ML Aggarwal Class 8 Maths Solutions Chapter 02 Exponents and Powers

Exercise 2.1

Question 1. Evaluate:
(i) \( (3/5)^{-2} \)
(ii) \( (-3)^{-3} \)
(iii) \( (2/7)^{-4} \)
Answer:
(i) To work out \( (3/5)^{-2} \), use the rule that \( a^{-n} = 1/a^{n} \). Flip the fraction and remove the negative sign:
\( (3/5)^{-2} = (5/3)^{2} = (5/3) \times (5/3) = 25/9 \)

(ii) For \( (-3)^{-3} \), apply the same rule:
\( (-3)^{-3} = (-1/3)^{3} = (-1/3) \times (-1/3) \times (-1/3) = -1/27 \)

(iii) For \( (2/7)^{-4} \), flip and simplify:
\( (2/7)^{-4} = (7/2)^{4} = (7/2) \times (7/2) \times (7/2) \times (7/2) = 2401/16 \)
In simple words: When you see a negative exponent, flip the fraction upside down and change the exponent to positive. Then multiply as usual.

Exam Tip: Always flip the base when you see a negative exponent - this is the key step. Make sure you count the multiplications carefully to get the right final answer.

Question 2. Simplify:
(i) \( [(2)^{-1} + (4)^{-1} + (3)^{-1}]^{-1} \)
(ii) \( [(4)^{-1} - (5)^{-1}]^{2} \times (5/8)^{-1} \)
(iii) \( [4^{0} + 4^{2} - 2^{3}] \times 3^{-2} \)
(iv) \( [(5)^{2} - (1/4)^{-2}] \times (3/4)^{-2} \)
Answer:
(i) Start by finding each reciprocal inside the brackets:
\( [(2)^{-1} + (4)^{-1} + (3)^{-1}]^{-1} = [(1/2) + (1/4) + (1/3)]^{-1} = [(6 + 3 + 4)/12]^{-1} = [13/12]^{-1} = 12/13 \)

(ii) Work out the inner brackets first, then square the result, and multiply:
\( [(4)^{-1} - (5)^{-1}]^{2} \times (5/8)^{-1} = [(1/4) - (1/5)]^{2} \times (8/5) = [(5 - 4)/20]^{2} \times (8/5) = [1/20]^{2} \times (8/5) = (1/400) \times (8/5) = 1/250 \)

(iii) Calculate the values inside brackets: any number to the power 0 equals 1, so \( 4^0 = 1 \). Then \( 4^2 = 16 \) and \( 2^3 = 8 \):
\( [1 + 16 - 8] \times (1/9) = 9 \times (1/9) = 1 \)

(iv) Simplify each term: \( (5)^2 = 25 \) and \( (1/4)^{-2} = (4)^{2} = 16 \):
\( [25 - 16] \times (4/3)^{2} = 9 \times (16/9) = 16 \)
In simple words: Work step by step. First handle the negative exponents by flipping. Then do the operations in the right order - brackets first, then power, then multiply.

Exam Tip: Follow the order of operations carefully. Pay special attention to negative exponents within brackets - flip them before adding or subtracting.

Question 3. Find the multiplicative inverse of the following:
(i) \( (81/16)^{-3/4} \)
(ii) \( \{(-3/2)^{-4}\}^{1/2} \)
(iii) \( (5/7)^{-2} \times (5/7)^{4} \div (5/7)^{3} \)
Answer:
(i) Start by simplifying using the rules of exponents. Flip the fraction and work with the exponent:
\( (81/16)^{-3/4} = (16/81)^{3/4} = (2^4/3^4)^{3/4} = (2/3)^{4 \times 3/4} = (2/3)^{3} = (2/3) \times (2/3) \times (2/3) = 8/27 \)
So the multiplicative inverse of 8/27 is 27/8.

(ii) Apply the power rule: multiply the exponents together:
\( \{(-3/2)^{-4}\}^{1/2} = (-3/2)^{-4 \times 1/2} = (-3/2)^{-2} = (2/-3)^{2} = (2/-3) \times (2/-3) = 4/9 \)
So the multiplicative inverse of 4/9 is 9/4.

(iii) Use the rules for multiplying and dividing powers with the same base - add or subtract the exponents:
\( (5/7)^{-2} \times (5/7)^{4} \div (5/7)^{3} = (7/5)^{2} \times (5/7)^{4-3} = (7/5)^{2} \times (5/7) = (7/5) \times (7/5) \times (5/7) = 7/5 \)
So the multiplicative inverse of 7/5 is 5/7.
In simple words: To find the multiplicative inverse (or reciprocal), you flip the number upside down. If the answer is 8/27, flip it to get 27/8.

Exam Tip: Remember that the multiplicative inverse of a/b is b/a. Always simplify the expression first, then find its reciprocal - do not try to do both steps at once.

Question 4. (i) Express \( 16^{-2} \) as a power with base 2.
Answer: First rewrite 16 as a power of 2. Since \( 16 = 2^4 \):
\( (16)^{-2} = (2^4)^{-2} = 2^{-8} = 1/2^{8} \)
In simple words: Change 16 to \( 2^4 \). Then apply the power rule: multiply -2 by 4 to get -8.

Exam Tip: Always rewrite composite numbers as powers of the requested base. For base 2, remember: 4 = 2², 8 = 2³, 16 = 2⁴, 32 = 2⁵.

Question 4. (ii) Express \( 125^{-4} \) as a power with base 5.
Answer: Rewrite 125 as a power of 5. Since \( 125 = 5^3 \):
\( (125)^{-4} = (5^3)^{-4} = 5^{-12} = 1/5^{12} \)
In simple words: Change 125 to \( 5^3 \). Then multiply the exponents: 3 times -4 gives -12.

Exam Tip: Know your perfect cubes: 8 = 2³, 27 = 3³, 125 = 5³. This helps you rewrite expressions quickly with the requested base.

Question 5. Write the following numbers in expanded form using exponents:
(i) 2789.453
(ii) 3007.805
Answer:
(i) Each digit has a place value that is a power of 10. The digits to the left of the decimal point use positive exponents, and those to the right use negative exponents:
\( 2789.453 = 2 \times 10^{3} + 7 \times 10^{2} + 8 \times 10^{1} + 9 \times 10^{0} + 4 \times 10^{-1} + 5 \times 10^{-2} + 3 \times 10^{-3} \)

(ii) For 3007.805, note that the tens and hundreds places are zero, so we skip them:
\( 3007.805 = 3 \times 10^{3} + 7 \times 10^{0} + 8 \times 10^{-1} + 5 \times 10^{-3} \)
In simple words: Write each digit multiplied by its place value as a power of 10. The first digit after the decimal point gets \( 10^{-1} \), the second gets \( 10^{-2} \), and so on.

Exam Tip: Digits before the decimal point (from right to left) use exponents 0, 1, 2, 3... Digits after the decimal point use exponents -1, -2, -3... Skip any place that has a 0 digit.

Question 6. Simplify and write in exponential form with positive exponent:
(i) \( [\{(5/7)^{2}\}^{-1}]^{-3} \)
(ii) \( (2/7)^{2} \times (7/2)^{-3} \div \{(7/5)^{-2}\}^{-4} \)
(iii) \( (4/5)^{2} \times 5^{4} \times (2/5)^{-2} \div (5/2)^{-3} \)
(iv) \( [8^{-1} \times 5^{3}]/2^{-4} \)
Answer:
(i) Multiply the exponents step by step, working from the innermost brackets outward:
\( [\{(5/7)^{2}\}^{-1}]^{-3} = \{(5/7)^{2}\}^{-1 \times -3} = (5/7)^{2 \times 3} = (5/7)^{6} \)

(ii) Convert all terms to have the same base if possible, then add or subtract exponents:
\( (2/7)^{2} \times (7/2)^{-3} \div \{(7/5)^{-2}\}^{-4} = (2/7)^{2} \times (2/7)^{3} \div (7/5)^{8} = (2/7)^{2+3} \times (5/7)^{8} = (2/7)^{5} \times (5/7)^{8} = (2^5 \times 5^8)/(7^{5+8}) = (2^5 \times 5^8)/7^{13} \)

(iii) Rewrite all fractions by separating numerators and denominators, then combine like bases:
\( (4/5)^{2} \times 5^{4} \times (2/5)^{-2} \div (5/2)^{-3} = (2^4 \times 5^4 \times 2^{-2} \times 2^{-3})/(5^2 \times 5^{-2} \times 5^{-3}) = 2^{4-2-3} \times 5^{4-2+2+3} = 2^{-1} \times 5^{7} = 5^{7}/2 \)

(iv) Rewrite 8 as \( 2^3 \), then simplify the fraction:
\( [8^{-1} \times 5^{3}]/2^{-4} = [(2^{3})^{-1} \times 5^{3}]/2^{-4} = [2^{-3} \times 5^{3}]/2^{-4} = 2^{-3+4} \times 5^{3} = 2^{1} \times 5^{3} = 2 \times 125 = 250 \)
In simple words: When you have powers of powers, multiply the exponents. When multiplying same bases, add exponents. When dividing, subtract exponents.

Exam Tip: Keep track of which bases you are working with - separate them by base before combining exponents. Write fractions with numerator powers and denominator powers clearly separated.

Question 7. Simplify and write the following in exponential form:
(i) \( ((-2)^{3})^{2} + 5^{-3} \div 5^{-5} - (-1/2)^{0} \)
(ii) \( 3^{-5} \times 3^{2} \div 3^{-6} + (2^{2} \times 3)^{2} + (2/3)^{-1} + 2^{-1} + (1/19)^{-1} \)
Answer:
(i) Work out each part separately:
\( ((-2)^{3})^{2} = (-2)^{6} = 64 \)
\( 5^{-3} \div 5^{-5} = 5^{-3+5} = 5^{2} = 25 \)
\( (-1/2)^{0} = 1 \)
So the full expression becomes: \( 64 + 25 - 1 = 88 \)

(ii) Break this into manageable pieces:
\( 3^{-5} \times 3^{2} \div 3^{-6} = 3^{-5+2+6} = 3^{3} = 27 \)
\( (2^{2} \times 3)^{2} = (4 \times 3)^{2} = 12^{2} = 144 \) or work as \( 2^{4} \times 3^{2} = 16 \times 9 = 144 \)
\( (2/3)^{-1} = 3/2 = 1.5 \)
\( 2^{-1} = 1/2 = 0.5 \)
\( (1/19)^{-1} = 19 \)
Adding: \( 27 + 144 + 1.5 + 0.5 + 19 = 192 \)
In simple words: Do each term one at a time. Use exponent rules to simplify powers first. Then add or subtract all the final numbers together.

Exam Tip: When you have a long expression with mixed operations, calculate each piece (exponent term, division, multiplication) separately before combining with addition or subtraction.

Question 8. Simplify and write in exponential form with negative exponent:
(i) \( 5^{3} \times (4/5)^{3} \)
(ii) \( [(3/7)^{-2}]^{-3} \)
(iii) \( (5/9)^{-2} \times (5/3)^{2} \div (1/5)^{-2} \)
(iv) \( 2^{-1} [(5/3)^{4} + (3/5)^{-2}] \div (17/9) \)
(v) \( (-7)^{3} \times (1/-7)^{-9} \div (-7)^{10} \)
Answer:
(i) Separate the power across the fraction:
\( 5^{3} \times (4/5)^{3} = 5^{3} \times (4^{3}/5^{3}) = 5^{3-3} \times 4^{3} = 1 \times 64 = 64 = (1/4)^{-3} \)

(ii) Multiply the exponents:
\( [(3/7)^{-2}]^{-3} = (3/7)^{-2 \times -3} = (3/7)^{6} = (7/3)^{-6} \)

(iii) Rewrite with fractional powers separated, then combine:
\( (5/9)^{-2} \times (5/3)^{2} \div (1/5)^{-2} = (5^{-2}/9^{-2}) \times (5^{2}/3^{2}) \div (1/5^{-2}) = (5^{-2} \times 5^{2} \times 5^{-2})/(9^{-2} \times 3^{2}) = 5^{-2}/(3^{-4} \times 3^{2}) = 5^{-2}/3^{-2} = (5/3)^{-2} \)

(iv) Calculate the bracket first, then divide:
\( 2^{-1} [(5/3)^{4} + (3/5)^{-2}] \div (17/9) = (1/2) [(625/81) + (25/9)] \div (17/9) = (1/2) [(625 + 225)/81] \times (9/17) = (1/2) \times (850/81) \times (9/17) = (1/2) \times (50 \times 9)/(81) \times (9/17) \)
After cancellation: \( (1/2) \times (850/81) \times (9/17) = 25/9 = (5/3)^{2} = (3/5)^{-2} \)

(v) Convert all to the same base and combine exponents:
\( (-7)^{3} \times (1/-7)^{-9} \div (-7)^{10} = (-7)^{3} \times (-7)^{9} \div (-7)^{10} = (-7)^{3+9-10} = (-7)^{2} = (1/-7)^{-2} \)
In simple words: To write the answer with a negative exponent, flip the fraction and change the sign of the exponent. For example, 64 = \( 4^3 \) = \( (1/4)^{-3} \).

Exam Tip: After simplifying, if your answer is a whole number or simple fraction, express it as a power with a negative exponent. This often means flipping a fraction and making the exponent negative.

Question 9. Simplify:
(i) \( (49 \times z^{-3}) / (7^{-3} \times 10 \times z^{-5}) \) (z ≠ 0)
(ii) \( (9^{3} \times 27 \times t^{4}) / (3^{2} \times 3^{4} \times t^{2}) \)
(iii) \( [(3^{-2})^{2} \times (5^{2})^{-3} \times (-t^{-3})^{2}] / [(3^{-2})^{5} \times (5^{3})^{-2} \times (t^{-4})^{3}] \)
(iv) \( (2^{-5} \times 15^{-5} \times 500) / (5^{-6} \times 6^{-5}) \)
Answer:
(i) Rewrite 49 as \( 7^2 \), then combine powers of 7 and z separately:
\( (7^{2} \times z^{-3}) / (7^{-3} \times 10 \times z^{-5}) = (7^{2+3} \times z^{-3+5}) / 10 = (7^{5} \times z^{2}) / 10 \)

(ii) Express 9 as \( 3^2 \) and 27 as \( 3^3 \), then combine all powers of 3 and t:
\( [(3^{2})^{3} \times 3^{3} \times t^{4}] / (3^{2} \times 3^{4} \times t^{2}) = (3^{6+3} \times t^{4}) / (3^{2+4} \times t^{2}) = 3^{6+3-2-4} \times t^{4-2} = 3^{3} \times t^{2} = 27t^{2} \)

(iii) Use power rules to simplify exponents, then combine:
\( [3^{-4} \times 5^{-6} \times t^{-6}] / [3^{-10} \times 5^{-6} \times t^{-12}] = 3^{-4+10} \times 5^{-6+6} \times t^{-6+12} = 3^{6} \times 5^{0} \times t^{6} = 729 \times 1 \times t^{6} = 729t^{6} \)

(iv) Rewrite 15 as \( 3 \times 5 \), 500 as \( 2^2 \times 5^3 \), and 6 as \( 2 \times 3 \), then combine by base:
\( (2^{-5} \times (3 \times 5)^{-5} \times 2^{2} \times 5^{3}) / (5^{-6} \times (2 \times 3)^{-5}) = (2^{-5+2+5} \times 3^{-5+5} \times 5^{-5+3+6}) = 2^{2} \times 3^{0} \times 5^{4} = 4 \times 1 \times 625 = 2500 \)
In simple words: Break down composite numbers into their prime factors. Then group by base and add or subtract exponents as needed.

Exam Tip: Always factor large numbers into primes (2, 3, 5, 7, etc.). This makes it easy to combine exponents correctly.

Question 10. By what number should \( (3/-2)^{-3} \) be divided to get \( (2/3)^{2} \)?
Answer: The required number is \( (3/-2)^{-3} \div (2/3)^{2} \). Simplify this division:
\( (3/-2)^{-3} \div (2/3)^{2} = (3^{-3})/(-2)^{-3} \times (3^{2})/(2^{2}) = [3^{-3} \times 3^{2}]/[(-2)^{-3} \times 2^{2}] = 3^{-1}/[(-2)^{-3} \times 2^{2}] \)
Since \( (-2)^{-3} = -1/8 \):
\( = 3^{-1} / [(-1/8) \times 4] = (1/3) / (-1/2) = (1/3) \times (-2) = -2/3 \)
In simple words: To find what number you need to divide by, divide the first number by the second. Simplify using exponent rules.

Exam Tip: When asked "by what number should A be divided to get B", set up the problem as A ÷ ? = B, which means ? = A ÷ B.

Question 11. Find the value of m for which \( 9^{m} \div 3^{-2} = 9^{4} \).
Answer: Rewrite 9 as a power of 3 to work with a single base:
\( (3^{2})^{m} \div 3^{-2} = (3^{2})^{4} \)
\( 3^{2m} \div 3^{-2} = 3^{8} \)
\( 3^{2m+2} = 3^{8} \)
Now compare the exponents on both sides:
\( 2m + 2 = 8 \)
\( 2m = 6 \)
\( m = 3 \)
In simple words: Rewrite everything using the same base. Then the exponents must be equal on both sides. Solve for the unknown.

Exam Tip: Always convert to a common base before comparing exponents. If bases are equal, their exponents must be equal too.

Question 12. If \( (-5/7)^{-4} \times (-5/7)^{12} = \{(-5/7)^{3}\}^{x} \times (-5/7)^{-1} \), find the value of x.
Answer: Simplify both sides using exponent rules:
Left side: \( (-5/7)^{-4+12} = (-5/7)^{8} \)
Right side: \( (-5/7)^{3x} \times (-5/7)^{-1} = (-5/7)^{3x-1} \)
Since both sides have the same base, equate the exponents:
\( 8 = 3x - 1 \)
\( 3x = 9 \)
\( x = 3 \)
In simple words: When adding exponents with the same base, the bases on both sides must be the same. Set exponents equal and solve for x.

Exam Tip: Add exponents when multiplying same bases. If \( a^m \times a^n = a^p \), then m + n = p. Use this to set up your equation.

Question 13. Find x, if \( (-2/3)^{-13} \times (3/-2)^{8} = (-2/3)^{-2x+1} \)
Answer: Rewrite \( (3/-2)^{8} \) to match the base \( (-2/3) \). Note that \( 3/-2 = -2/3 \):
\( (-2/3)^{-13} \times (-2/3)^{8} = (-2/3)^{-2x+1} \)
Combine the left side using exponent rules:
\( (-2/3)^{-13+8} = (-2/3)^{-2x+1} \)
\( (-2/3)^{-21} = (-2/3)^{-2x+1} \)
Equate exponents:
\( -21 = -2x + 1 \)
\( -2x = -22 \)
\( x = 11 \)
In simple words: Make sure both sides have the same base. Notice that \( 3/-2 \) is the same as \( -2/3 \) (just written differently). Then match exponents.

Exam Tip: Watch for fractions that look different but are equal - like \( 3/-2 \) and \( -2/3 \). Rewrite them to match before combining.

Question 14. (i) If \( 5^{2x-1} = 1/(125)^{x-3} \), find x.
Answer: Rewrite the right side using a single base. Since \( 125 = 5^3 \):
\( 5^{2x-1} = 1/(5^{3})^{x-3} \)
\( 5^{2x-1} = 1/5^{3x-9} \)
\( 5^{2x-1} = 5^{-3x+9} \)
Equate exponents:
\( 2x - 1 = -3x + 9 \)
\( 5x = 10 \)
\( x = 2 \)
In simple words: Rewrite 125 as \( 5^3 \). Then use the fact that \( 1/a = a^{-1} \) to flip the right side. Match exponents and solve.

Exam Tip: Know your powers: \( 2^3 = 8, 3^3 = 27, 5^3 = 125 \). Being able to recognize and write these powers quickly will save time.

Question 14. (ii) If \( (9^{n} \times 3^{5} \times 27^{3})/(3 \times 81^{4}) = 27 \), find n.
Answer: Rewrite all terms as powers of 3:
\( 9 = 3^2, 27 = 3^3, 81 = 3^4, 27 = 3^3 \)
Substitute:
\( [(3^{2})^{n} \times 3^{5} \times (3^{3})^{3}] / (3 \times (3^{4})^{4}) = 3^{3} \)
\( [3^{2n} \times 3^{5} \times 3^{9}] / (3 \times 3^{16}) = 3^{3} \)
Combine exponents in numerator and denominator:
\( 3^{2n+5+9} / 3^{1+16} = 3^{3} \)
\( 3^{2n+14} / 3^{17} = 3^{3} \)
\( 3^{2n+14-17} = 3^{3} \)
\( 3^{2n-3} = 3^{3} \)
Equate exponents:
\( 2n - 3 = 3 \)
\( 2n = 6 \)
\( n = 3 \)
In simple words: Change everything to base 3. Add exponents in the numerator. Subtract the denominator exponent. Then match the exponents on both sides.

Exam Tip: When working with powers in a fraction, simplify the numerator first, then subtract the denominator exponent. This avoids confusion with signs.

Exercise 2.2

Question 1. Express the following numbers in standard form:
(i) 0.0000000000085
(ii) 0.000000000000942
(iii) 6020000000000000
(iv) 0.00000000837
Answer:
(i) \( 0.0000000000085 = 8.5 \times 10^{-12} \)
(ii) \( 0.000000000000942 = 9.42 \times 10^{-13} \)
(iii) \( 6020000000000000 = 6.02 \times 10^{15} \)
(iv) \( 0.00000000837 = 8.37 \times 10^{-9} \)
In simple words: In standard form, you write a decimal between 1 and 10, then multiply by a power of 10. Move the decimal point until one digit is before it. Count how many places you moved.

Exam Tip: For numbers less than 1, the exponent is negative. For numbers greater than 1, the exponent is positive. The exponent tells you how many places the decimal point moved.

Question 2. Express the following numbers in usual form:
(i) \( 3.02 \times 10^{-6} \)
(ii) \( 1.007 \times 10^{11} \)
(iii) \( 5.375 \times 10^{14} \)
(iv) \( 7.579 \times 10^{-14} \)
Answer:
(i) \( 3.02 \times 10^{-6} = 0.00000302 \)
(ii) \( 1.007 \times 10^{11} = 100700000000 \)
(iii) \( 5.375 \times 10^{14} = 537500000000000 \)
(iv) \( 7.579 \times 10^{-14} = 0.00000000000007579 \)
In simple words: Move the decimal point the number of places shown by the exponent. If the exponent is positive, move right. If negative, move left.

Exam Tip: A positive exponent means a large number (move decimal right). A negative exponent means a small number less than 1 (move decimal left).

Question 3. Express the number appearing in the following statements in standard form:
(i) The mass of a proton is 0.000000000000000000000001673 gram.
(ii) The thickness of a piece of paper is 0.0016 cm.
(iii) The diameter of a wire on a computer chip is 0.000003 m.
(iv) A helium atom has a diameter of 22/100000000000 m.
(v) Mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.
(vi) The human body has 1 trillion cells which vary in shapes and sizes.
(vii) The distance from the Earth of the Sun is 149,600,000,000 m.
(viii) The speed of light is 300,000,000 m/sec.
(ix) Mass of the Earth is 5,970,000,000,000,000,000,000,000 kg.
(x) Express 3 years in seconds.
(xi) Express 7 hectares in cm².
(xii) A sugar factory has annual sales of 3 billion 720 million kilograms of sugar.
Answer:
(i) \( 0.000000000000000000000001673 \text{ gram} = 1.673 \times 10^{-24} \text{ gram} \)

(ii) \( 0.0016 \text{ cm} = 1.6 \times 10^{-3} \text{ cm} \)

(iii) \( 0.000003 \text{ m} = 3.0 \times 10^{-6} \text{ m} \)

(iv) \( 22/100000000000 \text{ m} = 2.2 \times 10^{-10} \text{ m} \)

(v) \( 0.00000000000000000000334 \text{ tons} = 3.34 \times 10^{-21} \text{ tons} \)

(vi) \( 1 \text{ trillion cells} = 1,000,000,000,000 = 10^{12} \)

(vii) \( 149,600,000,000 \text{ m} = 1.496 \times 10^{11} \text{ m} \)

(viii) \( 300,000,000 \text{ m/sec} = 3.0 \times 10^{8} \text{ m/sec} \)

(ix) \( 5,970,000,000,000,000,000,000,000 \text{ kg} = 5.97 \times 10^{24} \text{ kg} \)

(x) 3 years expressed in seconds:
\( 3 \text{ years} = 3 \times 365 \times 24 \times 3600 \text{ seconds} = 94608000 \text{ seconds} = 9.4608 \times 10^{7} \text{ seconds} \)

(xi) 7 hectares expressed in cm²:
\( 7 \text{ hectares} = 7 \times 10000 \text{ m}^{2} = 7 \times 10000 \times 100 \times 100 \text{ cm}^{2} = 700000000 \text{ cm}^{2} = 7.0 \times 10^{8} \text{ cm}^{2} \)

(xii) Annual sugar sales:
\( 3 \text{ billion } 720 \text{ million kg} = 3,720,000,000 \text{ kg} = 3.72 \times 10^{9} \text{ kg} \)
In simple words: Count carefully from right to left (for large numbers) or left to right (for small numbers) to place the decimal point correctly in standard form.

Exam Tip: For conversion problems (years to seconds, hectares to cm²), do the multiplication step by step. Then write the final answer in standard form.

Question 4. Compare the following:
(i) Size of a plant cell to the thickness of a piece of paper.
(ii) Size of a plant cell to the diameter of a wire on a computer chip.
(iii) The thickness of a piece of paper to the diameter of a wire on a computer chip.

Given: size of plant cell = 0.00001275 m, thickness of a piece of paper = 0.0016 cm, diameter of a wire on a computer chip = 0.000003 m
Answer:
First, express each measurement in standard form for comparison:
Size of plant cell = 0.00001275 m = 1.275 × 10⁻⁵ m
Thickness of paper = 0.0016 cm = 1.6 × 10⁻³ cm
Diameter of wire = 0.000003 m = 3.0 × 10⁻⁶ m

(i) Comparing plant cell to paper thickness:
\( 1.275 \times 10^{-5} : 1.6 \times 10^{-3} \)
Divide: \( 1.275 / 1.6 = 0.797 \approx 3/4 \)
The plant cell is about 3/4 times the thickness of the paper, or the paper is about 1.3 times thicker than the plant cell.

(ii) Comparing plant cell to wire diameter:
\( 1.275 \times 10^{-5} : 3.0 \times 10^{-6} \)
Divide: \( (1.275 / 3.0) \times 10^{1} = 0.425 \times 10 = 4.25 \)
The plant cell is about 4 times the diameter of the wire.

(iii) Comparing paper thickness to wire diameter (convert to same units first):
Paper = 1.6 × 10⁻³ cm = 1.6 × 10⁻⁵ m
\( 1.6 \times 10^{-5} : 3.0 \times 10^{-6} \)
Divide: \( (1.6 / 3.0) \times 10^{1} = 0.533 \times 10 = 5.33 \)
The paper is approximately 5 times thicker than the wire diameter.
In simple words: Convert all measurements to standard form and the same unit. Then divide to find how many times larger one is than the other.

Exam Tip: Always check units before comparing. Convert to the same units and standard form for clear comparison.

Question 5. The number of red blood cells per cubic millimeter of blood is approximately 5.5 million. If the average body contains 5 liters of blood, what is the total number of red cells in the body? (1 liter = 1,00,000 mm³)
Answer:
Given information:
Red blood cells per cubic millimeter = 5.5 million = 5.5 × 10⁶
Total blood in body = 5 liters = 5 × 100,000 mm³ = 5 × 10⁵ mm³

Total red blood cells = (Red cells per mm³) × (Total mm³ in body)
\( = 5.5 \times 10^{6} \times 5 \times 10^{5} \)
\( = 27.5 \times 10^{11} \)
\( = 2.75 \times 10^{12} \)

The total number of red blood cells in the body is 2.75 × 10¹².
In simple words: Multiply the number of cells in each mm³ by the total number of mm³ in the body. Express your answer in standard form.

Exam Tip: When multiplying numbers in standard form, multiply the coefficients and add the exponents: \( (a \times 10^m) \times (b \times 10^n) = (a \times b) \times 10^{m+n} \).

Question 6. Mass of Mars is 6.42 × 10²⁹ kg and the mass of the sun is 1.99 × 10³⁰ kg. What is the total mass?
Answer:
Mass of Mars = 6.42 × 10²⁹ kg
Mass of Sun = 1.99 × 10³⁰ kg

To add these, express both with the same exponent:
1.99 × 10³⁰ = 19.9 × 10²⁹

Total mass = 6.42 × 10²⁹ + 19.9 × 10²⁹
\( = (6.42 + 19.9) \times 10^{29} \)
\( = 26.32 \times 10^{29} \)
\( = 2.632 \times 10^{30} \) kg

The total mass is approximately 2.632 × 10³⁰ kg.
In simple words: When adding standard form numbers, first make the exponents the same. Then add the coefficients. Finally, adjust back to standard form if needed.

Exam Tip: To add or subtract numbers in standard form, always convert to the same power of 10 first, then add the coefficients.

Question 7. A particular star is at a distance of about 8.1 × 10¹³ km from the Earth. Assuming that the light travels at 3 × 10⁸ m/sec, find how long does light take from that star to reach the Earth.
Answer:
Given:
Distance from star to Earth = 8.1 × 10¹³ km = 8.1 × 10¹⁶ m (convert km to m by multiplying by 1000 = 10³)
Speed of light = 3 × 10⁸ m/sec

Time = Distance ÷ Speed
\( = (8.1 \times 10^{16}) / (3 \times 10^{8}) \)
\( = (8.1 / 3) \times 10^{16-8} \)
\( = 2.7 \times 10^{8} \text{ seconds} \)

Light takes 2.7 × 10⁸ seconds from the star to reach Earth.
In simple words: To divide numbers in standard form, divide the coefficients and subtract the exponents: \( (a \times 10^m) \div (b \times 10^n) = (a \div b) \times 10^{m-n} \).

Exam Tip: Watch your units - if distance is in km and speed is in m/sec, convert distance to meters first before dividing.