Ex 4.2, 6 - Chapter 4 Class 10 Quadratic Equations

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 4.2, 6
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 Rs 90, find the number of articles produced and the cost of each article.
We know that
Total Cost of Production = Number of articles produced × Cost of article
Given
Total cost of production = Rs 90
Also,
Cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day
Let Number of articles = x
Hence,
Cost of article = 2x + 3
Now,
Total Cost = Number of articles produced × Cost of article
90 = x (2x + 3)
x (2x + 3) = 90
x(2x) + 3x = 90
2x2 + 3x = 90
2x2 + 3x – 90 = 0
We factorize using
splitting the middle term method
2x2 + 15x – 12x – 90 = 0
x (2x + 15) – 6 (2x + 15) = 0
(x – 6) (2x + 15) = 0
So, x = 6 & x = (−15)/2 are the roots of the equation
Since x cannot be negative as number of articles is not negative
Hence, x = 6
x – 6 = 0
x = 6
2x + 15 = 0
2x = –15
x = (−𝟏𝟓)/𝟐
Therfore,
Number of articles = x = 6
& Cost of articles = 2x + 3
= 2(6) + 3
= 12 + 3
= Rs 15

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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