Represent the following situations mathematically: (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. We would like to find out how many marbles they had to start with. (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. We would like to find out the number of toys produced on that day.
Step-by-Step Solution
Key Concept: Translate the verbal information into algebraic equations. Use the sum and product of two numbers for part (i) and the relation ‘cost per unit × number of units = total cost’ for part (ii). Both lead to quadratic equations of the form $ax^2+bx+c=0$, which are solved using factorisation or the quadratic formula.
### Part (i)
1. Let $x$ = number of marbles John originally had, $y$ = number of marbles Jivanti originally had.
2. From the statement "together have 45 marbles":
$$x + y = 45 \tag{1}$$
3. After each loses 5 marbles, the numbers become $x-5$ and $y-5$. Their product is 124:
$$(x-5)(y-5) = 124 \tag{2}$$
4. Using (1), express $y = 45 - x$ and substitute in (2):
$$(x-5)(45 - x - 5) = 124$$
$$ (x-5)(40 - x) = 124$$
5. Expand and bring all terms to one side:
$$ -x^{2} + 45x - 200 = 124$$
$$ -x^{2} + 45x - 324 = 0$$
Multiply by $-1$:
$$ x^{2} - 45x + 324 = 0$$
6. Solve the quadratic. Discriminant $D = 45^{2} - 4\times1\times324 = 2025 - 1296 = 729 = 27^{2}$.
7. Roots:
$$ x = \frac{45 \pm 27}{2}$$
Hence $x = 36$ or $x = 9$.
8. Using (1), the corresponding $y$ values are $9$ or $36$ respectively.
9. Result: John and Jivanti originally had $36$ and $9$ marbles (order can be interchanged).
### Part (ii)
1. Let $n$ = number of toys produced in a day.
2. Cost of production of each toy = $55 - n$ rupees.
3. Total cost = $750$ rupees, so:
$$ n\,(55 - n) = 750 \tag{3}$$
4. Expand and rearrange:
$$ -n^{2} + 55n - 750 = 0$$
Multiply by $-1$:
$$ n^{2} - 55n + 750 = 0$$
5. Solve the quadratic. Discriminant $D = 55^{2} - 4\times1\times750 = 3025 - 3000 = 25 = 5^{2}$.
6. Roots:
$$ n = \frac{55 \pm 5}{2}$$
Hence $n = 30$ or $n = 25$.
7. Both values give a positive cost per toy ( $55-30 = 25$ rupees and $55-25 = 30$ rupees ), so both are admissible.
8. Result: The cottage industry could have produced either $30$ toys (cost per toy = Rs. 25) or $25$ toys (cost per toy = Rs. 30) on that day.
Correct Answer: Part (i): The two numbers are 36 and 9 (order may be interchanged).\nPart (ii): Number of toys = 30 or 25 (both satisfy the given condition).