10th cbse maths solution for exercise 4.2 part 2

This page 10th cbse maths solution for exercise 4.2 part 2 is going to provide you solution for every problems that you find in the exercise no 4.2

10th CBSE maths solution for Exercise 4.2 part 2

(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, the total cost of production was Rs. 750. We would like to find the number of toys produced on that day.

Solution:

Let 'x' be the number of toys produced in a particular day

cost of production of one toy = 55 - x

Total cost = Number of toys produced on that day x cost of one toy

 Total cost = x (55 - x)

      750 = 55 x - x²

x² - 55 x + 750 = 0

x² - 30 x - 25 x + 750 = 0

x (x - 30) - 25 (x - 30) = 0

(x - 30) (x - 25) = 0

  x - 30 = 0     x - 25 = 0

   x = 30          x = 25


(3) Find two numbers whose sum is 27 and product is 182.

Solution:

Let 'x' and '27 - x' are two numbers

Sum of those numbers = 27

product of those numbers = 182

x (27 - x) = 182

27 x - x² = 182

x² - 27 x + 182 = 0

x² - 14 x - 13 x + 182 = 0

x (x - 14) - 13 (x - 14) = 0

(x - 14) (x - 13) = 0

x - 14 = 0          x - 13 = 0
 
    x = 14                x = 13


(4) Find two consecutive positive integers,sum of whose squares is 365.

Solution:

Let 'x' and 'x + 1' are two consecutive integers

sum of whose squares = 365

x² + (x + 1)² = 365

x² + x²+ 2 x + 1 = 365

2 x² + 2 x + 1 - 365 = 0

2 x² + 2 x - 364 = 0

÷ by 2 = > x² + x - 182 = 0

                 x² + 13 x - 12 x - 182 = 0

           x (x + 13) - 12(x + 13) = 0

           (x - 12) (x + 13) = 0

 x - 12 = 0      x + 13 = 0

 x = 12           x = -13

 Since it is positive integer -13 is not admissible

x = 12

x + 1 = 13

 Therefore the two consecutive integers are 12 and 13.

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