Algebra — Complex Numbers, Quadratics, Sequences & Binomial Theorem — Practice

Question 1

The value of i¹³ + i¹⁶ + i¹⁹ + i²² is:
(a) 0  (b) -1  (c) 2i  (d) 2

Accepted Answer

Reduce each exponent mod 4: i¹³ (13 mod 4=1) = i; i¹⁶ (16 mod 4=0) = 1; i¹⁹ (19 mod 4=3) = −i; i²² (22 mod 4=2) = −1.

Sum = i + 1 − i − 1 = 0.

Answer: (a) 0. Trap (b) -1 comes from wrongly adding the exponents first (13+16+19+22=70) and computing i⁷⁰ instead of summing the individual terms (70 mod 4=2, giving i⁷⁰=−1). Trap (c) 2i comes from misremembering i³=+i instead of −i (sign slip), giving i+1+i−1=2i. Trap (d) 2 comes from misremembering i²² as +1 instead of −1 (confusing it with i²⁰), giving i+1−i+1=2.

Question 2

If ω is a complex cube root of unity, the value of (1+ω−ω²)(1−ω+ω²) is:
(a) 4  (b) 0  (c) -4  (d) 1

Accepted Answer

Since 1+ω+ω²=0, we have ω² = −1−ω. Substitute into each bracket:
1+ω−ω² = 1+ω−(−1−ω) = 2+2ω = 2(1+ω) = 2(−ω²) = −2ω² (using 1+ω=−ω²).
1−ω+ω² = 1−ω+(−1−ω) = −2ω.

Product = (−2ω²)(−2ω) = 4ω³ = 4(1) = 4 (since ω³=1).

Answer: (a) 4. Trap (b) 0 comes from wrongly assuming one of the brackets itself equals zero, confusing it with the identity 1+ω+ω²=0 (neither bracket is that expression). Trap (c) -4 comes from a sign slip treating ω³ as −1 instead of 1. Trap (d) 1 comes from correctly reducing to 4ω³ but then forgetting the factor of 4 and reporting only ω³=1.

Question 3

For what value of k does x² − 4x + k = 0 have real and equal roots?
(a) 4  (b) 16  (c) -4  (d) 1

Accepted Answer

Equal roots require D = b²−4ac = 0. Here D = (−4)² − 4(1)(k) = 16 − 4k = 0 ⇒ k = 4.

Answer: (a) 4. Trap (b) 16 comes from forgetting the factor of 4 in '4ac', wrongly computing D as b²−ac = 16−k = 0 ⇒ k=16. Trap (c) -4 comes from a sign error, setting D = b²+4ac = 16+4k = 0 ⇒ k=−4. Trap (d) 1 comes from using b instead of b² (D = b−4ac = 4−4k = 0 ⇒ k=1).

Question 4

If α and β are the roots of x² − 6x + 4 = 0, the value of α² + β² is:
(a) 28  (b) 36  (c) 44  (d) 20

Accepted Answer

α+β = 6, αβ = 4. α²+β² = (α+β)² − 2αβ = 36 − 8 = 28.

Answer: (a) 28. Trap (b) 36 comes from forgetting to subtract 2αβ at all, just squaring the sum. Trap (c) 44 comes from adding instead of subtracting: 36+2(4)=44. Trap (d) 20 comes from confusing this with (α−β)² = (α+β)²−4αβ = 36−16=20, a different (though related) identity.

Question 5

If x is a root of x² − 5x + 1 = 0 (x ≠ 0), the value of x³ + 1/x³ is:
(a) 110  (b) 125  (c) 140  (d) 23

Accepted Answer

Divide the equation by x: x − 5 + 1/x = 0 ⇒ x + 1/x = 5.

x³+1/x³ = (x+1/x)³ − 3(x+1/x) = 5³ − 3(5) = 125 − 15 = 110.

Answer: (a) 110. Trap (b) 125 comes from forgetting to subtract the 3k term, just cubing k. Trap (c) 140 comes from a sign error using k³+3k instead of k³−3k (125+15=140). Trap (d) 23 comes from computing the wrong identity altogether — x²+1/x² = k²−2 = 25−2 = 23 — confusing the square identity with the cube identity that was actually asked for.

Question 6

In a class of 60 students, 25 play cricket, 20 play football, and 8 play both. How many students play neither?
(a) 23  (b) 15  (c) 31  (d) 37

Accepted Answer

n(A∪B) = n(A)+n(B)−n(A∩B) = 25+20−8 = 37. Neither = Total − n(A∪B) = 60−37 = 23.

Answer: (a) 23. Trap (b) 15 comes from forgetting to subtract the intersection when finding the union (wrongly using 25+20=45), giving neither = 60−45=15. Trap (c) 31 comes from double-subtracting the intersection (25+20−2×8=29), giving neither=60−29=31. Trap (d) 37 comes from correctly finding n(A∪B) but stopping there, forgetting the final step of subtracting from the total to get 'neither'.

Question 7

The 15th term of the AP 7, 11, 15, 19, ... is:
(a) 63  (b) 67  (c) 56  (d) 71

Accepted Answer

a=7, d=4. a₁₅ = a+(n−1)d = 7+14(4) = 7+56 = 63.

Answer: (a) 63. Trap (b) 67 comes from the classic aₙ=a+nd slip: 7+15(4)=67. Trap (c) 56 comes from computing only (n−1)d and forgetting to add a: 14(4)=56. Trap (d) 71 comes from the opposite off-by-one error, using a+(n+1)d: 7+16(4)=71.

Question 8

The sum of the infinite GP 9, 3, 1, 1/3, ... is:
(a) 27/2  (b) 27  (c) 6  (d) 27/4

Accepted Answer

a=9, r=1/3 (|r|<1, so S∞ exists). S∞ = a/(1−r) = 9/(1−1/3) = 9/(2/3) = 9×3/2 = 27/2.

Answer: (a) 27/2. Trap (b) 27 comes from miscalculating 1−r as 1/3 instead of 2/3, giving 9/(1/3)=27. Trap (c) 6 comes from multiplying instead of dividing: 9×(2/3)=6. Trap (d) 27/4 comes from using the wrong-sign formula a/(1+r) instead of a/(1−r): 9/(1+1/3)=9/(4/3)=27/4.

Question 9

Two positive numbers have AM = 10 and GM = 8. Their HM is:
(a) 32/5  (b) 6  (c) 25/2  (d) 8

Accepted Answer

AM × HM = GM² ⇒ HM = GM²/AM = 64/10 = 32/5 (= 6.4).

Answer: (a) 32/5. Trap (b) 6 comes from guessing a linear relation like HM=2×GM−AM (16−10=6) instead of using AM×HM=GM² — exactly the 'guess a pair instead of solving' trap this chapter warns about. Trap (c) 25/2 comes from inverting the formula (using AM²/GM=100/8=12.5 instead of GM²/AM). Trap (d) 8 comes from wrongly assuming HM equals GM.

Question 10

The coefficient of x⁵ in the expansion of (x+2)⁸ is:
(a) 448  (b) 56  (c) 1792  (d) 224

Accepted Answer

General term T_(r+1) = C(8,r) x^(8−r) 2^r. Need 8−r=5 ⇒ r=3. T₄ = C(8,3) x⁵ 2³ = 56 × 8 × x⁵ = 448x⁵.

Answer: (a) 448. Trap (b) 56 comes from forgetting the 2^r factor entirely and reporting only C(8,3). Trap (c) 1792 comes from using 2^(n−r) instead of 2^r, i.e. C(8,3)×2⁵=56×32=1792 — mixing up which term the power of 2 attaches to. Trap (d) 224 comes from an off-by-one on the power of 2, using 2^(r−1) instead of 2^r: 56×4=224.

Question 11

The 4th term in the expansion of (x − 2/x)⁷ is:
(a) -280x  (b) 280x  (c) -280/x  (d) -35x

Accepted Answer

T_(r+1) = C(7,r) x^(7−r) (−2/x)^r = C(7,r)(−2)^r x^(7−2r). The 4th term has r=3: T₄ = C(7,3)(−2)³x^(7−6) = 35×(−8)×x¹ = −280x.

Answer: (a) -280x. Trap (b) 280x comes from a sign error, forgetting (−2)³ is negative and using +2³=8 instead. Trap (c) -280/x comes from an exponent sign error, computing 2r−7 instead of 7−2r (giving x⁻¹ instead of x¹). Trap (d) -35x comes from forgetting to raise (−2) to the power 3 at all, using just the coefficient C(7,3)=35 with the correct sign but no magnitude from the (−2/x)³ factor.

Question 12

Given log₁₀2 = 0.3010, the number of digits in 2⁵⁰ is:
(a) 16  (b) 15  (c) 17  (d) 4

Accepted Answer

Digits in N = ⌊log₁₀N⌋+1. log₁₀(2⁵⁰) = 50×0.3010 = 15.05. Digits = ⌊15.05⌋+1 = 15+1 = 16.

Answer: (a) 16. Trap (b) 15 comes from forgetting the '+1' in the digit-count formula and reporting just the floor. Trap (c) 17 comes from rounding 15.05 up to 16 (using ceiling instead of floor) and then still adding 1, double-counting. Trap (d) 4 comes from misreading 2⁵⁰ as 50² = 2500 (flipping base and exponent under time pressure), which has only 4 digits — a misreading error, not a calculation one.

Question 13

In how many ways can 3 letters be chosen from the word 'FLOWER' (all 6 letters distinct) if the order of selection does not matter?
(a) 20  (b) 120  (c) 15  (d) 216

Accepted Answer

Since order doesn't matter, this is a combination: ⁶C₃ = 6!/(3!3!) = 20.

Answer: (a) 20. Trap (b) 120 comes from using permutations instead — ⁶P₃=6×5×4=120 — the classic order-vs-selection confusion. Trap (c) 15 comes from using ⁶C₂=15 instead of ⁶C₃ (an off-by-one slip in r). Trap (d) 216 comes from treating this as 3 independent choices with repetition allowed (6×6×6), ignoring both the 'no repeats' condition and the combination formula entirely.

Question 14

A committee of 5 is to be formed from 6 men and 4 women. In how many ways can this be done if at least 2 women must be included?
(a) 186  (b) 252  (c) 66  (d) 246

Accepted Answer

Total ways to choose 5 from 10 = ¹⁰C₅ = 252. Subtract the cases with fewer than 2 women (complement method):
0 women: ⁴C₀×⁶C₅ = 1×6 = 6.
1 woman: ⁴C₁×⁶C₄ = 4×15 = 60.
Excluded = 6+60 = 66. Answer = 252−66 = 186.

Answer: (a) 186. Trap (b) 252 comes from computing the unrestricted total and forgetting to apply the 'at least 2 women' condition at all. Trap (c) 66 comes from correctly finding the excluded (complement) count but reporting that instead of subtracting it from the total. Trap (d) 246 comes from subtracting only the '0 women' case and forgetting the '1 woman' case: 252−6=246.

Practice — Algebra — Complex Numbers, Quadratics, Sequences & Binomial Theorem

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