UGCNET Questions

The average of three numbers is the sum of the numbers divided by 3. If the average is 25, then the total sum must be 25 × 3 = 75. The two known numbers are 20 and 30, whose sum is 50. Subtracting 50 from 75 gives 25, so the third number must be 25.

Option A:
Option A, 25, exactly matches the required third number, because 20 + 30 + 25 = 75 and 75 ÷ 3 = 25. It is therefore the correct choice.

Option B:
Option B, 20, would give a total of 20 + 20 + 30 = 70, and 70 ÷ 3 does not equal the required average of 25, so this choice fails the given condition.

Option C:
Option C, 30, leads to 20 + 30 + 30 = 80, giving an average of 80 ÷ 3, which is greater than 25 and inconsistent with the stated mean.

Option D:
Option D, 35, produces a total of 20 + 30 + 35 = 85, resulting in an average greater than 25, again contradicting the problem’s information.

Let the present ages of the father and son be 7x and 3x respectively. After 10 years, their ages will be 7x + 10 and 3x + 10. The future ratio is (7x + 10) : (3x + 10) = 13 : 7. Cross-multiplying gives 7(7x + 10) = 13(3x + 10), i.e., 49x + 70 = 39x + 130. This simplifies to 10x = 60, so x = 6. Therefore, the father’s present age is 7x = 7 × 6 = 42 years.

Option A:
Option A, 35 years, would make the son 15 years old (ratio 7 : 3), and after 10 years their ages would be 45 and 25, giving a ratio of 9 : 5, not 13 : 7.

Option B:
Option B, 42 years, makes the son 18 years old. After 10 years they would be 52 and 28, and 52 : 28 simplifies to 13 : 7, satisfying both the given present ratio and the future ratio.

Option C:
Option C, 49 years, would imply a son aged 21, and after 10 years their ages become 59 and 31, which do not reduce to 13 : 7.

Option D:
Option D, 56 years, would give the son 24 years at present and 66 and 34 after 10 years, with ratio 33 : 17, again not 13 : 7.

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.

Let the unknown number be N. If 25% of N is 60, then 0.25N = 60. Solving this equation gives N = 60 ÷ 0.25 = 240. To find 40% of this number, we compute 0.40 × 240, which equals 96. Thus, 40% of the number is 96.

Option A:
Option A, 80, corresponds to one third of 240, not 40%, and would arise from mixing up the percentage multipliers. It does not follow from the given condition.

Option B:
Option B, 90, is 37.5% of 240 and again has no direct link to the required 40% rate. It reflects an incorrect multiplication step.

Option C:
Option C is obtained by first recovering the full value of the number from the 25% information and then correctly applying the 40% factor to it. The arithmetic is consistent at each stage and produces 96.

Option D:
Option D, 100, is slightly higher than the correct value and might come from rounding or a mistaken assumption that 40% is close to half, but it is not the exact result.

Five consecutive odd integers can be written as n − 4, n − 2, n, n + 2 and n + 4, where n is the middle term. The average of such a symmetric set is always the middle term, so the average being 37 means the middle integer is 37. The largest of the five is therefore 37 + 4, which is 41. This uses both the definition of average and the structure of consecutive odd numbers.

Option A:
Option A, 39, is the second largest of the five integers, not the largest. It is 2 greater than the middle term instead of 4 and therefore does not represent the end of the sequence.

Option B:
Option B correctly identifies that with an average of 37, the five odd numbers are 33, 35, 37, 39 and 41. Among these, 41 is clearly the largest and fits the pattern of consecutive odd integers.

Option C:
Option C, 43, would require shifting the sequence so that the average is larger than 37. That would contradict the given mean and lead to a different central value.

Option D:
Option D, 45, is even further from the centre and again cannot be part of the exact sequence whose average is 37.

To find x and y, we solve the system of linear equations using elimination or substitution. Solving 3x + 2y = 26 and 2x + 3y = 19 gives x = 8 and y = 1. Their difference x − y is therefore 8 − 1 = 7, a clean integer result consistent with both equations.

Option A:
Option A, 5, would be the difference if the values of x and y were closer together, but solving the system shows a larger gap between them. It does not match the algebraic relationships specified.

Option B:
Option B, 6, again does not arise from any correct elimination or substitution procedure. When checked against the solved values of x and y, it fails to match their actual difference.

Option C:
Option C reflects the correct outcome after systematically eliminating one variable and solving for the other. Substituting back gives consistent values of x and y whose difference is exactly 7, agreeing with all the given equations.

Option D:
Option D, 8, overshoots the true difference and would imply values of x and y that do not satisfy both equations simultaneously.

To find the LCM, we factorize the numbers into primes. The number 12 is 2² × 3 and 18 is 2 × 3². The LCM takes the highest power of each prime appearing in either factorization, giving 2² × 3². This product equals 4 × 9 = 36. Hence, 36 is the smallest number that is a multiple of both 12 and 18.

Option A:
Option A, 36, includes both enough factors of 2 and enough factors of 3 to be divisible by the two given numbers. It is also the smallest such number and so satisfies the definition of the LCM.

Option B:
Option B, 24, is divisible by 12 but not by 18 because 24 ÷ 18 is not an integer. It lacks a second factor of 3.

Option C:
Option C, 30, is divisible by neither 12 nor 18 in an exact sense, as it does not contain the required combination of prime powers.

Option D:
Option D, 42, is divisible by 6 and 7, but not by 12 or 18 without leaving a remainder. It is larger than necessary and not a common multiple in this context.

Distance is the product of speed and time. In 3 hours, the slower train at 40 km/h covers 40 × 3 = 120 km. The faster train at 60 km/h covers 60 × 3 = 180 km in the same time. The difference in distances is 180 − 120 = 60 km, so the faster train travels 60 km more than the slower one.

Option A:
Option A, 40 km, would be correct if the speed difference were 40 km/h for 1 hour, but here the speed difference is 20 km/h and the time is 3 hours.

Option B:
Option B, 50 km, has no direct basis in the given speeds and time and thus reflects an incorrect computation.

Option C:
Option C, 60 km, matches the product of the speed difference (60 − 40 = 20 km/h) and the common time of 3 hours. It properly accounts for how much farther the faster train travels in the same interval.

Option D:
Option D, 80 km, overstates the difference and would require either a larger gap in speed or a longer time interval.

The ratio 3 : 2 means that out of every 5 parts of the mixture, 3 parts are milk and 2 parts are water. The total mixture is 30 litres, which corresponds to 5 parts. One part is therefore 30 ÷ 5 = 6 litres. Milk occupies 3 parts, so the amount of milk is 3 × 6 = 18 litres. Thus, 18 litres of the mixture are milk.

Option A:
Option A, 12 litres, would correspond to only 2 parts of the mixture at 6 litres per part, undercounting the milk portion relative to the given ratio.

Option B:
Option B, 15 litres, arises if one mistakenly assumes equal parts of milk and water, ignoring the 3 : 2 ratio that favours milk.

Option C:
Option C uses the ratio correctly by finding the size of one part and then scaling by the number of milk parts. This fully respects the structure of the ratio and total volume.

Option D:
Option D, 20 litres, would represent 4 parts, which contradicts the defined ratio and implies more milk than the 3 parts specified.

To combine the ratios A : B and B : C, we first express them with a common value for B. The ratio A : B = 2 : 3 and B : C = 4 : 5. Making the B terms equal involves scaling the first ratio by 4 and the second by 3, giving A : B = 8 : 12 and B : C = 12 : 15. Now B has the same value in both, allowing us to read off A : C as 8 : 15. This is the required compound ratio.

Option A:
Option A, 8 : 15, emerges naturally once B is expressed with the same multiplier in both original ratios. It preserves the structure of each pairwise relation and gives a consistent comparison between A and C.

Option B:
Option B, 3 : 5, retains the original A : B pattern but ignores the connection to C through the second ratio. It cannot represent A : C directly.

Option C:
Option C, 5 : 8, inverts the relationship and does not follow from any consistent scaling of the given ratios. It would imply C is relatively smaller than A in a way that contradicts the conditions.

Option D:
Option D, 10 : 9, is arbitrary and not produced by any straightforward manipulation of the initial pair of ratios.

The arithmetic mean of five numbers is 18, so their total sum is 18 × 5 = 90. One of the numbers is given as 24. To find the sum of the remaining four numbers, we subtract this known value from the total: 90 − 24 = 66. Thus, the combined total of the four remaining numbers is 66.

Option A:
Option A directly applies the relationship Total = Average × Number of items and then subtracts the known element to obtain the sum of the others. This method is standard in average problems and yields 66 consistently.

Option B:
Option B, 72, would correspond to a larger total of 96 if 24 were added back, which would produce an average of 19.2 instead of 18. This contradicts the given mean.

Option C:
Option C, 80, similarly gives a total of 104 when 24 is added back, leading to an average of 20.8, which is inconsistent with the initial information.

Option D:
Option D, 84, implies a total of 108 for five numbers, resulting in an average of 21.6, again conflicting with the given average of 18.

When two objects move in opposite directions, their relative speed is the sum of their individual speeds. Here, the relative speed is 15 + 17 = 32 km/h. After 3 hours, the distance between them is relative speed multiplied by time, which is 32 × 3 = 96 km. This distance represents how far apart they are from the starting point after 3 hours.

Option A:
Option A, 32 km, is only the distance they would be apart after 1 hour at the relative speed of 32 km/h and fails to account for the full 3 hours of travel.

Option B:
Option B, 90 km, underestimates the separation and could come from incorrectly adding or averaging speeds, but it does not match 32 × 3.

Option C:
Option C correctly multiplies the relative speed by time, applying the standard formula distance = speed × time for relative motion. This yields 96 km as the accurate separation.

Option D:
Option D, 102 km, overstates the distance and suggests a relative speed higher than 32 km/h, which is not given.

Pipe A fills 1/10 of the tank per hour, pipe B fills 1/15 per hour, and pipe C empties 1/30 per hour. The net hourly work rate when all three are open is 1/10 + 1/15 − 1/30. Using 30 as the common denominator, this becomes 3/30 + 2/30 − 1/30 = 4/30 = 2/15 of the tank per hour. The time to fill the tank is therefore 1 ÷ (2/15) = 15/2 = 7.5 hours, which matches Option B.

Option A:
Option A, 5 hours, would require a net rate of 1/5 per hour, which is much faster than the computed 2/15 and not achievable with the given pipes.

Option B:
Option B, 7.5 hours, is exactly the reciprocal of the net rate 2/15 and correctly represents the time taken to fill the tank with all three pipes open.

Option C:
Option C, 8 hours, corresponds to a net rate of 1/8 per hour, slower than 2/15, and inconsistent with the detailed rate computation.

Option D:
Option D, 10 hours, duplicates the time taken by pipe A alone and ignores the additional inflow from B and the outflow from C, so it cannot be correct.

The word BANANA has 6 letters, with A repeated 3 times, N repeated 2 times and B appearing once. The formula for permutations of n objects with repeated groups is n! divided by the product of factorials of the counts of repeated items. Here this gives 6! ÷ (3! × 2!). Evaluating this gives 720 ÷ (6 × 2) = 720 ÷ 12 = 60. So there are 60 distinct arrangements of the letters of BANANA.

Option A:
Option A, 30, underestimates the total arrangements and would correspond to a different combination of repeats or a smaller word. It does not follow the factorial formula for this word.

Option B:
Option B, 36, arises from miscalculating either 6! or one of the factorials in the denominator. It does not equal 720 ÷ 12 and so fails to match the correct combinatorial reasoning.

Option C:
Option C, 48, again does not match the computed result and suggests partial adjustment for repetition without applying the full divisor 3! × 2!.

Option D:
Option D carefully applies the repetition-adjusted permutation formula. It fully accounts for indistinguishable arrangements caused by repeated letters and yields the correct count of 60.

The pattern in this series is based on increasing differences that themselves double each time. The differences between successive terms are 3, 6, 12 and 24. Each new difference is twice the previous one, giving the next difference as 48. When this difference of 48 is added to the last term 46, the next term becomes 94. Thus, 94 continues the established pattern of doubling increments.

Option A:
Option A, 88, would correspond to adding 42 to 46, which does not match the doubling pattern of 3, 6, 12 and 24. It ignores the rule that each difference is twice the previous difference.

Option B:
Option B, 90, arises if we add 44 to 46, but there is no consistent reason in the earlier terms to justify a difference of 44. It breaks the strict doubling of the gaps.

Option C:
Option C, 94, uses a difference of 48, which is double the previous difference of 24. This respects the second order structure of the series and produces a term fully consistent with the pattern.

Option D:
Option D, 98, assumes an increase of 52, again not related to the repeated doubling of the differences. It introduces a new rule that is not supported by the earlier terms.