UGC NET Questions (Paper – 1)

Statements A, B and C state the standard time–work model: one-day work as 1/n, combined rates added and inverse relation between time and rate. Statement E is also true since many problems can be set up and solved using simple linear equations in the unknown rates. Statement D is false because if A is twice as fast as B, A will take half the time, not twice. Therefore, the combination A, B, C and E only is correct.

Option A:
Option A is incomplete because it leaves out E and thus fails to mention the useful algebraic viewpoint that rates can be handled by linear equations. Without E, the picture of modelling is not fully captured.

Option B:
Option B is correct since it collects all the true statements and excludes D, which reverses the relationship between speed and time. It reflects exactly how NET aspirants are expected to reason about time and work problems.

Option C:
Option C is incorrect because it includes D, the wrong claim about time for a faster worker, even though it also lists some true statements. Including a false statement invalidates the combination.

Option D:
Option D is also wrong as it retains D and omits A, failing to state the fundamental definition of one-day work while endorsing an incorrect comparison of times.

For a fixed distance d, time is given by t = d/speed, so it is inversely proportional to speed. The first train’s time is d/90 and the second’s time is d/60. Their time ratio is (d/90):(d/60) = 1/90:1/60 = 60:90, which simplifies to 2:3. Thus, the first train takes 2 parts of time for every 3 parts taken by the second.

Option A:
Option A, 3:2, reverses the correct ratio and would mean the faster train takes more time, which contradicts the fundamental relationship between speed and time for the same distance. A higher speed must correspond to a shorter time, not a longer one.

Option B:
Option B is correct because it arises from properly comparing the reciprocals of the speeds. When the distance is constant, the time ratio is the inverse of the speed ratio. Since the speed ratio is 90:60 = 3:2, the time ratio must be 2:3, matching the computed result.

Option C:
Option C, 1:2, would imply that the first train takes half the time of the second, meaning its speed would have to be double the second’s speed. However, 90 km/h is only 1.5 times 60 km/h, not twice, so 1:2 overstates the difference.

Option D:
Option D, 5:6, bears no direct relationship to 90 and 60 when we follow the inverse proportionality rule. It cannot be obtained by simplifying 60:90 or any equivalent fraction, so it does not represent the correct time ratio.

A is correct because time = distance/speed for fixed distance, so time varies inversely with speed. C is true since distance = speed × time at constant speed. D is true because halving speed doubles time for the same distance. B is false because doubling speed at the same time doubles distance, not halves it. E is false; starting time is not forced by speed alone. Therefore A, C and D only are correct.

Option A:
Option A is correct as it captures both inverse and direct proportion relationships along with a key consequence for halving speed.

Option B:
Option B is incomplete because it excludes D, missing an explicit and important inverse-proportion consequence.

Option C:
Option C is wrong since it includes B, which misstates the effect of doubling speed at fixed time.

Option D:
Option D is incorrect because it adds E (false) and omits A (true), so it cannot be correct.

For the same distance, speed is inversely proportional to time. Let the common distance be d. The first cyclist’s speed is d/4 and the second’s speed is d/6. Their speed ratio is (d/4):(d/6) = 1/4:1/6 = 6:4, which simplifies to 3:2. Thus, the ratio of their speeds is 3:2.

Option A:
Option A, 2:3, corresponds to the ratio of their times, not their speeds. It would imply that the faster cyclist rides more slowly than the slower one, which contradicts the idea that less time for the same distance means higher speed.

Option B:
Option B is correct because it correctly applies the inverse proportionality between speed and time. The cyclist who takes 4 hours must be faster, and the computed ratio 3:2 quantifies exactly how much faster he is than the one who takes 6 hours.

Option C:
Option C, 4:3, might arise from directly comparing 4 and 6 but simplifying incorrectly. However, when we compute speeds rather than times, we get 3:2, not 4:3. Thus 4:3 is not compatible with the relationship between speed and time here.

Option D:
Option D, 5:4, has no clear basis in the given numbers and does not result from any correct manipulation of the times or speeds. It therefore does not represent the true speed ratio.

Statements A and B give the usual representation of individual and combined work rates in time and work problems. Statement D is also correct because increasing the number of equally efficient workers generally reduces the time for a fixed job when total work is unchanged. Statement C is false, since a faster worker takes less time, and E is false because alternate-day work patterns depend on how many days are counted and cannot always be reduced to a simple average. Hence the correct set of statements is A, B and D only, matching option B.

Option A:
Option A is incomplete as it leaves out D, thereby ignoring the inverse relation between number of workers and time, which is a standard pattern tested in exam questions.

Option B:
Option B is correct because it collects the ideas of one-day work, combined rate and inverse proportionality while excluding C and E, which misstate how speed affects time and oversimplify alternate working-day scenarios.

Option C:
Option C is wrong since it brings in E, which incorrectly claims a simple average works in all alternate-day situations, and it omits B, so it both adds an error and loses a fundamental rule.

Option D:
Option D is incorrect because it includes C, which reverses the comparison of fast and slow workers, and it omits A, dropping the basic 1/x representation of one-day work.

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.

For a fixed distance, time is inversely proportional to speed. Let the distance between P and Q be d. The onward time is d/60 and the return time is d/40. Their ratio is (d/60):(d/40) = 1/60:1/40 = 40:60, which simplifies to 2:3. Hence, the ratio of onward time to return time is 2:3.

Option A:
Option A, 3:2, reverses the correct ratio. This would imply that the onward journey at higher speed takes more time than the return journey at lower speed, which contradicts the basic principle of inverse proportionality between time and speed for the same distance.

Option B:
Option B, 4:5, does not match the fractional comparison 1/60:1/40. If we attempted to equate 4:5 to 40:60, they would not simplify to the same reduced form. Therefore 4:5 is not aligned with the correct calculation.

Option C:
Option C is correct because it results from cancelling the common factor d and then simplifying the fraction 40:60. This ratio correctly reflects that the car spends less time on the onward trip at 60 km/h than on the return trip at 40 km/h by a factor of 2 to 3.

Option D:
Option D, 5:4, suggests only a mild difference between the times and in the wrong direction. It cannot be derived from the exact speeds given and leads to an incorrect relationship between the onward and return travel times.