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 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 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

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.1 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.1 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Construct the following quadrilaterals of given measures.

Question 1.
In ∆MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90°.
Solution:

Question 2.
Construct ∆DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.
Solution:

Question 3.
In ∆ABCD, l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70°, m∠B = 50°, m∠C = 140°.
Solution:

Question 4.
Construct ₹LMNO such that
l(LM) = l(LO) = 6 cm,
l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.
Solution:

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.1 Intext Questions and Activities

Question 1.
Construction of a triangle:
Construct the triangles with given measures. (Textbook pg. no. 41)
i. ∆ABC: l(AB) = 5 cm, l(BC) = 5.5, l(AC) = 6 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg BC of length 5.5 cm as the base.
Step 2 : By taking a distance of 5 cm on the compass and placing the metal tip of the compass on point B, draw an arc on one side of BC.
Step 3 : By taking a distance 6 cm on the compass and placing the metal tip of the t compass on point C and draw an arc ’ such that it intersects the previous arc. Name the point as A.
Step 4 : Draw segments AB and AC to get the triangle. ∆ABC is the required triangle.

ii. ∆DEF: m∠D = 35°, m∠F = 100°, l(DF) = 4.8 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg DF of length 4.8 cm as the base.
Step 2 : Placing the centre of the protractor at point D, mark point P such that m∠PDF = 35°.
Step 3 : Placing the centre of the protractor at point F, mark point Q such that m∠QFD = 100°.
Step 4 : Draw ray DP and ray FQ. Name their point of intersection as E.
∆DEF is required triangle.

iii. ∆MNP: l(MP) = 6.2 cm, l(NP) = 4.5 cm, m∠P = 75°.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg PN of length 4.5 cm as the base.
Step 2 : Placing the centre of the protractor at point P, mark point Q such that m∠QPN = 75°.
Step 3 : By taking a distance of 6.2 cm on the compass and placing the metal tip at point P, draw an arc on ray PQ. Name the point as M.
Step 4 : Draw seg MN to get the triangle. ∆MNP is the required triangle.

iv. ∆XYZ: m∠Y = 90°, l(XY) = 4.2 cm, l(XZ) = 7 cm.
Solution:

Steps of construction:
Step 1 : As shown in the rough figure, draw seg XY of 4.2 cm as the base.
Step 2 : Placing the centre of the protractor at point Y, mark point Q such that m∠QYX = 90°.
Step 3 : By taking a distance of 7 cm on the compass and placing the metal tip on point X, draw an arc on ray YQ. Name the point as Z.
Step 4 : Draw seg XZ to get the triangle. ∆XYZ is the required triangle.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.3 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.3 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Question 1.
Measures of opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of its each angle.
Solution:
Let ₹PQRS be the parallelogram.
m∠Q = (3x – 2)° and m∠S = (50 – x)°

m∠Q = m∠S
…..(i)
[Opposite angles of a parallelogram are congruent]
∴ 3x – 2 = 50 – x
∴ 3x + x = 50 + 2
∴ 4x = 52
∴ x = \(\frac { 52 }{ 4 }\)
∴ x = 13
Now, m∠Q = (3x – 2)°
= (3 x 13 – 2)° = (39 – 2)° = 37°
∴ m∠S = m∠Q = 37° …[From(i)]
m∠P + m∠Q = 180°
….[Adjacent angles of a parallelogram are supplementary]
∴ m∠P + 37° = 180°
∴ m∠P = 180° – 37° = 143°
∴ m∠R = m∠P = 143°
…..[Opposite angles of a parallelogram are congruent]
∴ The measures of the angles of the parallelogram are 37°, 143°, 37° and 143°.

Question 2.
Referring the given figure of a parallelogram, write the answers of questions given below.
i. If l(WZ) = 4.5 cm, then l(XY) = ?
ii. If l(YZ) = 8.2 cm, then l(XW) = ?
iii. If l(OX) = 2.5 cm, then l(OZ) = ?
iv. If l(WO) = 3.3 cm, then l(WY) = ?
v. If m∠WZY = 120°, then m∠WXY = ? and m∠XWZ = ?

Solution:
i. l(WZ) = 4.5 cm … [Given]
l(X Y) = l(WZ) ….[Opposite sides of a parallelogram are congrument ]
∴ l(X Y) = 4.5cm

ii. l(YZ) = 8.2 cm …[Given]
l(XW) = l(YZ)
…[Opposite sides of a parallelogram are congruent]
∴ l(XW) = 8.2cm … [Given]

iii. l(OX) = 2.5 cm …[Given]
l(OZ) = l(OX)
….[Diagonals of a parallelogram bisect each other]
∴ l(OZ) = 2.5cm

iv. l(WO) = 3.3 cm … [Given]
l(WO) = \(\frac { 1 }{ 2 }\) l(WY)
….[Diagonals of a parallelogram bisect each other]
∴ 3.3 = \(\frac { 1 }{ 2 }\) l(WY)
∴ 3.3 x 2 = l(WY)
∴ l(WY) = 6.6cm

v. m∠WZY =120° … [Given]
m∠WXY = m∠WZY
…..[Opposite angles of a parallelogram are congrument]
∴ m∠WXY = 120° …(i)
m∠XWZ + m∠WXY = 180°
….[Adjacent angles of a parallelogram are supplementary]
∴ m∠XWZ + 120° = 180° … [From (i)]
∴ m∠XWZ = 180°- 120°
∴ m∠XWZ = 60°

Question 3.
Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40°, l(AB) = 3 cm
Solution:

Opposite sides of a parallelogram are congruent.
∴ l(AB) = l(CD) = 3cm
l(BC) = l(AD) = 7 cm

Question 4.
Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4. Find the measure of its each angle. Write with reason, what type of a quadrilateral it is.
Solution:
Let ₹PQRS be the quadrilateral.
Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4.
Let the common multiple be x.
∴m∠P = x°, m∠Q = 2x°, m∠R = 3x° and m∠S = 4x°
In ₹PQRS,
m∠P + m∠Q + m∠R + m∠S = 360°
…[Sum of the measures of the angles of a quadrilateral is 360°]
∴x° + 2x° + 3x° + 4x° = 360°
∴10 x° = 360°
∴x° = \(\frac { 360 }{ 10 }\)
∴x° = 36°
∴m∠P = x° = 36°

m∠Q = 2x° = 2 × 36° = 72°
m∠R = 3x° = 3 × 36° = 108° and
m∠S = 4x° = 4 × 36° = 144°
∴The measures of the angles of the quadrilateral are 36°, 72°, 108°, 144°.
Here, m∠P + m∠S = 36° + 144° = 180°
Since, interior angles are supplementary,
∴side PQ || side SR
m∠P + m∠Q = 36° + 72° = 108° ≠ 180°
∴side PS is not parallel to side QR.
Since, one pair of opposite sides of the given quadrilateral is parallel.
∴The given quadrilateral is a trapezium.

Question 5.
Construct ₹BARC such that
l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm
Solution:

Question 6.
Construct ₹PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110°, m∠R = 70°. If it is given that ₹PQRS is a parallelogram, which of the given information is unnecessary?
Solution:

1. Since, the opposite sides of a parallelogram are congruent.
   ∴ Either l(PQ) or l(SR) is required.
2. To construct a parallelogram lengths of adjacent sides and measure of one angle is required.
   ∴ Either l(PQ) and m∠Q or l(SR) and m∠R is the unnecessary information given in the question.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.3 Intext Questions and Activities

Question 1.
Draw a parallelogram PQRS. Take two rulers of different widths, place one ruler horizontally and draw lines along its edges. Now place the other ruler in slant position over the lines drawn and draw lines along its edges. We get a parallelogram. Draw the diagonals of it and name the point of intersection as T.

1. Measure the opposite angles of the parallelogram.
2. Measure the lengths of opposite sides.
3. Measure the lengths of diagonals.
4. Measure the lengths of parts of the diagonals made by point T. (Textbook pg. no. 47)

Solution:
[Students should attempt the above activities on their own.]

Question 2.
In the given figure of ₹ABCD, verify with a divider that seg AB ≅ seg CB and seg AD ≅ seg CD. Similarly measure ∠BAD and ∠BCD and verify that they are congruent. (Textbook pg. no. 48)

Solution:
[Students should attempt the above activities on their own.]

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 7.3 8th Std Maths Answers Solutions Chapter 7 Variation.

Practice Set 7.3 8th Std Maths Answers Chapter 7 Variation

Question 1.
Which of the following statements are of inverse variation?
i. Number of workers on a job and time taken by them to complete the job.
ii. Number of pipes of same size to fill a tank and the time taken by them to fill the tank.
iii. Petrol filled in the tank of a vehicle and its cost.
iv. Area of circle and its radius.
Solution:
i. Let, x represent number of workers on a job, and y represent time taken by workers to complete the job.
As the number of workers increases, the time required to complete the job decreases.
∴ \(x \propto \frac{1}{y}\)

ii. Let, n represent number of pipes of same size to fill a tank and t represent time taken by the pipes to fill the tank.
As the number of pipes increases, the time required to fill the tank decreases.
∴ \(\mathrm{n} \propto \frac{1}{\mathrm{t}}\)

iii. Let, p represent the quantity of petrol filled in a tank and c represent the cost of the petrol.
As the quantity of petrol in the tank increases, its cost increases.
∴ p ∝ c

iv. Let, A represent the area of the circle and r represent its radius.
As the area of circle increases, its radius increases.
∴ A ∝ r
∴ Statements (i) and (ii) are examples of inverse variation.

Question 2.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Let, n represent the number of workers building the wall and t represent the time required.
Since, the number of workers varies inversely with the time required to build the wall.
∴ \(\mathrm{n} \propto \frac{1}{\mathrm{t}}\)
∴ \(\mathrm{n}=\mathrm{k} \times \frac{1}{\mathrm{t}}\)
where k is the constant of variation
∴ n × t = k …(i)
15 workers can build a wall in 48 hours,
i.e., when n = 15, t = 48
∴ Substituting n = 15 and t = 48 in (i), we get
n × t = k
∴ 15 × 48 = k
∴ k = 720
Substituting k = 720 in (i), we get
n × t = k
∴ n × t = 720 …(ii)
This is the equation of variation.
Now, we have to find number of workers required to do the same work in 30 hours.
i.e., when t = 30, n = ?
∴ Substituting t = 30 in (ii), we get
n × t = 720
∴ n × 30 = 720
∴ n = \(\frac { 720 }{ 30 }\)
∴ n = 24
∴ 24 workers will be required to build the wall in 30 hours.

Question 3.
120 bags of half litre milk can be filled by a machine within 3 minutes find the time to fill such 1800 bags?
Solution:
Let b represent the number of bags of half litre milk and t represent the time required to fill the bags.
Since, the number of bags and time required to fill the bags varies directly.
∴ b ∝ t
∴ b = kt …(i)
where k is the constant of variation.
Since, 120 bags can be filled in 3 minutes
i.e., when b = 120, t = 3
∴ Substituting b = 120 and t = 3 in (i), we get
b = kt
∴ 120 = k × 3
∴ k = \(\frac { 120 }{ 3 }\)
∴ k = 40
Substituting k = 40 in (i), we get
b = kt
∴ b = 40 t …(ii)
This is the equation of variation.
Now, we have to find time required to fill 1800 bags
∴ Substituting b = 1800 in (ii), we get
b = 40 t
∴ 1800 = 40 t
∴ t = \(\frac { 1800 }{ 40 }\)
∴ t = 45
∴ 1800 bags of half litre milk can be filled by the machine in 45 minutes.

Question 4.
A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is
to be covered in \(7\frac { 1 }{ 2 }\) hours?
Solution:
Let v represent the speed of car in km/hr and t represent the time required.
Since, speed of a car varies inversely as the time required to cover a distance.
∴ \(v \propto \frac{1}{t}\)
∴ \(\mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\)
where, k is the constant of variation.
∴ v × t = k …(i)
Since, a car with speed 60 km/hr takes 8 hours to travel some distance.
i.e., when v = 60, t = 8
∴ Substituting v = 60 and t = 8 in (i), we get
v × t = k
∴ 60 × 8 = t
∴ k = 480
Substituting k = 480 in (i), we get
v × t = k
∴ v × t = 480 …(ii)
This is the equation of variation.
Now, we have to find speed of car if the same distance is to be covered in \(7\frac { 1 }{ 2 }\) hours.
i.e., when t = \(7\frac { 1 }{ 2 }\) = 7.5 , v = ?
∴ Substituting, t = 7.5 in (ii), we get
v × t = 480
∴ v × 7.5 = 480
\(v=\frac{480}{7.5}=\frac{4800}{75}\)
∴ v = 64
The speed of vehicle should be 64 km/hr to cover the same distance in 7.5 hours.
∴ The increase in speed = 64 – 60
= 4km/hr
∴ The increase in speed of the car is 4 km/hr.