number be $$x$$
Then unit's digit $$= \dfrac x4$$
and ten's digit $$\displaystyle=\frac{3x}{4}$$
Given,
The sum of the digits is $$16$$
$$\Rightarrow \displaystyle x +\frac{x}{4}+\frac{3x}{4}=16$$
$$\Rightarrow \dfrac{4x+x+3x}{4}=16$$
$$\displaystyle \Rightarrow 2x = 16$$
$$\Rightarrow x=8$$
ten's digit $$=\dfrac{3x}{4} = \dfrac{3 \times 8}{4} = 6$$
unit's digit $$\displaystyle =\dfrac{x}{4}=\dfrac{8}{4}=2$$
$$\displaystyle \therefore $$ Required number $$= 100 \times 8 + 10 \times 6 + 2$$
$$= 800 + 60 + 2= 862$$
The required number is $$862$$
The ratio of a two digit number to the sum of its digits is $$7:1.$$ If the digit in the ten's place is $$1$$ more than the digit in the one's place, then the number is
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0%
$$65$$
0%
$$43$$
0%
$$32$$
0%
$$21$$
Explanation
Let the digit at the unit's place in a $$2$$ digit number be $$x$$.
Then, the digit at the ten's place $$= x + 1$$.
$$\therefore$$ The number $$= 10(x+1) + x = 11x +10$$
Sum of digits $$= x+ 1 + x = 2x + 1$$
Given, $$ \displaystyle \frac {11x+10}{2x+1} = \frac {7}{1}$$
$$ 11x + 10 = 14x + 7 $$
$$ 3x =3$$
$$ \Rightarrow x = 1$$
$$\therefore$$ The number $$= 11 \times 1 + 10= 21$$
Given that $$\displaystyle \frac {-6p - 9}{3}\, =\, \displaystyle \frac {2p + 9}{5}$$, find the value of $$p$$.
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0%
$$-4$$
0%
$$-2$$
0%
$$3$$
0%
$$5$$
Explanation
Given, $$\displaystyle \frac {-6p - 9}{3}\, =\, \displaystyle \frac {2p + 9}{5}$$
$$\Rightarrow -2p-3=\cfrac {2p + 9}{5}$$
$$\Rightarrow -10p-15=2p+9$$
$$\Rightarrow -12p=24$$
$$\Rightarrow p=-2$$
If $$2x + 3 = x - 4,$$ then the value of $$x$$ is
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0%
$$-4$$
0%
$$-7$$
0%
$$-9$$
0%
$$-10$$
Explanation
$${\textbf{Step -1: Solve linnear equation in variable with variable on both sides.}}$$
$${\text{Given}}$$
$$2x + 3 = x - 4$$
$${\text{put x terms on one side and constants on another side.}}$$
$$\Rightarrow$$ $$2x-x=-4-3$$
$$\Rightarrow$$ $$x=-7$$
$${\text{Hence, the value of x is -7.}} $$
$${\textbf{Hence , option B is correct. The value of x is -7.}}$$
If the sum of three consecutive even numbers is $$234$$, then the smallest among them is
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0%
$$76$$
0%
$$78$$
0%
$$80$$
0%
None of these
Explanation
Let the three consecutive even numbers be $$2x-2, 2x, 2x+2$$.
We have,
$$(2x - 2) + 2x + (2x + 2) = 234$$
$$\Rightarrow 6x = 234$$
$$\Rightarrow x = 39$$
$$\therefore$$ Least even number is
$$ 2x - 2 = 2(39) - 2 = 76$$.
$$2n$$ is an even number. What are the odd numbers on each side of it? The sum of two consecutive odd numbers is $$96$$. What are they?
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0%
$$41,43$$
0%
$$57,59$$
0%
$$47,49$$
0%
$$31,33$$
Explanation
Given: Even number $$=2n$$
The odd number preceding $$2n =2n-1$$
The odd number succeeding $$2n=2n+1$$
A.T.Q,
$$2n-1+2n+1=96$$
$$\implies 4n=96$$
$$\implies n=\dfrac{96}{4}=24$$
$$\therefore$$ the consecutive odd numbers are $$2\times 24-1=47$$ and $$2\times 47+1=49$$
They are $$47,49$$
$$3x =2x + 20$$
What is the value of the $$x$$?
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0%
16
0%
18
0%
20
0%
22
Explanation
We have, $$3x = 2x + 20$$
$$\Rightarrow$$ $$3x - 2x = 20$$ [Transposing $$2x$$ to LHS]
$$\Rightarrow $$ $$x=20$$
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