NCERT Class X Chapter 4: Quadratic Equation Example 1 (ii)

NCERT Class X Chapter 4: Quadratic Equation Example 1 (ii)

Question:

Represent the following situations mathematically: 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.

Given:

Cost of production of each toy = 55 - (number of toys produced)
Total cost of production on a particular day = 750 rupees

To Find:

The number of toys produced on that day.

Formula:

Total cost of production = (Number of toys) × (Cost of production of each toy)

Solution:

Let,the number of toys produced in a day = x

Then,the cost of production of each toy = 55 - x
Total cost of production = (Number of toys) × (Cost of each toy)

Total cost of production = x(55 - x)
On the given day, Total cost of production = 750
⇒ x(55 - x) = 750
⇒ 55x - x2 = 750
Subtracting on both sides by 750 to make right side of '=' 0
⇒ -x2 + 55x - 750 = 0
Multiplying by -1:
⇒ x2 - 55x + 750 = 0
WKT,
The quadratic formula, x = -b ± √(Δ) 2a
Here, a = 1, b = -55, c = 750
Discriminant (Δ) = b2 - 4ac
⇒ Δ = (-55)2 - 4(1)(750)
⇒ Δ = 3025 - 3000
⇒ Δ = 25
√(Δ) = √(25) = 5
Now, substitute values into the quadratic formula:
The quadratic formula, x = (55 ± 5) 2
Two possible values for x:

x1 = (55 + 5) 2  = 60 2  = 30
x2 = (55 - 5) 2  = 50 2  = 25
Case 1:

If the number of toys produced (x) = 30

Cost of each toy = 55 - 30 = 25 rupees

Total cost = 30 × 25 = 750 rupees (This matches the given condition)

Case 2:

If the number of toys produced (x) = 25

Cost of each toy = 55 - 25 = 30 rupees

Total cost = 25 × 30 = 750 rupees (This also matches the given condition)

Result:

The number of toys produced on that day could be either 25 or 30.
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