Surds and Indices Aptitude Questions and Answers (Solved MCQs)

Surds and indices frighten students only because of the symbols. Strip the symbols away and it is one idea, that a root is just a power written differently. The moment you accept that va is a to the power half, every root question turns into an index question and you already know the index laws from school. So do that conversion first, always. The second thing is rationalising, which sounds like a big word but is just flipping the sign in the middle and multiplying. And when you have to compare two surds with different roots, do not calculate them, bring them to a common index using the LCM and compare the numbers inside. Learn the eight index laws below properly, and this chapter becomes the quickest set of marks in the arithmetic section.

Surds and Indices Formulas

The laws of indices
aᵐ × aⁿ = a⁽ᵐ ⁺ ⁿ⁾
aᵐ ÷ aⁿ = a⁽ᵐ ⁻ ⁿ⁾
(aᵐ)ⁿ = a⁽ᵐ ˣ ⁿ⁾
(a × b)ᵐ = aᵐ × bᵐ
a⁰ = 1, for any a other than 0
a⁻ᵐ = 1 ÷ aᵐ
(a ÷ b)⁻ᵐ = (b ÷ a)ᵐ

Roots as powers
√a = a^(1/2), ³√a = a^(1/3), ⁿ√a = a^(1/n)
ⁿ√(aᵐ) = a^(m/n)
Do this conversion at the start of every question and the rest is just index laws.

Surd rules
√a × √b = √(a × b)
√a ÷ √b = √(a ÷ b)
(√a)² = a
Remember √a + √b is never √(a + b). This mistake costs more marks than anything else in the chapter.

Simplifying and adding
Break the number inside into a perfect square times something else, then take the square out. √72 = 6√2
Only like surds can be added. √50 − √18 = 5√2 − 3√2 = 2√2
If they are unlike after simplifying, leave them. √2 + √3 stays as it is.

Rationalising the denominator
1 ÷ √a = √a ÷ a
1 ÷ (√a + √b) = (√a − √b) ÷ (a − b)
1 ÷ (√a − √b) = (√a + √b) ÷ (a − b)
1 ÷ (a + √b) = (a − √b) ÷ (a² − b)
The rule is one line, flip the sign in the middle and multiply top and bottom by it.

Comparing surds of different orders
Take the LCM of the orders and bring both to that common order, then compare the numbers inside.
For √3 and ³√5, the LCM is 6, so √3 = ⁶√27 and ³√5 = ⁶√25. So √3 is bigger.
Never find the decimal value.

Nested surds
√(a + 2√b) = √x + √y, where x + y = a and x × y = b
So √(7 + 2√12) = 2 + √3, because 4 + 3 = 7 and 4 × 3 = 12.

Infinite surds
√(a × √(a × √(a…))) = a
√(a + √(a + √(a…))) = the positive root of x² − x − a = 0
So √(6 + √(6 + √(6…))) = 3

Solving index equations
Bring both sides to the same base, then equate the powers. For 4ˣ = 8, write 2^(2x) = 2³, so x = 3/2.
For surd equations, isolate the root and square, then always check your answer in the original equation, because squaring can create a false root.

60 Surds and Indices Aptitude Questions and Answers (Solved MCQs)

Question 1. What is the value of 2⁵×2⁻³?

a) 2
b) 4
c) 8
d) 16

Answer:

b) 4 — When powers having the same base are multiplied, their exponents are added. Therefore, 2⁵×2⁻³ = 2⁵⁻³ = 2² = 4.

Question 2. What is the value of (3⁴)³/3¹⁰?

a) 3
b) 6
c) 9
d) 27

Answer:

c) 9 — Using the power-of-a-power rule, (3⁴)³ = 3¹². Therefore, 3¹²/3¹⁰ = 3¹²⁻¹⁰ = 3² = 9.

Question 3. What is the value of 16^3/4?

a) 4
b) 6
c) 8
d) 12

Answer:

c) 8 — Since 16 = 2⁴, we have 16^3/4 = (2⁴)^3/4 = 2³ = 8.

Question 4. What is the value of 27^−2/3?

a) 1/27
b) 1/9
c) 1/3
d) 9

Answer:

b) 1/9 — Since 27^1/3 = 3, we get 27^2/3 = 3² = 9. A negative exponent gives the reciprocal, so 27^−2/3 = 1/9.

Question 5. Which expression is equivalent to [(x³y⁻²)²]/(x⁻¹y³), where x and y are non-zero?

a) (x/y)⁵
b) (x/y)⁶
c) (x/y)⁷
d) (y/x)⁷

Answer:

c) (x/y)⁷ — The numerator is x⁶y⁻⁴. Dividing by x⁻¹y³ gives x⁶⁻⁽⁻¹⁾y⁻⁴⁻³ = x⁷y⁻⁷ = (x/y)⁷.

Question 6. If 4^x = 8^x−1, what is the value of x?

a) 2
b) 3
c) 4
d) 5

Answer:

b) 3 — Expressing both sides with base 2 gives 2^2x = 2^3(x−1). Therefore, 2x = 3x−3, so x = 3.

Question 7. If 9^x+1 = 27^x−1, what is the value of x?

a) 3
b) 4
c) 5
d) 6

Answer:

c) 5 — Writing both sides as powers of 3 gives 3^2x+2 = 3^3x−3. Therefore, 2x+2 = 3x−3, giving x = 5.

Question 8. If 2^x+2^x+1 = 96, what is the value of x?

a) 4
b) 5
c) 6
d) 7

Answer:

b) 5 — Taking 2ˣ as a common factor gives 2ˣ(1+2) = 96. Therefore, 3×2ˣ = 96, so 2ˣ = 32 = 2⁵. Hence, x = 5.

Question 9. What is the value of (81/16)^−3/4?

a) 8/27
b) 16/27
c) 27/8
d) 81/64

Answer:

a) 8/27 — Since 81/16 = (3/2)⁴, we have [(3/2)⁴]^−3/4 = (3/2)⁻³ = (2/3)³ = 8/27.

Question 10. Which statement correctly compares 2³⁰ and 8¹⁰?

a) 2³⁰ > 8¹⁰
b) 2³⁰ < 8¹⁰
c) 2³⁰ = 8¹⁰
d) Their relationship cannot be determined

Answer:

c) 2³⁰ = 8¹⁰ — Since 8 = 2³, we have 8¹⁰ = (2³)¹⁰ = 2³⁰. Therefore, the two quantities are equal.

Question 11. What is the value of (6×10⁸)(3×10⁻⁵) in scientific notation?

a) 1.8×10³
b) 1.8×10⁴
c) 18×10⁴
d) 1.8×10¹³

Answer:

b) 1.8×10⁴ — Multiplying gives 18×10⁸⁻⁵ = 18×10³. In standard scientific notation, this is 1.8×10⁴.

Question 12. Which expression is equivalent to (a^1/2b^1/3)⁶?

a) a²b³
b) a³b²
c) a⁶b⁶
d) a^7/2b^8/3

Answer:

b) a³b² — Multiplying the exponents by 6 gives a^(1/2)×6b^(1/3)×6 = a³b².

Question 13. If 5^2x−1 = 125, what is the value of x?

a) 1
b) 2
c) 3
d) 4

Answer:

b) 2 — Since 125 = 5³, we have 5^2x−1 = 5³. Therefore, 2x−1 = 3, giving 2x = 4 and x = 2.

Question 14. What is the value of 32^2/5×8^1/3?

a) 4
b) 6
c) 8
d) 16

Answer:

c) 8 — Since 32 = 2⁵, we get 32^2/5 = 2² = 4. Also, 8^1/3 = 2. Therefore, the product is 4×2 = 8.

Question 15. What is the value of (2⁻²+4⁻¹)⁻¹?

a) 1
b) 2
c) 4
d) 8

Answer:

b) 2 — We have 2⁻² = 1/4 and 4⁻¹ = 1/4. Their sum is 1/2. Therefore, (1/2)⁻¹ = 2.

Question 16. What is the simplest form of √72?

a) 4√2
b) 6√2
c) 8√2
d) 12√2

Answer:

b) 6√2 — Since 72 = 36×2, we have √72 = √36×√2 = 6√2.

Question 17. What is the simplified value of 3√50−2√8+√18?

a) 10√2
b) 12√2
c) 14√2
d) 16√2

Answer:

c) 14√2 — We have √50 = 5√2, √8 = 2√2 and √18 = 3√2. Therefore, 3√50−2√8+√18 = 15√2−4√2+3√2 = 14√2.

Question 18. What is the value of √12×√27?

a) 9
b) 12
c) 18
d) 24

Answer:

c) 18 — Using √a×√b = √(ab), we get √12×√27 = √324 = 18.

Question 19. What is the value of √75/√3?

a) 3
b) 5
c) 15
d) 25

Answer:

b) 5 — Using √a/√b = √(a/b), we get √75/√3 = √(75/3) = √25 = 5.

Question 20. What is the rationalised form of 5/√3?

a) √15/3
b) 3√5/5
c) 5√3/3
d) 5√3

Answer:

c) 5√3/3 — Multiplying the numerator and denominator by √3 gives 5√3/(√3×√3) = 5√3/3.

Question 21. What is the rationalised form of 1/(√5−2)?

a) √5−2
b) √5+2
c) (√5+2)/2
d) 5+2√5

Answer:

b) √5+2 — Multiplying the numerator and denominator by the conjugate √5+2 gives (√5+2)/(5−4) = √5+2.

Question 22. What is the rationalised form of 3/(√7+√2)?

a) 3(√7+√2)/5
b) 3(√7−√2)/5
c) (√7−√2)/3
d) 3√14/5

Answer:

b) 3(√7−√2)/5 — Multiplying by the conjugate √7−√2 gives 3(√7−√2)/[(√7+√2)(√7−√2)]. The denominator is 7−2 = 5. Hence, the rationalised form is 3(√7−√2)/5.

Question 23. What is the value of (√5+√3)²?

a) 8+√15
b) 8+2√15
c) 15+2√8
d) 16+√15

Answer:

b) 8+2√15 — Expanding gives (√5)²+(√3)²+2√5√3 = 5+3+2√15 = 8+2√15.

Question 24. What is the value of (√7+√5)(√7−√5)?

a) 2
b) 7
c) 12
d) √2

Answer:

a) 2 — Using the identity (a+b)(a−b) = a²−b², the expression becomes (√7)²−(√5)² = 7−5 = 2.

Question 25. What is the simplified form of √(9+4√5)?

a) √5+1
b) √5+2
c) 2√5+1
d) 3+√5

Answer:

b) √5+2 — Suppose √(9+4√5) = √a+√b. Squaring gives a+b+2√(ab) = 9+4√5. Thus, a+b = 9 and ab = 20. The numbers are 5 and 4. Therefore, √(9+4√5) = √5+√4 = √5+2.

Question 26. What is the simplified form of √(7−4√3)?

a) √3−1
b) 2−√3
c) 3−√2
d) √6−√2

Answer:

b) 2−√3 — We observe that (2−√3)² = 4+3−4√3 = 7−4√3. Since 2−√3 is positive, √(7−4√3) = 2−√3.

Question 27. What is the value of 1/(√3+√2)+1/(√3−√2)?

a) 2√2
b) 2√3
c) √5
d) 2√6

Answer:

b) 2√3 — Rationalising separately, 1/(√3+√2) = √3−√2 and 1/(√3−√2) = √3+√2 because 3−2 = 1. Adding gives 2√3.

Question 28. Which statement correctly compares √48 and 4√3?

a) √48 > 4√3
b) √48 < 4√3
c) √48 = 4√3
d) The comparison cannot be determined

Answer:

c) √48 = 4√3 — Since 48 = 16×3, we have √48 = √16×√3 = 4√3. Therefore, the two quantities are equal.

Question 29. What is the simplified value of ∛54+∛16?

a) 3∛2
b) 4∛2
c) 5∛2
d) 6∛2

Answer:

c) 5∛2 — Since 54 = 27×2, ∛54 = 3∛2. Also, 16 = 8×2, so ∛16 = 2∛2. Therefore, ∛54+∛16 = 3∛2+2∛2 = 5∛2.

Question 30. What is the rationalised value of (√2+1)/(√2−1)?

a) 1+2√2
b) 2+√2
c) 3+2√2
d) 3−2√2

Answer:

c) 3+2√2 — Multiplying the numerator and denominator by √2+1 gives (√2+1)²/(2−1). Expanding the numerator gives 2+1+2√2 = 3+2√2.

Question 31. If 2^x+2 = 8^x−1, what is the value of x?

a) 3/2
b) 2
c) 5/2
d) 3

Answer:

c) 5/2 — Since 8 = 2³, the equation becomes 2^x+2 = 2^3x−3. Therefore, x+2 = 3x−3, giving 2x = 5 and x = 5/2.

Question 32. What is the sum of all real solutions of 9^x−10(3^x)+9 = 0?

a) 0
b) 1
c) 2
d) 3

Answer:

c) 2 — Let y = 3ˣ. Then 9ˣ = 3^2x = y². The equation becomes y²−10y+9 = 0, which factors as (y−1)(y−9) = 0. Thus, 3ˣ = 1 or 3ˣ = 9, giving x = 0 or x = 2. Their sum is 2.

Question 33. If x is non-negative and x^3/2 = 64, what is the value of x?

a) 8
b) 12
c) 16
d) 32

Answer:

c) 16 — We have x^3/2 = (√x)³ = 64. Therefore, √x = 4, giving x = 16.

Question 34. What is the value of x if √(x+9)−√x = 1?

a) 9
b) 12
c) 16
d) 25

Answer:

c) 16 — Rearranging gives √(x+9) = √x+1. Squaring both sides gives x+9 = x+1+2√x. Therefore, 2√x = 8, so √x = 4 and x = 16. Substitution confirms the solution.

Question 35. What is the value of x if √(x+5)+√(x−4) = 3?

a) 4
b) 5
c) 6
d) 9

Answer:

a) 4 — The domain requires x ≥ 4. Substituting x = 4 gives √9+√0 = 3. Since both square-root terms increase as x increases, the left side is strictly greater than 3 for x > 4. Therefore, x = 4 is the unique solution.

Question 36. If a = √3+√2, what is the value of a²+1/a²?

a) 8
b) 10
c) 12
d) 14

Answer:

b) 10 — Since (√3+√2)(√3−√2) = 1, we have 1/a = √3−√2. Therefore, a+1/a = 2√3. Using a²+1/a² = (a+1/a)²−2, the value is (2√3)²−2 = 12−2 = 10.

Question 37. If a = √5+2, what is the value of a+1/a?

a) 2√3
b) 2√5
c) 4√5
d) 5

Answer:

b) 2√5 — Since (√5+2)(√5−2) = 5−4 = 1, we have 1/a = √5−2. Therefore, a+1/a = (√5+2)+(√5−2) = 2√5.

Question 38. What is the value of (√2+1)⁴+(√2−1)⁴?

a) 30
b) 32
c) 34
d) 36

Answer:

c) 34 — Let a = √2+1 and b = √2−1. Then a+b = 2√2 and ab = 1. Therefore, a²+b² = (a+b)²−2ab = 8−2 = 6. Hence, a⁴+b⁴ = (a²+b²)²−2a²b² = 36−2 = 34.

Question 39. If √(x+√(x+√(x+⋯))) = 2, what is the value of x?

a) 1
b) 2
c) 3
d) 4

Answer:

b) 2 — Let the entire recurring radical be y. Then y = √(x+y). Given y = 2, we get 2 = √(x+2). Squaring gives 4 = x+2, so x = 2.

Question 40. The area of a square is 72 cm². What is the length of its diagonal?

a) 6√2 cm
b) 8 cm
c) 12 cm
d) 12√2 cm

Answer:

c) 12 cm — The side of the square is √72 = 6√2 cm. Its diagonal is side×√2 = 6√2×√2 = 12 cm.

Question 41. A bacterial culture initially contains 500 bacteria and triples every 2 hours. How many bacteria will there be after 8 hours?

a) 13,500
b) 27,000
c) 40,500
d) 45,000

Answer:

c) 40,500 — Eight hours contain four periods of 2 hours. Therefore, the population becomes 500×3⁴ = 500×81 = 40,500.

Question 42. What is the value of (4.8×10⁻³)/(1.2×10⁻⁷) in scientific notation?

a) 4×10³
b) 4×10⁴
c) 4×10⁵
d) 4×10⁻⁴

Answer:

b) 4×10⁴ — Dividing the coefficients gives 4.8/1.2 = 4. Subtracting the exponents gives −3−(−7) = 4. Therefore, the value is 4×10⁴.

Question 43. What is the value of x if √(x+6)+√(x−3) = 3√3?

a) 3
b) 4
c) 6
d) 9

Answer:

c) 6 — The domain requires x ≥ 3. Substituting x = 6 gives √12+√3 = 2√3+√3 = 3√3. Since the left side is strictly increasing over its domain, x = 6 is the unique solution.

Question 44. What is the sum of all real solutions of 5^x+5^−x = 26/5?

a) −1
b) 0
c) 1
d) 2

Answer:

b) 0 — Let y = 5ˣ, so 5^−x = 1/y. Then y+1/y = 26/5. Multiplying by 5y gives 5y²−26y+5 = 0, which factors as (5y−1)(y−5) = 0. Thus, y = 1/5 or y = 5, giving x = −1 or x = 1. Their sum is 0.

Question 45. A square has a side of √7+√3 centimetres. What is its area?

a) 10+√21 cm²
b) 10+2√21 cm²
c) 14+2√21 cm²
d) 21+2√10 cm²

Answer:

b) 10+2√21 cm² — The area is (√7+√3)² = 7+3+2√21 = 10+2√21 cm².

Question 46. What is the simplified form of √(5+2√6)?

a) √3+√2
b) √6+1
c) 2√2+1
d) 3+√2

Answer:

a) √3+√2 — Suppose √(5+2√6) = √a+√b. Squaring gives a+b+2√(ab) = 5+2√6. Therefore, a+b = 5 and ab = 6. The required numbers are 3 and 2. Hence, √(5+2√6) = √3+√2.

Question 47. What is the value of (√3+√2)⁶+(√3−√2)⁶?

a) 866
b) 898
c) 970
d) 1,024

Answer:

c) 970 — Let a = √3+√2 and b = √3−√2. Then a+b = 2√3 and ab = 1. Define S_n = aⁿ+bⁿ. The recurrence is S_n = (a+b)S_n−1−abS_n−2. Starting with S₀ = 2 and S₁ = 2√3, we obtain S₂ = 10, S₃ = 18√3, S₄ = 98, S₅ = 178√3 and S₆ = 970.

Question 48. If x = ∛2+1/∛2, what is the value of x³−3x?

a) 3/2
b) 2
c) 5/2
d) 3

Answer:

c) 5/2 — Let a = ∛2 and b = 1/∛2. Then ab = 1. Using (a+b)³ = a³+b³+3ab(a+b), we get x³ = 2+1/2+3x. Therefore, x³−3x = 5/2.

Question 49. What is the rationalised value of (√3+√2)/(√3−√2)?

a) 3+2√6
b) 5+2√6
c) 5−2√6
d) 6+√6

Answer:

b) 5+2√6 — Multiplying the numerator and denominator by √3+√2 gives (√3+√2)²/(3−2). Expanding the numerator gives 3+2+2√6 = 5+2√6.

Question 50. What is the sum of all real solutions of 2^x+2^1−x = 3?

a) 0
b) 1
c) 2
d) 3

Answer:

b) 1 — Let y = 2ˣ, so 2^1−x = 2/y. The equation becomes y+2/y = 3. Multiplying by y gives y²−3y+2 = 0, or (y−1)(y−2) = 0. Therefore, y = 1 or 2, giving x = 0 or 1. Their sum is 1.

Question 51. Which interval represents all real solutions of 3^2x−10(3^x)+9 ≤ 0?

a) x ≤ 0
b) 0 ≤ x ≤ 2
c) 1 ≤ x ≤ 3
d) x ≥ 2

Answer:

b) 0 ≤ x ≤ 2 — Let y = 3ˣ, where y > 0. The inequality becomes y²−10y+9 ≤ 0, or (y−1)(y−9) ≤ 0. Thus, 1 ≤ y ≤ 9. Since y = 3ˣ, this gives 3⁰ ≤ 3ˣ ≤ 3², so 0 ≤ x ≤ 2.

Question 52. What is the value of √(4+√15)+√(4−√15)?

a) √6
b) 2√2
c) √10
d) 2√3

Answer:

c) √10 — Let the expression be S. Squaring gives S² = 4+√15+4−√15+2√[(4+√15)(4−√15)]. This simplifies to 8+2√(16−15) = 8+2 = 10. Since S is positive, S = √10.

Question 53. If a = √3+√2, what is the value of a⁴+1/a⁴?

a) 82
b) 90
c) 96
d) 98

Answer:

d) 98 — Since (√3+√2)(√3−√2) = 1, we have 1/a = √3−√2. First, a²+1/a² = [(a+1/a)²−2] = (2√3)²−2 = 10. Therefore, a⁴+1/a⁴ = (a²+1/a²)²−2 = 10²−2 = 98.

Question 54. If x is positive and √x+1/√x = 5/2, what is the product of all possible values of x?

a) 1/4
b) 1
c) 4
d) 17/4

Answer:

b) 1 — Let y = √x. Then y+1/y = 5/2. Multiplying by 2y gives 2y²−5y+2 = 0, which factors as (2y−1)(y−2) = 0. Thus, y = 1/2 or 2, giving x = 1/4 or 4. Their product is 1.

Question 55. What is the value of [64^2/3×32^3/5]/8^4/3?

a) 4
b) 8
c) 16
d) 32

Answer:

b) 8 — We have 64^2/3 = (2⁶)^2/3 = 2⁴ = 16, and 32^3/5 = (2⁵)^3/5 = 2³ = 8. Also, 8^4/3 = (2³)^4/3 = 2⁴ = 16. Therefore, the value is (16×8)/16 = 8.

Question 56. What is the value of [(2+√3)⁵−(2−√3)⁵]/(2√3)?

a) 181
b) 199
c) 209
d) 221

Answer:

c) 209 — In the difference (2+√3)⁵−(2−√3)⁵, all terms containing even powers of √3 cancel. Using the binomial expansion, the difference is 2[5(2⁴)√3+10(2²)(√3)³+(√3)⁵]. This equals 2√3[80+120+9] = 418√3. Dividing by 2√3 gives 209.

Question 57. What is the simplified value of ∛(20+14√2)?

a) 1+√2
b) 2+√2
c) 2√2+1
d) 3+√2

Answer:

b) 2+√2 — Expanding (2+√2)³ gives 8+12√2+12+2√2 = 20+14√2. Therefore, ∛(20+14√2) = 2+√2.

Question 58. What is the minimum value of x+9/x for x > 0?

a) 3
b) 6
c) 9
d) 12

Answer:

b) 6 — By the AM-GM inequality, x+9/x ≥ 2√[x×(9/x)] = 2√9 = 6. Equality occurs when x = 9/x, giving x = 3. Therefore, the minimum value is 6.

Question 59. If a = ∛2+∛4, what is the value of a³−6a?

a) 2
b) 4
c) 6
d) 8

Answer:

c) 6 — Let u = ∛2, so ∛4 = u² and u³ = 2. Then a = u+u². Using (u+u²)³ = u³+u⁶+3u³(u+u²), we get a³ = 2+4+6a = 6+6a. Therefore, a³−6a = 6.

Question 60. What is the value of 1/(√5−√3)²+1/(√5+√3)²?

a) 2
b) 3
c) 4
d) 5

Answer:

c) 4 — Since (√5−√3)(√5+√3) = 2, we have 1/(√5−√3) = (√5+√3)/2 and 1/(√5+√3) = (√5−√3)/2. Therefore, the expression becomes [(√5+√3)²+(√5−√3)²]/4. The numerator is 2(5+3) = 16, so the value is 16/4 = 4.

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