Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 30

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Question 1
1 / -0

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:

A
$$29$$

B
$$31$$

C
$$25$$

D
$$27$$

Solution

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 $$27$$.
Question 2
1 / -0

One number is 3 less than the two times of the other; if their sum is increased by 7 the result is 37. Find the numbers.

A
$$9 , 11$$

B
$$11 , 13$$

C
$$11 , 19$$

D
$$9 , 13$$

Solution

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
Question 3
1 / -0

The solution of the the equation
$$5(3x+5)=2(7x-4)$$ is

A
$$x=17$$

B
$$x=33$$

C
$$x=-33$$

D
$$x=-17$$

Solution

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.
Question 4
1 / -0

If $$\displaystyle \frac{7x+2}{6}-\frac{x-4}{3}=\frac{x+3}{3}$$, then $$x=$$

A
$$\displaystyle -\frac{4}{3}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{-1}{3}$$

D
$$\displaystyle 1\frac{1}{3}$$

Solution

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
$$3x=-4$$
$$\therefore$$ $$x=\dfrac {-4}{3}$$
Question 5
1 / -0

The sum of 4 consecutive integers is 70. Then the greatest among them is:

A
19

B
23

C
17

D
16

Solution

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 $$19$$
Question 6
1 / -0

The total cost of three prizes is Rs. 2550. 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.

A
Rs. $$900$$

B
Rs. $$1500$$

C
Rs. $$1200$$

D
Rs. $$450$$

Solution

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 $$Rs.1200$$
Question 7
1 / -0

Solve for $$x$$: $$\displaystyle \frac{1}{4}(5x+4)=\frac{1}{3}(2x-1)$$

A
$$\displaystyle \frac{-16}{7}$$

B
$$\displaystyle \frac{10}{7}$$

C
$$\displaystyle \frac{8}{7}$$

D
$$\displaystyle \frac{-15}{7}$$

Solution

Given, $$\cfrac14(5x+4)=\cfrac13(2x-1)$$
$$\Rightarrow 3(5x+4)=4(2x-1)$$ ($$\text{cross multiply}$$)
$$\Rightarrow 15x+12=8x-4$$
$$\Rightarrow 15x-8x=-4-12$$
$$\Rightarrow 7x=-16$$
$$\Rightarrow x=-\cfrac{16}7$$
Question 8
1 / -0

$$\displaystyle \frac{1}{5}$$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

A
$$\displaystyle 5\frac{5}{11}m$$

B
$$\displaystyle \frac{60}{11}cm$$

C
$$5 km$$

D
None

Solution

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 $$=5\dfrac{5}{11}m$$
Question 9
1 / -0

$$\displaystyle \frac{1}{2}$$ 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 to 10. Find the number

A
$$\displaystyle \frac{3}{5}$$

B
$$\displaystyle \frac{7}{4}$$

C
$$\displaystyle \frac{6}{7}$$

D
$$\displaystyle \frac{2}{3}$$

Solution

Let the number be $$x$$
Given $$\displaystyle \left \{ \left ( x-\frac{1}{2} \right )\times4+25\right \}\div 3=10$$
$$\displaystyle \left \{ 4x-2+25 \right \}\times\frac{1}{3}=10$$
$$\displaystyle \left ( \frac{4x+23}{3} \right )=10$$
$$ 4x+23=30$$
$$\displaystyle 4x=7$$
$$\Rightarrow x=\frac{7}{4}$$
Question 10
1 / -0

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 ?

A
Rs. $$25$$

B
Rs. $$10$$

C
Rs. $$35$$

D
Rs. $$40$$

Solution

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. $$35$$.

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Linear Equations in One Variable Test - 30

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Linear Equations in One Variable Test - 30

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Question 1

$$29$$

$$31$$

$$25$$

$$27$$

Solution

Question 2

$$9 , 11$$

$$11 , 13$$

$$11 , 19$$

$$9 , 13$$

Solution

Question 3

$$x=17$$

$$x=33$$

$$x=-33$$

$$x=-17$$

Solution

Question 4

$$\displaystyle -\frac{4}{3}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{-1}{3}$$

$$\displaystyle 1\frac{1}{3}$$

Solution

Question 5

19

23

17

16

Solution

Question 6

Rs. $$900$$

Rs. $$1500$$

Rs. $$1200$$

Rs. $$450$$

Solution

Question 7

$$\displaystyle \frac{-16}{7}$$

$$\displaystyle \frac{10}{7}$$

$$\displaystyle \frac{8}{7}$$

$$\displaystyle \frac{-15}{7}$$

Solution

Question 8

$$\displaystyle 5\frac{5}{11}m$$

$$\displaystyle \frac{60}{11}cm$$

$$5 km$$

None

Solution

Question 9

$$\displaystyle \frac{3}{5}$$

$$\displaystyle \frac{7}{4}$$

$$\displaystyle \frac{6}{7}$$

$$\displaystyle \frac{2}{3}$$

Solution

Question 10

Rs. $$25$$

Rs. $$10$$

Rs. $$35$$

Rs. $$40$$

Solution

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