Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

1. Complete the following :

Solution:

(i) No

Reason : For x = 5

L.H.S. = x + 5 = 5 + 5 = 10

RHS = 0

Since L.H.S. ≠ R.H.S.

Therefore, given equation is not satisfied for x = 5

(ii) Yes

Reason : For x = -5

L.H.S. = x + 5

= -5 + 5 = 0

R.H.S. = 0

Since L.H.S. = R.H.S.

Therefore, given equation is satisfied for

x = – 5

(iii) NO

Reason : For x = 3

L.H.S. = x – 3

= 3 – 3 = 0

R.H.S. = 1

Since L.H.S. ≠ R.H.S.

Therefore, given equation is not satisfied for x = 3

(iv) No

Reason : x = – 3

L.H.S. = x – 3

= – 3 – 3 = -6

R.H.S. = 1

Since L.H.S. ≠ R.H.S.

Therefore, given equation is not satisfied for x = – 3

(v) Yes

Reason : For x = 5

L.H.S. = 2x

= 2 × 5 = 10

R.H.S. = 10

Since L.H.S. = R.H.S.

Therefore given equation is satisfied for

x = 5

(vi) No

Reason : For x = – 6

L.H.S. = \(\frac{x}{3}\)

= \(\frac {-6}{3}\)

= -2

R.H.S. = 2

Since L.H.S. ≠ R.H.S.

Therefore given equation is not satisfied for x = – 6

(vii) No,

Reason : For x = 0

L.H.S. = \(\frac{x}{2}\)

= \(\frac {0}{2}\) = 0

R.H.S. = 2

Since L.H.S. ≠ R.H.S.

Therefore given equation is not satisfied for x = 0

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

Question (i).

x + 4 = 11 (x = 7)

Answer:

Yes

Check : For x = 1

L.H.S. = x + 4

= 7 + 4 = 11

R.H.S. = 11

Since L.H.S. = R.H.S.

Therefore x = 7 is the solution to the given equation.

Question (ii).

8x + 4 = 28 (x = 4)

Answer:

No

Check: Forx-4

L.H.S. = 8x + 4

= 8 × 4 + 4

= 32 + 4

= 36

R.H.S. = 28

Since L.H.S. ≠ R.H.S.

Therefore x = 4 is not solution to the given equation.

Question (iii).

3m – 3 = 0 (m = 1)

Answer:

Yes

Check : For m = 1

L.H.S. = 3m – 3

= 3 × 1 – 3 = 3 – 3 = 0

R.H.S.= 0

Since L.H.S. = R.H.S.

Therefore, m = 1 is the solution to the given equation.

Question (iv).

\(\frac{x}{5}\) – 4 = -1 (x = 15)

Answer:

Yes

Check : For x = 15

L.H.S. = \(\frac{x}{5}\) – 4

= \(\frac {15}{5}\) – 4 = 3 – 4 = – 1

R.H.S. = – 1

Since L.H.S. = R.H.S.

Therefore, x = 15 is the solution to the given equation.

Question (v).

4x – 3 = 13 (x = 0)

Answer:

No

Check : For x = 0

L.H.S. = 4x – 3

= 4 × 0 – 3

= 0 – 3

= -3

R.H.S. = 13

Since L.H.S. ≠ R.H.S.

Therefore x = 0 is not the solution to the given equation.

3. Solve the following equations by trial and error method

Question (i).

5x + 2 = 17

Answer:

Value of x | L.H.S | R.H.S |

1 | 5 × 1 + 2 = 5 + 2 = 7 | 17 |

2 | 5 × 2 + 2 = 10 + 2 = 12 | 17 |

3 | 5 × 3 + 2 = 15 + 2 = 17 | 17 |

We observe that for x = 3, L.H.S. = R.H.S.

Hence x = 3 is the solution of the given equation.

Question (ii).

3p – 14 = 4

Answer:

Value of p | L.H.S. | R.H.S. |

1 | 3 × 1 – 14 = 3 – 14 = -11 | 4 |

2 | 3 × 2 – 14 = 6 – 14 = -8 | 4 |

3 | 3 × 3- 14 = 9 – 14 = -5 | 4 |

4 | 3 × 4 – 14 = 12 – 14 = -2 | 4 |

5 | 3 × 5 – 14 = 15 – 14 = 1 | 4 |

6 | 3 × 6 – 14 = 18 – 14 = 4 | 4 |

We observe that for p = 6, L.H.S. = R.H.S.

Hence p = 6 is the solution of the given equation

4. Write equations for the following statements.

Question (i).

The sum of numbers x and 4 is 9

Answer:

x + 4 = 9

Question (ii).

3 subtracted from y gives 9

Answer:

y – 3 = 9

Question (iii).

Ten times x is 50

Answer:

10x = 50

Question (iv).

Nine times x plus 6 is 87

Answer:

9x + 6 = 87

Question (v).

One fifth of a number y minus 6 gives 3.

Answer:

\(\frac {1}{5}\)x – 6 = 3

5. Write the following equations in statement form :

Question (i).

x – 2 = 6

Answer:

2 substracted from x is 6

Question (ii).

3y – 2 = 10

Answer:

2 subtracted from 3 times a number y is 10.

Question (iii).

\(\frac{x}{6}\) = 6

Answer:

One sixth of a number x is 6

Question (iv).

7x – 15 = 34

Answer:

15 subtracted 7 times a number x is 34.

Question (v).

\(\frac{x}{2}\) + 2 = 8

Answer:

add 2 to half of a number x to get 8.

6. Write an equation for the following statements :

Question (i).

Raju’s father’s age is 4 years more than five times Raju’s age. Raju’s father is 54 years old.

Answer:

Let x years be Raju’s age

Five times Raju’s age is 5x years

His father’s age will be 4 years more than five times

Raju’s age = 5x + 4

But 4 years more han five times Raju’s age = Raju’s father’s age

Therefore 5x + 4 = 54

Question (ii).

A teacher tells that the highest marks obtained by a student in his class is twice the lowest marks plus 6. The highest score is 86. (Take the lowest score to be x).

Answer:

Let the lowest score to be x

Twice the lowest score plus 6 = 2x + 6

The highest score obtained by a student is twice the lowest score plus 6 = 2x + 6

But highest score = 86

Hence 2x + 6 = 86

Question (iii).

In an isosceles triangle, the vertex angle is twice either base angle (Let the base angle be x in degrees. Remember that the sum of angles of a triangle is 180 degrees).

Answer:

Let the base angle be x (in degrees)

Therefore vertex angle is twice the base angle = 2x (in degrees)

Sum of three angles of a triangle = 180°

∴ x + x + 2x = 180° = 4x = 180°

Question (iv).

A shopkeeper sells mangoes in two types of boxes. One small and one large. The large box contains as many as 8 small boxes plus 4 loose mangoes. The number of mangoes in a large box is given to be 100.

Answer:

Let the mangoes in small box be x.

Large box contains mangoes

= 8 small box + 4

= 8x + 4

But mangoes in large box = 100

∴ 8x + 4 = 100