Lesson 3 Scientific Notation of Numbers 
Exercise 3.2
1. Write the following decimal numbers in the scientific notation:
(a) 45
= 4.5 × 10
= 4.5 × 10¹
(b) 3400
= 3.4 × 1000
= 3.4 × 10³
(c) 0.000023
= 23 / 1000000
= 2.3 × 10⁻⁵
(d) 101000
= 101000 / 1
= 1.01 × 10⁵
(e) 0.010
= 1 / 100
= 1.0 × 10⁻²
(f) 45.01
= 4501 / 100
= 4.501 × 10¹
(g) 7000000
= 7000000 / 1
= 7.0 × 10⁶
(h) 0.00671
= 671 / 100000
= 6.71 × 10⁻³
(i) 625.6
= 6256 / 10
= 6.256 × 10²
(j) 0.07882
= 7882 / 100000
= 7.882 × 10⁻⁴
(k) 118000
= 1.18 × 100000
= 1.18 × 10⁵
(l) 87200
= 8.72 × 10000
= 8.72 × 10⁴
(m) 0.00000272
= 272 / 100000000
= 2.72 × 10⁻⁶
(n) 0.000037
= 37 / 1000000
= 3.7 × 10⁻⁵
(o) 74171.7
= 741717 / 10
= 7.41717 × 10⁴
(p) 3456.78
= 345678 / 100
= 3.45678 × 10³
2. Reduce the following scientific notation of numbers to the decimal numbers:
(a) 2.30 × 10⁴
= 2.30 × 10000
= 23000
(b) 5.40 × 10¹
= 5.40 × 10
= 54
(c) 1.76 × 10⁰
= 1.76 × 1
= 1.76
(d) 1.76 × 10⁻³
= 1.76 × 10 / 1000
= 1.76 × 0.001
= 0.00176
(e) 7.4 × 10⁻⁵
= 7.4 × 10 / 100000
= 7.4 × 0.00001
= 0.000074
(f) 1.901 × 10⁻⁷
= 1.901 × 10 / 100000000
= 1.901 × 0.00000001
= 0.0000001901
(g) 1.525 × 10⁶
= 1.525 × 1000000
= 1525000
(h) 6.58157 × 10⁷
= 6.58157 × 10000000
= 65815700
(i) 5.256 × 10⁸
= 5.256 × 100000000
= 525600000
(j) 5.23 × 10⁻⁷
= 5.23 × 10 / 10000000
= 5.23 × 0.0000001
= 0.000000523
(k) 8.71 × 10⁻⁸
= 8.71 × 10 / 100000000
= 8.71 × 0.00000001
= 0.0000000871
(l) 7.75763 × 10⁻⁹
= 7.75763 × 10 / 1000000000
= 7.75763 × 0.000000001
= 0.00000000775763
3. The weight of a loaded truck is 12,000 kg. Express it in scientific notation:
= 12,000
= 1.2 x 10000
= 1.2 x 10⁴
4. The radius of an atom of argon is 0.000000000039 meter. Write it in scientific notation:
= 0.000000000039
= 3.9 x 10 / 1000000000
= 3.9 x 10⁻¹¹
5. If 3 × 10⁸ m/s is the speed of light in air, what is this value in the decimal number?
= 3 × 10⁸
= 3 x 100000000
= 300000000
6. There are 2,592,000 seconds in a month of 30 days. Express this number in scientific notation:
= 2,592,000
= 2.592 x 1000000
= 2.592 x 10⁶
7. The quantity of petrol stored by Nepal Oil Corporation, Kathmandu is 1.75 × 10⁶ liters. How much kiloliters of petrol is there?
= 1.75 × 10⁶ liters
= 1.75 × 10³ kiloliters
= 1750 kiloliters
8. If the speed of light is 300,000,000 m/s, express it in scientific notation:
= 300,000,000
= 3 x 100000000
= 3.0 x 10⁸
Exercise 3.3
1. Simplify and write the answer in scientific notation:
(a) (3.4 × 10²) × (4.57 × 10³)
Solution:
(3.4 × 10²) × (4.57 × 10³)
= (3.4 × 4.57) × 10²⁺³
= 15.538 × 10⁵
= 1.5538 × 10⁶
(b) (6.9 × 10²) ÷ (2.4 × 10³)
Solution:
(6.9 × 10²) ÷ (2.4 × 10³)
= (6.9 ÷ 2.4) × 10²⁻³
= 2.875 × 10⁻¹
(c) (9.7 × 10⁶) × (8.3 × 10⁵)
Solution:
(9.7 × 10⁶) × (8.3 × 10⁵)
= (9.7 × 8.3) × 10⁶⁺⁵
= 80.51 × 10¹¹
= 8.051 × 10¹²
(d) (3.6 × 10²) × (1.6 × 10⁴)
Solution:
(3.6 × 10²) × (1.6 × 10⁴)
= (3.6 × 1.6) × 10²⁺⁴
= 5.76 × 10⁶
(e) (8.4 × 10⁵) ÷ (7.0 × 10⁶)
Solution:
(8.4 × 10⁵) ÷ (7.0 × 10⁶)
= (8.4 ÷ 7.0) × 10⁵⁻⁶
= 1.2 × 10⁻¹
(f) (1.3 × 10⁵) × (4.9 × 10⁴)
Solution:
= (1.3 × 4.9) × 10⁵⁺⁴
= 6.37 × 10⁹
2. Simplify and write the answer in scientific notation:
(a) (4.3 × 10⁸)(2.0 × 10⁶)
Solution:
(4.3 × 10⁸)(2.0 × 10⁶)
= (4.3 × 2.0) × 10⁸⁺⁶
= 8.6 × 10¹⁴
= 8.6 × 10¹⁴
(b) (6.0 × 10³) / (1.5 × 10²)
Solution:
(6.0 × 10³) / (1.5 × 10²)
= (6.0 ÷ 1.5) × 10³⁻²
= 4.0 × 10¹
= 4.0 × 10¹
(c) (1.5 × 10²) / (8.0 × 10¹)
Solution:
(1.5 × 10²) / (8.0 × 10¹)
= (1.5 ÷ 8.0) × 10²⁻¹
= 0.1875 × 10¹
= 1.875 × 10⁰
= 1.875 × 10⁰
(d) (5.2 × 10¹¹) / (3.0 × 10¹⁰)
Solution:
(5.2 × 10¹¹) / (3.0 × 10¹⁰)
= (5.2 ÷ 3.0) × 10¹¹⁻¹⁰
= 1.733 × 10¹
= 1.733 × 10¹
(e) (1.2 × 10⁸) / (3.0 × 10³)
Solution:
(1.2 × 10⁸) / (3.0 × 10³)
= (1.2 ÷ 3.0) × 10⁸⁻³
= 0.4 × 10⁵
= 4.0 × 10⁴
= 4.0 × 10⁴
(f) (7.8 × 10¹²) / (1.3 × 10¹³)
Solution:
(7.8 × 10¹²) / (1.3 × 10¹³)
= (7.8 ÷ 1.3) × 10¹²⁻¹³
= 6.0 × 10⁻¹
= 6.0 × 10⁻¹
(g) (8.4 × 10⁴) / (1.2 × 10³)
Solution:
(8.4 × 10⁴) / (1.2 × 10³)
= (8.4 ÷ 1.2) × 10⁴⁻³
= 7.0 × 10¹
= 7.0 × 10¹
(h) (5.6 × 10¹⁸) / (1.4 × 10⁸)
Solution:
(5.6 × 10¹⁸) / (1.4 × 10⁸)
= (5.6 ÷ 1.4) × 10¹⁸⁻⁸
= 4.0 × 10¹⁰
= 4.0 × 10¹⁰
(i) (8.1 × 10⁹) / (9.0 × 10⁸)
Solution:
(8.1 × 10⁹) / (9.0 × 10⁸)
= (8.1 ÷ 9.0) × 10⁹⁻⁸
= 0.9 × 10¹
= 9.0 × 10⁰
= 9.0 × 10⁰
(j) (3.2 × 10¹⁰) / (1.6 × 10¹⁵)
Solution:
(3.2 × 10¹⁰) / (1.6 × 10¹⁵)
= (3.2 ÷ 1.6) × 10¹⁰⁻¹⁵
= 2.0 × 10⁻⁵
= 2.0 × 10⁻⁵
(c) (2.1 × 10⁶)(4.0 × 10⁻³) / (4.2 × 10⁴)
Solution:
(2.1 × 10⁶)(4.0 × 10⁻³) / (4.2 × 10⁴)
= (2.1 × 4.0) × (10⁶ × 10⁻³) / (4.2 × 10⁴)
= 8.4 × 10³ / 4.2 × 10⁴
= (8.4 ÷ 4.2) × 10³⁻⁴
= 2.0 × 10⁻¹
= 2.0 × 10⁻¹
(d) (6.48 × 10⁵) / ((2.4 × 10⁴)(1.8 × 10²)
Solution:
(6.48 × 10⁵) / ((2.4 × 10⁴)(1.8 × 10²)
= (6.48 × 10⁵) / (2.4 × 1.8 × 10⁴ × 10²)
= (6.48 × 10⁵) / (4.32 × 10⁶)
= (6.48 ÷ 4.32) × 10⁵⁻⁶
= 1.5 × 10⁻¹
= 1.5 × 10⁻¹
4. There are 3.2 × 10⁴ liters of water in a tank and 1.3 × 10³ liters of water in a second tank. How much water is there in two tanks?
Solution:
(3.2 × 10⁴) + (1.3 × 10³)
= 3.2 × 10⁴ + 0.13 × 10⁴
= (3.2 + 0.13) × 10⁴
= 3.33 × 10⁴
= 3.33 × 10⁴ liters
5. A rocket is projected to travel a distance of 2.7 × 10⁹ km. How much distance is left after traveling the distance of 1.35 × 10⁹ km?
Solution:
(2.7 × 10⁹) − (1.35 × 10⁹)
= (2.7 − 1.35) × 10⁹
= 1.35 × 10⁹
= 1.35 × 10⁹ km
6. How many tanks of capacity 1.6 × 10³ liters are needed to store 9.6 × 10⁶ liters of petrol?
Solution:
(9.6 × 10⁶) ÷ (1.6 × 10³)
= (9.6 ÷ 1.6) × 10⁶⁻³
= 6.0 × 10³
= 6.0 × 10³ tanks
7. There is 1.8 × 10⁸ liters of water in a pond. If 1.6 × 10⁵ liters of water are flowed outside, how much water is left?
Solution:
(1.8 × 10⁸) − (1.6 × 10⁵)
= 1.8 × 10⁸ − 0.0016 × 10⁸
= (1.8 − 0.0016) × 10⁸
= 1.7984 × 10⁸
= 1.7984 × 10⁸ liters
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.