In the following question, select the related letters from the given alternatives. AMG : WIC :: NZT : ?
JVP
Understanding Letter Analogy Questions
This question is a letter analogy, which falls under verbal reasoning. In letter analogies, a specific relationship or pattern exists between the first pair of letters (AMG : WIC), and you need to identify this pattern and apply it to the third term (NZT) to find the missing fourth term.
To solve letter analogies, it's helpful to know the position of each letter in the English alphabet. We can number the letters from 1 (A) to 26 (Z).
Analyzing the Pattern: AMG to WIC
Let's look at the position of each letter in the alphabet for the first pair:
| Word 1 | Letter | Position |
|---|---|---|
| AMG | A | \(1\) |
| M | \(13\) | |
| G | \(7\) |
| Word 2 | Letter | Position |
|---|---|---|
| WIC | W | \(23\) |
| I | \(9\) | |
| C | \(3\) |
Now let's find the relationship between the corresponding letters:
- First letter: A (1) to W (23). The difference is \(23 - 1 = 22\). Alternatively, moving backwards from A, 4 steps would be A → Z → Y → X → W. This is a shift of \(-4\) positions (wrapping around the alphabet). \(1 - 4 = -3\), which is \(26 - 3 = 23\) in alphabet position.
- Second letter: M (13) to I (9). The difference is \(9 - 13 = -4\). This is a shift of \(-4\) positions.
- Third letter: G (7) to C (3). The difference is \(3 - 7 = -4\). This is a shift of \(-4\) positions.
The consistent pattern observed is subtracting 4 from the alphabetical position of each letter. For the first letter, this involves wrapping around from A to Z.
Applying the Pattern to NZT
Now, let's apply the same \(-4\) shift pattern to the letters in NZT.
First, find the positions of the letters in NZT:
- N is the \(14\)th letter.
- Z is the \(26\)th letter.
- T is the \(20\)th letter.
Apply the \(-4\) shift to each position:
- For N (14): \(14 - 4 = 10\). The \(10\)th letter is J.
- For Z (26): \(26 - 4 = 22\). The \(22\)nd letter is V.
- For T (20): \(20 - 4 = 16\). The \(16\)th letter is P.
The resulting letters are JVP.
Comparing with Options
Let's compare our result (JVP) with the given options:
- Option 1: JVP - This matches our calculated result.
- Option 2: MOP - Does not match.
- Option 3: JVS - The last letter S does not match P.
- Option 4: JKR - Does not match.
Therefore, the correct alternative is JVP.
| Original Letter | Position | Shift (-4) | New Position | Resulting Letter |
|---|---|---|---|---|
| A | \(1\) | \(1 - 4 = -3\) | \(26 - 3 = 23\) | W |
| M | \(13\) | \(13 - 4\) | \(9\) | I |
| G | \(7\) | \(7 - 4\) | \(3\) | C |
| N | \(14\) | \(14 - 4\) | \(10\) | J |
| Z | \(26\) | \(26 - 4\) | \(22\) | V |
| T | \(20\) | \(20 - 4\) | \(16\) | P |
Additional Information on Letter Analogy Reasoning
Letter analogy problems test your ability to find patterns in sequences of letters. Common patterns include:
- Position Shift: Each letter is shifted by a fixed number of positions forward or backward in the alphabet (as seen in this problem).
- Reverse Order: Letters might be in reverse alphabetical order within a word.
- Skipping Letters: Letters might follow a pattern of skipping a fixed number of letters (e.g., A, C, E...).
- Vowel/Consonant Patterns: The pattern might relate to vowels and consonants.
- Reverse Alphabetical Order: A is related to Z, B to Y, and so on (A=1st from start, Z=1st from end).
- Combination of Patterns: Sometimes, different patterns are applied to different letters within the same set.
Practicing with different types of letter analogy and letter series questions helps improve your logical reasoning and problem-solving skills for competitive exams.
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
