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CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 5 - MCQExams.com
CBSE
Class 8 Maths
Linear Equations In One Variable
Quiz 5
The sum of two consecutive odd natural numbers is $$140$$. Find the bigger number.
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0%
$$85$$
0%
$$71$$
0%
$$91$$
0%
$$69$$
Explanation
Let the integer be $$x$$.
The two consecutive odd number $$= x, x+2$$
Sum $$= x+x+2 = 140$$
$$2x = 138\\ x = 69$$
The length of larger number $$= x+2 = 71$$.
In a shooting competition a marksman receives $$50$$ paise if he hits the mark and pays $$20$$ paise if he misses it. He tried $$60$$ shots and was paid Rs. $$1.30$$. How many times did he hit the mark?
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0%
$$10$$
0%
$$19$$
0%
$$14$$
0%
$$12$$
Explanation
Let the Number of hits be $$ x$$
Number of misses will be $$ 60 - x$$
Then according to the problem, we have
$$0.50x -(60-x)0.20 = 1.30$$
$$\Rightarrow 0.50x -12.0 +0.2x = 1.30$$
$$\Rightarrow 0.7x = 13.30$$
$$\Rightarrow x = 19$$
Number of times he hit the mark $$= 19$$.
Solve: $$8x+\displaystyle \frac{21}{4}= 3x+7$$ , then $$x=$$
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0%
$$\displaystyle \frac{2}{9}$$
0%
$$\displaystyle \frac{7}{20}$$
0%
$$\displaystyle \frac{6}{17}$$
0%
$$\displaystyle \frac{4}{21}$$
Explanation
Given, $$
8x+\cfrac { 21 }{ 4 } = 3x+7$$
Multiply $$4$$ on both the sides, we get
$$ 32x +21 =12x +28$$
Taking $$x$$ to one side and constants to another side.
$$ \\ 20x = 7$$
$$ \therefore x = \cfrac { 7 }{ 20 }
$$
Some part of a journey of $$555\ km$$ was completed by a car with speed $$60\ km/h$$. Then the speed is increased by $$15\ km/h$$ and the journey is completed. If it takes $$8$$ hours to reach, find the time taken by $$60\ km/h$$ speed.
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0%
$$11$$
0%
$$8$$
0%
$$13$$
0%
$$3$$
Explanation
Let time taken to cover some distance with $$60$$ km/hr be $$t$$.
Then, $$t (60) + (8-t)(75) = 555$$
$$\Rightarrow 75\times 8 - 15t = 555$$
$$\Rightarrow 600 - 555 = 15t$$
$$\Rightarrow t = 3$$ hours
Distance $$=$$ $$3\times60= 180$$ km
Therefore, the time taken by $$60$$ km/hr speed is $$3$$ hours.
If $$\displaystyle x=m+1$$ and $$\displaystyle \frac{1}{3}(6x-3)-(8-3x)=11;$$ find the value of $$m$$.
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0%
$$m=7$$
0%
$$m=6$$
0%
$$m=2$$
0%
$$m=3$$
Explanation
Given, $$\dfrac {1}{3}(6x-3)-(8-3x)=11$$
As L.C.M. is $$3$$, we have
$$ 6x-3-24+9x=33$$
$$\Rightarrow 15x-27=33$$
$$\Rightarrow 15x=60$$
$$\Rightarrow x=4=3+1$$
Thus comparing this with $$x=m+1$$, we get value of $$m$$ as $$3$$.
The angles of a triangle are $$2\left ( x-7 \right )$$ , $$\displaystyle \frac{3}{2}\left ( x-1 \right )$$ and $$3\left ( x+11 \right )$$ . Find $$x$$ and then show that the triangle is isosceles.
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0%
$$15$$
0%
$$18$$
0%
$$22$$
0%
$$25$$
Explanation
Sum of the angles $$= 180$$
$$2(x-7) + \cfrac { 3 }{ 2 } (x-1) + 3(x+11) = 180\\ LCM = 2\\ 4(x-7) + 3(x-1) + 6(x+11) = 360\\ 4x - 28 + 3x - 3 + 6x +66 = 360\\ 13x +35 = 360\\ 13x = 325\\ x = 25$$
Angle $$1 = 2(x-7) = 2(18) = 36$$
Angle $$2 = \cfrac { 3 }{ 2 } (x-1) = \cfrac { 3 }{ 2 } (24) = 36$$
Since two angles are equal, it is an isosceles triangle.
The measures of the angles of a triangle are $$2x,x+25$$ and $$3x+5$$. Find $$x$$ and then show that the triangle is isosceles.
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0%
$$25$$
0%
$$15$$
0%
$$28$$
0%
$$12$$
Explanation
Measures of the angle $$= 2x, x+25, 3x+5$$
Sum of the angles $$= 180$$
$$2x + x+25 +3x + 5 = 180\\ 6x + 30 = 180\\ 6x = 150\\ x = 25$$
So, the angles of the triangle are
1) $$2x = 50$$
2) $$x+25 = 50$$
3) $$3x +5 = 80$$
Since two angles are equal, the triangle is isosceles.
The difference between $$ \displaystyle \frac{3}{4} $$ of a line and $$ \displaystyle \frac{2}{5} $$ of the same line is $$28$$ cm. Find the length of the line.
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0%
$$39$$cm
0%
$$45$$cm
0%
$$80$$cm
0%
$$40$$cm
Explanation
Let the Length of the line is $$x$$.
Therefore, $$\dfrac {3}{4}x-\dfrac {2}{5}x=28$$
$$\Rightarrow \dfrac {7x}{20}=28$$
$$\Rightarrow 7x=20\times 28$$
$$\Rightarrow x=80$$
Therefore, the length of a line is $$80$$ cm.
A boy has $$x$$ coins of $$50$$ paise each$$,\: 2x$$ coins of $$25$$ paise each, $$4x$$ coins of $$10$$ paise each and $$8x$$ coins of $$5$$ paise each. Find the number of $$5$$ paise coins if the value of all the coins is Rs. $$9$$.
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0%
$$40$$
0%
$$20$$
0%
$$500$$
0%
$$300$$
Explanation
According to the problem, we have
$$x(0.50) + 2x (0.25) + 4x (0.10) + 8x (0.05) = 9.00$$
$$\Rightarrow\ 0.5x + 0.5x +0.4x +0.4x = 9.0$$
$$ \Rightarrow\ 1.8x = 9$$
$$\Rightarrow x = 5$$
Number of $$5$$ paise coins $$= 8x = 40$$.
If $$(2ax + 1) (3x + 1) = 6a (x + 1)$$ and $$x = 1$$, find the value of $$a$$.
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0%
$$1$$
0%
$$4$$
0%
$$3$$
0%
$$2$$
Explanation
$$(2ax + 1) (3x + 1) = 6a (x + 1) $$
$$\implies[2a(1)+1][3(1)+1]=6a[(1+1)]$$
$$\implies (2a+1)(3+1)=6a(2)$$
$$\implies (2a+1)(4)=12a$$
$$\implies 8a+4=12a$$
$$\implies 8a-12a=-4$$
$$\implies -4a=-4$$
$$\implies a=1$$
Two numbers are in the ratio $$4 : 7$$. If thrice the larger be added to twice the smaller, the sum is $$59$$. Find the numbers.
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$$8\displaystyle \frac{4}{29}$$ and $$14\displaystyle \frac{7}{29}$$
0%
$$8\displaystyle \frac{3}{34}$$ and $$14\displaystyle \frac{7}{5}$$
0%
$$8\displaystyle \frac{3}{34}$$ and $$14\displaystyle \frac{6}{25}$$
0%
$$8\displaystyle \frac{4}{29}$$ and $$14\displaystyle \frac{2}{19}$$
Explanation
Let the numbers be $$ 4x ; 7x $$
Given, $$ 3(7x) + 2(4x) = 59 $$
$$ => 21x + 8x = 59 $$
$$ 29x = 59 $$
$$ => x = \cfrac {59}{29} $$
Thus the numbers are $$ 4x = 4 \times \cfrac {59}{29} = \cfrac {236}{29} = 8 \cfrac {4}{29} $$
and $$ 7x = 7 \times \cfrac {59}{29} = \cfrac {413}{29} = 14 \cfrac {7}{29} $$
In an examination, a student scores $$4$$ marks for every correct answer and loses $$1$$ mark for every wrong answer. If he attempts in all $$60$$ questions and secures $$130$$ marks, the number of questions he attempts correctly, is ___
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0%
$$35$$
0%
$$38$$
0%
$$40$$
0%
$$42$$
Explanation
Let the number of questions attempted correctly be $$x$$.
Then, number of incorrect ones $$= (60-x)$$.
$$\therefore 4x-1\left( 60-x \right) =130$$
$$\Rightarrow 5x=190$$
$$\Rightarrow x=38$$
So, number of questions student attempts correctly are $$38$$.
Amit is now $$6$$ times as old as his son. Four years from now, the sum of their ages will be $$43$$ years. Determine Amit's present age:
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$$30$$ years
0%
$$32$$ years
0%
$$34$$ years
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$$28$$ years
Explanation
Let present age of Amit's son be $$x$$ yrs and age of Amit be $$6x$$ yrs.
Four yrs from now, $$(6x+4)+(x+4)=43$$.....(Given condition)
$$\therefore 7x=35$$
$$\therefore x=5$$,
i.e., present age of Amit $$=6x=6\times 5=30$$yrs.
Hence, option $$A$$ is the correct answer.
If
$$\displaystyle \frac{1}{3}\left ( 7-4x \right )-\frac{1}{4}\left ( 11-5x \right )+1=x-\frac{1}{2}\left ( 3x-7 \right )$$, then $$x=$$
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0%
$$1$$
0%
$$9$$
0%
$$6$$
0%
$$7$$
Explanation
$$\cfrac { 1 }{ 3 } (7-4x) -\cfrac { 1 }{ 4 } (11-5x) + 1 =x - \cfrac { 1 }{ 2 } (3x-7)\\ LCM = 12\\ 4(7-4x) - 3(11-5x) +12 = 12x - 6(3x-7)\\ 28 - 16x -33 +15x +12 = 12x -18x +42\\ 7 -x = 42 -6x\\ 5x = 35\\ x = 7$$
The ten's digit of a two digit number exceeds its unit's digit by $$4$$. If the ten's digit and the unit's digit are in the ratio $$3 : 1$$, find the number.
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0%
$$62$$
0%
$$51$$
0%
$$65$$
0%
$$85$$
Explanation
Let the units digit $$= x$$
Let the tens digit $$= x+4$$
Tens digit : One digit :: $$3:1$$
$$x+4 : x :: 3:1$$
$$\Rightarrow x+4 = 3x$$
$$\Rightarrow 2x = 4$$
$$\Rightarrow x = 2$$
Tens digit $$= 6$$
Therefore, the number is $$ 62$$.
Solve the following equation:
$$\displaystyle \frac{4}{5}\, \left(x\, +\, \frac{5}{6} \right)\, +\, \frac{2}{3} \left(x\, -\, \frac{1}{4}\right)\, =\, {1}\frac{1}{9}$$
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0%
$$\displaystyle \frac{5}{19}$$
0%
$$\displaystyle \frac{5}{16}$$
0%
$$\displaystyle \frac{5}{14}$$
0%
$$\displaystyle \frac{5}{12}$$
Explanation
$$\displaystyle \frac{4}{5}\left ( x+\cfrac{5}{6} \right )+\cfrac{2}{3}\left ( x-\cfrac{1}{4} \right )=1\cfrac{1}{9}$$
$$=>\cfrac { 4x }{ 5 } +\cfrac { 20 }{ 30 } +\cfrac { 2x }{ 3 } -\cfrac { 2 }{ 12 } =\cfrac { 10 }{ 9 } $$
$$=>\cfrac { 4x }{ 5 } +\cfrac { 2 }{ 3 } +\cfrac { 2x }{ 3 } -\cfrac { 1 }{ 6 } -\cfrac { 10 }{ 9 } =0$$
$$=>\cfrac { 72x+60+60x-15-100 }{ 90 } =0$$
$$=>72x+60x=-60+15+100$$
$$=>132x=55$$
$$=>x=\cfrac{55}{132}$$
$$=>x=\cfrac{5}{12}$$
A farmer divides his herd of $$x$$ cows among his $$4$$ sons so, that first son gets-one-half of the herd, the second son get, one-fourth, the third son gets one-fifth, and the fourth son gets $$7$$ cows, then the value of $$x$$ is :
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0%
$$100$$
0%
$$140$$
0%
$$160$$
0%
$$180$$
Explanation
Total number of cows $$= x$$
According to the question, we have
A $$= \displaystyle{\frac{x}{2}}$$, $$B = \displaystyle{\frac{x}{4}}$$, C $$= \displaystyle{\frac{x}{5}}$$, $$D = 7$$
$$\therefore$$$$\displaystyle{\frac{x}{2} + \frac{x}{4} + \frac{x}{5}} + 7 = x$$
$$\Rightarrow$$$$\displaystyle{\frac{10x + 5x + 4x}{20}} = x - 7$$
$$\Rightarrow19x = 20x - 140$$
$$\Rightarrow 140 = x$$
Hence, option 'B' is correct.
Out of six consecutive numbers the sum of first three is $$27$$. What is the sum of next three?
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0%
$$30$$
0%
$$40$$
0%
$$36$$
0%
$$45$$
Explanation
Let the first three consecutive numbers be $$x, x+1$$
and $$x + 2$$.
Then, $$x + x + 1 + x + 2 = 27$$
$$\Rightarrow 3x = 24$$
$$\displaystyle \Rightarrow x = 8$$
$$\displaystyle \therefore $$ The first three numbers are $$8,9,10$$.
$$\displaystyle \Rightarrow $$ The next three numbers are $$11, 12, 13$$.
Hence, the required sum $$= 11 + 12 + 13 = 36$$.
I bought a number of books one day, four times of them on the next day, and eight times on the third day, In all I bought $$78$$ books. How many books did I buy on the second
day?
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0%
$$6$$
0%
$$8$$
0%
$$16$$
0%
$$24$$
0%
$$36$$
Explanation
Let the number of books bought of the first day be $$a$$
Thus, number of books on second day $$=4a$$
and the number of books on third day $$=8a$$
Now, given $$a+4a+8a=78$$
$$\Rightarrow 13a=78$$
$$\Rightarrow a=\dfrac{78}{13}$$
$$\Rightarrow a=6$$
Thus, the number of books bought of second day $$=4\times 6=24$$
The difference of the squares of two consecutive even natural numbers isTaking x as the smaller of the two numbers, form an equation in x, and hence find the larger of the two numbers.
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0%
$$18$$
0%
$$24$$
0%
$$23$$
0%
$$21$$
Explanation
Given that,
The difference of squares of two consecutive even natural numbers is $$92$$.
The smaller number is $$x$$.
To find out,
$$(i)$$ The equation in $$x$$ representing the given information.
$$(ii)$$ The larger of the two numbers.
$$(i)$$ If the smaller number is $$x$$, then the larger number will be $$x+2$$
According to the question,
$$(x+2)^2-x^2=92$$
Hence, $$(x+2)^2-x^2=92$$ is the equation that represents the given information.
$$(ii)$$ Solving $$(x+2)^2-x^2=92$$
We know that, $$(a+b)^2=a^2+2ab+b^2$$
Applying the same identity, we get:
$$(x^2+4x+4)-x^2=92$$
$$\Rightarrow x^2+4x+4-x^2=92$$
$$\Rightarrow 4x+4=92$$
Dividing both sides by $$4$$, we get:
$$x+1=23$$
$$x=23-1$$
$$=22$$
Hence, $$x=22$$
$$\therefore\ x+2=22+2$$
$$=24$$
Hence, the larger of the two numbers is $$24$$.
The difference between two numbers is $$8$$ and their ratio is $$1:5$$. Which of the following is the smaller number?
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0%
$$24$$
0%
$$8$$
0%
$$10$$
0%
$$2$$
Explanation
Let the numbers be $$x$$ and $$5x$$.
Given, $$5x-x=8$$
$$\Rightarrow 4x=8$$
$$\Rightarrow \displaystyle x=\frac{8}{4}=2$$
Hence, smaller number is $$2$$.
The solution of the equation
$$\displaystyle \frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$$ is
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0%
$$a=2$$
0%
$$\displaystyle a=\frac{1}{3}$$
0%
$$a=-2$$
0%
$$\displaystyle a=\frac{1}{2}$$
Explanation
Given, $$\displaystyle \frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$$
L.C.M. of denominators is $$6$$.
Multiplying each term by $$6$$, we get
$$\displaystyle 6\times \left ( \frac{3a-2}{3} \right )+6\times \left(\frac{2a+3}{2}\right)=6\times a+6\times \frac{7}{6}$$
$$\Rightarrow 2(3a-2)+3(2a+3)=6a+7$$
$$\Rightarrow 6a-4+6a+9=6a+7$$
$$\Rightarrow 12a+5=6a+7$$
$$\Rightarrow 6a=2$$
$$\Rightarrow \displaystyle a=\frac{2}{6}=\frac{1}{3}$$
Hence, required solution is $$a=\cfrac{1}{3}$$.
The sum of one-half, one third and one-fourth of a number exceeds the number by $$12$$. What is the number?
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0%
$$72$$
0%
$$144$$
0%
$$98$$
0%
$$156$$
Explanation
Let the number be $$x$$.
Then,
$$\displaystyle \left ( \frac{x}{2}+\frac{x}{3}+\frac{x}{4} \right )-x=12$$,
L.C.M of $$2, 3, 4=12$$,
Thus multiplying above equation with $$12$$ both sides,
$$\Rightarrow 12 \times \dfrac {x}{2}+12 \times \dfrac {x}{3}+12 \times \dfrac {x}{4}-12 \times x-12 \times 12$$
$$\Rightarrow 6x+4x+3x-12x=144$$
$$\Rightarrow 13x-12x=144\Rightarrow x=144$$.
Hence, the required number is $$144.$$
Find the value of $$y$$ in the equation :
$$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$$
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0%
$$18$$
0%
$$9$$
0%
$$-38$$
0%
$$-32$$
Explanation
Given, $$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$$
$$\Rightarrow \displaystyle \frac{2+y}{y-7}=\frac{4}{5}$$
$$\Rightarrow 5(2+y)=4(y-7) $$ ....( cross multiplying )
$$\Rightarrow 10+5y=4y-28$$
$$\Rightarrow y=-38$$
Hence,. the solution is $$y=-38$$
Solve for $$x$$: $$\displaystyle \frac{2x+3}{2}=5-\frac{2(x-4)}{3}$$.
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0%
$$\dfrac{5}{3}$$
0%
$$\dfrac{12}{29}$$
0%
$$\dfrac{37}{10}$$
0%
None of these
Explanation
Given, $$\displaystyle \frac{2x+3}{2}=5-\frac{2(x-4)}{3}$$
Multiplying all the terms by the LCM of $$2, 3$$, i.e., $$6$$, we get
$$\displaystyle 6\times \frac{(2x+3)}{2}=6\times 5-\frac{6\times 2(x-4)}{3}$$
$$\Rightarrow 3(2x+3)=30-4(x-4)$$
$$\Rightarrow \displaystyle 6x+9=30-4x+16$$
$$\Rightarrow 6x+9=-4x+46$$
$$\Rightarrow \displaystyle 6x+4x=46-9$$
$$\Rightarrow 10x=37$$
$$\Rightarrow x=\dfrac{37}{10}$$
Hence, required solution is $$x=\cfrac{37}{10}$$.
X buys pens and pencils at Rs. 5 and Rs. 1 per piece respectively. For every two pens, he buys three pencils. He sold pens and pencils at 12 $$\%$$ and 10 $$\%$$ profit respectively. If his total sale is Rs. 725, then the number of pencils exceeds the number of pens by
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0%
$$25$$
0%
$$50$$
0%
$$75$$
0%
$$90$$
Explanation
Let number of pens be $$ x$$
Hence, number of pencils will be $$ \dfrac{3}{2}x$$
Cost of one pen is $$Rs.\ 5$$
Hence, total C.P. of pens $$= 5x$$
Also, cost of one pencil is $$Rs.\ 1$$
Hence, total C.P. of pencils $$= \dfrac{3}{2}x$$
It is also given that, pens are sold at $$12\%$$ profit,
Hence, $$S.P. = 112\% \ of\ 5x $$
$$= \dfrac{112}{100} \times 5x$$
$$=\dfrac{28x}{5}$$
Also, pencils are sold at $$10\%$$ profit,
Hence, $$S.P. = 110\% \ of\ \dfrac{3}{2}x$$
$$ = \dfrac{330}{200}x$$
$$ = \dfrac{33x}{20}$$
Now, total sale is $$Rs.\ 725$$.
Hence, $$\dfrac{28x}{5} + \dfrac{33x}{20} = 725$$
$$112 x + 33x = 725 \times 20$$
$$145 x = 14500$$
$$x = 100$$
Thus, number of pens $$= 100$$
Number of pencils $$= \dfrac{3}{2} \times 100$$
$$ = 150$$
So, the number of pencils exceeds the number of pens by $$50$$.
Hence, option B is correct.
A candidate in an examination was asked to find $$\displaystyle \dfrac{5}{14}$$ of a certain number. By mistake he found $$\displaystyle \dfrac{5}{4}$$ of it. Thus his answer was $$25$$ more than the correct answer. What was the number?
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0%
$$28$$
0%
$$56$$
0%
$$65$$
0%
$$20$$
Explanation
Let the number be x
Then $$\displaystyle x\times \dfrac{5}{4}-x\times \dfrac{5}{14}=25$$
$$\displaystyle \Rightarrow \dfrac{35x-10x}{28}=25$$
$$\Rightarrow \dfrac{25x}{28}=25$$
$$\displaystyle \Rightarrow x=\dfrac{25\times 28}{25}=28$$
Solve for $$x$$: $$\displaystyle \frac{3x}{4}-\frac{(x-1)}{3}=\frac{(x-2)}{2}$$.
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0%
$$9$$
0%
$$14$$
0%
$$16$$
0%
$$11$$
Explanation
Given, $$\displaystyle \frac{3x}{4}-\frac{(x-1)}{3}=\frac{(x-2)}{2}$$
L.C.M. of denominator $$=12$$
$$\therefore $$ Multiplying each term by $$12$$, we get
$$\displaystyle 12\times \frac{3x}{4}-12\times \frac{(x-1)}{3}=12\times \frac{(x-2)}{2}$$
$$\Rightarrow 9x-4(x-1)=6(x-2)$$
$$\Rightarrow 9x-4x+4=6x-12$$
$$\Rightarrow 5x+4=6x-12$$
$$\Rightarrow x=16$$
Hence, required value of $$x$$ is $$16$$.
The solution of the the equation
$$5(3x+5)=2(7x-4)$$ is
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0%
$$x=17$$
0%
$$x=33$$
0%
$$x=-33$$
0%
$$x=-17$$
Explanation
Given, $$5(3x+5)=2(7x-4)$$
$$\therefore 15x+25=14x-8$$
Putting $$x$$ terms on one side and constants on another side.
$$\therefore 15x-14x=-8-25$$
$$\therefore x=-33$$
Hence, option C is correct.
If two third, one half and one seventh of a number is added to itself the result is $$37$$. Find the number.
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0%
$$\displaystyle 14\frac{2}{97}$$
0%
$$\displaystyle 16\frac{2}{97}$$
0%
$$\displaystyle 18\frac{2}{97}$$
0%
None
Explanation
Let,
The number be $$x$$
Two-third of a number$$=\dfrac{2}{3}x$$
One-half of a number $$=\dfrac{1}{2}x$$
One-seventh of a number $$=\dfrac{1}{7}x$$
A.T.Q,
$$\dfrac{2}{3}x+\dfrac{1}{2}x+\dfrac{1}{7}x+x=37$$
$$\implies \dfrac{28x+21x+6x+42x}{42}=37$$
$$\implies \dfrac{97}{42}x=37$$
$$\implies x=\dfrac{37 \times 42}{97}$$
$$\implies x=\dfrac{1554}{97}=16\dfrac{2}{97}$$
Answer : B
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