Class 9 Indices (Exponents) Mock Test (ICSE)

Topics in Indices (Exponents)

• Laws of indices
• Zero, negative and fractional indices
• Simplifying expressions with indices
• Solving exponential equations

Indices (Exponents) practice questions with answers

1. Q1Multiple choice·1 mark
   In a water treatment plant, the number of micro-organisms $N$ in a settling tank after $x$ hours is given by $N = 2^{3x-1}$. If $N = 32$, the value of $x$ is:

   1. (a)$1$
   2. (b)$2$
   3. (c)$3$
   4. (d)$\frac{11}{3}$
   Show answer and step marks

   1. Correct option: (b) $2$+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   Simplify $\frac{(2^{3})^{2} \times 2^{-1}}{2^{4}}$. The value is:

   1. (a)$2^{3}$
   2. (b)$2$
   3. (c)$2^{5}$
   4. (d)$\frac{1}{2}$
   Show answer and step marks

   1. Correct option: (b) $2$+1

   Answer: (b)
3. Q3Assertion and reason·1 mark
   Assertion (A): $5^{-2} = \dfrac{1}{25}$. Reason (R): $a^{-m} = a^{m}$ for every non-zero number $a$ and every positive integer $m$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (c) Assertion (A) is true but Reason (R) is false, since $a^{-m} = \dfrac{1}{a^{m}}$, not $a^{m}$+1

   Answer: (c)
4. Q4Assertion and reason·1 mark
   Assertion (A): The solution of $4^{x} = 16$ is $x = 2$. Reason (R): Since $16 = 4^{2}$, the equation $4^{x} = 16$ gives $x = 3$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (c) Assertion (A) is true but Reason (R) is false.+1

   Answer: (c)
5. Q5Short answer·2 marks
   Let $P = 16^{-\frac{3}{4}}$ and $Q = 16^{\frac{3}{4}}$. (a) Find the value of $P$. [1] (b) State how many times greater $Q$ is than $P$. [1]
   Show answer and step marks

   1. $16^{-\frac{3}{4}} = \dfrac{1}{16^{\frac{3}{4}}} = \dfrac{1}{(16^{\frac{1}{4}})^3} = \dfrac{1}{2^3} = \dfrac{1}{8}$, so $P = \dfrac{1}{8}$+1
   2. $Q = 16^{\frac{3}{4}} = 8$, and $\dfrac{Q}{P} = 8 \div \dfrac{1}{8} = 64$, so $Q$ is 64 times $P$+1

   Answer: (a) $P = \dfrac{1}{8}$; (b) $Q$ is 64 times $P$
6. Q6Long answer·4 marks
   Evaluate each of the following expressions using the laws of indices: (a) $\left(\frac{216}{343}\right)^{-\frac{2}{3}} \div \left(\frac{25}{49}\right)^{-\frac{1}{2}}$ (b) Given $x = 2$, find the exact value of $\left( 3x^{-1} + 4x^{-2} \right)^{-1}$
   Show answer and step marks

   1. For part (a), express the bases in exponential form: $\left(\frac{6^3}{7^3}\right)^{-\frac{2}{3}} \div \left(\frac{5^2}{7^2}\right)^{-\frac{1}{2}}$+0.5
   2. Apply the power rule to get $\left(\frac{6}{7}\right)^{-2} \div \left(\frac{5}{7}\right)^{-1}$ and convert negative indices: $\left(\frac{7}{6}\right)^{2} \div \frac{7}{5}$+0.5
   3. Simplify: $\frac{49}{36} \div \frac{7}{5} = \frac{49}{36} \times \frac{5}{7} = \frac{35}{36}$+1
   4. For part (b), substitute $x = 2$ into the expression: $\left(3(2)^{-1} + 4(2)^{-2}\right)^{-1}$+0.5
   5. Simplify the inner terms: $\left(3\left(\frac{1}{2}\right) + 4\left(\frac{1}{4}\right)\right)^{-1} = \left(\frac{3}{2} + 1\right)^{-1} = \left(\frac{5}{2}\right)^{-1}$+1
   6. Apply the negative index to get the final value: $\frac{2}{5}$+0.5

   Answer: (a) $\frac{35}{36}$, (b) $\frac{2}{5}$

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