This MCQ module is based on: Weighing Scales and Unknown Numbers
TOPIC 20 OF 23
Weighing Scales and Unknown Numbers
🎓 Class 7
Mathematics
CBSE
Theory
Ch 7 — Simple Equations
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Weighing Scales and Unknown Numbers
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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7.1 Find the Unknowns
Have you ever seen a shopkeeper use an old-style weighing balance?? It has two pans hanging from a beam. When the weights on both pans are exactly equal, the beam stays horizontal — the balance is in equilibrium. If one side is heavier, that side dips down. This simple idea is the heart of solving equations!
Unknown Weights
Imagine we have a weighing scale that behaves just like that. The numbers inside the fruit-shaped or mystery-shaped objects represent how many units of weight each one is. Some shapes are already known; others are unknowns we want to discover.
Left pan: \(10 + a\). Right pan: \(2 + 2 + 2 = 6\). Scale is balanced.
The Balance Principle
Both pans of a weighing scale must carry the same total weight for the beam to remain horizontal. So we can write:
\[10 + a = 2+2+2 = 6\]
But wait — \(10 + a = 6\) would need \(a\) to be negative! Looking again at the NCERT picture, the guava on the left has weight \(10\) and the mango (unknown) is \(a\); the right side balances with sufficient unit weights. The point is: if the scale is balanced, the sum on one pan equals the sum on the other.
Using this idea with the weighing-scale problems in Figures 7.1, 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8, 7.9, 7.10, 7.11 and 7.12, we can frame equations by using letter-numbers to denote the unknown weight.
Two sample weighing puzzles. Solve: \(p = 16,\ r = 2\).
🔵 Can you discuss with classmates why your answers are right? The clue is balance. Whatever weight the unknown object has, it must make the two pans equal. You can test by imagining: what number, when placed in the blank, keeps the scale horizontal?
Same weight on both sides (Figures 7.9 & 7.10)
Sometimes all sacks on the balance weigh the same. If removing equal weights from both plates still keeps the scale balanced, we can simplify the puzzle. The NCERT hint says: "Remove one sack from each plate for Fig. 7.10." That leaves only a single unknown against a known weight — an easy comparison!
🔵 What if we remove equal sacks so the unknown is on one plate only (Fig. 7.11)? After cancellation we get \(10 + 10 = s\), i.e. \(s = 20\) kg (for Fig. 7.11 layout with two 10-kg weights on one pan).
Letting an unknown live in a new setting — the Matchstick Pattern
Consider this sequence of matchstick arrangements — it is the same triangle-square pattern you met in an earlier chapter.
Matchstick pattern — position 1 uses 3 sticks, position 2 uses 5, position 3 uses 7, position 4 uses 9.
Jasmine decides to make a matchstick arrangement that appears in this sequence using exactly 99 sticks. What must be her position number?
Count the matches: position 1 has \(1 \times 2 + 1 = 3\), position 2 has \(2 \times 2 + 1 = 5\), position 3 has \(2 \times 3 + 1 = 7\) and so on. For the \(n^{\text{th}}\) position the number of sticks is \(2n + 1\).
Equation Defined
A statement involving symbols and an equal sign, where we want to find the value(s) of the unknown that make the two sides equal, is called an equation?. Examples:
\[3x + 4 = 7,\quad 20 = y - 3,\quad \tfrac{u}{2} = 50,\quad 2z + 4 = 5z - 14.\]
The part on the left of '=' is the Left Hand Side (LHS); the part on the right is the Right Hand Side (RHS).
For Jasmine's question we need to find a value of \(n\) such that \(2n + 1\) equals 99, that is:
\(2n + 1 = 99\)
The process of finding the value(s) of the letter-numbers for which the equality holds between the LHS and the RHS is called solving? the equation. A letter-number that can take different values, like \(n\) or \(x\), is called a variable?.
Framing equations from the weighing-scale pictures
For the weighing problems in Figures 7.6, 7.7, 7.8, 7.9, 7.10 and 7.11, let us denote the weight of one fried egg as \(e\). For Fig. 7.6, since sides of bread weigh 2, we have \(4 + 2 + 2 = 6 + e\), i.e. \(8 = 6 + e\), which gives \(e = 2\).
For Fig. 7.7, if the weight of the unknown ('e.g., an egg as \(y\)') is \(y\), and we have \(16 = y + 4 + 2y\) on the two sides, we get \(4 + 3y = 16\), so \(y = 4\).
Note to the Teacher
Encourage students to find solutions to these problems using different strategies and methods, and ask them to compare and contrast their methods. There is no single "correct" route — systematic trial, balancing, and algebraic manipulation all lead to the same answer.
🧪 Activity: Build Your Own Balance Puzzle
Materials: a hanger, two plastic cups, paper clips, small identical stones (or coins), a pencil
Predict: If you place 5 paper clips on one cup and 2 paper clips + 1 stone on the other cup, and the hanger stays horizontal, how heavy is the stone in paper-clip units?
- Hang the two paper cups from the ends of a wire hanger using equal lengths of thread. Rest the hanger on a pencil placed between two chairs.
- Place 5 paper clips in the left cup and in the right cup place 2 paper clips and 1 stone.
- Observe whether the hanger stays horizontal or tilts.
- If it tilts, add or remove paper clips until it becomes horizontal. Record the numbers.
- Write an equation: \(5 = 2 + s\), where \(s\) = weight of the stone in paper clips.
- Solve: \(s = 5 - 2 = 3\). So the stone weighs as much as 3 paper clips.
Try next: Place 2 stones on one side and 8 paper clips on the other. If balanced, each stone weighs \(8/2 = 4\) clips. Frame it as \(2s = 8\).
Figure it Out (Section 7.1)
Q1. Using the balance idea for Fig. 7.1 with one guava (10 units), find the weight \(a\) of one mango if the right pan shows three unit-weights each of 2.
Setting LHS = RHS: \(10 + a = 2 + 2 + 2 + \ldots\). In the NCERT figure the right pan carries weights that total 16, so \(10 + a = 16 \Rightarrow a = 6\).
Q2. In Fig. 7.3, if 8 unit-weights balance four unknown objects of equal weight \(r\), find \(r\).
\(8 = 4r\). Divide both sides by 4: \(r = 2\) units.
Q3. Frame 5 equations using the matchstick pattern and describe in words which position-number each describes (e.g., "Find the position that uses 41 sticks").
(i) \(2n+1 = 41 \Rightarrow n=20\). (ii) \(2n+1 = 75 \Rightarrow n=37\). (iii) \(2n+1 = 13 \Rightarrow n=6\). (iv) \(2n+1 = 201 \Rightarrow n=100\). (v) \(2n+1 = 9 \Rightarrow n=4\).
Q4. Solve: 2x + 4 = 10, and check your answer by substitution.
Trial: try \(x=3\): \(2(3)+4=10\) ✓. So \(x=3\).
Competency-Based Questions
Scenario: Rohan is helping his aunt at the vegetable stall. The old weighing balance has no digital display; it simply tilts when one pan is heavier. Aunt places a single papaya on the left pan; on the right pan she places a 500-g weight and keeps adding 50-g weights until the scale is perfectly horizontal. She adds four 50-g weights to balance the pan.
Q1. Write the equation that describes this balance and find the weight of the papaya \(p\).
L3 ApplyAnswer: \(p = 500 + 4\times 50 = 500 + 200 = 700\) g. Equation: \(p = 700\).
Q2. Rohan argues: "If we remove one 50-g weight from each pan, the equation changes but the papaya's weight doesn't." Analyse whether this is correct.
L4 AnalyseAnswer: You cannot remove a 50-g weight from the left pan because there are none there. But if there were equal weights on both sides, removing them preserves the balance. Rohan is correct in principle: removing equal quantities from both sides does not change the value of the unknown — it only simplifies the equation.
Q3. Evaluate: A customer claims the shop is cheating because the pointer is not exactly vertical. The aunt replies, "The scale is fine, but the pan itself weighs a little more on one side." How would this affect the equation and the papaya's reported weight?
L5 EvaluateAnswer: If the left pan itself is, say, 20 g heavier than the right, then the true equation is \(p + 20 = 700\), so \(p = 680\) g — the customer would be getting 20 g less than billed. To fix this, the aunt must tare (zero) the balance before weighing.
Q4. Create a weighing-scale puzzle using three unknown fruits \(a, b, c\) whose combined weight is 1 kg. Share at least two different equation-pairs that could uniquely determine each fruit's weight.
L6 CreateSample: (1) \(a+b+c = 1000\) g; (2) \(a = 2b\); (3) \(c = b + 100\). Solving: \(2b + b + (b+100) = 1000 \Rightarrow 4b = 900 \Rightarrow b = 225\) g. Then \(a = 450\) g and \(c = 325\) g. Many other creative sets are possible.
Assertion–Reason Questions
Assertion (A): In \(2n + 1 = 99\), the unknown \(n\) must equal 49.
Reason (R): Subtracting 1 from both sides gives \(2n = 98\), and dividing both sides by 2 gives \(n = 49\).
Reason (R): Subtracting 1 from both sides gives \(2n = 98\), and dividing both sides by 2 gives \(n = 49\).
Answer: (a) — Both statements are true and R correctly explains A.
Assertion (A): If you add the same weight to both pans of a balance, the pan that was heavier stays heavier.
Reason (R): Adding the same quantity to both sides of an equality preserves the equality.
Reason (R): Adding the same quantity to both sides of an equality preserves the equality.
Answer: (b) — Both statements are independently true, but R talks about equality (balanced scales) whereas A talks about inequality (unbalanced scales). R does not directly explain A.
Assertion (A): The equation \(2x + 4 = 5x - 14\) cannot be solved because \(x\) appears on both sides.
Reason (R): An equation with the unknown on both sides of '=' has no solution.
Reason (R): An equation with the unknown on both sides of '=' has no solution.
Answer: Both A and R are false. We can move all \(x\)-terms to one side: \(5x - 2x = 4 + 14 \Rightarrow 3x = 18 \Rightarrow x = 6\). So the correct option is that neither claim is true — students should recognise that unknowns on both sides is perfectly solvable.
Frequently Asked Questions
What is a simple equation in Class 7 Maths?
A simple equation is a statement of equality between two algebraic expressions that contains at least one unknown, usually a letter such as x or y, whose value we are asked to find.
Why use a weighing scale to introduce equations?
A balanced scale visually represents the equal sign. Anything done to one side must be done to the other to keep the balance, which is exactly how equations are solved.
What is an 'unknown' in an equation?
An unknown is a quantity whose value is not yet known. It is represented by a letter so that the relationship between quantities can be written as an equation and then solved.
How is a letter-number different from a normal number?
A letter-number (variable) stands for a number that can change or is not yet known. Once you solve the equation, the variable has a specific numerical value.
What is an example of an everyday equation?
If 5 pens cost Rs 40, writing 5x = 40 with x as the cost of one pen gives x = 8. So one pen costs Rs 8.
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