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.

To find how many students like at least one of the two subjects, we use the inclusion–exclusion principle. We add those who like each subject and subtract those counted twice in the overlap: 22 + 18 − 10 = 30 students. These 30 students like Mathematics, Science or both.

Option A:
Option A, 30, is the direct result of applying inclusion–exclusion and correctly represents the number who like at least one of the two subjects.

Option B:
Option B, 32, would correspond to assuming an overlap of 8 instead of 10, which contradicts the given overlap and misuses the data.

Option C:
Option C, 40, assumes that every student likes at least one of the subjects, which is not stated and need not be true.

Option D:
Option D, 50, exceeds the total number of students in the group, so it is impossible.

For a conditional statement p → q, the contrapositive ¬q → ¬p is logically equivalent to the original. Here p is “the number is divisible by 4” and q is “the number is even.” The contrapositive is “if a number is not even, then it is not divisible by 4.” Both the original statement and its contrapositive are true or false together in all possible cases, so they are logically equivalent.

Option A:
Option A reverses the direction of the implication to q → p, which is the converse. The truth of the original statement does not guarantee the truth of its converse; being even does not ensure divisibility by 4.

Option B:
Option B states “If a number is not divisible by 4, then it is not even,” which is the inverse. The inverse is not logically equivalent to the original because many numbers are not divisible by 4 but are still even.

Option C:
Option C correctly expresses the contrapositive ¬q → ¬p: if a number is not even, then it is not divisible by 4. This preserves the logical content of the original implication and is therefore equivalent to it.

Option D:
Option D, “If a number is not divisible by 4, then it is even,” contradicts the original pattern and misstates the relationship between being even and being divisible by 4.

Distance is calculated using the formula Distance = Speed × Time. The given speed is 60 km/h, but the time is 30 minutes, which is half an hour or 0.5 hours. Substituting into the formula gives Distance = 60 × 0.5 = 30 km. Hence, the train covers 30 kilometres in 30 minutes at this speed.

Option A:
Option A, 20 km, would correspond to a speed of 40 km/h for 30 minutes, not the given 60 km/h. It reflects a misalignment between speed and time.

Option B:
Option B, 25 km, does not correspond to any simple fraction of 60 using half an hour and indicates an incorrect multiplication or conversion.

Option C:
Option C, 40 km, would be the distance travelled in 40 minutes at 60 km/h, or in 2/3 of an hour, not in 30 minutes. Thus, it overestimates the distance for the time given.

Option D:
Option D correctly uses 0.5 hours as the time and multiplies by 60 km/h to get 30 km. This matches both the formula and the unit conversion from minutes to hours.

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.

The given series consists of perfect squares of natural numbers. Specifically, 1² = 1, 2² = 4, 3² = 9, 4² = 16 and 5² = 25. The next term should therefore be 6², which is 36. This continues the clear pattern of successive squares.

Option A:
Option A, 30, does not correspond to any perfect square and breaks the pattern of square numbers. It would require a different rule that is not evident in the series.

Option B:
Option B, 32, is also not a perfect square and falls between 25 and 36 without a clear basis from squaring integers.

Option C:
Option C, 35, is again not a perfect square and therefore cannot follow the pattern of 1², 2², 3², 4² and 5².

Option D:
Option D, 36, is 6² and naturally follows 5² = 25 in the sequence of square numbers. It maintains the structure of the series and is therefore the correct next term.

The word LEVEL has five letters in total, with L appearing twice and E appearing twice, while V appears once. The formula for permutations of n objects with repetitions is n factorial divided by the product of factorials of repeated counts. Thus, the number of distinct permutations is 5! ÷ (2! × 2!). Calculating this gives 120 ÷ 4 = 30. Hence, there are 30 different arrangements of the letters in LEVEL.

Option A:
Option A, 20, underestimates the actual number and cannot be obtained by the correct factorial expression 5! ÷ (2! × 2!). It likely comes from an incorrect guess or misapplied formula.

Option B:
Option B accurately uses the repetition-adjusted permutation formula, recognizing two repeated L’s and two repeated E’s. By evaluating 120 ÷ 4, it yields 30, which counts all unique arrangements without overcounting identical ones.

Option C:
Option C, 40, exceeds the true count and would require a different numerator or denominator in the formula. It suggests that the effect of repeated letters has not been properly accounted for.

Option D:
Option D, 60, is exactly half of 120 and may arise from dividing by only one 2! instead of both 2!. This partially correct adjustment still leads to overcounting because both L and E are repeated.

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.

The equation (x − 3)(x + 5) = 0 is satisfied when either factor is zero. Setting x − 3 = 0 gives x = 3, and setting x + 5 = 0 gives x = −5. The sum of the roots is therefore 3 + (−5) = −2. This also agrees with the general property that for x² + bx + c = 0, the sum of the roots is −b, which is consistent if the equation is expanded.

Option A:
Option A correctly adds the two root values, 3 and −5, to obtain −2. It captures the essence of the relationship between factorized form and the numerical roots.

Option B:
Option B, 2, would arise if one mistakenly subtracted −5 from 3 or ignored the sign of one root, but that does not reflect the actual arithmetic.

Option C:
Option C, −8, is the product of the roots, not their sum. It reflects a different relation in quadratics and confuses two separate concepts.

Option D:
Option D, 8, would require both roots to be positive or both negative with different magnitudes, which is not the case for this factorization.

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.

Using the principle of inclusion–exclusion, the number of students who like at least one of Mathematics or English is 18 + 16 − 8 = 26. The total number of students in the class is 30. Therefore, those who like neither subject are 30 − 26 = 4. This calculation separates students who fall outside both preference sets.

Option A:
Option A, 2, would imply that 28 students like at least one of the two subjects, but the inclusion–exclusion result clearly shows only 26 students in that category. Hence, 2 underestimates the count of those liking neither.

Option B:
Option B, 3, similarly assumes 27 students like at least one subject, conflicting with the properly derived figure of 26. It reflects a minor miscalculation in subtracting from the total.

Option C:
Option C correctly applies inclusion–exclusion to avoid double counting those who like both subjects, then subtracts from the total class strength. This yields 4 students who like neither Mathematics nor English.

Option D:
Option D, 6, would require at least one count to be different because it implies only 24 students like at least one subject, which contradicts the overlap-adjusted total of 26.

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.

Average speed is defined as total distance divided by total time, not the average of the two speeds. The car travels 120 km each way, so the total distance is 240 km. The time taken in the first leg is 120 ÷ 40 = 3 hours and in the return leg is 120 ÷ 60 = 2 hours, giving a total of 5 hours. Dividing 240 by 5 yields 48 km/h as the average speed for the entire journey.

Option A:
Option A, 45 km/h, might be guessed by taking a simple average of the two speeds or by an approximate calculation, but it does not correspond to the correct total distance over total time.

Option B:
Option B uses the exact definition of average speed and calculates both distances and times explicitly. It also matches the harmonic mean formula for a round trip with equal distances, which simplifies to 2ab/(a + b).

Option C:
Option C, 50 km/h, exceeds the true average because more time is spent at the lower speed, pulling the overall average below the midpoint between 40 and 60.

Option D:
Option D, 52 km/h, is even higher and clearly inconsistent with spending a substantial part of the journey at just 40 km/h.

According to the BODMAS rule, division and multiplication are performed before addition and subtraction and are evaluated from left to right. Starting with 18 ÷ 3, we get 6, then multiplying 6 × 2 gives 12. The expression now becomes 12 + 12 − 4. Next, we perform addition and subtraction from left to right: 12 + 12 = 24 and then 24 − 4 = 20. Thus, the value of the expression is 20.

Option A:
Option A, 16, might result from an incorrect grouping of operations, such as performing addition or subtraction too early, but it does not respect the proper order dictated by BODMAS.

Option B:
Option B, 18, could arise from partial simplification without completing all operations in the correct sequence, but it fails to match the fully evaluated expression.

Option C:
Option C consistently applies the order of operations, handling division and multiplication first, then addition and subtraction. This careful sequencing leads to the accurate result of 20.

Option D:
Option D, 22, again does not correspond to the correct application of BODMAS and suggests an arithmetic error in combining the intermediate results.