UGC NET Questions (Paper – 1)

From a:b = 2:3, let a = 2k and b = 3k. From b:c = 4:5, let b = 4m and c = 5m. Equating b gives 3k = 4m, so we choose k = 4 and m = 3 to get a = 8, b = 12 and c = 15. Then (a + b) = 20 and (b + c) = 27, so the required ratio is 20:27. This ratio is unique after we bring the two given ratios to a common middle term.

Option A:
Option A, 14:19, does not arise from any consistent choice of a, b and c that satisfies both 2:3 and 4:5 as the given ratios. If we attempt to back-calculate from 14:19, the implied values of a, b and c will violate at least one of the original relationships. Hence 14:19 is not compatible with the data.

Option B:
Option B is correct because it is computed after properly equalising the middle term b in both ratios. The method ensures that b plays the same role in both proportional relationships before forming (a + b) and (b + c). This leads precisely to 20:27 as the resulting ratio of the sums.

Option C:
Option C, 7:9, looks like a simplified ratio but does not match the actual numerical values 20 and 27 obtained from solving. Any attempt to assign a, b and c consistent with 7:9 will contradict the starting ratios 2:3 and 4:5. Therefore 7:9 cannot represent (a + b):(b + c).

Option D:
Option D, 3:4, is unrelated to the sums and instead resembles a typical simple ratio. When we compute (a + b) and (b + c) using the correct method, their ratio is not reducible to 3:4. Thus, 3:4 is not the correct resultant ratio.

Solving 2x + 3 ≤ 11, we subtract 3 from both sides to get 2x ≤ 8. Dividing both sides by 2 yields x ≤ 4. On the real number line, this corresponds to all numbers less than or equal to 4. In interval notation, that set is written as (−∞, 4], with a square bracket at 4 indicating that 4 is included.

Option A:
Option A, (−∞, 4), excludes 4 itself and would represent x < 4, which is a strict inequality, not the given ≤ relation. Option B: Option B correctly captures x ≤ 4 by including all values up to and including 4, thus satisfying the inequality exactly. Option C: Option C, (−∞, 7], includes many values greater than 4, violating the upper bound imposed by 2x ≤ 8. Option D: Option D, [4, ∞), represents x ≥ 4, the opposite direction of the inequality, and only includes 4 and greater values.

When multiplying powers of the same base, we keep the base unchanged and add the exponents. The expression 3⁴ × 3² therefore becomes 3^(4 + 2). Adding the exponents gives 3⁶. This rule arises directly from expanding each power into repeated multiplication and then counting the total number of factors of 3.

Option A:
Option A, 3⁸, results from incorrectly multiplying the exponents instead of adding them. There is no exponent law that justifies multiplying 4 and 2 in this context.

Option B:
Option B correctly applies the exponent rule aᵐ × aⁿ = aᵐ⁺ⁿ. By adding 4 and 2, it expresses the product as a single power of 3 with exponent 6.

Option C:
Option C, 3², keeps only one of the original exponents and discards the contribution of the other factor. This contradicts the idea of combining powers.

Option D:
Option D, 3⁴, similarly ignores the factor 3² and so cannot be equivalent to the full product.

From a:b = 7:9, we can take a = 7k and b = 9k. From b:c = 3:4, take b = 3m and c = 4m. Equating b gives 9k = 3m, so m = 3k and c = 4m = 12k. Now a + b = 7k + 9k = 16k and b + c = 9k + 12k = 21k. Therefore, (a + b):(b + c) = 16:21.

Option A:
Option A, 7:9, simply restates the original a:b ratio and does not consider the contributions from c in b + c. Because the question asks for the ratio of sums, not of individual terms, 7:9 fails to capture the required combined relationships.

Option B:
Option B is correct because it arises from expressing a, b and c in terms of a single parameter k that satisfies both given ratios. Only after this alignment can we reliably compute a + b and b + c. The final ratio 16:21 cannot be obtained by any shortcut that ignores this common scaling step.

Option C:
Option C, 4:7, might be guessed by incorrectly mixing the numerators of the given ratios, but it is not supported by the actual expressions for a + b and b + c. Substituting 4:7 back into the linked equations fails to reproduce the original ratios.

Option D:
Option D, 5:7, is another arbitrary-looking pair that does not emerge from the algebraic derivation. Computing a + b and b + c explicitly shows that their ratio is 16:21, not 5:7, so this option is inconsistent with the calculation.

Let 3a = 4b = 6c = k for some positive constant k. Then a = k/3, b = k/4 and c = k/6. To eliminate denominators, multiply each term by 12, the LCM of 3, 4 and 6. This gives a:b:c = 4k:3k:2k, which simplifies to 4:3:2. Thus, the required ratio a:b:c is 4:3:2.

Option A:
Option A, 2:3:4, reverses the correct ordering and suggests that c is the largest term. But since 6c = k and 3a = k, we have c = k/6 and a = k/3, so a must be larger than c. Therefore, 2:3:4 is inconsistent with the given equalities.

Option B:
Option B, 3:4:2, places b as the largest value, which conflicts with 3a = 4b = 6c. From 3a = 4b we deduce a = (4/3)b, so a is larger than b. This means a should have the largest coefficient, which is not the case in 3:4:2.

Option C:
Option C, 4:2:3, misorders b and c relative to a. If we normalise using 3a = 4b = 6c, c must be the smallest because its divisor (6) is the largest. In 4:2:3, c’s coefficient is 3, not the smallest, so the pattern does not respect the initial conditions.

Option D:
Option D is correct because the derivation from 3a = 4b = 6c leads directly to a = k/3, b = k/4 and c = k/6, and scaling by 12 produces 4:3:2. This ordering matches the relative sizes implied by the denominators 3, 4 and 6.

Equal real roots occur when the discriminant of a quadratic equation is zero. For the equation 2x² − kx + 8 = 0, the discriminant is b² − 4ac, which becomes k² − 4(2)(8). Setting this equal to zero gives k² − 64 = 0. Solving, we get k² = 64 and hence k = ±8, but the given options only include 8 as a valid choice. Therefore, k = 8 satisfies the equal roots condition using the standard discriminant rule.

Option A:
Option A, 4√2, when squared gives 32, and substituting into the discriminant k² − 64 yields 32 − 64 = −32, which is negative. A negative discriminant would produce complex, not equal real, roots.

Option B:
Option B leads to k² = 64, so the discriminant becomes 64 − 64 = 0. This matches the precise condition for equal real roots in a quadratic and is consistent with the formula b² − 4ac = 0.

Option C:
Option C, 16, gives k² = 256 and a discriminant of 256 − 64 = 192, which is positive and corresponds to two distinct real roots rather than equal ones.

Option D:
Option D, 16√2, squares to 512, making the discriminant 512 − 64 = 448. This again yields unequal real roots, not the required equal roots.

To solve the inequality 2x − 5 > 7, we first add 5 to both sides, giving 2x > 12. Then we divide both sides by 2, which is positive, so the inequality direction remains the same, yielding x > 6. Thus, the solution set consists of all real numbers greater than 6, and this is the equivalent inequality in terms of x.

Option A:
Option A captures the full result of correctly isolating x. It incorporates both the addition and division steps and expresses the final constraint directly as x > 6.

Option B:
Option B, x > 5, would follow only if the inequality were 2x − 5 > 5, not 2x − 5 > 7. It represents a weaker condition than required.

Option C:
Option C, x > 4, ignores more of the numerical shift and significantly broadens the solution set beyond what the original inequality allows.

Option D:
Option D, x > 3, is even looser and does not align with the operations needed to isolate x in the given inequality.