Simple Equations Important Question Paper

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Questions:

Generic 2

10 x 2 = 20

Find the mean of 6, 15, 120, 50, 100, 80, 10, 158, 10, 15.

Solve for y : \(\frac { y }{ 5 } +3=2\)

Solve the following equation: 6x + 18 = 8x + 12.

Solve: 4q-12 = 0

Set up equations and solve them to find the unknown numbers in the following cases: Anwar thinks of a number. If he takes away 7 from \(\frac { 5 }{ 2 } \) of the number, the result is \(\frac { 11 }{ 2 } \).

Solve the equation. \(\frac { 3p }{ 4 } =6\)

Solve the following equations by trial and error method. 6m + 2 = 14.

Solve the following equation. 10p + 10 = 100.

Write the following equations in statement form.
m - 7 = 3.

10.

Write at least one other form
5p = 20

Generic 3

5 x 3 = 15

11.

Solve the following equations:
16 = 4 + 3(t + 2)

12.

Solve the following equation:
0.6x + 0.8 = 0.56x/+ 2.32

13.

Solve for x: \(\frac { 2x-1 }{ 3 } -\frac { 6x-2 }{ 5 } =\frac { 1 }{ 3 } \)

14.

Solve the following- Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

15.

Complete the last column of the table.

S.No	Equation	Value
(i)	x+3 =0	x=3
(ii)	x+3 =0	x=0
(iii)	x+3 =0	x= -3
(iv)	x-7 =1	x=7
(v)	x-7 =1	x=8
(vi)	5x= 25	x=0
(vii)	5x= 25	x=5
(viii)	5x= 25	x= -5
(ix)	\(\frac { m }{ 3 } =2\)	m= -6
(x)	\(\frac { m }{ 3 } =2\)	m=0
(xi)	\(\frac { m }{ 3 } =2\)	m=6

Generic 5

3 x 5 = 15

16.

The three scales shown below are perfectly balanced if • = 3. What are the values of \(\Delta\) and *?

17.

In a Mathematics quiz, 30 prizes consisting of 1st and 2nd prizes only are to be given. 1st and 2nd prizes are worth Rs 2000 and Rs 1000, respectively. If the total prize money is Rs 52000 then show that
(a) If 1st prizes are x in number, then the number of 2nd prizes are _______.
(b) The total value of prizes in terms of x are ________.
(c) The equation formed is ______.
(d) The solution of the equation is _______.
(e) The number of 1st prizes are ________ and the number of 2nd prizes are ________.

18.

If 45 is added to half a number, then result is triple the number. Find the number.

Multiple Choice Question

10 x 1 = 10

19.

The solution of the equation 2s = 0 is

(a)

(b)

-2

(c)

(d)

\(1\over2\)

20.

The solution of the equation 2m = 4 is

(a)

(b)

(c)

-1

(d)

-2

21.

Write the following statement in the form of an equation:
Four times a number p is 8.

(a)

4p = 8

(b)

p+4=8

(c)

p-4=8

(d)

p\(\div\)4=8

22.

The product of a fraction and its reciprocal is:

(a)

(b)

-1

(c)

(d)

the fraction itself

23.

The value of the variable that we get on solving an equation is:

(a)

degree of the equation

(b)

root of the equation

(c)

co-efficient of the variable

(d)

none of these

24.

If 7x + 4 = 25, then x is equal to

(a)

\(\frac { 29 }{ 7 } \)

(b)

\(\frac { 100 }{ 7 } \)

(c)

(d)

25.

Five times of a number minus 4 gives 6. Find which of the following is the number?

(a)

(b)

(c)

(d)

26.

If sum of p and 4 is 15, P is :

(a)

(b)

(c)

(d)

27.

Which of the following numbers satisfies the equation -6 + x = -18?

(a)

(b)

-13

(c)

-12

(d)

-16

28.

If a and b are positive integers, then the solution of the equation ax = b will always be a

(a)

positive number

(b)

negative number

(c)

(d)

Fill up / 1 Marks

3 x 1 = 3

29.

If \(\frac{1}{x}\) + 1 = 2, then x =________________.

30.

If \(x-\frac { 7 }{ 8 } =\frac { 7 }{ 8 } \), then x = ___________

31.

x - 0 = ________, when 3x = 12.

True or False

1 x 1 = 1

32.

\(\frac { 3 }{ 2 } \) is the solution of the equation 8x - 5 =7.

(a)

False

(b)

True

Generic 4

2 x 4 = 8

33.

Divide 184 into two parts such that one third of one part may exceed one seventh of other part by 8.

34.

Ram's father is 49 years old. He is 4 years older than three times Ram's age. What is Ram's age.

1 Marks

1 x 1 = 1

35.

What happens to the equality of an equation, if the same quantity is removed from both sides?

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Answers Key:

Generic 2

10 x 2 = 20

56.4

\(\frac { y }{ 5 } +3=2\)
\(\Rightarrow \frac { y }{ 5 } =2-3\)
\(\Rightarrow \quad \frac { y }{ 5 } =-1\)
\(\Rightarrow \quad \frac { y }{ 5 } =-1\times 5\)
y = -5

Since,
6x + 18 = 8x + 12
\(\therefore\) 6x - 8x + 18 = 12
[On transposing 8x to LHS]
\(\Rightarrow\) 6x - 8x = 12 - 18
[On transposing 18 to RHS]
\(\Rightarrow\) -2x = -6
\(\Rightarrow \quad \frac { -2x }{ -2 } =\frac { -6 }{ -2 } \)
\(\Rightarrow\) x = 3

4q-12 = 0
\(\Rightarrow\) 4q = 12
Thus, q = 3

Let the number be x. Then, the required equation is \(\frac { 5n }{ 2 } -7=\frac { 11 }{ 2 } \)
Multiplying both sides by 2, we get
5n - 14 = 11
\(\Rightarrow\) 5n - 14 + 14 = 11 + 14
\(\Rightarrow\) 5n = 25 \(\Rightarrow \frac { 5n }{ 5 } =\frac { 25 }{ 5 } \)
\(\Rightarrow\) n = 5

We have, \(\frac { 3p }{ 4 } =6\)
\(\Rightarrow \quad \frac { 3p }{ 4 } \times \frac { 4 }{ 3 } =6\times \frac { 4 }{ 3 } \)
\(\Rightarrow \quad p=2\times 7=8\)
So, p = 8 is the solution of the given equation.

Given, 6m + 2 =14
Put m = 0 in 6m + 2 = 14, we get
(6 x 0) + 2 = 14
\(\Rightarrow\) 0 + 2 = 14
\(\Rightarrow\) 2 = 14
Now, put m = 1 in 6m + 2 = 14, we get
(6 x 1) + 2 = 14 \(\Rightarrow\) 6 + 2 = 14 \(\Rightarrow\) 8 = 14
Now. put m = 2 in 6m + 2 = 14, we get
(6 x 2) + 2 = 14 \(\Rightarrow\) 12 + 2 = 14 \(\Rightarrow\) 14 = 14
So, m = 2 is the solution of the given equation 6m + 2 = 14.

We have, 10p + 10 = 100
On subtracting 10 from both sides, we get
10p + 10 - 10 = 100 - 10 \(\Rightarrow\) 10p = 90
On dividing both sides by 10, we get
\(\frac{10p}{10}=\frac{90}{10}\quad \Rightarrow\) p = 9
Hence, p = 9 is the solution of the given equation.

The given equations in statement form are as follows:
The difference of m and 7 is 3.

10.

Other forms for equation 5p = 20 are as follows:
(a) Five times a number p is equal to 20.
(b) Multiply a number p by 5 to get 20.

Generic 3

5 x 3 = 15

11.

Interchanging the sides, we have
4 + 3(t + 2) = 16
or 3(t+2)=16-4=12
[Transposing 4 to RH.S.]
or \(\frac{3(t+2)}{3}=\frac{12}{3}\)
[Dividing both sides by 3]
or t + 2 = 4
or t= 4 - 2
[Transposing 2 to RH.S.]
or t = 2
Thus, t = 2 is the required solution.

12.

We have:
0.6x + 0.8 = 0.56x + 2.32
Transposing 0.8 to R.H.S. and 0.56 to L.H.S.,
we have
0.6x - 0.56x = 2.32 - 0.8
\(\Rightarrow \) 0.60x - 0.56x = 2.32 - 0.80
\(\Rightarrow \)(0.60 - 0.56)x = (2.32 - 0.80)
\(\Rightarrow \)0.04x = 1.52
\(\Rightarrow \) x=\(\frac{1.52}{0.04}\)
\(\Rightarrow \) x=38
Thus, the required solution of the given equation is x = 38.

13.

Since,
\(\frac { 2x-1 }{ 3 } -\frac { 6x-2 }{ 5 } =\frac { 1 }{ 3 } \)
\(\therefore \quad \frac { 5(2x-1) }{ 3\times 5 } -\frac { 3(6x-2) }{ 3\times 5 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 10x-5 }{ 15 } -\frac { (18x-6) }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 10x-5-18x-6 }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { -18x+1 }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \) -18x+1=\(\frac { 15 }{ 3 } =5\)
\(\Rightarrow\) -18x = 5 - 1 = 4
\(\Rightarrow \quad x=\frac { 4 }{ -18 } \)
Thus, x=\(\frac { -2 }{ 9 } \)

14.

Let runs scored by Rahul be x.
Then, runs scored by Sachin = Twice of runs scored by Rahul = 2x
\(\therefore\) Sum of their runs = x + 2x = 3x
Since the sum of their runs be two short of a double century.
Therefore, we get the equation
3x + 2 = 200
To solve this equation, transposing (+ 2) from LHS to RHS, we get
3x = 200 - 2 \(\Rightarrow\) 3x = 198
On dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{198}{3}\)
\(\Rightarrow\) x = 66
Hence, the runs scored by Rahul is 66 and by Sachin = 2 x 66 = 132.

15.

S.No	Equation	Value	Value of LHS	Result
(i)	x+3 =0	x=3	x+3=3+3=6≠RHS	No
(ii)	x+3 =0	x=0	x+3=0+3=3≠RHS	No
(iii)	x+3 =0	x= -3	x+3=-3+3=0=RHS	Yes
(iv)	x-7 =1	x=7	x-7=7-7=0≠RHS	No
(v)	x-7 =1	x=8	x-7=-8-7=1=RHS	Yes
(vi)	5x= 25	x=0	5x=5\(\times\)0=0 ≠RHS	No
(vii)	5x= 25	x=5	5x=5\(\times\)5=25≠RHS	Yes
(viii)	5x= 25	x= -5	5x=5\(\times\)(-5)=-25≠RHS	No
(ix)	\(\frac { m }{ 3 } =2\)	m= -6	=\(\frac { m }{ 3 } =\frac { -6 }{ 3 } \) -2 ≠RHS	No
(x)	\(\frac { m }{ 3 } =2\)	m=0	\(\frac { m }{ 3 } =\frac { -6 }{ 3 } \)=0 ≠RHS	No
(xi)	\(\frac { m }{ 3 } =2\)	m=6	\(\frac { m }{ 3 } =\frac { 6 }{ 3 } \)= 2=RHS	Yes

Generic 5

3 x 5 = 15

16.

and given. = 3
From (a), y + y + y + y + y = x + x + 3 + 3
\(\Rightarrow\) 5y = 2x + 6 \(\Rightarrow\) 5y - 2x = 6
\(\Rightarrow\) 2x - 5y = - 6 ...(i)
From (b), x + x = y + y + 3 + 3
\(\Rightarrow\) 2x = 2y + 6 \(\Rightarrow\) 2x - 2y = 6
\(\Rightarrow\) x - y = 3 [dividing both sides by 2] ...(ii)
From (c), y + y + y + 3 + 3 + 3 = x + x + x
\(\Rightarrow\) 3 y + 9 = 3x \(\Rightarrow\) 3x - 3y = 9
\(\Rightarrow\) x - y = 3 [dividing both sides by 3] ...(iii)
From Eq. (iii), x - y = 3 \(\Rightarrow\) x = y + 3
On putting x = y + 3 in Eq. (i). we get
2(Y + 3) - 5Y = -6 \(\Rightarrow\) 2y + 6 - 5Y = -6
-3y + 6 = - 6 \(\Rightarrow\) -3 y = - 6 - 6 = -12
\(y=\frac{12}{3}=4\)
On putting y = 4 in Eq. (ii), we get
x - y = 3 \(\Rightarrow\) x - 4 = 3
\(\Rightarrow\) x = 3 + 4 = 7 \(\Rightarrow\) x = 7
:. The value of \(\Delta\) = x = 7 and the value of • = y = 4.

17.

Given, number of prizes = 30
Total prize money = Rs 52000
and 1st and 2nd prizes are worth Rs 2000 and Rs 1000 respectively.
(a) If 1stprizes are x in number, the number of 2nd prizes are 30 - x because total number of prizes are 30.
(b) The total value of prizes in terms of x are 2000 x + 1000 (30 - x).
(c) The equation formed is
1000x + 30000 = 52000
\(\therefore\) From (b) 2000x + 1000 (30 - x) = 52000
2000x + 30000 -1000x = 52000
1000x + 30000 = 52000
(d) The solution of the equation is 52.
\(\therefore\) From (c), 1000x + 30000 = 52000
1000x = 52000 - 30000 = 22000
\(x=\frac{22000}{1000}=22\)
(e) The number of 1st prizes are 22 and the number of 2nd prizes are 8.
\(\therefore\) From (b), 2000x + 1000(30 - x) = 52000
2 x + 30 - x = 52 [dividing both sides by 1000]
x + 30 = 52 \(\Rightarrow\) x = 52 - 30 = 22
\(\therefore\) Number of 2nd prizes = 30 - 22 = 8.

18.

Let x be the number. So, half of x is \(\frac{x}{2}\)
Then, \(\frac{x}{2}+45=3x\Rightarrow \frac{x+90}{2}=3x\)
x + 90 = 6x \(\Rightarrow\) 90 = 5x \(\Rightarrow \quad x=\frac{90}{5}=18\)
Hence, the number is 18.

Multiple Choice Question

10 x 1 = 10

19.

\(2s=0 \Rightarrow s={0\over2}=0\)

20.

2m = 4\(\Rightarrow m={4\over2}=2\)

21.

(a)
4p = 8

22.

23.

(b)
root of the equation

24.

(d)
3

25.

(b)
2

26.

27.

28.

(a)
positive number

Fill up / 1 Marks

3 x 1 = 3

29.

()
1

30.

()
\(\frac { 7 }{ 4 } \)

31.

If 3x = 12 \(\Rightarrow x=\frac{12}{3}=4\)
x - 0 = 4, when 3x = 12.

True or False

1 x 1 = 1

32.

(a)
False

Generic 4

2 x 4 = 8

33.

Let one part of 184 be x.
\(\therefore\) Other part be (184 - x)
Now, according to question,
\(\Rightarrow \quad \frac { 1 }{ 3 } x-\frac { 1 }{ 7 } (184-x)=8\)
\(\Rightarrow \quad \frac { x }{ 3 } +\frac { x }{ 7 } -\frac { 184 }{ 7 } =8\)
\(\Rightarrow \quad \frac { 7x+3x }{ 21 } =8+\frac { 184 }{ 7 } \)
\(\Rightarrow \quad \frac { 10x }{ 21 } =\frac { 56+184 }{ 7 } \)
\(\Rightarrow \quad \frac { 10x }{ 21 } =\frac { 240 }{ 7 } \)
\(\Rightarrow \quad x=\frac { 21\times 240 }{ 7\times 10 } \)
= 72
Hence, parts are 72 and 184 - 72 = 112.

34.

Age of Ram's father = 49 years
Let the age of Ram be x years
\(\therefore\) 3x + 4 = 49
\(\Rightarrow\) 3x = 49 - 4
\(\Rightarrow\) 3x = 45
\(\Rightarrow\) \(x=\frac { 45 }{ 3 } \)
x = 15
\(\Rightarrow\) Ram's age = 15 years