Category: Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.2 8th Std Maths Answers Solutions Chapter 11 Statistics.

Practice Set 11.2 8th Std Maths Answers Chapter 11 Statistics

practice set 11.2 8th class Question 1.
Observe the following graph and answer the questions.

i. State the type of the graph.
ii. How much is the savings of Vaishali in the month of April?
iii. How much is the total of savings of Saroj in the months March and April?
iv. How much more is the total savings of Savita than the total savings of Megha?
v. Whose savings in the month of April is the least?
Solution:
i. The given graph is a subdivided bar graph.
ii. Vaishali’s savings in the month of April is Rs 600.
iii. Total savings of Saroj in the months of March and April is Rs 800.
iv. Savita’s total saving = Rs 1000, Megha’s total saving = Rs 500
∴ difference in their savings = 1000 – 500 = Rs 500.
Savita’s saving is Rs 500 more than Megha.
v. Megha’s savings in the month of April is the least.

practice set 11.2 Question 2.
The number of boys and girls, in std 5 to std 8 in a Z.P. School is given in the table. Draw a subdivided bar graph to show the data. (Scale : On Y axis, 1cm = 10 students)

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20

Solution:

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20
Total	51	40	35	45

Statistics class 8 practice set 11.1 Question 3.
In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

Year\Town	karjat	Wadgaon	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150

Solution:

Year\Town	karjat	Wadgaon	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150
Total	350	550	450	250

Statistics class 8 Question 4.
In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale: On Y axis: 1 cm = 500 students)

Means of commutation\Town	Paithan	Yeola	Shahapur
Cycle	3250	1500	1250
Bus and auto	750	500	500
On foot	1000	1000	500

Solution:

Means of commutation\Town	Paithan	Yeola	Shahapur
Cycle	3250	1500	1250
Bus and auto	750	500	500
On foot	1000	1000	500
Total	5000	3000	2250

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.4 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Practice Set 5.4 8th Std Maths Answers Chapter 5 Expansion Formulae

Question 1.
Expand:
i. (2p + q + 5)²
ii. (m + 2n + 3r)²
iii. (3x + 4y – 5p)²
iv. (7m – 3n – 4k)²
Solution:
i. (2p + q + 5)² = (2p)² + (q)² + (5)² + 2(2p) (q) + 2(q) (5) + 2(2p) (5)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 4p² + q² + 25 + 4pq + 10q + 20p

ii. (m + 2n + 3r)² = (m)² + (2n)² + (3r)² + 2(m) (2n) + 2(2n) (3r) + 2(m) (3r)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= m² + 4n² + 9r² + 4mn + 12nr + 6mr

iii. (3x + 4y – 5p)² = (3x)² + (4y)² + (- 5p)² + 2(3x) (4y) + 2(4y) (- 5p) + 2(3x) (- 5p)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 9x + 16y² + 25p² + 24xy – 40py – 30px

iv. (7m – 3n – 4k)² = (7m)² + (- 3n)² + (- 4k)² + 2(7m) (- 3n) + 2 (- 3n) (- 4k) + 2 (7m) (- 4k)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49m² + 9n² + 16k² – 42mn + 24nk – 56km

Question 2.
Simplify:
i. (x – 2y + 3)² + (x + 2y – 3)²
ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
Solution:
i. (x – 2y + 3)² + (x + 2y – 3)²
= [(x)² + (- 2y)² + (3)² + 2 (x) (- 2y) + 2 (- 2y) (3) + 2 (x) (3)] + [(x)² + (2y)² + (- 3)² + 2 (x) (2y) + 2 (2y) (- 3) + 2 (x) (- 3)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= x² + 4y² + 9 – 4xy – 12y + 6x + x² + 4y² + 9 + 4xy – 12y – 6x
= x + x² + 4y² + 4y² + 9 + 9 – 4xy + 4xy – 12y – 12y + 6x – 6x
= 2x² + 8y² + 18 – 24y

ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
= [(3k)² + (- 4r)² + (- 2m)² + 2 (3k) (- 4r) + 2 (- 4r) (- 2m) + 2 (3k) (- 2m)] – [(3k)² + (4r)² + (- 2m)² + 2 (3k) (4r) + 2 (4r) (- 2m) + 2 (3k) (- 2m)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (9k² + 16r² + 4m² – 24kr + 16rm – 12km) – (9k² + 16r² + 4m² + 24kr – 16rm – 12km)
= 9k² + 16r² + 4m² – 24kr + 16rm – 12km – 9k² – 16r² – 4m² – 24kr + 16rm + 12km
= 9k² – 9k² + 16r² – 16r² + 4m² – 4m² – 24kr – 24kr + 16rm + 16rm – 12km + 12km
= 32rm – 48kr

iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
= [(7a)² + (- 6b)² + (5c)² + 2(7a) (-6b) + 2(-6b) (5c) + 2(7a) (5c)] + [(7a)² + (6b)² + (- 5c)² + 2 (7a) (6b) + 2 (6b) (- 5c) + 2 (7a) (- 5c)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49a² + 36b² + 25c² – 84ab – 60bc + 70ac + 49a² + 36b² + 25c² + 84ab – 60bc – 70ac
= 49a² + 49a² + 36b² + 36b² + 25c² + 25c² – 84ab + 84ab – 60bc – 60bc + 70ac – 70ac
= 98a² + 72b² + 50c² – 120bc

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.4 Intext Questions and Activities

Question 1.
Fill in the boxes with appropriate terms in the steps of expansion. (Textbook pg. no. 27)
(2p + 3m + 4n)²
= (2p)² + (3m)² + __ + 2 × 2p × 3m + 2 × __ × 4n + 2 × 2p × __
= __ + 9m² + __ + 12pm + __ + __
Solution:
(2p + 3m + 4n)²
= (2p)² + (3m)² + (4n)² + 2 x 2p x 3m + 2 x 3m x 4n + 2 x 2p x 4n
= 4p² + 9m² + 16n² + 12pm + 24mn + 16pn

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.

Practice Set 3.3 8th Std Maths Answers Chapter 3 Indices and Cube Root

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
∴ \(\sqrt[3]{8000}=20\)

ii. 729
= (3 × 3) × (3 × 3) × (3 × 3)
= (3 × 3)³
= 9³
∴ \(\sqrt[3]{729}=9\)

iii. 343
= 7 × 7 × 7
= 7³
∴ \(\sqrt[3]{343}=7\)

iv. -512
= 2 × 2 × 2 × 4 × 4 × 4
= (2 × 4) × (2 × 4) × (2 × 4)
= (2 × 4)³
= 8³
∴ – 512 = (- 8) × (- 8) × (- 8)
= (-8)³
∴ \(\sqrt[3]{-512}=-8\)

v. -2744
= 2 × 2 × 2 × 7 × 7 × 7
= (2 × 7) × (2 × 7) × (2 × 7)
= (2 × 7)³
= 14³
∴ -2744 = (-14) × (-14) × (-14)
= (-14)³
∴ \(\sqrt[3]{-2744}=-14\)

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
∴ \(\sqrt[3]{32768}=32\)

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. \(\sqrt[3]{\frac{27}{125}}\)

ii. \(\sqrt[3]{\frac{16}{54}}\)

iii. \(\sqrt[3]{0.000729}\)


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.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 2.2 8th Std Maths Answers Solutions Chapter 2 Parallel Lines and Transversals.

Practice Set 2.2 8th Std Maths Answers Chapter 2 Parallel Lines and Transversals

Question 1.
Choose the correct alternative.
i. In the given figure, if line m || line n and line p is a transversal, then find x.

(A) 135°
(B) 90°
(C) 45°
(D) 40°
Solution:
(C) 45°

Hint:

line m || line n and line p is a transversal.
∴ m∠BFG + m∠FGD = 180°
…[Interior angles]
∴ 3x + x = 180°
∴ 4x = 180°
∴ x = \(\frac { 180 }{ 4 }\)
∴ x = 45°

ii. In the given figure, if line a || line b and line l is a transversal, then find x.

(A) 90°
(B) 60°
(C) 45°
(D) 30°
Solution:
(D) 30°

Hint:

line a || line b and line l is a transversal.
∴ m∠UVS = m∠PUV
…[Alternate angles]
= 4x
m∠UVS + m∠WVS = 180°
… [Angles in a linear pair]
∴ 4x + 2x = 180°
∴ 6x = 180°
∴ x = \(\frac { 180 }{ 6 }\)
∴ x = 30°

Question 2.
In the given figure, line p || line q. Line t and line s are transversals. Find measure of ∠x and ∠y using the measures of angles given in the figure.

Solution:

i. Consider ∠z as shown in figure.
line p || line q and line t is a transversal.
∴ m∠z = 40° …(i) [Corresponding angles]
m∠x + m∠z = 180° …[Angles in a linear pair]
∴ m∠x + 40o = 180° …[From(i)]
∴ m∠x= 180° – 40°
∴ m∠x = 140°

ii. Consider ∠w as shown in the figure.
m∠w + 70° = 180° …[Angles in a linear pair]
∴ m∠w = 180° – 70°
∴ m∠w = 110° …(ii)
line p || line q and line s is a transversal.
∴ m∠y = m∠w …[Alternate angles]
∴ m∠y =110° …[From (ii)]
∴ m∠x = 140°, m∠y = 110°

Question 3.
In the given figure, line p || line q, line l || line m. Find measures of ∠a, ∠b and ∠c, using the measures of given angles. Justify your answers.

Solution:
i. line p || line q and line l is a transversal.
∴ m∠a + 80° = 180° …[Interior angles]
∴ m∠a= 180° – 80°
∴ m∠a= 100°

ii. line l || line m and line p is a transversal.
∴ m∠c = 80° …(i) [Exterior alternate angles]

iii. line p || line q and line m is a transversal.
∴ m∠b = m∠c … [Corresponding angles]
m∠b = 80° …[From (i)]
∴ m∠a = 100°, m∠b = 80°, m∠c = 80°

Question 4.
In the given figure, line a || line b, line l is a transversal. Find the measures of ∠x, ∠y, ∠z using the given information.

Solution:
line a || line b and line l is a transversal.
∴ m∠x = 105° …(i) [Corresponding angles]

ii. m∠y = m∠x … [Vertically opposite angles]
∴ m∠y = 105° …[From (i)]

iii. m∠z + 105° = 180° …[Angles in a linear pair]
∴ m∠z = 180°- 105°
∴ m∠z = 75°
∴ m∠x = 105°, m∠y = 105°, m∠z = 75°

Question 5.
In the given figure, line p || line l || line q. Find ∠x with the help of the measures given in the figure.

Solution:

line p || line l and line IJ is a transversal.
m∠IJN = m∠JIH … [Alternate angles]
∴ m∠IJN = 40° …(i)
line l || line q and line MJ is a transversal.
m∠MJN = m∠JMK … [Alternate angles]
∴ m∠MJN = 30° …(ii)
Now, m∠x = m∠IJN + m∠MJN
…[Angle addition property]
= 40° + 30° …[From (i) and (ii)]
∴ m∠x = 70°

Maharashtra Board Class 8 Maths Chapter 2 Parallel Lines and Transversals Practice Set 2.2 Intext Questions and Activities

Question 1.
When two parallel lines are intersected by a transversal eight angles are formed. If the measure of one of these eight angles is given, can we find measures of remaining seven angles? (Textbook pg, no. 9)
Solution:
Yes, we can find the measures of the remaining angles.

In the given figure, line m || line n and line l is a transversal.
m∠a = 60°(say) …(i)
i. m∠a + m∠b = 180° …[Angles in a linear pair]
∴ 60° + m∠b =180° … [From (i)]
∴ m∠b = 180° – 60°
∴ m∠b = 120° …(ii)

ii. m∠c = m∠b …[Vertically opposite angles]
∴ m∠c = 120° .. .(iii) [From (ii)]

iii. m∠d = m∠a …[Vertically opposite angles]
∴ m∠d = 60° …(iv) [From (i)]

iv. m∠e = m∠d …[Alternate angles]
∴ m∠e = 60° … [From (iv)]

v. m∠f = m∠c …[Alternate angles]
∴ m∠f = 120° …[From (iii)]

vi. m∠g = m∠d …[Corresponding angles]
∴ m∠g = 60° … [From (iv)]

vii. m∠h = m∠c … [Corresponding angles]
∴ m∠h = 120° …[From (iii)]

Question 2.
As shown in the figure (A), draw two parallel lines and their transversal on a paper. Draw a copy of the figure on another blank sheet using a trace paper, as shown in the figure (B). Colour part Land part II with different colours. Cut out the two parts with a pair of scissors. Place, part I and part II on each angle in the figure A and answer the following questions. (Textbook pg. no. 9)

1. Which angles coincide with part I?
2. Which angles coincide with part II?

Solution:

1. ∠d, ∠f and ∠h coincide with part I.
2. ∠c, ∠e and ∠g coincide with part II.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 10 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 10 Answers Solutions Chapter 3

Question 1.
Which number is neither a prime number nor a composite number?
Solution:
1

Question 2.
Which of the following are pairs of co-primes?
i. 8,14
ii. 4,5
iii. 17,19
iv. 27,15
Solution:
i. Factors of 8: 1, 2, 4, 8
Factors of 14: 1, 2, 7, 14
∴ Common factors of 8 and 14: 1,2
∴ 8 and 14 are not a pair of co-prime numbers.

ii. Factors of 4: 1, 4
Factors of 5: 1, 5
∴ Common factors of 4 and 5: 1
∴ 4 and 5 are a pair of co-prime numbers.

iii. Factors of 17: 1, 17
Factors of 19: 1, 19
∴ Common factors of 17 and 19: 1
∴ 17 and 19 are a pair of co-prime numbers.

iv. Factors of 27: 1, 3, 9, 27
Factors of 15: 1, 3, 5, 15 .
∴ Common factors of 27 and 15 : 1,3
∴ 27 and 15 are not a pair of co-prime numbers.

Question 3.
List the prime numbers from 25 to 100 and say how many they are.
Solution:
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
There are 16 prime numbers from 25 to 100.

Question 4.
Write all the twin prime numbers from 51 to 100.
Solution:

1. 59 and 61
2. 71 and 73

Question 5.
Write 5 pairs of twin prime numbers from 1 to 50.
Solution:

1. 3,5
2. 5,7
3. 11,13
4. 17,19
5. 29,31
6. 41,43

Question 6.
Which are the even prime numbers?
Solution:
2

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 10 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 15)
i. Which is the smallest prime number?
ii. List the prime numbers from 1 to 50. How many are they?
iii. Identify the prime numbers in the list below.
17, 15 ,4, 3, 1, 2, 12, 23, 27, 35, 41, 43, 58, 51, 72, 79, 91, 97.
Solution:
i. 2 is the smallest prime number.
ii. There are 15 prime numbers from 1 to 50.
They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
iii. [17], 15 ,4, [3], 1, [2], 12, [23], 27, 35, [41], [43], 58, 51, 72, [79], 91, [97].

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Practice Set 10.1 8th Std Maths Answers Chapter 10 Division of Polynomials

Question 1.
Divide and write the quotient and the remainder.
i. 21m² ÷ 7m
ii. 40a³ ÷ (-10a)
iii. (- 48p⁴) ÷ (- 9p²)
iv. 40m⁵ ÷ 30m³
v. (5x³ – 3x²) ÷ x²
vi. (8p³ – 4p²) ÷ 2p²
vii. (2y³ + 4y² + 3 ) ÷ 2y²
viii. (21x⁴ – 14x² + 7x) ÷ 7x³
ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²
x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³
Solution:
i. 21m² ÷ 7m

∴ Quotient = 3m
Remainder = 0

ii. 40a³ ÷ (-10a)

∴ Quotient = -4a²
Remainder = 0

iii. (- 48p⁴) ÷ (- 9p²)

∴ Quotient = \(\frac { 16 }{ 3 }\) p²
Remainder = 0

iv. 40m⁵ ÷ 30m³

∴ Quotient = \(\frac { 4 }{ 3 }\) m²
Remainder = 0

v. (5x³ – 3x²) ÷ x²

∴ Quotient = 5x – 3
Remainder = 0

vi. (8p³ – 4p²) ÷ 2p²

∴ Quotient = 4p – 2
Remainder = 0

vii. (2y³ + 4y² + 3 ) ÷ 2y²

∴ Quotient = y + 2
Remainder = 3

viii. (21x⁴ – 14x² + 7x) ÷ 7x³

∴ Quotient = 3x
Remainder = -14x² + 7x

ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²

∴ Quotient = 3x³ – 2x² + 4x + 1
Remainder = 0

x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³

∴ Quotient = 5m – 3
Remainder = 10m + 8

Maharashtra Board Class 8 Maths Chapter 10 Division of Polynomials Practice Set 10.1 Intext Questions and Activities

Question 1.
Fill in the blanks in the following examples. (Textbook pg. no. 61)

1. 2a + 3a = __
2. 7b – 4b = __
3. 3p × p² = __
4. 5m² × 3m² = __
5. (2x + 5y) × \(\frac { 3 }{ x }\) = __
6. (3x² + 4y) × (2x + 3y) = __

Solution:

1. 2a + 3a = 5a
2. 7b – 4b = 3b
3. 3p × p² = 3p³
4. 5m² × 3m² = 15m⁴
5. (2x + 5y) × \(\frac { 3 }{ x }\) = \(6+\frac { 15y }{ x }\)
6. (3x² + 4y) × (2x + 3y) = 6x³ + 9x²y + 8xy + 12y²

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.1 8th Std Maths Answers Solutions Chapter 11 Statistics.

Practice Set 11.1 8th Std Maths Answers Chapter 11 Statistics

Question 1.
The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

No. of saplings (Scores) xi	No. of students (frequency) fi	fi × xi
1	4	4
2	6	__
3	12	__
4	8	__
	N = __	∑ fi × xi = __

Solution:

No. of saplings (Scores) xi	No. of students (frequency) fi	fi × xi
1	4	4
2	6	12
3	12	36
4	8	32
	N = __	∑ fi × xi = 84

Question 2.
The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

Electricity used (Units) xi	No. of families (frequency) fi	fi × xi
30	4
45	6
60	12
75	8
90	3
	N = __	∑ fi × xi =

i. How many families use 45 units electricity?
ii. State the score, the frequency of which is 5.
iii. Find N, and ∑ f_i × x_i .
iv. Find the mean of electricity used by each family in the month of May.
Solution:

Electricity used (Units) xi	No. of families (frequency) fi	fi × xi
30	7	210
45	2	90
60	8	480
75	5	375
90	3	270
	N = 25	∑ fi × xi = 1425

i. 2 families used 45 units of electricity.
ii. The score for which the frequency is 5 is 75
iii. N = 25 and ∑ f_i × x_i = 1425
iv.

The mean of electricity used by each.

Question 3.
The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2,3, 5, 6, 4, 2. Prepare a frequency table and And the mean of members of 40 families.
Solution:


∴ The mean of the members of 40 families is 3.9.

Question 4.
The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exhibition is:
2, 3 ,4, 1, 2, 3, 1, 5, 4, 2, 3, 1, 3, 5, 4, 3, 2, 2, 3, 2. Prepare a frequency table and find the mean of the data.
Solution:


∴ The mean of the given data is 2.75.

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.1 Intext Questions and Activities

Question 1.
The number of pages of a book Ninad read for five consecutive days were 60, 50, 54, 46, 50. Find the average number of pages he read everyday. (Textbook pg. no. 67)
Solution:
\(\frac{60+[50]+[54]+[46]+50}{[5]}=\frac{260}{[5]}=[52]\)
∴ Average number of pages read daily is 52

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

Practice Set 5.2 8th Std Maths Answers Chapter 5 Expansion Formulae

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 = \(\frac { 1 }{ x }\)

vii. Here, a = 2m and b = \(\frac { 1 }{ 5 }\)

viii. Here, a = \(\frac { 5x }{ y }\) and b = \(\frac { y }{ 5x }\)

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.

Practice Set 4.1 8th Std Maths Answers Chapter 4 Altitudes and Medians of a Triangle

Question 1.
In ∆LMN, ___ is an altitude and __ is a median, (write the names of appropriate segments.)

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:

Question 3.
Draw an obtuse angled ∆STV. Draw its medians and show the centroid.
Solution:

Question 4.
Draw an obtuse angled ∆LMN. Draw its altitudes and denote the ortho centre by ‘O’.
Solution:

Question 5.
Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.
Solution:

Question 6.
Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.
Solution:

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) = ___.

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.

Step 2: By placing a set-square on line l, draw a perpendicular to the line from point P.

Question 2.
Draw an acute angled ∆ABC and all its altitudes. Observe the location of the orthocentre. (Textbook pg. no. 20)
Solution:

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:

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.

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:

ii. Obtuse angled triangle:

iii. Acute angled triangle:

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:

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)

Solution:
Draw seg AP ⊥ seg PE and seg EQ ⊥ seg QC.

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:

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:

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.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.2 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Practice Set 9.2 8th Std Maths Answers Chapter 9 Discount and Commission

Question 1.
John sold books worth Rs 4500 for a publisher. For this he received 15% commission. Complete the following activity to find the total commission John obtained.
Solution:
Selling price of the books = Rs 4500
Rate of commission = 15%
Commission obtained = 15% of selling price
\(=\frac{[15]}{[100]} \times[4500]\)
= 15 × 45
∴ Commission obtained = 675 Rupees.
∴ The total commission obtained by John is Rs 675.

Question 2.
Rafique sold flowers worth Rs 15,000 by giving 4% commission to the agent. Find the commission he paid. Find the amount received by Rafique.
Solution:
Here, selling price of flowers = Rs 15,000,
Rate of commission = 4%
i. Commission = 4% of selling price
= \(\frac { 4 }{ 100 }\) × 15,000
= 4 x 150
∴ Commission = Rs 600

ii. Amount received by Rafique = selling price – commission
= 15,000 – 600
= Rs 14,400
∴ Rafique paid Rs 600 as commission and the amount received by him was Rs 14,400.

Question 3.
A farmer sold food grains for Rs 9200 through an agent. The rate of commission was 2%. How much amount did the agent get ?
Solution:
Here, selling price of food grains = Rs 9200,
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 9200
= 2 × 92
= Rs 184
∴ The agent got a commission of Rs 184.

Question 4.
Umatai purchased following items from a Khadi – Bhandar.
i. 3 sarees for Rs 560 each.
ii. 6 bottles of honey for Rs 90 each.
On the purchase, she received a rebate of 12%. How much total amount did Umatai pay?
Solution:
Here, number of sarees = 3,
Price of each saree = Rs 560
∴ Cost of 3 sarees = 560 × 3
= Rs 1680 …(i)
Also, number of honey bottles = 6,
Price of each bottle = Rs 90
∴ Cost of 6 honey bottles = 90 × 6
= Rs 540
Total amount of purchase
= cost of 3 sarees + cost of 6 honey bottles
= 1680 + 540 … [From (i) and (ii)]
= Rs 2220 …(iii)
Rate of rebate = 12%
Rebate = 12% of total amount of purchase
= \(\frac { 12 }{ 100 }\) × 2220
= 12 × 22.20
= Rs 266.40 ..(iv)
Amount paid by Umatai
= Total amount of purchase – Rebate
= 2,220 – 266.40 … [From (iii) and (iv)]
= Rs 1953.60
∴ The total amount paid by Umatai is Rs 1953.60.

Question 5.
Use the given information and fill in the boxes with suitable numbers.
Smt. Deepanjali purchased a house for Rs 7,50,000 from Smt. Leelaben through an agent. Agent has charged 2 % brokerage from both of them.
Solution:
i. Smt. Deepanjali paid 7,50,000 × \(\frac { 2 }{ 100 }\)
= 7,500 × 2 = Rs 15,000 brokerage for purchasing the house.

ii. Smt. Leelaben paid brokerage of Rs 15,000

iii. Total brokerage received by the agent is = 15,000 + 15,000 = Rs 30,000

iv. The cost of house Smt. Deepanjali paid is = 7,50,000 + 15,000 = Rs 7,65,000

v. The selling price of house for Smt.Leelaben is = 7,50,000 – 15,000
= Rs 7,35,000