RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents

Exponents

Rules of Exponents

Exercise 5A

Question 1. Express each of the following as a power:
(i) \( \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \)
(ii) \( \frac{-4}{3} \times \frac{-4}{3} \times \frac{-4}{3} \times \frac{-4}{3} \times \frac{-4}{3} \)
(iii) \( \frac{-1}{6} \times \frac{-1}{6} \times \frac{-1}{6} \)
(iv) \( (-8) \times (-8) \times (-8) \times (-8) \times (-8) \)
Answer:
(i) \( \left(\frac{5}{7}\right)^4 \)
(ii) \( \left(\frac{-4}{3}\right)^5 \)
(iii) \( \left(\frac{-1}{6}\right)^3 \)
(iv) \( (-8)^5 \)

Exam Tip: Count how many times the number is repeated in the multiplication - that count becomes the exponent.

Question 2. Express each of the following as a rational number raised to a power:
(i) \( \frac{25}{36} \)
(ii) \( \frac{-27}{64} \)
(iii) \( \frac{-32}{243} \)
(iv) \( \frac{-1}{128} \)
Answer:
(i) \( \frac{25}{36} = \frac{5^2}{6^2} = \left(\frac{5}{6}\right)^2 \)
(ii) \( \frac{-27}{64} = \frac{(-3)^3}{4^3} = \left(\frac{-3}{4}\right)^3 \)
(iii) \( \frac{-32}{243} = \frac{(-2)^5}{3^5} = \left(\frac{-2}{3}\right)^5 \)
(iv) \( \frac{-1}{128} = \frac{(-1)^7}{2^7} = \left(\frac{-1}{2}\right)^7 \)

Exam Tip: Recognise the numerator and denominator as powers of smaller numbers first, then combine them into a single power of a fraction.

Question 3. Evaluate:
(i) \( \left(\frac{2}{3}\right)^5 \)
(ii) \( \left(\frac{-8}{5}\right)^3 \)
(iii) \( \left(\frac{-13}{11}\right)^2 \)
(iv) \( \left(\frac{1}{6}\right)^3 \)
(v) \( \left(\frac{-1}{2}\right)^5 \)
(vi) \( \left(\frac{-3}{2}\right)^4 \)
(vii) \( \left(\frac{-4}{7}\right)^3 \)
(viii) \( (-1)^9 \)
Answer:
(i) \( \frac{2^5}{3^5} = \frac{32}{243} \)
(ii) \( \frac{(-8)^3}{5^3} = \frac{-512}{125} \)
(iii) \( \frac{(-13)^2}{11^2} = \frac{169}{121} \)
(iv) \( \frac{1^3}{6^3} = \frac{1}{216} \)
(v) \( \frac{(-1)^5}{2^5} = \frac{-1}{32} \)
(vi) \( \frac{(-3)^4}{2^4} = \frac{81}{16} \)
(vii) \( \frac{(-4)^3}{7^3} = \frac{-64}{343} \)
(viii) \( -1 \), since -1 raised to any odd power stays -1

Exam Tip: A negative base raised to an even power always gives a positive result, while an odd power keeps the negative sign.

Question 4. Evaluate:
(i) \( 4^{-1} \)
(ii) \( (-6)^{-1} \)
(iii) \( \left(\frac{1}{3}\right)^{-1} \)
(iv) \( \left(\frac{-2}{3}\right)^{-1} \)
Answer:
(i) \( 4^{-1} = \left(\frac{4}{1}\right)^{-1} = \frac{1}{4} \)
(ii) \( (-6)^{-1} = \left(\frac{-6}{1}\right)^{-1} = \frac{-1}{6} \)
(iii) \( \left(\frac{1}{3}\right)^{-1} = \frac{3}{1} = 3 \)
(iv) \( \left(\frac{-2}{3}\right)^{-1} = \frac{-3}{2} \)

Exam Tip: A negative exponent of -1 always means "flip the fraction" - numerator and denominator swap places.

Question 5. Write the reciprocal of each of the following:
(i) \( \left(\frac{3}{8}\right)^4 \)
(ii) \( \left(\frac{-5}{6}\right)^{11} \)
(iii) \( 6^7 \)
(iv) \( (-4)^3 \)
Answer:
(i) Reciprocal of \( \left(\frac{3}{8}\right)^4 \) is \( \left(\frac{8}{3}\right)^4 \)
(ii) Reciprocal of \( \left(\frac{-5}{6}\right)^{11} \) is \( \left(\frac{-6}{5}\right)^{11} \)
(iii) Reciprocal of \( 6^7 \), written as \( \left(\frac{6}{1}\right)^7 \), is \( \left(\frac{1}{6}\right)^7 \)
(iv) Reciprocal of \( (-4)^3 \), written as \( \left(\frac{-4}{1}\right)^3 \), is \( \left(\frac{-1}{4}\right)^3 \)

Exam Tip: The reciprocal of a power flips the base fraction while leaving the exponent itself unchanged.

Question 6. Evaluate:
(i) \( 8^0 \)
(ii) \( (-3)^0 \)
(iii) \( 4^0 + 5^0 \)
(iv) \( 6^0 \times 7^0 \)
Answer:
(i) \( 8^0 = 1 \)
(ii) \( (-3)^0 = 1 \)
(iii) \( 4^0 + 5^0 = 1 + 1 = 2 \)
(iv) \( 6^0 \times 7^0 = 1 \times 1 = 1 \)

Exam Tip: Remember that any nonzero number raised to the power 0 equals exactly 1, regardless of how large or small the base is.

Question 7. Evaluate:
(i) \( \left(\frac{3}{2}\right)^4 \times \left(\frac{1}{5}\right)^2 \)
(ii) \( \left(\frac{-2}{3}\right)^5 \times \left(\frac{-3}{7}\right)^3 \)
(iii) \( \left(\frac{-1}{2}\right)^5 \times 2^3 \times \left(\frac{3}{4}\right)^2 \)
(iv) \( \left(\frac{2}{3}\right)^2 \times \left(\frac{-3}{5}\right)^3 \times \left(\frac{7}{2}\right)^2 \)
(v) \( \left\{\left(\frac{-3}{4}\right)^3 - \left(\frac{-5}{2}\right)^3\right\} \times 4^2 \)
Answer:
(i) \( \frac{3^4}{2^4} \times \frac{1^2}{5^2} = \frac{81}{16} \times \frac{1}{25} = \frac{81}{400} \)
(ii) \( \frac{(-2)^5}{3^5} \times \frac{(-3)^3}{7^3} = \frac{-32}{243} \times \frac{-27}{343} = \frac{32}{3087} \)
(iii) \( \frac{(-1)^5}{2^5} \times 2^3 \times \frac{3^2}{4^2} = \frac{-1 \times 2^3 \times 3^2}{2^5 \times 4^2} = \frac{-9}{64} \)
(iv) \( \frac{2^2}{3^2} \times \frac{(-3)^3}{5^3} \times \frac{7^2}{2^2} = \frac{-1 \times 3 \times 49}{125} = \frac{-147}{125} \)
(v) \( \left(\frac{-27}{64} - \frac{-125}{8}\right) \times 16 = \left(\frac{-27+1000}{64}\right) \times 16 = \frac{973}{64} \times 16 = \frac{973}{4} \)

Exam Tip: Simplify each power separately before multiplying or subtracting them - trying to combine unsimplified powers directly leads to mistakes.

Question 8. Simplify:
(i) \( \left(\frac{4}{9}\right)^6 \times \left(\frac{4}{9}\right)^{-4} \)
(ii) \( \left(\frac{-7}{8}\right)^{-3} \times \left(\frac{-7}{8}\right)^2 \)
(iii) \( \left(\frac{4}{3}\right)^{-3} \times \left(\frac{4}{3}\right)^{-2} \)
Answer:
(i) \( \left(\frac{4}{9}\right)^{6+(-4)} = \left(\frac{4}{9}\right)^2 = \frac{16}{81} \)
(ii) \( \left(\frac{-7}{8}\right)^{-3+2} = \left(\frac{-7}{8}\right)^{-1} = \frac{-8}{7} \)
(iii) \( \left(\frac{4}{3}\right)^{-3+(-2)} = \left(\frac{4}{3}\right)^{-5} = \left(\frac{3}{4}\right)^5 = \frac{243}{1024} \)

Exam Tip: When multiplying powers of the same base, add the exponents rather than multiplying the bases together.

Question 9. Evaluate:
(i) \( 5^{-3} \)
(ii) \( (-2)^{-5} \)
(iii) \( \left(\frac{1}{4}\right)^{-4} \)
(iv) \( \left(\frac{-3}{4}\right)^{-3} \)
(v) \( (-3)^{-1} \times \left(\frac{1}{3}\right)^{-1} \)
(vi) \( \left(\frac{5}{7}\right)^{-1} \times \left(\frac{7}{4}\right)^{-1} \)
(vii) \( (5^{-1} - 7^{-1})^{-1} \)
(viii) \( \left\{\left(\frac{4}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1} \)
(ix) \( \left(\frac{3}{2}\right)^{-1} \div \left(\frac{-2}{5}\right)^{-1} \)
(x) \( \left(\frac{23}{25}\right)^0 \)
Answer:
(i) \( \left(\frac{1}{5}\right)^3 = \frac{1}{125} \)
(ii) \( \left(\frac{-1}{2}\right)^5 = \frac{-1}{32} \)
(iii) \( 4^4 = 256 \)
(iv) \( \left(\frac{-4}{3}\right)^3 = \frac{-64}{27} \)
(v) \( \frac{-1}{3} \times 3 = -1 \)
(vi) \( \frac{7}{5} \times \frac{4}{7} = \frac{4}{5} \)
(vii) \( \left(\frac{1}{5} - \frac{1}{7}\right)^{-1} = \left(\frac{7-5}{35}\right)^{-1} = \left(\frac{2}{35}\right)^{-1} = \frac{35}{2} \)
(viii) \( \left(\frac{3}{4} - 4\right)^{-1} = \left(\frac{3-16}{4}\right)^{-1} = \left(\frac{-13}{4}\right)^{-1} = \frac{-4}{13} \)
(ix) \( \frac{2}{3} \div \frac{-5}{2} = \frac{2}{3} \times \frac{-2}{5} = \frac{-4}{15} \)
(x) \( \left(\frac{23}{25}\right)^0 = 1 \), since any nonzero number raised to the power 0 is 1

Exam Tip: Turn each negative exponent into a positive one by flipping the base first - this makes the rest of the calculation much simpler.

Question 10. Simplify:
(i) \( \left[\left\{\left(\frac{-1}{4}\right)^2\right\}^{-2}\right]^{-1} \)
(ii) \( \left\{\left(\frac{-2}{3}\right)^2\right\}^3 \)
(iii) \( \left(\frac{-3}{2}\right)^3 \div \left(\frac{-3}{2}\right)^6 \)
(iv) \( \left(\frac{-2}{3}\right)^7 \div \left(\frac{-2}{3}\right)^4 \)
Answer:
(i) Multiplying the exponents step by step: \( \left(\frac{-1}{4}\right)^{2 \times -2 \times -1} = \left(\frac{-1}{4}\right)^4 = \frac{1}{256} \)
(ii) \( \left(\frac{-2}{3}\right)^{2 \times 3} = \left(\frac{-2}{3}\right)^6 = \frac{(-2)^6}{3^6} = \frac{64}{729} \)
(iii) \( \left(\frac{-3}{2}\right)^{3-6} = \left(\frac{-3}{2}\right)^{-3} = \left(\frac{2}{-3}\right)^3 = \frac{-8}{27} \)
(iv) \( \left(\frac{-2}{3}\right)^{7-4} = \left(\frac{-2}{3}\right)^3 = \frac{-8}{27} \)

Exam Tip: When a power is raised to another power, multiply the exponents together; when powers of the same base are divided, subtract the exponents.

Question 11. Find the value of x such that \( (-5)^{-1} \times x = (8)^{-1} \).
Answer:
\( \frac{-1}{5} \times x = \frac{1}{8} \), so \( x = \frac{1}{8} \times (-5) = \frac{-5}{8} \)

Exam Tip: Convert both negative exponents into fraction form first, then solve the resulting simple equation for x.

Question 12. Find the value of x such that \( (3)^{-3} \times x = 4 \).
Answer:
\( \frac{1}{27} \times x = 4 \), so \( x = 4 \times 27 = 108 \)

Exam Tip: Isolating x here just means multiplying both sides by the reciprocal of the fractional coefficient.

Question 13. Find the value of x such that \( (-30)^{-1} \div x = 6^{-1} \).
Answer:
\( \frac{-1}{30} \div x = \frac{1}{6} \), so \( \frac{-1}{30x} = \frac{1}{6} \), giving \( x = \frac{6}{-30} = \frac{-1}{5} \)

Exam Tip: When the unknown sits in the denominator after a division, cross-multiply carefully to bring it into the numerator before solving.

Question 14. Find the value of x such that \( \left(\frac{3}{5}\right)^3 \times \left(\frac{3}{5}\right)^{-6} = \left(\frac{3}{5}\right)^{2x-1} \).
Answer:
\( \left(\frac{3}{5}\right)^{3+(-6)} = \left(\frac{3}{5}\right)^{2x-1} \), so \( \left(\frac{3}{5}\right)^{-3} = \left(\frac{3}{5}\right)^{2x-1} \)
Matching the exponents: \( -3 = 2x - 1 \), so \( 2x = -2 \), giving \( x = -1 \)

Exam Tip: Once both sides share the same base, the two exponents must be equal - this turns the problem into a simple linear equation.

Question 15. Simplify: \( \frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5} \)
Answer:
Writing \( 10 = 2 \times 5 \) and \( 25 = 5^2 \), \( 6 = 2 \times 3 \): \( \frac{3^5 \times 2^5 \times 5^5 \times 5^2}{5^7 \times 2^5 \times 3^5} \)
\( = 3^{5-5} \times 2^{5-5} \times 5^{7-7} = 3^0 \times 2^0 \times 5^0 = 1 \times 1 \times 1 = 1 \)

Exam Tip: Breaking composite bases like 10, 25 and 6 into their prime factors first often reveals matching bases that can be cancelled directly.

Question 16. Simplify: \( \frac{16 \times 2^{n+1} - 4 \times 2^n}{16 \times 2^{n+2} - 2 \times 2^{n+2}} \)
Answer:
Writing \( 16 = 2^4 \) and \( 4 = 2^2 \): \( \frac{2^4 \times 2^{n+1} - 2^2 \times 2^n}{2^4 \times 2^{n+2} - 2 \times 2^{n+2}} = \frac{2^{n+5} - 2^{n+2}}{2^{n+6} - 2^{n+3}} \)
\( = \frac{2^{n+2}(2^3 - 1)}{2^{n+3}(2^3 - 1)} = \frac{2^{n+2}}{2^{n+3}} = 2^{-1} = \frac{1}{2} \)

Exam Tip: Factor out the smallest matching power from both the top and bottom of the fraction, then cancel the identical bracket that remains.

Question 17. Find n such that:
(i) \( 5^{2n} \times 5^3 = 5^9 \)
(ii) \( 8 \times 2^{n+2} = 32 \)
(iii) \( 6^{2n+1} \div 36 = 6^3 \)
Answer:
(i) \( 5^{2n+3} = 5^9 \), so \( 2n+3 = 9 \), giving \( n = 3 \)
(ii) \( 2^3 \times 2^{n+2} = 2^5 \), so \( 2^{n+5} = 2^5 \), giving \( n = 0 \)
(iii) \( 6^{2n+1} \div 6^2 = 6^3 \), so \( 6^{2n-1} = 6^3 \), giving \( 2n-1 = 3 \), so \( n = 2 \)

Exam Tip: Rewrite every term with the same base first, then equate the exponents to turn the problem into ordinary algebra.

Question 18. Find n such that \( 2^{n-7} \times 5^{n-4} = 1250 \).
Answer:
Writing \( 1250 = 2 \times 5^4 \): \( \frac{2^n}{2^7} \times \frac{5^n}{5^4} = 2 \times 5^4 \)
Cross-multiplying: \( 2^n \times 5^n = 2^{1+7} \times 5^{4+4} = 2^8 \times 5^8 = (2 \times 5)^8 = 10^8 \)
So, \( (2 \times 5)^n = 10^8 \), giving \( 10^n = 10^8 \), so \( n = 8 \)

Exam Tip: When two different bases both need the same exponent to reach the target, combine them into a single base (like 10 = 2×5) to solve cleanly.

Exercise 5B

Question 1. Express each of the following in standard form:
(i) 538
(ii) 6428000
(iii) 82934000000
(iv) 940000000000
(v) 23000000
Answer:
(i) \( 538 = 5.38 \times 10^2 \), since the decimal point shifts 2 places to the left
(ii) \( 6428000 = 6.428 \times 10^6 \), since the decimal point shifts 6 places to the left
(iii) \( 82934000000 = 8.2934 \times 10^{10} \), since the decimal point shifts 10 places to the left
(iv) \( 940000000000 = 9.4 \times 10^{11} \), since the decimal point shifts 11 places to the left
(v) \( 23000000 = 2.3 \times 10^7 \), since the decimal point shifts 7 places to the left

Exam Tip: Count exactly how many places the decimal point moves to land just after the first non-zero digit - that count becomes the power of 10.

Question 2. Express the following facts in standard form:
(i) The diameter of the Earth is 12,756,000 m.
(ii) The distance between the Earth and the Moon is 384,000,000 m.
(iii) The population of India in March 2001 was 1,027,000,000.
(iv) The number of stars in a galaxy is about 100,000,000,000.
(v) The present age of the universe is about 12,000,000,000 years.
Answer:
(i) Diameter of the Earth \( = 1.2756 \times 10^7 \) m
(ii) Distance between the Earth and the Moon \( = 3.84 \times 10^8 \) m
(iii) Population of India in March 2001 \( = 1.027 \times 10^9 \)
(iv) Number of stars in a galaxy \( = 1.0 \times 10^{11} \)
(v) Present age of the universe \( = 1.2 \times 10^{10} \) years

Exam Tip: Very large real-world figures like these are exactly what standard form is designed for - it keeps the numbers compact and easy to compare.

Question 3. Write each of the following numbers in expanded exponential form:
(i) 684502
(ii) 4007185
(iii) 5807294
(iv) 50074
Answer:
(i) \( 684502 = 6 \times 10^5 + 8 \times 10^4 + 4 \times 10^3 + 5 \times 10^2 + 0 \times 10^1 + 2 \times 10^0 \)
(ii) \( 4007185 = 4 \times 10^6 + 0 \times 10^5 + 0 \times 10^4 + 7 \times 10^3 + 1 \times 10^2 + 8 \times 10^1 + 5 \times 10^0 \)
(iii) \( 5807294 = 5 \times 10^6 + 8 \times 10^5 + 0 \times 10^4 + 7 \times 10^3 + 2 \times 10^2 + 9 \times 10^1 + 4 \times 10^0 \)
(iv) \( 50074 = 5 \times 10^4 + 0 \times 10^3 + 0 \times 10^2 + 7 \times 10^1 + 4 \times 10^0 \)

Exam Tip: Each digit of the number gets multiplied by the power of 10 matching its place value, working from the highest place down to the units.

Question 4. Find the number whose expanded exponential form is given:
(i) \( 6 \times 10^4 + 3 \times 10^3 + 0 \times 10^2 + 7 \times 10^1 + 8 \times 10^0 \)
(ii) \( 9 \times 10^6 + 7 \times 10^5 + 0 \times 10^4 + 3 \times 10^3 + 4 \times 10^2 + 6 \times 10^1 + 2 \times 10^0 \)
(iii) \( 8 \times 10^5 + 6 \times 10^4 + 4 \times 10^3 + 2 \times 10^2 + 9 \times 10^1 + 6 \times 10^0 \)
Answer:
(i) \( 60000 + 3000 + 0 + 70 + 8 = 63078 \)
(ii) \( 9000000 + 700000 + 0 + 3000 + 400 + 60 + 2 = 9703462 \)
(iii) \( 800000 + 60000 + 4000 + 200 + 90 + 6 = 864296 \)

Exam Tip: Work out each term's actual value first, then add all the terms together to rebuild the original number.

Exercise 5C

Question 1. \( (6^{-1} - 8^{-1})^{-1} = \)
(a) \( \frac{1}{24} \)
(b) \( \frac{-1}{24} \)
(c) -24
(d) 24
Using the LCM of 6 and 8 (24): \( \left(\frac{4-3}{24}\right)^{-1} = \left(\frac{1}{24}\right)^{-1} = 24 \)
Answer: (d) 24

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 2. \( (5^{-1} \times 3^{-1})^{-1} = \)
(a) \( \frac{1}{15} \)
(b) \( \frac{-1}{15} \)
(c) 15
(d) -15
\( \left(\frac{1}{5} \times \frac{1}{3}\right)^{-1} = \left(\frac{1}{15}\right)^{-1} = 15 \)
Answer: (c) 15

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 3. \( (2^{-1} - 4^{-1})^2 = \)
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{8} \)
(c) \( \frac{1}{16} \)
(d) \( \frac{1}{2} \)
Using the LCM of 2 and 4 (4): \( \left(\frac{2-1}{4}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)
Answer: (c) \( \frac{1}{16} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 4. \( \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} = \)
(a) 19
(b) 29
(c) 39
(d) 9
\( 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29 \)
Answer: (b) 29

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 5. \( \left\{6^{-1} + \left(\frac{3}{2}\right)^{-1}\right\}^{-1} = \)
(a) \( \frac{5}{6} \)
(b) \( \frac{5}{4} \)
(c) \( \frac{6}{5} \)
(d) \( \frac{4}{5} \)
Using the LCM of 6 and 3 (6): \( \left(\frac{1+4}{6}\right)^{-1} = \left(\frac{5}{6}\right)^{-1} = \frac{6}{5} \)
Answer: (c) \( \frac{6}{5} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 6. \( \left(\frac{-1}{2}\right)^{-6} = \)
(a) -64
(b) 64
(c) \( \frac{-1}{64} \)
(d) \( \frac{1}{64} \)
\( \left(\frac{-1}{2}\right)^{-6} = (-2)^6 = 64 \)
Answer: (b) 64

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 7. \( \left\{\left(\frac{3}{4}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1} = \)
(a) \( \frac{3}{8} \)
(b) \( \frac{-3}{8} \)
(c) \( \frac{8}{3} \)
(d) \( \frac{-8}{3} \)
\( \left(\frac{4}{3} - 4\right)^{-1} = \left(\frac{4-12}{3}\right)^{-1} = \left(\frac{-8}{3}\right)^{-1} = \frac{-3}{8} \)
Answer: (b) \( \frac{-3}{8} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 8. \( \left[\left\{\left(\frac{-1}{2}\right)^2\right\}^{-2}\right]^{-1} = \)
(a) \( \frac{1}{16} \)
(b) 16
(c) \( \frac{-1}{16} \)
(d) -16
Multiplying the exponents: \( \left(\frac{-1}{2}\right)^{2 \times -2 \times -1} = \left(\frac{-1}{2}\right)^4 = \frac{1}{16} \)
Answer: (a) \( \frac{1}{16} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 9. \( \left(\frac{5}{6}\right)^0 = \)
(a) 0
(b) \( \frac{5}{6} \)
(c) 1
(d) \( \frac{6}{5} \)
Any nonzero number raised to the power 0 equals 1.
Answer: (c) 1

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 10. \( \left(\frac{2}{3}\right)^{-5} = \)
(a) \( \frac{32}{243} \)
(b) \( \frac{243}{32} \)
(c) \( \frac{-32}{243} \)
(d) \( \frac{-243}{32} \)
\( \left(\frac{2}{3}\right)^{-5} = \left(\frac{3}{2}\right)^5 = \frac{243}{32} \)
Answer: (b) \( \frac{243}{32} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 11. \( \left\{\left(\frac{1}{3}\right)^2\right\}^4 = \)
(a) \( \left(\frac{1}{3}\right)^6 \)
(b) \( \left(\frac{1}{3}\right)^8 \)
(c) \( \left(\frac{1}{3}\right)^2 \)
(d) \( \left(\frac{1}{3}\right)^{16} \)
Multiplying the exponents: \( \left(\frac{1}{3}\right)^{2 \times 4} = \left(\frac{1}{3}\right)^8 \)
Answer: (b) \( \left(\frac{1}{3}\right)^8 \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 12. \( \left(\frac{-3}{2}\right)^{-1} = \)
(a) \( \frac{3}{2} \)
(b) \( \frac{-2}{3} \)
(c) \( \frac{2}{3} \)
(d) \( \frac{-3}{2} \)
Flipping the fraction gives \( \left(\frac{-3}{2}\right)^{-1} = \frac{2}{-3} = \frac{-2}{3} \)
Answer: (b) \( \frac{-2}{3} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 13. \( (3^2 - 2^2) \times \left(\frac{2}{3}\right)^{-3} = \)
(a) \( \frac{8}{135} \)
(b) \( \frac{40}{27} \)
(c) \( \frac{27}{40} \)
(d) \( \frac{135}{8} \)
\( (9-4) \times \left(\frac{3}{2}\right)^3 = 5 \times \frac{27}{8} = \frac{135}{8} \)
Answer: (d) \( \frac{135}{8} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 14. \( \left\{\left(\frac{1}{3}\right)^{-3} - \left(\frac{1}{2}\right)^{-3}\right\} \div \left(\frac{1}{4}\right)^{-3} = \)
(a) \( \frac{19}{64} \)
(b) \( \frac{-19}{64} \)
(c) \( \frac{64}{19} \)
(d) \( \frac{35}{64} \)
\( (3^3 - 2^3) \div 4^3 = (27-8) \div 64 = 19 \div 64 = \frac{19}{64} \)
Answer: (a) \( \frac{19}{64} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 15. \( \left(\frac{-1}{5}\right)^3 \div \left(\frac{-1}{5}\right)^8 = \)
(a) \( 5^5 \)
(b) \( \frac{1}{5^5} \)
(c) \( (-5)^5 \)
(d) \( \frac{-1}{5^5} \)
\( \left(\frac{-1}{5}\right)^{3-8} = \left(\frac{-1}{5}\right)^{-5} = (-5)^5 \)
Answer: (c) \( (-5)^5 \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 16. \( \left(\frac{-2}{5}\right)^7 \div \left(\frac{-2}{5}\right)^5 = \)
(a) \( \frac{4}{25} \)
(b) \( \frac{-4}{25} \)
(c) \( \frac{25}{4} \)
(d) \( \frac{-25}{4} \)
\( \left(\frac{-2}{5}\right)^{7-5} = \left(\frac{-2}{5}\right)^2 = \frac{4}{25} \)
Answer: (a) \( \frac{4}{25} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 17. \( \left(\frac{-2}{3}\right)^2 = \)
(a) \( \frac{-4}{9} \)
(b) \( \frac{-4}{6} \)
(c) \( \frac{4}{9} \)
(d) \( \frac{4}{6} \)
\( \frac{-2}{3} \times \frac{-2}{3} = \frac{4}{9} \)
Answer: (c) \( \frac{4}{9} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 18. \( \left(\frac{-1}{2}\right)^3 = \)
(a) \( \frac{1}{8} \)
(b) \( \frac{-1}{8} \)
(c) \( \frac{3}{2} \)
(d) \( \frac{-3}{2} \)
\( \frac{-1}{2} \times \frac{-1}{2} \times \frac{-1}{2} = \frac{-1}{8} \)
Answer: (b) \( \frac{-1}{8} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 20. \( \left(\frac{5}{3}\right)^{-5} \times \left(\frac{5}{3}\right)^{11} = \left(\frac{5}{3}\right)^{8x} \). The value of x is:
(a) \( \frac{4}{3} \)
(b) \( \frac{3}{8} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{8}{3} \)
\( \left(\frac{5}{3}\right)^{-5+11} = \left(\frac{5}{3}\right)^{8x} \), so \( 6 = 8x \), giving \( x = \frac{6}{8} = \frac{3}{4} \)
Answer: (c) \( \frac{3}{4} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.

Question 21. \( (-8)^{-1} \times x = (10)^{-1} \). The value of x is:
(a) \( \frac{4}{5} \)
(b) \( \frac{-4}{5} \)
(c) \( \frac{-5}{4} \)
(d) \( \frac{5}{4} \)
\( \frac{-1}{8} \times x = \frac{1}{10} \), so \( x = \frac{1}{10} \times (-8) = \frac{-8}{10} = \frac{-4}{5} \)
Answer: (b) \( \frac{-4}{5} \)

Exam Tip: Work the value out first, then match it to the closest-looking option - the distractors are built from common slip-ups.