CBSE Class 7 Mathematics Finding the Unknown Worksheet Set 01

Very Short Answer Questions

Question 1. Define an equation.
Answer:
An equation is a mathematical statement that says two algebraic expressions are equal, for example 2n + 1 = 99.

Teacher's Note:
1. An equation must contain an equal sign.
2. Give an example along with the definition.

Question 2. What do the short forms LHS and RHS stand for in an equation?
Answer:
LHS means the Left Hand Side, the expression to the left of the equal sign.
RHS means the Right Hand Side, the expression to the right of the equal sign.

Teacher's Note:
1. The equal sign divides an equation into two sides.
2. Each side is a whole expression.

Question 3. A weighing scale is balanced. One pan carries two identical boxes, each of weight w units, and the other pan carries a weight of 14 units. Write an equation for this situation.
Answer:
Both pans carry the same total weight, so 2w = 14 (which gives w = 7 units).

Teacher's Note:
1. w + w is written as 2w.
2. A balanced scale gives an equation directly.

Question 4. Solve the equation y − 8 = 5.
Answer:
Add 8 to both sides: y = 5 + 8 = 13.
Check: 13 − 8 = 5.

Teacher's Note:
1. Undo subtraction by adding.
2. Always check by substitution.

Question 5. In a matchstick sequence, the arrangement at position n needs 2n + 1 sticks. How many sticks are needed for the arrangement at position 10?
Answer:
Put n = 10: 2 × 10 + 1 = 21 sticks.

Teacher's Note:
1. Replace n by the position number.
2. Multiply first, then add.

Question 6. What do we mean when we say that we have solved an equation?
Answer:
It means we have found the value of the letter-number that makes the LHS equal to the RHS.

Teacher's Note:
1. This value is called the solution.
2. Substitution can confirm it.

Question 7. Name the method in which different values are put in place of the unknown until the LHS becomes equal to the RHS.
Answer:
It is the trial and error method.

Teacher's Note:
1. It works, but can be slow for large numbers.
2. The balance method is faster.

Question 8. What happens to a balanced weighing scale when equal weights are removed from both of its pans?
Answer:
The scale stays balanced, because the two pans still carry equal total weights.

Teacher's Note:
1. This is the idea behind subtracting the same number from both sides.
2. Equal minus equal is equal.

Question 9. Solve the equation \( \frac{m}{4} \) = 7.
Answer:
Multiply both sides by 4: m = 7 × 4 = 28.
Check: \( \frac{28}{4} \) = 7.

Teacher's Note:
1. Undo division by multiplying.
2. Multiply the whole other side by the divisor.

Question 10. What was algebra called in ancient India, and what does the word bīja mean?
Answer:
Algebra was called bījagaṇita. The word bīja means seed: the answer lies hidden in the unknown just as a tree lies hidden in a seed.

Teacher's Note:
1. Answer both parts of the question.
2. The seed idea explains the name.

Question 11. A number x is multiplied by 5 and then 3 is added to the product, giving 23. Write this as an equation.
Answer:
Multiplying by 5 gives 5x; adding 3 gives 5x + 3. So the equation is 5x + 3 = 23.

Teacher's Note:
1. Write the steps in the order they happen.
2. (It gives x = 4.)

Question 12. Is 3x + 7 an equation? Give a reason for your answer.
Answer:
No. 3x + 7 has no equal sign, so it is only an expression. An equation must show two expressions as equal.

Teacher's Note:
1. No '=' means no equation.
2. An expression can become part of an equation.

Question 13. Solve 3x − 10 = 38.
Answer:
Add 10 to both sides: 3x = 48. Divide by 3: x = 16.
Check: 3 × 16 − 10 = 48 − 10 = 38.

Teacher's Note:
1. Remove the constant before the coefficient.
2. Check by substitution.

Question 14. Solve 5z = 3z + 12.
Answer:
Subtract 3z from both sides: 2z = 12, so z = 6.
Check: 5 × 6 = 30 and 3 × 6 + 12 = 30.

Teacher's Note:
1. Collect the unknown terms on one side first.
2. Subtract the smaller unknown term.

Question 15. Check whether x = 4 is a solution of 2x + 5 = 13.
Answer:
Put x = 4: LHS = 2 × 4 + 5 = 13 = RHS. So yes, x = 4 is a solution.

Teacher's Note:
1. Substitute and compare both sides.
2. Equal sides mean it is a solution.

Question 16. Solve 2(p + 3) = 16.
Answer:
Divide both sides by 2: p + 3 = 8. Subtract 3: p = 5.
Check: 2 × (5 + 3) = 16.

Teacher's Note:
1. You may keep the bracket and divide first.
2. Opening the bracket gives the same answer.

Question 17. A bag costs ₹120 and each notebook costs ₹35. If n notebooks and one bag together cost ₹400, write an equation and find n.
Answer:
Equation: 35n + 120 = 400.
35n = 280, so n = 280 ÷ 35 = 8 notebooks.

Teacher's Note:
1. The bag cost is the fixed part.
2. Check: 8 × 35 + 120 = 400.

Question 18. Solve 13 − z = 8.
Answer:
Add z to both sides: 13 = 8 + z. Subtract 8: z = 5.
Check: 13 − 5 = 8.

Teacher's Note:
1. Adding z keeps the unknown positive.
2. Then the equation is easy to finish.

Question 19. State the rule that allows us to change \( \frac{u}{12} \) = 5 into u = 60.
Answer:
When a divisor is removed from one side, the other side is multiplied by it: u = 5 × 12 = 60.

Teacher's Note:
1. This is multiplying both sides by 12.
2. Division is undone by multiplication.

Question 20. Write two different equations whose solution is x = 7.
Answer:
For example, x + 3 = 10 and 2x = 14; both are true when x = 7.

Teacher's Note:
1. Start from x = 7 and do the same thing to both sides.
2. Many correct answers are possible.

Question 21. Solve 7m = m − 18.
Answer:
Subtract m from both sides: 6m = −18, so m = −3.
Check: 7 × (−3) = −21 and −3 − 18 = −21.

Teacher's Note:
1. Solutions can be negative.
2. Take care with signs when checking.

Question 22. From 2v − 4 = 6, a student writes v − 4 = 6 − 2. Name the mistake.
Answer:
The student subtracted 2 instead of dividing by the factor 2. Correctly: 2v = 6 + 4 = 10, so v = 5.

Teacher's Note:
1. A factor is removed by division, not subtraction.
2. Remove the −4 first to avoid errors.

Question 23. It is given that 4k + 1 = 13. Without finding k, write the value of 8k + 2.
Answer:
8k + 2 is double 4k + 1, so doubling both sides gives 8k + 2 = 26.

Teacher's Note:
1. Spot that the new expression is a multiple of the old.
2. The same operation keeps the equality.

Question 24. Using Brahmagupta’s rule x = \( \frac{D - B}{A - C} \) for Ax + B = Cx + D, solve 5x + 4 = 3x + 8.
Answer:
A = 5, B = 4, C = 3, D = 8, so x = \( \frac{8 - 4}{5 - 3} \) = \( \frac{4}{2} \) = 2.
Check: 5 × 2 + 4 = 14 and 3 × 2 + 8 = 14.

Teacher's Note:
1. Read A, B, C, D from the equation carefully.
2. The rule is a shortcut for the balance steps.

Question 25. The sum of two numbers is 84 and one number is five times the other. Write an equation and find the smaller number.
Answer:
Let the smaller number be n; the other is 5n. Equation: n + 5n = 84, so 6n = 84 and n = 14.
Check: 14 + 70 = 84.

Teacher's Note:
1. Write the second number in terms of the first.
2. One unknown is then enough.

Question 26. One-fifth of a number increased by 8 gives the same number. Find the number.
Answer:
Equation: \( \frac{x}{5} \) + 8 = x. Multiply by 5: x + 40 = 5x, so 4x = 40 and x = 10.
Check: 10 ÷ 5 + 8 = 2 + 8 = 10.

Teacher's Note:
1. Clear the fraction first.
2. Check by putting the answer back.

Question 27. Explain in one or two lines why the equation k + 5 = k + 9 has no solution.
Answer:
Removing k from both sides leaves 5 = 9, which is never true. So no value of k can make the two sides equal.

Teacher's Note:
1. The RHS is always 4 more than the LHS.
2. Not every equation has a solution.

Question 28. Solve 3 − 7s = 7 − 3s.
Answer:
Add 7s to both sides: 3 = 7 + 4s. Subtract 7: 4s = −4, so s = −1.
Check: 3 − 7 × (−1) = 10 and 7 − 3 × (−1) = 10.

Teacher's Note:
1. Adding 7s keeps the unknown term positive.
2. Watch the signs when s is negative.

Question 29. From −5x = 3, a student writes x = \( \frac{-5}{3} \). Correct the step and write the right solution.
Answer:
Removing the factor −5 means dividing the other side by −5, not the other way round.
So x = \( \frac{3}{-5} \) = \( \frac{-3}{5} \). Check: −5 × (\( \frac{-3}{5} \)) = 3.

Teacher's Note:
1. The RHS goes on top, the coefficient goes below.
2. The sign of the coefficient goes with it.

Question 30. In a number trick you think of a number, subtract 2 from it, multiply the result by 5 and then add 10. The final answer is 40. Write an equation and find the starting number.
Answer:
Equation: 5(x − 2) + 10 = 40, that is 5x − 10 + 10 = 40, so 5x = 40 and x = 8.
Check: 8 − 2 = 6, 6 × 5 = 30, 30 + 10 = 40.

Teacher's Note:
1. Follow the trick step by step with x.
2. Simplify before solving.