This question is based on coding and decoding using opposite alphabetical letters and letter shifting. Each letter is first replaced by its opposite in the English alphabet and then shifted two positions forward. We need to verify this rule using the given examples and apply it to RANGE.
We need to identify the common rule used in both examples.
In the English alphabet, opposite letters are paired so that their alphabetical positions add up to 27.
After replacing each letter with its opposite, move two positions forward in the alphabet. If the shift goes beyond Z, continue from A.
Original word: SCOPE
Replace each letter with its opposite:
Combining the final letters gives:
JZNMX
This matches the given code.
Original word: GAMUT
Combining the final letters gives:
VBPHI
This also matches the given code.
Original word: RANGE
Combining the final letters gives:
K + B + O + V + X = KBOVX
The same opposite-letter and forward-shifting rule works consistently for all three words.
Correct Option: A) KBOVX
Each letter of RANGE is replaced by its opposite alphabetical letter and then shifted two positions forward. The resulting letters are K, B, O, V, and X. Therefore, the correct option is A) KBOVX.
The question asks about the specific location in animals like horses and rabbits where cellulose digestion occurs through bacterial action. This points to a specialized digestive organ found in herbivores.
A) Caecum — The caecum is a large, pouch-like organ in the digestive system of many herbivores, including horses and rabbits. It houses a rich microbial population that ferments cellulose, breaking it down into absorbable nutrients. This makes it the primary site for cellulose digestion in these animals.
Eight friends, Hari, Ganpati, Falak, Ekta, Janak, Rohit, Mangesh and Tony, are sitting around a round table, facing the centre of the table but not necessarily in the same order. Janak is sitting third to the right of Mangesh, who is not an immediate neighbour of either Falak or Tony. Falak is sitting second to the left of Tony, who is sitting third to the left of Hari. Ganpati is sitting fourth to the right of Tony. Ekta is sitting second to the left of Hari.
Who is second to the right of Janak?
This question is based on circular seating arrangements. Eight friends, Hari, Ganpati, Falak, Ekta, Janak, Rohit, Mangesh, and Tony, are sitting around a circular table facing the centre. We need to arrange them according to the given conditions and identify the person sitting second to the right of Janak.
Since all eight friends face the centre:
We can fix Hari at any position because rotating a circular arrangement does not change the relative positions.
Tony sits third to the left of Hari.
Starting from Hari, count three positions clockwise:
Therefore, Tony occupies the third clockwise position from Hari.
Ekta sits second to the left of Hari.
Therefore, Ekta occupies the second clockwise position from Hari.
The partial clockwise arrangement becomes:
Hari → Vacant → Ekta → Tony → Vacant → Vacant → Vacant → Vacant
Falak sits second to the left of Tony.
Count two positions clockwise from Tony:
The arrangement becomes:
Hari → Vacant → Ekta → Tony → Vacant → Falak → Vacant → Vacant
Ganpati sits fourth to the right of Tony.
Since everyone faces the centre, count four positions anticlockwise from Tony.
This places Ganpati immediately anticlockwise from Hari, in the eighth position.
The arrangement becomes:
Hari → Vacant → Ekta → Tony → Vacant → Falak → Vacant → Ganpati
Janak sits third to the right of Mangesh.
Also, Mangesh cannot be an immediate neighbour of Falak or Tony.
The remaining vacant positions are the second, fifth, and seventh positions.
The fifth position is adjacent to both Tony and Falak, so Mangesh cannot occupy it.
The seventh position is adjacent to Falak, so Mangesh cannot occupy it either.
Therefore, Mangesh must occupy the second position.
Counting three positions anticlockwise from Mangesh places Janak in the seventh position.
The only remaining vacant position is the fifth position.
Therefore, Rohit occupies this position.
The complete clockwise arrangement is:
Hari → Mangesh → Ekta → Tony → Rohit → Falak → Janak → Ganpati
All the given conditions are satisfied.
Since everyone faces the centre, move anticlockwise from Janak.
Therefore, Rohit sits second to the right of Janak.
Correct Option: B) Rohit
The clockwise arrangement is Hari, Mangesh, Ekta, Tony, Rohit, Falak, Janak, and Ganpati. Counting anticlockwise from Janak, Falak is first and Rohit is second. Therefore, the correct option is B) Rohit.
To find the sum of several decimal numbers, align them vertically by their decimal points. Then, add the numbers column by column, starting from the rightmost digit, carrying over when necessary.
The numbers are:
\[ \begin{array}{r} 5432.5678 \\ 435.6920 \\ 29.5400 \\ 6.3000 \\ +\quad 0.4321 \\ \hline \end{array} \]The sum is:
\[ \begin{array}{r} 5432.5678 \\ 435.6920 \\ 29.5400 \\ 6.3000 \\ +\quad 0.4321 \\ \hline 5904.5319 \end{array} \]Correct Option: B) 5904.5319
The question asks about a specific decision made by an Indian state government in January 2022 regarding the retirement age of its employees. This is a factual question related to current affairs and state government policies.
Correct Option: C) Andhra Pradesh
Eight people live on different floors of the same building. The lowermost floor in the building is numbered 1, the floor above it is numbered 2 and so on till the topmost floor is numbered 8. Only 4 people live above Aditya. Anu lives on the sixth floor. Only two people live between Anu and Shubham. Ravi lives immediately above Deep. Diksha lives immediately below Sakshi. Sakshi lives on the topmost floor.
On which floor does Deep live?
This is a floor-based puzzle. We need to deduce the living arrangement of eight people on eight different floors based on the given clues. We will start by placing the individuals whose positions are directly given or can be easily inferred, and then use relative positioning clues to fill in the rest.
Correct Option: A) 1
The question asks about a major event in plants during the daytime. This immediately brings to mind photosynthesis, the primary process plants undertake when sunlight is available.
Correct Option: C) Oâ‚‚ release
During the daytime, plants actively perform photosynthesis, a process that utilizes carbon dioxide and water in the presence of sunlight to produce glucose and oxygen. The oxygen is released into the atmosphere as a byproduct. While plants also respire and release \(\text{CO}_2\), the rate of photosynthesis during the day is significantly higher, leading to a net release of \(\text{O}_2\).
We need to find the initial total number of boys and girls. We are given two conditions: initially, the number of boys and girls is equal, and after 8 girls leave, the number of boys becomes twice the number of girls. We can set up algebraic equations based on these conditions to solve for the initial number of boys and girls.
B) 32 — The initial number of boys was 16 and girls was 16, making a total of 32. After 8 girls left, there were 16 boys and \(16 - 8 = 8\) girls. Indeed, 16 boys is twice 8 girls, satisfying the condition.
This problem involves arranging six people (P, Q, R, S, T, U) based on their weights using given relational statements. We will deduce the weight order step-by-step and identify the heaviest person.
Correct Option: C) R is the heaviest amongst them, as determined by the combined weight order.
Lines \( \overleftrightarrow{ℓ} \) ∥ \( \overleftrightarrow{m} \) ∥ \( \overleftrightarrow{n} \) and \( \overleftrightarrow{p} \), \( \overleftrightarrow{q} \) are transversals of \( \overleftrightarrow{ℓ} \), \( \overleftrightarrow{m} \) and \( \overleftrightarrow{n} \).
\( \overleftrightarrow{p} \) intersects lines \( \overleftrightarrow{â„“} \), \( \overleftrightarrow{m} \) and \( \overleftrightarrow{n} \) at the points A, B and C respectively, while \( \overleftrightarrow{q} \) intersects lines \( \overleftrightarrow{â„“} \), \( \overleftrightarrow{m} \) and \( \overleftrightarrow{n} \) at the points D, E and F respectively.
If m(\(\overline{AB}\)) = m(\(\overline{DE}\)), then which of the following is correct?
m(\(\overline{BC}\)) = m(\(\overline{EF}\))
m(\(\overline{AD}\)) = m(\(\overline{DF}\))
m(\(\overline{CE}\)) = m(\(\overline{DF}\))
m(\(\overline{BD}\)) = m(\(\overline{CE}\))
The problem describes three parallel lines intersected by two transversals. We are given that the segment lengths between the first two parallel lines on one transversal are equal to the segment lengths between the first two parallel lines on the other transversal. We need to determine the correct relationship between other segments.
This scenario directly relates to the Intercept Theorem (also known as Thales's Theorem or Basic Proportionality Theorem extension for parallel lines and transversals). A special case of this theorem states that if three or more parallel lines cut off equal intercepts on one transversal, they cut off equal intercepts on any other transversal.
A) \( m(\overline{BC}) = m(\overline{EF}) \). This is derived directly from the Intercept Theorem and the given condition \( m(\overline{AB}) = m(\overline{DE}) \).