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Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 2 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Question 1.
Angela deposited Rs 15000 in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to Rs 5400. For how many years had she deposited the amount?
Solution:
Here, P = Rs 15000, R = 9 p.c.p.a., I = Rs 5400

∴ T = 4
∴ Angela had deposited the amount for 4 years.

Question 2.
Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?
Solution:
Let us suppose that 8 men require x days to tar the road.
Number of days required by 10 men to tar the road = 4
The number of men and the number of days required to tar the road are in inverse proportion.
∴ 8 × x = 10 x 4
∴ \(x=\frac{10 \times 4}{8}\)
∴ x = 5
∴ 8 men will require 5 days to tar the road.

Question 3.
Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?
Solution:
Total amount invested = Rs 40,000 + Rs 60,000
= Rs 1,00,000
Profit earned = 30%
∴ Total profit = 30% of 1,00,000
= \(\frac { 30 }{ 100 }\) × 100000
= Rs 30000
Proportion of investment = 40000 : 60000
= 2:3 …. (Dividing by 20000)
Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.
∴ 2x + 3x = 30000
∴ 5x = 30000
∴ x = \(\frac { 30000 }{ 5 }\).
∴ x = 6000
∴ Nasruddin’s profit = 2x = 2 × 6000 = Rs 12000
Mahesh’s profit = 3x = 3 × 6000 = Rs 18000
∴ The profits of Nasruddin and Mahesh are Rs 12000 and Rs 18000 respectively.

Question 4.
The diameter of a circle is 5.6 cm. Find its circumference.
Solution:
Diameter of the circle (d) = 5.6 cm
Circumference = πd
= \(\frac{22}{7} \times 5.6\)
= \(\frac{22}{7} \times \frac{56}{10}\)
= 17.6 cm
∴ The circumference of the circle is 17.6 cm.

Question 5.
Expand:
i. (2a – 3b)²
ii. (10 + y)²
iii. \(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}\)
iv. \(\left(y-\frac{3}{y}\right)^{2}\)
Solution:
i. Here, A = 2a and B = 3b
∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²
…. [(A – B)² = A² – 2AB + B²]
= 4a² – 12ab + 9b²

ii. Here, a = 10 and b = y
(10 + y)² = 102 + 2 × 10xy + y²
…. [(a + b)² = a² + 2ab + b²]
= 100 + 20y + y²

iii. Here, a = \(\frac { p }{ 3 }\) and b = \(\frac { q }{ 4 }\)
\(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}=\left(\frac{p}{3}\right)^{2}+2 \times \frac{p}{3} \times \frac{q}{4}+\left(\frac{q}{4}\right)^{2}\)
…. [(a + b)² = a² + 2ab + b²]
\(\frac{p^{2}}{9}+\frac{p q}{6}+\frac{q^{2}}{16}\)

iv. Here, a = y and b = \(\frac { 3 }{ y }\)
\(\left(y-\frac{3}{y}\right)^{2}=y^{2}-2 \times y \times \frac{3}{y}+\left(\frac{3}{y}\right)^{2}\)
…. [(a – b)² = a² – 2ab + b²
= \(y^{2}-6+\frac{9}{y^{2}}\)

Question 6.
Use a formula to multiply:
i. (x – 5)(x + 5)
ii. (2a – 13)(2a + 13)
iii. (4z – 5y)(4z + 5y)
iv. (2t – 5)(2t + 5)
Solution:
i. Here, a = x and b = 5
(x – 5)(x + 5) = (x)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= x² – 25

ii. Here, A = 2a and B = 13
(2a – 13)(2a + 13) = (2a)² – (13)²
…. [(A + B)(A – B) = A² – B²]
= 4a² – 169

iii. Here, a = 4z and b = 5y
(4z – 5y)(4z + 5y) = (4z)² – (5y)²
…. [(a + b)(a – b) = a² – b²]
= 16z² – 25y²

iv. Here, a = 2t and b = 5
(2t – 5)(2t + 5) = (2t)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= 4t² – 25

Question 7.
The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?
Solution:
Diameter of the wheel (d) = 1.05 m
∴ Distance covered in 1 rotation of wheel = Circumference of the wheel
= πd
= \(\frac{22}{7} \times 1.05\)
= 3.3 m
∴ Distance covered in 1000 rotations = 1000 x 3.3 m
= 3300 m
= \(\frac { 3300 }{ 1000 }\) km …[1m = \(\frac { 1 }{ 1000 }\)km]
= 3.3 km
∴ The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

Question 8.
The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of Rs 250 per metre.
Solution:
Length of the rectangular garden = 40 m
Area of the rectangular garden = 1000 sq. m.
∴ length × breadth = 1000
∴ 40 × breadth = 1000
∴ breadth = \(\frac { 1000 }{ 40 }\)
= 25 m
Now, perimeter of the rectangular garden = 2 × (length + breadth)
= 2 (40 + 25)
= 2 × 65
= 130 m
Length of one round of fence = circumference of garden – width of the entrance
= 130 – 4
= 126 m
∴ Total length of fencing = length of one round of wire × number of rounds = 126 × 3
= 378 m
∴ Total cost of fencing = Total length of fencing × cost per metre of fencing
= 378 × 250
= 94500
∴ The cost of fencing the garden is Rs 94500.

Question 9.
From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
Solution:

In ∆ABC, ∠B = 90°, and l(BC) = 21, and l(AB) = 20
∴ According to Pythagoras’ theorem,
∴ l(AC)² = l(BC)² + l(AB)²
∴ l(AC)² = 21² + 20²
∴ l(AC)² = 441 + 400
∴ l(AC)² = 841
∴ l(AC)² = 29²
∴ l(AC) = 29
Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)
= 20 + 21 + 29
= 70
∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.

Question 10.
If the edge of a cube is 8 cm long, find its total surface area.
Solution: ,
Total surface area of the cube = 6 × (side)²
= 6 × (8)²
= 6 × 64
= 384 sq. cm
The total surface area of the cube is 384 sq.cm.

Question 11.
Factorize: 365y⁴z³ – 146y²z⁴
Solution:
= 365y⁴z³ – 146y²z⁴
= 73 (5y⁴z³ – 2y²z⁴)
= 73y² (5y²z³ – 2z⁴)
= 73y²z³(5y² – 2z)

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 7 Maths Solutions covers the 7th Std Maths Practice Set 12 Answers Solutions Chapter 3 HCF and LCM.

HCF and LCM Class 7 Practice Set 12 Answers Solutions Chapter 3

Question 1.
i. 25, 40
ii. 56, 32
iii. 40, 60, 75
iv. 16, 27
v. 18, 32,48
vi. 105, 154
vii. 42, 45, 48
viii. 57, 75, 102
ix. 56, 57
x. 777, 315, 588
Solution:
i. 25, 40

∴ 25 = 5 × 5
40 = 2 × 2 × 2 × 5
∴ HCF of 25 and 40 = 5

ii. 56, 32

∴ 56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8

iii. 40, 60, 75

∴ 40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5

iv. 16, 27

∴ 16 = 2 × 2 × 2 × 2 × 1
27 = 3 × 3 × 3 × 1
∴ HCF of 16 and 27 = 1

v. 18, 32,48

∴ 18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2

vi. 105, 154

∴ 105 = 3 × 5 × 7
154 = 2 × 2 × 11
∴ HCF of 105 and 154 = 7

vii. 42, 45, 48

∴ 42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 42,45 and 48 =3

viii. 57, 75, 102

∴ 57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
∴ HCF of 57, 75 and 102 = 3

ix. 56, 57

∴ 56 = 2 × 2 × 2 × 7 × 1
57 = 3 × 19 × 1
∴ HCF of 56 and 57 = 1

x. 777, 315, 588

∴ 777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
HCF of 777, 315 and 588 = 21

Question 2.
Find the HCF by the division method and reduce to the simplest form:
i. \(\frac { 275 }{ 525 }\)
ii. \(\frac { 76 }{ 133 }\)
iii. \(\frac { 161 }{ 69 }\)
Solution:
i. \(\frac { 275 }{ 525 }\)

ii. \(\frac { 76 }{ 133 }\)

iii. \(\frac { 161 }{ 69 }\)

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 12 Intext Questions and Activities

Question 1.
In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (Textbook pg. no. 17)
i. 28, 42
ii. 51, 27
iii. 25, 15, 35
Solution:
i. Factors of 28 = 1,2,4, 7, 14, 28
Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42
∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51
Factors of 27 = 1, 3, 9, 27
∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25
Factors of 15 = 1, 3, 5, 15
Factors of 35 = 1, 5, 7, 35
∴ HCF of 25, 15 and 35 = 5

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 9 HCF-LCM Class 6 Practice Set 23 Answers Solutions.

6th Standard Maths Practice Set 23 Answers Chapter 9 HCF-LCM

Question 1.
Write all the factors of the given numbers and list their common factors:
i. 12, 16
ii. 21, 24
iii. 25, 30
iv. 24, 25
v. 56, 72
Solution:
i. Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 16 = 1, 2, 4, 8, 16
∴ Common factors of 12 and 16 = 1, 2, 4

ii. Factors of 21 = 1, 3, 7, 21
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
∴ Common factors of 21 and 24 = 1, 3

iii. Factors of 25 = 1, 5, 25
Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
∴ Common factors of 25 and 30 = 1, 5

iv. Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 25 = 1,5, 25
∴ Common factor of 24 and 25 = 1

v. Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
∴ Common factors of 56 and 72 = 1, 2, 4, 8

Maharashtra Board Class 6 Maths Chapter 9 HCF-LCM Practice Set 23 Intext Questions and Activities

Question 1.
In the empty boxes, write the proper words: dividend, divisor, quotient, remainder. (Textbook pg. no. 46)

Solution:

When we divide 36 by 4, the remainder is zero. Therefore, 4 is a factor of 36 and 36 is a multiple of 4. But, when we divide 65 by 9, the remainder is not zero. Therefore, 9 is not a factor of 65. Also, 65 is not a multiple of 9.

Question 2.
Write all the factors of the numbers 36 and 48. Also, list their common factors. (Textbook pg. no. 46)
Solution:
36 = 1 × 36
= 2 × 18
= 3 × 12
= 4 × 9
= 6 × 6

48 = 1 × 48
= 2 × 24
= 3 × 16
= 4 × 12
= 6 × 8

∴ Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Common factors of 36 and 48: [1] ,[2], [3], [4], [6], [12]

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 13 Profit-Loss Class 6 Practice Set 32 Answers Solutions.

6th Standard Maths Practice Set 32 Answers Chapter 13 Profit-Loss

Question 1.
From a wholesaler, Santosh bought 400 eggs for Rs 1500 and spent Rs 300 on transport. 50 eggs fell down and broke. He sold the rest at Rs 5 each. Did he make a profit or a loss? How much?
Solution:
Cost price of 400 eggs = Rs 1500
Transportation cost = Rs 300
∴ Total cost price of 400 eggs = Cost price of 400 eggs + Transportation cost
= 1500 + 300 = Rs 1800
50 eggs fell and broke
∴ Remaining eggs = 400 – 50 = 350
Selling price of 1 egg = Rs 5
∴ Selling price of 350 eggs = 5 x 350 = Rs 1750
Total cost price is greater than the selling price.
∴ Santosh suffered a loss.
Loss = Total cost price – Selling price
= 1800 – 1750
= Rs 50
∴ Santosh incurred a loss of Rs 50.

Question 2.
Abraham bought goods worth Rs 50000 and spent Rs 7000 on transport and octroi. If he sold the goods for Rs 65000, did he make a profit or a loss? How much?
Solution:
Cost price of goods = Rs 50000
Transportation cost and octroi = Rs 7000
∴ Total cost price for buying goods = Cost price of goods + Transportation cost and octroi
= 50000 + 7000 = Rs 57000
Selling price of goods = Rs 65000
Selling price is greater than the total cost price
∴ Abraham made a profit.
Profit = Selling price – Total cost price
= 65000 – 57000
= Rs 8000
∴ Abraham made a profit of Rs 8000.

Question 3.
Ajit Kaur bought a 50 kg sack of sugar for Rs 1750, but as sugar prices fell, she had to sell it at Rs 32 per kg. How much loss did she incur?
Solution:
Cost price of 50 kg sugar = Rs 1750
Selling price of 1 kg sugar = Rs 32
∴ Selling price of 50 kg sugar = 50 x 32 = Rs 1600
Loss = Total cost price – Selling price
= 1750 – 1600 = Rs 150
∴ Ajit Kaur incurred a loss of Rs 150.

Question 4.
Kusumtai bought 80 cookers at Rs 700 each. Transport cost her Rs 1280. If she wants a profit of Rs 18000, what should be the selling price per cooker?
Solution:
Cost price of one cooker = Rs 700
∴ Cost price of 80 cookers = 700 x 80 = Rs 56000
Transportation cost = Rs 1280
∴ Total cost price = Cost price of 80 cookers + Transportation cost
= 56000 + 1280
= Rs 57280
Profit = Rs 18000
Profit = Selling Price – Total Cost Price
∴ Required selling price = Total cost price + profit
= 57280 + 18000
= Rs 75280
∴ Selling price of 80 cookers = Rs 75280
∴ Selling price of 1 cooker = \(\frac { 75280 }{ 80 }\) = Rs 941
∴ The selling price per cooker should be Rs 941.

Question 5.
Indrajit bought 10 refrigerators at Rs 12000 each and spent Rs 5000 on transport. For how much should he sell each refrigerator in order to make a profit of Rs 20000?
Solution:
Cost price of 1 refrigerator = Rs 12000
Cost price of 10 refrigerator = 10 x 12000 = Rs 120000
Transportation cost = Rs 5000
∴ Total cost price of 10 refrigerators = Cost price of 10 refrigerators + Transportation cost
= 120000 + 5000 = Rs 125000
Profit = Rs 20000
Profit = Selling Price – Total Cost Price
∴ Required selling price = Total cost price + Profit
= 125000 + 20000 = Rs 145000
∴ Selling price of 10 refrigerators = Rs 145000
∴ Selling price of 1 refrigerator = \(\frac { 145000 }{ 10 }\) = Rs 14500
∴ Indrajit must sell each refrigerator at Rs 14500 to make a profit of Rs 20000.

Question 6.
Lalitabai sowed seeds worth Rs 13700 in her field. She had to spend Rs 5300 on fertilizers and spraying pesticides and Rs 7160 on labor. If, on selling her produce, she earned Rs 35400 what was her profit or her loss?
Solution:
Cost price of seeds = Rs 13700
Cost of fertilizers and pesticides = Rs 5300
Labor cost = Rs 7160
∴ Total cost price = Cost price of seeds + Cost of fertilizers and pesticides + Labor cost
= 13700 + 5300 + 7160
= Rs 26160
Selling price = Rs 35400
Selling price is greater than the total cost price.
∴ Lalitabai made a profit.
Profit = Selling price – Cost price
= 35400 – 26160
= Rs 9240
∴ Lalitabai made a profit of Rs 9240.

Maharashtra Board Class 6 Maths Chapter 13 Profit-Loss Practice Set 32 Intext Questions and Activities

Question 1.
At Diwali, in a certain school, the students undertook a Design a Diya project. They bought 1000 diyas for Rs 1000 and some paint for Rs 200. To bring the diyas to the school, they spent Rs 100 on transport. They sold the painted lamps at Rs 2 each. Did they make a profit or incur a loss? (Textbook pg. no. 67 and 68)

i. Is Anju right?
ii. What about the money spent on paints and transport?
iii. How much money was actually spent before the diyas could be sold?
iv. How much actual profit was made in this project of colouring the diyas and selling them?

Ans:
i. No, Anju is wrong.
Cost price of diyas also includes the painting and transportation cost.
∴ Total cost price of diyas = Cost of diyas + Cost of paint + Transportation cost
= 1000 + 200+ 100
= Rs 1300
ii. The cost of paint was Rs 200 and that for transportation was Rs 100. These costs are also to be added to the cost price of diyas.
iii. Rs 1300 was actually spent before the diyas could be sold.
iv. Total Cost Price of 1000 Diyas = Rs 1300
Selling Price of 1 Diya = Rs 2
∴ Selling Price of 1000 Diyas = 2 x 1000 = Rs 2000
∴ Profit = Selling Price – Total Cost Price
= 2000 – 1300
= Rs 700
∴ The profit made by coloring the diyas and selling them was Rs 700.

Question 2.
A farmer sells what he grows in his fields. How is the total cost price calculated? What does a farmer spend on his produce before he can sell it? What are the other expenses besides seeds, fertilizers and transport? (Textbook pg. no. 68)
Solution:
The farmer, in order to calculate the total cost price of his produce, needs to consider all the expenses associated with the growing and selling of his produce.

Following are the things on which farmer spends money before he can sell it.

1. Time and energy
2. Ploughing and tilling
3. Irrigation and electricity cost
4. Harvesting and cleaning
5. Packing

As given above, there are a multiple of costs to be included besides seeds, fertilizers and transport for the farmer to price its produce appropriately.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 4 Operations on Fractions Class 6 Practice Set 10 Answers Solutions.

6th Standard Maths Practice Set 10 Answers Chapter 4 Operations on Fractions

Question 1.
Add:
i. \(6 \frac{1}{3}+2 \frac{1}{3}\)
ii. \(1 \frac{1}{4}+3 \frac{1}{2}\)
iii. \(5 \frac{1}{5}+2 \frac{1}{7}\)
iv. \(3 \frac{1}{5}+2 \frac{1}{3}\)
Solution:
i. \(6 \frac{1}{3}+2 \frac{1}{3}\)

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

iii. \(5 \frac{1}{5}+2 \frac{1}{7}\)

iv. \(3 \frac{1}{5}+2 \frac{1}{3}\)

Question 2.
Subtract:
i. \(3 \frac{1}{3}-1 \frac{1}{4}\)
ii. \(5 \frac{1}{2}-3 \frac{1}{3}\)
iii. \(7 \frac{1}{8}-6 \frac{1}{10}\)
iv. \(7 \frac{1}{2}-3 \frac{1}{5}\)
Solution:
i. \(3 \frac{1}{3}-1 \frac{1}{4}\)

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

iii. \(7 \frac{1}{8}-6 \frac{1}{10}\)

iv. \(7 \frac{1}{2}-3 \frac{1}{5}\)

Question 3.
Solve:
i. Suyash bought \(2\frac { 1 }{ 2 }\) kg of sugar and Ashish bought \(3\frac { 1 }{ 2 }\) kg. How much sugar did they buy altogether? If sugar costs Rs 32 per kg, how much did they spend on the sugar they bought?

ii. Aradhana grows potatoes in \(\frac { 2 }{ 5 }\) part of her garden, greens in \(\frac { 1 }{ 3 }\) part and brinjals in the remaining part. On how much of her plot did she plant brinjals?

iii. Sandeep filled water in \(\frac { 4 }{ 7 }\) of an empty tank. After that, Ramakant filled \(\frac { 1 }{ 4 }\) part more of the same tank. Then Umesh used \(\frac { 3 }{ 14 }\) part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
Solution:
i. Sugar bought by Suyash = \(2\frac { 1 }{ 2 }\) kg
Sugar bought by Ashish = \(3\frac { 1 }{ 2 }\) kg
∴ Total sugar bought by both

Cost of 1 kg of sugar = Rs 32
∴ Cost of 6 kg of sugar = 32 x 6
= Rs 192
∴ They bought 6 kg sugar altogether and the total money spent on sugar is Rs 192.

ii. Part of garden occupied by potatoes = \(\frac { 2 }{ 5 }\)
Part of garden occupied by greens = \(\frac { 1 }{ 3 }\)
Since brinjals are planted in the remaining part,
∴ (Part occupied by potatoes) + (part occupied by greens) + (part occupied by brinjals) = 1 entire garden.
∴ Part of garden occupied by brinjals = 1 – (part of garden occupied by potatoes + part of garden occupied by greens)

∴ Aradhana planted brinjals on \(\frac { 4 }{ 15 }\) part of her plot.

iii. Part of tank filled by Sandeep = \(\frac { 4 }{ 7 }\)
Part of tank filled by Ramakant = \(\frac { 1 }{ 4 }\)

Since maximum capacity of tank is 560 litres
∴ Quantity of water left in tank = \(\frac { 17 }{ 28 }\times560\) = 340 litres
∴ The quantity of water left in the tank is 340 litres.

Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Practice Set 10 Intext Questions and Activities

Question 1.
How to do this subtraction: \(4 \frac{1}{4}-2 \frac{1}{2}\) ? Is it same as \(\left[4-2+\frac{1}{4}-\frac{1}{2}\right]\) ? (Textbook pg. no. 23)
Solution:
\(4 \frac{1}{4}-2 \frac{1}{2}\)

\(\left[4-2+\frac{1}{4}-\frac{1}{2}\right]\)

The subtraction \(4 \frac{1}{4}-2 \frac{1}{2}\) is the same as \(\left[4-2+\frac{1}{4}-\frac{1}{2}\right]\).

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 13 Profit-Loss Class 6 Practice Set 31 Answers Solutions.

6th Standard Maths Practice Set 31 Answers Chapter 13 Profit-Loss

Question 1.
The cost price and selling price are given in the following table. Find out whether there was a profit or a loss and how much it was.

Ex.	Cost price (in Rs)	Selling price (in Rs)	Profit or Loss	How much?
i.	4500	5000
ii.	4100	4090
iii.	700	799
iv.	1000	920

Solution:

i. Cost price = Rs 4500
Selling price = Rs 5000
Selling price is greater than cost price.
∴ There is a profit.
∴ Profit = Selling price – Cost price
= 5000 – 4500
Profit = Rs 500

ii. Cost price = Rs 4100
Selling price = Rs 4090
Cost price is greater than selling price.
∴ There is a loss.
∴ Loss = Cost price – Selling price
= 4100 – 4090
∴ Loss = Rs 10

iii. Cost price = Rs 700
Selling price = Rs 799
Selling price is greater than cost price.
∴ There is a profit.
∴ Profit = Selling price – Cost price
= 799 – 700
∴ Profit = Rs 99

iv. Cost price = Rs 1000
Selling price = Rs 920
Cost price is greater than selling price.
∴ There is a loss.
∴ Loss = Cost price – Selling price
= 1000 – 920
∴ Loss = Rs 80

Ex.	Cost price (in Rs)	Selling price (in Rs)	Profit or Loss	How much?
i.	4500	5000	Profit	Rs 500
ii.	4100	4090	Loss	Rs 10
iii.	700	799	Profit	Rs 99
iv.	1000	920	Loss	Rs 80

Question 2.
A shopkeeper bought a bicycle for Rs 3000 and sold the same for Rs 3400. How much was his profit?
Solution:
Cost price = Rs 3000, Selling price = Rs 3400
∴ Profit = Selling price – Cost price
= 3400 – 3000
= Rs 400
The shopkeeper’s profit was Rs 400.

Question 3.
Sunandabai bought milk for Rs 475. She converted it into yogurt and sold it for Rs 700. How much profit did she make?
Solution:
∴ Cost price = Rs 475, Selling price = Rs 700
∴ Profit = Selling price – Cost price
= 700 – 475
= Rs 225
∴ Sunandabai made a profit of Rs 225.

Question 4.
The Jijamata Women’s Saving Group bought raw materials worth Rs 15000 for making chakalis.
They sold the chakalis for Rs 22050. How much profit did the WSG make?
Solution:
Cost price of raw materials = Rs 15000
Selling price of chakalis = Rs 22050
∴ Profit = Selling price – Cost price
= 22050 – 15000
= Rs 7050
∴ The Women’s Saving Group made a profit of Rs 7050.

Question 5.
Pramod bought 100 bunches of methi greens for Rs 400. In a sudden downpour, 30 of the bunches on his handcart got spoil. He sold the rest at the rate of Rs 5 each. Did he make a profit or a loss? How much?
Solution:
Cost price of 100 bunches of methi green = Rs 400
Since, 30 bunches got spoil,
∴ Remaining bunches of methi green = 100 – 30 = 70
Selling price of 1 bunch of methi green = Rs 5
∴ Selling price of 70 bunches of methi green = 5 x 70 = Rs 350
Cost price is greater than selling price
∴ Pramod suffered a loss.
Loss = Cost price – Selling price
= 400 – 350
= Rs 50
∴ Pramod suffered a loss of Rs 50.

Question 6.
Sharad bought one quintal of onions for Rs 2000. Later he sold them all at the rate of Rs 18 per kg. Did he make a profit or incur a loss? How much was it?
Solution:
Cost price of one quintal onions = Rs 2000
Selling price of 1 kg onions = Rs 18
Since, 1 quintal = 100 kg
∴ Selling price of 1 quintal (100 kg) onions = 18 x 100 = Rs 1800
Cost price is greater than selling price
∴ Sharad suffered a loss.
∴ Loss = Cost price – Selling price
= Rs 2000 – Rs 1800
= Rs 200
∴ Sharad incurred a loss of Rs 200.

Question 7.
Kantabai bought 25 saris from a wholesale merchant for Rs 10000 and sold them all at Rs 460 each. How much profit did Kantabai get in this transaction?
Solution:
Cost price of 25 saris = Rs 10000
Selling price of 1 sari = Rs 460
∴ Selling price of 25 saris = 460 x 25 = Rs 11500
Selling price is greater than cost price.
∴ Profit = Selling price – Cost price
= 11500 – 10000
= Rs 1500
∴ Kantabai made a profit of Rs 1500.

Maharashtra Board Class 6 Maths Chapter 13 Profit-Loss Practice Set 31 Intext Questions and Activities

Question 1.
Pranav and sarita had set up stalls in a fun fair. Study the data given below and answer the questions. (Textbook pg. no. 65)


Solution:
Total amount invested by Pranav = 70 + 25 + 45 + 14 + 20 = Rs 174
Amount gained through sale = Rs 160
∴ Selling price is less than invested price.
∴ Pranav incurred a loss in his Pav Bhaji business. Hence, he is disappointed.

Total amount invested by Sarita = 20 + 10 + 30 + 50 + 20 + 60 = Rs 190
Amount gained by selling = Rs 230
∴ Selling price is more than invested price.
∴ Sarita made profit in her business. Hence, she is happy.

Question 2.
For the above example,

1. If Sarita had bought twice as much, would she have gained twice as much?
2. What should Pranav do the next time he sets up a stall to sell more pav bhaji and make more gains? (Textbook pg. no. 66)

Solution:

1. If Sarita would have bought twice as much, she would have prepared double quantity of food items. Hence, she would have gained twice as much.
2. Next time Pranav sets a stall, he must sell pav bhaji at a higher cost than he had sold earlier in order to make more gains.

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 31 Answers Solutions Chapter 7 Joint Bar Graph.

Joint Bar Graph Class 7 Practice Set 31 Answers Solutions Chapter 7

Question 1.
The number of saplings planted by schools on World Tree Day is given in the table below. Draw a joint bar graph to show these figures.

School Name\Name of Sapling	Almond	Karanj	Neem	Ashok	Gulmohar
Nutan Vidyalaya	40	60	72	15	42
Bharat Vidyalaya	42	38	60	25	40

Solution:

Question 2.
The table below shows the number of people who had the different juices at a juice bar on a Saturday and a Sunday. Draw a joint bar graph for this data.

Days\Fruits	Sweet Lime	Orange	Apple	Pineapple
Saturday	43	30	56	40
Sunday	59	65	78	67

Solution:

Question 3.
The following numbers of votes were cast at 5 polling booths during the Gram Panchayat elections. Draw a joint bar graph for this data.

Persons\Booth No.	1	2	3	4	5
Men	200	270	560	820	850
Women	700	240	340	640	470

Solution:

Question 4.
The maximum and minimum temperatures of five Indian cities are given in °C. Draw a joint bar graph for this data.

City\Temperature	Delhi	Mumbai	Kolkata	Nagpur	Kapurthala
Maximum temperature	35	32	37	41	37
Minimum temperature	26	25	26	29	26

Solution:

Question 5.
The numbers of children vaccinated in one day at the government hospitals in Solapur and Pune are given in the table. Draw a joint bar graph for this data:

City\Vaccine	D.P.T. (Booster)	Polio (Booster)	Measles	Hepatitis
Solapur	65	60	65	63
Pune	89	87	88	86

Solution:

Question 6.
The percentage of literate people in the states of Maharashtra and Gujarat are given below. Draw a joint bar graph for this data.

State\Year	1971	1981	1991	2001	2011
Maharashtra	46	57	65	77	83
Gujarat	40	45	61	69	79

Solution:

Maharashtra Board Class 7 Maths Chapter 7 Joint Bar Graph Practice Set 31 Intext Questions and Activities

Question 1.
Observe the graph shown below and answer the following questions. (Textbook pg. no. 51)

1. In which year did Ajay and Vijay both produce equal quantities of wheat?
2. In year 2014, who produced more wheat?
3. In year 2013, how much wheat did Ajay and Vijay each produce?

Solution:

1. Both produced equal quantities of wheat in the year 2011.
2. Ajay produced more wheat in the year 2014.
3. Ajay’s wheat production in 2013 = 40 quintal.
   Vijay’s wheat production in 2013 = 30 quintal.

Question 2.
The minimum and maximum temperature in Pune for five days is given. Read the joint bar graph and answer the questions below: (Textbook pg. no. 52)

1. What data is shown on X- axis?
2. What data is shown on Y- axis?
3. Which day had the highest temperature?
4. On which day is the minimum temperature the highest?
5. On Thursday, what is the difference between the minimum and maximum temperature?
6. On which day is the difference between the minimum and maximum temperature the greatest?

Solution:

1. Five days of a week are shown on X – axis.
2. Temperature in the city of Pune is shown on Y – axis.
3. Monday had the highest temperature.
4. The minimum temperature was highest on Wednesday.
5. Maximum temperature = 29.5° C
   Minimum temperature = 15° C
   ∴ Difference in temperature = 29.5° C – 15° C = 14.5 ° C
6. The difference in minimum and maximum temperature is greatest on Thursday.

Question 3.
Collect various kinds of graphs from newspapers and discuss them. (Textbook pg. no. 53)
i. Histogram

ii. Line graph

iii. Pie chart

Solution:
(Students should attempt the above activities on their own.)

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 8 Divisibility Class 6 Practice Set 22 Answers Solutions.

6th Standard Maths Practice Set 22 Answers Chapter 8 Divisibility

Question 1.
There are some flowering trees in a garden. Each tree bears many flowers with the same number printed on it. Three children took a basket each to pick flowers. Each basket has one of the numbers, 3, 4 or 9 on it. Each child picks those flowers which have numbers divisible by the number on his or her basket. If He / She takes only 1 flower from each tree. Can you tell which numbers the flowers in each basket will have?

Solution:
Each child will have flowers bearing the following numbers:
Girl with basket number 3: 111, 369, 435, 249, 666, 450, 960, 432, 999, 72, 336, 90, 123, 108
Boy with basket number 4: 356, 220, 432, 960, 72, 336, 108
Girl with basket number 9: 369, 666, 450, 432, 999, 72, 90, 108

Maharashtra Board Class 6 Maths Chapter 8 Divisibility Practice Set 22 Intext Questions and Activities

Question 1.
Read the numbers given below. Which of these numbers are divisible by 2, by 5, or by 10? Write them in the empty boxes. 125,364,475,750,800,628,206,508,7009,5345,8710. (Textbook pg. no. 43)

Divisible by 2	Divisible by 5	Divisible by 10

Solution:

Divisible by 2	Divisible by 5	Divisible by 10
364,750, 800, 628, 206, 508, 8710	125,475, 750, 800, 5345, 8710	750, 800, 8710

Question 2.
Complete the following table: (Textbook pg. no. 43)

Number	Sum of digits in the number	Is the sum divisible by 3?	Is the given number divisible by 3?
63	6 + 3 = 9	✓	✓
872	17	X	X
91
552
9336
4527

Solution:

Number	Sum of digits in the number	Is the sum divisible by 3?	Is the given number divisible by 3?
63	6 + 3 = 9	✓	✓
872	8 + 7 + 2 = 17	X	X
91	9 + 1 = 10	X	X
552	5 + 5 + 2 = 12	✓	✓
9336	9 + 3 + 3 + 6 = 21	✓	✓
4527	4 + 5 + 2 + 7 = 18	✓	✓

Question 3.
Complete the following table: (Textbook pg. no. 44)

Number	Divide the number by 4. Is it completely divisible?	The number formed by the digits in the tens and units places.	Is this number divisible by 4?
992	✓	92	✓
7314
6448
8116
7773
3024

Solution:

Number	Divide the number by 4. Is it completely divisible?	The number formed by the digits in the tens and units places.	Is this number divisible by 4?
992	✓	92	✓
7314	X	14	X
6448	✓	48	✓
8116	✓	16	✓
7773	X	73	X
3024	✓	24	✓

Question 4.
Complete the following table: (Textbook pg. no. 44)

Number	Divide the number by 9. Is it completely divisible?	Sum of the digits in the number.	Is the sum divisible by 9?
1980	✓	1 + 9 + 8 + 0 = 18	✓
2999	X	2 + 9 + 9 + 9 = 29	X
5004
13389
7578
69993

Solution:

Number	Divide the number by 9. Is it completely divisible?	Sum of the digits in the number.	Is the sum divisible by 9?
1980	✓	1 + 9 + 8 + 0 = 18	✓
2999	X	2 + 9 + 9 + 9 = 29	X
5004	✓	5 + 0 + 0 + 4 = 9	✓
13389	X	1 + 3 + 3 + 8 + 9 = 24	X
7578	✓	7 + 5 + 7 + 8 = 27	✓
69993	✓	6 + 9 + 9 + 9 + 3 = 36	✓

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 7 Symmetry Class 6 Practice Set 21 Answers Solutions.

6th Standard Maths Practice Set 21 Answers Chapter 7 Symmetry

Question 1.
Along each figure shown below, a line l has been drawn. Complete the symmetrical figures by drawing a figure on the other side such that the line l becomes the line of symmetry.

Solution:

Maharashtra Board Class 6 Maths Chapter 7 Symmetry Practice Set 21 Intext Questions and Activities

Question 1.
In the figures below, the line l divides the figure in two parts. Do these parts fall on each other? Verify? (Textbook pg. no. 42)

Solution:
Yes, the two parts of both the figures fall on each other on folding along the line l.