4. Simple Equations

Solutions to Ex. 4.1

Q1. Complete the last column of the table.

Ans:

Q2. Check whether the value given in the brackets is a solution to the given equation or not:

Ans:

(a) n + 5 = 19 (n = 1)

When n = 1, 

LHS = 1 + 5 = 6

6 \( \neq \) 19

Hence, n = 1 is not the solution to this equation.

(b) 7n + 5 = 19 (n = – 2) 

When n = -2,   

LHS = (7 x -2) + 5

       = -14 + 5 = -9

-9 \( \neq \) 19

Hence, n = -2 is not the solution to this equation.

(c) 7n + 5 = 19 (n = 2)

When n = 2,

LHS = (7 x 2) + 5

       = 14 + 5 = 19

19 = 19

Hence, n = 2 is the solution to this equation.

(d) 4p – 3 = 13 (p = 1)

When p = 1

LHS = 4 - 3 = 1

1 \( \neq \) 13

Hence, p = 1 is not the solution to this equation.

(e) 4p – 3 = 13 (p = – 4) 

When p = -4,      

LHS = (4 x -4) - 3

       = -16 - 3 = -19

-19 \( \neq \) 13

Hence, p = -4 is not the solution to this equation.

(f) 4p – 3 = 13 (p = 0) 

When p = 0,     

LHS = 0 - 3 = -3

-3 \( \neq \) 13

Hence, p = 0 is not the solution to this equation.

Q3. Solve the following equations by trial and error method.

Ans:

(i) 5p + 2 = 17

Let's take p = 0

LHS = 5 x 0 + 2 = 2

2 \( \neq \) 17

Let's take p = 1

LHS = 5 x 1 + 2

       = 5 + 2 = 7

7 \( \neq \) 17

Let's take p = 2

LHS = 5 x 2 + 2

       = 10 + 2 = 12

12 \( \neq \) 17

Let's take p = 3

LHS = 5 x 3 + 2

       = 15 + 2 = 17

17 = 17

Hence, p = 3 is the solution of the equation.

(ii) 3m – 14 = 4

Let's take m = 4,

LHS = 3 x 4 - 14

       = 12 - 14 = -2

-2 \( \neq \) 4

Let's take m = 5,

LHS = 3 x 5 - 14

       = 15 - 14 = 1

1 \( \neq \) 4

Let's take m = 6,

LHS = 3 x 6 - 14

       = 18 - 14 = 4

4 = 4

Hence, m = 6 is the solution of the equation.

Q4. Write equations for the following statements:

Ans:

(i) x + 4 = 9

(ii) y-2 = 8

(iii) 10a = 70

(iv) \( \frac{b}{5} \) = 6

(v) \( \frac{3t}{4} \) = 15

(vi) 7m + 7 = 77

(vii) \( \frac{x}{4} \) - 4 =4

(viii) 6y - 6 = 60

(ix)  \( \frac{z}{3} \) + 3 = 30

Q5. Write the following equations in statement forms:

Ans:

(i) The sum of numbers p & 4 is 15.

(ii) 7 subtracted from m is 3

(iii) Two times m is 7.

(iv) One-fifth m gives 3.

(v) Three-fifth m is 6.

(vi) If you add 4 to 3 times p, you get 25.

(vii) If you take away 2 from 4 times p, you get 18.

(viii) If you add 2 to half of p, you get 8.

Q6. Set up an equation in the following cases:

Ans:

(i) 5m + 7 = 37

(ii) 3y + 4 = 49

(iii) 2l + 7 = 87

(iv) b + b + 2b = 180

      4b = 180

Solutions to Ex. 4.2

Q1. Give first the step you will use to separate the variable and then solve the equation:

Ans:

(a) x – 1 = 0

Add 1 to both sides,

x - 1 + 1 = 0 + 1

x = 1

(b) x + 1 = 0

Subtract 1 from both sides,

x + 1 - 1 = 0 - 1

x = -1

(c) x – 1 = 5

Add 1 to both sides,

x - 1 +1 = 5 + 1

x = 6

(d) x + 6 = 2

Subtract 6 from both sides,

x + 6 - 6 = 2 - 6

x = -4

(e) y – 4 = – 7

Add 4 to both sides,

y - 4 + 4 = -7 + 4

y = -3

(f) y – 4 = 4

Add 4 to both sides,

y - 4 + 4 = 4 + 4

y = 8

(g) y + 4 = 4

Subtract 4 from both sides,

y + 4 - 4 = 4 - 4

y = 0

(h) y + 4 = – 4

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:

Ans:

(a) 3l = 42

Divide both sides by 3

\( \frac{3l}{3} \) = \( \frac{42}{3} \)

l = 14

(b) \( \frac{b}{2} \) = 6

Multiply both sides by 2

 \( \frac{b}{2} \) x 2 = 6 x 2

b = 12

(c) \( \frac{p}{7} \) = 4

Multiply both sides by 7

 \( \frac{p}{7} \) x 7 = 4 x 7

p = 28

(d) 4x = 25

Divide both sides by 4

\( \frac{4x}{4} \) = \( \frac{25}{4} \)

x = \( \frac{25}{4} \) = \( 6\frac{1}{4} \)

(e) 8y = 36 

Divide both sides by 4

\( \frac{8y}{8} \) = \( \frac{36}{8} \)

y = \( \frac{36}{8} \) = \( 4\frac{1}{4} \)

(f) \( \frac{z}{3} \) = \( \frac{5}{4} \) 

Multiply both sides by 3

\( \frac{z}{3} \) x 3 = \( \frac{5}{4} \) x 3

z = \( \frac{15}{4} \) = \( 3\frac{3}{4} \)

(g) \( \frac{a}{5} \) = \( \frac{7}{15} \) 

Multiply both sides by 5

\( \frac{a}{5} \) x 5 = \( \frac{7}{15} \) x 5

a = \( \frac{7}{3} \) = \( 2\frac{1}{3} \)

(h) 20t = – 10

Divide both sides by 20

\( \frac{20t}{20} \) = \( \frac{-10}{20} \)

t = \( \frac{-1}{2} \) 

Q3. Give the steps you will use to separate the variable and then solve the equation:

Ans:

(a) 3n – 2 = 46

Add 2 to both sides

3n - 2 + 2 = 46 + 2

3n = 48

Divide both sides by 3

\( \frac{3n}{3} \) = \( \frac{48}{3} \)

n = 16

(b) 5m + 7 = 17

Subtract 7 from both sides

5m + 7 - 7 = 17 - 7

5m = 10

Divide both sides by 5

\( \frac{5m}{5} \) = \( \frac{10}{5} \)

m = 2

(c) \( \frac{20p}{3} \) = 40

Multiply both sides by 3

\( \frac{20p}{3} \) x 3 = 40 x 3

20p = 120

Divide both sides by 20

\( \frac{20p}{20} \) = \( \frac{120}{20} \)

p = 6

(d) \( \frac{3p}{10} \) = 6

Multiply both sides by 10

 \( \frac{3p}{10} \) x 10 = 6 x 10

3p = 60

Divide both sides by 3

\( \frac{3p}{3} \) = \( \frac{60}{3} \)

p = 20

Q4. Solve the following equations:

Ans:

(a) 10p = 100

Divide both sides by 10

\( \frac{p}{10} \) = \( \frac{100}{10} \)

p = 10

(b) 10p + 10 = 100

10p = 100 - 10

10p = 90

Divide both sides by 10

\( \frac{p}{10} \) = \( \frac{90}{10} \)

p = 9

(c) \( \frac{p}{4} \) = 5

Multiply both sides by 4

\( \frac{p}{4} \) x 4 = 5 x 4

p = 20

(d) \( \frac{-p}{3} \) = 5

Multiply both sides by -3

\( \frac{-p}{3} \) x -3 = 5 x -3

p = -15

(e) \( \frac{3p}{4} \) = 6

Multiply both sides by 4

\( \frac{3p}{4} \) x 4 = 6 x 4

3p = 24

Divide both sides by 3

\( \frac{3p}{3} \) = \( \frac{24}{3} \)

p = 8

(f) 3s = –9

Divide both sides by 3

\( \frac{3s}{3} \) = \( \frac{-9}{3} \)

s = -3

(g) 3s + 12 = 0 

Subtract 12 from both sides

3s + 12 - 12 = -12

3s = -12

Divide both sides by 3

\( \frac{3s}{3} \) = \( \frac{-12}{3} \)

s = -4

(h) 3s = 0

Divide both sides by 3

\( \frac{3s}{3} \) = \( \frac{0}{3} \)

s = 0

(i) 2q = 6

Divide both sides by 2

\( \frac{2q}{2} \) = \( \frac{6}{2} \)

q = 3

(j) 2q – 6 = 0

Add 6 to both sides

2q - 6 + 6 = 0 + 6

2q = 6

Divide both sides by 2

\( \frac{2q}{2} \) = \( \frac{6}{2} \)

q = 3

(k) 2q + 6 = 0

Subtract 6 from both sides

2q + 6 - 6 = 0 - 6

2q = -6

Divide both sides by 2

\( \frac{2q}{2} \) = \( \frac{-6}{2} \)

q = -3

(l) 2q + 6 = 12

Subtract 6 from both sides

2q + 6 - 6 = 12 - 6

2q = 6

Divide both sides by 2

\( \frac{2q}{2} \) = \( \frac{6}{2} \)

q = 3

Solutions to Ex. 4.3

Q1. Set up equations and solve them to find the unknown numbers in the following cases:

Ans:

(a) Let the number be x.

Eight times a number is 8x.

Add 4 to 8x, you get 60.

8x + 4 = 60

Transpose 4 to RHS, 

8x = 60 - 4 = 56

Divide both sides by 8,

x = 7

The number is 7.

(b) Let the number be x.

One- fifth of number is x/5.

x/5 minus 4 gives 3

\( \frac{x}{5} \) - 4 = 3

Transpose 4 to RHS,

\( \frac{x}{5} \) = 3 + 4 = 7

Multiply both sides by 5,

x = 35

The number is 35.

(c) Let the number be x.

Three- fourth of number is 3x/4.

Add 3  to 3x/4 to get 21

\( \frac{3x}{4} \) + 3 = 21

Transpose 3 to RHS,

\( \frac{3x}{4} \) = 21 - 3 = 18

Multiply both sides by 4,

3x = 72

Divide both sides by 3,

x = 24

The number is 24.

(d) Let the number be x

Twice the number is 2x.

Subtract 11 from 2x to get 15,

2x - 11 = 15

Transpose 11 to RHS, 

2x = 15 + 11 = 26

Divide both sides by 2, 

x = 13

The number is 13.

(e) Let number of notebooks be x.

Thrice the notebooks is 3x.

Subtract 3x from 50 to get 8,

50 - 3x = 8

Transpose 50 to RHS,

-3x = 8 - 50 = -42

Divide both sides by -3,

x = 14

Number of notebooks Munna has is 14.

(f) Let the number be x.

Add 19 to x and divide the sum by 5 to get 8,

\( \frac{x + 19}{5}\)= 8

Multiply both sides by 5,

x + 19 = 40

Transpose 19 to RHS,

x = 40 - 19 = 21

The number is 21.

(g) Let the number be x. 

5/2 of x is \( \frac{5x}{2} \)

Take away 7 from \( \frac{5x}{2} \) to get 23,

\( \frac{5x}{2} \) - 7 = 23

Transpose 7 to RHS,

\( \frac{5x}{2} \) = 23 + 7 = 30

Multiply both sides by 2,

5x = 60

Divide both sides by 5,

x = 12

The number is 12.

Q2. Solve the following:

Ans:

(a) Let the lowest score be x.

Twice the lowest marks is 2x.

Add 7 to 2x to get 87,

2x + 7 = 87

Transpose 7 to RHS,

2x = 87 - 7 = 80

Divide both sides by 2,

x = 40

The lowest score obtained in the class is 40.

(b) Let the base angle be x.

Vertex angle = 40^o

Base angle + base angle + vertex angle = 180^o (sum interior angles of triangle property)

As base angles are equal,

x + x + 40 = 180

2x + 40 = 180

Transpose 40 to RHS

2x = 180 - 40 = 140

Divide both sides by 2,

x = 70^o

The base angles of the triangle are 70^o.

(c) Let the runs scored by Rahul be x.

Runs scored by Sachin = 2x

Sum of their runs = 198

x + 2x = 198

3x = 198

Divide both sides by 3,

x = 66

Rahul scored 66 runs

Sachin scored = 66 x 2 = 132 runs.

Q3. Solve the following:

Ans:

(i) Let Parmit have x marbles.

5 times the marbles is 5x.

Irfan has 7 more than 5x.

Number of marbles Irfan has is 37

5x + 7 = 37

Transpose 7 to RHS,

5x = 37 - 7 = 30

Divide both sides by 5,

x = 6

Parmit has 6 marbles.

(ii) Let Laxmi's age by x years.

Laxmi's fathers age = 49 years

Three times Laxmi's age is 3x, 

3x + 4 = 49

Transpose 4 to RHS,

3x = 49 - 4 = 45

Divide both sides by 3,

x = 15

Laxmi's age is 15 years.

(iii) Let the number of fruit trees be x.

Number of non-fruit trees = 77

Three times fruit trees = 3x

2 more than 3x is 3x + 2

3x + 2 = 77

Transpose 2 to RHS,

3x = 77 - 2 = 75

Divide both sides by 3, 

x = 25

The number of fruit trees is 25 trees.

Q4. Solve the following riddle:

Ans:

Let the number be x.

7 times the number = 7x,

Add 50 = 7x + 50, 

To reach triple century you need 40 means 260,

7x + 50 = 260

Transpose 50 to RHS,

7x = 260 - 50 = 210

Divide both sides by 7,

x = 30

The number is 30.