Solve the problems given in Example 1.
Example 1: Represent the following situations mathematically and solve them:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. How many marbles did each of them have originally?
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. How many toys were produced on that day?
(i) \(9, 36\)
(ii) \(25, 30\)
Key Idea: Convert each situation into an equation by assigning a variable, translating the conditions into algebra, forming a quadratic equation, and then solving by factorisation.
1. Let John originally have \(x\) marbles. Since the total is 45, Jivanti originally has \(45 - x\) marbles.
2. Both lose 5 marbles each, so their new counts become:
John: \(x - 5\), Jivanti: \(45 - x - 5 = 40 - x\).
3. We are told the product of the marbles they now have is 124:
\[(x - 5)(40 - x) = 124\]
4. Expand the expression:
\[(x - 5)(40 - x) = 40x - x^2 - 200 + 5x = -x^2 + 45x - 200\]
So the equation becomes:
\[-x^2 + 45x - 200 = 124\]
5. Move 124 to the left side:
\[-x^2 + 45x - 324 = 0\]
6. Multiply by \(-1\) to simplify:
\[x^2 - 45x + 324 = 0\]
7. Factorise. We need two numbers whose product is 324 and whose sum is 45. These are 9 and 36.
Thus:
\[(x - 9)(x - 36) = 0\]
8. Solving:
\[x = 9 \quad ext{or} \quad x = 36\]
Hence, the original number of marbles was 9 and 36.
1. Let the number of toys produced on that day be \(x\).
2. Cost of production per toy is given as:
\[55 - x\]
3. Total production cost is ₹750. Total cost = (number of toys) × (cost per toy), so:
\[x(55 - x) = 750\]
4. Expand the expression:
\[55x - x^2 = 750\]
5. Rearrange to form a quadratic equation:
\[-x^2 + 55x - 750 = 0\]
6. Multiply by \(-1\) to simplify:
\[x^2 - 55x + 750 = 0\]
7. Factorise the quadratic. We need two numbers whose product is 750 and whose sum is 55. These numbers are 25 and 30.
So:
\[(x - 25)(x - 30) = 0\]
8. Solving:
\[x = 25 \quad ext{or} \quad x = 30\]
Both values satisfy the conditions. So the industry produced 25 or 30 toys on that day (depending on the production-cost model).
Conclusion: Real-life word problems involving sum, product, or cost constraints often lead naturally to quadratic equations, which can be solved efficiently by factorisation.