CBSE Class 10 Mathematics Surds and Indices MCQs Set 25

Answer: (a) 4

Explanation:
1. \( 3^{3k}=3^{2-k} \)
2. So \( 3k = 2 - k \), giving \( k = 1/2 \)
3. \( \dfrac1{k^2}=4 \).

Teacher's Note:
1. Write both sides with the same base and equate the exponents.
2. \( 27 = 3^{3} \) and \( 9 = 3^{2} \).

Answer: (d) \( \dfrac{xy}{x+y} \)

Explanation:
1. Let each equal \( t \). Then \( 4=t^{1/x},\ 5=t^{1/y},\ 20=t^{1/z} \)
2. Since \( 20 = 4 \times 5 \): \( \dfrac1z=\dfrac1x+\dfrac1y \)
3. So \( z=\dfrac{xy}{x+y} \).

Teacher's Note:
1. Express each base as a power of the common value \( t \).
2. Then use the product relationship between the bases.

Answer: (c) 0

Explanation:
1. \( 2^{6x}=2^{-8y} \)
2. \( 6x = -8y \), so \( 3x + 4y \) = 0.

Teacher's Note:
1. Reduce both sides to base 2.
2. Then equate the exponents.

Answer: (b) 5

Explanation:
1. \( 625^{(-1/2)(-1/4)(2)}=625^{1/4} \)
2. \( 625 = 5^{4} \), so the value is 5.

Teacher's Note:
1. Multiply the exponents when raising a power to a power.
2. \( (-1/2) \times (-1/4) \times 2 = 1/4 \).

Answer: (c) 5

Explanation:
1. \( 8^{1/3} + 27^{1/3} = 2 + 3 = 5 \)
2. \( [5 \times 125]^{1/4} = 625^{1/4} \) = 5.

Teacher's Note:
1. Evaluate the cube roots first.
2. Then \( 5 \times 5^{3} = 5^{4} \).

Answer: (a) \( 10^{-3} \)

Explanation:
1. \( 1 + 8 + 27 + 64 = 100 \)
2. \( 100^{-3/2}=(10^2)^{-3/2}= \)\( 10^{-3} \).

Teacher's Note:
1. Add the cubes first.
2. Then apply the fractional negative exponent.

Answer: (d) \( \dfrac{10^6}{3^6} \)

Explanation:
1. \( 0.000729 = 0.09^{3} \), so the product is \( (0.09)^{-9/4-3/4}=(0.09)^{-3} \)
2. \( (0.09)^{-3}=\left(\dfrac{100}{9}\right)^3=\dfrac{10^6}{3^6} \).

Teacher's Note:
1. Notice that 0.000729 is the cube of 0.09.
2. Then combine the exponents.

Answer: (a) \( \dfrac1{x^2} \)

Explanation:
1. Inner part: \( (1/x)^{1/4\cdot(-12)}=x^3 \)
2. \( (x^3)^{-2/3}=x^{-2}=\dfrac1{x^2} \).

Teacher's Note:
1. A fourth root is the power 1/4.
2. Multiply exponents step by step.

Answer: (c) \( 10^4 \)

Explanation:
1. \( x^{-3/2}=\dfrac{100}{10^8}=10^{-6} \)
2. \( x=10^{6\cdot2/3}=10^4 \).

Teacher's Note:
1. Isolate the power of \( x \).
2. Raise both sides to the reciprocal exponent −2/3.

Answer: (a) \( 5^{5/2} \)

Explanation:
1. \( 4^{x-1}(4-1)=24\Rightarrow 4^{x-1}=8 \)
2. \( 2^{2x-2}=2^3 \) gives \( x = 5/2 \)
3. \( (2x)^x=5^{5/2} \).

Teacher's Note:
1. Take \( 4^{x-1} \) as a common factor.
2. Then solve the simple exponent equation.

Answer: (b) \( 2^{-3} \)

Explanation:
1. \( a=b^{2x}=c^{4xy}=a^{8xyz} \)
2. So 8xyz = 1 and \( xyz=\dfrac18=2^{-3} \).

Teacher's Note:
1. Substitute the three relations one after another.
2. Compare the power of a on both sides.

Answer: (c) \( \dfrac38 \)

Explanation:
1. \( 27^{2/3} = 9 \) and \( 16^{-3/2} = 1/64 \)
2. \( m^2=\dfrac9{64} \), so \( m \) = 3/8 (positive root).

Teacher's Note:
1. Evaluate each fractional power separately.
2. Take the positive root for \( m \).

Answer: (a) \( \dfrac57 \)

Explanation:
1. Numerator: \( 7^{-1}\times5^4=\dfrac{625}{7} \)
2. \( \sqrt{25^3}=125 \), so the value is \( \dfrac{625}{7\times125}=\dfrac57 \).

Teacher's Note:
1. Square the bracket first.
2. \( \sqrt{25^{3}} = 5^{3} = 125 \).

Answer: (b) \( \dfrac89 \)

Explanation:
1. \( 27^{-2/3} = 3^{-2} = 1/9 \)
2. \( \sqrt{4^3} = 8 \), so the product is 8/9.

Teacher's Note:
1. \( 27 = 3^{3} \) gives \( 27^{-2/3} = 3^{-2} \).
2. \( \sqrt{4^{3}} = 2^{3} = 8 \).

Answer: (d) \( \dfrac12 \)

Explanation:
1. \( 3^{1+m}=3^{3m} \)
2. So \( 1 + m = 3m \) and \( m \) = 1/2.

Teacher's Note:
1. \( 9^{1/2} = 3 \).
2. Equate exponents with base 3.

Answer: (b) \( \dfrac43 \)

Explanation:
1. \( 9^{x/2}=9^{2/3} \)
2. So \( x/2 = 2/3 \) and \( x \) = 4/3.

Teacher's Note:
1. Both sides have base 9.
2. Equate exponents.

Answer: (d) 0.008

Explanation:
1. \( a^{3/2} = (\sqrt{a})^{3} = 0.2^{3} \)
2. = 0.008.

Teacher's Note:
1. \( a^{3/2} \) is the cube of \( \sqrt{a} \).
2. \( 0.2^{3} = 0.008 \).

Answer: (b) \( a^3-3a \)

Explanation:
1. Cube a: \( a^3=x^3+\dfrac1{x^3}+3\left(x+\dfrac1x\right) \)
2. So \( x^3+x^{-3}=a^3-3a \).

Teacher's Note:
1. Expand \( (x + 1/x)^{3} \).
2. The middle term is \( 3(x + 1/x) \).

Answer: (c) \( 9a^2b^2c^2 \)

Explanation:
1. When \( a + b + c = 0, a^{3} + b^{3} + c^{3} \) = 3abc
2. Squaring: \( 9a^2b^2c^2 \).

Teacher's Note:
1. Use \( a^{3} + b^{3} + c^{3} \) − 3abc = (\( a + b \) + c)(…).
2. The bracket is multiplied by zero.

Answer: (b) x = 12, y = 6, z = 4

Explanation:
1. \( x = 2y = 3z \). Let \( x = 6k, y = 3k, z = 2k \)
2. xyz = \( 36k^{3} = 288 \), so \( k = 2 \)
3. Hence x = \( 12, y = 6, z \) = 4.

Teacher's Note:
1. Convert to a common base and relate the exponents.
2. Then use the product condition.

Answer: (a) 1

Explanation:
1. Let each equal \( t \): \( 2=t^{1/x},\ 3=t^{1/y},\ 12=t^{1/z} \)
2. \( 12 = 2^{2} \times 3 \), so \( \dfrac1z=\dfrac2x+\dfrac1y=\dfrac{x+2y}{xy} \)
3. So \( \dfrac{z(x+2y)}{xy}=1 \).

Teacher's Note:
1. Express 12 as 2² × 3.
2. Reciprocals of the exponents add.

Answer: (a) x = 3/4

Explanation:
1. \( \sqrt{p/q}=(q/p)^{-1/2} \)
2. \( 1 - 2x = -1/2 \), so \( x \) = 3/4.

Teacher's Note:
1. Rewrite the RHS with base \( q/p \).
2. Equate exponents.

Answer: (a) \( \dfrac57 \)

Explanation:
1. Base 2: \( 2^{2x-2}\cdot2^{2x-3}=2^{-3x} \)
2. \( 4x - 5 = -3x \), so \( x \) = 5/7.

Teacher's Note:
1. Write \( 4, 1/2 \) and 0.125 as powers of 2.
2. Add exponents on the left.

Answer: (b) \( \dfrac{11}{50} \)

Explanation:
1. \( 6^0 = 1 \), so \( 100x - 22 = 0 \)
2. \( x = 22/100 \) = 11/50.

Teacher's Note:
1. Any non-zero base to the power 0 equals 1.
2. Solve the linear equation.