Std 6 Mathematics · Chapter 9

9. Equations

Solutions to Practice Set 9.1, 9.2, 9.3, 9.4 and Exercise 9. Maharashtra State Board (Balbharati), Grade Six Mathematics.

About the chapter

In this chapter, we learn the order of operations called BODMAS. We learn what an equality is and its properties. We use letters in place of numbers. Then we write equations and solve them.

Key points

In an expression, first solve the brackets. Then do division and multiplication. Last, do addition and subtraction.
Solve the innermost bracket ( ) first, then [ ] and then { }.
If we add, subtract, multiply or divide both sides of an equality by the same number, the equality stays.
A letter used in place of a number is called a variable.
The value of the variable which makes both sides equal is the solution of the equation.

Formulas and terms

BODMAS – Brackets, Of, Division, Multiplication, Addition, Subtraction.
Equation – an equality with a variable, like x + 4 = 21.
a + b = b + a, a + 0 = a
a × 0 = 0, a × 1 = a
Profit = Selling price − Cost price, Loss = Cost price − Selling price

Practice Set 9.1

1Complete the following steps.

(i) 32 × 4 ÷ 8 − 6
= 128 ÷ 8 − 6
= 16 − 6
= 10

(ii) 65 ÷ 5 × 2 − 20
= 13 × 2 − 20
= 26 − 20
= 6

(iii) 80 × (15 + 8 − 3) + 5
= 80 × (23 − 3) + 5
= 80 × 20 + 5
= 1600 + 5
= 1605

(iv) 100 ÷ (18 + 5 − 3) + 7
= 100 ÷ (23 − 3) + 7
= 100 ÷ 20 + 7
= 5 + 7
= 12

(v) 4 + {100 ÷ [3(7 − 5) + 4]}
= 4 + {100 ÷ [3 × 2 + 4]}
= 4 + {100 ÷ [6 + 4]}
= 4 + {100 ÷ 10}
= 4 + 10
= 14

(vi) 210 ÷ {125 − [5 + 3(14 − 9)]}
= 210 ÷ {125 − [5 + 3 × 5]}
= 210 ÷ {125 − [5 + 15]}
= 210 ÷ {125 − 20}
= 210 ÷ 105
= 2

2Find the values of the following expressions.

(i) 28 + 12 − 25
= 40 − 25
= 15

(ii) 8 × 4 − 3 × 6
= 32 − 18
= 14

(iii) 112 − 18 × 2 ÷ 9 + 2
= 112 − 36 ÷ 9 + 2
= 112 − 4 + 2
= 110

(iv) 13 − [4 + (32 ÷ 4 − 5)]
= 13 − [4 + (8 − 5)]
= 13 − [4 + 3]
= 13 − 7
= 6

(v) {100 ÷ [3(18 − 12) + 7]} + 6
= {100 ÷ [3 × 6 + 7]} + 6
= {100 ÷ [18 + 7]} + 6
= {100 ÷ 25} + 6
= 4 + 6
= 10

3Sharvari made expressions using the numbers 1 to 4 whose value is 21. Help her make more such expressions.

Some more expressions whose value is 21:

(4 + 3) × (2 + 1) = 7 × 3 = 21
4 × 4 + 4 + 1 = 16 + 4 + 1 = 21
3 × (4 + 2) + 3 = 18 + 3 = 21
2 × 4 × 3 − 3 = 24 − 3 = 21

4Obtain a verbal example from each of the expressions given below.

(i) (5 + 4) − 8
There were 5 mangoes in a basket. 4 more mangoes were put in it. Then 8 mangoes were eaten. How many mangoes are left?
Answer: (5 + 4) − 8 = 9 − 8 = 1 mango

(ii) (6 × 12) ÷ 4
There are 6 boxes. Each box has 12 pencils. The pencils are shared equally among 4 friends. How many pencils does each friend get?
Answer: 72 ÷ 4 = 18 pencils

(iii) 2 × 6 + 3 × 4 + 1
Rahul bought 2 pens at ₹6 each and 3 notebooks at ₹4 each. He also bought an eraser for ₹1. How much did he spend?
Answer: 12 + 12 + 1 = ₹25

Practice Set 9.2

1Find the values on the left and right and write '=' or '≠' in the box.

(i) 5 + 9 = 14 and 2 × 7 = 14
∴ 5 + 9 = 2 × 7

(ii) 18 ÷ 3 = 6 and 9 × 3 = 27
∴ 18 ÷ 3 ≠ 9 × 3

(iii) 6 ÷ 2 + 4 = 3 + 4 = 7 and 13 − 6 = 7
∴ 6 ÷ 2 + 4 = 13 − 6

2Write the correct number in the blank so that both sides are equal.

(i) 16 + 4 = 20
∴ 16 + 4 = 12 + 8

(ii) 5 + 8 = 13
∴ 26 ÷ 2 = 5 + 8

(iii) 21 ÷ 3 = 7
∴ 38 − 31 = 21 ÷ 3

3Create your own examples of equality using the given numbers and symbols.

1) 13 + 7 = 4 × 5 (20 = 20)
2) 12 + 6 = 3 × 6 (18 = 18)
3) 8 × 5 = 4 × 10 (40 = 40)
4) 12 ÷ 4 = 6 − 3 (3 = 3)
5) 10 − 4 = 13 − 7 (6 = 6)

✦Properties of equality — complete the table for (6 + 4) = (5 × 2).

Property	Left side	Right side
Addition	(6 + 4) + 5 = 10 + 5 = 15	(5 × 2) + 5 = 10 + 5 = 15
Subtraction	(6 + 4) − 2 = 10 − 2 = 8	(5 × 2) − 2 = 10 − 2 = 8
Multiplication	(6 + 4) × 3 = 10 × 3 = 30	(5 × 2) × 3 = 10 × 3 = 30
Division	(6 + 4) ÷ 2 = 10 ÷ 2 = 5	(5 × 2) ÷ 2 = 10 ÷ 2 = 5

When we do the same operation with the same number on both sides, the equality does not change.

Practice Set 9.3 – using letters

1Write mathematical expressions using letters.

(i) The sum of a number and 9 – a + 9
(ii) A number divided by 8 – a ÷ 8, that is a/8
(iii) Multiplication of a number and 0 – a × 0 = 0
(iv) 12 is subtracted from a number – a − 12
(v) 4 times a number – 4a
(vi) Multiply a number by 1 – a × 1 = a

2Rewrite the rules and formulas using the letters given in brackets.

(i) Profit = Selling price − Cost price → P = S − C
(ii) Loss = Cost price − Selling price → L = C − S

✦Complete the table of rules for addition and multiplication.

Example	Rule	In words
6 + 3 = 9; 3 + 6 = 9	a + b = b + a	The sum does not change if the order of numbers is changed.
3 + 0 = 3 5 + 0 = 5	a + 0 = a	When zero is added to a number, the sum is the same number.
7 × 0 = 0; 25 × 0 = 0	a × 0 = 0	The product of any number and 0 is zero.
8 × 1 = 8	a × 1 = a	When a number is multiplied by 1, the product is the same number.

Practice Set 9.4

1Write equations from the information given below.

(i) 4 times a number is 24. – 4x = 24
(ii) If one number is subtracted from 37, 12 remains. – 37 − x = 12
(iii) Shrimayi gave 8 of her fruits to Sai and now has 15 left. – x − 8 = 15
(iv) Samarth divides a number by 7 and gets 4 as the quotient. – x/7 = 4

2Determine which values of the variables are solutions of the equation.

(i) 7x = 14
If x = 1, 7 × 1 = 7 ≠ 14
If x = 2, 7 × 2 = 14
∴ x = 2 is the solution.

(ii) p + 6 = 9
If p = 1, 1 + 6 = 7; if p = 2, 2 + 6 = 8
If p = 3, 3 + 6 = 9
∴ p = 3 is the solution.

(iii) 4 = y/8
If y = 31, 31/8 is not 4
If y = 32, 32/8 = 4
∴ y = 32 is the solution.

(iv) 8 − m = 5
If m = 1, 8 − 1 = 7; if m = 2, 8 − 2 = 6
If m = 3, 8 − 3 = 5
∴ m = 3 is the solution.

3Solve the following equations using the properties of equality.

(i) 3x = 12
∴ 3x/3 = 12/3 … dividing both sides by 3
∴ x = 4

(ii) x + 4 = 21
∴ x + 4 − 4 = 21 − 4 … subtracting 4 from both sides
∴ x = 17

(iii) 17 − p = 8
∴ 17 − p + p = 8 + p … adding p to both sides
∴ 17 = 8 + p
∴ 17 − 8 = 8 + p − 8 … subtracting 8 from both sides
∴ p = 9

(iv) k/8 = 9
∴ k/8 × 8 = 9 × 8 … multiplying both sides by 8
∴ k = 72

4Complete the following activity.

(i) The sum of two consecutive natural numbers is 131. Find those numbers.

Let the two consecutive numbers be m and m + 1.
∴ m + (m + 1) = 131
∴ 2m + 1 = 131
∴ 2m + 1 − 1 = 131 − 1 … subtracting 1 from both sides
∴ 2m = 130
∴ 2m/2 = 130/2 … dividing both sides by 2
∴ m = 65
∴ m + 1 = 65 + 1 = 66

The two numbers are 65 and 66.

(ii) A pant costs ₹250 more than a shirt, and together they cost ₹1250. Find the price of both.

Let the price of the shirt be x.
∴ The price of the pant is x + 250.
∴ x + (x + 250) = 1250
∴ 2x + 250 = 1250
∴ 2x = 1000 … subtracting 250 from both sides
∴ x = 500 … dividing both sides by 2
∴ Price of the shirt = ₹500
∴ Price of the pant = 500 + 250 = ₹750

5Write and solve equations from the following verbal examples.

(i) A garden has 36 Amla trees and the rest are Eucalyptus. There are 150 trees in all. How many Eucalyptus trees are there?

Let the number of Eucalyptus trees be x.
36 + x = 150
36 + x − 36 = 150 − 36
x = 114

There are 114 Eucalyptus trees in the garden.

(ii) Of the food grains supplied, 34 quintals were distributed and 48 quintals remained. How many quintals were available in total?

Let the total food grains be x quintals.
x − 34 = 48
x − 34 + 34 = 48 + 34
x = 82

82 quintals of food grains were available in total.

Exercise 9

1Choose the correct option.

(i) If x + 12 = 27 then x = ? — (b) 15
x = 27 − 12 = 15
(ii) If 23 − n = 12 then n = ? — (a) 11
n = 23 − 12 = 11
(iii) If 12a = 96 then a = ? — (b) 8
a = 96 ÷ 12 = 8
(iv) 24 ÷ (3 + 5) − 2 = ? — (d) 1
24 ÷ 8 − 2 = 3 − 2 = 1
(v) In which pair is there an equality? — (c) 5 × 3; 14 + 1
5 × 3 = 15 and 14 + 1 = 15
(vi) The sum of a number and its triple is 24. Which is that number? — (b) 6
x + 3x = 24, so 4x = 24 and x = 6.

2Complete the equation by placing the correct value in the box.

(i) 12 ÷ 3 = 4
∴ 12 ÷ 3 = 9 − 5

(ii) 15 + 7 = 22
∴ 15 + 7 = 11 × 2

(iii) 15 − 12 = 3
∴ 48 ÷ 16 = 15 − 12

3Solve the following expressions.

(i) 8 × 4 − 17
= 32 − 17
= 15

(ii) 9 ÷ (13 − 10) + 5
= 9 ÷ 3 + 5
= 3 + 5
= 8

(iii) 100 ÷ {6 + [12 − 4 + (5 × 3 − 4)]}
= 100 ÷ {6 + [12 − 4 + (15 − 4)]}
= 100 ÷ {6 + [12 − 4 + 11]}
= 100 ÷ {6 + 19}
= 100 ÷ 25
= 4

4Solve the following equations using the properties of equality.

(i) x + 12 = 25
∴ x + 12 − 12 = 25 − 12
∴ x = 13

(ii) 6p = 48
∴ 6p/6 = 48/6
∴ p = 8

(iii) m/7 = 7
∴ m/7 × 7 = 7 × 7
∴ m = 49

(iv) 27 − t = 14
∴ 27 − t + t = 14 + t
∴ 27 = 14 + t
∴ 27 − 14 = t
∴ t = 13

5Sonal thought of a number and multiplied it by five; the answer was 30. What is that number?

Let the number be x.
5x = 30
5x/5 = 30/5
x = 6

The number is 6.

6Deepali's age is 5 years less than Asma's age. The sum of their ages is 27 years. What are their ages?

Let Asma's age be x years.
∴ Deepali's age = (x − 5) years
x + (x − 5) = 27
2x − 5 = 27
2x = 27 + 5 = 32
x = 16

Asma's age is 16 years. Deepali's age is 16 − 5 = 11 years.

7Students collected funds equal to four times their number. Teachers added ₹1500 and the total became ₹5140. How many students are there?

Let the number of students be x.
4x + 1500 = 5140
4x = 5140 − 1500 = 3640
x = 3640 ÷ 4 = 910

There are 910 students in the school.

8The number of Neem trees is 52 more than the number of Peepal trees. The total number of both is 170. How many Neem trees were planted?

Let the number of Peepal trees be x.
∴ Neem trees = x + 52
x + (x + 52) = 170
2x + 52 = 170
2x = 170 − 52 = 118
x = 59
Neem trees = 59 + 52 = 111

111 Neem trees were planted.

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