Page 128 - Maths Skills - 8
P. 128

126                                                                                                  Maths


              Let’s Attempt


        Example 1:  The population of a town is 35000 in a certain year. It increases at the rate of 4% per annum. What
                       will be its population after 2 years? Also find its net growth.
        Solution:      Let ‘P’ be the population at the beginning of a certain year. Then, the population after 2 years (A)
                       is given by


                                  R    n               4   2
                       A = P 1         = 35000    1

                                 100                 100
                                     1   2          26   2          26     6
                       = 35000   1       = 35000         = 35000 ×           = 37856
                                    25                25              25  25
                       Hence, the population of the town after 2 years will be 37856.
                       Net growth = A – P = 37856 – 35000 = 2856

                       Hence, the net growth in population will be 2856.

        Example 2:  The population of a town is 10000 in a certain year. It increases at the rate of 5% in the first year
                       and 10% in the second year. Find the net growth in two years.

        Solution:      Let the population at the beginning of a certain year be P  = 10000.
                       R  = 5% for the first year                           R  = 10% for the second year.
                        1                                                    2
                       Total population after 2 years


                                  R         R                   5        10              105 110
                       A = P  1+    1      1+  2    = 10000 1+        1+      = 10000 ×      ×     = 11550


                                 100       100                100       100                100 100
                       Net growth = 11550 – 10000 = 1550
                       Hence, the net growth in two years is 1550.

        Example 3:  The cost of a production plant is ` 90000. If its value depreciates at 10% annually, find the value
                       of the plant after 4 years.

        Solution:      Let ‘P’ be the initial value, i.e., P = `  90000
                       Depreciation  R = 10%                                    n = 4 years
                       Final value (A) of the plant after 4 years is given by


                                  R    n                10    4              9   4
                       A = P 1         = ` 90000    1        = ` 90000 ×

                                 100                   100                   10
                                      9       9        9       9
                       = ` 90000 ×       ×       ×        ×       = ` 59049
                                      10         10         10         10
                       Hence, the value of the production plant after 4 years will be ` 59049.
   123   124   125   126   127   128   129   130   131   132   133