CBSE Class 7 Mathematics Finding the Unknown Assignment Set 01

Short Answer Questions

Question 1. The weighing scale shown below is balanced. Every sack has the same weight s units, and the number on a block is its weight in the same units.
(i) Frame an equation for this balance.
(ii) Find s by removing equal weights from both pans.
Answer:
The two pans carry equal weights, so their totals can be equated.
(i) Left pan = s + s + s + 4 = 3s + 4; right pan = 19. Equation: 3s + 4 = 19.
(ii) Remove a weight of 4 from both pans; the scale stays balanced: 3s = 15.
Three equal sacks weigh 15 units, so one sack weighs s = 5 units.
Check: 3 × 5 + 4 = 19, the weight on the right pan.
In simple words: Take the 4 off both sides, then share 15 equally among 3 sacks.

Teacher's Note:
1. Write each pan as an expression first.
2. Removing equal weights is subtracting the same number from both sides.
3. Always check with the original picture.

Question 2. What is an equation? In the equation 2z + 4 = 5z − 14, write the LHS and the RHS, and state what is meant by solving this equation.
Answer:
An equation is a statement that two algebraic expressions are equal, with an “=” sign between them.
LHS (Left Hand Side) = 2z + 4; RHS (Right Hand Side) = 5z − 14.
Solving the equation means finding the value of z that makes the LHS equal to the RHS. (Here z = 6, since both sides become 16.)
In simple words: An equation sets two expressions equal; solving it means finding z that makes both sides the same.

Teacher's Note:
1. The whole expression on each side is the LHS or RHS.
2. Answer all three parts.
3. Checking the value of z is a bonus.

Question 3. Solve 4x − 9 = 23 step by step. Write beside each step the operation you have done to both sides, and check the solution.
Answer:
4x − 9 = 23
Add 9 to both sides: 4x − 9 + 9 = 23 + 9, so 4x = 32.
Divide both sides by 4: 4x ÷ 4 = 32 ÷ 4, so x = 8.
Check: LHS = 4 × 8 − 9 = 32 − 9 = 23 = RHS. So x = 8 is the solution.
In simple words: Add 9 to both sides, then divide both sides by 4.

Teacher's Note:
1. Name the operation at each step.
2. Undo subtraction first, then multiplication.
3. Substitute back to check.

Question 4. In a sequence of rangoli designs, the design at position n is made with 4n + 3 petals. Which position of the sequence uses exactly 43 petals? Write the equation you solve.
Answer:
Let the position be n. Equation: 4n + 3 = 43.
Subtract 3 from both sides: 4n = 40.
Divide both sides by 4: n = 10.
So the design at position 10 uses 43 petals. Check: 4 × 10 + 3 = 43.
In simple words: Set the petal rule equal to 43 and solve for n.

Teacher's Note:
1. The rule gives the number of petals at any position.
2. The answer must be a counting number.
3. Check by putting n = 10 in the rule.

Question 5. Solve 2m + 5 = 17 by the trial and error method, showing at least three trials. Why can this method be slow for larger numbers?
Answer:
Try values of m in 2m + 5:
m = 3 gives LHS = 11, too small.
m = 5 gives LHS = 15, still less than 17.
m = 6 gives LHS = 17 = RHS. So m = 6.
For larger numbers we may have to try a great many values before the LHS equals the RHS, so the method becomes slow.
In simple words: Keep trying numbers until both sides match; with big numbers this takes many tries.

Teacher's Note:
1. Use each trial to decide whether to go higher or lower.
2. Trial and error is fine for small answers.
3. The balance method is faster in general.

Question 6. It is given that 12 + 9 = 21. Show that the equality still holds when
(i) 7 is added to both sides,
(ii) both sides are multiplied by 4.
Which property of an equation does this illustrate?
Answer:
(i) LHS = 12 + 9 + 7 = 28 and RHS = 21 + 7 = 28. The equality holds.
(ii) LHS = (12 + 9) × 4 = 84 and RHS = 21 × 4 = 84. The equality holds.
This shows that doing the same operation on both sides of an equation keeps the equality, just as a balanced scale stays balanced.
In simple words: Doing the same thing to both equal sides keeps them equal.

Teacher's Note:
1. In (ii), multiply the whole LHS, not just one number.
2. This is the balance property.
3. It is the basis of solving equations.

Question 7. Ancient Indian mathematicians wrote 2x + 1 as “yā 2 rū 1”.
(i) What did “yā” and “rū” stand for?
(ii) How was a negative number shown in this notation?
Answer:
(i) “yā” (short for yāvat-tāvat) stood for the unknown number, like our x. “rū” (short for rūpa) stood for the known quantity, the constant term.
(ii) A negative number was shown by putting a dot over the number. So 2x − 8 was written as “yā 2 rū 8” with a dot over the 8.
In simple words: yā meant the unknown, rū the known number, and a dot over a number made it negative.

Teacher's Note:
1. yā is like x and rū is like the constant.
2. An example makes part (ii) clear.
3. This is from the 'A Pinch of History' section.

Question 8. (i) Check by substitution whether y = 3 is a solution of 6y − 5 = 4y + 2.
(ii) If it is not, find the correct solution.
Answer:
(i) Put y = 3: LHS = 6 × 3 − 5 = 13 and RHS = 4 × 3 + 2 = 14. Since LHS ≠ RHS, y = 3 is not a solution.
(ii) Subtract 4y from both sides: 2y − 5 = 2. Add 5: 2y = 7, so y = \( \frac{7}{2} \) = 3.5.
Check: LHS = 21 − 5 = 16 and RHS = 14 + 2 = 16.
In simple words: y = 3 does not balance the sides; solving properly gives y = 3.5.

Teacher's Note:
1. Unequal sides mean the value is not a solution.
2. Solutions need not be whole numbers.
3. Check the new answer too.

Question 9. In the balanced scale below, all the sacks have the same weight s units and the numbers on the blocks are their weights.
(i) Frame an equation for the balance.
(ii) Remove equal numbers of sacks and equal weights from the two pans to find s.
Answer:
(i) Left pan = 5s + 4; right pan = 2s + 22. Equation: 5s + 4 = 2s + 22.
(ii) Remove 2 sacks from each pan: 3s + 4 = 22.
Now remove a weight of 4 from each pan: 3s = 18.
Three equal sacks weigh 18 units, so s = 6 units.
Check: 5 × 6 + 4 = 34 and 2 × 6 + 22 = 34.
In simple words: Take 2 sacks and then 4 units off both pans; 3 sacks are left weighing 18.

Teacher's Note:
1. Count the sacks on each pan carefully.
2. Remove sacks first, then the small block.
3. Both pans must give the same total in the check.

Question 10. It is known that 7452 − 318 + 96 + 45 = 7275. Without working out the big numbers again, find the value of 7452 − 318 + 96. Which fact about operations allows you to do this?
Answer:
The second expression is the first one without the term + 45.
Subtract 45 from both sides of the given equality: 7452 − 318 + 96 + 45 − 45 = 7275 − 45.
So 7452 − 318 + 96 = 7230.
This works because addition and subtraction are inverse operations, and doing the same subtraction on both sides keeps the equality.
In simple words: Take 45 away from both sides of the known equality.

Teacher's Note:
1. Compare the two expressions to find what is missing.
2. Subtracting 45 undoes adding 45.
3. No long calculation is needed.

Question 11. An auto-rickshaw charges a fixed amount of ₹30 for a trip and ₹12 for every kilometre travelled. One trip cost ₹174. Taking the distance as d kilometres, frame an equation and find d.
Answer:
Cost of a trip of d km = 30 + 12d rupees.
Equation: 30 + 12d = 174.
Subtract 30: 12d = 144. Divide by 12: d = 12.
So the trip was 12 km long. Check: ₹30 + ₹12 × 12 = ₹30 + ₹144 = ₹174.
In simple words: Take off the fixed ₹30, then divide the rest by ₹12 per km.

Teacher's Note:
1. Fixed charge + rate × distance.
2. Remove the fixed part first.
3. Check with the total fare.

Question 12. Solve 3(p + 5) − 4 = 26 in two different ways and check that both give the same value of p.
Answer:
Method 1 (keep the bracket): 3(p + 5) − 4 = 26 ⇒ 3(p + 5) = 30 ⇒ p + 5 = 10 ⇒ p = 5.
Method 2 (open the bracket first): 3p + 15 − 4 = 26 ⇒ 3p + 11 = 26 ⇒ 3p = 15 ⇒ p = 5.
Check: 3 × (5 + 5) − 4 = 30 − 4 = 26 = RHS. Both methods give p = 5.
In simple words: Whether you open the bracket or not, p comes out as 5.

Teacher's Note:
1. When opening the bracket, multiply both terms by 3.
2. Different routes lead to the same solution.
3. Check with the original equation.

Question 13. Solve \( \frac{u}{7} \) + 4 = 10. State the rule used when the divisor 7 is removed from the LHS, and verify your answer.
Answer:
Subtract 4 from both sides: \( \frac{u}{7} \) = 10 − 4 = 6.
Rule: when a divisor is removed from one side, the other side is multiplied by that divisor. So u = 6 × 7 = 42.
Check: \( \frac{42}{7} \) + 4 = 6 + 4 = 10 = RHS.
In simple words: Remove the 4, then multiply by 7.

Teacher's Note:
1. Remove the added 4 before the divisor.
2. Removing a divisor means multiplying.
3. Verify by substituting.

Question 14. The sum of two numbers is 84, and the larger number is 5 times the smaller one. Taking the smaller number as x, frame an equation in one unknown and find both numbers.
Answer:
Smaller number = x, larger number = 5x.
Equation: x + 5x = 84, that is 6x = 84.
x = 84 ÷ 6 = 14 and the larger number = 5 × 14 = 70.
The numbers are 14 and 70. Check: 14 + 70 = 84 and 70 = 5 × 14.
In simple words: Call the smaller one x, the larger 5x, and use their sum.

Teacher's Note:
1. Write the second number using the first.
2. Find both numbers, as asked.
3. Check both conditions.

Question 15. Priya solved 5 − 3z = 11 as shown: 3z = 11 − 5, so 3z = 6 and z = 2. Describe her mistake, mend it, and give the correct solution with a check.
Answer:
Mistake: the unknown term is −3z, not 3z. Priya moved 5 correctly but also changed −3z into 3z, which is not allowed.
Correct working: 5 − 3z = 11 ⇒ −3z = 11 − 5 = 6 ⇒ z = 6 ÷ (−3) = −2.
Check: 5 − 3 × (−2) = 5 + 6 = 11 = RHS.
In simple words: She dropped the minus sign of −3z; the right answer is z = −2.

Teacher's Note:
1. The sign belongs to the term that follows it.
2. Only the moved term changes sign.
3. A negative answer is fine.

Question 16. Given that 6k − 7 = 29, find the values of (i) 6k, (ii) k and (iii) 12k − 14. For part (iii), get the answer without substituting the value of k.
Answer:
(i) Add 7 to both sides: 6k = 36.
(ii) Divide both sides by 6: k = 6.
(iii) 12k − 14 = 2 × (6k − 7), so multiplying both sides of the given equation by 2 gives 12k − 14 = 2 × 29 = 58.
Check: 12 × 6 − 14 = 72 − 14 = 58.
In simple words: Add 7, divide by 6, and double the whole equation for part (iii).

Teacher's Note:
1. Notice that 12k − 14 is double 6k − 7.
2. Doubling both sides keeps the equality.
3. The check uses k = 6.

Question 17. For an equation of the form Ax + B = Cx + D, Brahmagupta gives the rule x = \( \frac{D - B}{A - C} \). Use this rule to solve 7x + 5 = 3x + 29, and then solve the same equation step by step to show that both methods agree.
Answer:
By the rule: A = 7, B = 5, C = 3, D = 29, so x = \( \frac{29 - 5}{7 - 3} \) = \( \frac{24}{4} \) = 6.
Step by step: 7x + 5 = 3x + 29 ⇒ 7x − 3x = 29 − 5 ⇒ 4x = 24 ⇒ x = 6.
Both methods agree: x = 6. Check: LHS = 42 + 5 = 47 and RHS = 18 + 29 = 47.
In simple words: The rule and the usual steps both give x = 6.

Teacher's Note:
1. The rule is the balance method done once in general.
2. Match A, B, C, D carefully.
3. Check with both sides.

Question 18. Begin with the equation z = 4 and build a chain of three new equations, each obtained from the previous one by doing the same operation to both sides. Explain why all the equations in your chain have the same solution.
Answer:
One possible chain, starting from z = 4:
Multiply both sides by 5: 5z = 20.
Add 3 to both sides: 5z + 3 = 23.
Subtract z from both sides: 4z + 3 = 23 − z.
Each step does the same thing to both sides, so the equality is never disturbed. A value that makes one equation true makes the next one true too, so all of them have the solution z = 4.
Check: 4 × 4 + 3 = 19 and 23 − 4 = 19.
In simple words: Doing the same thing to both sides at every step keeps the solution z = 4.

Teacher's Note:
1. Many different chains are correct.
2. Each step must act on both sides.
3. Check the last equation with z = 4.

Question 19. The weight of a water jug is 2 kg more than one-third of its weight. Taking the weight of the jug as w kg, frame an equation, find w and check your answer.
Answer:
One-third of the weight is \( \frac{w}{3} \), so w = \( \frac{w}{3} \) + 2.
Multiply both sides by 3: 3w = w + 6.
Subtract w from both sides: 2w = 6.
Divide both sides by 2: w = 3. The jug weighs 3 kg.
Check: one-third of 3 kg is 1 kg, and 1 + 2 = 3 kg, which is the weight itself.
In simple words: Weight = one-third of weight + 2 gives 3 kg.

Teacher's Note:
1. Multiply by 3 to clear the fraction.
2. Multiply every term, including the 2.
3. Check against the wording.

Question 20. (i) Show that the equation 4t + 7 = 4t + 10 has no solution.
(ii) Write one equation of your own that is satisfied by every number, and explain why it is satisfied by every number.
Answer:
(i) Subtract 4t from both sides: 7 = 10, which is false. Whatever t is, the LHS is always 3 less than the RHS, so no value of t works.
(ii) Example: 5n + 3 = 3 + 5n. Both sides are the same expression written in a different order, so for every value of n the LHS equals the RHS. Every number is a solution.
In simple words: The first equation can never balance; the second balances for every number.

Teacher's Note:
1. Removing the same term from both sides exposes the problem.
2. Same expression on both sides means every number works.
3. Other examples such as 2(n + 1) = 2n + 2 are also correct.