Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Twenty years ago, my age was $$\left (\displaystyle \frac{1}{3}\right)$$rd of what it is now. What is my present age?

A
$$66$$ years

B
$$30$$ years

C
$$33$$ years

D
$$36$$ years

Solution

Let the present age be $$x$$ years
$$20$$ years ago, age will be $$x-20$$
As per the given condition, we have
$$x\, -\, 20\, =\, \displaystyle \frac{x}{3}$$
$$x-\dfrac {x}{3}=20$$
$$\displaystyle \frac{2x}{3}\, =\, 20$$
$$x = 30$$ years
Therefore, present age is $$30$$ years.
Question 2
1 / -0

The number of variables in a simple linear equation is

A
Two

B
One

C
0

D
None

Solution

A linear equation can comprise of many variables. The most simple linear equation is in $$one$$ variable ,i.e, $$ax+b=0$$.
Question 3
1 / -0

In $$\displaystyle \frac {a}{8}\, +\, \displaystyle \frac {a}{4}\, =\, 6$$, the value of $$a$$ is:

A
$$122$$

B
$$-16$$

C
$$16$$

D
$$0$$

Solution

$$\displaystyle \frac {a+2a}{8}\, =\, 6$$
$$\Rightarrow \displaystyle \frac {3a}{8}\, =\, 6$$
$$\Rightarrow 3a= 48 \Rightarrow a = 16$$
Hence the solution is $$a=16$$
Question 4
1 / -0

The length of a rectangle is $$12$$ m and its area is $$72 m^2$$. Then the breadth equals to

A
$$72$$ m

B
$$60$$ m

C
$$6$$ m

D
$$9$$ m

Solution

Let $$l,b$$ and $$A$$ are length, breadth and area of the rectangle respectively.
Then given , $$l= 12$$ m, $$A = 72$$ sq.m
Now we know, area of rectangle $$=lb$$
$$\Rightarrow 12b=72$$
$$\Rightarrow b = \cfrac {72}{12}\, =\, 6\,$$ m
Hence, breadth of the given rectangle is $$6$$ m.
Question 5
1 / -0

A person was asked to state his age in years. His reply was, take my age three years hence multiply it by $$3$$ and then subtract three times my age three years ago and you will know how old I am. What was the age of the person?

A
$$24$$ years

B
$$20$$ years

C
$$32 $$ years

D
$$18$$ years

Solution

Let the present age of the person be $$x$$ years
His age after $$3$$ years hence $$= (x + 3)$$ years
His age $$3$$ years ago $$= (x - 3)$$ years
Then according to given condition, his present age will be $$ (x+3) 3 - 3(x - 3)$$
$$= (3x + 9 - 3x + 9)=18$$ years.
Question 6
1 / -0

A farmer divided his herd of x cows among his 4 sons so that one son gets one half of the herd, the second gets one-fourth, the third son gets one-fifth and the fourth gets 7 crows . Then x is

A
100

B
140

C
0

D
1

Solution

Given that:
No. of cows $$=x$$
According to the question,
$$\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{5}+7=x$$

$$\dfrac{10x+5x+4x}{20}+7=x$$
$$19x+140=20x$$
$$x=140.$$
Hence, no. of cows are $$140.$$
Question 7
1 / -0

If the sum of $$\displaystyle\frac{1}{3}$$ and $$\displaystyle\frac{1}{4}$$ is x times the difference of $$\displaystyle\frac{1}{3}$$ and $$\displaystyle\frac{1}{4}$$, then the value of x is equal to

A
4

B
5

C
6

D
7

Solution

Sum of $$\frac { 1 }{ 3 } $$ and $$\frac { 1 }{ 4 } $$ is x times the difference of $$\frac { 1 }{ 3 } $$ and $$\frac { 1 }{ 4 } $$
As per problem,
$$\Rightarrow \left( \frac { 1 }{ 3 } +\frac { 1 }{ 4 } \right) =\left( \frac { 1 }{ 3 } -\frac { 1 }{ 4 } \right)x $$
$$\Rightarrow \left( \frac { 4+3 }{ 12 } \right) =\left( \frac { 4-3 }{ 12 } \right) x$$
$$\Rightarrow \frac { 7 }{ 12 } =\frac { 1 }{ 12 } x$$
$$x=7$$
Question 8
1 / -0

Anita had to do a multiplication. Instead of taking 35 as one of the multipliers she took 53. As a result the product went up by 540. What is the new product?

A
1050

B
540

C
1440

D
1590

Solution

Let the number that Anita wanted to multiply be 'X'.
She wanted to find the value of 35X.
Instead, she found the value of 53X.
The difference between the value that she got (53X) and what she was expected to get (35X) is 540.
Then $$53x-35x=540$$
$$18 x=540$$
$$x=30$$
$$\therefore $$ correct product is $$53\times 30=1590$$
Question 9
1 / -0

A man has 480 rupees in the denominations of one-rupee, five rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?

A
90

B
75

C
45

D
60

Solution

Let there be x notes of each kind
$$\implies1.x+5.x+10.x=480$$
$$\therefore x=30$$
Total number of notes $$=x+x+x=3x=90$$
Question 10
1 / -0

The sum of a number and its half is $$84$$. Number will be-

A
$$65$$

B
$$56$$

C
$$66$$

D
None of these

Solution

Let the number be $$x$$
According to the question, the number$$+\cfrac{1}{2}$$ of the number $$=84$$
$$\Rightarrow$$ $$x+\cfrac{x}{2}=84$$
$$\Rightarrow$$ $$\cfrac{2x+x}{2}=84$$
$$\Rightarrow$$ $$3x=84\times 2$$
$$\Rightarrow$$ $$x=\cfrac{84\times 2}{3}$$
$$\Rightarrow$$ $$x=56$$
Hence the number is $$56$$

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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 19

Result

Linear Equations in One Variable Test - 19

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SHARING IS CARING

Question 1

$$66$$ years

$$30$$ years

$$33$$ years

$$36$$ years

Solution

Question 2

Two

One

0

None

Solution

Question 3

$$122$$

$$-16$$

$$16$$

$$0$$

Solution

Question 4

$$72$$ m

$$60$$ m

$$6$$ m

$$9$$ m

Solution

Question 5

$$24$$ years

$$20$$ years

$$32 $$ years

$$18$$ years

Solution

Question 6

100

140

0

1

Solution

Question 7

4

5

6

7

Solution

Question 8

1050

540

1440

1590

Solution

Question 9

90

75

45

60

Solution

Question 10

$$65$$

$$56$$

$$66$$

None of these

Solution

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