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Simple Equations Questions For Class 7
Mathematics is a very important subject that you study in school. It is not a mere subject in school, but we need Mathematics to solve our reallife problems. In your early classes, you have learned about variables.
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ToggleIn this chapter, you will learn about simple equations. Simple equations include only one variable and the value of the variable is calculated by solving the equations. In higher classes, equations are largely used with multiple variables, and students must build the concepts. They must practice questions regularly to clear their doubts.
Extramarks is a wellknown company providing study materials related to CBSE and NCERT. Our experts believe in practice, and for this purpose, they have made the Important Questions Class 7 Mathematics Chapter 4. They collected different types of questions from various sources. They solved the questions like the textbook exercises, CBSE sample papers, CBSE question papers of past years, NCERT Exemplars and important reference books. Experienced professionals have further checked the answers to ensure the best quality of the content.
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Important Questions Class 7 Mathematics Chapter 4 with Solutions
Practice is very important for students to score better in exams. The experts of Extramarks understand the importance of practice and they have made this question series. They have collected the question from various sources such as the textbook exercises, CBSE sample papers, CBSE question papers of past years, NCERT Exemplars and important reference books. Thus, the important questions Class 7 Mathematics Chapter 4 will help students to practice the question regularly. The important questions are given below
Question 1. Check whether the given value in the brackets is the correct solution to the belowgiven equation or not:
(a) n + 5 = 19 (n = 1)
Answer 1.:
LHS = n + 5
By substituting the value of n = 1
Then we get,
LHS = n + 5
= 1 + 5
= 6
By comparing the LHS and RHS
6 ≠ 19
LHS ≠ RHS
Hence we get that the value of n = 1 is not the solution to the given equation above n + 5 = 19.
(b) 7n + 5 = 19 (n = – 2)
Answer :
LHS = 7n + 5
Now, by substituting the value of n = 2
Then we get,
LHS = 7n + 5
= (7 × (2)) + 5
= – 14 + 5
= – 9
By comparing the LHS and RHS
9 ≠ 19
LHS ≠ RHS
Hence we know that the value of n = 2 is not the solution to the abovegiven equation 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
Answer:
LHS = 7n + 5
By substituting the value of n as = 2
Then,
LHS = 7n + 5
= (7 × (2)) + 5
= 14 + 5
= 19
By comparing both the LHS and the RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is the solution to the given equation is 7n + 5 = 19.
(d) 4p – 3 = 13 (p = 1)
Answer:
LHS = 4p – 3
By substituting the value of p = 1
Then,
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing both the LHS and the RHS
1 ≠ 13
LHS ≠ RHS
Hence then, the value of p = 1 is not the solution to the abovegiven equation 4p – 3 = 13.
(e) 4p – 3 = 13 (p = – 4)
Answer:
LHS = 4p – 3
By substituting the value of p as = – 4
Then,
LHS = 4p – 3
= (4 × (4)) – 3
= 16 – 3
= 19
By comparing both the LHS and the RHS
19 ≠ 13
LHS ≠ RHS
Hence then, the value of p = 4 is not the solution to the abovegiven equation 4p – 3 = 13.
(f) 4p – 3 = 13 (p = 0)
Answer:
LHS = 4p – 3
By substituting the value of p = 0
Then,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= 3
By comparing the LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence then, the value of p = 0 is not the solution to the abovegiven equation 4p – 3 = 13.
Question 2. Solve the following equations by the trial and error method:
(i) 5p + 2 = 17
Answer:
LHS = 5p + 2
By substituting the value of p as = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing both the LHS and RHS
2 ≠ 17
LHS ≠ RHS
Hence then, the value of p = 0 is not the solution to the abovegiven equation.
Let, p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing the LHS and the RHS
7 ≠ 17
LHS ≠ RHS
Hence then, the value of p = 1 is not the solution to the abovegiven equation.
Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing both the LHS and the RHS
12 ≠ 17
LHS ≠ RHS
Hence then, the value of p = 2 is not the solution to the abovegiven equation.
Let, p = 3
LHS = 5p + 2
= (5 × 3) + 2
= 15 + 2
= 17
By comparing both the LHS and the RHS
17 = 17
LHS = RHS
Hence, the value of p = 3 is the solution given to the given equation.
(ii) 3m – 14 = 4
Answer:
LHS = 3m – 14
By substituting the value of m as = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing both the LHS and the RHS
5 ≠ 4
LHS ≠ RHS
Hence then, the value of m = 3 is not the solution to the abovegiven equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing both the LHS and the RHS
2 ≠ 4
LHS ≠ RHS
Hence then, the value of m = 4 is not the solution to the abovegiven equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing both the LHS and the RHS
1 ≠ 4
LHS ≠ RHS
Hence then, the value of m = 5 is not the solution to the abovegiven equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing both the LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is the solution to the abovegiven equation.
Question 3. Write the equations for the following statements:
(i) The sum of the numbers x and 4 is 9.
Answer:
The above statement can also be written in the equation form as,
= x + 4 = 9
(ii) 2 subtracted from y is 8.
Answer:
The above statement can also be written in the equation form as,
= y – 2 = 8
(iii) Ten times a is 70.
Answer:
The above statement can also be written in the equation form,
= 10a = 70
(iv) The number b, when divided by 5, gives 6.
Answer:
The above statement can also be written in the equation form as,
= (b/5) = 6
(v) Threefourths of t is 15.
Answer:
The above statement can also be written in the equation form,
= ¾t = 15
(vi) Seven times m plus seven will get you 77.
Answer:
The above statement can also be written in the equation form,
Seven times m will be 7m
= 7m + 7 = 77
(vii) Onefourth of the number x minus 4 gives 4.
Answer:
The above statement can also be written in the equation form,
Onefourth of the number x is x/4.
= x/4 – 4 = 4
(viii) If you take away six from 6 times y, you get 60.
Answer:
The above statement can also be written in the equation form as,
Six times y is 6y.
= 6y – 6 = 60
(ix) If you add 3 to the onethird of z, you get 30.
Answer:
The above statement can also be written in the equation form as,
Onethird of z is z/3.
= 3 + z/3 = 30
Question 4. Set up an equation form in the following cases given below:
(i) Irfan says that he has seven marbles, more than five times the marbles Permit has. Irfan has 37 marbles. (Take m to be the number of Permit’s marbles.)
Answer :
From the question, it is given that,
Number of Permit’s marbles = m
Then,
Irfan has seven marbles, more than five times the marbles Permit has
= 5 × Number of Permit’s marbles + 7 = Total number of the marbles Irfan having
= (5 × m) + 7 = 37
= 5m + 7 = 37
(ii) Laxmi’s father is 49 years old. He is four years older than Laxmi three times. (Take Laxmi’s age to be y years.)
Answer :
From the question, it is given that,
Let Laxmi’s age be = y years old.
Then,
Lakshmi’s father is four years older than three times her age.
= 3 × Laxmi’s age + 4 = Age of her father
= (3 × y) + 4 = 49
= 3y + 4 = 49
(iii) A teacher tells her class that the highest marks obtained by a student in the class are twice the lowest marks obtained plus 7. The highest score is 87. (Take the lowest score to be denoted as l.)
Answer:
From the question, it is given that,
Highest score in the class = 87
Let the lowest score be equal to the l
= 2 × Lowest score + 7 = Highest score in class
= (2 × l) + 7 = 87
= 2l + 7 = 87
(iv) In the isosceles triangle, the vertex angle is twice the base angle. (Let the base angle be denoted as b in degrees. Remember that the sum of the angles of the triangle is 180 degrees).
Answer:
From the above question, it is given that,
We know that the total sum of angles of a triangle is 180o.
Let the base angle be b
Then,
Vertex angle = 2 × base angle = 2b
= b + b + 2b = 180o
= 4b = 180o
Question 5. Write the following equations in statement forms:
(i) p + 4 = 15
Answer:
The sum of the numbers p and 4 is 15.
(ii) m – 7 = 3
Answer:
Seven subtracted from m is 3.
(iii) 2m = 7
Answer:
Twice of number m is 7.
(iv) m/5 = 3
Answer:
The number m divided by the number 5 gives 3.
(v) (3m)/5 = 6
Answer:
Threefifth of m is 6.
(vi) 3p + 4 = 25
Answer:
Three times p plus four will give you 25.
(vii) 4p – 2 = 18
Answer:
Four times p minus 2 gives you the number 18.
(viii) p/2 + 2 = 8
Answer:
If you add half of the number p to 2, you get 8.
Question 6. Give the steps which are used to separate the variable and then solve the equation:
(a) 3n – 2 = 46
Answer :
First, we have to add two to both the side of the equation,
Then, we receive,
= 3n – 2 + 2 = 46 + 2
= 3n = 48
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3n/3 = 48/3
= n = 16
(b) 5m + 7 = 17
Answer :
First, we have to subtract seven from the both sides of the equation,
Then, we receive,
= 5m + 7 – 7 = 17 – 7
= 5m = 10
Now,
We have to divide both these sides of the equation by 5,
Then, we receive,
= 5m/5 = 10/5
= m = 2
(c) 20p/3 = 40
Answer:
First, we have to multiply both of these sides of the equation by 3,
Then, we get,
= (20p/3) × 3 = 40 × 3
= 20p = 120
Now,
We have to divide both of these sides of the equation by 20,
Then, we receive,
= 20p/20 = 120/20
= p = 6
(d) 3p/10 = 6
Answer:
First, we have to multiply both of these sides of the equation by 10,
Then, we receive,
= (3p/10) × 10 = 6 × 10
= 3p = 60
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3p/3 = 60/3
= p = 20
Question 7. Solve the following equations:
(a) 10p = 100
Answer:
Now,
We have to divide both of these sides of the equation by 10,
Then, we get,
= 10p/10 = 100/10
= p = 10
(b) 10p + 10 = 100
Answer:
Firstly we have to subtract 10 from both sides of the following equation,
Then, we get,
= 10p + 10 – 10 = 100 – 10
= 10p = 90
Now,
We have to divide both of these sides of the equation by 10,
Then, we get,
= 10p/10 = 90/10
= p = 9
(c) p/4 = 5
Answer:
Now,
We have to multiply both of these sides of the equation by 4,
Then, we get,
= p/4 × 4 = 5 × 4
= p = 20
(d) – p/3 = 5
Answer:
Now,
We have to multiply both of these sides of the equation by – 3,
Then, we receive,
= – p/3 × ( 3) = 5 × ( 3)
= p = – 15
(e) 3p/4 = 6
Answer:
First, we have to multiply both of these sides of the equation by 4,
Then, we get,
= (3p/4) × (4) = 6 × 4
= 3p = 24
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3p/3 = 24/3
= p = 8
(f) 3s = – 9
Answer:
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = 9/3
= s = 3
(g) 3s + 12 = 0
Answer:
First, we have to subtract the number 12 from the both sides of the equation,
Then, we get,
= 3s + 12 – 12 = 0 – 12
= 3s = 12
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = 12/3
= s = – 4
(h) 3s = 0
Answer:
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = 0/3
= s = 0
(i) 2q = 6
Answer:
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3
(j) 2q – 6 = 0
Answer:
Firstly we have to add 6 to both the sides of the following equation,
Then, we get,
= 2q – 6 + 6 = 0 + 6
= 2q = 6
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3
(k) 2q + 6 = 0
Answer :
First, we have to subtract the number 6 from the both sides of the following equation,
Then, we get,
= 2q + 6 – 6 = 0 – 6
= 2q = – 6
Now,
We have to divide both of these sides of the equation by the number 2,
Then, we get,
= 2q/2 = – 6/2
= q = – 3
(l) 2q + 6 = 12
Answer:
First, we have to subtract the number 6 to the both sides of the following equation,
Then, we get,
= 2q + 6 – 6 = 12 – 6
= 2q = 6
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3
Question 8. Solve the following equations:
(a) 2y + (5/2) = (37/2)
Answer:
By transposing the fraction (5/2) from LHS to RHS, so it becomes 5/2
Then,
= 2y = (37/2) – (5/2)
= 2y = (375)/2
= 2y = 32/2
Now,
Divide both the side by the number 2,
= 2y/2 = (32/2)/2
= y = (32/2) × (1/2)
= y = 32/4
= y = 8
(b) 5t + 28 = 10
Answer :
By transposing 28 from the LHS to RHS, it becomes 28
Then,
= 5t = 10 – 28
= 5t = – 18
Now,
Dividing both these sides by 5,
= 5t/5= 18/5
= t = 18/5
(c) (a/5) + 3 = 2
Answer:
By transposing three from the LHS to RHS, it becomes 3
Then,
= a/5 = 2 – 3
= a/5 = – 1
Now,
Multiply both these sides by 5,
= (a/5) × 5= 1 × 5
= a = 5
(d) (q/4) + 7 = 5
Answer:
By transposing the number 7 from the LHS to RHS, it becomes 7
Then,
= q/4 = 5 – 7
= q/4 = – 2
Now,
Multiply both these sides by 4,
= (q/4) × 4= 2 × 4
= a = 8
(e) (5/2) x = 5
Answer:
First, we have to multiply both these sides by 2,
= (5x/2) × 2 = – 5 × 2
= 5x = – 10
Now,
We have to divide both these sides by 5,
Then we get,
= 5x/5 = 10/5
= x = 2
(f) (5/2) x = 25/4
Answer:
First, we have to multiply both these sides by 2,
= (5x/2) × 2 = (25/4) × 2
= 5x = (25/2)
Now,
We have to divide both these sides by 5,
Then we get,
= 5x/5 = (25/2)/5
= x = (25/2) × (1/5)
= x = (5/2)
Question 9. Solve the following equations:
(a) 2(x + 4) = 12
Answer:
Let us divide both these sides by 2,
= (2(x + 4))/2 = 12/2
= x + 4 = 6
By transposing four from the LHS to RHS, it becomes 4
= x = 6 – 4
= x = 2
(b) 3(n – 5) = 21
Answer:
Let us divide both these sides by 3,
= (3(n – 5))/3 = 21/3
= n – 5 = 7
By transposing 5 from the LHS to the RHS, it becomes 5
= n = 7 + 5
= n = 12
(c) 3(n – 5) = – 21
Answer:
Let us divide both these sides by 3,
= (3(n – 5))/3 = – 21/3
= n – 5 = 7
By transposing 5 from the LHS to the RHS, it becomes 5
= n = – 7 + 5
= n = – 2
(d) – 4(2 + x) = 8
Answer:
Let us divide both these sides by 4,
= (4(2 + x))/ (4) = 8/ (4)
= 2 + x = 2
By transposing two from the LHS to the RHS, it becomes – 2
= x = 2 – 2
= x = – 4
(e) 4(2 – x) = 8
Answer:
Let us divide both these sides by 4,
= (4(2 – x))/ 4 = 8/ 4
= 2 – x = 2
By transposing the number 2 from the LHS to the RHS, it becomes – 2
= – x = 2 – 2
= – x = 0
= x = 0
Question 10. Solve the following equations:
(a) 4 = 5(p – 2)
Answer:
Let us divide both these sides by 5,
= 4/5 = (5(p – 2))/5
= 4/5 = p 2
By transposing – 2 from the RHS to the LHS, it becomes 2
= (4/5) + 2 = p
= (4 + 10)/ 5 = p
= p = 14/5
(b) – 4 = 5(p – 2)
Answer:
Let us divide both these sides by 5,
= – 4/5 = (5(p – 2))/5
= – 4/5 = p 2
Now by transposing – 2 from the RHS to the LHS, it becomes 2
= – (4/5) + 2 = p
= ( 4 + 10)/ 5 = p
= p = 6/5
(c) 16 = 4 + 3(t + 2)
Answer:
By transposing four from the RHS to the LHS, it becomes – 4
= 16 – 4 = 3(t + 2)
= 12 = 3(t + 2)
Let us divide both these side by 3,
= 12/3 = (3(t + 2))/ 3
= 4 = t + 2
By transposing 2 from the RHS to the LHS it becomes – 2
= 4 – 2 = t
= t = 2
(d) 4 + 5(p – 1) =34
Answer:
By transposing 4 from the LHS to the RHS it becomes – 4
= 5(p – 1) = 34 – 4
= 5(p – 1) = 30
Let us divide both these side by 5,
= (5(p – 1))/ 5 = 30/5
= p – 1 = 6
By transposing – 1 from the RHS to the LHS it becomes 1
= p = 6 + 1
= p = 7
(e) 0 = 16 + 4(m – 6)
Answer:
By transposing 16 from the RHS to the LHS it becomes – 16
= 0 – 16 = 4(m – 6)
= – 16 = 4(m – 6)
Let us divide both these side by the number 4,
= – 16/4 = (4(m – 6))/ 4
= – 4 = m – 6
Now by transposing – 6 from the RHS to the LHS it becomes 6
= – 4 + 6 = m
= m = 2
Question 11. (a) Construct 3 equations starting with x = 2
Answer:
First equation is,
Multiply both these side by 6
= 6x = 12 … [equation 1]
Second equation is,
Subtracting the number 4 from both side,
= 6x – 4 = 12 4
= 6x – 4 = 8 … [equation 2]
Third equation is,
Divide both these sides by 6
= (6x/6) – (4/6) = (8/6)
= x – (4/6) = (8/6) … [equation 3]
(b) Construct three equations that start with x = – 2
Answer:
First equation is,
Multiply both these side by 5
= 5x = 10 … [equation 1]
Second equation is,
Subtracting the number 3 from both side,
= 5x – 3 = – 10 – 3
= 5x – 3 = – 13 … [equation 2]
Third equation is,
Dividing both these side by 2
= (5x/2) – (3/2) = (13/2) … [equation 3]
Benefits of Solving Important Questions Class 7 Mathematics Chapter 4
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Q.1 10 more than twice a number x is 83, it is represented as ___________.
Marks:1
1. 10 – 2x = 83
2. 10 + 2x = 83
3. 10 + x = 83
4. 20 + x = 83
Ans
2. 10 + 2x = 83
Explanation
Twice a number x = 2x
10 more than twice a number x = 10 + 2x
It is given that
10 + 2x = 83, which is the required equation.
Q.2 The sum of two consecutive even numbers is 114.
Which one of the following equations shows the given situation?
Marks:1
1. x + 2 = 114
2. 2x + 1 = 114
3. 2x + 2 = 114
4. 2x + 3 = 114
Ans
3. 2x + 2 = 114
Explanation
Let the two consecutive even numbers be x and (x + 2).
Given,
x + (x + 2) = 114
or, 2x + 2 = 114, which is the required equation.
Q.3 If an integer is 5 more than the other integer and their sum is 71. What are the values of these integers?
Marks:1
1. 30 and 41
2. 33 and 38
3. 35 and 36
4. 31 and 40
Ans
2. 33 and 38
Explanation
Let the first integer be x.
Then, other integer = x + 5
Now, x + (x + 5) = 71
or, 2x + 5 = 33
or, x = 33
The other integer is
x + 5 = 38
Hence, required integers are 33 and 38.
Q.4 Give one example of an equation whose solution is 5.
Marks:1
Ans
Let x = 5
On multiplying both sides by 2, we get
2x = 10
On adding 3 to both sides, we get
2x + 3 = 10 + 3
Or 2x + 3 = 13 which is one of the equations whose solution is 5.
Q.5 Set up an equation for the following cases:
1. The age of Kanika is four times that of her daughter. The difference between their ages is 27 years.
2. The interest received by Karim is 30 more than that of Ramesh. The total interest received by them is 70.
Marks:3
Ans
1. Let the age of Kanika’s daughter be x years
So, age of Kanika = 4x years
According to the statement, the difference between their ages is 27 years,
i.e., 4x – x = 27, which is the required equation.
2. Let interest received by Ramesh = x
So, interest received by Karim = (x + 30)
According to the statement, the total interest received is 70
i.e., x + (x+30) = 70, which is the required equation.
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FAQs (Frequently Asked Questions)
1. What is the content of Class 7 Mathematics Chapter 4?
Class 7 Mathematics Chapter 4 is about simple equations. In simple equations, the value of one variable generally has to be calculated. It is the most basic type of equation. In an equation, there are some known values and some unknown values. The unknown values are calculated by determining the relationship between the known and unknown values. Variables present unknown values. Equations are an integral part of mathematics; students will solve more complicated equations in their higher classes. So, they must build a strong concept of equations, how they can be built and different ways of finding solutions. They should practice as much as possible to score better in exams. They can take help from the Important Questions Class 7 Mathematics Chapter 4 for practice. It will help them to achieve better marks in exams.
2. How can the Important Questions Class 7 Mathematics Chapter 4 help students?
Extramarks is a reputed educational company in India. Our experts have made chapterwise question series for students to help them in practice. They have collected the questions from CBSE sample papers, CBSE question papers of past years, NCERT textbooks and important reference books. They have also solved the questions so that students can follow their answers if they cannot solve the questions. Thus, the Important Questions Class 7 Mathematics Chapter 4 will help them in practice, and it will help them to score better in exams.