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MathematicsClass 7Finding The

Finding The | Class 7 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

Finding The – this guide gives you a concise, exam-ready overview of Finding The from Class 7 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Properties of Equality

This section introduces the fundamental properties of equality that help in solving equations. The two main properties are the Addition Property of Equality and the Multiplication Property of Equality. The Addition Property states that if the same number is added to both sides of an equation, the equality remains true. Similarly, the Multiplication Property states that if both sides of an equation are multiplied by the same non-zero number, the equality remains true. These properties allow us to manipulate equations to isolate the unknown variable. The chapter explains that these properties maintain the balance of the equation, similar to a balance scale. For example, if x + 3 = 7, subtracting 3 from both sides gives x = 4, maintaining equality. The section also cautions that division by zero is undefined and cannot be used. Understanding these properties is essential for solving linear equations efficiently.

📊 Diagram: Diagram showing a balance scale with weights being added or multiplied equally on both sides to maintain balance.

🧪 Activity: No specific activity in this section.

🔗 Connection: Leads to the next section where students apply these properties to solve simple equations.

Frequently asked questions

We have the expression 3k + 1 which gives the number of tiles needed to make an arrangement in Step k. To check whether an arrangement is possible using 100 tiles at some Step k, we can solve the equation: 3k + 1 = 100. Find the value of k.

Given the equation 3k + 1 = 100, Subtract 1 from both sides: 3k = 100 - 1 3k = 99 Divide both sides by 3: k = 99 ÷ 3 k = 33 Therefore, the arrangement using 100 tiles corresponds to Step 33.

Madhubanti wants to organise a party. She decides to buy snacks for the party from the chaat shop in town. Each plate of snacks costs ₹25. The shop charges an additional fixed amount of ₹50 to deliver the snacks to Madhubanti’s house. There are 5 members in Madhubanti’s family, including herself. Her parents tell her she can spend ₹500 on this party. How many friends can she invite to the party if she wants to give a plate of snacks to each person, including her family and friends?

Total money available = ₹500 Delivery charge = ₹50 Money left for snacks = 500 - 50 = ₹450 Cost per plate = ₹25 Number of plates that can be bought = 450 ÷ 25 = 18 Number of family members = 5 Number of friends invited = 18 - 5 = 13 Therefore, Madhubanti can invite 13 friends.

Two friends want to save money. Jahnavi starts with an initial amount of ₹4000, and in addition, saves ₹650 per month. Sunita starts with ₹5050 and saves ₹500 per month. After how many months will they have the same amount of money?

Let m be the number of months after which their savings are equal. Jahnavi's savings after m months = 4000 + 650m Sunita's savings after m months = 5050 + 500m Set equal: 4000 + 650m = 5050 + 500m Subtract 500m from both sides: 4000 + 150m = 5050 Subtract 4000 from both sides: 150m = 1050 Divide both sides by 150: m = 1050 ÷ 150 = 7 Therefore, after 7 months, both will have the same amount of money.

Solve 28 (x + 4) + 300 = 1000.

Given equation: 28(x + 4) + 300 = 1000 Step 1: Subtract 300 from both sides: 28(x + 4) = 1000 - 300 28(x + 4) = 700 Step 2: Divide both sides by 28: x + 4 = 700 ÷ 28 x + 4 = 25 Step 3: Subtract 4 from both sides: x = 25 - 4 x = 21 Therefore, the solution is x = 21.

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