Andhra Pradesh SCERT Class 7 Maths Simple Equations Question and Answers Solutions
Andhra Pradesh SCERT 7th Class Maths Solutions Chapter 4 Simple Equations Question and answers. Students who are searching for Andhra Pradesh Class 7 Maths Chapter 4 can find here Solution of this chapter.
Board |
Andhra Pradesh (AP Board) |
Class |
7th |
Subject |
Maths |
Topic |
Solution |
EXERCISE 4.1
1.) Complete the last column of the table.
ANSWER:
(i) x + 3 = 0
Here, we have to satisfy x = 3 for given equation.
x + 3 = 0 , 3 + 3 = 6
x = 3 is not satisfy.
(ii) x + 3 = 0
Here, we have to satisfy x = 0 for given equation.
x + 3 = 0 , 0 + 3 = 3
x = 0 is not satisfy.
(iii) x + 3 = 0
Here, we have to satisfy x = -3 for given equation.
x + 3 = 0 , -3 + 3 = 0
x = -3 is satisfy.
(iv) x – 7 = 1
Here, we have to satisfy x = 7 for given equation.
x – 7 = 1, 7 – 7 = 0
x = 7 is not satisfy.
(v) x – 7 = 1
Here, we have to satisfy x = 8 for given equation.
x – 7 = 1, 8 – 7 = 1
x = 8 is satisfy.
(vi) 5x = 25
Here, we have to satisfy x = 0 for given equation.
5x = 25, 5 x 0 = 0
x = 0 is not satisfy.
vii) 5x = 25
Here, we have to satisfy x = 5 for given equation.
5x = 25, 5 x 5 = 25
x = 5 is satisfy.
(viii) 5x = 25
Here, we have to satisfy x = -5 for given equation.
5x = 25, 5 x -5 = -25
x = -5 is not satisfy.
(ix) m/3 = 2
Here, we have to satisfy m = – 6 for given equation.
m/3 = 2 = – 6/3 = -2
m = – 6 is not satisfy.
(x) m/3 = 2
Here, we have to satisfy m = 0 for given equation.
m/3 = 2 = 0 /3 = 0
m = 0 is not satisfy.
(xi) m/3 = 2
Here, we have to satisfy m = 6 for given equation.
m/3 = 2 = 6/3 = 2
m = 6 is satisfy.
2.) Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
ANSWER:
We have to put n = 1 in n + 5 = 19.
We get,
1 + 5 = 6
Which is not equal to 19.
n = 1 is not a solution of equation n + 5 = 19
(b) 7n + 5 = 19 (n = – 2)
ANSWER:
We have to put n = -2 in 7n + 5 = 19
We get,
7 x (-2) + 5 = -9
Which is not equal to 19.
n = -2 is not a solution of equation 7n + 5 = 19
(c) 7n + 5 = 19 (n = 2)
ANSWER:
We have to put n = 2 in 7n + 5 = 19
We get,
7 x 2 + 5 = 19
Which is equal to 19.
n = 2 is a solution of equation 7n + 5 = 19
(d) 4p – 3 = 13 (p = 1)
ANSWER:
We have to put p = 1 in 4p – 3 = 13
We get,
4 x 1 – 3 = 1
Which is not equal to 13.
p = 1 is a not solution of equation 4p – 3 = 13
(e) 4p – 3 = 13 (p = – 4)
ANSWER:
We have to put p = – 4 in 4p – 3 = 13
We get,
4 x (– 4) – 3 = -19
Which is not equal to 13.
p = – 4 is a not solution of equation 4p – 3 = 13
(f) 4p – 3 = 13 (p = 0)
ANSWER:
We have to put p =0 1 in 4p – 3 = 13
We get,
4 x 0 – 3 = -3
Which is not equal to 13.
p = 0 is a not solution of equation 4p – 3 = 13
3.) Solve the following equations by trial and error method:
(i) 5p + 2 = 17
ANSWER:
We have to solve equation by trial and error method.
5p + 2 = 17
We put p = 1 in 5p + 2 = 17, we get
5 x 1 + 2 = 7
p = 1 not satisfy.
We put p = 2 in 5p + 2 = 17, we get
5 x 2 + 2 = 12
p = 2 not satisfy.
We put p = 3 in 5p + 2 = 17, we get
5 x 3 + 2 = 17
p = 3 satisfy.
p = 3 is solution of equation 5p + 2 = 17
(ii) 3m – 14 = 4
ANSWER:
We have to solve equation by trial and error method.
3m – 14 = 4
3m = 14 + 4
3m = 18
We put m= 4 in 3m = 18 we get,
3 x 4 = 12
m= 4 not satisfy.
We put m= 5 in 3m = 18 we get,
3 x 5 = 15
m= 5 not satisfy.
We put m= 6 in 3m = 18 we get,
3 x 6 = 18
m= 6 satisfy.
m= 6 is solution of equation 3m – 14 = 4
4.) Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
ANSWER:
We have to write equation for given statement.
The sum of numbers x and 4 is 9 = x + 4 = 9
(ii) 2 subtracted from y is 8.
ANSWER:
We have to write equation for given statement.
2 subtracted from y is 8 = y – 2 = 8
(iii) Ten times a is 70.
ANSWER:
We have to write equation for given statement.
Ten times a is 70 = 10a = 70
(iv) The number b divided by 5 gives 6.
ANSWER:
We have to write equation for given statement.
The number b divided by 5 gives 6 = b/5 = 6
(v) Three-fourth of t is 15.
ANSWER:
We have to write equation for given statement.
Three-fourth of t is 15 = 3t/4 = 15
(vi) Seven times m plus 7 gets you 77.
ANSWER:
We have to write equation for given statement.
Seven times m plus 7 gets you 77 = 7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
ANSWER:
We have to write equation for given statement.
One-fourth of a number x minus 4 gives 4 = x/4 – 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
ANSWER:
We have to write equation for given statement.
If you take away 6 from 6 times y, you get 60 = 6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30.
ANSWER:
We have to write equation for given statement.
you add 3 to one-third of z, you get 30 = z/3 + 3 = 30
5.) Write the following equations in statement forms:
(i) p + 4 = 15
ANSWER:
We have to write given equation in statement.
p + 4 = 15 = The sum of p and 4 is 15
(ii) m – 7 = 3
ANSWER:
We have to write given equation in statement.
m – 7 = 3 = 7 subtracted from m is 3
(iii) 2m = 7
ANSWER:
We have to write given equation in statement.
2m = 7 = Twice a number m is 7
(iv)m/5 = 3
ANSWER:
We have to write given equation in statement.
m/5 = 3 = One-fifth of a number m is 3
(v)3m/5= 6
ANSWER:
We have to write given equation in statement.
3m/5= 6 =Three-fifth of a number m is 6
(vi) 3p + 4 = 25
ANSWER:
We have to write given equation in statement.
3p + 4 = 25 = Three times a number p when added to 4 gives 25
(vii) 4p – 2 = 18
ANSWER:
We have to write given equation in statement.
4p – 2 = 18 = 2 subtracted from four times a number p is 18
(viii) p/2 + 2 = 8
ANSWER:
We have to write given equation in statement.
p/2 + 2 = 8 = Add 2 to half of a number p to get 8
6.) Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
ANSWER:
We have to setup equation for given information.
Let, the number of Parmit’s marbles is m.
Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles.
Equation is 5m + 7 = 37
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
ANSWER:
We have to setup equation for given information.
Let, Laxmi’s age to be y years.
Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age.
Equation is 3y + 4 = 49
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
ANSWER:
We have to setup equation for given information.
Let, lowest score to be ‘l’.
The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7.
Equation is 2l + 7 = 87
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
ANSWER:
We have to setup equation for given information.
Let, the base angle be ‘b’ in degrees.
We know,
The sum of angles of a triangle is 1800
In an isosceles triangle, the vertex angle is twice either base angle.
Equation is 4b = 180°
EXERCISE 4.2
1.) Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
ANSWER:
Here we have to give first the step which use to separate the variable.
x – 1 = 0
We add 1 to both sides
x – 1 + 1 = 0 + 1
x = 1
x = 1 is solution of equation x – 1 = 0
(b) x + 1 = 0
ANSWER:
Here we have to give first the step which use to separate the variable.
x + 1 = 0
We subtract 1 from both sides
x + 1 – 1 = 0 – 1
x = –1
x = -1 is solution of equation x + 1 = 0.
(c) x – 1 = 5
ANSWER:
Here we have to give first the step which use to separate the variable.
x – 1 = 5
We add 1 to both sides
x – 1 + 1 = 5 + 1
x = 6
x = 6 is solution of equation x – 1 = 5
(d) x + 6 = 2
ANSWER:
Here we have to give first the step which use to separate the variable.
x + 6 = 2
Subtract 6 from both sides
x + 6 – 6 = 2 – 6
x = – 4
x = -4 is solution of equation x + 6 = 2.
(e) y – 4 = – 7
ANSWER:
Here we have to give first the step which use to separate the variable.
y – 4 = – 7
We add 4 to both sides
y – 4 + 4 = – 7 + 4
y = –3
y = –3 is solution of equation y – 4 = – 7.
(f) y – 4 = 4
ANSWER:
Here we have to give first the step which use to separate the variable.
y – 4 = 4
We add 4 to both sides
y – 4 + 4 = 4 + 4
y = 8
y = 8 is solution of equation y – 4 = 4
(g) y + 4 = 4
ANSWER:
Here we have to give first the step which use to separate the variable.
y + 4 = 4
Subtract 4 from both sides
y + 4 – 4 = 4 – 4
y = 0
y = 0 is solution of equation y + 4 = 4
(h) y + 4 = – 4
ANSWER:
Here we have to give first the step which use to separate the variable.
y + 4 = – 4
Subtract 4 from both sides
y + 4 – 4 = – 4 – 4
y = – 8
y = – 8 is solution of equation y + 4 = – 4
2.) Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
ANSWER:
Here we have to give first the step which use to separate the variable.
3l = 42
Divide both sides by 3
3l/3 = 42/3
l = 14
l = 14 is solution of equation 3l = 42
(b) b/2= 6
ANSWER:
Here we have to give first the step which use to separate the variable.
b/2= 6
Multiply both sides by 2
b/2 x 2 = 6 x 2
b = 12
b = 12 is solution of equation b/2= 6.
(c) p/7= 4
ANSWER:
Here we have to give first the step which use to separate the variable.
p/7= 4
Multiply both sides by 7
p/7 x 7 = 4 x 7
p = 28
p = 28 is solution of equation p/7= 4.
(d) 4x = 25
ANSWER:
Here we have to give first the step which use to separate the variable.
4x = 25
Divide both sides by 4
4x/4 = 25/4
x = 25/ 4
x = 25/ 4 is solution of equation 4x = 25
(e) 8y = 36
ANSWER:
Here we have to give first the step which use to separate the variable.
8y = 36
Divide both sides by 8
8y/8 = 36/8
y = 36/8
y = 36/8 is solution of equation 8y = 36
(f) z/3 = 5/4
ANSWER:
Here we have to give first the step which use to separate the variable.
z/3 = 5/4
Multiply both sides by 3
z/3 x 3 = 5/4 x 3
z = 15/4
z = 15/4 is solution of equation z/3 = 5/4.
(g) a / 5 = 7/15
ANSWER:
Here we have to give first the step which use to separate the variable.
a / 5 = 7/15
Multiply both sides by 5
a / 5 x 5 = 7/15 x 5
a = 7/3
a = 7/3 is solution of equation a / 5 = 7/15
(h) 20t = – 10
ANSWER:
Here we have to give first the step which use to separate the variable.
20t = – 10
Divide both sides by 20
20t/20 = – 10/20
t = – 1/2
t = – ½ is solution of equation 20t = – 10
- Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
ANSWER:
Here we have to give first the step which use to separate the variable.
3n – 2 = 46
Step 1: Add 2 to both sides
3n – 2 + 2 = 46 + 2
3n = 48
Step 2: Divide both sides by 3
3n/3 = 48/3
n = 16
n = 16 is solution of equation 3n – 2 = 46
(b) 5m + 7 = 17
ANSWER:
Here we have to give first the step which use to separate the variable.
5m + 7 = 17
Step 1: Subtract 7 from both sides
5m + 7 – 7 = 17 – 7
5m = 10
Step 2: Divide both sides by 5
5m/5 = 10/5
m = 2
m = 2 is solution of equation 5m + 7 = 17
(c) 20 p /3 = 40
ANSWER:
Here we have to give first the step which use to separate the variable.
20 p /3 = 40
Step 1: Multiply both sides by 3
20 p /3 x 3 = 40 x 3
20 p = 120
Step 2: Divide both sides by 20
20 p / 20 = 120 / 20
p = 6
p = 6 is solution of equation 20 p /3 = 40
(d) 3 p/10 = 6
ANSWER:
Here we have to give first the step which use to separate the variable.
3 p/10 = 6
Multiply both sides 10
3 p/10 x 10 = 6 x 10
3 p = 60
Step 2: Divide both sides by 3
3 p/3 = 60/3
p = 20
p = 20 is solution of equation 3 p/10 = 6
4.) Solve the following equations:
(a) 10p = 100
ANSWER:
10p = 100
Step 1: Divide both sides by 10
10p/10 = 100/10
p = 10
p = 10 is solution of equation 10p = 100
(b) 10p + 10 = 100
ANSWER:
10p + 10 = 100
Step 1: subtract 10 on both sides
10p + 10 – 10 = 100 – 10
10p = 90
Step 2: Divide both sides by 10
10p/10 = 90/10
p = 9
p = 9 is solution of equation 10p + 10 = 100.
(c) p / 4 = 5
ANSWER:
p / 4 = 5
Multiply both sides 4
p / 4 x 4 = 5 x 4
p = 20
p = 20 is solution of equation p / 4 = 5
(d) –p/3 =5
ANSWER:
–p/3 =5
Multiply both sides -3
–p/3 x -3 =5 x -3
p = -15
p = -15 is solution of equation –p/3 = 5
(e) 3p/ 4 = 6
ANSWER:
3p/ 4 = 6
Multiply both sides 4
3p/ 4 x 4 = 6 x 4
3p = 24
Divide both sides by 3
3p/3 = 24/3
p = 8
p = 8 is solution of equation 3p/ 4 = 6
(f) 3s = –9
ANSWER:
3s = –9
Divide both sides by 3
3s/3 = –9/3
S = -3
S = -3 is solution of equation 3s = –9.
(g) 3s + 12 = 0
ANSWER:
3s + 12 = 0
Subtract 12 on both sides
3s + 12 – 12 = 0 – 12
3s = -12
Divide both sides by 3
3s/3 = -12/3
S = -4
S = -4 is solution of equation 3s + 12 = 0
(h) 3s = 0
ANSWER:
3s = 0
Divide both sides by 3
3s/3 = 0/3
S = 0
S = 0 is solution of equation 3s = 0
(i) 2q = 6
ANSWER:
2q = 6
Divide both sides by 2
2q/2 = 6/2
q = 3
q = 3 is solution of equation 2q = 6
(j) 2q – 6 = 0
ANSWER:
2q – 6 = 0
Add 6 on both sides
2q – 6 + 6 = 0 + 6
2q = 6
Divide both sides by 2
2q/2 = 6/2
q = 3
q = 3 is solution of equation 2q – 6 = 0
(k) 2q + 6 = 0
ANSWER:
2q + 6 = 0
Subtract 6 on both sides
2q + 6 – 6 = 0 – 6
2q = -6
Divide both sides by 2
2q/2 = -6/2
q = -3
q = -3 is solution of equation 2q + 6 = 0
(l) 2q + 6 = 12
ANSWER:
2q + 6 = 12
Subtract 6 on both sides
2q + 6 – 6 = 12 – 6
2q = 6
Divide both sides by 2
2q/2 = 6/2
q = 3
q = 3 is solution of equation 2q + 6 = 12
EXERCISE 4.3
1.) Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
ANSWER:
We have to set equation.
Let, x is the given number.
Equation is 8x + 4 = 60
8x + 4 = 60
Subtract 4 on both sides
8x + 4 – 4 = 60 – 4
8x = 56
Divide both sides by 8
8x/8 = 56/8
X = 7is solution of equation.
(b) One-fifth of a number minus 4 gives 3.
ANSWER:
We have to set equation.
Let, x is the given number.
Equation is x/5– 4 = 3
x/5 – 4 = 3
Add 4 on both sides
x/5 – 4 + 4 = 3 + 4
x/5 = 7
Multiply 5 on both sides
x/5 x 5 = 7 x 5
x = 35 is solution of equation
(c) If I take three-fourths of a number and add 3 to it, I get 21.
ANSWER:
We have to set equation.
Let, y is the given number.
Equation is 3/4 y + 3 = 21
3/4 y + 3 = 21
Subtract 3 on both sides
3/4 y + 3 – 3 = 21 – 3
3/4 y = 18
Multiply by 4 on both sides
3/4 y x 4 = 18 x 4
3y = 72
Divide by 3 on both sides
y = 24
y = 24 is solution of equation
(d) When I subtracted 11 from twice a number, the result was 15.
ANSWER:
We have to set equation.
Let, m is the given number.
Equation is 2m – 11 = 15
Add 11 on both sides
2m – 11 + 11 = 15 + 11
2m = 26
Divide by 2 on both sides
m = 13
m = 13 is solution of equation.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
ANSWER:
We have to set equation.
Let, x is the given number.
Equation is 50 – 3x = 8
50 – 3x = 8
Subtract 50 on both sides
-50 + 50 – 3x = 8 – 50
– 3x = -42
Divide by -3 on both sides
– 3x/-3 = -42/-3
x = 14
x = 14 is solution of equation.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
ANSWER:
We have to set equation.
Let, x is the given number.
Equation isx +19/ 5 = 8
x +19/ 5 = 8
Multiply by 5 on both sides
x +19/ 5 x 5 = 8 x 5
x +19 = 40
Subtract 19 on both sides
x +19 – 19 = 40 – 19
x = 21
x = 21 is solution of equation.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
ANSWER:
We have to set equation.
Let, n is the given number.
Equation is5n/2–7 = 23
Add 7 on both sides
5n/2 –7 + 7 = 23 + 7
5n/2 = 30
Multiply by 2 on both sides
5n/2 x 2 = 30 x 2
5n = 60
Divide by 5 on both sides
5n/5 = 60/5
n = 12
n = 12 is solution of equation.
2.) Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
ANSWER:
Given that,
The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7
The highest score is 87.
We have to find lowest score.
Let, lowest score be x.
Equation is 2x + 7 = 87
2x + 7 = 87
Subtract 7 on both sides
2x + 7 – 7 = 87 – 7
2x = 80
Divide by 2 on both sides
2x/2 = 80/2
X = 40
Lowest score is 40.
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
ANSWER:
Given that,
In an isosceles triangle, the base angles are equal. The vertex angle is 40°.
We have to find the base angles of the triangle
Let, the base angles of the triangle be x.
Equation is 2x + 40° = 180°
2x + 40° = 180°
Subtract 40° on both sides
2x + 40° – 40° = 180°- 40°
2x = 140°
Divide by 2 on both sides
2x/2 = 140°/2
X = 70°
Base angles of the triangle is 70°
(c) 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?
ANSWER:
Given that,
Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century.
Let, Sachin scored x runs.
Rahul scored y runs
Equation is x = 2y and x + y = 198
We put x = 2y in x + y = 198
2y + y = 198
3y = 198
Y = 66
But y is Rahul score.
Rahul score 66 runs.
x = 2y
x = 2 x 66
x = 132
Sachin score 132 runs.
4.) Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
ANSWER:
Let, the number is x.
Equation is 7x + 50 = 260
7x + 50 = 260
Subtract 50 on both sides
7x + 50 – 50 = 260 – 50
7x = 210
Divide by 7 on both sides
7x/7 = 210/7
X = 30
The number is 30.