Class 7 Maths - Simple Equations NCERT Solutions

Chapter 4: Simple Equations (NCERT Solutions)

Exercise 4.1

Q1. Complete the last column of the table.
(S.No., Equation, Value, Say whether the Equation is Satisfied. (Yes/No))

S.No.	Equation	Value of Variable	Equation Satisfied? (Yes/No)
(i)	x + 3 = 0	x = 3	No (3+3=6, not 0)
(ii)	x + 3 = 0	x = 0	No (0+3=3, not 0)
(iii)	x + 3 = 0	x = -3	Yes (-3+3=0)
(iv)	x - 7 = 1	x = 7	No (7-7=0, not 1)
(v)	x - 7 = 1	x = 8	Yes (8-7=1)
(vi)	5x = 25	x = 0	No (5(0)=0, not 25)
(vii)	5x = 25	x = 5	Yes (5(5)=25)
(viii)	5x = 25	x = -5	No (5(-5)=-25, not 25)
(ix)	m / 3 = 2	m = -6	No (-6/3 = -2, not 2)
(x)	m / 3 = 2	m = 0	No (0/3 = 0, not 2)
(xi)	m / 3 = 2	m = 6	Yes (6/3 = 2)

Q2. Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (n = -2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p - 3 = 13 (p = 1)
(e) 4p - 3 = 13 (p = -4)
(f) 4p - 3 = 13 (p = 0)

(a) n + 5 = 19. Put n = 1: 1 + 5 = 6 ≠ 19. So, it is not a solution.

(b) 7n + 5 = 19. Put n = -2: 7(-2) + 5 = -14 + 5 = -9 ≠ 19. So, it is not a solution.

(c) 7n + 5 = 19. Put n = 2: 7(2) + 5 = 14 + 5 = 19. LHS = RHS. So, it is a solution.

(d) 4p - 3 = 13. Put p = 1: 4(1) - 3 = 4 - 3 = 1 ≠ 13. So, it is not a solution.

(e) 4p - 3 = 13. Put p = -4: 4(-4) - 3 = -16 - 3 = -19 ≠ 13. So, it is not a solution.

(f) 4p - 3 = 13. Put p = 0: 4(0) - 3 = 0 - 3 = -3 ≠ 13. So, it is not a solution.

Q3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m - 14 = 4

(i) 5p + 2 = 17
Try p = 1: 5(1) + 2 = 7 ≠ 17
Try p = 2: 5(2) + 2 = 12 ≠ 17
Try p = 3: 5(3) + 2 = 15 + 2 = 17. LHS = RHS.
Therefore, p = 3 is the solution.

(ii) 3m - 14 = 4
Try m = 4: 3(4) - 14 = 12 - 14 = -2 ≠ 4
Try m = 5: 3(5) - 14 = 15 - 14 = 1 ≠ 4
Try m = 6: 3(6) - 14 = 18 - 14 = 4. LHS = RHS.
Therefore, m = 6 is the solution.

Q4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.

(i) x + 4 = 9

(ii) y - 2 = 8

(iii) 10a = 70

(iv) b / 5 = 6

(v) (3/4)t = 15 OR 3t/4 = 15

(vi) 7m + 7 = 77

(vii) (x / 4) - 4 = 4

(viii) 6y - 6 = 60

(ix) (z / 3) + 3 = 30

Q5. Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m - 7 = 3
(iii) 2m = 7
(iv) m / 5 = 3
(v) (3m) / 5 = 6
(vi) 3p + 4 = 25
(vii) 4p - 2 = 18
(viii) (p / 2) + 2 = 8

(i) The sum of p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Twice of a number m is 7.

(iv) One-fifth of a number m is 3.

(v) Three-fifth of a number m is 6.

(vi) Three times a number p, when added to 4, gives 25.

(vii) 2 subtracted from 4 times a number p is 18.

(viii) Add 2 to half of a number p to get 8.

Q6. 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.)
(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.)
(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.)
(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).

(i) Let Parmit's marbles = m.
Five times Parmit's marbles = 5m.
7 more than 5m = 5m + 7.
Irfan has 37 marbles. So, equation is: 5m + 7 = 37

(ii) Let Laxmi's age = y.
Three times her age = 3y.
4 years older than 3y = 3y + 4.
Father's age is 49. So, equation is: 3y + 4 = 49

(iii) Let the lowest score = l.
Twice the lowest score = 2l.
Twice lowest plus 7 = 2l + 7.
Highest score is 87. So, equation is: 2l + 7 = 87

(iv) Let each base angle = b.
Vertex angle is twice base angle = 2b.
Sum of angles = b + b + 2b = 4b.
Since sum is 180°, equation is: 4b = 180

Exercise 4.2

Q1. Give first the step you will use to separate the variable and then solve the equation:
(a) x - 1 = 0
(b) x + 1 = 0
(c) x - 1 = 5
(d) x + 6 = 2
(e) y - 4 = -7
(f) y - 4 = 4
(g) y + 4 = 4
(h) y + 4 = -4

(a) x - 1 = 0
Step: Add 1 to both sides.
x - 1 + 1 = 0 + 1 ⇒ x = 1

(b) x + 1 = 0
Step: Subtract 1 from both sides.
x + 1 - 1 = 0 - 1 ⇒ x = -1

(c) x - 1 = 5
Step: Add 1 to both sides.
x - 1 + 1 = 5 + 1 ⇒ x = 6

(d) x + 6 = 2
Step: Subtract 6 from both sides.
x + 6 - 6 = 2 - 6 ⇒ x = -4

(e) y - 4 = -7
Step: Add 4 to both sides.
y - 4 + 4 = -7 + 4 ⇒ y = -3

(f) y - 4 = 4
Step: Add 4 to both sides.
y - 4 + 4 = 4 + 4 ⇒ y = 8

(g) y + 4 = 4
Step: Subtract 4 from both sides.
y + 4 - 4 = 4 - 4 ⇒ y = 0

(h) y + 4 = -4
Step: Subtract 4 from both sides.
y + 4 - 4 = -4 - 4 ⇒ y = -8

Q2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) b / 2 = 6
(c) p / 7 = 4
(d) 4x = 25
(e) 8y = 36
(f) z / 3 = 5 / 4
(g) a / 5 = 7 / 15
(h) 20t = -10

(a) 3l = 42
Step: Divide both sides by 3.
3l / 3 = 42 / 3 ⇒ l = 14

(b) b / 2 = 6
Step: Multiply both sides by 2.
(b / 2) × 2 = 6 × 2 ⇒ b = 12

(c) p / 7 = 4
Step: Multiply both sides by 7.
(p / 7) × 7 = 4 × 7 ⇒ p = 28

(d) 4x = 25
Step: Divide both sides by 4.
4x / 4 = 25 / 4 ⇒ x = 25/4

(e) 8y = 36
Step: Divide both sides by 8.
8y / 8 = 36 / 8 ⇒ y = 36/8 = 9/2

(f) z / 3 = 5 / 4
Step: Multiply both sides by 3.
(z / 3) × 3 = (5 / 4) × 3 ⇒ z = 15/4

(g) a / 5 = 7 / 15
Step: Multiply both sides by 5.
(a / 5) × 5 = (7 / 15) × 5 ⇒ a = 35 / 15 = 7/3

(h) 20t = -10
Step: Divide both sides by 20.
20t / 20 = -10 / 20 ⇒ t = -1/2

Q3. Give the steps you will use to separate the variable and then solve the equation:
(a) 3n - 2 = 46
(b) 5m + 7 = 17
(c) 20p / 3 = 40
(d) 3p / 10 = 6

(a) 3n - 2 = 46
Step 1: Add 2 to both sides. ⇒ 3n = 48
Step 2: Divide both sides by 3. ⇒ 3n / 3 = 48 / 3 ⇒ n = 16

(b) 5m + 7 = 17
Step 1: Subtract 7 from both sides. ⇒ 5m = 10
Step 2: Divide both sides by 5. ⇒ 5m / 5 = 10 / 5 ⇒ m = 2

(c) 20p / 3 = 40
Step 1: Multiply both sides by 3. ⇒ 20p = 120
Step 2: Divide both sides by 20. ⇒ 20p / 20 = 120 / 20 ⇒ p = 6

(d) 3p / 10 = 6
Step 1: Multiply both sides by 10. ⇒ 3p = 60
Step 2: Divide both sides by 3. ⇒ 3p / 3 = 60 / 3 ⇒ p = 20

Q4. Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) p / 4 = 5
(d) -p / 3 = 5
(e) 3p / 4 = 6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q - 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12

(a) 10p = 100 ⇒ p = 100 / 10 ⇒ p = 10

(b) 10p + 10 = 100 ⇒ 10p = 90 ⇒ p = 90 / 10 ⇒ p = 9

(c) p / 4 = 5 ⇒ p = 5 × 4 ⇒ p = 20

(d) -p / 3 = 5 ⇒ -p = 15 ⇒ p = -15

(e) 3p / 4 = 6 ⇒ 3p = 24 ⇒ p = 24 / 3 ⇒ p = 8

(f) 3s = -9 ⇒ s = -9 / 3 ⇒ s = -3

(g) 3s + 12 = 0 ⇒ 3s = -12 ⇒ s = -12 / 3 ⇒ s = -4

(h) 3s = 0 ⇒ s = 0 / 3 ⇒ s = 0

(i) 2q = 6 ⇒ q = 6 / 2 ⇒ q = 3

(j) 2q - 6 = 0 ⇒ 2q = 6 ⇒ q = 6 / 2 ⇒ q = 3

(k) 2q + 6 = 0 ⇒ 2q = -6 ⇒ q = -6 / 2 ⇒ q = -3

(l) 2q + 6 = 12 ⇒ 2q = 6 ⇒ q = 6 / 2 ⇒ q = 3

Exercise 4.3

Q1. Solve the following equations:
(a) 2y + 5/2 = 37/2
(b) 5t + 28 = 10
(c) a/5 + 3 = 2
(d) q/4 + 7 = 5
(e) (5/2)x = -10
(f) (5/2)x = 25/4
(g) 7m + 19/2 = 13
(h) 6z + 10 = -2
(i) 3l/2 = 2/3
(j) 2b/3 - 5 = 3

(a) 2y + 5/2 = 37/2
Transposing 5/2 to RHS: 2y = 37/2 - 5/2
2y = 32/2 ⇒ 2y = 16
Dividing by 2: y = 16/2 ⇒ y = 8

(b) 5t + 28 = 10
Transposing 28 to RHS: 5t = 10 - 28
5t = -18 ⇒ t = -18/5

(c) a/5 + 3 = 2
Transposing 3 to RHS: a/5 = 2 - 3
a/5 = -1
Multiplying by 5: a = -1 × 5 ⇒ a = -5

(d) q/4 + 7 = 5
Transposing 7 to RHS: q/4 = 5 - 7
q/4 = -2
Multiplying by 4: q = -2 × 4 ⇒ q = -8

(e) (5/2)x = -10
Multiplying by 2: 5x = -20
Dividing by 5: x = -20 / 5 ⇒ x = -4

(f) (5/2)x = 25/4
Multiplying by 2: 5x = (25/4) × 2 = 25/2
Dividing by 5: x = (25/2) / 5 = 5/2 ⇒ x = 5/2

(g) 7m + 19/2 = 13
Transposing 19/2 to RHS: 7m = 13 - 19/2
7m = (26 - 19)/2 ⇒ 7m = 7/2
Dividing by 7: m = (7/2) / 7 ⇒ m = 1/2

(h) 6z + 10 = -2
Transposing 10 to RHS: 6z = -2 - 10
6z = -12 ⇒ z = -12 / 6 ⇒ z = -2

(i) 3l/2 = 2/3
Multiplying by 2: 3l = 4/3
Dividing by 3: l = (4/3) / 3 ⇒ l = 4/9

(j) 2b/3 - 5 = 3
Transposing 5 to RHS: 2b/3 = 3 + 5
2b/3 = 8
Multiplying by 3: 2b = 24
Dividing by 2: b = 24 / 2 ⇒ b = 12

Q2. Solve the following equations:
(a) 2(x + 4) = 12
(b) 3(n - 5) = 21
(c) 3(n - 5) = -21
(d) -4(2 + x) = 8
(e) 4(2 - x) = 8

(a) 2(x + 4) = 12
Divide by 2: x + 4 = 6
Transpose 4: x = 6 - 4 ⇒ x = 2

(b) 3(n - 5) = 21
Divide by 3: n - 5 = 7
Transpose 5: n = 7 + 5 ⇒ n = 12

(c) 3(n - 5) = -21
Divide by 3: n - 5 = -7
Transpose 5: n = -7 + 5 ⇒ n = -2

(d) -4(2 + x) = 8
Divide by -4: 2 + x = 8 / -4 ⇒ 2 + x = -2
Transpose 2: x = -2 - 2 ⇒ x = -4

(e) 4(2 - x) = 8
Divide by 4: 2 - x = 8 / 4 ⇒ 2 - x = 2
Transpose 2: -x = 2 - 2 ⇒ -x = 0 ⇒ x = 0

Q3. Solve the following equations:
(a) 4 = 5(p - 2)
(b) -4 = 5(p - 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p - 1) = 34
(e) 0 = 16 + 4(m - 6)

(a) 4 = 5(p - 2)
Divide by 5: 4/5 = p - 2
Transpose 2: p = 4/5 + 2 = (4 + 10)/5 = 14/5

(b) -4 = 5(p - 2)
Divide by 5: -4/5 = p - 2
Transpose 2: p = -4/5 + 2 = (-4 + 10)/5 = 6/5

(c) 16 = 4 + 3(t + 2)
Transpose 4: 16 - 4 = 3(t + 2) ⇒ 12 = 3(t + 2)
Divide by 3: 12 / 3 = t + 2 ⇒ 4 = t + 2
Transpose 2: t = 4 - 2 ⇒ t = 2

(d) 4 + 5(p - 1) = 34
Transpose 4: 5(p - 1) = 34 - 4 ⇒ 5(p - 1) = 30
Divide by 5: p - 1 = 30 / 5 ⇒ p - 1 = 6
Transpose 1: p = 6 + 1 ⇒ p = 7

(e) 0 = 16 + 4(m - 6)
Transpose 16: -16 = 4(m - 6)
Divide by 4: -16 / 4 = m - 6 ⇒ -4 = m - 6
Transpose 6: m = -4 + 6 ⇒ m = 2

Q4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = -2

(a) Starting with x = 2:
1. Multiply both sides by 10: 10x = 20
2. Add 2 to both sides: x + 2 = 4
3. Multiply by 5 and add 3: 5x = 10 ⇒ 5x + 3 = 13

(b) Starting with x = -2:
1. Multiply both sides by 3: 3x = -6
2. Subtract 2 from both sides: x - 2 = -4
3. Multiply by 4 and add 10: 4x = -8 ⇒ 4x + 10 = 2

Exercise 4.4

Q1. 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.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

(a) Let the number be x.
Equation: 8x + 4 = 60
8x = 60 - 4 ⇒ 8x = 56 ⇒ x = 56/8 ⇒ x = 7

(b) Let the number be y.
Equation: (y / 5) - 4 = 3
y / 5 = 3 + 4 ⇒ y / 5 = 7 ⇒ y = 7 × 5 ⇒ y = 35

(c) Let the number be z.
Equation: (3/4)z + 3 = 21
(3/4)z = 21 - 3 ⇒ (3/4)z = 18
z = 18 × (4/3) ⇒ z = 6 × 4 ⇒ z = 24

(d) Let the number be p.
Equation: 2p - 11 = 15
2p = 15 + 11 ⇒ 2p = 26 ⇒ p = 26/2 ⇒ p = 13

(e) Let the number of notebooks be m.
Equation: 50 - 3m = 8
-3m = 8 - 50 ⇒ -3m = -42 ⇒ m = -42 / -3 ⇒ m = 14

(f) Let the number be n.
Equation: (n + 19) / 5 = 8
n + 19 = 8 × 5 ⇒ n + 19 = 40
n = 40 - 19 ⇒ n = 21

(g) Let the number be x.
Equation: (5/2)x - 7 = 23
(5/2)x = 23 + 7 ⇒ (5/2)x = 30
x = 30 × (2/5) ⇒ x = 6 × 2 ⇒ x = 12

Q2. 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?
(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°).
(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?

(a) Let the lowest marks be x.
Highest marks = 2x + 7
Given highest score = 87.
So, 2x + 7 = 87 ⇒ 2x = 87 - 7 ⇒ 2x = 80 ⇒ x = 40
Therefore, the lowest score is 40.

(b) Let each base angle be b degrees.
Vertex angle = 40°.
Sum of angles = b + b + 40 = 180
2b + 40 = 180 ⇒ 2b = 180 - 40 ⇒ 2b = 140 ⇒ b = 70
Therefore, the base angles are 70° each.

(c) Let Rahul's score be R.
Sachin's score = 2R.
Together: R + 2R = 3R.
Given sum is two short of a double century (200 - 2 = 198).
3R = 198 ⇒ R = 198 / 3 ⇒ R = 66
Sachin's score = 2 × 66 = 132.
Therefore, Rahul scored 66 runs and Sachin scored 132 runs.

Q3. Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi's father is 49 years old. He is 4 years older than three times Laxmi's age. What is Laxmi's age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

(i) Let Parmit's marbles be m.
Irfan's marbles = 5m + 7
Given Irfan has 37 marbles: 5m + 7 = 37
5m = 37 - 7 ⇒ 5m = 30 ⇒ m = 6
Parmit has 6 marbles.

(ii) Let Laxmi's age be y years.
Father's age = 3y + 4
Given father's age is 49: 3y + 4 = 49
3y = 49 - 4 ⇒ 3y = 45 ⇒ y = 15
Laxmi is 15 years old.

(iii) Let the number of fruit trees be f.
Number of non-fruit trees = 3f + 2
Given non-fruit trees = 77: 3f + 2 = 77
3f = 77 - 2 ⇒ 3f = 75 ⇒ f = 25
The number of fruit trees planted was 25.

Q4. 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!

Let the number be n.
Take it seven times over: 7n
Add a fifty: 7n + 50
To reach a triple century (300), you still need forty, which means this sum plus 40 equals 300.
7n + 50 + 40 = 300
7n + 90 = 300
7n = 300 - 90
7n = 210
n = 210 / 7 = 30
Therefore, the number is 30.

Class 7 Maths - Simple Equations Practice

Chapter 4: Simple Equations (Practice Questions)

RD Sharma / Extra Practice

Q1. Solve for x: 3x - 5 = 13

3x = 13 + 5
3x = 18
x = 18 / 3 = 6

Q2. Solve the equation: (2y + 5) / 3 = y - 1

Multiply both sides by 3:
2y + 5 = 3(y - 1)
2y + 5 = 3y - 3
5 + 3 = 3y - 2y
y = 8

Q3. A number increased by 8 equals 26. Find the number.

Let the number be n.
n + 8 = 26
n = 26 - 8 = 18

Q4. Three-fourths of a number is 15. Find the number.

Let the number be x.
(3/4)x = 15
x = 15 × 4/3
x = 5 × 4 = 20

Q5. The sum of three consecutive integers is 51. What are the integers?

Let the integers be x, x+1, x+2.
x + (x+1) + (x+2) = 51
3x + 3 = 51
3x = 48 ⇒ x = 16
The integers are 16, 17, and 18.

Q6. If 1/3 of a number subtracted from 1/2 of the same number gives 5, find the number.

Let the number be x.
x/2 - x/3 = 5
(3x - 2x)/6 = 5
x/6 = 5 ⇒ x = 5 × 6 = 30

Q7. Find a number whose double is 45 greater than its half.

Let the number be x.
2x = (x/2) + 45
2x - x/2 = 45
(4x - x)/2 = 45
3x/2 = 45 ⇒ 3x = 90 ⇒ x = 30
The number is 30.

Q8. A man's age is three times his son's age. Ten years ago, he was five times as old as his son. Find their present ages.

Let son's age = x, Father's age = 3x.
Ten years ago: Father = 3x - 10, Son = x - 10
3x - 10 = 5(x - 10)
3x - 10 = 5x - 50
40 = 2x ⇒ x = 20
Son is 20 years old, Father is 60 years old.

Q9. Solve for m: 4(m - 3) = 2(m + 5)

4m - 12 = 2m + 10
4m - 2m = 10 + 12
2m = 22 ⇒ m = 11

Q10. The sum of two consecutive even numbers is 74. Find them.

Let numbers be 2n and 2n+2.
2n + 2n + 2 = 74
4n = 72 ⇒ n = 18
The numbers are 2(18) = 36 and 36+2 = 38.
36 and 38.

Q11. The length of a rectangle is twice its breadth. If the perimeter is 60 cm, find its length and breadth.

Let breadth = b, length = 2b.
Perimeter = 2(L + B) = 60
2(2b + b) = 60
3b = 30 ⇒ b = 10
Breadth = 10 cm, Length = 20 cm.

Q12. The ages of A and B are in the ratio 5:7. Eight years ago, their ages were in the ratio 7:13. Find their present ages.

Let present ages be 5x and 7x.
(5x - 8) / (7x - 8) = 7/13
13(5x - 8) = 7(7x - 8)
65x - 104 = 49x - 56
16x = 48 ⇒ x = 3
A's age = 15 years, B's age = 21 years.

Q13. I have Rs 1000 in ten and twenty rupee notes. If the number of ten rupee notes that I have is ten more than the number of twenty rupee notes, how many of each do I have?

Let number of Rs 20 notes = x.
Number of Rs 10 notes = x + 10.
Total value: 20x + 10(x + 10) = 1000
20x + 10x + 100 = 1000
30x = 900 ⇒ x = 30
30 notes of Rs 20, and 40 notes of Rs 10.

Q14. Solve for p: (p-2)/5 + (p+1)/2 = 8

Multiply by LCM (10):
2(p-2) + 5(p+1) = 80
2p - 4 + 5p + 5 = 80
7p + 1 = 80
7p = 79 ⇒ p = 79/7

Q15. The difference between two positive numbers is 40 and the ratio of these integers is 1 : 3. Find the integers.

Let numbers be x and 3x.
3x - x = 40
2x = 40 ⇒ x = 20
The numbers are 20 and 60.

Q16. Solve: 0.25(4f - 3) = 0.05(10f - 9)

f(0.25×4) - 0.75 = f(0.05×10) - 0.45
1f - 0.75 = 0.5f - 0.45
0.5f = 0.30 ⇒ f = 0.30 / 0.50 = 3/5 = 0.6

Q17. In a fraction, twice the numerator is 2 more than the denominator. If 3 is added to the numerator and to the denominator, the new fraction is 2/3. Find the original fraction.

Let numerator be n. Then denominator = 2n - 2.
(n + 3) / (2n - 2 + 3) = 2/3
(n + 3) / (2n + 1) = 2/3
3(n + 3) = 2(2n + 1)
3n + 9 = 4n + 2
n = 7. Denominator = 2(7)-2=12.
Original fraction is 7/12.

Q18. Two supplementary angles differ by 44 degrees. Find the angles.

Let angles be x and x+44.
x + (x+44) = 180
2x = 136 ⇒ x = 68
The angles are 68° and 112°.

Q19. Three times a number increased by 6 gives 24. Find the number.

Let number be n.
3n + 6 = 24
3n = 18 ⇒ n = 6

Q20. Solve for y: 4/5 * y = 12

Multiply by 5/4 on both sides:
y = 12 × (5/4)
y = 3 × 5 = 15

Class 7 Maths - Simple Equations Summary

Chapter 4: Simple Equations (Concepts & Summary)

1. What is an Equation?

An equation is a condition on a variable such that two expressions in the variable have equal value.
It must always contain an equality sign (`=`).
The expression on the left of the equality sign is the Left Hand Side (LHS).
The expression on the right of the equality sign is the Right Hand Side (RHS).
In an equation, the LHS is strictly equal to the RHS.

2. Variable

The word variable means something that can vary, i.e., change.
A variable takes on different numerical values; its value is not fixed.
Variables are commonly denoted by letters such as x, y, z, l, m, n, p, etc.

3. Solving an Equation

The value of the variable for which the equation is satisfied is called the solution of the equation.
An equation remains unchanged (balanced) if:
- The same number is added to both sides.
- The same number is subtracted from both sides.
- Both sides are multiplied by the same non-zero number.
- Both sides are divided by the same non-zero number.

4. Transposing Method

Transposing a number (i.e., changing its side) is the same as adding or subtracting the number from both sides.
When you transpose a number from one side of the equation to the other side, you change its sign.
- `+` becomes `-`
- `-` becomes `+`
- `×` becomes `÷`
- `÷` becomes `×`

5. Applications (Word Problems)

Equations are extremely useful in dealing with practical situations.
Steps:
1. Read the problem carefully to understand what is given and what is to be found.
2. Choose a variable, say x, to stand for the unknown quantity.
3. Form an equation using the given conditions.
4. Solve the equation to find the unknown value.

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