UGEE REAP Mock Test- 6 Free Online Test 2026

∴ IFSHJW is the code for DANCER

E and A play Football.
To find out which pair of friends play football:
1. Identify who doesn't play football:

• The tallest (unknown) does not play football.
• B does not play football.
• G does not play football.

2. Identify football players:

• There are two football players.
• E and A are not excluded from playing football.

Thus, the pair of friends playing football is: EA

The original sentence is “Increased productivity necessary reflects greater efforts made by the employees,” which is grammatically incorrect because “necessary” is an adjective and cannot modify the verb “reflects.”
An adverb, such as “necessarily,” is required. Option C, “Increase in productivity necessarily,” replaces the underlined part, resulting in “Increase in productivity necessarily reflects greater efforts made by the employees,” which is grammatically correct.
Thus, the correct option is C.

To determine which statement must be true, we analyze all possible configurations of A, B, and C as knights or knaves, ensuring their statements are consistent.

• Case 1: A is a knight (tells the truth).
  • A says "B is a knave," so B is a knave (lies).
  • B says "C is a knave," but since B lies, this is false, so C is not a knave, meaning C is a knight.
  • C says "A is a knight," and since C is a knight, this is true, confirming A is a knight.
  • Result: A is a knight, B is a knave, C is a knight.
  • Check options:
    • A (knight) and B (knave): different types.
    • B (knave) and C (knight): different types.
    • A (knight) and C (knight): same type.
• Case 2: A is a knave (lies).
  • A says "B is a knave," but since A lies, this is false, so B is not a knave, meaning B is a knight.
  • B says "C is a knave," and since B is a knight, this is true, so C is a knave.
  • C says "A is a knight," but since C lies, this is false, confirming A is a knave.
  • Result: A is a knave, B is a knight, C is a knave.
  • Check options:
    • A (knave) and B (knight): different types.
    • B (knight) and C (knave): different types.
    • A (knave) and C (knave): same type.
• Conclusion:
  • In both possible cases:
    • Case 1: A knight, B knave, C knight → A and C are the same type (knights).
    • Case 2: A knave, B knight, C knave → A and C are the same type (knaves).
  • Option C ("A and C are the same type") holds true in all scenarios, while A, B, and D do not.

Thus, the correct answer is C.

Initially, 8 people: (8/2) = 28 handshakes.
2 leave, leaving 6 (3 males, 3 females, e.g., from 4M+4F, 1M+1F leave).
At end: Females (3/2) = 3, males (3/2) = 3, total 6 handshakes.
Total: 28 + 6 = 34.

The total earnings are Rs 84, meaning there are 84 people below. The structure is as follows:

• Level 1: 4 people (earn Re 1 each for those below)
• Level 2: Each of the 4 has 4, totaling 16
• Level 3: 16¹4 = 64

People at Level 3 earn zero as they have no one under them. Thus, 64 is correct.

The answer matches option B, so no changes are needed to the question or options.

Since the dealer must purchase at least one machine, the total number of machines, x + y, must be greater than zero.
Given that x ≥ 0 and y ≥ 0, the condition x + y > 0 ensures that at least some positive quantity of machines is purchased, excluding the case where x = y = 0.
Thus, the correct option is C: x + y > 0.

At (0, 0), z = 0 
At (16, 0), z = 352
At (8, 12), z = 392
At (0, 20), z = 360
At (10,10),
Budget: 360(10) + 240(10) = 6000 (exceeds 5760 → not feasible)
So actually, (10,10) is not even in the feasible region.

It can be observed that max z occur at (8, 12). Thus, z will attain its optimal value at (8, 12).

The optimal solution occurs at every point on the line joining these two points.

• If an LPP (Linear Programming Problem) has optimal solutions at two consecutive vertices, it means the entire line segment between these vertices is optimal.
• This happens because the objective function has the same value at both vertices, making every point along the line equally optimal.

The code is obtained by rearranging the letters in the order of positions 2, 4, 6, 1, 3, 5, 7 of the original word.
For CORRECT (C, O, R, R, E, C, T), this gives O (2), R (4), C (6), C (1), R (3), E (5), T (7) → ORCCRET.
For KINGDOM (K, I, N, G, D, O, M), this gives I (2), G (4), O (6), K (1), N (3), D (5), M (7) → IGOKNDM, which is option A.

The sentence is looking for a contrast as it is joined by the conjunction ‘despite’. The best pair of words that can fit the context of the sentence is ‘effectiveness – prescribed’. Though the medicine is ‘effective’ in treating diabetes, it is not being ‘prescribed’ widely. A new medicine cannot have a ‘prescription’ or ‘availability’ for treating a disease. ‘Proscribed’ means forbidden by law. In case we use ‘acceptance – proscribed’, the sentence will not make any sense because it will mean that though the medicine is accepted widely, it is not forbidden by law.

Felicitated is correct because it means to express congratulations or good wishes. This fits the context of honouring the teacher on Teacher's Day.

Each of the 10 questions has 4 possible answers.

• Each question can be answered in one of 4 ways.
• Thus, the total number of ways to answer all questions is 4¹⁰.

The problem is about a square where, in each step, we draw a new square by joining the midpoints of the sides of the previous square. The goal is to find the sum of the areas of all these squares.

Step 1: Area of the outermost square

The diagonal of the outermost square is given as 6√2 cm. In a square, the diagonal (d) is related to the side length (s) by the formula:

d = s√2

Here, d = 6√2, so:

6√2 = s√2

Cancel out the √2 on both sides:

s = 6 cm

Now, the area of the outermost square is:

Area = side × side = 6 × 6 = 36 cm²

Step 2: Area of the next square

The next square is formed by joining the midpoints of the sides of the outer square. The side of this new square will be s / √2 (since the diagonal of this square is the same as the side of the previous square).

So, the side length of the next square is:

Side of new square = 6 / √2 = 6 × (1/√2) = 6 / √2 = 3√2 cm

The area of the next square is:

Area = (3√2)² = 9 × 2 = 18 cm²

Step 3: Sum of areas

We can see that the area of each successive square is half of the previous one. So, we have:

• First square: 36 cm²

• Second square: 18 cm²

• Third square: 9 cm²

• Fourth square: 4.5 cm²

• And so on...

This forms a geometric series where the first term (a) is 36, and the common ratio (r) is 1/2.

The sum of an infinite geometric series is given by:

Sum = a / (1 - r)

Here, a = 36 and r = 1/2, so:

Sum = 36 / (1 - 1/2) = 36 / (1/2) = 36 × 2 = 72 cm²

Final Answer:

The sum of the areas of all the squares is 72 cm².

Solar radiation consists of electromagnetic waves, which are transverse in nature. This means their electric and magnetic fields:

• Oscillate perpendicular to the direction of propagation.

Therefore, the correct answer is A.

The ozone layer is primarily located in the stratosphere, approximately 15-30 kilometres above Earth's surface. It is formed when ultraviolet radiation causes oxygen molecules to split and recombine into ozone (O₃), playing a crucial role in absorbing harmful UV-B and UV-C rays, thus protecting life on Earth.

While small amounts of ozone exist in the troposphere, the stratospheric ozone layer is:

• The densest
• The most significant

The large semicircle has a radius R (diameter AB = 2R) with an area:

Point S, at (0, 7R/8), is the center of a circle with radius R/8, grazing an area:

Horses at points P and R are tied at the tangency points of this circle with semicircles of diameters AO and OB.
Centers at 
Radius R/2
Approximate positions:

Each horse grazes a semicircle with radius:

Area of each grazed semicircle:

The total grazed area, after adjusting for overlaps, is approximately 72% of:

So, the un-grazed area is 28% of  which corresponds to option B.

M is 13^th letter in the English alphabetical order.
E is 5^th letter, T is 20^th letter, and R is 18^th letter.

13 + 18 = 31
5 + 20 + 5 = 30
31 - 30 = 1
Similarly,
In ROGER,
R is 18th letter, O is 15^th letter, G is 7^th letter and E is 5^th letter.

18 + 18 = 36
15 + 7 + 5 = 27
36 - 27 = 9
Hence, code for "ROGER" is 9.

Option B is correct.

Let the radius of each inscribed sphere be r. For a regular hexagon of side a, the apothem (distance from centre to a side) is r = (√3/2) a.

The volume of one sphere is (4/3)π r³. Substituting r = (√3/2) a gives V_sphere = (π√3 / 2) a³.

The prism over the hexagon has base area A = (3√3 / 2) a² and height equal to the sphere diameter 2r = a√3. Thus the prism volume is V_prism = A × height = (3√3 / 2) a² × a√3 = (9/2) a³.

The shaded volume in one hexagonal prism (prism minus sphere) is V_shaded,1 = V_prism - V_sphere = (9/2 - π√3 / 2) a³ = (a³/2)(9 - π√3).

The total number of hexagons in the triangular arrangement with n hexagons in the last row is N = n(n+1)/2. Therefore the total shaded volume is

V_total = N × V_shaded,1 = (n(n+1)/2) × (a³/2)(9 - π√3) = (a³/4) n(n+1) (9 - π√3).

Equivalently, this expression can be written as (√3 / 4) a³ (3√3 - π) n(n+1), which matches Option B.

The correct word to complete the sentence is 'hanged' because it correctly uses the past participle form in passive voice for executing a person.

• Safety Stock for March = 20% × 400 = 80 units

• Opening Stock for March = 100 units

• Required Production for March = (400 + 80) - 100 = 380 units

Based on the table provided, the production requirement for each month is calculated as Demand + Safety Stock - Opening Stock. Using the data:

• January: 300 + 60 - 50 = 310
• February: 500 + 100 - 60 = 540
• March: 400 + 80 - 100 = 380
• April: 100 + 20 - 80 = 40
• May: 200 + 40 - 20 = 220
• June: 300 + 60 - 40 = 320

Thus, the highest production requirement is in February.

The opening stock for each month is the safety stock from the previous month:

• January: 50
• February: 60 (20% of 300)
• March: 100 (20% of 500)
• April: 80 (20% of 400)
• May: 20 (20% of 100)
• June: 40 (20% of 200)

Therefore, the lowest opening stock is in May.

The ninth and the tenth of this month are Monday and Tuesday, respectively.

With 1000 possible codes and no repetition, the correct code is equally likely to be any of the 1000 attempts.
Thus, the probability of success exactly on the s^th trial is 1/1000 for 1 ≤ s ≤ 1000.

The Earth's magnetic field strength typically ranges between 25 and 65 microteslas. This is equivalent to approximately 2.5 × 10^-5 T to 6.5 × 10^-5 T.

Among the given options, the correct order of magnitude for the Earth's magnetic field is:

• 10^-5 T (option C)

At equator vertical component of magnetic fields is zero.

Solution:

• The horizontal and vertical components of the Earth's magnetic field are equal.
• This means that tan δ (the tangent of the angle of dip) is equal to 1.
• Thus, the angle of dip δ is 45°.

Here consonants of the words are coded according to their positional values in English alphabetical series.

• Consonants → same positional value

The Vowels are coded according to the opposite letter place value × 2.

• Vowels → Opposite letter positional value × 2.

For AROUND = 52182412144

Similarly, 
For PLASTIC = ?

Hence, '1612521920363' is the correct answer.

To determine the probability that four points on a circle lie within some semicircle, we consider the geometric condition: all four points must be contained within a 180-degree arc. Fix one point at angle 0 (due to rotational symmetry). The other three points are chosen uniformly, and all four must fit within a semicircle. If the convex hull contains the center, the points span more than 180 degrees, making it impossible. The probability that the center lies inside the convex quadrilateral formed by four random points is 1/2 (a known geometric probability result). Thus, the probability that all points lie within a semicircle is the complement: 1 - 1/2 = 1/2.