Practice Set 2.2 Algebra 9th Standard Maths Part 1 Chapter 2 Real Numbers Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Practice Set 2.2 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 2.2 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
Show that 4√2 is an irrational number.
Solution:
Let us assume that 4√2 is a rational number .
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
4√2 = \(\frac { a }{ b }\)
∴ √2 = \(\frac { a }{ 4b }\)
Since, a and b are integers, \(\frac { a }{ 4b }\) is a rational number and so √2 is a rational number.

Alternate Proof:
Let us assume that 4√2 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.2 1
Since, 32 divides a2, so 32 divides ‘a’ as well.
So, we write a = 32c, where c is an integer.
∴ a2 = (32c)2 … [Squaring both the sides]
∴ 32b2 = 32 x 32c2 …[From(i)]
∴ b2 = 32c2
∴ c2 = \(\frac { { b }^{ 2 } }{ 32 }\)
Since, 32 divides b2, so 32 divides ‘b’.
∴ 32 divides both a and b.
a and b have at least 32 as a common factor.
But this contradicts the fact that a and b have no common factor other than 1.
∴ Our assumption that 4√2 is a rational number is wrong.
∴ 4√2 is an irrational number.

Question 2.
Prove that 3 + √5 is an irrational number.
Solution:
Let us assume that 3 + √5 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.2 2
Since, a and b are integers, \(\frac { a }{ b }\) – 3 is a rational
number and so √5 is a rational number.
But this contradicts the fact that √5 is an irrational number.
∴ Our assumption that 3 – √5 is a rational number is wrong.
3 + √5 is an irrational number.

Question 3.
Represent the numbers √5 and √10 on a number line.
Solution:
i. Draw a number line and take point A at 2.
Draw AB perpendicular to the number line such that AB = 1 unit.
In ∆OAB, m∠OAB = 90°
∴ (OB)2 = (OA)2 + (AB)2 … [Pythagoras theorem]
= (2)2 + (1)2
∴ (OB)2 = 5
∴ OB = √5 units. … [Taking square root of both sides]
With O as centre and radius equal to OB, draw an arc to intersect the number line at C.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.2 3
The coordinate of the point C is √5 .

ii. Draw a number line and take point Pat 3.
Draw PR perpendicular to the number line such that PR = 1 unit.
In ∆OPR, m∠OPR = 90°
∴ (OR)2 = (OP)2 + (PR)2 … [Pythagoras theorem]
= (3)2 + (1)2
∴ (OR)2 = 10
∴ OR= √10units. … [Taking square root of both sides]
With O as centre and radius equal to OR, draw an arc to intersect the number line at Q.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.2 4
The coordinate of the point Q is √10 .

Question 4.
Write any three rational numbers between the two numbers given below.
i. 0.3 and – 0.5
ii. – 2.3 and – 2.33
iii. 5.2 and 5.3
iv. – 4.5 and – 4.6
Solution:
i. 0.3 = 0.30 and -0.5 = -0.50
We know that,
0. 30 >0.29 >….. >0.10>.. > – 0.10>…. > -0.30>…> -0.50
∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

Alternate Method:
A rational number between two rational numbers a and b
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.2 5
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.2 6
∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

ii. -2.3 = -2.300 and -2.33 = -2.330
We know that,
-2.300 > -2.301>… > -2.310>…> -2.320>…> -2.330
∴ the three rational numbers between -2.3 and -2.33 are -2.310, -2.320 and -2.325.

iii. 5.2 = 5.20 and 5.3 = 5.30
We know that,
5.20 < 5.21 < 5.22 < 5.23 < … < 5.30
∴ the three rational numbers between 5.2 and 5.3 are 5.21, 5.22 and 5.23.

iv. -4.5 = -4.50 and -4.6 = -4.60 We know that,
-4.50 > -4.51 > -4.52 >… > – 4.55 >…>- 4.60
∴ the three rational numbers between -4.5 and -4.6 are -4.51, -4.52 and -4.55.

Class 9 Maths Digest

Practice Set 2.3 Algebra 9th Standard Maths Part 1 Chapter 2 Real Numbers Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Practice Set 2.3 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 2.3 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
State the order of the surds given below.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 1
Answer:
i. 3, ii. 2, iii. 4, iv. 2, v. 3

Question 2.
State which of the following are surds Justify. [2 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 2
Answer:
i. \(\sqrt [ 3 ]{ 51 }\) is a surd because 51 is a positive rational number, 3 is a positive integer greater than 1 and \(\sqrt [ 3 ]{ 51 }\) is irrational.

ii. \(\sqrt [ 4 ]{ 16 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 3
= 2, which is not an irrational number.

iii. \(\sqrt [ 5 ]{ 81 }\) is a surd because 81 is a positive rational number, 5 is a positive integer greater than 1 and \(\sqrt [ 5 ]{ 81 }\) is irrational.

iv. \(\sqrt { 256 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 4
= 16, which is not an irrational number.

v. \(\sqrt [ 3 ]{ 64 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 5
= 4, which is not an irrational number.

vi. \(\sqrt { \frac { 22 }{ 7 } }\) is a surd because \(\frac { 22 }{ 7 }\) is a positive rational number, 2 is a positive integer greater than 1 and \(\sqrt { \frac { 22 }{ 7 } }\) is irrational.

Question 3.
Classify the given pair of surds into like surds and unlike surds. [2 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 6
Solution:
If the order of the surds and the radicands are same, then the surds are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 7
Here, the order of 2\(\sqrt { 13 }\) and 5\(\sqrt { 13 }\) is same and their radicands are also same.
∴ \(\sqrt { 52 }\) and 5\(\sqrt { 13 }\) are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 8
Here, the order of 2\(\sqrt { 17 }\) and 5\(\sqrt { 3 }\) is same but their radicands are not.
∴ \(\sqrt { 68 }\) and 5\(\sqrt { 3 }\) are unlike surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 9
Here, the order of 12\(\sqrt { 2 }\) and 7\(\sqrt { 2 }\) is same and their radicands are also same.
∴ 4\(\sqrt { 18 }\) and 7\(\sqrt { 2 }\) are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 10
Here, the order of 38\(\sqrt { 3 }\) and 6\(\sqrt { 3 }\) is same and their radicands are also same.
∴ 19\(\sqrt { 12 }\) and 6\(\sqrt { 3 }\) are like surds.

v. 5\(\sqrt { 22 }\), 7\(\sqrt { 33 }\)
Here, the order of 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) is same but their radicands are not.
∴ 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) are unlike surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 11
Here, the order of 5√5 and 5√3 is same but their radicands are not.
∴ 5√5 and √75 are unlike surds.

Question 4.
Simplify the following surds.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 12
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 13

Question 5.
Compare the following pair of surds.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 14
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 15
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 16
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 17

Question 6.
Simplify.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 18
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 19

Question 7.
Multiply and write the answer in the simplest form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 20
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 21

Question 8.
Divide and write form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 22
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 23

Question 9.
Rationalize the denominator.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 24
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 25
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 26

Question 1.
\(\sqrt { 9+16 }\) ? + \(\sqrt { 9 }\) + \(\sqrt { 16 }\) (Texbookpg. no. 28)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 27

Question 2.
\(\sqrt { 100+36 }\) ? \(\sqrt { 100 }\) + \(\sqrt { 36 }\) (Textbook pg. no. 28)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 28

Question 3.
Follow the arrows and complete the chart by doing the operations given. (Textbook pg. no. 34)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 29
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 30

Question 4.
There are some real numbers written on a card sheet. Use these numbers and construct two examples each of addition, subtraction, multiplication and division. Solve these examples. (Textbook pg. no. 34)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 31
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 32

Class 9 Maths Digest

Practice Set 9.1 Class 8 Answers Chapter 9 Discount and Commission Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.1 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Discount and Commission Class 8 Maths Chapter 9 Practice Set 9.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 9.1 Chapter 9 Solutions Answers

8th Standard Maths Practice Set 9.1 Question 1. If marked price = Rs 1700, selling price = Rs 1540, then find the discount.
Solution:
Here, Marked price = Rs 1700,
selling price = Rs 1540
Selling price = Marked price – Discount
∴ 1540 = 1700 – Discount
∴ Discount = 1700 – 1540
= Rs 160
∴ The amount of discount is Rs 160.

Discount and Commission Practice Set 9.1 Question 2. If marked price Rs 990 and percentage of discount is 10, then find the selling price.
Solution:
Here, marked price = Rs 990,
discount = 10%
Let the percentage of discount be x
∴ x = 10%
i. Discount
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 1
= Rs 99

ii. Selling price = Marked price – Discount
= 990 – 99
= Rs 891
∴ The selling price is Rs 891.

Practice Set 9.1 Question 3. If selling price Rs 900, discount is 20%, then find the marked price.
Solution:
Here, selling price = Rs 900, discount = 20%
Let the marked price be Rs 100
Since, the discount given = 20%
∴ Amount of discount = Rs 20
∴ Selling price = 100 – 20 – Rs 80
Let actual marked price be Rs x
∴ For marked price of Rs x, selling price is Rs 900
\(\frac{80}{100}=\frac{900}{x}\)
∴ 80 × x = 100 × 900
∴ \(x=\frac{100 \times 900}{80}\)
∴ x = Rs 1125
∴ The marked price is Rs 1125.

Discount and Commission Std 8 Question 4. The marked price of the fan is Rs 3000. Shopkeeper gave 12% discount on it. Find the total discount and selling price of the fan.
Solution:
Here, Marked price = Rs 3000, discount = 12%
Let the percentage of discount be x.
∴ x = 12%
i. Discount
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 2
= 30 × 12
= Rs 360

ii. Selling price = Marked price – Discount
= 3000 – 360
= Rs 2640
∴ The total discount is Rs 360 and the selling price of the fan is Rs 2640.

Discount and Commission 8th Standard Question 5. The marked price of a mixer is Rs 2300. A customer purchased it for Rs 1955. Find percentage of discount offered to the customer.
Solution:
Here, marked price = Rs 2300,
selling price = Rs 1955
i. Selling price = Marked price – Discount
∴ 1955 = 2300 – Discount
∴ Discount = 2300 – 1955
= Rs 345

ii. Let the percentage of discount be x
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 3
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 4
∴ x = 15%
∴ The percentage of discount offered to the customer is 15%.

Question 6.
A shopkeeper gives 11% discount on a television set, hence the cost price of it is Rs 22,250. Then find the marked price of the television set.
Solution:
Here, selling price = Rs 22,250, discount = 11%
Let marked price be Rs 100
Since, the discount given = 11%
∴ Amount of discount = Rs 11
∴ Selling price = 100 – 11 = Rs 89
∴ Let actual marked price be Rs x
∴ For marked price of Rs x, selling price is Rs 22,250
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 5
∴ x = Rs 25,000
∴ The marked price of the television set is Rs 25,000.

8th Std Maths Discount and Commission Question 7. After offering discount of 10% on marked price, a customer gets total discount of Rs 17. To find the cost price for the customer, fill in the following boxes with appropriate numbers and complete the activity.
Solution:
Suppose, marked price of the item = 100 rupees Therefore, for customer that item costs 100 – 10 = 90 rupees.
Hence, when the discount is [10] then the selling price is [90] rupees.
Suppose when the discount is [17] rupees, the selling price is x rupees.
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 6
∴ The customer will get the item for Rs 153.

Question 8.
A shopkeeper decides to sell a certain item at a certain price. He tags the price on the item by increasing the decided price by 25%. While selling the item, he offers 20% discount. Find how many more or less percent he gets on the decided price.
Solution:
Here, price increase = 25%,
discount offered = 20%
Let the decided price be Rs 100
∴ Increase in price = Rs 25
∴ Shopkeeper marks the price = 100 + 25
= Rs 125
∴ marked price = Rs 125
Let the percentage of discount be x
∴ x = 20%
Maharashtra Board Class 8 Maths Solutions Chapter 9 Discount and Commission Practice Set 9.1 7
∴ Selling price = Marked price – Discount
= 125 – 25
= Rs 100
∴ If the decided price is Rs 100, then shopkeeper gets Rs 100.
∴ The shopkeeper gets neither more nor less than the decided price i.e. he gets 0% more / less.

Maharashtra Board Class 8 Maths Chapter 9 Discount and Commission Practice Set 9.1 Intext Questions and Activities

Question 1.
Write the appropriate numbers in the following boxes. (Textbook pg. no. 51)

  1. \(\frac { 12 }{ 100 }=\) __ percent = __%
  2. 47% = __
  3. 86% = __
  4. 4% of 300 = 300 × __ = __
  5. 15% of 1700 = 1700 × __= __

Solution:

  1. \(\frac { 12 }{ 100 }=\) 12 percent = 12%
  2. 47% = \(\frac { 47 }{ 100 }\)
  3. 86% = \(\frac { 86 }{ 100 }\)
  4. 4% of 300 = 300 × \(\frac { 4 }{ 100 }\) = 12
  5. 15% of 1700 = 1700 × \(\frac { 15 }{ 100 }\) = 255

Question 2.
You may have seen advertisements like ‘Monsoon Sale’, ‘Stock Clearance Sale’ etc offering different discount. In such a sale, a discount is offered on various goods. Generally in the month of July, sales of clothes are declared. Find and discuss the purpose of such sales. (Textbook pg. no. 51)
Solution:
(Students should attempt the above activity on their own)

Maharashtra Board Class 8 Maths Solutions

Std 8 Maths Digest

Practice Set 5.1 Class 8 Answers Chapter 5 Expansion Formulae Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.1 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.1 Chapter 5 Solutions Answers

Question 1.
Expand :
i. (a + 2)(a – 1)
ii. (m – 4)(m + 6)
iii. (p + 8) (p – 3)
iv. (13 + x)(13 – x)
v. (3x + 4y) (3x + 5y)
vi. (9x – 5t) (9x + 3t)
vii. \(\left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right)\)
viii. \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
ix. \(\left(\frac{1}{y}+4\right)\left(\frac{1}{y}-9\right)\)
Solution:
i. (a + 2)(a – 1)
= a² + (2 – 1) a + 2 × (-1)
..[∵ (x + A) (x + B) = x² + (A + B)x + AB]
= a² + a – 2

ii. (m – 4)(m + 6)
= m² + (- 4 + 6) m + (-4) × 6
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= m² + 2m – 24

iii. (p + 8) (p – 3)
= p² + (8 – 3) p + 8 x (-3)
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= p² + 5p – 24

iv. (13 + x) (13 – x)
= (13)² + (x – x) 13 + x × (-x)
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 169 + 0 × 13 – x²
= 169 – x²

v. (3x + 4y) (3x + 5y)
= (3x)² + (4y + 5y) 3x + 4y × 5y
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 9x² + 9y × 3x + 20y²
= 9x² + 27xy + 20y²

vi. (9x – 5t) (9x + 3t)
= (9x)² + [(-5t) + 3t] 9x + (-5t) × 3t
…[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 81x² + (-2t) × 9x – 15t²
= 81x² – 18xt – 15t²

vii. \(\left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right)\)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 1

viii. \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 2

ix. \(\left(\frac{1}{y}+4\right)\left(\frac{1}{y}-9\right)\)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 3

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.1 Intext Questions and Activities

Question 1.
Use the above formulae to fill proper terms in the following boxes. (Textbook pg. no. 23)

  1. (x + 2y)² = x² + ___ + 4y²
  2. (2x – 5y)² = __ – 20xy + __
  3. (101)² = (100 + 1)² = ___+ ___ + 1² = ___
  4. (98)² = (100 – 2)² = 10000 – ___ + ___ = ___
  5. (5m + 3n) (5m – 3n) = ___ – ___ = ___ – ___

Solution:

  1. (x + 2y)² = x² + 4xy + 4y²
  2. (2x – 5y)² = 4x² – 20xy + 25y²
  3. (101)² = (100 + 1)² = 10000 + 200 + 1² = 10201
  4. (98)² = (100 – 2)² = 10000 – 400 + 4 = 9604
  5. (5m + 3n) (5m – 3n) = (5m)² – (3n)² = 25m² – 9n²

Question 2.
Expand (x + a) (x + b) using formulae for areas of a square and a rectangle. (Textbook pg. no. 23)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 4
(x + a) (x + b) = x² + ax + bx + ab
(x + a) (x + b) = x² + (a + b) x + ab
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 5
∴ (x + a) (x + b) = x² + ax + bx + ab
∴ (x + a) (x + b) = x² + (a + b) x + ab

Std 8 Maths Digest

Practice Set 5.2 Class 8 Answers Chapter 5 Expansion Formulae Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.2 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.2 Chapter 5 Solutions Answers

Question 1.
Expand:
i. (k + 4)³
ii. (7x + 8y)³
iii. (7x + m)³
iv. (52)³
v. (101)³
vi. \(\left(x+\frac{1}{x}\right)^{3}\)
vii. \(\left(2 m+\frac{1}{5}\right)^{3}\)
viii. \(\left(\frac{5 x}{y}+\frac{y}{5 x}\right)^{3}\)
Solution:
i. Here, a = k and b = 4
(k + 4)³ = (k)³ + 3(k)² (4) + 3(k)(4)² + (4)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= k³ + 12k² + 3(k)(16) + 64
= k³ + 12k² + 48k + 64

ii. Here, a = 7x and b = 8y
(7x + 8y)³
= (7x)³ + 3(7x)² (8y) + 3(7x) (8y)² + (8y)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 343x³ + 3(49x²)(8y) + 3(7x)(64y²) + 512y³
= 343x³ + 1176x²y + 1344xy² + 512y³

iii. Here, a = 7 and b = m
(7 + m)³ = (7)³ + 3(7)²(m) + 3(7)(m)² + (m)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 343 + 3(49)(m) + 3(7)(m²) + m³
= 343 + 147m + 21m² + m³

iv. (52)³ = (50 + 3)³
Here, a = 50 and b = 2
(52)³ = (50)³ + 3(50)² (2) + 3(50)(2)² + (2)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 125000 + 3(2500)(2) + 3(50)(4) + 8
= 125000 + 15000 + 600 + 8
=140608

v. (101)³ = (100 + 1)³
Here, a = 100 and b = 1
(101)³
= (100)³ + 3(100)²(1) + 3(100)(1)² + (1)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1000000 + 3(10000) + 3(100) (1) + 1
= 1000000 + 30000 + 300 + 1
= 1030301

vi. Here, a = x and b = \(\frac { 1 }{ x }\)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.2 1

vii. Here, a = 2m and b = \(\frac { 1 }{ 5 }\)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.2 2

viii. Here, a = \(\frac { 5x }{ y }\) and b = \(\frac { y }{ 5x }\)
Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.2 3

Std 8 Maths Digest

Practice Set 4.1 Class 8 Answers Chapter 4 Altitudes and Medians of a Triangle Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 4.1 8th Std Maths Answers Solutions Chapter 4 Altitudes and Medians of a Triangle.

Altitudes and Medians of a Triangle Class 8 Maths Chapter 4 Practice Set 4.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 4.1 Chapter 4 Solutions Answers

Question 1.
In ∆LMN, ___ is an altitude and __ is a median, (write the names of appropriate segments.)
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 1
Solution:
In ∆LMN, seg LX is an altitude and seg LY is a median.

Question 2.
Draw an acute angled ∆PQR. Draw all of its altitudes. Name the point of concurrence as ‘O’.
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 2

Question 3.
Draw an obtuse angled ∆STV. Draw its medians and show the centroid.
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 3

Question 4.
Draw an obtuse angled ∆LMN. Draw its altitudes and denote the ortho centre by ‘O’.
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 4

Question 5.
Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 5

Question 6.
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 6
The point of concurrence of medians i.e. G and that of altitudes i.e. O lie on the same line PS which is the perpendicular bisector of seg QR.

Question 7.
Fill in the blanks.
Point G is the centroid of ∆ABC.
i. If l(RG) = 2.5, then l(GC) = ___
ii. If l(BG) = 6, then l(BQ) = ____
iii. If l(AP) = 6, then l(AG) = ___ and l(GP) = ___.
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 7
Solution:
The centroid of a triangle divides each median in the ratio 2:1.
i. Point G is the centroid and seg CR is the median.
∴ \(\frac{l(\mathrm{GC})}{l(\mathrm{RG})}=\frac{2}{1}\)
∴ \(\frac{l(\mathrm{GC})}{2.5}=\frac{2}{1}\) ……[∵ l(RG) = 2.5]
∴ l(GC) × 1 = 2 × 2.5
∴ l(GC) = 5

ii. Point G is the centroid and seg BQ is the median.
∴ \(\frac{l(\mathrm{BG})}{l(\mathrm{GQ})}=\frac{2}{1}\)
∴ \(\frac{6}{l(\mathrm{GQ})}=\frac{2}{1}\) …..[∵ l(BG) = 6]
∴ 6 × 1 = 2 × l(GQ)
∴ \(\frac { 6 }{ 2 }\) = l(GQ)
∴ 3 = l(GQ)
i.e. l(GQ) = 3
Now, l (BQ) = l(BG) + l(GQ)
∴ l(BQ) = 6 + 3
∴ l(BQ) = 9

iii. Point G is the centroid and seg AP is the median.
∴ \(\frac{l(\mathrm{AG})}{l(\mathrm{GP})}=\frac{2}{1}\)
∴ l(AG) = 2 l(GP) …..(i)
Now, l(AP) = l(AG) + l(GP) … (ii)
∴ l(AP) = 2l(GP) + l(GP) … [From (i)]
∴ l(AP) = 3l(GP)
∴ 6 = 3l(GP) ..[∵ l(AP) = 6]
∴ \(\frac { 6 }{ 3 }\) = l(GP)
∴ 2 = l(GP)
i.e. l(GP) = 2
l(AP) = l(AG) + l(GP) …[from (ii)]
∴ 6 = l(AG) + 2
∴ l(AG) = 6 – 2
∴ l(AG) = 4

Maharashtra Board Class 8 Maths Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 Intext Questions and Activities

Question 1.
Draw a line. Take a point outside the line. Draw a perpendicular from the point to the line with the help of a set-square (Textbook pg. no, 19)
Solution:
Step 1: Draw a line l and a point P lying outside it.
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 8
Step 2: By placing a set-square on line l, draw a perpendicular to the line from point P.
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 9

Question 2.
Draw an acute angled ∆ABC and all its altitudes. Observe the location of the orthocentre. (Textbook pg. no. 20)
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 10
Point O is the orthocentre.
Orthocentre lies in the interior of ∆ABC.

Question 3.
Draw a right angled triangle and draw all its altitudes. Write the point of concurrence. (Textbook: pg, no. 20)
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 11
Point Q is the orthocentre.
The point of concurrence of altitudes PQ, QR and QS is Q.

Question 4.
i. Draw an obtuse angled triangle and all its altitudes.
ii. Do they intersect each other?
Draw the lines containing the altitudes. Observe that these lines are concurrent. (Textbook pg. no. 20)
Solution:
i.
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 12
Point O is the orthocentre.

ii. Yes, all the altitudes intersect at point O in the exterior of ∆PQR.

Question 5.
Draw three different triangles; a right angled triangle, an obtuse angled triangle and an acute angled triangle. Draw the medians of the triangles. Note that the centroid of each of them is in the interior of the triangle. (Textbook pg. no. 21)
Solution:
i. Right angled triangle:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 13
ii. Obtuse angled triangle:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 14
iii. Acute angled triangle:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 15

Question 6.
Draw a sufficiently large ∆ABC.
Draw medians; seg AR, seg BQ and seg CP of ∆ABC.
Name the point of concurrence as G.
Measure the lengths of segments from the figure and fill in the boxes in the following table.

l(AG) = l(GR) = l(AG): l(GR) =
l(BG) = l(GQ) = l(BG): l(GQ) =
l(CG) = l(GP) = l(CG): l(GP) =

Observe that all of these ratios are nearly 2 : 1 (Textbook pg. no. 21)
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 16

l(AG) = 2.9 l(GR) = 1.4 l(AG): l(GR) = \(\frac{2.8}{1.4}=\frac{2}{1}\)
l(BG) = 2.4 l(GQ) = 1.2 l(BG): l(GQ) = \(\frac{2.4}{1.2}=\frac{2}{1}\)
l(CG) = 2.8 l(GP) = 1.4 l(CG): l(GP) = \(\frac{2.8}{1.4}=\frac{2}{1}\)

Question 7.
As shown in the given figure, a student drew ∆ABC using five parallel lines of a notebook. Then he found the centroid G of the triangle. How will you decide whether the location of G he found, is correct. (Textbook pg. no. 21)
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 17
Solution:
Draw seg AP ⊥ seg PE and seg EQ ⊥ seg QC.
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 18
Side AP || side EQ and AC is their transversal.
∴ ∠PAE ≅ ∠QEC …(i) [Corresponding angles]
In ∆ APE and ∆ EQC,
∠PAE ≅ ∠QEC …[From (i)]
∠APE ≅ ∠EQC
… [Each angle is of measure 90°]
side PE ≅ side QC
…. [Perpendicular distance between parallel lines]
∴ ∆ APE ≅ ∆ EQC … [By AAS test]
∴ AE = EC
… [Corresponding sides of congruent triangles]
∴ E is the midpoint of AC.
∴ seg BE is the median.
Similarly, seg CF is the median.
Since, the medians of a triangle are concurrent.
∴ G is the centroid of ∆ABC.

Question 8.
Draw an equilateral triangle. Find its circumcentre (C), incentre (I), centroid (G) and orthocentre (O). Write your observation. (Textbook pg. no. 22)
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 19
From the figure, circumcentre (C), incentre (I), centroid (G) and orthocentre (O) of an equilateral triangle are the same.

Question 9.
Draw an isosceles triangle. Locate its centroid, orthocentre, circumcentre and incentre. Verify that they are collinear. (Textbook pg. no. 22)
Solution:
Maharashtra Board Class 8 Maths Solutions Chapter 4 Altitudes and Medians of a Triangle Practice Set 4.1 20
From the figure, centroid (G), orthocentre (O), circumcentre (C) and incentre (I) of an isosceles triangle lie on the same line AD.
∴ they are collinear.

Std 8 Maths Digest

Problem Set 1 Algebra 9th Standard Maths Part 1 Chapter 1 Sets Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 1 Sets.

9th Standard Maths 1 Problem Set 1 Chapter 1 Sets Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Problem Set 1 Chapter 1 Sets Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for each of the following questions.
i. M= {1, 3, 5}, N= {2, 4, 6}, then M ∩ N = ?
(A) {1, 2, 3, 4, 5, 6}
(B) {1, 3, 5}
(C) φ
(D) {2, 4, 6}
Answer:
(C) φ

ii. P = {x | x is an odd natural number, 1< x ≤ 5}. How to write this set in roster form?
(A) {1, 3, 5}
(B) {1, 2, 3, 4, 5}
(C) {1, 3}
(D) {3, 5}
Answer:
(D) {3, 5}

iii. P= {1, 2, ………. , 10}. What type of set Pis?
(A) Null set
(B) Infinite set
(C) Finite set
(D) None of these
Answer:
(C) Finite set

iv. M ∪ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4}, then which of the following represent set N ?
(A) {1, 2, 3}
(B) {3, 4, 5, 6}
(C) {2, 5, 6}
(D) {4, 5, 6}
Answer:
(B) {3, 4, 5, 6}

v. If P ⊆ M, then which of the following set represent P ∩ (P ∪ M)?
(A) P
(B) M
(C) P ∪ M
(D) P’ ∩ M
Answer:
(A) P

vi. Which of the following sets are empty sets?
(A) Set of intersecting points of parallel lines.
(B) Set of even prime numbers.
(C) Month of an english calendar having less than 30 days.
(D) P = {x | x ∈ I , – 1 < x < 1}
Answer:
(A) Set of intersecting points of parallel lines.

Hints:
v. Here, P ⊆ M
∴ P ∪ M = M
∴ P ∩ (P ∪ M) = P ∩ M
= P … [∵ P ⊆M]

Question 2.
Find the correct option for the given question.
i. Which of the following collections is a set ?
(A) Colors of the rainbow
(B) Tall trees in the school campus.
(C) Rich people in the village
(D) Easy examples in the book
Answer:
(A) Colors of the rainbow

ii. Which of the following set represent N ∩W?
(A) {1, 2, 3,….}
(B) {0, 1, 2, 3,….}
(C) {0}
(D) { }
Answer:
(A) {1, 2, 3,….}

iii. P = {x | x is an odd natural number, 1< x < 5}. How to write this set in roster form?
(A) {1, 3, 5}
(B) {1, 2, 3, 4, 5}
(C) {1, 3}
(D) {3, 5}
Answer:
(B) {1, 2, 3, 4, 5}

iv. If T = {1, 2, 3, 4, 5} and M = {3,4, 7, 8}, then T ∪ M = ?
(A) {1, 2, 3, 4, 5,7}
(B) {1, 2, 3, 7, 8}
(C) {1, 2, 3, 4, 5, 7, 8}
(D) {3, 4}
Answer:
(C) {1, 2, 3, 4, 5, 7, 8}

Hints:
i. The elements of options B, C and D cannot be definitely and clearly decided.
ii. The common elements of N and W are 1 2, 3,….

Question 3.
Out of 100 persons in a group, 72 persons speak English and 43 persons speak French. Each one out of 100 persons speak at least one language. Then how many speak only English? How many speak only French ? How many of them speak English and French both?
Solution:
i. Let U be the set of all the persons,
E be the set of persons who speak English and
F be the set of persons who speak French.
∴ n(E) = 72, n(F) = 43
Since, each one out of 100 persons speak at least one language
∴ n(U) = n(E ∪ F)= 100,

ii. n (E ∪ F) = n (E) + n (F) – n(E ∩ F)
100 = 72 + 43 – n (E ∩ F)
n (E ∩ F) = 72 + 43 – 100
∴ n(E ∩ F) = 15
Number of people who speak English and French = 15

iii. Number of people who speak only English = n(E) – n(E ∩ F)
= 72 – 15 = 57

iv. Number of
people who speak only French = n(F) – n(E ∩ F)
= 43 – 15 = 28

Alternate Method:
Let U be the set of all the persons,
E be the set of persons who speak English,
F be the set of persons who speak French and x people speak both the languages.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 1
Since, each one out of 100 persons speak at least one language.
∴ n(U) = n(E ∪ F) = 100
∴ 72 – x + x + 43 – x = 100
∴ 115 – x= 100
∴ x = 115 – 100= 15.
Number of people who speak English and French = 15
Number of people who speak only English = 72 – x = 72 – 15 = 57
Number of people who speak only French = 43 – x = 43 – 15 = 28

Question 4.
70 trees were planted by Parth and 90 trees were planted by Pradnya on the occasion of Tree Plantation Week. Out of these 25 trees were planted by both of them together. How many trees were planted by Parth or Pradnya?
Solution:
i. Let P be the trees planted by Parth and Q be the trees planted by Pradnya
∴ n(P) = 70 and n(Q) = 90
Total number of trees planted by Parth and Pradnya = n(P ∩ Q) = 25

ii. Number of trees planted by Parth or Pradnya = n(P ∪ Q)
= n(P) + n(Q) – n(P ∩ Q)
= 70 + 90 – 25 = 135
∴ A total of 135 trees were planted by Parth or Pradnya.

Alternate Method:
Let P be the trees planted by Parth and Q be the trees planted by Pradnya
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 2
From Venn diagram
∴ Total trees planted by parth or pradnya = n(P ∪ Q)
= 45 + 25 + 65
= 135
A total of 135 trees were planted by Parth or Pradnya.

Question 5.
If n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, then n(A ∩ B) = ?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 36 = 20 + 28 – n(A ∩ B)
∴ n(A ∩ B) = 20 + 28 – 36
∴ n(A ∩ B) = 12

Question 6.
In a class, 8 students out of 28 have a dog as their pet animal at home, 6 students have a cat as their pet animal, 10 students have dog and cat both, then how many students do not have dog or cat as their pet animal at home?
Solution:
i. Let U be the set of all the students, then n(U) = 28
Let D be the set of students who have dog as pet and C be the set of students who have cat as pet.
10 students have dog and cat as their pet animal
n(D ∩ C) = 10
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 3
From venn diagram,

ii. Number of students who have cat or dog as pet
= n(D ∪ C)
= 8 + 10 + 6
= 24

iii. Number of students who do not have dog or cat as pet = n (U) – n(D ∪ C)
= 28 – 24
= 4

Question 7.
Represent the union of two sets by Venn diagram for each of the following.
i. A = {3, 4, 5, 7},B = {1, 4, 8} l Marks
ii. P {a, b, c, e, f, Q = {l, m, n, e, b) I Markj
iii. X = {x x is a prime number between 80 and 100}
Y = { y | y is an odd number between 90 and 100}
Solution:
i. A = {3, 4, 5, 7}, B = {1, 4, 8}
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 4

ii. P = {a, b, c, e, f}, Q = {l, m, n, e, b}
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 5

iii. X = {x | x is a prime number between 80 and 100}
∴ X = {83, 89, 97}
Y = {y | y is an odd number between 90 and 100}
∴ Y = {91, 93, 95, 97, 99}
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 6

Question 8.
Write the subset relations between the following sets.
X = set of all quadrilaterals.
Y = set of all rhombuses.
S = set of all squares.
T = set of all parallelograms.
V = set of all rectangles. [3 Marks]
Solution:
i. Rhombus, square, parallelogram and rectangle all are quadrilaterals.
∴ Y ⊆ X,S ⊆ X,T ⊆ X,V ⊆ X

ii. Every square is a rhombus, parallelogram and rectangle.
∴ S ⊆ Y, S ⊆ T, S ⊆ V

iii. Every rhombus and rectangle is a parallelogram.
∴ Y ⊆ T, V ⊆ T

Question 9.
If M is any set, then write M ∪Φ and M ∩ Φ.
Solution:
Let M = {2, 3, 4, 8} and Φ = { }
∴ M ∪ Φ = {2, 3, 4, 8}
∴ M ∪ Φ = M Also, M ∩ Φ = { }
∴ M ∩ Φ = i(i

Question 10.
Observe the Venn diagram and write the given sets U, A, B, A ∪ B and A ∩ B.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 7
U = {1,2, 3,4, 5, 7, 8, 9, 10, 11, 13}
A = {1, 2, 3, 5,7}
B = {1, 5, 8, 9, 10}
A ∪ B = {1,2, 3, 5, 7, 8, 9, 10}
A ∩ B = {1, 5}

Question 11.
If n(A) = 7, n(B) = 13, n(A ∩ B) = 4, then n(A ∪ B) = ?
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 7 + 13 – 4
n(A ∪ B) = 16

Question 1.
Set of students in a class and set of students in the same class who can swim, are shown by the Venn diagram.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 8
Observe the diagram and draw Venn diagrams for the following subsets.
i. a. Set of students in a class
b. Set of students who can ride bicycles in the same class

ii. A set of fruits is given as follows.
U = {guava, orange, mango, jackfruit, chickoo, jamun, custard apple, papaya, plum}
Show these subsets.
A = fruit with one seed
B = fruit with more than one seed. (Textbook pg. no. 8)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 9

ii. A = {mango, jamun, plum}
B = {guava, orange, jackfruit, chickoo, custard apple, papaya}
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 10

Question 2.
Every student should take 9 triangular sheets of paper and one plate. Numbers from 1 to 9 should, be written on each triangle. Everyone should keep some numbered triangles in the plate. Now the triangles in each plate form a subset of the set of numbers from 1 to 9.
Maharashtra Board Class 9 Maths Solutions Chapter 1 Sets Problem Set 1 11
Look at the plates of Sujata, Hameed, Mukta, Nandini, Joseph with the numbered triangles. Guess the thinking behind selecting these numbers. Hence write the subsets in set builder form. (Textbook pg, no. 9)
Solution:
Sujata:
S = {x | x = 2n- 1, n ∈ N, x < 9}
Hameed:
f H = {x | x = 2n, n ∈ N, x < 9}
Mukta:
M = {x | x = n2, n ∈ N, x ≤ 9}
Nandini:
N = {x | x ∈ N, x ≤ 9}
Joseph:
J = {x | x is a prime number between 1 and 9}

Question 3.
Collect the following information from 20 families nearby your house.
i. Number of families subscribing for Marathi Newspaper.
ii. Number of families subscribing for English Newspaper.
iii. Number of families subscribing for both English as well as Marathi Newspaper.
Show the collected information using Venn diagram. (Textbook pg.no. 18)
[Students should attempt the above activity on their own.]

Class 9 Maths Digest

Practice Set 2.1 Algebra 9th Standard Maths Part 1 Chapter 2 Real Numbers Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Practice Set 2.1 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 2.1 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 1
Solution:
i. Denominator = 5 = 1 x 5
Since, 5 is the only prime factor denominator.
the decimal form of the rational number \(\frac { 13 }{ 5 }\) will be terminating type.

ii. Denominator = 11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number \(\frac { 2 }{ 11 }\) will be non-terminating recurring type.

iii. Denominator = 16
= 2 x 2 x 2 x 2
Since, 2 is the only prime factor in the denominator.
∴ the decimal form of the rational number \(\frac { 29 }{ 16 }\) will be terminating type.

iv. Denominator = 125
= 5 x 5 x 5
Since, 5 is the only prime factor in the denominator.
the decimal form of the rational number \(\frac { 17 }{ 125 }\) will be terminating type.

v. Denominator = 6
= 2 x 3
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number \(\frac { 11 }{ 6 }\) will be non-terminating recurring type.

Question 2.
Write the following rational numbers in decimal form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 2
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 3
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 4
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 5
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 6
Solution:
i. \(\frac { -5 }{ 7 }\)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 7

ii. \(\frac { 9 }{ 11 }\)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 8

iii. √5
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 9

iv. \(\frac { 121 }{ 13 }\)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 10

v. \(\frac { 29 }{ 8 }\)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 11

Question 3.
Write the following rational numbers in \(\frac { p }{ q }\) form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 12
Solution:
i. Let x = \(0.\dot { 6 }\) …(i)
∴ x = 0.666…
Since, one number i.e. 6 is repeating after the decimal point.
Thus, multiplying both sides by 10,
10x = 6.666…
∴ 10 x 6.6 …(ii)
Subtracting (i) from (ii),
10x – x = 6.6 – 0.6
∴ 9x = 6
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 13

ii. Let x = \(0.\overline { 37 }\)
∴ x = 0.3737…
Since, two numbers i.e. 3 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 37.3737……
∴ 100x = \(37.\overline { 37 }\) ……(ii)
Subtracting (i) from (ii),
100x – x = \(37.\overline { 37 }\) – \(0.\overline { 37 }\)
∴ 99x = 37
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 14

iii. Letx = \(3.\overline { 17 }\) …(i)
∴ x = 3.1717…
Since, two numbers i.e. 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 317.1717…
∴ 100x= 317.17 …(ii)
Subtracting (i) from (ii),
100x – x = \(317.\overline { 17 }\) – \(3.\overline { 17 }\)
∴ 99x = 314
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 15

iv. Let x = \(15.\overline { 89 }\) …….. (i)
∴ x = 15.8989…
Since, two numbers i.e. 8 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x= 1589.8989…
∴ 100x = \(1589.\overline { 89 }\) …(ii)
Subtracting (i) from (ii),
100x – x = \(1589.\overline { 89 }\) – \(15.\overline { 89 }\)
∴ 99x = 1574
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 16

v. Let x = \(2.\overline { 514 }\)
∴ x = 2.514514…
Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 2514.514514…
1000x = \(2514.\overline { 514 }\) ….(ii)
Subtracting (i) from (ii),
1000x – x = \(2514.\overline { 514 }\) – \(2.\overline { 514 }\)
∴ 999x = 2512
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 17

Question 1.
How to convert 2.43 in \(\frac { p }{ q }\) form ? (Textbook pg. no. 20)
Solution:
Let x = 2.43
In 2.43, the number 4 on the right side of the decimal point is not recurring.
So, in order to get only recurring digits on the right side of the decimal point, we will multiply 2.43 by 10.
∴ 10x = 24.3 …(i)
∴ 10x = 24.333…
Here, digit 3 is the only recurring digit. Thus, by multiplying both sides by 10, 100x = 243.333…
∴ 100x= 243.3 …(ii)
Subtracting (i) from (ii),
100x – 10x = 243.3 – 24.3
∴ 90x = 219
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 18

Class 9 Maths Digest

Practice Set 3.3 Class 8 Answers Chapter 3 Indices and Cube Root Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 3.3 8th Std Maths Answers Solutions Chapter 3 Indices and Cube Root.

Indices and Cube Root Class 8 Maths Chapter 3 Practice Set 3.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 3.3 Chapter 3 Solutions Answers

Question 1.
Find the cube root of the following numbers.
i. 8000
ii. 729
iii. 343
iv. -512
v. -2744
vi. 32768
Solution:
i. 8000
= 2 × 2 × 2 × 10 × 10 × 10
= (2 × 10) × (2 × 10) × (2 × 10)
= (2 × 10)³
= 20³
∴ \(\sqrt[3]{8000}=20\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 1

ii. 729
= (3 × 3) × (3 × 3) × (3 × 3)
= (3 × 3)³
= 9³
∴ \(\sqrt[3]{729}=9\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 2

iii. 343
= 7 × 7 × 7
= 7³
∴ \(\sqrt[3]{343}=7\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 3

iv. -512
= 2 × 2 × 2 × 4 × 4 × 4
= (2 × 4) × (2 × 4) × (2 × 4)
= (2 × 4)³
= 8³
∴ – 512 = (- 8) × (- 8) × (- 8)
= (-8)³
∴ \(\sqrt[3]{-512}=-8\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 4

v. -2744
= 2 × 2 × 2 × 7 × 7 × 7
= (2 × 7) × (2 × 7) × (2 × 7)
= (2 × 7)³
= 14³
∴ -2744 = (-14) × (-14) × (-14)
= (-14)³
∴ \(\sqrt[3]{-2744}=-14\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 5

vi. 32768
= 2 × 2 × 2 × 4 × 4 × 4 × 4 × 4 × 4
= (2 × 4 × 4) × (2 × 4 × 4) × (2 × 4 × 4)
= (2 × 4 × 4)³
= 32³
∴ \(\sqrt[3]{32768}=32\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 6

Question 2.
Simplify:
i. \(\sqrt[3]{\frac{27}{125}}\)
ii. \(\sqrt[3]{\frac{16}{54}}\)
iii. If \(\sqrt[3]{729}=9\) then \(\sqrt[3]{0.000729}\) = ?
Solution:
i. \(\sqrt[3]{\frac{27}{125}}\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 7

ii. \(\sqrt[3]{\frac{16}{54}}\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 8

iii. \(\sqrt[3]{0.000729}\)
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 9
Maharashtra Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root Practice Set 3.3 10
Note:
Here, number of decimal places in cube root = 6
∴ number of decimal places in cube of number = 2

Maharashtra Board Class 8 Maths Chapter 3 Indices and Cube Root Practice Set 3.3 Intext Questions and Activities

Question 1.
17 is a positive number. The cube of 17, which is 4913, is also a positive number. Cube of -6 is -216. Take some more positive and negative numbers and obtain their cubes. Find the relation between the sign of a number and the sign of its cube. (Textbook pg. no. 17)
Solution:
Consider, 6³ = 6 × 6 × 6 = 216 and (-4)³ = (- 4) × (- 4) × (- 4) = – 64
Thus, cube of a positive number is positive and cube of a negative number is negative.
∴ Sign of a number = sign of its cube.

Question 2.
In example 4 and 5 on textbook pg. no. 17, observe the number of decimal places in the number and number of decimal places in the cube of the number. Is there any relation between the two? (Textbook pg. no. 17)
Solution:
Yes, there is a relation between the number of decimal places in the number and its cube.
(1.2)³ = 1.728, (0.02)³ = 0.000008
No. of decimal places in 1.2 = 1
No. of decimal places in 1.728 = 3
No. of decimal places in 0.02 = 2
No. of decimal places in 0.000008 = 6
Thus, number of decimal places in cube of a number is three times the number of decimal places in that number.

Std 8 Maths Digest

Practice Set 3.1 Class 8 Answers Chapter 3 Indices and Cube Root Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 3.1 8th Std Maths Answers Solutions Chapter 3 Indices and Cube Root.

Indices and Cube Root Class 8 Maths Chapter 3 Practice Set 3.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 3.1 Chapter 3 Solutions Answers

Question 1.
Express the following numbers in index form.
i. Fifth root of 13
ii. Sixth root of 9
iii. Square root of 256
iv. Cube root of 17
v. Eighth root of 100
vi. Seventh root of 30
Solution:
i. \((13)^{\frac{1}{5}}\)
ii. \((9)^{\frac{1}{6}}\)
iii. \((256)^{\frac{1}{2}}\)
iv. \((17)^{\frac{1}{3}}\)
v. \((100)^{\frac{1}{8}}\)
vi. \((30)^{\frac{1}{7}}\)

Question 2.
Write in the form ‘nth root of a’ in each of the following numbers.
i. \((81)^{\frac{1}{4}}\)
ii. \((49)^{\frac{1}{2}}\)
iii. \((15)^{\frac{1}{5}}\)
iv. \((512)^{\frac{1}{9}}\)
v. \((100)^{\frac{1}{19}}\)
vi. \((6)^{\frac{1}{7}}\)
Solution:
i. Fourth root of 81.
ii. Square root of 49.
iii. Fifth root of 15.
iv. Ninth root of 512.
v. Nineteenth root of 100.
vi. Seventh root of 6.

Maharashtra Board Class 8 Maths Chapter 3 Indices and Cube Root Practice Set 3.1 Intext Questions and Activities

Question 1.
Using laws of indices, write proper numbers in the following boxes. (Textbook pg, no. 14)
i. \({ 3 }^{ 5 }\times { 3 }^{ 2 }={ 3 }^{ \left( \right) }\)
ii. \({ 3 }^{ 7 }\div { 3 }^{ 9 }={ 3 }^{ \left( \right) }\)
iii. \(({ 3 }^{ 4 })^{ 5 }={ 3 }^{ \left( \right) }\)
iv. \(5^{ -3 }=\frac { 1 }{ { 5 }^{ \left( \right) } }\)
v. \(5^{ 0 }=\left( \right) \)
vi. \(5^{ 1 }=\left( \right) \)
vii. \((5\times 7)^{ 2 }={ 5 }^{ \left( \right) }\times { 7 }^{ \left( \right) }\)
viii. \({ \left( \frac { 5 }{ 7 } \right) }^{ 3 }=\frac { { \left( \right) }^{ 3 } }{ { \left( \right) }^{ 3 } } \)
ix. \({ \left( \frac { 5 }{ 7 } \right) }^{ -3 }={ \left( \frac { \left( \right) }{ \left( \right) } \right) }^{ 3 }\)
Solution:
i. \({ 3 }^{ 5 }\times { 3 }^{ 2 }={ 3 }^{ 7 }\)
ii. \({ 3 }^{ 7 }\div { 3 }^{ 9 }={ 3 }^{ -2 }\)
iii. \(({ 3 }^{ 4 })^{ 5 }={ 3 }^{ 20 }\)
iv. \(5^{ -3 }=\frac { 1 }{ { 5 }^{ 3 } } \)
v. \(5^{ 0 }=1\)
vi. \(5^{ 1 }=5\)
vii. \((5\times 7)^{ 2 }={ 5 }^{ 2 }\times { 7 }^{ 2 }\)
viii. \({ \left( \frac { 5 }{ 7 } \right) }^{ 3 }=\frac { { 5 }^{ 3 } }{ { 7 }^{ 3 } } \)
ix. \({ \left( \frac { 5 }{ 7 } \right) }^{ -3 }={ \left( \frac { 7 }{ 5 } \right) }^{ 3 }\)

Std 8 Maths Digest