7 th maths chapter 3 || 3.3

7th Standard Mathematics — Chapter 3: Algebra

Exercise 3.3 Solutions

1. Fill in the blanks

1. An expression equated to another expression is called an equation.
2. If a = 5, the value of 2a + 5 is 15.
   Working: 2(5) + 5 = 10 + 5 = 15
3. The sum of twice and four times of the variable x is 6x.
   Working: 2x + 4x = 6x

2. Say True or False

1. Every algebraic expression is an equation
   Answer: False
   Reason: An expression becomes an equation only when it contains an equality sign (=).
2. The expression 7x + 1 can not be reduced without knowing the value of x.
   Answer: True
   Reason: 7x and 1 are unlike terms, so they cannot be added together into a single term without evaluating x.
3. To add two like terms, its coefficients can be added.
   Answer: True
   Reason: Adding like terms involves adding their numerical coefficients together.

3. Solve

1. x + 5 = 8
   x = 8 - 5
   x = 3
2. p - 3 = 7
   p = 7 + 3
   p = 10
3. 2x = 30
   x = 30 ÷ 2
   x = 15
4. m / 6 = 5
   m = 5 × 6
   m = 30
5. 7x + 10 = 80
   7x = 80 - 10
   7x = 70
   x = 70 ÷ 7
   x = 10

4. Algebraic Addition Problem

What should be added to 3x + 6y to get 5x + 8y?

Solution:

Required expression = (5x + 8y) - (3x + 6y)

= 5x + 8y - 3x - 6y

= (5x - 3x) + (8y - 6y)

= 2x + 2y

Answer: 2x + 2y should be added.

5. Number Puzzle Problem

Nine added to thrice a whole number gives 45. Find the number.

Solution:

• Let the required whole number be x.
• Thrice the number = 3x
• According to the question: 3x + 9 = 45

3x = 45 - 9

3x = 36

x = 36 ÷ 3 = 12

Answer: The required number is 12.

6. Consecutive Odd Numbers Problem

Find two consecutive odd numbers whose sum is 200.

Solution:

• Let the first odd number be x.
• Then the next consecutive odd number is x + 2.
• Given sum = 200

x + (x + 2) = 200

2x + 2 = 200

2x = 200 - 2

2x = 198

x = 198 ÷ 2 = 99

• First odd number (x) = 99

• Second odd number (x + 2) = 99 + 2 = 101

Answer: The two consecutive odd numbers are 99 and 101.

7. Taxi Fare Problem

The taxi charges in a city comprise of a fixed charge of ₹ 100 for 5 kms and ₹ 16 per km for every additional km. If the amount paid at the end of the trip was ₹ 740, find the distance travelled.

Solution:

• Fixed charge for initial 5 km = ₹ 100
• Rate for additional distance = ₹ 16 per km
• Total amount paid = ₹ 740
• Let the additional distance travelled beyond 5 km be x km.

Equation:

100 + 16x = 740

16x = 740 - 100

16x = 640

x = 640 ÷ 16 = 40 km (additional distance)

Total Distance Calculation:

Total distance = Initial 5 km + Additional distance (x)

Total distance = 5 + 40 = 45 km

Answer: The total distance travelled is 45 km.