Prasanna

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 9 Answers Solutions.

6th Standard Maths Practice Set 9 Answers Chapter 4 Operations on Fractions

Question 1.
Convert into improper fractions:
i. \(7 \frac{2}{5}\)
ii. \(5 \frac{1}{6}\)
iii. \(4 \frac{3}{4}\)
iv. \(2 \frac{5}{9}\)
v. \(1 \frac{5}{7}\)
Solution:
i. \(7 \frac{2}{5}\)

ii. \(5 \frac{1}{6}\)

iii. \(4 \frac{3}{4}\)

iv. \(2 \frac{5}{9}\)

v. \(1 \frac{5}{7}\)

Question 2.
Convert into mixed numbers:
i. \(\frac { 30 }{ 7 }\)
ii. \(\frac { 7 }{ 4 }\)
iii. \(\frac { 15 }{ 12 }\)
iv. \(\frac { 11 }{ 8 }\)
v. \(\frac { 21 }{ 4 }\)
v. \(\frac { 20 }{ 7 }\)
Solution:
i. \(\frac { 30 }{ 7 }\)

ii. \(\frac { 7 }{ 4 }\)

iii. \(\frac { 15 }{ 12 }\)

iv. \(\frac { 11 }{ 8 }\)

v. \(\frac { 21 }{ 4 }\)

v. \(\frac { 20 }{ 7 }\)

Question 3.
Write the following examples using fraction:
i. If 9 kg rice is shared among 5 people, how many kilograms of rice does each person get?
ii. To make 5 shirts of the same size, 11 meters of cloth is needed. How much cloth is needed for one shirt?
Solution:
i. Total quantity of rice = 9 kg
Number of people = 5
∴ Kilograms of rice received by each person = \(\frac { 9 }{ 5 }\)
∴ Each person will get \(\frac { 9 }{ 5 }\) kg of rice.

ii. Total meters of cloth = 11 meters
Number of shirts to be made = 5
Meters of cloth needed to make 1 shirt = \(\frac { 11 }{ 5 }\)
∴ Cloth needed to make 1 shirt is \(\frac { 11 }{ 5 }\) meters.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.3 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
Write sample space ‘S’ and number of sample points n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).

i. One die is rolled,
Event A: Even number on the upper face.
Event B: Odd number on the upper face.
Event C: Prime number on the upper face.

ii. Two dice are rolled simultaneously,
Event A: The sum of the digits on upper faces is a multiple of 6.
Event B: The sum of the digits on the upper faces is minimum 10.
Event C: The same digit on both the upper faces.

iii. Three coins are tossed simultaneously.
Condition for event A: To get at least two heads.
Condition for event B: To get no head.
Condition for event C: To get head on the second coin.

iv. Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A: The number formed is even.
Condition for event B: The number is divisible by 3.
Condition for event C: The number formed is greater than 50.

v. From three men and two women, environment committee of two persons is to be formed.
Condition for event A: There must be at least one woman member.
Condition for event B: One man, one woman committee to be formed.
Condition for event C: There should not be a woman member.

vi. One coin and one die are thrown simultaneously.
Condition for event A: To get head and an odd number.
Condition for event B: To get a head or tail and an even number.
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
Solution:
i. Sample space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Condition for event A: Even number on the upper face.
∴ A = {2,4,6}
∴ n(A) = 3
Condition for event B: Odd number on the upper face.
∴ B = {1, 3, 5}
∴ n(B) = 3
Condition for event C: Prime number on the upper face.
∴ C = {2, 3, 5}
∴ n(C) = 3

ii. Sample space,
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
∴ n(S) = 36
Condition for event A: The sum of the digits on the upper faces is a multiple of 6.
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
∴ n(A) = 6

Condition for event B: The sum of the digits on the upper faces is minimum 10.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 6

Condition for event C: The same digit on both the upper faces.
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(C) = 6

iii. Sample space,
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ n(S) = 8

Condition for event A: To get at least two heads.
∴ A = {HHT, HTH, THH, HHH}
∴ n(A) = 4

Condition for event B: To get no head.
∴ B = {TTT}
∴ n(B) = 1

Condition for event C: To get head on the second coin.
∴ C = {HHH, HHT, THH, THT}
∴ n(C) = 4

iv. Sample space (S) = {10, 12, 13, 14, 15,
20, 21, 23, 24, 25,
30, 31, 32, 34, 35,
40, 41, 42, 43,
45, 50, 51, 52, 53, 54}
∴ n(S) = 25
Condition for event A: The number formed is even
∴ A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54)
∴ n(A) = 13
Condition for event B: The number formed is divisible by 3.
∴ B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n(B) = 9
Condition for event C: The number formed is greater than 50.
∴ C = {51,52, 53,54}
∴ n(C) = 4

v. Let the three men be M₁, M₂, M₃ and the two women be W₁, W₂.
Out of these men and women, a environment committee of two persons is to be formed.
∴ Sample space,
S = {M₁M₂, M₁M₃, M₁W₁, M₁W₂, M₂M₃, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}
∴ n(S) = 10
Condition for event A: There must be at least one woman member.
∴ A = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}
∴ n(A) = 7
Condition for event B: One man, one woman committee to be formed.
∴ B = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₂, M₃W₂}
∴ n(B) = 6
Condition for event C: There should not be a woman member.
∴ C = {M₁M₂, M₁M₃, M₂M₃}
∴ n(C) = 3

vi. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) = 12
Condition for event A: To get head and an odd number.
∴ A = {(H, 1), (H, 3), (H, 5)}
∴ n(A) = 3
Condition for event B: To get a head or tail and an even number.
∴ B = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}
∴ n(B) = 6
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
The greatest number on the upper face of a die is 6.
∴ Event C is an impossible event.
∴ C = { }
∴ n(C) = 0

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 40 Answers Solutions Chapter 10 Bank and Simple Interest.

Bank and Simple Interest Class 7 Practice Set 40 Answers Solutions Chapter 10

Question 1.
If Rihanna deposits Rs 1500 in the school fund at 9 p.c.p.a for 2 years, what is the total amount she will get?
Solution:
Here, P = Rs 1500, R = 9 p.c.p.a , T = 2 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{1500 \times 9 \times 2}{100}\)
= 1500 x 9 x 2
= Rs 270
∴ Total amount = Principal + Interest
= 1500 + 270
= Rs 1770
∴ Rihanna will get a total amount of Rs 1770.

Question 2.
Jethalal took a housing loan of Rs 2,50,000 from a bank at 10 p.c.p.a. for 5 years. What is the yearly interest he must pay and the total amount he returns to the bank?
Solution:
Here, P = Rs 250000, R = 10 p.c.p.a., T = 5 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{250000 \times 10 \times 5}{100}\)
= 2500 x 10 x 5
= Rs 1,25,000
∴ Yearly interest = Total interest ÷ Time = 1,25,000 ÷ 5 = Rs 25000
Total amount to be returned = Principal + Total interest
= 250000 + 125000 = Rs 375000
∴ The yearly interest is Rs 25,000 and Jethalal will have to return Rs 3,75,000 to the bank.

Question 3.
Shrikant deposited Rs 85,000 for \(2\frac { 1 }{ 2 }\) years at 7 p.c.p.a. in a savings bank account. What is the total
interest he received at the end of the period?
Solution:
Here, P = Rs 85000, R = 7 p.c.p.a., T = \(2\frac { 1 }{ 2 }\) years = 2.5 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{85000 \times 7 \times 2.5}{100}\)
= \(\frac{85000 \times 7 \times 25}{100 \times 10}\)
= 85 x 7 x 25
= Rs 14875
∴ The total interest received by Shrikant at the end of the period is Rs 14875.

Question 4.
At a certain rate of interest, the interest after 4 years on Rs 5000 principal is Rs 1200. What would be the interest on Rs 15000 at the same rate of interest for the same period?
Solution:
The interest on Rs 5000 after 4 years is Rs 1200.
Let us suppose the interest on Rs 15000 at the same rate after 4 years is Rs x.
Taking the ratio of interest and principal, we get
∴ \(\frac{x}{15000}=\frac{1200}{5000}\)
∴ \(x=\frac{1200 \times 15000}{5000}\)
= Rs 3600
∴ The interest received on Rs 15000 is Rs 3600.

Question 5.
If Pankaj deposits Rs 1,50,000 in a bank at 10 p.c.p.a. for two years, what is the total amount he will get from the bank?
Solution:
Here, P = 150000, R = 10 p.c.p.a., T = 2 years
∴ Total interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{150000 \times 10 \times 2}{100}\)
= Rs 30000
∴ Total amount = Principal + Total Interest
= 150000 + 30000
= Rs 180000
∴ Pankaj will receive Rs 180000 from the bank.

Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 40 Intext Questions and Activities

Question 1.
Observe the entries made in the page of a passbook shown below and answer the following questions. (Textbook pg. no. 70)

1. On 2.2.16 the amount deposited was Rs__and the balance Rs__.
2. On 12.2.16, Rs__were withdrawn by cheque no. 243965. The balance was Rs__
3. On 26.2.2016 the bank paid an interest of Rs__

Solution:

1. 1500, 7000
2. 3000, 9000
3. 135

Practice Set 40 Class 7 Question 2.
Suvidya borrowed a sum of Rs 30000 at 8 p.c.p.a. interest for a year from her bank to buy a computer. At the end of the period, she had to pay back an amount of Rs 2400 over and above what she had borrowed.
Based on this information fill in the blanks below. (Textbook pg. no. 70)

1. Principal = Rs__
2. Rate of interest =__%
3. Interest = Rs__
4. Time =__year.
5. The total amount returned to the bank = 30,000 + 2,400 = Rs__

Solution:

1. 30000
2. 8
3. 2400
4. 1
5. Rs 32400

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 16 Quadrilaterals Class 6 Practice Set 37 Answers Solutions.

6th Standard Maths Practice Set 37 Answers Chapter 16 Quadrilaterals

Question 1.
Observe the figures below and find out their names:

Solution:
i. Pentagon (5 sides)
ii. Hexagon (6 sides)
iii. Heptagon (7 sides)
iv. Octagon (8 sides)

Maharashtra Board Class 6 Maths Chapter 16 Quadrilaterals Practice Set 37 Intext Questions and Activities

Question 1.
Observe the figures given below and say which of them are quadrilaterals. (Textbook pg. no. 81)

Solution:
Is a quadrilateral: (i)

Question 2.
Draw a quadrilateral. Draw one diagonal of this quadrilateral and divided it into two triangles. Measures all the angles in the figure. Is the sum of the measures of the four angles of the quadrilateral equal to the sum of the measures of the six angles of the two triangles? Verity that this is so with other quadrilaterals. (Textbook pg. no. 84)
Solution:

m∠PQR = 104°
m∠QRP = 26°
m∠RPQ = 50°
m∠PRS = 34°
m∠RSP = 106°
m∠SPR = 40°
∴ Sum of the measures of the angles of quadrilateral = m∠PQR + m∠QRP + m∠RPQ + m∠PRS + m∠RSP + m∠SPR
= 104° + 26° + 50° + 34° + 106° + 40°
= 360°
Also, we observe that
Sum of the measures of the angles of quadrilateral = Sum of the measures of angles of the two triangles (PQR and PRS)
= (104°+ 26°+ 50°)+ (34° + 106° + 40°)
= 180° + 180°
= 360°
[Note: Students should drew different quadrilaterals and verify the property.]

Question 3.
For the pentagon shown in the figure below, answer the following: (Textbook pg. no. 84)

1. Write the names of the five vertices of the pentagon.
2. Name the sides of the pentagon.
3. Name the angles of the pentagon.
4. See if you can sometimes find players on a field forming a pentagon.

Solution:

1. The vertices of the pentagon are points A, B, C, D and E.
2. The sides of the pentagon are segments AB, BC, CD, DE and EA.
3. The angles of the pentagon are ∠ABC, ∠BCD, ∠CDE, ∠DEA and ∠EAB.
4. The players shown in the above figure form a pentagon. The players are standing on the vertices of

Question 4.
Cut out a paper in the shape of a quadrilateral. Make folds in it that join the vertices of opposite angles. What can these folds be called? (Textbook pg. no. 83)

Solution:
The folds are called diagonals of the quadrilateral.

Question 5.
Take two triangular pieces of paper such that . one side of one triangle is equal to one side of the other. Let us suppose that in ∆ABC and ∆PQR, sides AC and PQ are the equal sides. Join the triangles so that their equal sides lie B side by side. What figure do we get? (Textbook pg. no. 83)

Solution:
If we place the triangles together such that the equal sides overlap, the two triangles form a quadrilateral.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 13 Profit-Loss Class 6 Practice Set 33 Answers Solutions.

6th Standard Maths Practice Set 33 Answers Chapter 13 Profit-Loss

Question 1.
Maganlal bought trousers for Rs 400 and a shirt for Rs 200 and sold them for Rs 448 and Rs 250 respectively. Which of these transactions was more profitable?
Solution:
Cost price of trousers = Rs 400
Selling price of trousers = Rs 448
Profit = Selling price – Cost price
= 448 – 400 = Rs 48
Let Maganlal make x % profit on selling trousers

∴ x = 12%
Cost price of shirt = Rs 200
Selling price of shirt = Rs 250
∴ Profit = Selling price – Cost price
= 250 – 200
= Rs 50
Let Maganlal make y% profit on selling shirt.

∴ y = 25%
∴ Transaction involving selling of shirt was more profitable.

Question 2.
Ramrao bought a cupboard for Rs 4500 and sold it for Rs 4950. Shamrao bought a sewing machine for Rs 3500 and sold it for Rs 3920. Whose transaction was more profitable?
Solution:
Cost price of cupboard = Rs 4500
Selling price of cupboard = Rs 4950
∴ Profit = Selling price – Cost price
= 4950 – 4500
= Rs 450
Let Ramrao make x% profit on selling cupboard

∴ x = 10%
Cost price of sewing machine = Rs 3500
Selling price of sewing machine = Rs 3920
∴Profit = Selling price – Cost price
= 3920 – 3500
= Rs 420
Shamrao make y% profit on selling sewing machine.

∴y = 12%
∴Shamrao’s transaction was more profitable.

Question 3.
Hanif bought one box of 50 apples for Rs 400. He sold all the apples at the rate of Rs 10 each. Was there a profit or loss? What was its percentage?
Solution:
Cost price of 50 apples = Rs 400
Selling price of one apple = Rs 10
∴ Selling price of 50 apples = 10 x 50 = Rs 500
Selling price is greater than the total cost price.
∴ Hanif made a profit.
∴ Profit = Selling price – Cost price
= 500 – 400
= Rs 100
Let Hanif make of x% profit on selling apples.

∴ x = 25%
∴ Hanif made a profit of 25%.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 9 HCF-LCM Class 6 Practice Set 24 Answers Solutions.

6th Standard Maths Practice Set 24 Answers Chapter 9 HCF-LCM

Question 1.
Find the HCF of the following numbers.
i. 45, 30
ii. 16, 48
iii. 39, 25
iv. 49, 56
v. 120, 144
vi. 81, 99
vii. 24, 36
viii. 25, 75
ix. 48, 54
x. 150, 225
Solution:
i. Factors of 45 = 1, 3, 5, 9,15, 45
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
∴ HCF of 45 and 30 = 15

ii. Factors of 16 = 1, 2, 4, 8, 16
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
∴ HCF of 16 and 48 = 16

iii. Factors of 39 = 1, 3, 13, 39
Factors of 25 = 1, 5, 25
∴ HCF of 39 and 25 = 1

iv. Factors of 49 = 1, 7, 49
Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
∴ HCF of 49 and 56 = 7

v. Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
Factors of 144 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
∴ HCF of 120 and 144 = 24

vi. Factors of 81 = 1, 3, 9, 27, 81
Factors of 99 = 1, 3, 9, 11, 33, 99
∴ HCF of 81 and 99 = 9

vii. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
∴ HCF of 24 and 36 = 12

viii. Factors of 25 = 1, 5, 25
Factors of 75 = 1, 3, 5, 15, 25, 75
∴ HCF of 25 and 75 = 25

ix. Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54
∴ HCF of 48 and 54 = 6

x. Factors of 150 = 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150
Factors of 225 = 1, 3, 5, 9, 15, 25, 45, 75, 225
∴ HCF of 150 and 225 = 75

Question 2.
If large square beds of equal size are to be made for planting vegetables on a plot of land 18 metres long and 15 metres wide, what is the maximum possible length of each bed?
Solution:
Length of the land = 18 m
Width of the land = 15 m
The maximum length of each bed will be the greatest common factor of 18 and 15.
Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 15 = 1, 3, 5, 15
∴ HCF of 18 and 15 = 3
∴ The maximum possible length of each bed is 3 metres.

Question 3.
Two ropes, one 8 metres long and the other 12 metres long are to be cut into pieces of the same length. What will the maximum possible length of each piece be?
Solution:
Length of first rope = 8 m
Length of second rope = 12 m
The maximum length of each piece will be the greatest common factor of 8 and 12.
Factors of 8 = 1, 2, 4, 8
Factors of 12 = 1, 2, 3, 4, 6, 12
∴ HCF of 8 and 12 = 4
∴ The maximum possible length of each piece is 4 metres.

Question 4.
The number of students of Std 6th and Std 7th who went to visit the Tadoba Tiger Project at Chandrapur was 140 and 196 respectively. The students of each class are to be divided into groups of the same number of students. Each group can have a paid guide. What is the maximum number of students that can be there in each group? Why do you think each group should have the maximum possible number of students?
Solution:
Number of students of Std 6th = 140
Number of students of Std 7th = 196
The maximum number of students in each group will be the greatest common factor of 140 and 196.
Factors of 140 = 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140
Factors of 196 = 1, 2, 4, 7, 14, 28, 49, 98, 196
∴ HCF of 140 and 196 = 28
∴ Maximum students in each group are 28.
Each group should have maximum number students so that there will be minimum number of groups and hence minimum number of paid guides.

Question 5.
At the Rice Research Centre at Tumsar there are 2610 kg of seeds of the basmati variety and 1980 kg of the indrayani variety. If the maximum possible weight of seeds has to be filled to make bags of equal weight what would be the weight of each bag? How many bags of each variety will there be?
Solution:
Weight of basmati rice = 2610 kg
Weight of indrayani rice = 1980 kg
The weight of each bag will be the greatest common factor of 2610 and 1980.
Factors of 2610 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 29, 30, 45, 58, 87, 90, 145, 174, 261, 290, 435, 522, 870, 1305, 2610
Factors of 1980 = 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, 22, 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990, 1980
∴ HCF of 2610 and 1980 = 90
Maximum weight of each bag = 90 kg
Number of bags of basmati rice = 2610 ÷ 90 = 29
Number of bags of indrayani rice = 1980 ÷ 90 = 22
Maximum weight of each bag is 90 kg.
The number of bags of basmati rice is 29, and the number of bags of indrayani rice is 22.

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 9 Maths Solutions covers the Practice Set 4.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.5 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let the number to be subtracted be x.
∴ (12 – x), (16 – x) and (21 – x) are in continued proportion.

∴ 84 – 4x = 80 – 5x
∴ 5x – 4x = 80 – 84
∴ x = -4
∴ -4 should be subtracted from 12,16 and 21 so that the resultant numbers in continued proportion.

Question 2.
If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.
Solution:
(28 – x) is the mean proportional of (23 – x) and (19-x). …[Given]

∴ -5(19 – x) = 9(28 – x)
∴ -95 + 5x = 252 – 9x
∴ 5x + 9x = 252 + 95
∴ 14x = 347
∴ x = \(\frac { 347 }{ 14 }\)

Question 3.
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let the first number be x.
∴ Third number = 26 – x
12 is the mean proportional of x and (26 – x).
∴ \(\frac { x }{ 12 }\) = \(\frac { 12 }{ 26 – x }\)
∴ x(26 – x) = 12 x 12
∴ 26x – x² = 144
∴ x² – 26x + 144 = 0
∴ x² – 18x – 8x + 144 = 0
∴ x(x – 18) – 8(x – 18) = 0
∴ (x – 18) (x – 8) = 0
∴ x = 18 or x = 8
∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18
∴ The numbers are 18, 12, 8 or 8, 12, 18.

Question 4.
If (a + b + c)(a – b + c) = a² + b² + c², show that a, b, c are in continued proportion.
Solution:
(a + b + c)(a – b + c) = a² + b² + c² …[Given]
∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2
∴ a² – ab + ac + ab – b² + be + ac – be + c² = a² + b² + c²
∴ a² + 2ac – b² + c² = a² + b² + c²
∴ 2ac – b² = b²
∴ 2ac = 2b²
∴ ac = b²
∴ b² = ac
∴ a, b, c are in continued proportion.

Question 5.
If \(\frac { a }{ b }\) = \(\frac { b }{ c }\) and a, b, c > 0, then show that,
i. (a + b + c)(b – c) = ab – c²
ii. (a² + b²)(b² + c²) = (ab + be)²
iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)
Solution:
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk =(ck)k
∴ a = ck² …(ii)

i. (a + b + c)(b – c) = ab – c²
L.H.S = (a + b + c) (b – c)
= [ck² + ck + c] [ck – c] … [From (i) and (ii)]
= c(k² + k + 1) c (k – 1)
= c² (k² + k + 1) (k – 1)
R.H.S = ab – c²
= (ck²) (ck) – c² … [From (i) and (ii)]
= c²k³ – c²
= c²(k³ – 1)
= c² (k – 1) (k² + k + 1) … [a³ – b³ = (a – b) (a² + ab + b²]
∴ L.H.S = R.H.S
∴ (a + b + c) (b – c) = ab – c²

ii. (a² + b²)(b² + c²) = (ab + bc)²
b = ck; a = ck²
L.H.S = (a² + b²) (b² + c²)
= [(ck²) + (ck)²] [(ck)² + c²] … [From (i) and (ii)]
= [c²k⁴ + c²k²] [c²k² + c²]
= c²k² (k² + 1) c² (k² + 1)
= c4k² (k² + 1)²
R.H.S = (ab + bc)²
= [(ck²) (ck) + (ck)c]² …[From (i) and (ii)]
= [c²k³ + c²k]²
= [c²k (k² + 1)]2 = c⁴(k² + 1)²
∴ L.H.S = R.H.S
∴ (a² + b²) (b² + c²) = (ab + bc)²

iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)

9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of \(\frac{x+y}{x-y}, \frac{x^{2}-y^{2}}{x^{2} y^{2}}\).
Solution:
Let a be the mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^{2}-y^{2}}{x^{2} y^{2}}\)

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 1 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Question 1.
Solve the following:
i. (-16) × (-5)
ii. (72) ÷ (-12)
iii. (-24) × (2)
iv. 125 ÷ 5
v. (-104) ÷ (-13)
vi. 25 × (-4)
Solution:
i. (-16) × (-5) = 80

ii. 72 ÷ (-12) = \(\frac { 72 }{ -12 }\)
= \(\frac{1}{(-1)} \times \frac{72}{12}\)
(-1) × 12
= -6

iii. (-24) × 2 = -48

iv. 125 ÷ 5 = \(\frac { 125 }{ 5 }\)
= 25

v. (-104) ÷ (-13) = \(\frac { -104 }{ -13 }\)
= \(\frac { 104 }{ 13 }\)
= 8

vi. 25 × (-4) = -100

Question 2.
Find the prime factors of the following numbers and find their LCM and HCF:
i. 75,135
ii. 114,76
iii. 153,187
iv. 32,24,48
Solution:
i. 75 = 3 × 25
= 3 × 5 × 5
135 = 3 × 45
= 3 × 3 × 15
= 3 × 3 × 3 × 5
∴ HCF of 75 and 135 = 3 × 5
= 15
LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3
= 675

ii. 114 = 2 × 57
= 2 × 3 × 19
76 = 2 × 38
= 2 × 2 × 19
∴ HCF of 114 and 76 = 2 × 19
= 38
LCM of 114 and 76 = 2 × 19 × 3 × 2
= 228

iii. 153 = 3 × 51
= 3 × 3 × 17
187 = 11 × 17
∴ HCF of 153 and 187 = 17
LCM of 153 and 187 = 17 × 3 × 3 × 11
= 1683

iv. 32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
∴ HCF of 32, 24 and 48 = 2 × 2 × 2
= 8
LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question 3.
Simplify:
i. \(\frac { 322 }{ 391 }\)
ii. \(\frac { 247 }{ 209 }\)
iii. \(\frac { 117 }{ 156 }\)
Solution:
i. \(\frac { 322 }{ 391 }\)

ii. \(\frac { 247 }{ 209 }\)

iii. \(\frac { 117 }{ 156 }\)

Question 4.
i. 784
ii. 225
iii. 1296
iv. 2025
v. 256
Solution:
i. 784

∴ 784 = 2 × 2 × 2 × 2 × 7 × 7
∴ √784 = 2 × 2 × 7
= 28

ii. 225

∴ 225 = 3 × 3 × 5 × 5
∴ √225 = 3 × 5
= 15

iii. 1296

∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ √1296 = 2 × 2 × 3 × 3
= 36

iv. 2025

∴ 2025 = 3 × 3 × 3 × 3 × 5 × 5
∴ √2025 = 3 × 3 × 5
= 45

v. 256

∴ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ √256 = 2 × 2 × 2 × 2
= 16

Question 5.
There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.

Polling Booths	Navodaya Vidyalaya	Vidyaniketan School	City High School	Eklavya School
Women	500	520	680	800
Men	440	640	760	600

Solution:

Question 6.
Simplify the expressions:
i. 45 ÷ 5 + 120 × 4 – 12
ii. (38 – 8) × 2 ÷ 5 + 13
iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
Solution:
i. 45 ÷ 5 + 120 × 4 – 12
= 9 + 80 – 12
= 89 – 12
= 77

ii. (38 – 8) × 2 ÷ 5 + 13
= 30 × 2 ÷ 5 + 13
= 60 ÷ 5 + 13
= 12 + 13
= 25

iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
\(\frac{5}{3}+\frac{4}{7} \times \frac{21}{32}\)
\(\frac{5}{3}+\frac{3}{8}=\frac{40}{24}+\frac{9}{24}\)
\(\frac{49}{24}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
= 3 × {4[90 – 5] + 2}
= 3 × {4 × 85 + 2}
= 3 × (340 + 2)
= 3 × 342
= 1026

Question 7.
Solve:
i. \(\frac{5}{12}+\frac{7}{16}\)
ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
Solution:
i. \(\frac{5}{12}+\frac{7}{16}\)

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
= 4 × (-2)
= -8

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
= \(\frac{7}{4} \times \frac{9}{5}\)
= \(\frac { 63 }{ 20 }\)

Question 8.
Construct ∆ABC such that m∠A = 55°, m∠B = and l(AB) = 5.9 cm.
Solution:

Question 9.
Construct ∆XYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Solution:

Question 10.
Construct ∆PQR such that, m∠P = 80°, m∠Q = 70°, l(QR) = 5.7 cm.
Ans:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° …. (Sum of the measures of the angles of a triangle is 180°)
∴ 80 + 70 + m∠R = 180
∴ 150 + m∠R = 180
∴ m∠R = 180 – 150
∴ m∠R = 30°

Question 11.
Construct ∆EFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.
Solution:

Question 12.
In ∆LMN, l(LM) = 6.2 cm, m∠LMN = 60°, l(MN) 4 cm. Construct ∆LMN.
Solution:

Question 13.
Find the measures of the complementary angles of the following angles:
i. 35°
ii. a°
iii. 22°
iv. (40 – x)°
Solution:
i. Let the measure of the complementary
angle be x°.
35 + x = 90
∴35 + x-35 = 90 – 35
….(Subtracting 35 from both sides)
∴x = 55
∴The complementary angle of 35° is 55°.

ii. Let the measure of the complementary angle be x°.
a + x = 90
∴a + x – a = 90 – a
….(Subtracting a from both sides)
∴x = (90 – a)
∴The complementary angle of a° is (90 – a)°.

iii. Let the measure of the complementary angle be x°.
22 + x = 90
∴22 + x – 22 = 90 – 22
….(Subtracting 22 from both sides)
∴x = 68
∴The complementary angle of 22° is 68°.

iv. Let the measure of the complementary angle be a°.
40 – x + a = 90
∴40 – x + a – 40 + x = 90 – 40 + x
….(Subtracting 40 and adding x on both sides)
∴a = (50 + x)
∴The complementary angle of (40 – x)° is (50 + x)°.

Question 14.
Find the measures of the supplements of the following angles:
i. 111°
ii. 47°
iii. 180°
iv. (90 – x)°
Solution:
i. Let the measure of the supplementary
angle be x°.
111 + x = 180
∴ 111 + x – 111 = 180 – 111
…..(Subtracting 111 from both sides)
∴ x = 69
∴ The supplementary angle of 111° is 69°.

ii. Let the measure of the supplementary angle be x°.
47 + x = 180
∴47 + x – 47 = 180 – 47
….(Subtracting 47 from both sides)
∴x = 133
∴The supplementary angle of 47° is 133°.

iii. Let the measure of the supplementary angle be x°.
180 + x = 180
∴180 + x – 180 = 180 – 180
….(Subtracting 180 from both sides)
∴x = 0
∴The supplementary angle of 180° is 0°.

iv. Let the measure of the supplementary angle be a°.
90 – x + a = 180
∴90 – x + a – 90 + x = 180 – 90+ x
….(Subtracting 90 and adding x on both sides)
∴a = 180 – 90 + x
∴a = (90 + x)
∴The supplementary angle of (90 – x)° is (90 + x)°.

Question 15.
Construct the following figures:
i. A pair of adjacent angles
ii. Two supplementary angles which are not adjacent angles.
iii. A pair of adjacent complementary angles.
Solution:
i.

ii.

iii.

Question 16.
In ∆PQR the measures of ∠P and ∠Q are equal and m∠PRQ = 70°, Find the measures of the following angles.

1. m∠PRT
2. m∠P
3. m∠Q

Solution:
Here, ∠PRQ and ∠PRT are angles in a linear pair.
m∠PRQ + m∠PRT = 180°
∴70 + m∠PRT = 180
∴m∠PRT = 180 – 70
∴m∠PRT = 110°
Now, ∠PRT is the exterior angle of ∆PQR.
∴m∠P + m∠Q = m∠PRT
∴m∠P + m∠P = m∠PRT ….(The measures of ∠P and ∠Q is same)
∴2m∠P = 110
∴m∠P = \(\frac { 110 }{ 2 }\)
∴m∠P = 55°
∴m∠Q =

Question 17.
Simplify
i. 5⁴ × 5³
ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Solution:
Simplify
i. 5⁴ × 5³
= 5⁴⁺³
= 5⁷

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)

Question 18.
Find the value:
i. 17¹⁶  ÷ 17¹⁶
ii. 10^-3
iii. (2³)²
iv. 4⁶ × 4^-4
Solution:
i. 17¹⁶  ÷ 17¹⁶
= 17⁰
= 1

ii. 10^-3
= \(\frac{1}{10^{3}}\)
= \(\frac{1}{1000}\)

iii. (2³)²
= 2^3×2
= 2⁶
= 2 × 2 × 2 × 2 × 2 × 2
= 64

iv. 4⁶ × 4^-4
= 4^6+(-4)
= 4²
= 4 × 4
= 16

Question 19.
Solve:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
ii. (3x + 2y) (7x – 8y)
iii. (7m – 5n) – (-4n – 11m)
iv. (11m – 12n + 3p) – (9m + 7n – 8p)
Solution:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
= (6a + 2a) + (-5b + 15b) + (-8c – 5c)
= 8a + 10b – 13c

ii. (3x + 2y) (7x – 8y)
= 3x × (7x – 8y) + 2yx (7x – 8y)
= 21x² – 24xy + 14xy – 16y²
= 21x² – 10xy – 16y²

iii. (7m – 5n) – (-4n – 11m)
= 7m – 5n + 4n + 11m
= (7m + 11m) + (-5n + 4n)
= 18m – n

iv. (11m – 12n + 3p) – (9m + 7n – 8p)
= 11m – 12n + 3p – 9m – 7n + 8p
= (11m – 9m) + (-12n – 7n) + (3p + 8p)
= 2m – 19n + 11p

Question 20.
Solve the following equations:
i 4(x + 12) = 8
ii. 3y + 4 = 5y – 6
Solution:
i. 4(x + 12) = 8
∴4x + 48 = 8
∴4x + 48 – 48 = 8 – 48
….(Subtracting 48 from both sides)
∴ 4x = -40
∴ x = \(\frac { -40 }{ 4 }\)
∴ x = -10

ii. 3y + 4 = 5y – 6
∴ 3y + 4 + 6 = 5y – 6 + 6
….(Adding 6 on both sides)
∴ 3y + 10 = 5y
∴ 3y + 10 – 3y = 5y – 3y
….(Subtracting 3y from both sides)
∴ 10 = 2y
∴ 2y = 10
∴ y = \(\frac { 10 }{ 2 }\)
∴ y = 5

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