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Q6:
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ $90$, find the number of articles produced and the cost of each article.
Solution :
Given:
1. The total cost of production on a particular day = ₹ $90$.
2. The cost of production of each article is $3$ more than twice the number of articles produced.
To Find:
1. The number of articles produced on that day.
2. The cost of each article.
Step 1: Defining Variables
Let the number of pottery articles produced on that day be $x$.
According to the problem, the cost of production of each article is $3$ more than twice the number of articles produced.
Therefore, the cost of each article = $(2x + 3)$ rupees.
Step 2: Formulating the Quadratic Equation
The total cost of production is calculated by multiplying the number of articles by the cost per article.
Total Cost = (Number of articles) $\times$ (Cost of each article)
$90 = x(2x + 3)$ [Given that total cost is ₹ 90]
$90 = 2x^2 + 3x$ [Distributive property of multiplication]
Rearranging the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:
$2x^2 + 3x - 90 = 0$
Step 3: Solving the Quadratic Equation by Factorization
To factorize $2x^2 + 3x - 90 = 0$, we need to find two numbers whose product is $a \times c = 2 \times (-90) = -180$ and whose sum is $b = 3$.
The factors of $-180$ that add up to $3$ are $15$ and $-12$.
Splitting the middle term:
$2x^2 + 15x - 12x - 90 = 0$
Grouping the terms:
$(2x^2 + 15x) - (12x + 90) = 0$
$x(2x + 15) - 6(2x + 15) = 0$ [Factoring out common terms]
$(2x + 15)(x - 6) = 0$
Step 4: Finding the Roots
Setting each factor to zero:
1. $2x + 15 = 0 \implies 2x = -15 \implies x = -7.5$
2. $x - 6 = 0 \implies x = 6$
Since the number of articles produced cannot be negative, we discard $x = -7.5$.
Therefore, the number of articles produced is $x = 6$.
Step 5: Calculating the Cost of Each Article
Cost of each article = $2x + 3$
Substitute $x = 6$ into the expression:
Cost = $2(6) + 3$
Cost = $12 + 3 = 15$
The cost of each article is ₹ $15$.
Final Answer: The number of articles produced is 6 and the cost of each article is ₹ 15.
More Questions from Class 10 Mathematics Quadratic Equations EXERCISE 4.2
- Q1(i): Find the roots of the following quadratic equations by factorisation: (i) $x^2 – 3x – 10 = 0$
- Q1(ii): Find the roots of the following quadratic equations by factorisation: (ii) $2x^2 + x – 6 = 0$
- Q1(iii): Find the roots of the following quadratic equations by factorisation: (iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
- Q1(iv): Find the roots of the following quadratic equations by factorisation: (iv) $2x^2 – x + \frac{1}{8} = 0$
- Q1(v): Find the roots of the following quadratic equations by factorisation: (v) $100x^2 – 20x + 1 = 0$
- Q2: Solve the problems given below. 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 Rs.750. We would like to find out the number of toys produced on that day.
- Q3: Find two numbers whose sum is $27$ and product is $182$.
- Q4: Find two consecutive positive integers, sum of whose squares is $365$.
- Q5: The altitude of a right triangle is $7$ cm less than its base. If the hypotenuse is $13$ cm, find the other two sides.
CBSE Solutions for Class 10 Mathematics Quadratic Equations
Chapters in CBSE - Class 10 Mathematics
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