Class 7

MP Board Class 7th Social Science Solutions Chapter 28 Asia: Physical Features

MP Board Class 7th Social Science Chapter 28 Text Book Questions

Fill in the blanks:

1. The largest continent in the world is …………….
2. ……………… is the highest mountain peak of the world.
3. The plains of Northern part of Asia is known as ……………….
4. The plateau which lies in the ……………… central part of Asia is known as the roof of the world.
5. High pressure area is formed in …………… the part of Asia during winters.
6. …………… receives the highest rainfall in the world.

Answer:

1. Asia
2. Mount Everest
3. Siberian Plain
4. Pamir
5. Central
6. Mansynram.

MP Board Class 7th Social Science Chapter 28 Short Answer Type Questions

Question 1.
Which oceans surround the continent of Asia?
Answer:
The continent of Asia is surrounded by oceans on three sides. The Arctic Ocean in the north, the Pacific Ocean in the east and the Indian Ocean lies to the south of the continent.

Question 2.
Write the names of any two mountains and any two plateaus of Asia.
Answer:
Name of two mountains:

1. Ural Mountains
2. The Central Mountains Name of two plateaus
3. The Pamir plateau
4. The Tibetan Plateau.

Question 3.
Into how many physical features do we divide the continent of Asia? Write their names.
Answer:
The continent of Asia can be divided into five physical features. Their names are:

1. The northern lowlands
2. The central mountains
3. The southern plateaus
4. The great river valleys
5. The island groups.

Question 4.
Write any two factors which influence the climate of Asia.
Answer:
The two factors which influence the climate of Asia:

• Its vast size.
• Great latitudinal extent in the central mountains and plateaus.

Question 5.
Write any two differences between the vegetation found in the Northern and Southern part of Asia.
Answer:
Two differences between the vegetation found in the Northern and Southern part of Asia is:
1. The Northern part of Asia experiences extreme cold. But its Southern part has evergreen climate.

2. The vegetation found in the North empart are lichens, jherberi, mosses etc. which can withstand the extreme cold. The vegetation found in die Southern part are evergreen trees like spruce, fir and pine.

MP Board Class 7th Social Science Chapter 28 Long Answer Type Questions

Question 1.
Describe in detail the relief of the continent of Asia.
Answer:
The continent of Asia consists of the highest mountain ranges, lowlands and plains. It maybe divided into five major.

physical divisions:
1. The North Lowlands:
It is a vast plain which extends between the Ural Mountains in die west, Lena river in the east and the Central Mountains in the south, known as the ‘Siberian Plain’, it is drained by rivers like Ob, Yenisei and Lena Lake Baikal, die deepest lake of die world is located in Siberia.

2. The Central Mountains:
The folded mountains and plateaus lie in the south of the northern lowlands. The Pamir plateau is a meeting place of several ranges, which form the Pamir knot This plateau is known as die ‘Roof of the World’, because it is the highest plateau in the world. The Tibetan plateau lies to the east of the Pamir plateau. Himalayas are the highest mountain ranges of the world. Mount Everest is situated in the Himalayas.

3. The Southern Plateaus:
To the south of the central mountain belt, there’ are some plateau which form the major part of the peninsula projecting southwards from the mainland of Asia. These plateaus are the plateau of Arabia, the Deccan plateau and the plateau of Yunnan.

4. The Great River Valleys:
To the east and south of Asia are found riverineplains. In the south are the Sindhu, Ganga, Brahmaputra plains, in the southeast are the plains formed by rivers Irrawady, Salvin, Yang-Tse-Kiang, Sikiang, Minang- Mekong and Amursar.

5. The Island Groups:
Some groups of islands are situated to the south-east and east of the continent of Asia. There are three major island groups-Indonesia, Philippines and Japan.

Question 2.
Describe the climate and wild life of the continent of Asia.
Answer:
Climate:
The southern part of the continent is situated near the equator and hence, remains hot and humid throughout the year. The northern part remains covered with snow as it is situated away from the equator.

During the summer season die rays of the sun falls vertically on the Northern Hemisphere. The temperature rises and a low – pressure area is developed over Central Asia. The moisture laden winds start blowing from the seas to the low pressure. These winds are known as southwestern monsoon winds.

Asia experiences the cold weather season during Oct-Dec. in the North  part of the Central mountains. High pressure builds over Central Asia The wind starts blowing from High pressure (Central Asia) to low pressure area (Southern Asia). Wild Life of Asia-In the northern part of Asia where die area is covered with snow throughout die year reindeer, polar bear, seal and whales are found.

In the south where die climate is hot and humid and deciduous forests are spread, lion, tiger, deer, elephants, wild buffaloes and rich rocuous etc. are die main wild life found. In die western dry and desert area sheep, goats, camel, ass and gazelle are found.

Map Work:

1. The Pamir plateau
2. The Deccan plateau
3. The Himalayan mountains
4. The Ural mountains
5. The Ganga river
6. Yang-Tie-kiang river
7. The Siberian plateau
8. The Ganga-Brahamaputra river
9. The Indonesian group of islands
10. Japan is group of islands.

Answer:

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.
Solution:
Length of garden (l) = 90 m
Breadth of garden (b) = 75 m
Area of garden = l × b = 90 × 75 = 6750 m²

From the figure, it can be observed that the new length and breadth of the garden, when path is also included, are (90 + 5 + 5) m i.e., 100 m and (75 + 5 + 5) m i.e., 85 m respectively.
Area of the garden including the path = 100 × 85 = 8500 m²
Area of path = Area of the garden including the path – Area of garden
= 8500 – 6750 = 1750 m²
1 hectare = 10000 m²
Therefore, area of garden in hectare = \(\frac{6750}{10000}\)
= 0.675 hectare

Question 2.
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Solution:
Length of park (I) = 125 m
Breadth of park (b) = 65 m
Area of park = l × b = 125 × 65 = 8125 m²

From the figure, it can be observed that the new length and breadth of the park, when path is also included, are (125 + 3 + 3) m i.e., 131 m and (65 + 3 + 3) m i.e., 71 m respectively.
Area of the park including the path = 131 × 71 = 9301 m²
Area of path = Area of the park including the path – Area of park
= 9301 – 8125 = 1176 m²

Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Length of cardboard (l) = 8 cm
Breadth of cardboard (b) = 5 cm
Area of cardboard = l × b = 8 × 5 = 40 cm²

From the figure, it can be observed that the new length and breadth of the cardboard, when margin is not included, are (8 – 1.5 – 1.5) cm i.e., 5 cm and (5 – 1.5 – 1.5) cm i.e., 2 cm respectively.
Area of the cardboard not including the margin = 5 × 2 = 10 cm²
Area of the margin = Area of cardboard – area of cardboard not including the margin
= 40 – 10 = 30 cm²

Question 4.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m².
Solution:
(i) Length of room (l) = 5.5 m
Breadth of room (b) = 4 m
Area of room = l × b = 5.5 × 4 = 22 m²

From the figure, it can be observed that the new length and breadth of the room, when verandah is also included, are (5.5 + 2.25 + 2.25) m i.e., 10 m and (4 + 2.25 + 2.25) m i.e., 8.5 m respectively.
Area of the room including the verandah = 8.5 × 10 = 85 m²
Area of verandah = Area of the room including the verandah – Area of room = 85 – 22 = 63 m²

(ii) Cost of cementing 1 m² area of the floor of the verandah = ₹ 200
Cost of cementing 63 m² area of the floor of the verandah = 200 × 63 = ₹ 12600

Question 5.
A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m².
Solution:
(i) Side of square garden (a) = 30 m
Area of square garden = a² = (30)² = 900 m²

From the figure, it can be observed that the side of the square garden, when path is not included, is (30 – 1 – 1) m i.e., 28 m.
Area of the square garden not including the path = (28)² = 784 m²
Area of path = Area of the square garden – Area of square garden not including the path = 900 – 784 = 116 m²

(ii) Cost of planting grass in 1 m² area of the garden = ₹ 40
Cost of planting grass in 784 m² area of the garden = 784 × 40 = ₹ 31360

Question 6.
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Solution:

Length of park (l) = 700 m
Breadth of park (b) = 300 m
Area of park = 700 × 300 = 210000 m²
Length of road PQRS = 700 m
Length of road ABCD = 300 m
Width of each road = 10 m.
KLMN is a square of side 10 m.
Area of two roads = area (PQRS) + area (ABCD) – area (KLMN)
= (700 × 10) + (300 × 10) – (10 × 10)
= 7000 + 3000 – 100 = 10000 – 100 = 9900 m²
= \(\frac{9900}{10000}\) hectare = 0.99 hectare (∵ 1 hectare = 10000 m²)
Area of park excluding roads = 210000 – 9900
= 200100 m² = \(\frac{200100}{10000}\) hectare = 20.01 hectare

Question 7.
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of ₹ 110 per m².
Solution:

Length of field (l) = 90 m
Breadth of field (b) = 60 m
Area of field = 90 × 60 = 5400 m²
Length of road PQRS = 90 m
Length of road ABCD = 60 m
Width of each road = 3 m
KLMN is a square of side 3 m.
(i) Area of the roads = area (PQRS) + area (ABCD) – area (KLMN)
= (90 × 3) + (60 × 3) – (3 × 3)
= 270 + 180 – 9 = 441 m²
(ii) Cost for constructing 1 m² road = ₹ 110
Cost for constructing 441 m² road
= 110 × 441 = ₹ 48510

Question 8.
Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)

Solution:
Perimeter of circular pipe = 2πr
= 2 × 3.14 × 4 = 25.12 cm
Perimeter of the square = 4 × Side of the square = 4 × 4 = 16 cm
Length of cord left withPfagya = 25.12 – 16 = 9.12 cm

Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:

(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed
(iv) the circumference of the flower bed.
Solution:
(i) Area of whole land = Length × Breadth
= 10 × 5 = 50 m²
(ii) Area of flower bed = πr² = 3.14 × 2 × 2
= 12.56 m²
(iii) Area of lawn excluding the flower bed = Area of whole land – Area of flower bed
= 50 – 12.56 = 37.44 m²
(iv) Circumference of the flower bed = 2πr
= 2 × 3.14 × 2 = 12.56 m

Question 10.
In the following figures( find the area of the shaded portions:


Solution:
(i) Area of EFDC
= area(ABCD) – area (BCE) – area (AFE)

Question 11.
Find the area of the quadrilateral ABCD.

Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM⊥AC, DN⊥AC
Solution:
Area (ABCD) = area (ABC) + area (ADC)
= \(\frac{1}{2}\) (3 × 22) + \(\frac{1}{2}\) (3 × 22)
= 33 + 33 = 66 cm²

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 25 Our Judicial System

MP Board Class 7th Social Science Chapter 25 Text Book Questions

Choose the correct alternatives:

Question 1.
The Judge of the Supreme Court remains in the office till the age of:
(a) 65 years
(b) 62 years
(c) 67 years
(d) 60 years
Answer:
(a) 65 years

Question 2.
The Chief Justice of the High Court is appointed by the President in consultation with the:
(a) Chief Minister
(b) Governor
(c) Education Minister
(d) None of these
Answer:
(b) Governor

Fill in the blanks

1. The Supreme Court of India is situated at ……………
2. To become a Judge one must be a citizen of ……………
3. The ………….. appoints die Chief Justice.
4. The High Court of Madhya Pradesh is situated at ………….

Answer:

1. Delhi
2. India
3. President
4. Jabalpur

MP Board Class 7th Social Science Chapter 25 Short Answer Type Questions

Question 1.
How many Judges are there in Supreme Court?
Answer:
There is one Chief Justice and 25 other judges in the Supreme Court.

Question 2.
What qualifications are required for the post of the Judge of a High Court?
Answer:
The following qualifications are essential for a person to be a judge of a High Court:

• He must be a citizen of India.
• He must have worked as a Judge in any court of the state for 10 years or practiced law for 10 years.

Question 3.
Write any two important functions of the High Court.
Answer:

1. The High Court has original and appellate jurisdiction. It has the power to hear for the first time civil and criminal cases and cases regarding the fundamental rights.

2. The High Court takes necessary action regarding the protection of fundamental rights granted in the Indian Constitution. It is binding on all the subordinate courts to obey the orders of the High Court regarding the protection of fundamental rights.

MP Board Class 7th Social Science Chapter 25 Long Answer Type Questions

Question 1.
Describe in detail the jurisdiction and function of the Supreme Court.
Answer:
Jurisdiction and functions of the Supreme Court –
1. Original Jurisdiction -Cases which come directly to the Supreme Court are under its original jurisdiction.
These are –

• Disputes between central and state government
• Disputes between two or more states.
• Disputes between the central government and one or more states on one side and one or more states on the other side.
• Those matters, which are related to the violation of Fundamental Rights of a person.

2. Appellate Jurisdiction:
Supreme Court is the highest court of appeal in the country. It hears appeals against decision of the High Courts in three types of cases:

• Civil
• Criminal
• Cases involving interpretation of the Constitution.
• Guardian of Fundamental Rights.
• Guardian of the Constitution.
• Court of Record.

3. Protection of the Constitution and the Fundamental Rights:
If any law enacted by the Government is against the spirit of the Constitution, the Supreme Court can declare it null and void. The Supreme Court also protects the Fundamental Rights of a person if they are violated.

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.3

Question 1.
Find the circumference of the circles with the following radius: (Take π = \(\frac{22}{7}\))
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:
(a) r = 14 cm
∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 14 =88 cm
(b) r = 28 mm
∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 28 = 176 mm
(c) r = 21 cm
∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 21 = 132 cm

Question 2.
Find the area of the following circles, given that: (Take π = \(\frac{22}{7}\))
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm
Solution:
(a) r = 14 mm

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = \(\frac{22}{7}\))
Solution:
Circumference = 2πr = 154 m

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = \(\frac{22}{7}\))
Solution:
Diameter (d) = 21 m 21
∴ Radius (r) = \(\frac{21}{2}\)m
Circumference = 2πr = 2 × \(\frac{22}{7} \times \frac{21}{2}\) = 66 m
Length of rope required for fencing = 2 × 66 m = 132 m
Cost of 1 m rope = ₹ 4
Cost of 132 m rope = 4 × 132 = ₹ 528

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:
Outer radius of circular sheet (R) = 4 cm
Inner radius of circular sheet (r) = 3 cm
Remaining area = πR² – πr²
= 3.14 × 4 × 4 – 3.14 × 3 × 3
= 50.24 – 28.26 = 21.98 cm²

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)
Solution:
The length of the lace required = circumference of circular table
Circumference = 2πr = 2 × 3.14 × \(\frac{d}{2}\)
= 2 × 3.14 × \(\frac{1.5}{2}\) = 4.71 m
Cost of 1 m lace = ₹ 15
Cost of 4.71 m lace = 4.71 × 15 = ₹ 70.65

Question 7.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution:
Diameter = 10 cm
Radius = \(\frac{10}{2}\) = 5 cm
Circumference of semicircle = \(\frac{2 \pi r}{2}\)
= 2 × \(\frac{1}{2} \times \frac{22}{7}\) × 5 = 15.71 cm
Total perimeter = Circumference of semicircle + Length of diameter
= 15.71 + 10 = 25.71 cm

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
Solution:
Diameter = 1.6 m
∴ Radius = \(\frac{1.6}{2}\) = 0.8 m
Area = πr² = 3.14 × 0.8 × 0.8 = 2.0096 m²
Cost for polishing 1 m² area = ₹ 15
Cost for polishing 2.0096 m² area
= 15 × 2.0096 = ₹ 30.14
Therefore, it will cost ₹ 30.14 for polishing circular table.

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = \(\frac{22}{7}\))
Solution:
If the wire is bent into a circle, then the length of wire = circumference of the circle
⇒ 2πr = 44 cm
⇒ 2 × \(\frac{22}{7}\) × r = 44
⇒ r = 7 cm
Area = πr²= \(\frac{22}{7}\) × 7 × 7 = 154 cm²
If the wire is bent into a square, then the length of the wire = perimeter of the square
⇒ 4 × side = 44cm ⇒ side = \(\frac{44}{4}\) = 11 cm
Area of square = (11)² = 121 cm²
As 154 > 121,
Therefore, circle encloses more area.

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the following figure). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))

Solution:
Area of bigger circle = \(\frac{22}{7}\) × 14 × 14 = 616 cm²
Area of 2 small circles = 2 × πr²
= 2 × \(\frac{22}{7}\) × 3.5 × 3.5 = 77 cm²
Area of rectangle = Length × Breadth = 3 × 1
= 3 cm²
Area of remaining sheet = Area of bigger circle – (Area of 2 small circles + Area of rectangle)
= 616 – (77 + 3) = 536 cm²

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Solution:
Area of square-shaped sheet = (Side)²
= (6)² = 36 cm²
Area of circle = 3.14 × 2 × 2= 12.56 cm²
Area of remaining sheet = Area of square sheet – area of circle
= 36 – 12.56 = 23.44 cm²

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:
Let r be the radius of circle. Circumference = 2πr = 31.4 cm
⇒ 2 × 3.14 × r = 31.4 cm
⇒ r = 5 cm
Area = 3.14 × 5 × 5 = 78.50 cm²

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Solution:
Radius of flower bed = \(\frac{66}{2}\) = 33 m
Width of the path = 4 m
Radius of flower bed and path together = 33 + 4 = 37 m
Area of flower bed and path together
= 3.14 × 37 × 37 = 4298.66 m²
Area of flower bed = 3.14 × 33 × 33 = 3419.46 m²
Area of path = Area of flower bed and path together – Area of flower bed
= 4298.66 – 3419.46 = 879.20 m²

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)
Solution:
Area = πr² = 314 m²
3.14 × r² = 314 ⇒ r² = 100 ⇒ r = 10 m
Yes, the sprinkler will water the whole garden.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Solution:
Radius of outer circle = 19 m
Circumference = 2πr =2 × 3.14 × 19 = 119.32 m
Radius of inner circle = 19 – 10 = 9 m
Circumference = 2πr = 2 × 3.14 × 9 = 56.52 m

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \(\frac{22}{7}\))
Solution:
r = 78 cm

Therefore, it will rotate 200 times.

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
Distance travelled by the tip of minute hand = Circumference of the clock
= 2πr = 2 × 3.14 × 15 = 94.2 cm

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.
(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Solution:
(a) Cost price = ₹ 250,
Selling price = ₹ 325
Profit = 325 – 250 = ₹ 75

(b) Cost price = ₹ 12000,
Selling price = ₹ 13500
Profit = 13500 – 12000 = ₹ 1500
Profit % = \(\frac{1500}{12000} \times 100=12.5 \%\)

(c) Cost price = ₹ 2500,
Selling price = ₹ 3000
Profit = 3000 – 2500 = ₹ 500

(d) Cost price = ₹ 250,
Selling price = ₹ 150
Loss = 250 – 150 = ₹ 100

Question 2.
Convert each part of the ratio to percentage:
(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5
Solution:

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Initial population = 25000 and Final population = 24500
Decrease = 500

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase?
Solution:
Initial price = ₹ 350000
Final price = ₹ 370000
Increase = ₹ 20000

Question 5.
I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
Cost price = ₹ 10,000
Profit % = 20%
∴ Profit = 20% of 10000
Selling price = Profit + Cost price
\(=\frac{20}{100} \times 10000+10000\)
= 2000 + 10000 = ₹ 12,000

Question 6.
Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Solution:
Selling price = ₹ 13500,
Loss% = 20%
Let the cost price be x.
∴ Loss = 20% of x
Cost price – Loss = Selling price

Therefore, she bought it for ₹ 16875.

Question 7.
(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Solution:
(i) Ratio of calcium, carbon and oxygen = 10 : 3 : 12
Therefore, percentage of carbon = \(\frac{3}{25} \times 100 \%\)
= 12%

(ii) Let the weight of the chalk stick be x g.

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
Cost price = ₹ 275
Loss% = 15% or Loss = 15% of 275
Cost price – Loss = Selling price


⇒ 275 – 41.25 = Selling price
∴ Selling price = ₹ 233.75

Question 9.
Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹ 1,200 at 12% p.a.
(b) Principal = ₹ 7,500 at 5% p.a.
Solution:
(a) Principal (P) = ₹ 1200
Rate (R) = 12% p.a.
Time (T) = 3 years

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:

Therefore, 0.25% gives ₹ 280 as interest on the given sum.

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a„ What is the sum she has borrowed?
Solution:

Therefore, Meena has borrowed ₹ 500.

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 24 The Rise of the Sikh and Maratha Power

MP Board Class 7th Social Science Chapter 24 Text Book Questions

Choose the correct alternatives:

Question 1.
Khalsa group was organised by:
(a) Guru Govind Singh
(b) Guru Teg Bahadur
(c) Banda Bahadur
(d) Gum Hargovind.
Answer:
(a) Guru Govind Singh

Question 2.
The credit of organisation of Maratha power goes to:
(a) Sambaji
(b) Shahji
(c) Shivaji
(d) Peshwa.
Answer:
(c) Shivaji

Question 3.
To Supress the power of Shivaji the Sultan of Bijapur sent:
(a) Afzal Khan
(b) Adil Shah
(c) ShaistaKhan
(d) Hasan Khan
Answer:
(a) Afzal Khan

Question 4.
At the highest position of Shivaji’s Ashta Pradhan was:
(a) Amatya
(b) Secretary
(c) Panditrao
(d) Peshwa.
Answer:
(d) Peshwa.

Fill in the blanks:

1. The title of Sachcha Padshah was conferred on …………
2. …………… was fee first Guru of the Sikhs.
3. Shivaji adorned fee title of …………. after his coronation.
4. ………….. was the source of income of Shivaji’s Kingdom.

Answer:

1. Banda Bahadur
2. Guru Nanak
3. Chhatrapati
4. land revenue

MP Board Class 7th Social Science Chapter 24 Short Answer Type Questions

Question 1.
What efforts was made by Guru Govind Singh to make the Sikhs powerful?
Answer:
Gum Govind Singh transformed the Sikhs into a separate community and named them Khalsa. He prescribed the five K’s-Kara, Kripan, Kesh, Kachacha, and Kangha for the Sikhs. He transformed the Sikhs into a powerful military organization.

Question 2.
What education did Shivaji receive in his childhood?
Answer:
Shivaji was taught to be independent. His mother instilled inspiration and determination in him to defend his people and his country.

Question 3.
Why did Shivaji kill Afzal Khan?
Answer:
Afzal Khan was sent to capture Shivaji. He plotted to kill him. Shivaji came to know about his plan and killed him in order to save his own life.

Question 4.
Write short notes on:

1. Ashta Pradhan.
2. The military administration of Shivaji.

Answer:
1. Ashta Pradhan:
Shivaji had appointed a council of eightministers. It was called Ashta Pradhan. Their main function was to advise Shivaji in carrying out the administration of his territories. Each person was the head of his department. However all worked under the chairmanship of Shivaji.

These Ashta Pradhan were –

• Peshwa (Prime Minister)
• Amatya (Finance Minister)
• Sumant (External Affair Minister)
• Mari
• Sachiv (Secretary)
• Panditrao (Purohit)
• Senapati (army general)
• Nyayadhish (Judge)

2. The military administration of Shivaji:
Shivaji had maintained discipline in his army. His army comprised of cavalry, infantry, artillery and navy. The soldiers were under control. They never tried to break the rules of discipline. Beside other duties, they also protected the holy books and safeguarded the women, children or old people from abuse.

MP Board Class 7th Social Science Chapter 24 Long Answer Type Questions

Question 1.
Clarify the Mughal and the Sikh relations.
Answer:
The Skihs were the followers of Guru Nanak. By the seventeenth century, Sikhism (new religion) had become the religion of the peasants and artisans in many parts of the the Punjab. After Gum Nanak, there were other nine Sikh Gurus. The earlier Gums concentrated mainly on Sikhism But the later Gums became the military leaders of the Sikhs also. They did so because they had to defend themselves from the atrocities of the Mughals.

The fifth Gum Aijundev was accused by Jahangir for helping his son Khusro in the revolt against him and was killed. The confrontation and martyrdom of the gurus transformed die Sikhs into a military brotherhood. To curb the growing power and strength of the Sikhs, Aurangzeb ordered the execution of Gum Tegh Bahadur in 1675 A.D. This enraged the Sikhs.

As a result, the tenth and last Gum Govind Singh organised the Sikhs as soldiers and prepared them for a long battle against the Mughals. Like Maratha, the Sikhs carried out raids in various places, but unlike Maratha, they could not establish an independent state during the reign of Aurangzeb. Thus we see that the relations between the Mughals and the Sikhs were not friendly. They were always on fighting terms. Enmity was at its height between both the sects.

Question 2.
Shivaji had excellent administrative ability. Explain?
Answer:
Shivaji was not only a great general but also a good administrator of top order. Shivaji’s administration was of high order which inspired by ideals of public welfare. Though Shivaji was all in all, in all matters, he kept a committee of 8 persons to advise him on the affairs of the state. This committee came to be known as Ashta Pradhan. This was file main feature of Shivaji’s administration.

The main source of income was the tax on the land which amounted to two – fifths of file land produce. Chauth and Sardeshmukhi were also levied on those living outside Maratha kingdom. Chauth was one fourth of the tax which farmers paid such kingdoms by their peasants. Sardeshmukhi was over and above this tax. It was one tenth of the total revenue, from which these taxes were collected, remained free from the Maratha looting’s and attacks.

For the smooth and efficient administration, Shivaji divided his kingdom into a number of provinces known as prants, and each prant into districts and parganas. In this way Shivaji proved himself as an able administrator.

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

Question 1.
Find the area of each of the following parallelograms:

Solution:
Area of parallelogram = Base × Height
(a) Height = 4 cm, Base = 7 cm
Area of parallelogram = 7 × 4 = 28 cm²

(b) Height = 3 cm, Base = 5 cm
Area of parallelogram = 5 × 3 = 15 cm²

(c) Height = 3.5 cm, Base = 2.5 cm
Area of parallelogram = 2.5 × 3.5 = 8.75 cm²

(d) Height = 4.8 cm, Base = 5 cm
Area of parallelogram = 5 × 4.8 = 24 cm²

(e) Height = 4.4 cm, Base = 2 cm
Area of parallelogram = 2 × 4.4 = 8.8 cm²

Question 2.
Find the area of each of the following triangles:


Solution:
Area of triangle = \(\frac{1}{2}\) × Base × Height
(a) Base = 4 cm, height = 3 cm
Area = \(\frac{1}{2}\) × 4 × 3 = 6 cm²

(b) Base = 5 cm, height = 3.2 cm
Area = \(\frac{1}{2}\) × 5 × 3.2 = 8 cm²

(c) Base = 3 cm, height = 4 cm
Area = \(\frac{1}{2}\) × 3 × 4 = 6cm²

(d) Base = 3 cm, height = 2 cm
Area = \(\frac{1}{2}\) × 3 × 2 = 3 cm²

Question 3.
Find the missing values:

Solution:
Area of parallelogram = Base × Height
(a) Base = 20 cm
Let height = h
Area of parallelogram = 246 cm²
∴ 20 × h = 246
⇒ h = \(\frac{246}{20}\) = 12.3 cm
Therefore, the height of parallelogram is 12.3 cm.

(b) Let base = b
Height = 15 cm
Area of parallelogram = 154.5 cm²
∴ b × 15 = 154.5
⇒ b = \(\frac{154.5}{15}\) = 10.3 cm
Therefore, the base of parallelogram is 10.3 cm.

(c) Let base = b
Height = 8.4 cm
Area of parallelogram = 48.72 cm²
∴ b × 8.4 = 48.72
⇒ b = \(\frac{48.72}{8.4}\) = 5.8 cm
Therefore, the base of parallelogram is 5.8 cm.

(d) Base = 15.6 cm
Let height = h
Area of parallelogram = 16.38 cm²
∴15.6 × h = 16.38
⇒ h = \(\frac{16.38}{15.6}\) = 1.05 cm
Therefore, the height of parallelogram is 1.05 cm.

Question 4.
Find the missing values:

Solution:
Area of triangle = \(\frac{1}{2}\) × Base × Height
Let b be the base of triangle and h be the height of triangle.
(i) b = 15 cm

Therefore, the height of triangle is 11.6 cm.

(ii) h = 31.4 mm
Area = \(\frac{1}{2}\) × b × h = 1256 mm²

Therefore, the base of triangle is 80 mm.

(iii) b = 22 cm

Therefore, the height of triangle is 15.5 cm.

Question 5.
PQRS is a parallelogram (see the given figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm

Solution:
(a) Area of parallelogram = Base × Height
= SR × QM
= 12 × 7.6 = 91.2 cm²

(b) PS = 8 cm
Area of parallelogram = Base × Height
= PS × QN = 91.2 cm²
⇒ 8 × QN = 91.2
⇒ QN = \(\frac{91.2}{8}\) = 11.4 cm

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (see the given figure).

If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Solution:
Area of parallelogram = Base × Height
= AB × DL
⇒ 1470 = 35 × DL
⇒ DL = \(\frac{1470}{35}\) = 42 cm
Also, area of parallelogram = AD × BM
⇒ 1470 = 49 × BM
∴ BM = \(\frac{1470}{49}\) = 30 cm

Question 7.
∆ABC is right angled at A (see the given figure). AD is perpendicular to BC. If AB = 5 cm, BC – 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.

Solution:

Question 8.
∆ABC is isosceles with AB = AC= 7.5 cm and BC = 9 cm (see the given figure). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i. e., CE?

Solution:

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 8 Comparing Quantities Ex 8.2

Question 1.
Convert the given fractional numbers to percents.

Solution:

Question 2.
Convert the given decimal fractions to percents.
(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Solution:

Question 3.
Estimate what part of the figures is coloured and hence find the percent which is coloured.

Solution:
(i) Here, 1 part out of 4 equals parts is shaded which represents the fraction \(\frac{1}{4}\).

(ii) Here, 3 parts out of 5 equal parts are shaded which represents the fraction \(\frac{3}{5}\).

(iii) Here 3 parts out of 8 equal parts are shaded which represents the fraction \(\frac{3}{8}\).

Question 4.
Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg
Solution:

Question 5.
Find the whole quantity if
(a) 5% of it is 600
(b) 12% of it is? 1080
(c) 40% of it is 500 km
(d) 70% of it is 14 minutes
(e) 8% of it is 40 litres
Solution:
Let the whole quantity be x.

Question 6.
Convert given percents to decimal fractions and also to fractions in simplest forms:
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Solution:

Question 7.
In a city, 30% are females, 40% are males and remaining are children. What percent are children?
Solution:
It is given that 30% are females and 40% are males.
Children = 100% – (40% + 30%)
= 100% – 70% = 30%

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Percentage of voters who voted = 60%
Percentage of those who did not vote = 100% – 60%
= 40%
Number of people who did not vote = 40% of 15000
= 40% × 15000

Therefore, 6000 people did not vote.

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let Meeta’s salary be ₹ x.
Given that, 10% of x = 400

Therefore, Meeta’s salary is ₹ 4000.

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solution:
Number of games won = 25% of 20
\(=\frac{25}{100} \times 20=5\)
Therefore, the team won 5 matches.

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 23 The Administration of the Mughal and the Life of the People

MP Board Class 7th Social Science Chapter 23 Text Book Questions

Choose the correct alternatives:

Question 1.
During the Mughal period the Di-wan or Wazir looked after:
(a) The income & expenditure
(b) The army
(c) Home affairs
(d) Judicial affairs
Answer:
(a) The income & expenditure

Question 2.
The main occupation of the people during the Mughal rule was:
(a) Agriculture
(b) Timber industry
(c) Foreign trade
(d) None of the above
Answer:
(a) Agriculture

Question 3.
Which new faith was propagated by Akbar:
(a) Din-i-illahi
(b) Bhakti Movement
(c) Sikhism
(d) All the above
Answer:
(a) Din-i-illahi

Fill in the blanks:

1. The court language was …………. during the Mughal period. (Sanskrit, Arabic, Urdu and Pessian)
2. During the reign of Akbar …………. was the great singer of India, (Tansen, Tulsidas, Raidas, Mirabai).
3. The ………….. built by Shajahan is included in the world heritage site. (Jama Masjid, Redfort, Tajmahal, Hawaniahal)

Answer:

1. Urdus and Persian
2. Tansen
3. Tajmahal.

MP Board Class 7th Social Science Chapter 23 Short Answer Type Questions

Question 1.
Write a short account of the administration of the Mughal period.
Answer:
The Emperor was all powerful. He ruled with the help of his army. There were many Ministers to did and assist the Emperor, e.g. Wakil, Wazir or Diwan, Mirabakshi, Khan – i – sama, Qazi – ul – Qazat etc. Akbar introduced the provincial system of administration in his empire. He divided his empire in 18 Subas.

Question 2.
Name of the items of import and export during the Mughal period.
Answer:

1. Items of import:
Gold, Silver, Copper, Tin, Steel, Glass, Minors, Wines, Horses, Corals, Mercury etc.

2. Items of export:
Muslim, Spices, Turmeric, Gun powder, Indigo, opium, Sugar, Gum, Sugar candy, Precious stones etc.

Question 3.
Give an account of Akbar’s religious policy.
Answer:
Akbar followed a policy of broad religious toleration. He gave full religious freedom to die people. In 1594 he abolished the Jazia which was used by the Ulema to humiliate the non – Muslims. He abolished the pilgrim’s tax. Generally he removed himself from orthodoxy in Islam. He set up a new religion which was compounded of many existing religions Hinduism, Christianity, Zoroastrianism etc. This new religion was known as Din-i-Illahi.

MP Board Class 7th Social Science Chapter 23 Long Answer Type Questions

Question 1.
Describe with examples the development of architecture during the Mughal Period.
Answer:
The Mughal Emperors were great lovers of architecture. The buildings of this period reflect the fusion of Hindu – Muslim – style of architecture. The tomb of Humayun is an excellent piece of architecture. Akbar built the city of Fatehpur Sikri in which besides Buland Darwaza many beautiful buildings were also constructed. Noorjahan built the tomb of her father Itmad – ud – daulah that was decorated wife precious gems.

The Tajmahal built by Shahjahan in fee memory of his Begun Mumtaz is fee best example of architecture of Mughal period. It has been included in fee World Heritage Site. Shahjahan also built – Jama Masjid at Delhi and Agra, The Red Fort of Delhi, Shish Mahal etc. The Moti Masjid was built by Aurangzeb. Thus we can find magnificent buildings during fee Mughal period.

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land, if 1 m² of the land costs ₹ 10,000.
Solution:
Length (l) = 500 m
Breadth (b) = 300 m
(i) Area = Length × Breadth = 500 × 300 = 150000 m²
(ii) Cost of 1 m² land = ₹ 10000
∴ Cost of 150000 m² land
= 150000 × 10000 = ₹ 1500000000

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:
Perimeter of the square park = 320 m
∴ 4 × Length of the side of park = 320
Length of the side of park = \(\frac{320}{4}\) = 80 m
Area = (Length of the side of park)²
= (80)² = 6400 m²

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Solution:
Area of a rectangular plot = 440 m²
Length = 22 m
Area = Length x Breadth = 440 m²
∴ 22 × Breadth = 440
⇒ Breadth = \(\frac{440}{22}\) = 20 m
∴ Perimeter = 2 (Length + Breadth)
= 2 (22 + 20) = 2(42) = 84 m

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Solution:
Length = 35 cm
Perimeter = 100 cm
∴ 2 (35 + Breadth) = 100
⇒ 35 + Breadth = 50
⇒ Breadth = 50 – 35 = 15 cm
∴ Area = Length × Breadth
= 35 × 15 = 525 cm²

Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:
Side of the square park = 60 m
Length of the rectangular park = 90 m
Area of the square park = (side)² = (60)² = 3600 m²
Area of rectangular park = Length × Breadth
= 90 × Breadth
It is given that area of square park = area of rectangular park
∴ 3600 = 90 × Breadth
⇒ Breadth = 40 m

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Solution:
Length of rectangle = 40 cm
Breadth of rectangle = 22 cm
Perimeter of rectangle = Perimeter of square
∴ 2 (Length + Breadth) = 4 × Side of square
⇒ 2 (40 + 22) = 4 × Side of square
⇒ 2 × 62 = 4 × Side of square
∴ Side of square = \(\frac{124}{4}\) = 31 cm
Now, area of rectangle = 40 × 22 = 880 cm²
Area of square = (Side)² = 31 × 31 = 961 cm²
As 961 > 880.
Therefore, the square-shaped wire encloses more area than rectangle – shaped wire.

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Solution:
Breadth = 30 cm
Perimeter = 130 cm
∴ 2 (Length + 30) = 130
⇒ Length + 30 = 65
⇒ Length = 65 – 30 = 35 cm
Now, area = Length × Breadth
= 35 × 30 = 1050 cm²

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (see the given figure). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m².

Solution:
Length of wall = 4.5 m
Breadth of wall = 3.6 m
Area of wall = Length × Breadth
= 4.5 × 3.6
= 16.2 m²
Area of door = 2 × 1 = 2 m²
Area to be white-washed
= Area of wall – Area of door
= 16.2 – 2 = 14.2 m²
Cost of white-washing 1 m² area = ₹ 20 2.
∴ Cost of white-washing 14.2 m² area
= 14.2 × 20 = ₹ 284

MP Board Class 7th Maths Solutions