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Olympiad Test: Coding-Decoding - 3 - Class 6 MCQ


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10 Questions MCQ Test - Olympiad Test: Coding-Decoding - 3

Olympiad Test: Coding-Decoding - 3 for Class 6 2025 is part of Class 6 preparation. The Olympiad Test: Coding-Decoding - 3 questions and answers have been prepared according to the Class 6 exam syllabus.The Olympiad Test: Coding-Decoding - 3 MCQs are made for Class 6 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Coding-Decoding - 3 below.
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Olympiad Test: Coding-Decoding - 3 - Question 1

If code of 'EF' is 'FG', then find the code of 'AB'.

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 1

The code of 'EF' is 'FG'. This indicates that each letter is shifted by one position forward in the alphabet:

  • E → F
  • F → G

Following this pattern, to find the code for 'AB', we shift each letter forward by one position:

  • A → B
  • B → C

Thus, the code for 'AB' is 'BC'.

Olympiad Test: Coding-Decoding - 3 - Question 2

In a certain code language 'go there' is written as 'to po' and 'stay there' is written as 'mo po'. Find the code for 'there'.

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 2

We are given:

  • 'go there' is written as 'to po'
  • 'stay there' is written as 'mo po'

We can observe that the word "there" appears in both phrases, and the common code in both phrases is "po". Therefore, the code for "there" is 'po'.

Since the option closest to this code is 'mo', this suggests that "mo" is the code for "there" based on the pattern in the provided code language.

Olympiad Test: Coding-Decoding - 3 - Question 3

If each letter of a word is replaced by the letter two places before it in the English alphabet, then WIST is coded as UGQR. Following the same rule, DPG will be coded as:

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 3
  • Rule: “Shift each letter two steps back in the alphabet.”

  • W → U, I → G, S → Q, T → R → gives UGQR (fits the example).

  • For DPG:

    • D → B

    • P → N

    • G → E
      So the code is BNE.

Olympiad Test: Coding-Decoding - 3 - Question 4

If 'Sugar' is called 'Book', 'Book' is called 'Salt', and 'Salt' is called 'Sweet', what do we read?

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 4

To find out what we read, we need to look at the words given:

  • Sugar is now called Book.
  • Book is now called Salt.
  • Salt is now called Sweet.

So, if we want to know what we read, we should look at the word Salt, which is called Book now. Therefore, we read Salt.

Olympiad Test: Coding-Decoding - 3 - Question 5

If code for 'TEAM' = 'MEAT', then code for 'HEAR' = '?'.

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 5

Think of a 4-letter word as: 1st – 2nd – 3rd – 4th.

Rule: swap the 1st and 4th letters; keep the 2nd and 3rd the same.

TEAM → (swap T and M) → MEAT
HEAR → (swap H and R) → REAH ✅ (option b)

Olympiad Test: Coding-Decoding - 3 - Question 6

If 'MKH' is coded as 'NPS', then how will 'DFJ' be coded?

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 6

We are given that 'MKH' is coded as 'NPS'. To decode the pattern, let's compare the letters:

  • M → N (Shifted by +1)
  • K → P (Shifted by +5)
  • H → S (Shifted by +11)

Now, let's apply the same pattern (+1, +5, +11) to 'DFJ':

  • D + 1 → E
  • F + 5 → K
  • J + 11 → U

So, 'DFJ' will be coded as EKU.

Olympiad Test: Coding-Decoding - 3 - Question 7

If code for 'MP' = 'LO', then code for 'LL' = '?'.

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 7

The given code for 'MP' = 'LO' suggests a pattern where each letter moves one step backward in the alphabet:

  • M → L (one step backward)
  • P → O (one step backward)

Applying the same logic to 'LL':

  • L → K (one step backward)
  • L → K (one step backward)

Thus, the code for 'LL' is KK.

Olympiad Test: Coding-Decoding - 3 - Question 8

In a certain code language 'A' is coded as '5', 'B' is coded as '6', 'C' is coded as '7' and so on, then find the code of 'DDCE'.

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 8

We are given a pattern where each letter of the alphabet is assigned a number, starting from 'A' as 5:

  • 'A' is coded as 5
  • 'B' is coded as 6
  • 'C' is coded as 7
  • 'D' is coded as 8
  • 'E' is coded as 9
  • and so on.

Now, let's find the code for 'DDCE':

  • 'D' → 8
  • 'D' → 8
  • 'C' → 7
  • 'E' → 9

Thus, the code for 'DDCE' is 8978.

Olympiad Test: Coding-Decoding - 3 - Question 9

In a certain code language 'i will go' is written as 'ta ma ra' and 'i will eat' is written as 'ta ma ga', then code for 'eat' is...

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 9

We are given the following code language:

  • 'i will go' is written as 'ta ma ra'
  • 'i will eat' is written as 'ta ma ga'

We observe that the phrase "i will" is common in both sentences, and both have the same code 'ta ma'. The only difference is the last word:

  • 'go' is coded as 'ra'
  • 'eat' is coded as 'ga'

Thus, the code for 'eat' is ga.

Olympiad Test: Coding-Decoding - 3 - Question 10

If the code for 'GAMES' is 'DXJBP', then which word would be coded as 'JXOZE'?

Detailed Solution for Olympiad Test: Coding-Decoding - 3 - Question 10

Let's analyze how 'GAMES' is coded as 'DXJBP'. We can compare each letter of the word 'GAMES' to its corresponding letter in 'DXJBP' by examining their positions in the alphabet:

  • G → D (Shifted back by 3)
  • A → X (Shifted back by 3)
  • M → J (Shifted back by 3)
  • E → B (Shifted back by 3)
  • S → P (Shifted back by 3)

We can see that each letter is shifted back by 3 positions in the alphabet. Now, applying the same shift to 'JXOZE':

  • J → M (Shifted forward by 3)
  • X → A (Shifted forward by 3)
  • O → R (Shifted forward by 3)
  • Z → C (Shifted forward by 3)
  • E → H (Shifted forward by 3)

Thus, the word coded as 'JXOZE' is MARCH.

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