Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 27

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Question 1
1 / -0

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?

A
$$10$$

B
$$19$$

C
$$14$$

D
$$12$$

Solution

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 $$= 19$$.
Question 2
1 / -0

If $$\displaystyle x=m+1$$ and $$\displaystyle \frac{1}{3}(6x-3)-(8-3x)=11;$$ find the value of $$m$$.

A
$$m=7$$

B
$$m=6$$

C
$$m=2$$

D
$$m=3$$

Solution

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 $$3$$.
Question 3
1 / -0

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.

A
$$15$$

B
$$18$$

C
$$22$$

D
$$25$$

Solution

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.
Question 4
1 / -0

If the sum of three consecutive integers is $$63$$. Find the smallest number.

A
$$11$$

B
$$15$$

C
$$23$$

D
$$20$$

Solution

Let the smallest integer be $$x$$.
The three consecutive numbers $$= x, x+1, x+2$$
Given, $$x + x+1 + x +2 = 63$$
$$ 3x + 3 = 63$$
$$ 3x = 60$$
$$\therefore x =20$$
Question 5
1 / -0

Solve: $$8x+\displaystyle \frac{21}{4}= 3x+7$$ , then $$x=$$

A
$$\displaystyle \frac{2}{9}$$

B
$$\displaystyle \frac{7}{20}$$

C
$$\displaystyle \frac{6}{17}$$

D
$$\displaystyle \frac{4}{21}$$

Solution

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.
$$ \\ 20x = 7$$
$$ \therefore x = \cfrac { 7 }{ 20 }
$$
Question 6
1 / -0

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.

A
$$11$$

B
$$8$$

C
$$13$$

D
$$3$$

Solution

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.
Question 7
1 / -0

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.

A
$$25$$

B
$$15$$

C
$$28$$

D
$$12$$

Solution

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.
Question 8
1 / -0

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. $$9$$.

A
$$40$$

B
$$20$$

C
$$500$$

D
$$300$$

Solution

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 $$= 8x = 40$$.
Question 9
1 / -0

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.

A
$$39$$cm

B
$$45$$cm

C
$$80$$cm

D
$$40$$cm

Solution

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.
Question 10
1 / -0

If $$(2ax + 1) (3x + 1) = 6a (x + 1)$$ and $$x = 1$$, find the value of $$a$$.

A
$$1$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

$$(2ax + 1) (3x + 1) = 6a (x + 1) $$
$$\implies[2a(1)+1][3(1)+1]=6a[(1+1)]$$
$$\implies (2a+1)(3+1)=6a(2)$$
$$\implies (2a+1)(4)=12a$$
$$\implies 8a+4=12a$$
$$\implies 8a-12a=-4$$
$$\implies -4a=-4$$
$$\implies a=1$$

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Linear Equations in One Variable Test - 27

Result

Linear Equations in One Variable Test - 27

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SHARING IS CARING

Question 1

$$10$$

$$19$$

$$14$$

$$12$$

Solution

Question 2

$$m=7$$

$$m=6$$

$$m=2$$

$$m=3$$

Solution

Question 3

$$15$$

$$18$$

$$22$$

$$25$$

Solution

Question 4

$$11$$

$$15$$

$$23$$

$$20$$

Solution

Question 5

$$\displaystyle \frac{2}{9}$$

$$\displaystyle \frac{7}{20}$$

$$\displaystyle \frac{6}{17}$$

$$\displaystyle \frac{4}{21}$$

Solution

Question 6

$$11$$

$$8$$

$$13$$

$$3$$

Solution

Question 7

$$25$$

$$15$$

$$28$$

$$12$$

Solution

Question 8

$$40$$

$$20$$

$$500$$

$$300$$

Solution

Question 9

$$39$$cm

$$45$$cm

$$80$$cm

$$40$$cm

Solution

Question 10

$$1$$

$$4$$

$$3$$

$$2$$

Solution

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