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Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.6 Questions and Answers.

Std 12 Maths 1 Exercise 1.6 Solutions Commerce Maths

Question 1.
Prepare the truth tables for the following statement patterns:
(i) p → (~p ∨ q)
Solution:
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.

(ii) (~p ∨ q) ∧ (~p ∨ ~q)
Solution:

(iii) (p ∧ r) → (p ∨ ~q)
Solution:
Here are three statements and 4 connectives.
∴ there are 2 × 2 × 2 = 8 rows and 3 + 4 = 7 columns in the truth table.

(iv) (p ∧ q) ∨ ~r
Solution:

Question 2.
Examine, whether each of the following statement patterns is a tautology or a contradiction or a contingency:
(i) q ∨ [~(p ∧ q)]
Solution:

All the entries in the last column of the above truth table are T.
∴ q ∨ [~(p ∧ q)] is a tautology.

(ii) (~q ∧ p) ∧ (p ∧ ~p)
Solution:

All the entries in the last column of the above truth table are F.
∴ (~q ∧ p) ∧ (p ∧ ~p) is a contradiction.

(iii) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(iv) ~p → (p → ~q)
Solution:

All the entries in the last column of the truth table are T.
∴ p → (p → ~q) is a tautology.

Question 3.
Prove that each of the following statement pattern is a tautology:
(i) (p ∧ q) → q
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → q is a tautology.

(ii) (p → q) ↔ (~q → ~p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~q → ~p) is a tautology.

(iii) (~p ∧ ~q) → (p → q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∧ ~q) → (p → q) is a tautology.

(iv) (~p ∨ ~q) ↔ ~(p ∧ q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (~p ∨ ~q) ↔ ~(p ∧ q) is a tautology.

Question 4.
Prove that each of the following statement pattern is a contradiction:
(i) (p ∨ q) ∧ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.

(ii) (p ∧ q) ∧ ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∧ ~p is a contradiction.

(iii) (p ∧ q) ∧ (~p ∨ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p ∧ q) ∧ (~p ∨ ~q) is a contradiction.

(iv) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

Question 5.
Show that each of the following statement pattern is a contingency:
(i) (p ∧ ~q) → (~p ∧ ~q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p ∧ ~q) → (~p ∧ ~q) is a contingency.

(ii) (p → q) ↔ (~p ∧ q)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ↔ (~p ∧ q) is a contingency.

(iii) p ∧ [(p → ~q) → q]
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ p ∧ [(p → ~q) → q] is a contingency.

(iv) (p → q) ∧ (p → r)
Solution:

The entries in the last column of the above truth table are neither all T nor all F.
∴ (p → q) ∧ (p → r) is a contingency.

Question 6.
Using the truth table, verify:
(i) p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).

(ii) p → (p → q) ≡ ~q → (p → q)
Solution:

The entries in columns 5 and 6 are identical.
∴ p → (p → q) ≡ ~q → (p → q)

(iii) ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q
Solution:

The entries in columns 5, 7 and 8 are identical.
∴ ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q.

(iv) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:

The entries in columns 3 and 7 are identical.
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 7.
Prove that the following pairs of statement patterns are equivalent:
(i) p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r)
Solution:

The entries in columns 5 and 8 are identical.
∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

(ii) p ↔ q and (p → q) ∧ (q → p)
Solution:

The entries in columns 3 and 6 are identical.
∴ p ↔ q ≡ (p → q) ∧ (q → p)

(iii) p → q and ~q → ~p and ~p ∨ q
Solution:

The entries in columns 5, 6 and 7 are identical.
∴ p → q ≡ ~q → ~p ≡ ~p ∨ q.

(iv) ~(p ∧ q) and ~p ∨ ~q
Solution:

The entries in columns 6 and 7 are identical.
∴ ~(p ∧ q) ≡ ~p ∨ ~q.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 1.1 Solutions
• 12th Commerce Maths Exercise 1.2 Solutions
• 12th Commerce Maths Exercise 1.3 Solutions
• 12th Commerce Maths Exercise 1.4 Solutions
• 12th Commerce Maths Exercise 1.5 Solutions
• 12th Commerce Maths Exercise 1.6 Solutions
• 12th Commerce Maths Exercise 1.7 Solutions
• 12th Commerce Maths Exercise 1.8 Solutions
• 12th Commerce Maths Exercise 1.9 Solutions
• 12th Commerce Maths Exercise 1.10 Solutions
• 12th Commerce Maths Exercise Miscellaneous Exercise 1 Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

Std 12 Maths 1 Exercise 1.5 Solutions Commerce Maths

Question 1.
Use qualifiers to convert each of the following open sentences defined on N, into a true statement:
(i) x² + 3x – 10 = 0
Solution:
∃ x ∈ N, such that x² + 3x – 10 = 0 is a true statement
(x = 2 ∈ N satisfy x² + 3x – 10 = 0)

(ii) 3x – 4 < 9
Solution:
∃ x ∈ N, such that 3x – 4 < 9 is a true statement.
(x = 1, 2, 3, 4 ∈ N satisfy 3x – 4 < 9)

(iii) n² ≥ 1
Solution:
∀ n ∈ N, n² ≥ 1 is a true statement.
(All n ∈ N satisfy n² ≥ 1)

(iv) 2n – 1 = 5
Solution:
∃ x ∈ N, such that 2n – 1 = 5 is a true statement.
(n = 3 ∈ N satisfy 2n – 1 = 5)

(v) y + 4 > 6
Solution:
∃ y ∈ N, such that y + 4 > 6 is a true statement.
(y = 3, 4, 5, … ∈ N satisfy y + 4 > 6

(vi) 3y – 2 ≤ 9
Solution:
∃ y ∈ N, such that 2y ≤ 9 is a true statement.
(y = 1, 2, 3 ∈ N satisfy 3y – 2 ≤ 9).

Question 2.
If B = {2, 3, 5, 6, 7}, determine the truth value of each of the following:
(i) ∀ x ∈ B, x is a prime number.
Solution:
(i) x = 6 ∈ B does not satisfy x is a prime number.
So, the given statement is false, hence its truth value is F.

(ii) ∃ n ∈ B, such that n + 6 > 12.
Solution:
Clearly n = 7 ∈ B satisfies n + 6 > 12.
So, the given statement is true, hence its truth value is T.

(iii) ∃ n ∈ B, such that 2n + 2 < 4.
Solution:
No element n ∈ B satisfy 2n + 2 < 4.
So, the given statement is false, hence its truth value is F.

(iv) ∀ y ∈ B, y² is negative.
Solution:
No element y ∈ B satisfy y² is negative.
So, the given statement is false, hence its truth value is F.

(v) ∀ y ∈ B, (y – 5) ∈ N.
Solution:
y = 2 ∈ B, y = 3 ∈ B and y = 5 ∈ B do not satisfy (y – 5) ∈ N.
So, the given statement is false, hence its truth value is F.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 1.1 Solutions
• 12th Commerce Maths Exercise 1.2 Solutions
• 12th Commerce Maths Exercise 1.3 Solutions
• 12th Commerce Maths Exercise 1.4 Solutions
• 12th Commerce Maths Exercise 1.5 Solutions
• 12th Commerce Maths Exercise 1.6 Solutions
• 12th Commerce Maths Exercise 1.7 Solutions
• 12th Commerce Maths Exercise 1.8 Solutions
• 12th Commerce Maths Exercise 1.9 Solutions
• 12th Commerce Maths Exercise 1.10 Solutions
• 12th Commerce Maths Exercise Miscellaneous Exercise 1 Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

Std 12 Maths 1 Exercise 1.4 Solutions Commerce Maths

Question 1.
Write the following statements in symbolic form:
(i) If the triangle is equilateral, then it is equiangular.
Solution:
Let p : Triangle is equilateral.
q : It is equiangular.
Then the symbolic form of the given statement is p → q.

(ii) It is not true that ‘i’ is a real number.
Solution:
Let p : ‘i’ is a real number.
Then the symbolic form of the given statement is ~p.

(iii) Even though it is not cloudy, it is still raining.
Solution:
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is ~p ∧ q.

(iv) Milk is white if and only if the sky is not blue.
Solution:
Let p : Milk is white.
q : Sky is blue.
Then the symbolic form of the given statement is p ↔ (~q).

(v) Stock prices are high if and only if stocks are rising.
Solution:
Let p : Stock prices are high.
q : stocks are rising.
Then the symbolic form of the given statement is p ↔ q

(vi) If Kutub-Minar is in Delhi, then Taj Mahal is in Agra.
Solution:
Let p : Kutub-Minar is in Delhi.
q : Taj Mahal is in Agra.
Then the symbolic form of the given statement is p → q

Question 2.
Find the truth value of each of the following statements:
(i) It is not true that 3 – 7i is a real number.
Solution:
Let p : 3 – 7i be a real number.
Then the symbolic form of the given statement is ~p.
The truth value of p is F.
∴ the truth value of ~p is T. ….[~F ≡ T]

(ii) If a joint venture is a temporary partnership, then a discount on purchase is credited to the supplier.
Solution:
Let p : Joint venture is a temporary partnership.
q : Discount on purchases is credited to the supplier.
Then the symbolic form of the given statement is p → q.
The truth values of p and q are T and F respectively.
∴ the truth value of p → q is F. …..[T → F ≡ F]

(iii) Every accountant is free to apply his own accounting rules if and only if machinery is an asset.
Solution:
Let p : Every accountant is free to apply his own accounting rules.
q : Machinery is an asset.
Then the symbolic form of the given statement is p ↔ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ↔ q is F. ….[F ↔ T ≡ F]

(iv) Neither 27 is a prime number nor divisible by 4.
Solution:
Let p : 27 is a prime number.
q : 27 is divisible by 4.
Then the symbolic form of the given statement is ~p ∧ ~q.
The truth values of both p and q are F.
∴ the truth value of ~p ∧ ~q is T. …..[~F ∧ ~F ≡ T ∧ T ≡ T]

(v) 3 is a prime number and an odd number.
Solution:
Let p : 3 be a prime number.
q : 3 is an odd number.
Then the symbolic form of the given statement is p ∧ q
The truth values of both p and q are T.
∴ the truth value of p ∧ q is T. …..[T ∧ T ≡ T]

Question 3.
If p and q are true and r and s are false, find the true value of each of the following statements:
(i) p ∧ (q ∧ r)
Solution:
Truth values of p and q are T and truth values of r and s are F.
p ∧ (q ∧ r) ≡ T ∧ (T ∧ F)
≡ T ∧ F
≡ F
Hence, the truth value of the given statement is false.

(ii) (p → q) ∨ (r ∧ s)
Solution:
(p → q) ∨ (r ∧ s) ≡ (T → T) ∨ (F ∧ F)
≡ T ∨ F
≡ T
Hence, the truth value of the given statement is true.

(iii) ~[(~p ∨ s) ∧ (~q ∧ r)]
Solution:
~[(~p ∨ s) ∧ (~q ∧ r)] ≡ ~[(~ T ∨ F) ∧ (~T ∧ F)]
≡ ~[(F ∨ F) ∧ (F ∧ F)]
≡ ~(F ∧ F)
≡ ~F
≡ T
Hence, the truth value of the given statement is true.

(iv) (p → q) ↔ ~(p ∨ q)
Solution:
(p → q) ↔ ~(p ∨ q) = (T → T) ↔ ~(T ∨ T)
≡ T ↔ ~ (T)
≡ T ↔ F
≡ F
Hence, the truth value of the given statement is false.

(v) [(p ∨ s) → r] ∨ [~(p → q) ∨ s]
Solution:
[(p ∨ s) → r] ∨ ~[~(p → q) ∨ s]
≡ [(T ∨ F) → F] ∨ ~[ ~(T → T) ∨ F]
≡ (T → F) ∨ ~(~T ∨ F)
≡ F ∨ ~ (F ∨ F)
≡ F ∨ ~F
≡ F ∨ T
≡ T
Hence, the truth value of the given statement is true.

(vi) ~[p ∨ (r ∧ s)] ∧ ~[(r ∧ ~s) ∧ q]
Solution:
~[p ∨ (r ∧ s)] ∧ ~[(r ∧ ~s) ∧ q]
≡ ~[T ∨ (F ∧ F)] ∧ ~[(F ∧ ~F) ∧ T]
≡ ~[T ∨ F] ∧ ~[(F ∧ T) ∧ T]
≡ ~T ∧ ~(F ∧ T)
≡ F ∧ ~F
≡ F ∧ T
≡ F
Hence, the truth value of the given statement is false.

Question 4.
Assuming that the following statements are true:
p : Sunday is a holiday.
q : Ram does not study on holiday.
Find the truth values of the following statements:
(i) Sunday is not holiday or Ram studies on holiday.
Solution:
The symbolic form of the statement is ~p ∨ ~q.

∴ the truth value of the given statement is F.

(ii) If Sunday is not a holiday, then Ram studies on holiday.
Solution:
The symbolic form of the given statement is ~p → ~q.

∴ the truth value of the given statement is T.

(iii) Sunday is a holiday and Ram studies on holiday.
Solution:
The symbolic form of the given statement is p ∧ q.

∴ the truth value of the given statement is F.

Question 5.
If p : He swims.
q : Water is warm.
Give the verbal statements for the following symbolic statements:
(i) p ↔ ~q
Solution:
p ↔ ~ q
He swims if and only if the water is not warm.

(ii) ~(p ∨ q)
Solution:
~(p ∨ q)
It is not true that he swims or water is warm.

(iii) q → p
Solution:
q → p
If water is warm, then he swims.

(iv) q ∧ ~p
Solution:
q ∧ ~p
The water is warm and he does not swim.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 1.1 Solutions
• 12th Commerce Maths Exercise 1.2 Solutions
• 12th Commerce Maths Exercise 1.3 Solutions
• 12th Commerce Maths Exercise 1.4 Solutions
• 12th Commerce Maths Exercise 1.5 Solutions
• 12th Commerce Maths Exercise 1.6 Solutions
• 12th Commerce Maths Exercise 1.7 Solutions
• 12th Commerce Maths Exercise 1.8 Solutions
• 12th Commerce Maths Exercise 1.9 Solutions
• 12th Commerce Maths Exercise 1.10 Solutions
• 12th Commerce Maths Exercise Miscellaneous Exercise 1 Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

Std 12 Maths 1 Exercise 1.3 Solutions Commerce Maths

Question 1.
Write the negation of each of the following statements:
(i) All men are animals.
Solution:
Some men are not animals.

(ii) 3 is a natural number.
Solution:
-3 is not a natural number.

(iii) It is false that Nagpur is the capital of Maharashtra.
Solution:
Nagpur is the capital of Maharashtra.

(iv) 2 + 3 ≠ 5.
Solution:
2 + 3 = 5.

Question 2.
Write the truth value of the negation of each of the following statements:
(i) √5 is an irrational number.
Solution:
Let p : √5 is an irrational number.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(ii) London is in England.
Solution:
Let p : London is in England.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(iii) For every x ∈ N, x + 3 < 8.
Solution:
Let p : For every x ∈ N, x + 3 < 8.
The truth value of p is F.
Therefore, the truth value of ~p is T.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 1.1 Solutions
• 12th Commerce Maths Exercise 1.2 Solutions
• 12th Commerce Maths Exercise 1.3 Solutions
• 12th Commerce Maths Exercise 1.4 Solutions
• 12th Commerce Maths Exercise 1.5 Solutions
• 12th Commerce Maths Exercise 1.6 Solutions
• 12th Commerce Maths Exercise 1.7 Solutions
• 12th Commerce Maths Exercise 1.8 Solutions
• 12th Commerce Maths Exercise 1.9 Solutions
• 12th Commerce Maths Exercise 1.10 Solutions
• 12th Commerce Maths Exercise Miscellaneous Exercise 1 Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

Std 12 Maths 1 Exercise 1.2 Solutions Commerce Maths

Question 1.
Express the following statements in symbolic form:
(i) e is a vowel or 2 + 3 = 5.
Solution:
Let p : e is a vowel.
q: 2 + 3 = 5.
Then the symbolic form of the given statement is p ∨ q.

(ii) Mango is a fruit but potato is a vegetable.
Solution:
Let p : Mango is a fruit.
q : Potato is a vegetable.
Then the symbolic form of the given statement is p ∧ q.

(iii) Milk is white or grass is green.
Solution:
Let p : Milk is white.
q : Grass is green.
Then the symbolic form of the given statement is p ∨ q.

(iv) I like playing but not singing.
Solution:
Let p : I like playing.
q : I am not singing.
Then the symbolic form of the given statement is p ∧ q.

(v) Even though it is cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is cloudy and it is still raining.
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

Question 2.
Write the truth values of the following statements:
(I) Earth is a planet and Moon is a star.
Solution:
Let p : Earth is a planet.
q : Moon is a star.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. …[T ∧ F ≡ F]

(ii) 16 is an even number and 8 is a perfect square.
Solution:
Let p : 16 is an even number.
q : 8 is a perfect square.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. ….[T ∧ F ≡ F]

(iii) A quadratic equation has two distinct roots or 6 has three prime factors.
Solution:
Let p : A quadratic equation has two distinct roots.
q : 6 has three prime factors.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are F.
∴ the truth value of p ∨ q is F. …..[F ∨ F ≡ F]

(iv) The Himalayas are the highest mountains but they are part of India in the northeast.
Solution:
Let p : the Himalayas are the highest mountains.
q : They are part of India in the northeast.
Then the symbolic form of the given statement is p ∧ q.
The truth values of both p and q are T.
∴ the truth value of p ∧ q is T. …..[T ∧ T ≡ T]

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 1.1 Solutions
• 12th Commerce Maths Exercise 1.2 Solutions
• 12th Commerce Maths Exercise 1.3 Solutions
• 12th Commerce Maths Exercise 1.4 Solutions
• 12th Commerce Maths Exercise 1.5 Solutions
• 12th Commerce Maths Exercise 1.6 Solutions
• 12th Commerce Maths Exercise 1.7 Solutions
• 12th Commerce Maths Exercise 1.8 Solutions
• 12th Commerce Maths Exercise 1.9 Solutions
• 12th Commerce Maths Exercise 1.10 Solutions
• 12th Commerce Maths Exercise Miscellaneous Exercise 1 Solutions

Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

Std 12 Maths 1 Exercise 1.1 Solutions Commerce Maths

State which of the following sentences are statements. Justify your answer. In case of statements, write down the truth value:

Question (i).
A triangle has ‘ n’ sides.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (ii).
The sum of interior angles of a triangle is 180°.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (iii).
You are amazing!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question (iv).
Please grant me a loan.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (v).
√-4 is an irrational number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vi).
x² – 6x + 8 = 0 implies x = -4 or x = -2.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vii).
He is an actor.
Solution:
It is an open sentence, hence it is not a statement.

Question (viii).
Did you eat lunch yet?
Solution:
It is an interrogative sentence, hence it is not a statement.

Question (ix).
Have a cup of cappuccino.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (x).
(x + y)² = x² + 2xy + y² for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xi).
Every real number is a complex number.
Solution:
It is a statement that is true, hence its truth value is ‘T.

Question (xii).
1 is a prime number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xiii).
With the sunset, the day ends.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xiv).
1! = 0.
Solution:
It is a statement that is false, hence its truth value is

Question (xv).
3 + 5 > 11.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xvi).
The number π is an irrational number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xvii).
x² – y² = (x + y)(x – y) for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xviii).
The number 2 is only even a prime number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xix).
Two coplanar lines are either parallel or intersecting.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xx).
The number of arrangements of 7 girls in a row for a photograph is 7!
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxi).
Give me a compass box.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxii).
Bring the motor car here.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxiii).
It may rain today.
Solution:
It is an open sentence, hence it is not a statement.

Question (xxiv).
If a + b < 7, where a ≥ 0 and b ≥ 0, then a < 7 and b < 7.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxv).
Can you speak English?
Solution:
It is an interrogative sentence, hence it is not a statement.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 1.1 Solutions
• 12th Commerce Maths Exercise 1.2 Solutions
• 12th Commerce Maths Exercise 1.3 Solutions
• 12th Commerce Maths Exercise 1.4 Solutions
• 12th Commerce Maths Exercise 1.5 Solutions
• 12th Commerce Maths Exercise 1.6 Solutions
• 12th Commerce Maths Exercise 1.7 Solutions
• 12th Commerce Maths Exercise 1.8 Solutions
• 12th Commerce Maths Exercise 1.9 Solutions
• 12th Commerce Maths Exercise 1.10 Solutions
• 12th Commerce Maths Exercise Miscellaneous Exercise 1 Solutions

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.3 Questions and Answers.

Std 11 Maths 2 Exercise 6.3 Solutions Commerce Maths

Question 1.
Find n if ⁿP₆ : ⁿP₃ = 120 : 1
Solution:
ⁿP₆ : ⁿP₃ = 120 : 1
∴ \(\frac{n !}{(n-6) !} \div \frac{n !}{(n-3) !}=\frac{120}{1}\)
∴ \(\frac{\mathrm{n} !}{(\mathrm{n}-6) !} \times \frac{(\mathrm{n}-3) !}{\mathrm{n} !}\) = 120
∴ \(\frac{n !}{(n-6) !} \times \frac{(n-3)(n-4)(n-5)(n-6) !}{n !}\) = 120
∴ (n – 3) (n – 4) (n – 5) = 120
∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4
Comparing on both sides, we get
n – 3 = 6
∴ n = 9

Question 2.
Find m and n if ^(m+n)P₂ = 56 and ^(m-n)P₂ = 12
Solution:
^m+nP₂ = 56
∴ \(\frac{(\mathrm{m}+\mathrm{n}) !}{(\mathrm{m}+\mathrm{n}-2) !}\) = 56
∴ \(\frac{(m+n)(m+n-1)(m+n-2) !}{(m+n-2) !}\) = 56
∴ (m + n) (m + n – 1) = 8 × 7
Comparing on both sides, we get
m + n = 8 …..(i)
Also ^m-nP₂ = 12
∴ \(\frac{(m-n) !}{(m-n-2) !}\) = 12
∴ \(\frac{(m-n)(m-n-1)(m-n-2) !}{(m-n-2) !}\) = 12
∴ (m – n) (m – n – 1) = 4 × 3
Comparing on both sides, we get
m – n = 4 …..(ii)
Adding (i) and (ii), we get
2m = 12
∴ m = 6
Substituting m = 6 in (ii), we get
6 – n = 4
∴ n = 2

Question 3.
Find r if ¹²P_r-2 : ¹¹P_r-1 = 3 : 14
Solution:

∴ (14 – r)(13 – r) = 8 × 7
Comparing on both sides, we get
14 – r = 8
∴ r = 6

Question 4.
Show that (n + 1) ⁿP_r = (n – r + 1) ⁽ⁿ⁺¹⁾P_r.
Solution:

Question 5.
How many 4 letter words can be formed using letters in the word MADHURI if
(i) letters can be repeated?
(ii) letters cannot be repeated?
Solution:
There are 7 letters in the word MADHURI.
(i) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.
∴ 1st letter can be filled in 7 ways.
2nd letter can be filled in 7 ways.
3rd letter can be filled in 7 ways.
4th letter can be filled in 7 ways.
∴ Total no. of ways a 4-letter word can be formed = 7 × 7 × 7 × 7 = 2401
∴ 2401 four-lettered words can be formed when repetition of letters is allowed.

(ii) When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is
⁷P₄ = \(\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}\) = 840
∴ 840 four-letter words can be formed when repetition of letters is not allowed.

Alternate method:
There are 7 letters in the word MADHURI.
(i) Since letters can be repeated
∴ In all places of a four-letter word, any one of seven letters M, A, D, H, U, R, I can appear.
∴ Using the Multiplication theorem, we get
Number of four-letter words with repetition of letters M, A, D, H, U, R, I = 7 × 7 × 7 × 7 = 2401

(ii) Since the letters cannot be repeated therefore 1st, 2nd, 3rd, 4th places can be filled in 7, 6, 5, 4 ways respectively
∴ Using the multiplication theorem, we get
Number of four-letter words, with no repetition of letters M, A, D, H, U, R, I = 7 × 6 × 5 × 4 = 840

Question 6.
Determine the number of arrangements of letters of the word ALGORITHM if
(i) vowels are always together.
(ii) no two vowels are together.
(iii) Consonants are at even positions
(iv) O is first and T is last.
Solution:
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.
(i) When vowels are always together:
There are 3 vowels in the word ALGORITHM. (i.e, A, I, O)
Let us consider these 3 vowels as one unit.
This unit with 6 other letters is to be arranged.
∴ It becomes an arrangement of 7 things which can be done in ⁷P₇ i.e., 7! ways and 3 vowels can be arranged among themselves in ³P₃ i.e., 3! ways.
∴ the total number of ways in which the word can be formed = 7! × 3!
= 5040 × 6
= 30240
∴ 30240 words can be formed if vowels are always together.

(ii) When no two vowels are together:
There are 6 consonants in the word ALGORITHM.
They can be arranged among themselves in ⁶P₆ i.e., 6! ways.
Let consonants be denoted by C.
_C_C_ C_C_C_C_
6 consonants create 7 gaps in which 3 vowels are to arranged.
∴ 3 vowels can be filled in ⁷P₃
= \(\frac{7 !}{(7-3) !}\)
= \(\frac{7 \times 6 \times 5 \times 4 !}{4 !}\)
= 210 ways
∴ total number of ways in which the word can be formed = 6! × 210
= 720 × 210
= 151200
∴ 151200 words can be formed if no two vowels are together.

(iii) When consonants are at even positions:
There are 4 even places and 6 consonants in the word ALGORITHM.
1st, 2nd, 3rd, 4th even places are filled in 6, 5, 4, 3 way respectively.
∴ The number of ways to fill four even places by consonants = 6 × 5 × 4 × 3 = 360
The remaining 5 letters (3 vowels and 2 consonants) can be arranged among themselves in ⁵P₅ i.e., 5! ways.
∴ Total number of ways the words can be formed
In which even places are occupied by consonants = 360 × 5!
= 360 × 120
= 43200
∴ 43200 words can be formed if even positions are occupied by consonants.

(iv) When beginning with O and ends with T:
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
7 letters other than O and T can be filled between O and T in ⁷P₇ i.e., 7! ways = 5040 ways.
∴ 5040 words beginning with O and ending with T can be formed.

Question 7.
In a group photograph, 6 teachers and principals are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.
Solution:
In 1st row middle seat is fixed for the principal.
Also 1st row, 6 teachers can be arranged among themselves in ⁶P₆ i.e., 6! ways.
In the 2nd row, 12 boys can be arranged among themselves in ¹²P₁₂ i.e., 12! ways.
13 gaps are created by 12 boys, in which 6 girls are to be arranged.
together which can be done in ¹³P₆ ways.
∴ total number of arrangements = 6! × 12! × ¹³P₆ …..[using Multiplications Principle]
= 6! × 12! × \(\frac{13 !}{(13-6) !}\)
= 6! × 12! × \(\frac{13 !}{7 !}\)
= \(\frac{6 ! \times 12 ! \times 13 !}{7 \times 6 !}\)
= \(\frac{12 ! 13 !}{7}\)

Question 8.
Find the number of ways letters of the word HISTORY can be arranged if
(i) Y and T are together
(ii) Y is next to T.
Solution:
There are 7 letters in the word HISTORY
(i) When ‘Y’ and ‘T’ are together.
Let us consider ‘ Y’ and ‘T’ as one unit
This unit with the other 5 letters is to be arranged.
∴ The number of arrangements of one unit and 5 letters = ⁶P₆ = 6!
Also, ‘Y’ and ‘T’ can be arranged among themselves in ²P₂ i.e., 2! ways.
∴ a total number of arrangements when Y and T are always together = 6! × 2!
= 720 × 2
= 1440
∴ 1440 words can be formed if Y and T are together.

(ii) When ‘Y’ is next to ‘T’
Let us take this (‘Y’ next to ‘T’) as one unit.
This unit with 5 other letters is to be arranged.
∴ The number of arrangements of 6 letters and one unit = ⁶P₆ = 6!
Also ‘Y’ has to be always next to ‘T’.
So they can be arranged in 1 way.
∴ total number of arrangements possible when Y is next to T = 6! × 1 = 720
∴ 720 words can be formed if Y is next to T.

Question 9.
Find the number of arrangements of the letters in the word BERMUDA so that consonants and vowels are in the same relative positions.
Solution:
There are 7 letters in the word “BERMUDA” out of which 3 are vowels and 4 are consonants.
If relative positions of consonants and vowels are not changed.
3 vowels can be arranged among themselves in ³P₃ i.e., 3! ways.
4 consonants can be arranged among themselves in ⁴P₄ i.e., 4! ways.
∴ total no. of arrangements possible if relative positions of vowels and consonants are not changed = 3! × 4!
= 6 × 24
= 144

Question 10.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if
(i) digits can be repeated
(ii) digits cannot be repeated
Solution:
(i) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.
∴ The unit’s place digit can be filled in 6 ways.
10’s place digit can be filled in 6 ways.
100’s place digit can be filled in 6 ways.
1000’s place digit can be filled in 6 ways.
∴ total number of numbers = 6 × 6 × 6 × 6 = 6⁴ = 1296
∴ 1296 four-digit numbers can be formed if repetition of digits is allowed.

(ii) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.
∴ 4 different digits are to be arranged from 6 given digits which can be done in ⁶P₄
= \(\frac{6 !}{(6-4) !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360 ways
∴ 360 four-digit numbers can be formed, if repetition of digits is not allowed.

Question 11.
How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that the resulting numbers are between 100 and 1000?
Solution:
A number between 100 and 1000 is a 3 digit number and is to be formed from the digits 0, 1, 2, 3, 4, 5, without repetition of digits.
∴ 100’s place digit must be a non-zero number which can be filled in 5 ways.
10’s place digits can be filled in 5 ways.
Unit’s place digit can be filled in 4 ways.
∴ total number of ways the number can be formed = 5 × 5 × 4 = 100
∴ 100 numbers between 100 and 1000 can be formed.

Question 12.
Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are
(i) divisible by 5?
(ii) not divisible by 5?
Solution:
A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in ⁶P₆ i.e., 6! = 720 ways
(i) If the number is divisible by 5, then
The unit’s place digit must be 5, and hence unit’s place can be filled in 1 way
Other 5 digits can be arranged among themselves in ⁵P₅ i.e., 5! ways
∴ Total number of ways in which numbers divisible by 5 can be formed = 1 × 5! = 120

(ii) If the number is not divisible by 5, then
Unit’s place can be any digit from 3, 4, 6, 7, 8 which can be selected in 5 ways.
Other 5 digits can be arranged in ⁵P₅ i.e., 5! ways
∴ The total number of ways in which numbers not divisible by 5 can be formed = 5 × 5!
= 5 × 120
= 600

Question 13.
A code word is formed by two distinct English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.
Solution:
(i) There is a total of 26 alphabets.
A code word contains 2 English alphabets.
∴ 2 alphabets can be filled in ²⁶P₂
= \(\frac{26 !}{(26-2) !}\)
= \(\frac{26 \times 25 \times 24 !}{24 !}\)
= 650 ways
Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in ⁹P₂
= \(\frac{9 !}{(9-2) !}\)
= \(\frac{9 \times 8 \times 7 !}{7 !}\)
= 72 ways
∴ Total number of a code words = 650 × 72 = 46800

(ii) There are total 26 alphabets.
A code word contains 2 English alphabets.
∴ 2 alphabets can be filled in ²⁶P₂
= \(\frac{26}{(26-2) !}\)
= \(\frac{26 \times 25 \times 24 !}{24 !}\)
= 650 ways
For a code word to end with an even integer, the digit in the unit’s place should be an even number between 1 to 9 which can be filled in 4 ways.
Also, 10’s place can be filled in 8 ways.
∴ Total number of codewords = 650 × 4 × 8 = 20800 ways
∴ 20800 codewords end with an even integer.

Question 14.
Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
Solution:
There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.
∴ Each letter can be posted in 3 ways.
∴ Total number of ways in which 5 letters can be posted = 3 × 3 × 3 × 3 × 3 = 243

Question 15.
Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object
(i) always occurs
(ii) never occurs
Solution:
There are 11 distinct objects and 4 are to be taken at a time.
(i) The number of permutations of n distinct objects, taken r at a time, when one specified object will always occur is r × ^(n-1)P_(r-1)
Here, r = 4, n = 11
∴ The number of permutations of 4 out of 11 objects when a specified object occurs.
= 4 × ^(11-1)P_(4-1)
= 4 × ¹⁰P₃
= 4 × \(\frac{10 !}{(10-3) !}\)
= 4 × \(\frac{10 !}{7 !}\)
= 4 × \(\frac{10 \times 9 \times 8 \times 7 !}{7 !}\)
= 2880
∴ There are 2880 permutations of 11 distinct objects, taken 4 at a time, in which one specified object always occurs.

(ii) When one specified object does not occur then 4 things are to be arranged from the remaining 10 things, which can be done in ¹⁰P₄ ways
= 10 × 9 × 8 × 7 ways
= 5040 ways
∴ There are 5040 permutations of 11 distinct objects, taken 4 at a time, in which one specified object never occurs.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 6.1 Solutions
• 11th Commerce Maths Exercise 6.2 Solutions
• 11th Commerce Maths Exercise 6.3 Solutions
• 11th Commerce Maths Exercise 6.4 Solutions
• 11th Commerce Maths Exercise 6.5 Solutions
• 11th Commerce Maths Exercise 6.6 Solutions
• 11th Commerce Maths Exercise 6.7 Solutions
• 11th Commerce Maths  Miscellaneous Exercise 6 Solutions

Linear Inequations Class 11 Commerce Maths 2 Chapter 8 Exercise 8.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.1 Questions and Answers.

Std 11 Maths 2 Exercise 8.1 Solutions Commerce Maths

Question 1.
Write the inequations that represent the interval and state whether the interval is bounded or unbounded:
(i) [-4, \(\frac{7}{3}\)]
Solution:
[-4, \(\frac{7}{3}\)]
Here, x takes values between -4 and \(\frac{7}{3}\) including -4 and \(\frac{7}{3}\)
∴ the required inequation is -4 ≤ x ≤ \(\frac{7}{3}\)
∴ it is a bounded (closed) interval.

(ii) (0, 0.9]
Solution:
(0, 0.9]
Here, x takes values between 0 and 0.9, including 0.9 and excluding 0.
∴ the required inequation is 0 < x ≤ 0.9
∴ it is a bounded (semi-right closed) interval.

(iii) (-∞, ∞)
Solution:
(-∞, ∞)
Here, x takes values between -∞ and ∞
∴ the required inequation is -∞ < x < ∞
∴ it is an unbounded (open) interval.

(iv) [5, ∞)
Solution:
[5, ∞)
Here, x takes values between 5 and ∞ including 5.
∴ the required inequation is 5 ≤ x < ∞
∴ it is an unbounded (semi-left closed) interval.

(v) (-11, -2)
Solution:
(-11, -2)
Here, x takes values between -11 and -2
∴ the required inequation is -11 < x < -2
∴ it is a bounded (open) interval.

(vi) (-∞, 3)
Solution:
(-∞, 3)
Here, x takes values between -∞ and 3
∴ the required inequation is -∞ < x < 3
∴ it is an unbounded (open) interval.

Question 2.
Solve the following inequations
(i) 3x – 36 > 0
Solution:
3x – 36 > 0
Adding 36 both sides, we get
3x – 36 + 36 > 0 + 36
∴ 3x > 36
Dividing both sides by 3, we get
\(\frac{3 x}{3}>\frac{36}{3}\)
∴ x > 12
∴ x takes all real values more than 12
∴ Solution set = (12, ∞)

(ii) 7x – 25 ≤ -4
Solution:
7x – 25 ≤ -4
Adding 25 on both sides, we get
7x – 25 + 25 ≤ -4 + 25
∴ 7x ≤ 21
Dividing both sides by 7, we get
x ≤ 3
∴ x takes all real values less or equal to 3.
∴ Solution Set = (-∞, 3]

(iii) 0 < \(\frac{x-5}{4}\) < 3
Solution:
0 < \(\frac{x-5}{4}\) < 3
0 < x – 5 < 12
Adding 5 on both sides, we get
5 < x < 17
x takes all real values between 5 and 17.
∴ Solution set = (5, 17)

(iv) |7x – 4| < 10
Solution:
|7x – 4| < 10
-10 < 7x – 4 < 10 …….[|x| < k is same as -k < x < k]
Adding 4 on both sides, we get
-6 < 7x < 14
Dividing both sides by 7, we get
\(-\frac{6}{7}<x<\frac{14}{7}\)
∴ \(-\frac{6}{7}\) < x < 2
∴ x takes all real values between \(-\frac{6}{7}\) and 2.
∴ Solution set = (\(-\frac{6}{7}\), 2)

Question 3.
Sketch the graph which represents the solution set for the following inequations:
(i) x > 5
Solution:
x > 5
Here, x takes all real values that are greater than 5.
∴ Solution set represents the unbounded (open) interval (5, ∞)
∴ the required graph of the solution set is as follows:

(ii) x ≥ 5
Solution:
x ≥ 5
Here, x takes all real values that are greater than or equal to 5
∴ Solution set represents the unbounded (semi-left closed) interval [5, ∞)
∴ the required graph of the solution set is as follows:

(iii) x < 3
Solution:
x < 3
Here, x takes all real values that are less than 3.
∴ Solution set represents the unbounded (open) interval (-∞, 3)
∴ the required graph of the solution set is as follows:

(iv) x ≤ 3
Solution:
x ≤ 3
Here, x takes all real values less than and including 3
∴ Solution set represents the unbounded (semi-right closed) interval (-∞, 3]
∴ the required graph of the solution set is as follows:

(v) -4 < x < 3
Solution:
-4 < x < 3
Here, x takes all real values between -4 and 3.
∴ Solution set represents the bounded (open) interval (-4, 3)
∴ the required graph of the solution set is as follows:

(vi) -2 ≤ x < 2.5
Solution:
-2 ≤ x < 2.5
Here, x takes all values between -2 and 2.5 including -2.
∴ Solution set represents the bounded (semi-left closed) interval [-2, 2.5)
∴ the required graph of the solution set is as follows.

(vii) -3 ≤ x ≤ 1
Solution:
-3 ≤ x ≤ 1
Here, x takes all real values between -3 and 1 including -3 and 1
∴ Solution set represents the bounded (closed) interval [-3, 1]
∴ the required graph of the solution set is as follows:

(viii) |x| < 4
Solution:
|x| < 4 ⇒ -4 < x < 4
Here, x takes all real values between -4 and 4.
∴ Solution set represents bounded (open) interval (-4, 4)
∴ the required graph of the solution set is as follows:

(ix) |x| ≥ 3.5
Solution:
|x| ≥ 3.5 ⇒ x ≥ 3.5 or x ≤ -3.5
Here, x takes values greater than or equal to 3.5 or it takes values less than or equal to -3.5
∴ Solution set represents the unbounded (semi-left closed) interval [3.5, ∞) or the unbounded (semi-right closed) interval (-∞, -3.5]
∴ x ∈ (-∞, -3.5] ∪ [3.5, ∞)
∴ the required graph of the solution set is as follows:

Question 4.
Solve the inequations:
(i) 5x + 7 > 4 – 2x
Solution:
5x + 7 > 4 – 2x
Adding 2x on both sides, we get
7x + 7 > 4
Subtracting 7 from both sides, we get
7x > -3
Dividing by 7 on both sides, we get
∴ x > \(-\frac{3}{7}\)
i.e., x takes all real values greater than \(-\frac{3}{7}\)
∴ the solution set is (\(-\frac{3}{7}\), ∞)

(ii) 3x + 1 ≥ 6x – 4
Solution:
3x + 1 ≥ 6x – 4
Subtracting 3x from both sides, we get
1 ≥ 3x – 4
Adding 4 on both sides, we get
5 ≥ 3x
Dividing by 3 on both sides, we get
\(\frac{5}{3}\) ≥ x
i.e., x ≤ \(\frac{5}{3}\)
i.e., x takes all real values less than or equal to \(\frac{5}{3}\).
∴ the solution set is (-∞, \(\frac{5}{3}\)]

(iii) 4 – 2x < 3(3 – x)
Solution:
4 – 2x < 3(3 – x)
∴ 4 – 2x < 9 – 3x
Adding 3x on both sides, we get
4 + x < 9
Subtracting 4 from both sides, we get
x < 5
i.e., x takes all real values less than 5
∴ the solution set is (-∞, 5)

(iv) \(\frac{3}{4}\)x – 6 ≤ x – 7
Solution:
\(\frac{3}{4}\)x – 6 ≤ x – 7
Multiplying by 4 on both sides, we get
3x – 24 ≤ 4x – 28
Subtracting 3x from both sides, we get
-24 ≤ x – 28
Adding 28 on both the sides, we get
∴ 4 ≤ x i.e., x ≥ 4
i.e., x takes all real values greater or equal to 4.
∴ the solution set is [4, ∞)

(v) -8 ≤ -(3x – 5) < 13
Solution:
-8 < -(3x – 5) < 13 Multiplying by -1 throughout (so inequality sign changes) 8 ≥ 3x – 5 > -13
i.e., -13 < 3x – 5 ≤ 8
Adding 5 on both the sides, we get
-8 < 3x ≤ 13
Dividing, by 3 on both sides, we get
∴ \(-\frac{8}{3}\) < x ≤ \(\frac{13}{3}\)
i.e., x takes all real values between \(-\frac{8}{3}\) and \(\frac{13}{3}\) including \(\frac{13}{3}\).
∴ the solution set is \(\left(-\frac{8}{3}, \frac{13}{3}\right]\)

(vi) -1 < 3 – \(\frac{x}{5}\) ≤ 1
Solution:
-1 < 3 – \(\frac{x}{5}\) ≤ 1
Subtracting 3 from both sides, we get
-4 < –\(\frac{x}{5}\) < -2 Multiplying by -1 throughout (so inequality sign changes) ∴ 4 > \(\frac{x}{5}\) > 2
i.e., 2 < \(\frac{x}{5}\) < 4
Multiplying by 5 on both sides, we get
10 < x < 20
i.e., x takes all real values between 10 and 20.
∴ the solution set is (10, 20)

(vii) 2|4 – 5x| ≥ 9
Solution:
2|4 – 5x | ≥ 9
∴ |4 – 5x| ≥ \(\frac{9}{2}\)
∴ 4 – 5x ≥ \(\frac{9}{2}\) or 4 – 5x ≤ –\(\frac{9}{2}\) ……[|x| ≥ a implies x ≤ -a or x ≥ a]
Subtracting 4 from both sides, we get
-5x ≥ \(\frac{1}{2}\) or -5x ≤ \(\frac{-17}{2}\)
Divide by -5 (so inequality sign changes)
∴ x ≤ \(-\frac{1}{10}\) or x ≥ \(\frac{17}{10}\)
∴ x takes all real values less than or equal to \(-\frac{1}{10}\)
or it takes all real values greater or equal to \(\frac{17}{10}\).
∴ the solution set is (-∞, \(-\frac{1}{10}\)] or [\(\frac{17}{10}\), ∞)

(viii) |2x + 7| ≤ 25
Solution:
|2x + 7| < 25
∴ -25 ≤ 2x + 7 ≤ 25 …..[|x| ≤ a implies -a ≤ x ≤ a]
Subtracting 7 from both sides, we get
-32 ≤ 2x ≤ 18
Dividing by 2 on both sides, we get
-16 ≤ x ≤ 9
∴ x can take all real values between -16 and 9 including -16 and 9.
∴ the solution set is [-16, 9]

(ix) 2|x + 3| > 1
Solution:
2|x + 3| > 1
Dividing by 2 on both sides, we get
|x + 3| > \(\frac{1}{2}\)
∴ x + 3 < –\(\frac{1}{2}\) or x + 3 > \(\frac{1}{2}\) …..[|x| > a implies x < -a or x > a]
Subtracting 3 from both sides, we get
x < – 3 – \(\frac{1}{2}\) or x > -3 + \(\frac{1}{2}\)
∴ x < \(\frac{-7}{2}\) or x > \(\frac{-5}{2}\)
∴ x can take all real values less \(\frac{-7}{2}\) or it can take values greater than \(\frac{-5}{2}\).
∴ Solution set is (-∞, \(\frac{-7}{2}\)) ∪ (\(\frac{-7}{2}\), ∞)

(x) \(\frac{x+5}{x-3}\) < 0
Solution:
\(\frac{x+5}{x-3}\) < 0
Since \(\frac{a}{b}\) < 0, when a > 0 and b < 0 or a < 0 and b > 0
∴ either x + 5 > 0 and x – 3 < 0
or x + 5 < 0 and x – 3 > 0
Case I:
x + 5 > 0 and x – 3 < 0 ∴ x > -5 and x < 3
∴ -5 < x < 3
∴ solution set = (-5, 3)
Case II:
x + 5 < 0 and x – 3 > 0
∴ x < -5 and x > 3
which is not possible
∴ solution set = Φ
∴ solution set of the given inequation is (-5, 3)

(xi) \(\frac{x-2}{x+5}\) > 0
Solution:
\(\frac{x-2}{x+5}\) > 0
Since \(\frac{a}{b}\) > 0,
when a > 0 and b > 0 or a < 0 and b < 0 b
∴ either x – 2 > 0 and x + 5 > 0
or x – 2 < 0 and x + 5 < 0 Case I: x – 2 > 0 and x + 5 > 0
∴ x > 2 and x > -5
∴ x > 2
∴ solution set = (2, ∞)
Case II:
x – 2 < 0 and x + 5 < 0
∴ x < 2 and x < -5
∴ x < -5
∴ solution set = (-∞, -5)
∴ the solution set of the given inequation is (-∞, -5) ∪ (2, ∞)

Question 5.
Rajiv obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let x₁, x₂, x₃ denote the marks in 1st, 2nd and 3rd unit test respectively. Then
\(\frac{x_{1}+x_{2}+x_{3}}{3}\) ≥ 60
∴ \(\frac{70+75+x_{3}}{3}\) ≥ 60
∴ 145 + x₃ ≥ 3(60)
Subtracting 145 from both sides, we get
x₃ ≥ 180 – 145
∴ x₃ ≥ 35
Rajiv must obtain a minimum of 35 marks to maintain an average of at least 60 marks.

Question 6.
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in the first four examinations are 87, 92, 94, and 95, find the minimum marks that Sunita must obtain in the fifth examination to get a grade ‘A’ in the course.
Solution:
Let x₁, x₂, x₃, x₄, x₅ denote the marks in five examinations. Then
\(\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}\) ≥ 90
∴ \(\frac{87+92+94+95+x_{5}}{5}\) ≥ 90
∴ 368 + x₅ ≥ 450
Subtracting 368 from both sides, we get
∴ x₅ ≥ 82
Sunita must obtain a minimum of 82 marks in the 5th examination to get a grade of A.

Question 7.
Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let two consecutive positive integers be 2n – 1, 2n + 1 where n ≥ 1 ∈ Z,
Given that 2n – 1 < 10 and 2n + 1 < 10
∴ 2n < 11 and 2n < 9
∴ 2n < 9
∴ n < \(\frac{9}{2}\) …..(i) Also, (2n – 1) + (2n + 1) > 11
∴ 4n > 11
∴ n > \(\frac{11}{4}\) …….(ii)
From (i) and (ii)
\(\frac{11}{4}<n<\frac{9}{2}\) Since, n is an integer,
∴ n = 3, 4
n = 3 gives 2n – 1 = 5, 2n + 1 = 7
and n = 4 gives 2n – 1 = 7, 2n + 1 = 9
∴ The pairs of positive consecutive integers are (5, 7) and (7, 9).

Question 8.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Solution:
Let 2n, 2n + 2 be two positive consecutive integers where n ≥ 1 ∈ Z.
Given that 2n > 5 and 2n + 2 > 5
∴ n > \(\frac{5}{2}\) and 2n > 3
∴ n > \(\frac{5}{2}\) and n > \(\frac{3}{2}\)
∴ n > \(\frac{5}{2}\) ……(i)
Also (2n) + (2n + 2) < 23
∴ 4n + 2 < 23
∴ 4n < 21
∴ n < \(\frac{21}{4}\) ……(ii)
From (i) and (ii)
\(\frac{5}{2}<n<\frac{21}{4}\) and n is an integer.
∴ n = 3, 4, 5
n = 3 gives 2n = 6, 2n + 2 = 8
n = 4 gives 2n = 8, 2n + 2 = 10
n = 5 gives 2n = 10, 2n + 2 = 12
∴ The pairs of positive even consecutive integers are (6, 8) (8, 10), (10, 12)

Question 9.
The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum integer length of the shortest side.
Solution:
Let the shortest side be x.
Then longest side length = 2x
and third side length = x + 2
Perimeter = x + 2x + x + 2 = 4x +2
Given, perimeter > 166
∴ 4x + 2 > 166
∴ 4x > 164
∴ x > 41
∴ Minimum integer length of shortest side is 42 cm.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 8.1 Solutions
• 11th Commerce Maths Exercise 8.2 Solutions
• 11th Commerce Maths Exercise 8.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 8 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.2 Questions and Answers.

Std 11 Maths 2 Exercise 9.2 Solutions Commerce Maths

Question 1.
Mr. Sarad purchased a laptop for ₹ 24,000 and sold it for ₹ 30,000. What was the profit percentage?
Solution:
Cost price (C.P.) = ₹ 24000
Selling price (S. P.) = ₹ 30,000
Profit = S.P. – C.P.
= 30,000 – 24,000
= 6,000

∴ Profit Percentage = 25%

Question 2.
Shraddha purchased a mobile phone and refrigerator for ₹ 18,000 and ₹ 15,000 respectively. She sold the refrigerator at a loss of 20% and the mobile at a profit of 20%. What is her overall profit or loss?
Solution:
C.P. of mobile phone = ₹ 18,000
Profit percentage on mobile phone = 20%
Selling price (S.P.) of mobile phone = 18,000 (1 + \(\frac{20}{100}\))
= 18,000 (1 + \(\frac{1}{5}\))
= 18,000 × \(\frac{6}{5}\)
= ₹ 21,600
C.P. of refrigerator = 15,000
Loss percentage on refrigerator = 20%
∴ Selling price (S.P.) = 15,000(1 – \(\frac{20}{100}\))
= 15,000(1 – \(\frac{1}{5}\))
= 15,000 × \(\frac{4}{5}\)
= ₹ 12,000
∴ Total gelling price for the transaction = 21,600 + 12,000 = ₹ 33,600
Total cost price (purchase price) for the transaction = 18,000 + 15,000 = ₹ 33,000
∴ Overall profit made by Shraddha = Total S.P. – Total C.P.
= 33,600 – 33,000
= ₹ 600
Thus, Shraddha made on overall profit of ₹ 600.

Question 3.
A vendor bought toffees at 6 for ₹ 10. How many for ₹ 10 must he sell to gain 20%?
Solution:
Vendor bought toffees at the rate of 6 for ₹ 10
∴ Cost price of one toffee = \(\frac{10}{6}\)
i.e. C.P. = \(\frac{10}{6}\) …….(i)
Let x be the number of toffees he must sell in ₹ 10 to gain 20%
i.e. S.P. = \(\frac{10}{x}\) …….(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }}\)
Using (i) and (ii) we have

∴ 30(6 – x) = 6x
∴ 180 – 30x = 6x
∴ 36x = 180
∴ x = 5
The vendor must sell 5 toffees for ₹ 10 in order to gain 20%.

Question 4.
The percentage profit earned by selling an article for ₹ 2,880 is equal to the percentage loss incurred by selling the same article for ₹ 1,920. At what price the article should be sold to earn a 25% profit?
Solution:
Let x be C.P. of the article
Let y % be both, the gain and loss made when article is sold at ₹ 2,880 and ₹ 1,920 respectively. Then
x + \(\frac{y}{100}\) x = 2880 ……(i)
x – \(\frac{y}{100}\) x = 1920 …..(ii)
Adding (i) and (ii), we get
2x = 4800
∴ x = 2400
i.e. C.P. of the article = ₹ 2400
Required profit percentage = 25%

∴ The article should be sold at ₹ 3000 to earn 25% profit.

Question 5.
A cloth merchant advertises for selling cloth at a 4% loss. By using a faulty meter scale, he is earning a profit of 20%. What is the actual length of the scale?
Solution:
Let the cost price of the cloth be ₹ ‘x’ per meter
He claims a loss of 4%
∴ Selling price of the cloth
S.P. = C.P.(1 – \(\frac{loss%}{100}\))
= x(1 – \(\frac{4}{100}\))
= 0.96x …..(i)
The actual cost price of the cloth is lower as the cloth is measured by a faulty meter scale.
Given that shopkeeper’s profit = 20%

∴ The actual cost price is 0.8 times the cost price as advertised.
In other words, the meter scale used for the fraud is 0.8 times the meter scale that should have been used.
∴ The length of the faulty meter scale used = 0.8 × 1 = 0.8 meter
∴ The actual length of the scale is 0.8 meters.

Question 6.
Sunil sells his bike worth ₹ 25,000 to Rohit at a profit of 20%. After 6 months Rohit sells the bike back to Sunil at a loss of 20%. Find the total profit percent of Sunil considering both the transactions.
Solution:
Sunil sells his bike to Rohit at 20% profit.
So S.P. of the bike for Sunil
= 25000 + \(\frac{20}{100}\) × 25000
= 25000 + 5000
= 30000
∴ Cost price of bike to Rohit = ₹ 30000
Rohit sells the bike back to Sunil at 20% loss
∴ S.P. of the bike for Rohit = 30000 – \(\frac{20}{100}\) × 30000
= 30000 – 6000
= 24000
∴ In second transaction Sunil pays 24000 to Rohit
In the first transaction, he had received 30000 from Rohit
∴ Sunil made a profit of ₹ (30000 – 24000) = ₹ 6000
Sunil earned this profit on the bike which costed him ₹ 25000
∴ Total profit % that Sunil makes = \(\frac{6000}{25000}\) × 100
= \(\frac{600}{25}\)
= 24
∴ Sunil makes 24% profit considering both the transactions.

Question 7.
By selling a book at ₹ 405 bookseller incurs a loss of 25%. Find the cost price of the book.
Solution:
S.P. = ₹ 405
Loss% = 25
S.P. when there is a loss is given by
S.P. = C.P. × \(\frac{\text { Loss } \%}{100}\)

∴ The cost price of the book is ₹ 540.

Question 8.
A cloth costs ₹ 675. If it is sold at a loss of 20%, what is its cost price as a percentage of its selling price?
Solution:
C.P. = ₹ 675
Loss% = 20%
∴ Loss made in selling = \(\frac{20}{100}\) × 675 = ₹ 135
S.P. = C.P. – Loss
= 675 – 135
= ₹ 540
Let C.P. be x % S.P.,
Then 675 = \(\frac{x}{100}\) × 540
∴ x = \(\frac{675 \times 100}{540}\) = 125
∴ Cost price is 125% of the selling price.

Question 9.
Ashwin buys an article for ₹ 500. He marks it for sale at 75% more than the cost price. He offers a 25% discount on the marked price to his customer. Calculate the actual percentage of profit made by Ashwin.
Solution:
C.P. = ₹ 500
Marked price = C.P. + \(\frac{75}{100}\) × C.P.
= \(\frac{75}{100}\) × 500
= 500 + 75 × 5
= 500 + 375
= 875
25% discount was given on marked price
∴ Discount = \(\frac{25}{100}\) × 875 = \(\frac{875}{4}\)
Selling price = marked price – discount

∴ Ashwin makes 31.25% profit.

Question 10.
The combined cost price of a refrigerator and a mixer is ₹ 12,400. If the refrigerator costs 600% more than the mixer, find the cost price of the mixer.
Solution:
Let ₹ x be the cost price of the mixer.
The cost price of the refrigerator = x + \(\frac{600}{100}\) x
= x + 6x
= 7x
Total cost price =12400 …..[Given]
i.e. x + 7x = 12400
i.e. 8x = 12400
∴ x = \(\frac{12400}{8}\) = 1550
∴ The cost price of mixer is ₹ 1550.

Question 11.
Find the single discount equivalent to the discount series of 5%, 7%, and 9%.
Solution:
Let the marked price be ₹ 100
After 1st discount the price = 100(1 – \(\frac{5}{100}\)) = 95
After 2nd discount the price = 95(1 – \(\frac{7}{100}\)) = \(\frac{95 \times 93}{100}\)
After 3rd discount the price = \(\frac{95 \times 93}{100}\left(1-\frac{9}{100}\right)\)
= \(\frac{95 \times 93 \times 91}{100 \times 100}\)
= \(\frac{803985}{10000}\)
= 80.3985 ~ 80.4
Selling price after 3 discounts is ₹ 80.4
Single equivalent discount = Marked price – Selling price
= 100 – 80.4
= ₹ 19.6
∴ Single equivalent discount is 19.6%.

Question 12.
The printed price of a shirt is ₹ 390. Lokesh pays ₹ 175.50 for it after getting two successive discounts. If the first discount is 10%, find the second discount.
Solution:
Marked price = ₹ 390
After the first discount of 10%, the price of the shirt

∴ x = 50
∴ Second discount is 50%

Question 13.
Amar, a manufacturer, gives a discount of 25% on the list price to his distributor Akbar, Akbar sells at a 10% discount on the list price to his customer Anthony. Anthony paid ₹ 540 for the article. What is the profit percentage of Akbar on his cost price?
Solution:
Let ₹ ‘x’ be the list price of the article.
Amar gives a discount of 25% on the list price.
∴ Selling price for Amar = \(x\left(1-\frac{25}{100}\right)\)
= \(x\left(1-\frac{1}{4}\right)\)
= ₹ \(\frac{3 x}{4}\)
Amar sells the article to Akbar
Cost price of article for Akbar = ₹ \(\frac{3 x}{4}\) ……(i)
Akbar sells the article to Anthony at 10% discount on list price
∴ Selling price for Akbar = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(1-\frac{1}{10}\right)\)
= ₹ \(\frac{9 x}{10}\) …..(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }} \times 100\)
Using (i) and (ii), we have the profit percentage for Akbar as,

∴ Akbar gets a profit of 20% on his cost price.

Question 14.
A man sells an article at a profit of 25%. If he had bought it at a 10% loss and sold it for ₹ 7 less, he would have gained 35%. Find the cost price of the article.
Solution:
Let ₹ ‘x’ be the C.P. of the article
∴ Article was sold at 25% profit
∴ S.P. of the article = \(x\left(1+\frac{25}{100}\right)\)
= \(x\left(1+\frac{1}{4}\right)\)
= 1.25x
If the article was bought at 10% loss
i.e., the new C.P. = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(\frac{9}{10}\right)\)
= 0.9x
and sold at ₹ 7 less
∴ New S.P. = 1.25x – 7
Then, the profit would have been 35%
Using profit percentage = \(\frac{\text { S.P.-C.P. }}{\text { C.P. }} \times 100\)

∴ Cost price of the article is ₹ 200

Question 15.
Mr. Mehta sold his two luxury cars at ₹ 39,10,000 each. On one he gains 15% but on the other, he loses 15%. How much does he gain or lose in the whole transaction?
Solution:
Let x, y be the C.P. of two cars.
S.P. of both the cars = 39,10,000 …..[Given]
∴ One car is sold at 15% loss
∴ S.P. of the first car = x – \(\frac{15}{100}\)x
∴ \(\frac{85}{100}\)x = 39,10,000
∴ x = \(\frac{39,10,000 \times 100}{85}\)
∴ x = 46,000 × 100
∴ x = 46,00,000
Other car is sold at 15% gain
∴ S.P. of second car = y + \(\frac{15}{100}\) y
∴ y + \(\frac{15}{100}\) y = 39,10,000
∴ \(\frac{115}{100}\)y = 39,10,000
∴ y = \(\frac{39,10,000 \times 100}{115}\)
∴ y = 34,000 × 100
∴ y = 34,00,000
x + y = Total C.P. of two cars
= 46,00,000 + 34,00,000
= 80,00,000
Total S.P. = 39,10,000 + 39,10,000 = 78,20,000
∴ S.P. < C.P.
∴ There is a loss of ₹ (80,00,000 – 78,20,000) = ₹ 1,80,000
∴ Loss % = \(\frac{1,80,000}{80,00,000} \times 100\)
= \(\frac{18}{8}\)
= 2.25
∴ Mr. Mehta bears a 2.25% loss in the whole transaction.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Measures of Dispersion Class 11 Commerce Maths 2 Chapter 2 Exercise 2.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.3 Questions and Answers.

Std 11 Maths 2 Exercise 2.3 Solutions Commerce Maths

Question 1.
The mean and standard deviation of two distributions of 100 and 150 items are 50, 5, and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Solution:


∴ The mean and standard deviation of all 250 items taken together are 44 and √55.6 respectively.

Question 2.
For certain bivariate data, the following information is available.

Obtain the combined standard deviation.
Solution:

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are: 2, 4, 6, 8, 10. (Given: √8 = 2.8284)
Solution:

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:
Given, \(\bar{x}\) = 25, σ = 5
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
= 100 × \(\frac{5}{25}\)
= 20%

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variation of 2.5%. What is the standard deviation of their height?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
∴ 2.5 = 100 × \(\frac{\sigma}{150.4}\)
∴ \(\frac{2.5 \times 150.4}{100}\) = σ
∴ σ = 3.76
∴ the standard deviation of students’ height is 3.76.

Question 6.
Two workers on the same job show1 the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:

(i) Since, C.V. (P) < C.V.(Q)
∴ Worker P is more consistent regarding the time required to complete the job.

(ii) Since, \(\bar{p}\) > \(\bar{q}\)
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:

(i) Since, \(\bar{x}_{1}>\bar{x}_{2}\)
i.e., average weekly wages are more for the first department.
∴ the first department has a larger bill.
(ii) Since, C.V. (1) < C.V. (2)
∴ the second department is less consistent.
∴ the second department has larger variability in wages.

Question 8.
The following table gives the weights of the students of class A. Calculate the coefficient of variation (Given √8 = 0.8944)

Solution:

Question 9.
Compute coefficient of variation for team A and team B. (Given: √2.5162 = 1.5863, √2.244 = 1.4980)

Which team is more consistent?
Solution:
Let f₁ denote no. of goals of team A and f₂ denote no. of goals of team B.


Since C.V. of team A > C.V. of team B.
∴ team B is more consistent.

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Solution:

Since C.V. (S) > C.V. (M)
∴ The subject statistics show higher variability in marks.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 2.1 Solutions
• 11th Commerce Maths Exercise 2.2 Solutions
• 11th Commerce Maths Exercise 2.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 2 Solutions