Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.8

Question 1. Represent the following numbers in the scientific notation:
(i) 569430000000
(ii) 2000.57
(iii) 0.0000006000
(iv) 0.0009000002
Answer:
(i) \( 569430000000 = 5.6943 \times 10^{11} \)
(ii) \( 2000.57 = 2.00057 \times 10^3 \)
(iii) \( 0.0000006000 = 6.0 \times 10^{-7} \)
(iv) \( 0.0009000002 = 9.000002 \times 10^{-4} \) Scientific notation helps us write very large or very small numbers in a compact form, making them easier to read and work with in math and science.
In simple words: To write a number in scientific notation, move the decimal point so there is only one non-zero digit before it. Then, multiply by 10 raised to the power of how many places you moved the decimal.

🎯 Exam Tip: Remember to use a positive exponent for large numbers (decimal moved left) and a negative exponent for small numbers (decimal moved right).

Question 2. Write the following numbers in decimal form:
(i) \( 3.459 \times 10^6 \)
(ii) \( 5.678 \times 10^4 \)
(iii) \( 1.00005 \times 10^{-5} \)
(iv) \( 2.530009 \times 10^{-7} \)
Answer:
(i) \( 3.459 \times 10^6 = 3459000 \)
(ii) \( 5.678 \times 10^4 = 56780 \)
(iii) \( 1.00005 \times 10^{-5} = 0.0000100005 \)
(iv) \( 2.530009 \times 10^{-7} = 0.0000002530009 \) When converting from scientific notation, a positive exponent means you move the decimal point to the right, making the number larger, and a negative exponent means moving it to the left to make the number smaller.
In simple words: To change from scientific notation to a normal decimal number, move the decimal point. If the power of 10 is positive, move it right. If the power is negative, move it left.

🎯 Exam Tip: Make sure you add enough zeros as placeholders when moving the decimal point to get the correct number of places.

Question 3. Represent the following numbers in scientific notation:
(i) \( (300000)^2 \times (20000)^4 \)
(ii) \( (0.000001)^{11} \div (0.005)^3 \)
(iii) \( \{(0.00003)^6 \times (0.00005)^4\} \div \{(0.009)^3 \times (0.05)^2\} \)
Answer:
(i) \( (300000)^2 \times (20000)^4 \)
\( = (3 \times 10^5)^2 \times (2 \times 10^4)^4 \)
\( = 3^2 \times (10^5)^2 \times 2^4 \times (10^4)^4 \)
\( = 9 \times 10^{10} \times 16 \times 10^{16} \)
\( = 9 \times 16 \times 10^{10+16} \)
\( = 144 \times 10^{26} \)
\( = 1.44 \times 10^{28} \)

(ii) \( (0.000001)^{11} \div (0.005)^3 \)
\( = (1.0 \times 10^{-6})^{11} \div (5.0 \times 10^{-3})^3 \)
\( = \frac{(1.0)^{11} \times 10^{-6 \times 11}}{5^3 \times 10^{-3 \times 3}} \)
\( = \frac{1.0 \times 10^{-66}}{125 \times 10^{-9}} \)
\( = 0.008 \times 10^{-66+9} \)
\( = 8.0 \times 10^{-3} \times 10^{-57} \)
\( = 8.0 \times 10^{-3-57} \)
\( = 8.0 \times 10^{-60} \)

(iii) \( \{(0.00003)^6 \times (0.00005)^4\} \div \{(0.009)^3 \times (0.05)^2\} \)
\( = \frac{(3.0 \times 10^{-5})^6 \times (5.0 \times 10^{-5})^4}{(9.0 \times 10^{-3})^3 \times (5.0 \times 10^{-2})^2} \)
\( = \frac{(3^6 \times 10^{-5 \times 6}) \times (5^4 \times 10^{-5 \times 4})}{(9^3 \times 10^{-3 \times 3}) \times (5^2 \times 10^{-2 \times 2})} \)
\( = \frac{(729 \times 10^{-30}) \times (625 \times 10^{-20})}{(729 \times 10^{-9}) \times (25 \times 10^{-4})} \)
\( = \frac{729 \times 625 \times 10^{-30-20}}{729 \times 25 \times 10^{-9-4}} \)
\( = \frac{625 \times 10^{-50}}{25 \times 10^{-13}} \)
\( = (625 \div 25) \times 10^{-50-(-13)} \)
\( = 25 \times 10^{-50+13} \)
\( = 25 \times 10^{-37} \)
\( = 2.5 \times 10^1 \times 10^{-37} \)
\( = 2.5 \times 10^{-36} \) When you multiply numbers in scientific notation, you multiply the decimal parts and add the exponents of 10; when you divide, you divide the decimal parts and subtract the exponents of 10. These rules make complex calculations much simpler.
In simple words: First, change all numbers to scientific notation. Then, use the rules for powers: when multiplying, add the small numbers (exponents) on the 10s; when dividing, subtract them. When a power is raised to another power, you multiply the small numbers.

🎯 Exam Tip: Always convert the numbers into proper scientific notation before applying the exponent rules to avoid errors.

Question 4. Represent the following information in scientific notation:
(i) The world population is nearly 7000,000,000.
(ii) One light year means the distance 9460528400000000 km.
(iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg.
Answer:
(i) World population = \( 7.0 \times 10^9 \)
(ii) Distance = \( 9.4605 \times 10^{15} \) km.
(iii) Mass of an electron = \( 9.1093822 \times 10^{-31} \) kg. Scientific notation is especially useful for expressing quantities found in astronomy or physics, like the vast distances in space or the tiny mass of subatomic particles.
In simple words: Take big numbers like the world population or very small numbers like the electron's mass, and write them in the special short way using scientific notation. This makes them easy to read.

🎯 Exam Tip: When dealing with numbers from real-world data, identify the significant digits and then correctly count the decimal places for the exponent of 10.

Question 5. Simplify:
(i) \( (2.75 \times 10^7) + (1.23 \times 10^8) \)
(ii) \( (1.598 \times 10^{17}) - (4.58 \times 10^{15}) \)
(iii) \( (1.02 \times 10^{10}) \times (1.20 \times 10^{-3}) \)
(iv) \( (8.41 \times 10^4) \div (4.3 \times 10^5) \)
Answer:
(i) \( (2.75 \times 10^7) + (1.23 \times 10^8) \)
Convert to standard form for addition:
\( 2.75 \times 10^7 = 27,500,000 \)
\( 1.23 \times 10^8 = 123,000,000 \)
Adding them: \( 27,500,000 + 123,000,000 = 150,500,000 \)
Convert back to scientific notation: \( 150,500,000 = 1.505 \times 10^8 \)

(ii) \( (1.598 \times 10^{17}) - (4.58 \times 10^{15}) \)
Convert to standard form for subtraction:
\( 1.598 \times 10^{17} = 159,800,000,000,000,000 \)
\( 4.58 \times 10^{15} = 4,580,000,000,000,000 \)
Subtracting them:
\( 159,800,000,000,000,000 - 4,580,000,000,000,000 = 155,220,000,000,000,000 \)
Convert back to scientific notation: \( 155,220,000,000,000,000 = 1.5522 \times 10^{17} \)

(iii) \( (1.02 \times 10^{10}) \times (1.20 \times 10^{-3}) \)
\( = (1.02 \times 1.20) \times (10^{10} \times 10^{-3}) \)
\( = 1.224 \times 10^{10-3} \)
\( = 1.224 \times 10^7 \)

(iv) \( (8.41 \times 10^4) \div (4.3 \times 10^5) \)
\( = \frac{8.41 \times 10^4}{4.3 \times 10^5} \)
\( = \frac{8.41}{4.3} \times \frac{10^4}{10^5} \)
\( = 1.955813953... \times 10^{4-5} \)
\( = 1.9558139 \times 10^{-1} \) When adding or subtracting numbers in scientific notation, it is helpful to first adjust one of the numbers so that both have the same power of 10. For multiplication and division, the decimal parts are handled separately from the powers of 10.
In simple words: To add or subtract, make sure the powers of 10 are the same first, then add or subtract the main numbers. To multiply, multiply the main numbers and add the powers of 10. To divide, divide the main numbers and subtract the powers of 10.

🎯 Exam Tip: For addition and subtraction, convert the numbers to a common exponent (usually the larger one) or to standard decimal form to avoid mistakes. For multiplication and division, handle the coefficients and powers of 10 separately.