UGC NET Questions (Paper – 1)

The condition “x varies inversely as the square of y” means x = k/y² for some constant k. Using x = 9 when y = 2, we get 9 = k/4, so k = 36. For y = 3, we have x = 36/3² = 36/9 = 4. Therefore, the new value of x corresponding to y = 3 is 4.

Option A:
Option A, 3, would correspond to x = 36/y² giving 3 = 36/y², so y² = 12, which does not match y = 3. Hence, 3 cannot be the value of x when y is 3 under this variation law.

Option B:
Option B is correct because it uses the constant k = 36 derived from the first pair (x, y) = (9, 2) and substitutes y = 3 appropriately. The computation 36/9 is straightforward and yields 4, which fully respects the inverse-square relationship between x and y.

Option C:
Option C, 6, could result from treating x as inversely proportional to y instead of y², leading to x = 18/3. However, that would violate the stated condition involving the square of y, and it does not satisfy x = 9 when y = 2 if we consistently use the same rule.

Option D:
Option D, 8, might arise from doubling 4 under the mistaken belief that x halves when y increases from 2 to 3, but inverse-square variation is more rapid than simple inverse variation. The correct calculation yields 4, not 8, so this option is not consistent with the given dependency.

If A fills the tank in 12 hours, A’s one-hour work is 1/12; for B it is 1/18, and for C it is 1/36. The ratio A:B:C in one hour is (1/12):(1/18):(1/36). Taking the LCM of 12, 18 and 36 as 36, we get 3:2:1 as the simplified ratio. Thus, A, B and C complete the work in the ratio 3:2:1 per hour.

Option A:
Option A, 6:4:3, represents 6:4:3 which, when normalised, does not match 1/12:1/18:1/36. Dividing these by 3 gives 2:4/3:1, which is not proportional to 3:2:1, so it misstates the relative speeds of the taps.

Option B:
Option B, 3:2:2, suggests that B and C work at the same rate, which contradicts the given times. Since C takes 36 hours, which is double B’s 18 hours, C must work at half of B’s rate, not the same rate. Therefore, 3:2:2 is inconsistent with the time data.

Option C:
Option C, 2:3:6, would imply that C is working fastest, with the largest share of work per hour, while in reality C is the slowest tap. This directly contradicts the inverse relationship between time and rate, so 2:3:6 cannot be correct.

Option D:
Option D is correct because converting the one-hour work fractions to a common denominator and simplifying gives 3:2:1. This ratio accurately reflects that A is fastest, B is intermediate and C is slowest, in exact accordance with their filling times.

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.

The groups descend in blocks of three letters: XYZ (24–26), UVW (21–23), RST (18–20) and OPQ (15–17). Each starting letter reduces by 3 positions (24, 21, 18, 15). The next starting position should be 12, corresponding to L, and the next block is LMN (12–14). Hence LMN is the only group that continues the pattern of moving each block three letters earlier in the alphabet.

Option A:
Option A, LMN, is correct because it begins at L, which is 3 letters before O, and forms the next three-letter block in descending order. It respects the length of the groups and the step size of −3 for the starting letter. This makes LMN a perfect fit for the series.

Option B:
Option B, KLM, starts too early at K, corresponding to a step of −4 from O, not −3. This over-adjustment does not match the observed behaviour of the series. As such, KLM cannot be considered correct.

Option C:
Option C, NOP, reverses the direction by moving forward after OPQ instead of backward. It also overlaps with OPQ by reusing O and P in a way that contradicts the blockwise descent. Therefore NOP does not align with the series rule.

Option D:
Option D, MNO, starts at M, which is 2 letters before O, not 3 letters before O. It breaks the fixed interval between the initial letters of successive blocks. Because it ignores the −3 step pattern, MNO is not a valid continuation.

A’s marks are 72% of 750, which is (72/100) × 750 = 540. B’s marks are 84% of 750, which is (84/100) × 750 = 630. The ratio A:B is 540:630. Dividing both numbers by 90 gives 6:7. Hence, the ratio of A’s marks to B’s marks is 6:7.

Option A:
Option A is correct because it simplifies the raw scores 540 and 630 using their highest common factor. Dividing both by 90 yields exactly 6:7, which preserves the proportion while using smaller, more convenient numbers. This process directly matches standard ratio reduction techniques.

Option B:
Option B, 7:8, might be guessed from the numerical percentages 72% and 84%, but the percentages cannot be compared directly without considering the total marks. When we compute the actual scores, the ratio 540:630 does not simplify to 7:8, so this option is incorrect.

Option C:
Option C, 4:5, would require A’s marks to be four-fifths of B’s marks. If B had 630, then A would need to have 504, not 540. Since A’s actual score is higher than 504, the ratio 4:5 understates A’s relative performance and does not match the data.

Option D:
Option D, 9:10, suggests a nearly equal performance, but if we scale 540 and 630 to 9 and 10, we find that 540 × 10 ≠ 630 × 9. This shows that 9:10 is not equivalent to 540:630, so it cannot represent their actual mark ratio.

Each new term is obtained by doubling the previous term and then adding consecutive odd numbers. From 2 to 5 we apply 2×2+1, from 5 to 13 we use 5×2+3, from 13 to 31 we use 13×2+5, and from 31 to 69 we use 31×2+7. The series of added odd numbers is 1, 3, 5, 7, so the next odd number must be 9. Therefore the next term is 69×2+9 = 147.

Option A:
Option A would correspond to 69×2+5, reusing an earlier odd adjustment instead of moving to 9. This would make the added sequence 1, 3, 5, 7, 5, which does not show a clear progression. As such, 143 does not respect the steadily increasing odd-number adjustment rule.

Option B:
Option B equals 141, which would arise from 69×2+3. That uses an even smaller odd addition than in the previous step, reversing the pattern 1, 3, 5, 7. Hence 141 cannot be justified by the observed rule.

Option C:
Option C uses 69×2+9, giving 147 and continuing the series of added odd numbers as 1, 3, 5, 7, 9. This maintains both the doubling structure and the simple progression among the adjustments. Therefore 147 is the correct next term.

Option D:
Option D, 151, would require adding 13 to 69×2, jumping from an added 7 directly to 13 and skipping 9 and 11. This breaks the neat sequence of consecutive odd numbers that defines the pattern. Thus 151 is not a valid extension.

Let the original numbers be 7x and 4x. After subtracting 5 from each, they become 7x − 5 and 4x − 5, and their ratio is 3:1. Thus, (7x − 5)/(4x − 5) = 3/1. This gives 7x − 5 = 3(4x − 5) = 12x − 15, which simplifies to 10 = 5x and hence x = 2. Therefore, the original numbers are 14 and 8, and the larger is 14.

Option A:
Option A, 8, is the smaller of the two numbers in the correct solution. While it satisfies the original ratio when combined with 14, the question asks for the larger number. Choosing 8 would misinterpret which quantity the problem seeks.

Option B:
Option B is correct because 14 equals 7x with x = 2 and is the larger term in the ratio 7:4. Substituting 14 and 8 into the transformed condition, we get (14 − 5):(8 − 5) = 9:3 = 3:1, which exactly matches the new ratio, confirming that 14 is the correct larger original number.

Option C:
Option C, 16, would require the smaller number to be 16 × (4/7) ≈ 9.14 to preserve the ratio 7:4. This is not an integer and, even if approximated, would not yield a new ratio of 3:1 after subtracting 5. Thus, 16 cannot be the larger number that fits all conditions.

Option D:
Option D, 18, similarly leads to a non-integer partner of 18 × (4/7) ≈ 10.29 under the original ratio and fails to produce the required new ratio of 3:1 after the subtraction. Hence, 18 is inconsistent with the given information.

Statements A, B, C and E are correct as they capture the essence of mixture problems: combining components, summing component quantities, using the rule of alligation as a shortcut and representing scenarios algebraically or graphically. Statement D is false because mixture questions are naturally modelled by weighted averages where proportions act as weights. Thus, the combination A, B, C and E only is correct.

Option A:
Option A is incomplete as it leaves out E, not mentioning that visual or diagrammatic approaches are often used to make mixture relationships clearer in exams. Without E, the modelling variety is understated.

Option B:
Option B is also incomplete because it omits A and thereby fails to state directly that mixtures can involve various types of components like solutions or grains, which is basic contextual information.

Option C:
Option C is correct because it collects all the true statements about mixture structures, alligation shortcuts and representation methods, while excluding D, which incorrectly denies the connection with weighted averages. This matches the treatment in UGC NET aptitude.

Option D:
Option D is wrong as it includes D, which wrongly claims that weighted averages cannot be used for mixtures, and omits B, which accurately describes how component quantities combine. The presence of D invalidates the set.

The first and last letters P and T remain fixed in each word, while the middle vowel changes in alphabetical order: A, E, I, O. The next vowel after O in this sequence is U. Therefore the next word formed with the same consonants is PUT.

Option A:
Option A, PET, repeats the second term rather than moving to a new vowel. The pattern clearly shows a progression of different vowels in order. Since PET does not introduce the next vowel, it cannot be the correct continuation.

Option B:
Option B, PAT, is the first word of the series and would move the pattern backward. There is no indication that the sequence cycles back to earlier words. Hence PAT does not fit as the next term.

Option C:
Option C is correct because it uses U as the next vowel after O while keeping P and T unchanged. This preserves both the consonant framework and the alphabetical order of the vowels. As a result, PUT is the only word that logically follows POT.

Option D:
Option D, PUTT, adds an extra T at the end, changing the word length and structure. The series maintains three-letter words, so this alteration in length disrupts the pattern. Therefore PUTT is not an appropriate continuation.

Let the two numbers be 4x and 5x. Their sum is 4x + 5x = 9x, which is given as 81. Solving 9x = 81 gives x = 9. Therefore, the numbers are 4 × 9 = 36 and 5 × 9 = 45, and the larger number is clearly 45. This uses the standard technique of expressing numbers with a common factor and solving a single linear equation.

Option A:
Option A, 36, is the smaller of the two numbers in the pair 36 and 45. While 36 does fit the ratio 4:5 when paired with 45, the question specifically asks for the larger number, so choosing 36 would ignore the direction of comparison. It therefore does not answer what is being asked.

Option B:
Option B is correct because it corresponds to 5x with x = 9, the larger term in the ratio 4:5. When we check, 36 + 45 = 81 and 36:45 simplifies to 4:5 on dividing both numbers by 9. This double verification confirms that 45 is the only number that fits both the ratio and sum conditions as the larger value.

Option C:
Option C, 40, might seem plausible as a mid-range value, but if 40 were the larger number, the smaller would have to be 40 × (4/5) = 32 to maintain the ratio 4:5. The pair 32 and 40 has a sum of 72, not 81, so this option fails the sum condition.

Option D:
Option D, 48, would give a smaller number of 48 × (4/5) = 38.4 if the ratio 4:5 were preserved, which is not even an integer. Moreover, 38.4 + 48 ≠ 81, so this choice violates both the integer requirement and the sum requirement.

The positions of the letters are D(4), G(7), J(10), M(13) and P(16). Each step increases the position by 3 (4→7→10→13→16). Continuing this constant difference, the next position is 16 + 3 = 19, which corresponds to S. Therefore S is the only letter that preserves the fixed +3 pattern of the series.

Option A:
Option A is correct because S is exactly 3 positions after P in the alphabet. It keeps the same arithmetic progression of positions that has been followed throughout the series. By maintaining this constant increment, S fits perfectly as the next term.

Option B:
Option B, R, is only 2 positions after P(16→18), which breaks the consistent step size of 3. Choosing R would change the underlying rule from a fixed difference to a variable one. Hence R cannot logically continue the established sequence.

Option C:
Option C, T, is 4 positions after P(16→20), overshooting the required increment. This would increase the step size without any prior indication of such change. As the pattern clearly uses equal jumps, T is not a valid continuation.

Option D:
Option D, U, lies 5 positions after P(16→21), producing an even larger jump. This completely disrupts the constant difference structure of the series. Therefore U cannot be accepted as the next term.

The positions of the letters are B(2), D(4), G(7), K(11) and P(16). The differences are +2, +3, +4 and +5 respectively. The pattern shows that each gap increases by 1, so the next difference should be +6. Adding 6 to 16 gives 22, which corresponds to V. Therefore V is the correct continuation.

Option A:
Option A, U, is at position 21, which is only 5 steps after P. That would repeat the last gap rather than increase it to 6. Since the pattern clearly demands a growing difference, U does not fit the rule.

Option B:
Option B, W, lies at position 23, implying a jump of 7 from P. This overshoots the expected difference and would break the smooth progression of 2, 3, 4, 5, 6. Therefore W cannot be accepted as the next term.

Option C:
Option C is correct because V occupies position 22, giving exactly the required +6 increment. It continues the trend of increasing the gap by one at each step. This strict adherence to the numerical progression of differences makes V the logical answer.

Option D:
Option D, X, at position 24, corresponds to a jump of 8 from P. Such a jump is not consistent with the incremental pattern of the series. As it would disrupt the orderly growth of the differences, X is not a valid continuation.

Statements A and B summarise the basic relationship and unit harmony. Statement C is correct (m/s → km/h multiply by 18/5). Statement D is correct (km/h → m/s multiply by 5/18). Statement E is false because inconsistent units can cause major errors even when approximating. Therefore, A, B, C and D only are correct.

Option A:
Option A is incomplete as it omits D (km/h → m/s).

Option B:
Option B is incomplete because it omits B (minutes to hours conversion) even though it includes conversion factors.

Option C:
Option C is incorrect because it omits A (distance = speed × time).

Option D:
Option D is correct since it gathers all true statements and excludes E (false).

The groups are four letter consecutive blocks where the starting letters J, L, N and P occupy positions 10, 12, 14 and 16, increasing by 2 each time. Each block then takes the next three letters in order. Continuing this rule, the next starting letter is position 18, which is R, and the block from R is RSTU. Thus RSTU uniquely keeps both the plus two starting step and the four letter structure.

Option A:
Option A, STUV, is a valid four letter block but begins at S, which is three steps after P instead of two. This changes the size of the jump in the starting letter, contradicting the constant increment of two. Therefore STUV does not align with the governing rule of the series.

Option B:
Option B, QRST, starts at Q, which is only one position after P. That reduces the step from two to one and breaks the arithmetic progression of starting positions. Because the pattern clearly shows a consistent plus two movement, QRST cannot be considered correct.

Option C:
Option C is correct because R is exactly two positions ahead of P and RSTU forms a proper four letter consecutive group. It extends the same even step difference in the starting letter and preserves the internal consecutiveness. This match with both structural aspects of the series makes RSTU the right answer.

Option D:
Option D, NOPQ, repeats an earlier block instead of moving forward. The series never reuses a previous group, and each term has a strictly increasing starting letter. As a result, NOPQ cannot be the next term.

In partnership problems, profit is divided in the ratio of “capital × time”. Taking capitals 5k, 7k and 9k for A, B and C, and times 12, 10 and 8 months, their effective investments are 5k × 12 = 60k, 7k × 10 = 70k and 9k × 8 = 72k. These are in the ratio 60:70:72. The sum of the ratio parts is 202, so C’s share is (72/202) of ₹2,02,000, which equals ₹72,000.

Option A:
Option A, ₹60,000, corresponds to 60 parts out of 202 and would be A’s share if we followed the correct method. Assigning this to C would contradict both the given investment ratio and the time periods. It undervalues C’s contribution, which is actually the largest.

Option B:
Option B is correct because it matches the 72 parts assigned to C in the 60:70:72 ratio. Multiplying the total profit by 72/202 leads exactly to ₹72,000, and the remaining profits allocate correctly to A and B when their parts are used. This ensures the distribution honours both capital and time contributions.

Option C:
Option C, ₹84,000, would represent a different fraction of the total profit and would imply that C’s effective investment is larger than 72 parts out of 202. That would disturb the precise ratio derived from 5:7:9 and the corresponding time periods. Hence ₹84,000 does not reflect the correct proportional sharing.

Option D:
Option D, ₹90,000, exceeds even Option C and would require a much stronger claim for C’s investment than is supported by the data. If we tried to back-calculate the implied ratio, the figures would no longer match the product of capital and time given in the question. Thus, ₹90,000 cannot be C’s correct share.