10th CBSE maths solution for exercise 3.3 part 4

This page 10th CBSE maths solution for exercise 3.3 part 4 is going to provide you solution for every problems that you find in the exercise no 3.3

10th CBSE maths solution for Exercise 3.3 part 4

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later,she buys 3 bats and 5 balls for Rs.1750. Find the cost if each bat and each ball.

Solution:

Let "x" be the cost of each bat

Let "y" be the cost of  each ball

7 x + 6 y = 3800 ----- (1)

3 x + 5 y = 1750 ----- (2)

 6 y = 3800 - 7 x

   y = (3800 - 7 x)/6

 substitute  y = (3800 - 7 x)/6 in the second equation

  3 x + 5 (3800 - 7 x)/6 = 1750

 [18 x + 5 (3800 - 7 x)]/6 = 1750

 (18 x + 19000 - 35 x)/6 = 1750

 -17 x + 19000 = 1750 x 6

  -17x + 19000 = 10500

   - 17 x = 10500 - 19000

    - 17 x = -8500

           x = 8500/17

           x = 500

Now,we have to apply the value of x in the equation

y = (3800 - 7 x)/6

          y = [3800 - 7(500)]/6

           y = (3800 - 3500)/6

           y = 300/6

           y = 50

Therefore the cost of each bat = Rs.500

Cost of each ball = Rs.50


(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km,the charge paid is Rs.105 and for a journey of 15 km,the charge paid is Rs 155. What are the fixed charge and charge per km? How much does a person have to pay for traveling a distance of 25 km?

Solution:

Let "x" be the fixed charge

Let "y" be the charge  per km for the distance covered

 x + 10 y = 105   ------ (1)

 x + 15 y = 155   ------ (2)

x = 105 - 10 y

Substitute x = 105 - 10 y in the second equation

 105 - 10 y + 15 y = 155

   105 + 5 y = 155

               5 y = 155 - 105

               5 y = 50

                  y = 50/5

                  y = 10

   x = 105 -10 (10)

  x = 105 - 100

  x = 5

Therefore the fixed charge is Rs.5

the charge  per km for the distance covered = Rs.10




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