Class 7

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

MP Board Class 7th Maths Solutions Chapter 7 त्रिभुजों की सर्वांगसमता Ex 7.2

प्रश्न 1.
निम्न में आप कौन-से सर्वांगसम प्रतिबन्धों का प्रयोग करेंगे ?
(a) दिया है : AC = DF AB = DE, BC = EF

इसलिए, ∆ABC ≅ ∆DEF
(b) दिया है : ZX = RP RQ = ZY
∠PRQ = ∠XZY
इसलिए, ∆PQR = ∆XYZ
(c) दिया है: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
इसलिए, ∆LMN ≅ ∆GFH
(d) दिया है: EB = DB
AE = BC
∠A = ∠C
इसलिए, ∆ABE ≅ ∆CDB
उत्तर:
(a) S.S.S. सर्वांगसमता प्रतिबन्ध द्वारा,
∆ABC ≅ ∆DEE
(b) S.A.S. सर्वांगसमता प्रतिबन्ध द्वारा,
∆PQR ≅ ∆XYZ.
(c) A.S.A. सर्वांगसमता प्रतिबन्ध द्वारा,
∆LMN ≅ ∆GFH.
(d) R.H.S. सर्वांगसमता प्रतिबन्ध द्वारा,
∆ABE ≅ ∆CDB.

प्रश्न 2.
आप ∆ART ≅ ∆PEN दर्शाना चाहते हैं।
(a) यदि आप S.S.S. सर्वांगसमता प्रतिबन्ध का प्रयोग करें तो आपको दर्शाने की आवश्यकता है:
(i) AR =
(ii) RT =
(iii) AT =
(b) यदि यह दिया गया है कि ∠T = ∠N और आपको S.A.S. प्रतिबन्ध का प्रयोग करना है, तो आपको आवश्यकता होगी:
(i) RT = और (ii) PN =

(c) यदि यह दिया गया है कि AT = PN और आपको A.S.A. प्रतिबन्ध का प्रयोग करना है, तो आपको आवश्यकता होगी:
(i) ? =
(ii) ? =
हल:
(a) ∆ART ≅ ∆PEN को S.S.S. सर्वांगसमता प्रतिबन्ध द्वारा दर्शाने के लिए दर्शाना होगा –
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) ∴ ∠T = ∠N
∴ (i) RT = EN
(ii) PN = AT
(c) यदि AT = PN और A.S.A. सर्वांगसमता के लिए आवश्यकता होगी –
(i) ∠RAT = ∠EPN
(ii) ∠ATR = ∠PNE

प्रश्न 3.
आपको ∆AMP ≅ ∆AMQ दर्शाना है। निम्न चरणों में, रिक्त कारणों को भरिए:

उत्तर:
(i) दिया है
(ii) दिया है
(iii) उभयनिष्ठ
(iv) S.A.S. सर्वांगसमता प्रतिबन्ध।

प्रश्न 4.
∆ABC में ∠L = 30°, ∠B = 40° और ∠C = 110°, ∆PQR में, ∠P = 30° ∠Q = 40° और ∠R = 110°. एक विद्यार्थी कहता है कि A.A.A. सर्वांगसमता प्रतिबन्ध से ∆ABC ≅ ∆PQR है।
क्या यह कथन सत्य है ? क्यों या क्यों नहीं ?
हल:
यहाँ ∆MBC के तीनों कोण ∆PQR के तीनों कोणों के बराबर हैं। तो यह आवश्यक नहीं कि त्रिभुज सर्वांगसम हों क्योंकि यदि ∆ABC में, भुजा BC = 3.0-सेमी तथा ∆POR में, भुजा QR = 4.0 सेमी हो, तो इस दशा में त्रिभुज के संगत कोण तो बराबर हैं परन्तु यह सर्वांगसम नहीं हैं। क्योंकि BC ≠ QR अतः विद्यार्थी की A.A.A. सर्वांगसमता का प्रतिबन्ध तर्कसंगत नहीं है।

प्रश्न 5.
संलग्न आकृति में दो त्रिभुज ART तथा OWN सर्वांगसम हैं जिनके संगत भागों को अंकित किया गया है। हम लिख सकते हैं ∆RAT = ?

हल:
हम लिख सकते हैं ∆RAT ≅ ∆WON
(∴ O ↔ A, N ↔ T, W ↔ R)

प्रश्न 6.
कथनों को पूरा कीजिए –

∆BCA ≅ ? ∆QRS ≅ ?
उत्तर:
∆BCA ≅ ∆ABTA, ∆QRS = ∆TPQ

प्रश्न 7.
एक वर्गांकित शीट पर, बराबर क्षेत्रफलों वाले दो त्रिभुजों को इस प्रकार बनाइए कि
(i) त्रिभुज सर्वांगसम हों
(ii) त्रिभुज सर्वांगसम न हों। आप उनके परिमाप के बारे में क्या कह सकते हैं?
हल:

(i) चित्र 7.19 (1) में,
∆ ABC का क्षेत्रफल = ∆EDC का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × 3 × 4 = 6 cm²
∆ ABC का परिमाप = 3 + 4 + 5 = 12 cm
∆ EDE का परिमाप = 3 + 4 + 5 = 12 cm
∆ ABC का परिमाप = ∆EDC का परिमाप,
अतः चित्र 7.19 में, ∆ABC ≅ ∆EDC है।
(ii) चित्र 7.19 (ii) में,
∆ PQR का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × PQ × PR
= \(\frac { 1 }{ 2 } \) × 3 × 4 = 6 cm²
तथा ∆ PSR का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × ST × PR
\(\frac { 1 }{ 2 } \) × 3 × 4 = 6 cm²

∴ ∆ POR का क्षेत्रफल = ∆ PSR का क्षेत्रफल
अब, ∆ PQR का परिमाप = 3 + 4 + 5 = 12 cm
तथा ∆ PRS का परिमाप = 4 + 35 + 4 = 11’5 cm
∆ POR का परिमाप ≠ ∆PRS का परिमाप
अत: चित्र 7.19 (ii) में ∆POR व ∆PRS सर्वांगसम नहीं हैं क्योंकि इनके क्षेत्रफल तो समान हैं परन्तु परिमाप समान नहीं

प्रश्न 8.
संलग्न आकृति में एक सर्वांगसम भागों का एक अतिरिक्त युग्म बताइए जिससे ∆ABC और ∆PQR सर्वांगसम हो जाएँ। आपने किस प्रतिबन्ध का प्रयोग किया ?

हल:
यहाँ, ∆ABC ≅ ∆PQR
∴ ∠B = ∠Q IR ∠C = ∠R
∴ सर्वांगसम भागों का अतिरिक्त युग्म –
BC = QR
उत्तर हमने यहाँ A.S.A. सर्वांगसम प्रतिबन्ध का प्रयोग किया है।

प्रश्न 9.
चर्चा कीजिए, क्यों?
∆ABC ≅ ∆FED.

हल:
∠B = ∠E = 90°,
∠A = ∠F (दिया हुआ है)
∴ ∠C = ∠D (तीसरा कोण)
BC = DE (दिया हुआ है)
अत: ASA सर्वांगसम प्रतिबन्ध से ∆ ABC ≅ ∆ FED परिणाम प्राप्त होगा।

पाठ्य-पुस्तक पृष्ठ संख्या # 163

ज्ञानवर्धक क्रियाकलाप

प्रश्न 1.
अलग-अलग माप के वर्गों के कट-आउट सोचिए। अध्यारोपण विधि का प्रयोग वर्गों की सर्वांगसमता के लिए प्रतिबन्ध ज्ञात करने के लिए कीजिए। कैसे “सर्वांगसम भागों” की संकल्पना सर्वांगसम के अंतर्गत उपयोग होती है ? क्या यहाँ संगत भुजाएँ हैं ? क्या यहाँ संगत विकर्ण हैं ?
हल:
हम जानते हैं कि समतल आकृतियाँ सर्वांगसम होती हैं। जब आकृतियों के आकार समान होते हैं तो वे एक-दूसरे की ठीक-ठीक पूरा ढक लेती हैं। सभी वर्ग समान आकृति के होते हैं लेकिन वर्ग का आकार उनकी भुजाओं की लम्बाई पर निर्भर करता है।

ABCD व PQRS दो वर्ग हैं। वर्ग ABCD के कट-आउट को वर्ग PQRS के ऊपर इस प्रकार रखते हैं कि शीर्ष A, वर्ग PQRS के शीर्ष P पर और भुजा AB भुजा PQ पर आए।

स्पष्ट है कि ABCD वर्ग PQRS को पूर्णतया ढक लेता है।

यदि AB = PQ तो दो वर्ग सर्वागसम होंगे यदि उनकी भुजाओं की लम्बाइयाँ समान हों।

अत: वर्ग ABCD ≅ वर्ग PORS यदि AB = PQ

हम एक वर्ग की किसी भी भुजा को दूसरे वर्ग की किसी भुजा के संगत ले सकते हैं। दूसरी संगत भुजाओं के युग्म इसी प्रकार बदल जाएँगे। यह बात विकर्णों के लिए भी सत्य है।

प्रश्न 2.
यदि आप वृत्त लेते हैं तो क्या होता है ? दो वृत्तों की सर्वांगसमता के लिए प्रतिबन्ध क्या है ? क्या, आप फिर अध्यारोपण विधि का प्रयोग कर सकते हैं ? पता लगाइए।
हल:
सभी वृत्तों की समान आकृति होती है और वृत्त का आकार वृत्त की त्रिज्या पर निर्भर करता है। यहाँ दो वृत्त C₁ व C₂ हैं। इनमें से किसी एक वृत्त का कट-आउट (माना वृत्त C₂ का) वृत्त C₁ पर रखते हैं। वृत्त C₂ वृत्त C₁ को पूरी तरह ठीक-ठीक ढल लेता है। यदि दोनों वृत्तों की त्रिज्याएँ समान होंगी तो दोनों वृत्त सर्वांगसम होंगे।

वृत्त C₁ वृत्त C₂ जबकि C₁ वृत्त की त्रिज्या = C₂ वृत्त की त्रिज्या।

प्रश्न 3.
इस संकल्पना को बढ़ाकर तल की दूसरी आकृतियाँ जैसे समषद्भुज इत्यादि के लिए प्रयत्न कीजिए।
हल:
हम जानते हैं कि समतल आकृतियाँ सर्वांगसम होती हैं यदि वे एक-दूसरे को पूर्णतया ढक लेती हैं। सभी समषट्भुज समान आकृति के होते हैं और इनका आकार समषट्भुज की भुजा की लम्बाई पर निर्भर करता है। दो समषट्भुज ABCDEF व PQRSTU लेते हैं। इनके कट-आउट लेते हैं जिनमें से प्रत्येक की सभी भुजाएँ समान हों।

अब PQRSTU के कट-आउट को ABCDEF पर इस प्रकार रखते हैं कि PQRSTU का बिन्दु P बिन्दु A पर आए तथा भुजा PQ भुजा AB पर आए। यदि PQ = AB तो समषट्भुज PQRSTU, समषट्भुज ABCDEF को पूर्णतया ठीक-ठीक ढक लेता है। अत: दो समषट्भुज सर्वांगसम होते हैं यदि इनकी भुजाओं की लम्बाई समान हो।

अत: समषट्भुज ABCDEF = समषट्भुज PQRSTU.

प्रश्न 4.
एक त्रिभुज की दो सर्वांगसम प्रतिलिपियाँ लीजिए। कागज को मोड़कर पता लगाइए कि क्या उनके शीर्ष लम्ब बराबर हैं ? क्या उनकी माध्यिकाएँ समान हैं ? आप उनके परिमाप तथा क्षेत्रफल के बारे में क्या कह सकते हैं ?
हल:
माना ∆ABC ≅ ∆DEF
कागज को मोड़कर प्रत्येक त्रिभुज के शीर्ष बनाए। हम देखते हैं कि
AL = DP BM = EQ और CN = FR
अर्थात् संगत शीर्ष लम्ब समान हैं।

इसी प्रकार हम देख सकते हैं कि सर्वांगसम त्रिभुजों में संगत माध्यिकाएँ समान होती हैं और इनके परिमाप व क्षेत्रफल समान होते हैं।

MP Board Class 7th Maths Solutions

MP Board Class 7th Social Science Solutions Chapter 22 Aurangzeb and the Decline of the Mughal Empire

MP Board Class 7th Social Science Chapter 22 Text Book Questions

Choose the correct alternatives:

Question 1.
Aurangzeb ruled for:
(a) 30 years
(b) 40 years
(c) 50 years
(d) 60 years
Answer:
(b) 40 years

Question 2.
The Jats lived near:
(a) Delhi and Agra
(b) Agra and Mathura
(c) Mathura and Bharatpur
(d) Arga and Jhansi
Answer:
(b) Agra and Mathura

Question 3.
The third battle of Panipat was fought in:
(a) 1526 AD
(b) 1556 AD
(c) 1560 AD
(d) 1761 AD
Answer:
(d) 1761 AD

Question 4.
The Indian goods which were in great demand in the European markets were:
(a) Clothes and spices
(b) Clothes and silver
(c) Spices and gems
(d) Spices and horses
Answer:
(a) Clothes and spices

Fill in the blanks:

1. ……………. and …………. were die Bundela rulers who revolted against Aurangzeb.
2. Gurudwara Sheshganj was built at the Martyrdom place of ……………
3. Nadh  Shah invaded India in ………….
4. Due to Aurangzeb’s policy of …………….. the Mughal administration weakend.

Answer:

1. Compestral, Chhatrasal
2. Guru Teg Bahadur
3. 1739 AD
4. Deccan

MP Board Class 7th Social Science Chapter 22 Short Answer Type Questions

Question 1.
Which taxes were imposed on the Hindus by Aurangzeb?
Answer:
Aurangzeb imposed the Jazia tax on the Hindus.

Question 2.
Who do you know about the Satnamis?
Answer:
Satnamis were the followers of truth. They used to dress like the purohits orpriests. They revolted due to the religious oppression of Aurangzeb. Their main center was Namaul and Mewar. However Aurangzeb suppressed their revolts ruthlessly.

Question 3.
Due to which policy of Aurangzeb his successors remained weak?
Answer:
Due to Aurangzeb’s policy of expansion his successors remained weak. The Mughal Empire was very vast and it was difficult to manage such a vast empire. The incompetent successors were not able to protect it.

Question 4.
What were the outcomes of the invasions of Nadir shah and Ahmad Shah Abdali?
Answer:
As a result of the invasions of Nadir Shah and Ahmad Shah Abdali the Mughal Empire became weak and remained restricted to Delhi and surrounding area.

MP Board Class 7th Social Science Chapter 22 Long Answer Type Questions

Question 1.
Describe the causes of the decline of the Mughal Empire.
Answer:
After the death of Aurangzeb in 1707, the Mughal Empire broke up. Its fall had already begun during Aurangzeb’s reign. But during the eighteenth century the fall was nearly complete.

The factors that led to the down-break of the Mughal empire were as follows:
1. Weak and incapable successors:
The successors of Aurangzeb were very weak and incapable. They had no capacity to rule over such a vast empire. They remained puppets in the hands of their subedars and other officers.

2. Wars of succession:
After the death of Aurangzeb war of succession broke out. They reduced the strength of the Kingdom.

3. The Suspicious nature of Aurangzeb and his rigid religious policy:
Aurabgzeb’s nature was very suspicious. Due this nature he could not provide administrative and military training to his sons. As a result they became incapable of managing the empire. His religious policy was based on intolerance. This made the Sikhs, Jats, Satnamis, Rajputs and Marathas his great enemies. They revolted against him.

4. Aurangzeb’s Deccan Policy:
Aurangzeb spent the last 25 years of his life fighting battles in the south. This made the Mughal administration in the north very weak.

5. Luxurious life styles of the Mughal rulers and Aristocrats:
Resulted in the decline of the Mughal Empire.

6. Foreign invasion:
Nadirshah, the ruler of Iran invaded the Mughal Empire in 1739. He looted the wealth about seventy crores of rupees. After his assassination in 1747 AD his general Ahmadshah Abdali, became the ruler and fought third battle of Panipat with Maratha in which Maratha were defeated. Because of these invasions Mughal Empire became weak and remained restricted to Delhi and surrounding areas.

7. Lack of Naval Power:
Neglect of navy proved fatal for the Mughal Empire.

Project Work:
On an outline map of India mark the extent of the Mughal Empire. Also mark the areas of Bundelas, Jats and Satnamis.
Answer:

MP Board Class 7th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic
watches or calculators,

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind

Solution:
(a) It is given that the number of segments required to form n digits of the kind
is (5n + 1).
Number of segments required to form 5 digits = (5 × 5 + 1) = 25 + 1 = 26
Number of segments required to form 10 digits = (5 × 10 + 1) = 50 + 1 = 51
Number of segments required to form 100 digits = (5 × 100 + 1) = 500 + 1 = 501

(b) It is given that the number of segments required to form n digits of the kind
[ is (3n +1).
Number of segments required to form 5 digits = (3 × 5 + 1) = 15 + 1 = 16
Number of segments required to form 10 digits = (3 × 10 + 1) = 30 + 1 = 31
Number of segments required to form 100 digits = (3 × 100 + 1) = 300 + 1 = 301

(c) It is given that the number of segments required to form n digits of the kind
is (5n + 2).
Number of segments required to form 5 digits = (5 × 5 + 2) = 25 + 2 = 27
Number of segments required to form 10 digits = (5 × 10 + 2) = 50 + 2 = 52
Number of segments required to form 100 digits = (5 × 100 + 2) = 500 + 2 = 502

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution:
(i) Number pattern for expression 2n – 1
Put n = 1, 2, 3,…. and so on, we get

(ii) For expression 3 n + 2, 5^th, 10^th term and 100th term of the pattern are 3 × 5 + 2 = 17, 3 × 10 + 2 = 32 and 3 × 100 + 2 = 302 respectively.
(iii) For expression 4n +1, 5^th, 10^th and 100^th term of the pattern are 4 × 5 +1 = 21, 4 × 10 + 1 = 41 and 4 × 100 + 1 = 401 respectively.
(iv) For expression 7n + 20, 5^th, 10^th and 100^th term of the pattern are 7 × 5 + 1 = 36, 7 × 10 + 20 = 90 and 7 × 100 + 20 = 720 respectively.
(v) For expression n² + 1, 5^th and 10th term of the pattern are 5² + 1 = 26 and 10² + 1 = 101 respectively.

MP Board Class 7th Maths Solutions