NCERT Solution Class 8th Maths Chapter – 7 Comparing Quantities
Textbook | NCERT |
Class | 8th |
Subject | Mathematics |
Chapter | 7th |
Chapter Name | Comparing Quantities |
Category | Class 8th Maths |
Medium | English |
Source | Last Doubt |
NCERT Solution Class 8th Maths Chapter 7 Comparing Quantities
Chapter – 7
Comparing Quantities
Exercise – 7.3
Q1. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. (i) Find the population in 2001. Ans – Population in 2003 is P = 54000 (i) Let the population in 2001 (i.e. 2 years ago) = P ∴ Present population = p(1+5\100) or 54000 = p(21\20)2 or 54000 = p(441\400) or p = 54000 x 400 \441 =48979.59 = 48980 (approx.) Thus, the population in 2001 was about 48980. (ii) Initial population (in 2003), i.e. P = 54000 Rate of increment in population = 5% p.a. ∴ A = p(1+R\100)2 = 135 * 21 * 21 = 59535 |
Q2. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000. Ans- Initial count of bacteria (P) = 5,06,000 A =P(+R\100)n A = 506000(1+2.5\100)2= 506000 (1+25\1000)2 = 506000 (41\40)2 = 506000 x 41\40x 41\40 = 6325x41x41 \20 =10632325\20 = 531616.25 or 531616 (approx.) Thus, the count of bacteria after 2 hours will be 531616 (approx.). |
Q3. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year. Ans – Initial cost (value) of the scooter (P) = Rs 42000 Using A = P(1+R\100)n,We have A= Rs 42000 x(1-8\100)1 Thus, the value of the scooter after 1 year will be ₹ 38640. |