NMAT 2026 Quiz 7: Practice Questions with Solutions

Q1. Arrange the following in ascending order of their magnitude: 2³⁰, 3²⁰, 5¹⁵, 7¹⁰, 17⁵ 17⁵ < 7¹⁰ < 2³⁰ < 3²⁰ < 5¹⁵ 17⁵ < 7¹⁰ < 5¹⁵ < 3²⁰ < 2³⁰ 17⁵ < 7¹⁰ < 3²⁰ < 2³⁰ < 5¹⁵ 7¹⁰ < 17⁵ < 2³⁰ < 3²⁰ < 5¹⁵ 17⁵ < 3²⁰ < 5¹⁵ < 7¹⁰ < 2³⁰ Correct Option A … Explanation: All five exponents share the common factor 5, so take the fifth root of each term — the order is preserved. This gives 2⁶ = 64, 3⁴ = 81, 5³ = 125, 7² = 49 and 17¹ = 17. Arranging these bases in ascending order, 17 < 49 < 64 < 81 < 125, which translates back to 17⁵ < 7¹⁰ < 2³⁰ < 3²⁰ < 5¹⁵. Option B wrongly places 2³⁰ as the largest, Option C swaps 2³⁰ and 3²⁰, Option D wrongly places 7¹⁰ below 17⁵ (49 > 17), and Option E scatters the order entirely. Hence, option A. Q2. If 2^(x+1) × 4^(2x−3) = 8^(x+2), find x. 4 7/2 9/2 11/2 5 Correct Option D … Explanation: Express every term with base 2. The left side becomes 2^(x+1) × 2^(4x−6) = 2^(5x−5), and the right side becomes 2^(3x+6). Since the bases are equal, the exponents must be equal: 5x − 5 = 3x + 6, giving 2x = 11 and x = 11/2. Options A, B, C and E all arise from mishandling the multiplier 2 on (2x − 3) or the multiplier 3 on (x + 2) while converting to a common base. Hence, option D. Q3. In a 9-term arithmetic progression, the 5th term is the number obtained by reversing the digits of the first term, and its 9th term is 5/2 times the first term. If the first term is greater than 20 but less than 30, find the common difference. 2.5 3 4.5 5.5 6 Correct Option C … Explanation: Let the first term be a = 20 + u, so its reverse is 10u + 2. From the 9th term condition, a + 8d = 2.5a, giving 8d = 1.5a and d = 3a/16. From the 5th term condition, a + 4d = 10u + 2, and substituting d gives 7a/4 = 10u + 2. Putting a = 20 + u yields 140 + 7u = 40u + 8, so u = 4, a = 24 and d = 3(24)/16 = 4.5. The progression checks out: the 5th term is 24 + 18 = 42 (the reverse of 24) and the 9th term is 24 + 36 = 60 = 2.5 × 24. Options A, B, D and E fail one or both of these two conditions. Hence, option C. Q4. An arithmetic progression has 11 terms, a common difference of 5, and a mean of 30. If one term is deleted, the mean of the remaining 10 terms becomes 31. Which term, by its position from the beginning, was deleted? 2nd 3rd 4th 5th 6th Correct Option C … Explanation: For an AP with an odd number of terms, the mean equals the middle term, so the 6th term is 30 and the total sum is 11 × 30 = 330. After deletion the sum of the remaining 10 terms is 10 × 31 = 310, so the deleted term is 330 − 310 = 20. Working backwards from the 6th term with d = 5, the first term is 30 − 25 = 5, so the series runs 5, 10, 15, 20, 25 … and 20 occupies the 4th position. Options A, B, D and E correspond to the values 10, 15, 25 and 30, none of which restores the required sum of 310. Hence, option C. Q5. Find the positive value of y if y = √(90 + √(90 + √(90 + … ∞))) 9 10 15 18 12 Correct Option B … Explanation: Because the radical is infinitely nested, the expression inside the outermost root is identical to y itself, so y = √(90 + y). Squaring gives y² − y − 90 = 0, which factorises as (y − 10)(y + 9) = 0. The roots are y = 10 and y = −9, and since a square root yields a non-negative value, only y = 10 is admissible. A quick verification confirms it: 10² = 100 = 90 + 10. Options A, C, D and E do not satisfy y² = 90 + y. Hence, option B. Q6. How many integers x satisfy log₂(x − 1) + log₂(9 − x) > 3? 2 3 4 5 6 Correct Option D … Explanation: The domain requires x − 1 > 0 and 9 − x > 0, so 1 < x < 9. Combining the logarithms gives log₂[(x − 1)(9 − x)] > 3, that is (x − 1)(9 − x) > 8, which simplifies to x² − 10x + 17 < 0. The roots are 5 ± 2√2, so x must lie between roughly 2.17 and 7.83. The integers in this interval are 3, 4, 5, 6 and 7 — five values in all. Options A, B, C and E result from mis-solving the quadratic or from including the boundary integers 2 and 8, which give a product of 7 and fail the strict inequality. Hence, option D. Q7. If logₐ 8 = 3/2 and log_b a = 2, then b equals √2 2 4 8 ½ Correct Option B … Explanation: From logₐ 8 = 3/2 we get a^(3/2) = 8, so a = 8^(2/3) = 4. Substituting into log_b a = 2 gives b² = 4, and since a logarithmic base must be positive, b = 2. Option C mistakes the intermediate value a = 4 for the final answer, Option D simply repeats the number 8 from the question, and Options A and E arise from inverting the exponent 3/2 or the exponent 2 incorrectly. Hence, option B. Q8. If a = √3 + √2, find a⁴ + 1/a⁴ 23 27 48 52 98 Correct Option E … Explanation: Rationalising gives 1/a = 1/(√3 + √2) = √3 − √2. Then a² = 5 + 2√6 and 1/a² = 5 − 2√6, so a² + 1/a² = 10. Using the identity a⁴ + 1/a⁴ = (a² + 1/a²)² − 2, the answer is 10² − 2 = 98. Option D (52) is the value obtained by forgetting to subtract 2 and mis-squaring, Option C (48) drops the surd terms, and Options A and B do not follow from any consistent expansion. Hence, option E. Q9. Arrange the following in descending order: ∛4, √3, ⁶√36, ⁴√9 ⁶√36 > ⁴√9 > √3 > ∛4 ⁶√36 > ⁴√9 > ∛4 > √3 ⁶√36 > ⁴√9 = √3 > ∛4 ⁴√9 = √3 > ⁶√36 > ∛4 ⁶√36 > ⁴√9 > √3 > ∛4 Correct Option C … Explanation: Reduce each surd and bring all of them to a common index of 6. Note that ⁴√9 = 9^(1/4) = 3^(1/2) = √3, so these two are exactly equal. Writing the rest with index 6: ⁶√36 = 36^(1/6), √3 = 27^(1/6) and ∛4 = 16^(1/6). Since 36 > 27 > 16, the correct descending order is ⁶√36 > ⁴√9 = √3 > ∛4. Option A wrongly shows ⁴√9 as strictly greater than √3 when they are equal, Option B additionally places ∛4 above √3, and Option D wrongly demotes ⁶√36 below √3. Hence, option C. Q10. Find n: 3¹⁰ × 729 ÷ 19683 = 3ⁿ × 81 2 3 4 5 6 Correct Option B … Explanation: Convert every number to a power of 3: 729 = 3⁶, 19683 = 3⁹ and 81 = 3⁴. The left side becomes 3^(10 + 6 − 9) = 3⁷ and the right side becomes 3^(n + 4). Equating the exponents gives n + 4 = 7, so n = 3. Options A, C, D and E follow from misidentifying 19683 as 3⁸ or 3¹⁰, or from adding the exponent of 81 instead of subtracting it. Hence, option B. Q11. If 2ˣ · 3ʸ = 576 and 2ʸ · 3ˣ = 2916, find the value of x + y. 8 7 4 5 6 Correct Option A … Explanation: Factorise both numbers into primes: 576 = 2⁶ × 3² and 2916 = 2² × 3⁶. Matching the first equation gives x = 6 and y = 2, and the second equation confirms the same pair since 2² × 3⁶ = 2916. Therefore x + y = 8. A faster route is to multiply the two equations: 6^(x+y) = 576 × 2916 = 1679616 = 6⁸, giving x + y = 8 directly. Options B, C, D and E follow from incomplete factorisation of 576 or 2916. Hence, option A. Q12. Positive integers x, y, z satisfy log₂ x, log₄ y, log₈ z are in arithmetic progression. Also xyz = 16384 and z/x = 256. Find y. 8 16 32 64 2 Correct Option B … Explanation: Write x = 2ᵖ, y = 2^q and z = 2ʳ, so the three logarithms become p, q/2 and r/3. The AP condition gives 2(q/2) = p + r/3, that is q = p + r/3. From xyz = 16384 = 2¹⁴ we get p + q + r = 14, and from z/x = 256 = 2⁸ we get r − p = 8. Solving these three relations gives p = 1, q = 4 and r = 9, so y = 2⁴ = 16. Options A, C, D and E correspond to q = 3, 5, 6 and 1, none of which satisfies all three conditions simultaneously. Hence, option B. Q13. If log₃ x + log₉ x + log₂₇ x = 11, find x. 64 125 1331 11 729 Correct Option E … Explanation: Convert all three logarithms to base 3. Since 9 = 3² and 27 = 3³, we have log₉ x = (log₃ x)/2 and log₂₇ x = (log₃ x)/3. Letting t = log₃ x, the equation becomes t + t/2 + t/3 = 11, that is 11t/6 = 11, so t = 6 and x = 3⁶ = 729. Options A and B are powers of 4 and 5 rather than of 3, Option C is 11³ (a trap built on the number 11 in the question), and Option D simply repeats 11. Hence, option E. Q14. Seven distinct integers are arranged in increasing order. Their sum is 149, the smallest integer is 9, the largest integer is 36, and the median is 22. The 2nd, 3rd, 5th, and 6th integers, in that order, form an arithmetic progression. What is the greatest possible value of the 3rd integer? 21 20 19 18 17 Correct Option D … Explanation: The seven terms are 9, a, b, 22, c, d, 36, and since the total is 149 we get a + b + c + d = 82. As these four form an AP with common difference k, their sum is 4a + 6k = 82, so 2a + 3k = 41, which forces k to be odd. Testing odd values: k = 1 gives c = 21 and k = 3 gives c = 22, both violating c > 22; k = 5 gives the valid set 9, 13, 18, 22, 23, 28, 36; k = 7 gives 9, 10, 17, 22, 24, 31, 36; and k = 9 forces a = 7, which is below the minimum of 9. The largest admissible 3rd integer is therefore 18. Options A, B and C are ruled out because they push the 5th term to 22 or below, and Option E is valid but not the greatest. Hence, option D. Q15. log₉(3 log₂(1 + log₃(1 + 2 log₂ x))) = 1/2. Find x. 1/2 1 2 4 3 Correct Option C … Explanation: Unwrap the expression one layer at a time. Since log₉(N) = 1/2, N = 9^(1/2) = 3, so 3 log₂(…) = 3 and log₂(1 + log₃(…)) = 1. This gives 1 + log₃(1 + 2 log₂ x) = 2, so log₃(1 + 2 log₂ x) = 1 and 1 + 2 log₂ x = 3. Hence log₂ x = 1 and x = 2. Option B (x = 1) makes log₂ x = 0 and collapses the innermost bracket to 1, Option D (x = 4) makes the innermost bracket 5 rather than 3, and Options A and E do not yield integer values at any layer. Hence, option C. Q16. If log₂[3 + log₃{4 + log₄(x − 1)}] − 2 = 0 then x equals 4/5 3/4 3/2 2/3 5/4 Correct Option E … Explanation: Rearranging gives log₂[3 + log₃{4 + log₄(x − 1)}] = 2, so the bracket equals 2² = 4 and log₃{4 + log₄(x − 1)} = 1. The inner brace therefore equals 3, giving log₄(x − 1) = −1, so x − 1 = 4⁻¹ = 1/4 and x = 5/4. Note that x must exceed 1 for log₄(x − 1) to be defined, which immediately eliminates Options A, B and D, all of which are less than 1. Option C (3/2) would make log₄(x − 1) = log₄(1/2) = −1/2, which does not satisfy the equation. Hence, option E. Q17. The number of terms common to both the arithmetic progressions 2, 5, 8, 11, …, 419 and 3, 10, 17, 24, …, 381 is 17 18 20 21 22 Correct Option B … Explanation: The first series has common difference 3 and the second has common difference 7, so the common terms themselves form an AP with common difference equal to the LCM of 3 and 7, namely 21. Scanning the second series, the first term that also appears in the first series is 17. The common terms must not exceed the smaller of the two last terms, which is 381, so we need 17 + 21n ≤ 381, giving n ≤ 17.33 and hence n = 0 to 17 — a total of 18 terms. Option A (17) is the count obtained by forgetting the first term, and Options C, D and E overshoot by using 419 instead of 381 as the ceiling. Hence, option B. Q18. If x, 2x + 2, 3x + 3 are in G.P., then the fourth term is −13.5 −15 13.5 15 7.5 Correct Option A … Explanation: For three terms in GP the middle term squared equals the product of the outer two, so (2x + 2)² = x(3x + 3). This gives 4(x + 1)² = 3x(x + 1), and factoring out (x + 1) gives (x + 1)(x + 4) = 0. The root x = −1 is rejected because it makes every term zero, so x = −4 and the GP is −4, −6, −9 with common ratio 3/2. The fourth term is −9 × 3/2 = −13.5. Options C, D and E ignore the negative sign that follows from x = −4, and Option B does not arise from the ratio 3/2. Hence, option A. Q19. A bouncing tennis ball is dropped from a height of 32 metres. The ball rebounds each time to a height equal to half the height of the previous bounce. The approximate distance travelled by the ball till the time it hits the ground for the 10th time is 64 m 96 m 128 m 144 m 160 m Correct Option B … Explanation: The initial drop of 32 m accounts for the first contact with the ground. Between the first and tenth contacts there are 9 rebounds, each covering its height twice (up and down), with heights 16, 8, 4, … forming a GP of ratio 1/2. The sum of these 9 rebound heights is 16(1 − (1/2)⁹)/(1 − 1/2) ≈ 31.94 m, so the rebound travel is about 63.88 m. The total is therefore 32 + 63.88 ≈ 95.88, or approximately 96 m. Option A counts only the descents, Option C doubles the initial drop unnecessarily, and Options D and E ignore that the GP converges to a finite limit close to 96. Hence, option B. Q20. If Sₙ = 4n² + 5n, where Sₙ denotes the sum of the first n terms of a series, then the nᵗʰ term is 8n + 1 10n − 1 4n + 5 5n + 4 10n + 4 Correct Option A … Explanation: The nth term of any series is aₙ = Sₙ − Sₙ₋₁. Here Sₙ₋₁ = 4(n − 1)² + 5(n − 1) = 4n² − 3n − 1, so aₙ = (4n² + 5n) − (4n² − 3n − 1) = 8n + 1. A quick check confirms it: S₁ = 9 and a₁ = 8(1) + 1 = 9. Option C merely copies the coefficients of the given expression, Option D reverses them, and Options B and E give a₁ = 9 only by coincidence or not at all — substituting n = 2 shows S₂ = 26, so a₂ must be 17, which only 8n + 1 satisfies. Hence, option A.