Class 11 Applied Maths Chapter 8 (Ex – 8.1)

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Class 11 Applied Maths Chapter 8 Solutions

Logical Reasoning

EXERCISE- 8.1

Q.1 In a certain language ‘ORANGE’ is coded as ‘AEGNOR’, how ‘PINEAPPLE’ is coded?

Ans. In the given language, the word ‘ORANGE’ is coded by arranging its letters in alphabetical order to form ‘AEGNOR’. To code the word ‘PINEAPPLE’ in the same way, we need to arrange its letters in alphabetical order.

The letters in ‘PINEAPPLE’ are: P, I, N, E, A, P, P, L, E

Sorting these letters alphabetically gives: A, E, E, I, L, N, P, P, P

Therefore, ‘PINEAPPLE’ is coded as ‘AEEILNPPP’.

Q.2 In a certain language ‘MIXTURE’ is coded as ‘ERUTXIM’, how ‘SOLUTION’ is coded?

Ans. In the given language, the word ‘MIXTURE’ is coded by reversing its letters to form ‘ERUTXIM’. To code the word ‘SOLUTION’ in the same way, we need to reverse its letters.

The word ‘SOLUTION’ reversed is: ‘NOITULOS’

Therefore, ‘SOLUTION’ is coded as ‘NOITULOS’.

Q.3 In a certain language ‘REGISTRATION’ is coded as ‘TSIGERNOITAR’, how ‘PUBLICATIONS” is coded?

Ans. In the given language, the word ‘REGISTRATION’ is coded as ‘TSIGERNOITAR’. This indicates that the word is split into two equal parts, reversed individually, and then concatenated.

Let’s apply this to ‘PUBLICATIONS’:

1. Split ‘PUBLICATIONS’ into two parts:
   • First part: ‘PUBLIC’
   • Second part: ‘ATIONS’
2. Reverse each part:
   • ‘PUBLIC’ reversed is ‘CILBUP’
   • ‘ATIONS’ reversed is ‘SNOITA’
3. Concatenate the reversed parts:
   • ‘CILBUP’ + ‘SNOITA’ = ‘CILBUPSNOITA’

Therefore, ‘PUBLICATIONS’ is coded as ‘ACILBUPSNOIT’.

Q.4 In a certain language ‘START’ is coded as UQEMZ and ‘EXCEL’ is coded as GUGZR. How is ‘PANIC’ coded?

Ans. To determine the coding pattern, let’s examine how ‘START’ becomes ‘UQEMZ’ and ‘EXCEL’ becomes ‘GUGZR’.

START to UQEMZ:

• S to U: +2
• T to Q: -3
• A to E: +4
• R to M: -5
• T to Z: +6

EXCEL to GUGZR:

• E to G: +2
• X to U: -3
• C to G: +4
• E to Z: -5
• L to R: +6

It appears the transformations follow a repeating pattern of +2, -3, +4, -5, +6.

Let’s apply this pattern to ‘PANIC’:

• P to R: P + 2 = R
• A to X: A – 3 = X
• N to R: N + 4 = R
• I to D: I – 5 = D
• C to I: C + 6 = I

Thus, ‘PANIC’ is coded as ‘RXRDI’.

Q.5 In a certain language ‘SACHIN’ is coded as RZBGHM. How ‘ISHANT’ is coded?

Ans. To determine the coding pattern, let’s examine how ‘SACHIN’ becomes ‘RZBGHM’:

• S to R: -1
• A to Z: -1
• C to B: -1
• H to G: -1
• I to H: -1
• N to M: -1

Each letter in ‘SACHIN’ is decreased by 1 to form ‘RZBGHM’.

Let’s apply the same pattern to ‘ISHANT’:

• I to H: I – 1 = H
• S to R: S – 1 = R
• H to G: H – 1 = G
• A to Z: A – 1 = Z
• N to M: N – 1 = M
• T to S: T – 1 = S

Thus, ‘ISHANT’ is coded as ‘HRGZMS’.

Q.6 In a certain language ‘QUEEN’ is coded as OVCFL. How is ‘KING’ coded?

Ans. To determine the coding pattern, let’s examine how ‘QUEEN’ becomes ‘OVCFL’:

• Q to O: -2
• U to V: +1
• E to C: -2
• E to F: +1
• N to L: -2

The pattern alternates between -2 and +1.

Let’s apply the same pattern to ‘KING’:

• K to I: K – 2 = I
• I to J: I + 1 = J
• N to L: N – 2 = L
• G to H: G + 1 = H

Thus, ‘KING’ is coded as ‘IJLH’.

Q.7 In a certain language ‘TEACHER’ is coded as GVZXSVI. How is ‘STUDENT’ coded?

Ans. To determine how ‘STUDENT’ is coded based on the pattern where ‘TEACHER’ is coded as ‘GVZXSVI’, let’s analyze the encoding pattern:

Analyze the Encoding Pattern

Compare each letter in ‘TEACHER’ with ‘GVZXSVI’:

1. T (20th letter) -> G (7th letter)
2. E (5th letter) -> V (22nd letter)
3. A (1st letter) -> Z (26th letter)
4. C (3rd letter) -> X (24th letter)
5. H (8th letter) -> S (19th letter)
6. E (5th letter) -> V (22nd letter)
7. R (18th letter) -> I (9th letter)

The pattern is a substitution based on the alphabet in reverse:

• A -> Z
• B -> Y
• C -> X
• D -> W
• E -> V
• F -> U
• G -> T
• H -> S
• I -> R
• J -> Q
• K -> P
• L -> O
• M -> N
• N -> M
• O -> L
• P -> K
• Q -> J
• R -> I
• S -> H
• T -> G
• U -> F
• V -> E
• W -> D
• X -> C
• Y -> B
• Z -> A

Apply the Same Pattern to ‘STUDENT’

1. S -> H
2. T -> G
3. U -> F
4. D -> W
5. E -> V
6. N -> M
7. T -> G

Using this pattern, ‘STUDENT’ is coded as HGFWVMG.

Q.8 If ‘REPUBLIC’ is coded as TGRWDNKE taken what information can be obtained from the message FGOQETCVKE?

Ans. To decode ‘FGOQETCVKE’ based on the coding pattern where ‘REPUBLIC’ is coded as ‘TGRWDNKE’, we need to identify the pattern used in the transformation from ‘REPUBLIC’ to ‘TGRWDNKE’.

Analyze the Encoding Pattern

Let’s map each letter from ‘REPUBLIC’ to ‘TGRWDNKE’:

1. R (18th letter) -> T (20th letter): Shift of +2
2. E (5th letter) -> G (7th letter): Shift of +2
3. P (16th letter) -> R (18th letter): Shift of +2
4. U (21st letter) -> W (23rd letter): Shift of +2
5. B (2nd letter) -> D (4th letter): Shift of +2
6. L (12th letter) -> N (14th letter): Shift of +2
7. I (9th letter) -> K (11th letter): Shift of +2
8. C (3rd letter) -> E (5th letter): Shift of +2

The pattern is a simple shift of +2 positions in the alphabet.

Decode ‘FGOQETCVKE’

Apply the reverse shift (-2) to decode ‘FGOQETCVKE’:

1. F (6th letter) -> D (4th letter): Shift of -2
2. G (7th letter) -> E (5th letter): Shift of -2
3. O (15th letter) -> M (13th letter): Shift of -2
4. Q (17th letter) -> O (15th letter): Shift of -2
5. E (5th letter) -> C (3rd letter): Shift of -2
6. T (20th letter) -> R (18th letter): Shift of -2
7. C (3rd letter) -> A (1st letter): Shift of -2
8. V (22nd letter) -> T (20th letter): Shift of -2
9. K (11th letter) -> I (9th letter): Shift of -2
10. E (5th letter) -> C (3rd letter): Shift of -2

So, ‘FGOQETCVKE’ decoded is DEMOCRATIC.

Q.9 In a certain language ‘AGRA’ is coded as ‘CUBC’, how ‘JHANSI’ is coded?

Ans. To determine how ‘JHANSI’ is coded based on the pattern where ‘AGRA’ is coded as ‘CUBC’ using a multiplication method, we need to establish a pattern based on the given example.

Analyze the Encoding Pattern with Multiplication

Compare each letter in ‘AGRA’ with ‘CUBC’:

• A (1st letter) → C (3rd letter)
  • A (1) * 3 = C (3)
• G (7th letter) → U (21st letter)
  • G (7) * 3 = U (21)
• R (18th letter) → B (2nd letter)
  • R (18) * 3 = 54 (54 % 26 = 2) = B (2)
• A (1st letter) → C (3rd letter)
  • A (1) * 3 = C (3)

So the pattern is that each letter’s position in the alphabet is multiplied by 3, and if the result is greater than 26, it is taken modulo 26.

Apply the Same Pattern to ‘JHANSI’

1. J (10th letter) → Multiply by 3:
   • 10 * 3 = 30 → 30 % 26 = 4 → D
2. H (8th letter) → Multiply by 3:
   • 8 * 3 = 24 → X
3. A (1st letter) → Multiply by 3:
   • 1 * 3 = 3 → C
4. N (14th letter) → Multiply by 3:
   • 14 * 3 = 42 → 42 % 26 = 16 → P
5. S (19th letter) → Multiply by 3:
   • 19 * 3 = 57 → 57 % 26 = 5 → E
6. I (9th letter) → Multiply by 3:
   • 9 * 3 = 27 → 27 % 26 = 1 → A

Result

Using the multiplication method, ‘JHANSI’ is coded as DXCPEA.

Q.10 In a certain language ‘TABLE’ is coded as ‘NBDXJ’, how ‘CHAIR’ is coded?

Ans. To determine how ‘CHAIR’ is coded based on the pattern where ‘TABLE’ is coded as ‘NBDXJ’ using a doubling method, we need to analyze the pattern and apply it to the new word.

Analyze the Encoding Pattern

Given that ‘TABLE’ is coded as ‘NBDXJ’:

1. T (20th letter) -> N (14th letter)
2. A (1st letter) -> B (2nd letter)
3. B (2nd letter) -> D (4th letter)
4. L (12th letter) -> X (24th letter)
5. E (5th letter) -> J (10th letter)

If we use a doubling method (multiplying the position by 2 and taking modulo 26 if necessary), let’s check if it applies:

• T: 20 * 2 = 40, 40 % 26 = 14 -> N
• A: 1 * 2 = 2 -> B
• B: 2 * 2 = 4 -> D
• L: 12 * 2 = 24 -> X
• E: 5 * 2 = 10 -> J

Apply the Same Pattern to ‘CHAIR’

1. C (3rd letter) -> Multiply by 2:
   • 3 * 2 = 6 -> F
2. H (8th letter) -> Multiply by 2:
   • 8 * 2 = 16 -> P
3. A (1st letter) -> Multiply by 2:
   • 1 * 2 = 2 -> B
4. I (9th letter) -> Multiply by 2:
   • 9 * 2 = 18 -> R
5. R (18th letter) -> Multiply by 2:
   • 18 * 2 = 36, 36 % 26 = 10 -> J

Result

Using the doubling method, ‘CHAIR’ is coded as FPBRJ.

Q.11 If ‘MIRINDA’ is coded as 151120111663 then how is THUMSUP coded?

Ans. To determine the coding pattern, let’s analyze how ‘MIRINDA’ is coded as 151120111663.

First, let’s break down the word ‘MIRINDA’ and its corresponding code:

• M: 13
• I: 9
• R: 18
• I: 9
• N: 14
• D: 4
• A: 1

Now, let’s see how these letters map to the code 151120111663:

• M (13) -> 15
• I (9) -> 11
• R (18) -> 20
• I (9) -> 11
• N (14) -> 16
• D (4) -> 6
• A (1) -> 3

It appears each letter’s numerical position in the alphabet is increased by a certain number to get the code:

• M (13) + 2 = 15
• I (9) + 2 = 11
• R (18) + 2 = 20
• I (9) + 2 = 11
• N (14) + 2 = 16
• D (4) + 2 = 6
• A (1) + 2 = 3

The pattern is to add 2 to each letter’s position in the alphabet.

Let’s apply the same pattern to ‘THUMSUP’:

• T: 20 + 2 = 22
• H: 8 + 2 = 10
• U: 21 + 2 = 23
• M: 13 + 2 = 15
• S: 19 + 2 = 21
• U: 21 + 2 = 23
• P: 16 + 2 = 18

So, ‘THUMSUP’ is coded as 22102315212318.

Q.12 In a certain language ‘VIOLET’ is coded as 196129217. Decode 1215241142.

Ans. To decode ‘1215241142’ based on the pattern where ‘VIOLET’ is coded as ‘196129217’ using the subtraction method,

Analyze the Encoding Pattern

The code for ‘VIOLET’ is ‘196129217’. If we consider that each digit in the code represents the position in the alphabet after adding or subtracting a certain number, let’s decode ‘VIOLET’ by determining if a consistent pattern like adding or subtracting a certain number is applied:

1. V (22nd letter) -> 19
2. I (9th letter) -> 6
3. O (15th letter) -> 12
4. L (12th letter) -> 9
5. E (5th letter) -> 21
6. T (20th letter) -> 7

Decoding the Number ‘1215241142’:

Let’s decode ‘1215241142’ by adding 3 to each numbers:

1. 12 + 3 = 15 (15th letter) ->O
2. 15 + 3 = 18 (18th letter) -> R
3. 24 + 3 = 27 (27th letter) -> A
4. 11 + 3 = 14 (14th letter) -> N
5. 4 + 3 = 7 (7th letter) -> N
6. 2 + 3 = 5 (5th letter) -> E

Therefore, ‘1215241142’ decoded is ORANGE.

Q.13. If ‘POOR’ is coded as 46 and ‘BLOWER’ is coded as 57 then how is ‘TOWER’ coded?

Ans. To determine the coding pattern, let’s examine how ‘POOR’ is coded as 46 and ‘BLOWER’ is coded as 57.

Alphabetical no. of CORONA:

• P = 16
• O = 15
• O = 15
• R = 18

Total = 16 + 15 + 15 + 18 = 64

Here, 64 is taken as reversing the number, i.e. 46

Alphabetical number of BLOWER :

• B = 2
• L = 12
• O = 15
• W = 23
• E = 5
• R = 18

Total = 2 + 12 + 15 + 23 + 5 + 18 = 75

Here, 55 is taken as reversing the number, i.e, 57

Let’s apply this pattern to ‘TOWER’:

• T = 20
• O = 15
• W = 23
• E = 5
• R = 18

Total = 20 + 15 + 23 + 5 + 18 = 81

Thus, ‘TOWER’ is coded as 18.

Q.14 In a certain language if ‘CORONA’ is coded as 6 and ‘MALARIA’ is coded as 5 then how is ‘CANCER’ coded?

Ans. To determine the coding pattern, let’s examine how ‘CORONA’ is coded as 6 and ‘MALARIA’ is coded as 5.

Alphabetical no. of CORONA:

• C = 3
• O = 15
• R = 18
• O = 15
• N = 14
• A = 1

Total = 3 + 15 + 18 + 15 + 14 + 1 = 66

Here, 66 is taken as 6

Alphabetical number of MALARIA:

• M = 13
• A = 1
• L = 12
• A = 1
• R = 18
• I = 9
• A = 1

Total = 13 + 1 + 12 + 1 + 18 + 9 + 1 = 55

Here, 55 is taken as 5

Let’s apply this pattern to ‘CANCER’:

• C = 3
• A = 1
• N = 14
• C = 3
• E = 5
• R = 18

Total = 3 + 1 + 14+ 3 + 5 + 18 = 44

Thus, ‘CANCER’ is coded as 4.

Q.15 If ‘PETER’ is coded as 64841, ‘SERVENT’ is coded as 5412478 and ‘MASTER’ is coded as 935841, then how is ‘PERMANENT’ coded?

Ans. To determine how ‘PERMANENT’ is coded, let’s first analyze the pattern used in coding the given words:

1. PETER is coded as 64841
2. SERVENT is coded as 5412478
3. MASTER is coded as 935841

Step-by-Step Analysis:

Let’s map the letters to their corresponding numbers based on the given codes:

• PETER -> 64841
  • P -> 6
  • E -> 4
  • T -> 8
  • E -> 4
  • R -> 1
• SERVENT -> 5412478
  • S -> 5
  • E -> 4
  • R -> 1
  • V -> 2
  • E -> 4
  • N -> 7
  • T -> 8
• MASTER -> 935841
  • M -> 9
  • A -> 3
  • S -> 5
  • T -> 8
  • E -> 4
  • R -> 1

From these mappings, we can see a pattern for each letter:

• A -> 3
• E -> 4
• M -> 9
• N -> 7
• P -> 6
• R -> 1
• S -> 5
• T -> 8
• V -> 2

Now we can use this pattern to code ‘PERMANENT’:

• P -> 6
• E -> 4
• R -> 1
• M -> 9
• A -> 3
• N -> 7
• E -> 4
• N -> 7
• T -> 8

Therefore, ‘PERMANENT’ is coded as 641937478.

Q.16 In a certain language 458 means “All is well,” 876 means “Well done boys” and 689 m “Boys played well”. What does ‘played’ mean in this language?

Ans. To determine what ‘played’ means in this language, let’s analyze the given information:

1. 458 means “All is well”
2. 876 means “Well done boys”
3. 689 means “Boys played well”

We need to find the code for ‘played’. Let’s identify common elements in the sentences to find the corresponding numbers.

Step-by-Step Analysis:

1. Identify common words:
   • ‘well’ is common between “All is well” (458) and “Well done boys” (876).
   • ‘boys’ is common between “Well done boys” (876) and “Boys played well” (689).
2. Determine codes for common words:
   • Comparing 458 (“All is well”) and 876 (“Well done boys”), the common word is ‘well’.
     • The common number is ‘8’, so ‘well’ corresponds to ‘8’.
   • Comparing 876 (“Well done boys”) and 689 (“Boys played well”), the common word is ‘boys’.
     • The common number is ‘6’, so ‘boys’ corresponds to ‘6’.
3. Analyze remaining code elements:
   • For “Well done boys” (876):
     • ‘well’ is 8
     • ‘boys’ is 6
     • ‘done’ corresponds to ‘7’
   • For “Boys played well” (689):
     • ‘boys’ is 6
     • ‘well’ is 8
     • The remaining number ‘9’ must correspond to ‘played’

Therefore, in this language, ‘played’ is represented by the digit ‘9’.

Q.17 In a certain language 123 means “He is good”, 456 means “You are bad” and 472 means “Good and bad”. What does ‘and’ means in this language?

Ans. To determine what ‘and’ means in this language, let’s analyze the given information:

1. 123 means “He is good”
2. 456 means “You are bad”
3. 472 means “Good and bad”

We need to find the code for ‘and’. Let’s identify common elements in the sentences to find the corresponding numbers:

Step-by-Step Analysis:

1. Identify common words:
   • ‘good’ is common between “He is good” (123) and “Good and bad” (472).
   • ‘bad’ is common between “You are bad” (456) and “Good and bad” (472).
2. Determine codes for common words:
   • Comparing 123 (“He is good”) and 472 (“Good and bad”), the common number is ‘3’, so ‘good’ corresponds to ‘3’.
   • Comparing 456 (“You are bad”) and 472 (“Good and bad”), the common number is ‘4’, so ‘bad’ corresponds to ‘4’.
3. Analyze remaining code elements:
   • 472: Good (3), and (?), bad (4)
   • With 472 and knowing ‘good’ is ‘3’ and ‘bad’ is ‘4’, the remaining number ‘7’ must correspond to ‘and’.

Therefore, in this language, ‘and’ is represented by the digit ‘7’.

Q.18 In a certain language ‘fli cdo vas’ means ‘Doctors save life’, ‘erp fli si’ means ‘life is precious’ and ‘dgo can cdo’ means ‘Doctors are god. What word stand for ‘Save’ in this language?

Ans. To determine which word stands for ‘save’ in this language, let’s analyze the given sentences and their codes:

1. ‘fli cdo vas’ means ‘Doctors save life’
2. ‘erp fli si’ means ‘life is precious’
3. ‘dgo can cdo’ means ‘Doctors are god’

We need to identify which code corresponds to the word ‘save’.

Let’s first identify the common words and their corresponding codes:

• From ‘fli cdo vas’ (‘Doctors save life’) and ‘dgo can cdo’ (‘Doctors are god’), the common word is ‘Doctors’, and the common code is ‘cdo’. Therefore, ‘cdo’ means ‘Doctors’.
• From ‘fli cdo vas’ (‘Doctors save life’) and ‘erp fli si’ (‘life is precious’), the common word is ‘life’, and the common code is ‘fli’. Therefore, ‘fli’ means ‘life’.

Now we can break down the codes further:

• ‘fli cdo vas’ = ‘life Doctors save’
• ‘dgo can cdo’ = ‘Doctors are god’
• ‘erp fli si’ = ‘life is precious’

We know:

• ‘cdo’ means ‘Doctors’
• ‘fli’ means ‘life’

From ‘fli cdo vas’ (‘life Doctors save’):

• We have identified ‘fli’ as ‘life’ and ‘cdo’ as ‘Doctors’.
• Therefore, ‘vas’ must mean ‘save’.

So, the word for ‘save’ in this language is ‘vas’.

Q.19 In a certain code ‘mte ves rme’ means ‘Save more time’, ’emt rfo two’ means ‘Wait for me’ and ‘rme vge emt’ means ‘Give me more’. What does ‘Give’ mean in this language?

Ans. To determine what ‘Give’ means in this language, let’s analyze the given sentences and their codes:

1. ‘mte ves rme’ means ‘Save more time’
2. ’emt rfo two’ means ‘Wait for me’
3. ‘rme vge emt’ means ‘Give me more’

We need to identify which word corresponds to ‘Give’ by comparing the sentences and their codes.

Let’s look at the words and their codes:

• From ‘mte ves rme’ and ‘rme vge emt’, the common word is ‘rme’, which corresponds to ‘me’ because ‘me’ appears in both ‘Save more time’ and ‘Give me more’.
• ‘more’ is the word common between ‘Save more time’ and ‘Give me more’, so ‘more’ should be ‘rme’ as well.This leads us to understand that ‘rme’ is likely ‘more’ (as ‘more’ is common in both sentences).

Given that:

• ‘rme’ is ‘more’ (from ‘mte ves rme’ and ‘rme vge emt’)
• ’emt’ is ‘me’ (from ‘Wait for me’ and ‘Give me more’)

Now, in the sentence ‘rme vge emt’ means ‘Give me more’, we already know:

• ‘rme’ is ‘more’
• ’emt’ is ‘me’

Therefore, ‘vge’ must correspond to ‘Give’.

So, ‘Give’ is coded as ‘vge’ in this language.

FAQ’s related to Class 11 Applied Maths Chapter 8 on Logical Reasoning:

Q.1 What is logical reasoning?

Ans. Logical reasoning refers to the process of using a structured and methodical approach to arrive at a conclusion based on given premises or facts. It involves critical thinking and the ability to see the logical connections between ideas.

Q.2 What are the common mistakes to avoid in logical reasoning?

Ans. Common mistakes include:

• Assuming correlation implies causation.
• Overgeneralizing from insufficient evidence.
• Ignoring counterexamples.
• Misinterpreting logical connectives.

Q.3 What is a Venn diagram?

Ans. A Venn diagram is a graphical way to represent sets and their relationships using circles. It is used in logical reasoning to visually illustrate the logical relationships between different sets.

Q.4 How are Venn diagrams used in logical reasoning?

Ans. Venn diagrams are used to:

• Show logical relationships between different sets.
• Solve problems involving set operations such as union, intersection, and difference.
• Visualize logical propositions and their truth values.

Q.5 Can you give an example of a logical reasoning problem and solution?

Ans. If all dogs are mammals and some pets are dogs, can we conclude that some pets are mammals?

Solution: Yes, because if all dogs are mammals, and some pets are dogs, it logically follows that those pets which are dogs must also be mammals. Therefore, some pets are mammals.

These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 8

In Class 11 Applied Maths chapter 8, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 8 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 8, we delve deep into advanced mathematical concepts that are crucial for understanding.