7th Standard CBSE Syllabus & Materials
7th Standard CBSE
CBSE 7th Social Science Theme E - Understanding Market - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme E - From Barter to Money - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme D - The Constitution of India- An Introduction - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme D - From the Rulers to the Ruled : Types of Governments - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme B - The Age of Reorganisation - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme B - The Rise of Empires - New Sample Question Papers Study Material - QB365 Set A
Published on: 05/03/2019
Simple Equations Important Question Paper
Download CBSE Class 7th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 7th Standard CBSE Mathematics
Questions + Answers key
Take MCQ Mathematics Test
1.
Find the mean of 6, 15, 120, 50, 100, 80, 10, 158, 10, 15.
2.
Solve for y : \(\frac { y }{ 5 } +3=2\)
3.
Solve the following equation: 6x + 18 = 8x + 12.
4.
Solve: 4q-12 = 0
5.
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 \(\frac { 11 }{ 2 } \).
6.
Solve the equation. \(\frac { 3p }{ 4 } =6\)
7.
Solve the following equations by trial and error method. 6m + 2 = 14.
8.
Solve the following equation. 10p + 10 = 100.
9.
Write the following equations in statement form.
m - 7 = 3.
10.
Write at least one other form
5p = 20
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 | Say,whether the equation is satisfied (Yes/No) |
|---|---|---|---|
| (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 |
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.
19.
The solution of the equation 2s = 0 is
2
-2
0
\(1\over2\)
20.
The solution of the equation 2m = 4 is
1
2
-1
-2
21.
Write the following statement in the form of an equation:
Four times a number p is 8.
4p = 8
p+4=8
p-4=8
p\(\div\)4=8
22.
The product of a fraction and its reciprocal is:
0
-1
1
the fraction itself
23.
The value of the variable that we get on solving an equation is:
degree of the equation
root of the equation
co-efficient of the variable
none of these
24.
If 7x + 4 = 25, then x is equal to
\(\frac { 29 }{ 7 } \)
\(\frac { 100 }{ 7 } \)
2
3
25.
Five times of a number minus 4 gives 6. Find which of the following is the number?
20
2
15
35
26.
If sum of p and 4 is 15, P is :
10
9
11
0
27.
Which of the following numbers satisfies the equation -6 + x = -18?
10
-13
-12
-16
28.
If a and b are positive integers, then the solution of the equation ax = b will always be a
positive number
negative number
1
0
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.
32.
\(\frac { 3 }{ 2 } \) is the solution of the equation 8x - 5 =7.
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.
35.
What happens to the equality of an equation, if the same quantity is removed from both sides?
1.
56.4
2.
\(\frac { y }{ 5 } +3=2\)
\(\Rightarrow \frac { y }{ 5 } =2-3\)
\(\Rightarrow \quad \frac { y }{ 5 } =-1\)
\(\Rightarrow \quad \frac { y }{ 5 } =-1\times 5\)
y = -5
3.
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
4.
4q-12 = 0
\(\Rightarrow\) 4q = 12
Thus, q = 3
5.
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
6.
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.
7.
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.
8.
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.
9.
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.
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 |
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.
19.
\(2s=0 \Rightarrow s={0\over2}=0\)
20.
2m = 4\(\Rightarrow m={4\over2}=2\)
21.
(a)
4p = 8
22.
(c)
1
23.
(b)
root of the equation
24.
(d)
3
25.
(b)
2
26.
(c)
11
27.
(c)
-12
28.
(a)
positive number
29.
( )
1
30.
( )
\(\frac { 7 }{ 4 } \)
31.
If 3x = 12 \(\Rightarrow x=\frac{12}{3}=4\)
x - 0 = 4, when 3x = 12.
32.
(a)
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
35.
( )
The equality does not change.
7th Standard CBSE Syllabus & Materials
7th Standard CBSE
CBSE 7th Social Science Theme B - New Beginnings : Cities and States - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme A - Climates of India - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Social Science Theme A - Geographical Diversity of India - New Sample Question Papers Study Material - QB365 Set A
NEW7th Standard CBSE
CBSE 7th Science Earth, Moon and the Sun - New Sample Question Papers Study Material - QB365 Set A
CBSE 7th Standard CBSE Subjects
CBSE Standards
