Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of $$y$$ in the equation :
$$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$$

A
$$18$$

B
$$9$$

C
$$-38$$

D
$$-32$$

Solution

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 $$y=-38$$
Question 2
1 / -0

The sum of one-half, one third and one-fourth of a number exceeds the number by $$12$$. What is the number?

A
$$72$$

B
$$144$$

C
$$98$$

D
$$156$$

Solution

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 $$144.$$
Question 3
1 / -0

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

A
$$\dfrac{5}{3}$$

B
$$\dfrac{12}{29}$$

C
$$\dfrac{37}{10}$$

D
None of these

Solution

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 $$x=\cfrac{37}{10}$$.
Question 4
1 / -0

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

A
$$25$$

B
$$50$$

C
$$75$$

D
$$90$$

Solution

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.
Question 5
1 / -0

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?

A
$$28$$

B
$$56$$

C
$$65$$

D
$$20$$

Solution

Let the number be x
Then $$\displaystyle x\times \dfrac{5}{4}-x\times \dfrac{5}{14}=25$$

$$\displaystyle \Rightarrow \dfrac{35x-10x}{28}=25$$

$$\Rightarrow \dfrac{25x}{28}=25$$

$$\displaystyle \Rightarrow x=\dfrac{25\times 28}{25}=28$$
Question 6
1 / -0

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

A
$$9$$

B
$$14$$

C
$$16$$

D
$$11$$

Solution

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 $$16$$.
Question 7
1 / -0

Out of six consecutive numbers the sum of first three is $$27$$. What is the sum of next three?

A
$$30$$

B
$$40$$

C
$$36$$

D
$$45$$

Solution

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 $$= 11 + 12 + 13 = 36$$.
Question 8
1 / -0

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?

A
$$6$$

B
$$8$$

C
$$16$$

D
$$24$$

E
$$36$$

Solution

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 $$=4\times 6=24$$
Question 9
1 / -0

A person travelled $$\displaystyle \frac{5}{8}$$th of the distance by train $$\displaystyle \frac{1}{4}$$th by bus and the remaining 15 km by boat .The total distance travelled by him was

A
$$90$$ km

B
$$120$$ km

C
$$150$$ km

D
$$180 $$km

Solution

Let total distance be x km
$$\displaystyle x-\left ( \frac{5x}{8}+\frac{x}{4} \right )=15$$
$$\displaystyle x-\frac{7x}{8}=15$$
x = 120 km
Question 10
1 / -0

If two third, one half and one seventh of a number is added to itself the result is $$37$$. Find the number.

A
$$\displaystyle 14\frac{2}{97}$$

B
$$\displaystyle 16\frac{2}{97}$$

C
$$\displaystyle 18\frac{2}{97}$$

D
None

Solution

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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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 29

Result

Linear Equations in One Variable Test - 29

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SHARING IS CARING

Question 1

$$18$$

$$9$$

$$-38$$

$$-32$$

Solution

Question 2

$$72$$

$$144$$

$$98$$

$$156$$

Solution

Question 3

$$\dfrac{5}{3}$$

$$\dfrac{12}{29}$$

$$\dfrac{37}{10}$$

None of these

Solution

Question 4

$$25$$

$$50$$

$$75$$

$$90$$

Solution

Question 5

$$28$$

$$56$$

$$65$$

$$20$$

Solution

Question 6

$$9$$

$$14$$

$$16$$

$$11$$

Solution

Question 7

$$30$$

$$40$$

$$36$$

$$45$$

Solution

Question 8

$$6$$

$$8$$

$$16$$

$$24$$

$$36$$

Solution

Question 9

$$90$$ km

$$120$$ km

$$150$$ km

$$180 $$km

Solution

Question 10

$$\displaystyle 14\frac{2}{97}$$

$$\displaystyle 16\frac{2}{97}$$

$$\displaystyle 18\frac{2}{97}$$

None

Solution

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