Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

A piece of string is $$80$$ cm long. It is cut into three pieces. The longest piece is $$3$$ times as long as the middle sized piece and the shortest piece is $$46$$ cm shorter than the longest piece. Find the length of the shortest piece.

A
$$14$$ cm

B
$$10$$ cm

C
$$8$$ cm

D
$$18$$ cm

Solution

Let the length of middle piece $$=$$ $$x$$
Then length of longest piece $$=$$ $$3x$$
Then length of shortest piece $$=$$ $$3x - 46$$
Thus, $$x + 3x + 3x - 46 = 80$$
$$\Rightarrow 7x = 126$$
$$\Rightarrow x = 18$$
Hence, shortest piece $$=$$ $$18 \times 3 - 46 = 54 - 46 = 8$$ cm
Question 2
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The sum of one half, one third and one fourth of a number exceed the number itself by $$12$$. The number is

A
$$72$$

B
$$144$$

C
$$180$$

D
$$244$$

Solution

Let the number be $$x$$
$$\Rightarrow\;\begin{bmatrix}\displaystyle\frac{x}{2}+\displaystyle\frac{x}{3}+\displaystyle\frac{x}{4}\end{bmatrix}-x=12$$
$$\Rightarrow\;\displaystyle\frac{6x+4x+3x-12x}{12}=12$$
$$\Rightarrow\;x=144$$
Question 3
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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

A
a multiple of $$5$$

B
a multiple of $$6$$

C
a multiple of $$7$$

D
divisible by $$8$$

Solution

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.
Question 4
1 / -0

When $$19$$ is subtracted from the product of $$p$$ and $$4$$, the result is $$17$$.
Then, the value of $$p$$ is

A
$$\displaystyle \dfrac { -1 }{ 9 } $$

B
$$\displaystyle \dfrac { 1 }{ 9 } $$

C
$$\displaystyle -9$$

D
$$\displaystyle 9$$

Solution

According to the question, $$\displaystyle 4p-19=17$$
Adding $$19$$ to both sides we get,
$$\Longrightarrow \displaystyle 4p=17+19$$
$$\Longrightarrow\displaystyle 4p=36$$

$$\Longrightarrow \displaystyle p=\frac { 36 }{ 4 } $$ ...(Divide by $$4$$ on both sides)

$$\therefore \displaystyle p=9$$

So, option D is correct.
Question 5
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The perimeter of an isosceles triangle is $$\displaystyle 3\frac { 9 }{ 4 } $$ cm. The base of an isosceles triangle is $$\dfrac{3}{2}$$ cm. What is the length of either of the remaining equal sides?

A
$$\displaystyle \frac { 15 }{ 4 } $$ cm

B
$$\displaystyle \frac { 15 }{ 8 } $$ cm

C
$$\displaystyle \frac { 8 }{ 15 } $$ cm

D
$$\displaystyle \frac { 4 }{ 15 } $$ cm

Solution

Perimeter = $$\displaystyle 3\frac { 9 }{ 4 } =\frac { 21 }{ 4 } $$ cm

Property of an isosceles triangle = two sides are equal

Perimeter of triangle = Sum of three sides
$$\therefore \displaystyle \frac { 21 }{ 4 } =a+a+\frac { 3 }{ 2 } $$

$$\Longrightarrow\displaystyle 2a=\frac { 21 }{ 4 } -\frac { 3 }{ 2 } $$

$$\Longrightarrow 2a=\dfrac { 21-6 }{ 4 } $$

$$\therefore a= \dfrac { 15 }{ 8 } $$
Question 6
1 / -0

The sum of three consecutive multiples of $$9$$ is $$135$$, then the multiples are

A
$$36, 45, 54$$

B
$$45, 54, 63$$

C
$$27, 36, 45$$

D
None of the above.

Solution

If $$p$$ is a multiple of $$9$$, then next two multiples are $$\displaystyle p+9$$ and $$p+18$$
According to the question:
$$\Longrightarrow \displaystyle p+(p+9)+(p+18)=135$$
$$\Longrightarrow\displaystyle 3p+27=135$$
$$\Longrightarrow \displaystyle 3p=135-27$$
$$\Longrightarrow \displaystyle 3p=108$$
$$\Longrightarrow \displaystyle p=36$$
Thus,
$$p=36$$
$$p+9=45$$
$$p+18=54$$
Hence the required multiples are $$36,45,54$$
Question 7
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In a division, a student took $$63$$ as divisor instead of $$36$$. His answer was $$24$$. The correct answer is -

A
$$42$$

B
$$32$$

C
$$48$$

D
$$28$$

E
$$38$$

Solution

From the given condition we have,
$$\dfrac {x}{63} = 24$$
$$x = 24 \times 63$$
So, correct answer would be,

$$\dfrac {24 \times 63}{36} = 42$$
Question 8
1 / -0

When a number is divided by $$4$$, the result is $$-9$$. The number is

A
$$36$$

B
$$-36$$

C
$$\dfrac{1}{36}$$

D
$$\dfrac{-1}{36}$$

Solution

Let the required number be $$x$$

According to the question, $$\displaystyle \frac { x }{ 4 } =-9\\ $$
$$\Longrightarrow\displaystyle x=-9\times 4$$ ...(Multiply both sides by $$4$$)
$$\Longrightarrow\displaystyle x=-36$$

So, option B is correct.
Question 9
1 / -0

In a contest there were $$10$$ problems. Each correct answer is eligible for $$5$$ points while each incorrect answer is penalized with a deduction of $$2$$ points. Venkat answered all the questions but got only $$29$$ points. The number of correct answers of Venkat was

A
$$5$$

B
$$2$$

C
$$7$$

D
$$4$$

Solution

Let the correct answer of Venkat be x. Then
$$5x -2(10 -x) =29$$
or $$5x -20 + 2x = 29$$
or $$7x=49$$
or $$x=7$$
Question 10
1 / -0

In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 75 questions and secures 125 marks, the number of questions he attempted correctly, is

A
35

B
40

C
42

D
46

Solution

If the number of correct answer be x, then
the number of incorrect answer is $$\displaystyle (75- x)$$.
$$\displaystyle \therefore \quad 4x-\left( 75-x \right) =125\quad or\quad x=40$$

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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 35

Result

Linear Equations in One Variable Test - 35

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SHARING IS CARING

Question 1

$$14$$ cm

$$10$$ cm

$$8$$ cm

$$18$$ cm

Solution

Question 2

$$72$$

$$144$$

$$180$$

$$244$$

Solution

Question 3

a multiple of $$5$$

a multiple of $$6$$

a multiple of $$7$$

divisible by $$8$$

Solution

Question 4

$$\displaystyle \dfrac { -1 }{ 9 } $$

$$\displaystyle \dfrac { 1 }{ 9 } $$

$$\displaystyle -9$$

$$\displaystyle 9$$

Solution

Question 5

$$\displaystyle \frac { 15 }{ 4 } $$ cm

$$\displaystyle \frac { 15 }{ 8 } $$ cm

$$\displaystyle \frac { 8 }{ 15 } $$ cm

$$\displaystyle \frac { 4 }{ 15 } $$ cm

Solution

Question 6

$$36, 45, 54$$

$$45, 54, 63$$

$$27, 36, 45$$

None of the above.

Solution

Question 7

$$42$$

$$32$$

$$48$$

$$28$$

$$38$$

Solution

Question 8

$$36$$

$$-36$$

$$\dfrac{1}{36}$$

$$\dfrac{-1}{36}$$

Solution

Question 9

$$5$$

$$2$$

$$7$$

$$4$$

Solution

Question 10

35

40

42

46

Solution

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