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

MP Board Class 8 Maths Chapter 8 Exercise 8.2 Question 1.
A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
Solution:
We have,
Increase in salary = 10%
New salary = ₹ 154000
Let original salary be ₹ s.

∴ His original salary is ₹ 140000.

MP Board Class 8 Maths Chapter 8 Question 2.
On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday?
Solution:
Number of people went to Zoo on Sunday = 845
Number of people went to Zoo on Monday = 169
Decrease in number of people = 845 – 169
= 676

MP Board Class 8 Maths Solutions English Medium Chapter 8 Question 3.
A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
We have,
Cost price = ₹ 2400
Profit = 16%
Number of articles = 80

∴ Selling price of 80 articles

Thus, the selling price of one article = \(\frac{2784}{80}\)
= ₹ 34.80

Class 8 MP Board Maths Chapter 8 Question 4.
The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.
Solution:
Cost of an article = ₹ 15500
Money spent on repairs = ₹ 450
∴ Total cost = Cost + Repairs = ₹ 15950
Profit = 15 %

Class 8 Maths Exercise 8.2 Question 5.
A VCR and TV were bought for? 8000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
Solution:
Cost of VCR = ₹ 8000
Cost of TV = ₹ 8000
Total cost = 8000 + 8000 = ₹ 16000
Loss on VCR = 4 %

Class 8 Maths 8.2 English Medium Question 6.
During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?
Solution:
We have,
Cost of a pair of jeans = ₹ 1450
Cost of two shirts = 2 × 850 = ₹ 1700
∴ Total cost = 1450 + 1700 = ₹ 3150
Discount = 10 %

Comparing Quantities Class 8 Exercise 8.2 Question 7.
A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find C.P. of each).
Solution:
We have,
Selling price of each buffalo = ₹ 20000
Total selling price = ₹ 40000
On buffalo 1, profit = 5 %
Cost price of buffalo 1

Total cost price = 19047.62 + 22222.22
= ₹ 41269.84
Since cost price is greater than the selling price.
∴ Loss = 41269.84 – 40000 = ₹ 1269.84

Class 8 Maths Chapter 8 Exercise 8.2 Question 3 Question 8.
The price of a TV is ₹ 13000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution:
Cost price of TV = ₹ 13000
Sales tax = 12 %

Class 8 Maths Comparing Quantities Exercise 8.2 Question 9.
Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.
Solution:
Selling price = ₹ 1600
Discount = 20 %
We know,

Class 8 Maths Chapter 8 MP Board Question 10.
I purchased a hair dryer for? 5,400 including 8% VAT. Find the price before VAT was added.
Solution:
Cost price = ₹ 5400
VAT = 8%

MP Board Class 8th Maths Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Class 10 Maths Chapter 5 Exercise 5.3 MP Board Question 1.
Find the sum of the following APs:
(i) 2,7,12, ….. to 10 terms
(ii) -37, -33, -29, …. to 12 terms
(iii) 0.6, 1.7, 2.8, …. to 100 terms
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}\), ……
Solution:
(i) Here, a = 2, d = 7 – 2 = 5, n = 10

Thus, the sum of first 10 terms is 245.

(ii) Here, a = -37, d = -33 – (-37) = 4, n = 12

Thus, the sum of first 12 terms is -180.

(iii) Here, a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100

Thus, the required sum of first 100 terms is 5505.

(iv)

Thus, the required sum of first 11 terms is \(\frac{33}{20}\)

Maths Class 10 5.3 Question 2.
Find the sums given below:
(i) 7+10\(\frac{1}{2}\) + 14 + …… + 84
(ii) 34 + 32 + 30+ …….. + 10
(iii) -5 + (-8) + (-11) + ……. + (-230)
Solution:
(i) Here,

Let n be the number of terms.

(ii) Here, a = 34, d = 32 – 34 = -2, l = 10
Let n be the number of terms

(iii) Here, a = -5,d = -8- (-5) = -3,1 = -230
Let n be the number of terms.

Class 10 Maths 5.3 Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50,find n and S_n.
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) given a_n = 4, d = 2,S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) Here, a = 5, d = 3 and a_n = 50 = l

Arithmetic Progression Class 10 5.3 Question 4.
How many terms of the AP : 9,17, 25, ….. must be taken to give a sum of 636?
Solution:
Here, a = 9, d = 17 – 9 = 8, S_n = 636

Class 10 Ka Math 5.3 Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Ex 5.3 Class 10 Question 6.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
We have, first term (a) = 17, last term (l) = 350 = T_n and common difference (d) = 9
Let the number of terms be n.

Class 10 Maths Chapter 5 Exercise 5.3 Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.
Solution:
Here, n = 22, T₂₂ = 149 = l, d = 7
Let the first term of the AP be a.

Class 10 Maths Exercise 5.3 Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:

Ch 5 Maths Class 10 Ex 5.3 Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, we have S₇ = 49 and S₁₇ = 289
Let the first term of the AP be a and d be the common difference, then

Maths Exercise 5.3 Class 10 Question 10.
Show that a₁, a₂,…… a_n,………, form an AP where a_n is defined as below:
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:

5.3 Ka Question Number 3 Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3^rd, the 10^th and the n^th terms.
Solution:
We have, S_n = 4n – n²
S₁ = 4(1) – (1)² = 4 – 1 = 3 ⇒ First term (a) = 3

Math 10 Class Chapter 5.3 Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 are 6,12,18, ……. , (6 × 40)
And, these numbers are in AP, such that a = 6,
d = 12 – 6 = 6 and _n = 6 × 40 = 240 = l

= 20[12 + 39 × 6] = 20[12 + 234]
= 20 × 246 = 4920

Class 10 Maths Exercise 5.3 Solutions Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 are 8, 16, 24, 32, ….., 120.
These numbers are in AP, where, a = 8 and l = 120

Thus, the sum of first 15 multiples of 8 is 960.

Class 10 Maths Chapter 5 Exercise 5.3 Question 3 Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, …… , 49
These numbers are in AP such that a = 1 and l = 49

Thus, the sum of odd numbers between 0 and 50 is 625.

MP Board Class 10th Maths Solution Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
Here, penalty for delay on 1^st day = ₹ 200
2^nd day = ₹ 250
3^rd day = ₹ 300
……………………………………….
……………………………………….

MP Board Solution Class 10th Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Sum of all the prizes = ₹ 700
Let the first prize be a
2^nd prize = (a – 20)
3^rd prize = (a- 40)
4^th prize = (a – 60)
The above prizes form an AP
Now, we have, first term = a
Common difference = d = (a – 20) – a = -20

Class 10 Chapter 5 Exercise 5.3 Question Number 3 Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students ?
Solution:
Number of classes = 12
∵ Each class has 3 sections.
∴ Number of trees planted by class I = 1 × 3 = 3
Number of trees planted by class II =2 × 3 = 6
Number of trees planted by class III = 3 × 3 = 9
Number of trees planted by class IV = 4 × 3 = 12
………………………………………………………………….
Number of trees planted by class XII = 12 × 3 = 36
The numbers 3, 6, 9,12, ……. , 36 form an AP
Here, a = 3, d = 6 – 3 = 3 and n = 12

Class 10th Chapter 5.3 Question 18.
A spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles? (Take \(\pi=\frac{22}{7}\))

Solution:

Class 10th Chapter 5.3 Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

5.3 Class 10 Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Solution:
Here, number of potatoes = 10
The to and fro distance of the bucket:

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.3 Pdf, Class 10 Subject Maths Chapter 3 Exercise 3.3, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Find the value of x in each of the given triangles.

MP Board Class 10th Maths Chapter 3 Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14; x – y = 4
(ii) s – f = 3; \(\frac{s}{3}+\frac{t}{2}\) = 6
(iii) 3x – y = 3; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
(v) \(\sqrt{2} x+\sqrt{3} y\) = 0; \(\sqrt{3} x-\sqrt{8} y\) = 0
(vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2, \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
(i) x + y = 14 … (1),
x – y = 4 …. (2)
From (1) , we get x = (14 – y) …. (3)
Substituting value of x in (2) , we get
(14 – y) – y = 4 ⇒ 14 – 2y = 4 ⇒ -2y = -10 ⇒ y = 5
Substituting y = 5 in (3), we have
x = 14 – 5 ⇒ x = 9
Hence, x = 9, y = 5

(ii) s – t = 3 … (1)

From (1), we have s = (3 + t) … (2)
Substituting this value of s in (2), we get

Substituting, t = 6 in (3) we get,
S = 3 + 6 = 9
Thus, S = 9, f = 6

(iii) 3x – y = 3 … (1),
9x – 3y = 9 … (2)
From (1) , y = (3x – 3)
Substituting this value of y in (2),
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9 ⇒ 9 = 9 which is true,
Eq. (1) and eq. (2) have infinitely many solutions.

(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 …. (2)
From the equation (1),

Substituting the value of y in (2), we have


Substituting the value of x in (1), we have

Solve using Substitution Calculator is a free online tool that displays the solution of the pair of linear equations using the substitution method.

MP Board Solutions Class 10 Maths Chapter 3 Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:

Enter the equation A and B in the substitution calculator for solving the linear equations.

Class 10 Maths Chapter 3 MP Board Question 3.
Form the pair of linear equations for the following problems and find their solution by
substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for 13800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let the two numbers be X and y such that x > y
It is given that
Difference between two numbers = 26
∴ x – y = 26 … (1)
Also one number = 3 [the other number]
⇒ x = 3y … (2)
Substituting x = 3y in (1) , we get 3y – y = 26 ⇒ 2y = 26
Now, substituting y = 13 in (2) , we have
x = 3(13) ⇒ x = 39
Thus, two numbers are 39 and 13.

(ii) Let the two angles be x and y such that x > y
∵ The larger angle exceeds the smaller by 18° (Given)
∴ x = y + 18°…. (1)
Also, sum of two supplementary angles = 180°
∴ x + y = 180° … (2)
Substituting the value of x from (1) in (2) , we get,
(18° + y) + y = 180°

Substituting, y = 81° in (1) , we get
x = 18° + 81° = 99°
Thus, x = 99° and y = 81°

(iii) Let the cost of a bat = ₹ x
And the cost of a ball = ₹ y
∵ [cost of 7 bats] + [cost of 6 balls] = ₹ 3800
⇒ 7x + 6y = 3800 … (1)
Also, [cost of 3 bats] + [cost of 5 balls] = ₹ 1750
3x + 5y = 1750 …. (2)

Substituting this value of y in (1) , we have

(iv) Let fixed charges = ₹ x
and charges per km = ₹ y
∵ Charges for the journey of 10 km = ₹ 105 (Given)
∴ x + 10y = 105 … (1)
and charges for the journey of 15 km = ₹ 155
∴ x + 15y = 155 … (2)
From (1) , we have, x = 105 – 10y …. (3)
Putting the value of x in (2) , we get
(105 – 10y) + 15y = 155
⇒ 5y = 155 – 105 = 50 ⇒ y = 10
Substituting y = 10 in (3) , we get
x = 105 – 10(10) ⇒ x = 105 – 100 = 5
Thus, x = 5 and y = 10
⇒ Fixed charges = ₹ 5
and charges per km = ₹ 10
Now, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255
∴ The charges for 25 km journey = ₹ 255

(v) Let the numerator = x
and the denominator = y
∴ Fraction = \(\frac{x}{y}\)
According to the given condition,

⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y + 4 = 0 … (1)

Substituting this value of x in (1) , we have

(vi) Let the present age of Jacob = x years
and the present age of his son = y years
∴ 5 years hence: Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
According to given condition,
[Age of Jacob] = 3[Age of his son]
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x – 3y -10 = 0 … (1)
5years ago : Age of Jacob = (x – 5) years,
Age of his son = (y – 5) years
According to given condition,
[Age of Jacob] = 7[Age of his son]
∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35
⇒ x – 7y + 30 = 0 … (2)
From (1) , x = [10 + 3y] … (3)
Substituting this value of x in (2) , we get
(10 + 3y) – 7y + 30 = 0
⇒ -4y = -40 ⇒ y = 10
Now, substituting y = 10 in (3) ,
we get x = 10 + 3(10)
⇒ x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and present age of his son = 10 years

MP Board Class 8th Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.2

MP Board Class 8 Maths Chapter 7 Exercise 7.2 Question 1.
Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
(ii) 512
(iii) 10648
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
Solution:

Class 8 Subject Math Chapter 7 Exercise 7.2 Question 2.
State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Solution:
(i) False
Because the cube of any odd number is odd.

(ii) True
Because to make a perfect cube, we need each and every factor to occur three times in a group.
So, we need 3, 6, 9, …. zeros to make a perfect cube.

(iii) False
15² = 225, which ends with 5.
and 15³ = 3375, which ends with 75.

(iv) False
∵ 12³ = 1728, which ends with 8.

(v) False
∵ 10³ = 1000, where 10 is the smallest two digit number, and its cube contains 4 digits.

(vi) False
∵ 99³ = 970299, where 99 is the largest two digit number and its cube contains 6 digits.

(vii) True
∵ 1³ = 1, Which is a single digit number.

Class 8 Maths Exercise 7.2 Question Number 3 Question 3.
You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
The given number is 1331.
Step-1: Start making groups of three digits starting from the right most digit of the, number we get 331 and 1 as two groups of three and one digits.
Step-2: First group, i.e., 331 will give the one’s digit (unit’s) of the required cube root. The number 331 ends with 1. We know that 1 comes at the unit’s place of a number only
when its cube root ends in 1.
So, we get 1 at the unit’s place of the cube root.
Step-3: Now take another group, i.e., 1
We know that 1³ = 1 and 2³ = 8. Also 1 ≤ 1 ≤ 8. We take the smaller number 1 as the ten’s place of the required cube root. So, we get
\(\sqrt[3]{1331}=11\)
Cube root of 4913
Step-1: Forming groups of three digits starting ’ from the right most digit of 4913 we get two groups 913 and 4.
Step-2: One’s place digit of first group is 3.
∴ We take one’s place digit of required cube root is 7(∵ 7³ = 343)
Step-3: Now the second group is 4.
1³ = 1, 2³ = 8
1 < 4 < 8, hence 1 is the ten’s place digit of required cube root.
So, we get \(\sqrt[3]{4913}\) = 17.
Cube root of 12167.
Step-1 : Forming groups of three digits starting from the right most digit of 12167 we get two groups 167 and 12.
Step-2 : First group is 167.
∴ unit’s place digit of required cube root is 3. (∵ 3³ = 27)
Step-3: Second group is 12
2³ = 8 and 3³ = 27, 8 < 12 < 27
⇒ 23 < 12 < 33
∴ 2 is the ten’s place digit of required cube root. So, we get \(\sqrt[3]{12167}\) = 23.
Cube root of 32768
Step-1: Forming groups of three digits starting from the right most digit of 32768 we get two groups 768 and 32.
Step-2: First group is 768.
∴ unit’s place digit of required cube root is 2
(∵ 2³ = 8).
Step-3: Second group is 32
3³ = 27 and 4³ = 64, 27 < 32 < 64
⇒ 33 < 32 < 43
∴ 3 is the ten’s place digit of required cube root. So, we get \(\sqrt[3]{32768}\) = 32.

MP Board Class 8th Maths Solutions

MP Board Class 8th Science Solutions Chapter 5 कोयला और पेट्रोलियम

MP Board Class 8th Science Chapter 5 पाठ के अन्तर्गत के प्रश्नोत्तर

पाठ्य-पुस्तक पृष्ठ संख्या # 56

क्रियाकलाप 5.1

MP Board Class 8 Science Chapter 5 प्रश्न 1.
अपने दैनिक जीवन में उपयोग में लाये जाने वाले पदार्थों की एक सूची बनाइए और उनको प्राकृतिक तथा मानव निर्मित वर्गों में वर्गीकृत कीजिए –
उत्तर:

MP Board Class 8 Science Chapter 3 Coal And Petroleum प्रश्न 2.
क्या इस सूची में वायु, जल, मृदा और खनिज सम्मिलित हैं?
उत्तर:
हाँ, इस सूची में वायु, जल, मृदा और खनिज सम्मिलित हैं।

MP Board Class 8th Science Chapter 5 प्रश्न 3.
क्या हम अपने सभी प्राकृतिक संसाधनों का निरन्तर उपयोग कर सकते हैं?
उत्तर:
नहीं, हम अपने सभी प्राकृतिक संसाधनों का निरन्तर उपयोग नहीं कर सकते हैं। एक दिन वे समाप्त हो जायेंगे।

Class 8 Science Chapter 5 MP Board प्रश्न 4.
क्या वायु, जल और मृदा मानवीय क्रियाकलापों द्वारा समाप्त हो सकते हैं?
उत्तर:
नहीं, वायु, जल और मृदा मानवीय क्रियाकलापों द्वारा समाप्त नहीं हो सकते हैं।

MP Board Class 8 Science प्रश्न 5.
क्या जल एक असीमित संसाधन है?
उत्तर:
हाँ, जल एक असीमित संसाधन है। लेकिन इसके असीमित दोहन से यह संकट उत्पन्न कर सकता है।

क्रियाकलाप 5.2

MP Board Solutions Class 8 Science प्रश्न 1.
अब अन्तिम रूप से देखिए कि तीसरी पीढ़ी के सभी उपभोक्ताओं को खाने हेतु कुछ मिला या नहीं। यह भी देखिए कि क्या पात्रों में अब भी कुछ शेष बच गया है?
उत्तर:
हाँ, तीसरी पीढ़ी के उपभोक्ताओं को खाने हेतु कुछ मिला है, लेकिन उन्हें बहुत थोड़ी मात्रा में मिला है।’ नहीं, पात्रों में अब कुछ शेष नहीं बचा है।

MP Board 8th Class Science Book प्रश्न 2.
क्या किसी समूह की पहली पीढ़ी बहुत अधिक लालची है?
उत्तर:
हाँ, किसी समूह की पहली पीढ़ी बहुत अधिक लालची है।

पाठ्य-पुस्तक पृष्ठ संख्या # 57

MP Board Science Class 8 प्रश्न 3.
कोयला हमें कहाँ से प्राप्त होता है और यह कैसे बनता है?
उत्तर:
कोयला हमें जमीन के अन्दर से प्राप्त होता है। बाढ़ जैसे प्राकृतिक प्रक्रमों के कारण वन मृदा के नीचे दब जाते हैं। इनके ऊपर मृदा जम जाने के कारण ये सम्पीडित हो जाते हैं। जैसे-जैसे ये गहरे होते जाते हैं उनका ताप भी बढ़ता जाता है। उच्च ताप व उच्च दाब के कारण पृथ्वी के अन्दर मृत पेड़-पौधे धीरे-धीरे कोयले में परिवर्तित हो जाते हैं। कोयले में मुख्य रूप से कार्बन होता है। यह प्रक्रम कार्बनीकरण कहलाता है।

पाठ्य-पुस्तक पृष्ठ संख्या # 59

पेट्रोलियम

Class 8 Science MP Board प्रश्न 1.
क्या आप जानते हैं कि पेट्रोलियम कैसे बनता है?
उत्तर:
पेट्रोलियम का निर्माण समुद्र में रहने वाले जीवों से हुआ। जब ये जीव मृत हुए तो इनके शरीर समुद्र के तल में जाकर बैठ गए और फिर रेत तथा मिट्टी की तहों द्वारा ढक गए। लाखों वर्षों के बाद वायु की अनुपस्थिति, उच्च ताप और उच्च दाब के कारण मृत जीव पेट्रोलियम और प्राकृतिक गैस में परिवर्तित हो गए।

MP Board Class 8th Science प्रश्न 2.
पेट्रोलियम तेल और गैस की परत, जल की परत के ऊपर है। ऐसा क्यों है?
उत्तर:
पेट्रोलियम तेल और गैस जल से हल्के होते हैं, ये जल में मिश्रित नहीं होते। अतः ये जल की परत के ऊपर रहते हैं।

पाठ्य-पुस्तक पृष्ठ संख्या # 61

Class 8 MP Board Science प्रश्न 1.
क्या प्रयोगशाला में मृत जीवों से पेट्रोलियम और प्राकृतिक गैस बनाई जा सकती है?
उत्तर:
नहीं, पेट्रोलियम और प्राकृतिक गैस का बनना एक बहुत धीमा प्रक्रम है और इसके बनने की परिस्थितियाँ प्रयोगशाला में उत्पन्न नहीं की जा सकती। अतः इन्हें प्रयोगशाला में नहीं बनाया जा सकता।

MP Board Class 8th Science Chapter 5 पाठान्त अभ्यास के प्रश्नोत्तर

MP Board Class 8 Science Solution प्रश्न 1.
सीएनजी और एलपीजी का ईंधन के रूप में उपयोग करने के क्या लाभ हैं?
उत्तर:
सीएनजी और एलपीजी उपयोग करने के लाभ:

1. ये कम प्रदूषणकारी है।
2. ये स्वच्छ ईंधन है।
3. ये अन्य ईंधनों की अपेक्षा सस्ते हैं।
4. ये आसानी से उपलब्ध हैं।
5. इन्हें घरों और कारखानों में सीधा जलाया जा सकता है, जहाँ इनकी आपूर्ति पाइपों के द्वारा की जाती है।

MP Board Class 8 Science Solutions English Medium प्रश्न 2.
पेट्रोलियम का कौन-सा उत्पाद सड़क निर्माण हेतु उपयोग में लाया जाता है?
उत्तर:
सड़क निर्माण हेतु पेट्रोलियम उत्पाद बिटुमिन उपयोग में लाया जाता है।

MP Board 8th Science प्रश्न 3.
वर्णन कीजिए, मृत वनस्पति से कोयला किस प्रकार बनता है? यह प्रक्रम क्या कहलाता है?
उत्तर:
कोयला हमें जमीन के अन्दर से प्राप्त होता है। बाढ़ जैसे प्राकृतिक प्रक्रमों के कारण वन मृदा के नीचे दब जाते हैं। इनके ऊपर मृदा जम जाने के कारण ये सम्पीडित हो जाते हैं। जैसे-जैसे ये गहरे होते जाते हैं उनका ताप भी बढ़ता जाता है। उच्च ताप व उच्च दाब के कारण पृथ्वी के अन्दर मृत पेड़-पौधे धीरे-धीरे कोयले में परिवर्तित हो जाते हैं। कोयले में मुख्य रूप से कार्बन होता है। यह प्रक्रम कार्बनीकरण कहलाता है।

यूपी बोर्ड कक्षा 8 विज्ञान पाठ 5 प्रश्न 4.
रिक्त स्थानों की पूर्ति कीजिए –

1. …………. तथा ………… जीवाश्म ईंधन हैं।
2. पेट्रोलियम के विभिन्न संघटकों को पृथक करने का प्रक्रम ………… कहलाता है।
3. वाहनों के लिए सबसे कम प्रदूषक ईंधन ……..।

उत्तर:

1. कोयला, पेट्रोलियम
2. परिष्करण
3. सीएनजी (Compressed Natural Gas-CNG)

कक्षा 8 विज्ञान पाठ 5 के प्रश्न उत्तर प्रश्न 5.
निम्नलिखित कथनों के सामने सत्य/असत्य लिखिए –

1. जीवाश्म ईंधन प्रयोगशाला में बनाए जा सकते हैं। (सत्य/असत्य)
2. पेट्रोल की अपेक्षा सीएनजी अधिक प्रदूषक ईंधन (सत्य/असत्य)
3. कोक, कार्बन का लगभग शुद्ध रूप है। (सत्य/असत्य)
4. कोलतार विभिन्न पदार्थों का मिश्रण है। (सत्य/असत्य)
5. मिट्टी का तेल एक जीवाश्म ईंधन नहीं है। (सत्य/असत्य)

उत्तर:

1. असत्य।
2. असत्य।
3. सत्य।
4. सत्य।
5. असत्य।

एमपी बोर्ड क्लास 8 विज्ञान प्रश्न 6.
समझाइए, जीवाश्म ईंधन समाप्त होने वाले प्राकृतिक संसाधन क्यों हैं?
उत्तर:
जीवाश्म ईंधन समाप्त होने वाले संसाधन हैं क्योंकि इनके बनने में मृतजीवों को ईंधन में परिवर्तित होने के लिए लाखों वर्ष लगते हैं और उपलब्ध भण्डार सीमित हैं। इसलिए जीवाश्म ईंधन समाप्त होने वाले प्राकृतिक संसाधन हैं।

कोयला और पेट्रोलियम Class 8 प्रश्न 7.
कोक के अभिलक्षणों और उपयोगों का वर्णन कीजिए।
उत्तर:
कोक के अभिलक्षण: कोक एक कठोर, सरन्ध्र और काला पदार्थ है। यह कार्बन का लगभग शुद्ध रूप है।

कोक का उपयोग: कोक का उपयोग इस्पात के औद्योगिक निर्माण और बहुत से धातुओं के निष्कर्षण में किया जाता है।

कोयला और पेट्रोलियम के प्रश्न उत्तर प्रश्न 8.
पेट्रोलियम निर्माण के प्रक्रम को समझाइए।
उत्तर:
पेट्रोलियम का निर्माण समुद्र में रहने वाले जीवों से हुआ। जब ये जीव मृत हुए तो इनके शरीर समुद्र के तल में जाकर बैठ गए और फिर रेत तथा मिट्टी की तहों द्वारा ढक गए। लाखों वर्षों के बाद वायु की अनुपस्थिति, उच्च ताप और उच्च दाब के कारण मृत जीव पेट्रोलियम और प्राकृतिक गैस में परिवर्तित हो गए।

कक्षा 8 विज्ञान पाठ 3 कोयला और पेट्रोलियम प्रश्न 9.
निम्नलिखित सारणी में 2004 से 2010 तक भारत में विद्युत् की कुल कमी को दिखाया गया है। इन आँकड़ों को ग्राफ द्वारा आलेखित करिए। वर्ष में कमी प्रतिशतता को Y-अक्ष पर तथा वर्ष को X-अक्ष पर आलेखित करिए।

उत्तर:

MP Board Class 8th Science Solutions

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1

Solve the following equations.

MP Board Class 8 Maths Solutions English Medium Chapter 2 Question 1.
x – 2 = 7.
Solution:
We have, x – 2 = 7
Transposing -2 to R.H.S., we get
x = 7 + 2 ⇒ x = 9, which is the required solution.

Class 8 Maths Chapter 2 MP Board Question 2.
y + 3 = 10.
Solution:
We have, y + 3 = 10
Transposing 3 to R.H.S., we get y = 10 – 3
⇒ y = 7, which is the required solution.

MP Board Class 8 Maths Solution Chapter 2 Question 3.
6 = z + 2.
Solution:
We have, 6 = z + 2
Transposing 2 to L.H.S., we get
6 – 2 = z
⇒ z = 4, which is the required solution.

Class 8 MP Board Maths Chapter 2 Question 4.
\(\frac{3}{7}\) + x = \(\frac{17}{7}\).
Solution:

⇒ x = 2, which is the required solution.

MP Board Class 8 Maths Chapter 2 Exercise 2.1 Question 5.
6x = 12.
Solution:
We have, 6x = 12
Dividing both sides by 6, we get
\(\frac{6 x}{6}=\frac{12}{6}\)
⇒ x = 2, which is the required solution.

Class 8 Maths Chapter 2 Exercise 2.1 Solutions 2023 Question 6.
\(\frac{t}{5}=10\)
Solution:
We have, \(\frac{t}{5}\) = 10
Multiplying both sides by 5, we get

⇒ t = 50, which is the required solution.

Class 8 Maths Chapter 2 Exercise 2.1 Solutions Question 7.
\(\frac{2 x}{3}\) = 18
Solution:
We have,
\(\frac{2 x}{3}\) = 18
Dividing both sides by 2, we get

Now, multiplying both sides by 3, we get

⇒ x = 27, which is the required solution.

Class 8 Maths MP Board Chapter 2 Question 8.
\(1.6=\frac{y}{1.5}\)
Solution:
We have, \(1.6=\frac{y}{1.5}\)
Multiplying both sides by 1.5, we get

⇒ y = 2.4, which is the required solution.

Class 8 Maths Chapter 2 Exercise 2.1 Question 8 Question 9.
7x – 9 = 16.
Solution:
We have, 7x – 9 = 16
Transposing -9 to R.H.S., we get
7x = 16 + 9
⇒ 7x = 25
Now, dividing both sides by 7, we get

⇒ x = \(\frac{25}{7}\), which is the required solution.

8th English 2.1 Question Answe rQuestion 10.
14y – 8 = 13.
Solution:
We have, 14y -8 = 13
Transposing -8 to R.H.S., we get
14y = 13 + 8 ⇒ 14y = 21
Now, dividing both sides by 14, we get

MP Board 8th Class Maths Solution Question 11.
17 + 6p = 9.
Solution:
We have, 17 + 6p = 9
Transposing 17 to R.H.S., we get 6p = 9 – 17
⇒ 6p = – 8
Now, dividing both sides by 6, we get

MP Board Solution Class 8 Maths Question 12.
\(\frac{x}{3}+1=\frac{7}{15}\)
Solution:

Now, multiplying both sides by 3, we get

⇒ x = \(-\frac{8}{5}\), which is the required solution.

MP Board Class 8th Maths Solutions

In this article, we will share MP Board Class 7th Hindi Solutions Chapter 8 उत्साह Pdf, Class 7 Hindi Lesson 8, These solutions are solved subject experts from the latest edition books.

MP Board Class 7th Hindi Sugam Bharti Solutions Chapter 8 उत्साह

MP Board Class 7th Hindi Sugam Bharti Chapter 8 प्रश्न-अभ्यास

वस्तुनिष्ठ प्रश्न

MP Board Class 7th Hindi Chapter 8 प्रश्न 1.
(क) सही जोड़ी बनाइए
1. बुरी – (क) महूरत
2. बड़ी – (ख) शेर
3. बड़े – (ग) जिंदगी
4. यात्रा – (घ) शर्मिंदगी
उत्तर
1. (ग), 2. (घ) , 3. (ख) , 4. (क)

Class 7 Hindi Chapter 8 MP Board प्रश्न (ख)
दिए गए शब्दों में से उपयुक्त शब्द चुनकर रिक्त स्थानों की पूर्ति कीजिए
1. फिसड्डी रहे और जीते रहे। किसी का दिया पीते रहे। (दूध/पानी)
2. भला यात्रा का……….. भी है। (जरूरत/महूरत)
3. कभी तुम बड़े………….. ये ठीक है। (शेर/सियार)
4. जरा………… अपना बदलते चलो। (मार्ग/भाग)
उत्तर
1. पानी
2. महूरत
3. शेर
4. भाग।

MP Board Class 7th Hindi Sugam Bharti Chapter 8 अति लघु उत्तरीय प्रश्न

MP Board Class 7 Hindi Chapter 8 प्रश्न 2.
निम्नलिखित प्रश्नों के उत्तर एक वाक्य में लिखिए

(क)
इस कविता में फिसड्डी रहने का क्या अर्थ है?
उत्तर
इस कविता में फिसड्डी रहने का अर्थ है ‘हर कार्य में पीछे रहना।

(ख)
कवि के अनुसार उत्साह की आवश्यकता क्यों
उत्तर
कवि के अनुसार जीवन में किसी भी कार्य में उन्नति करने के लिए उत्साह की आवश्यकता होती

(ग)
दूसरों के सहारे जीने का भाव कविता की कौन-सी पंक्ति में आया है?
उत्तर
दूसरों के सहारे जीने का भाव कविता की ‘किसी का दिया दूध पीते रहे’ पंक्ति से लिया गया है।

(घ)
हमें मुहूर्त की प्रतीक्षा में रुककर क्यों नहीं बैठना चाहिए?
उत्तर
हमें मुहूर्त की प्रतिक्षा में रुककर इसलिए नहीं बैटना चाहिए क्योंकि समय कभी किसी के लिए रुकता नहीं है।

(ङ)
क्या अपने अतीत पर अभिमान करते रहना उचित है?
उत्तर
हमें मात्र अपने अतीत के गौरव का मान करते रहना नहीं चाहिए बल्कि जीवन में आगे बढ़ते रहना चाहिए।

MP Board Class 7th Hindi Sugam Bharti Chapter 8 लघु उत्तरीय प्रश्न

Class 7 Hindi Chapter 8 Question Answer Board प्रश्न 3.
निम्नलिखित प्रश्नों के उत्तर कम-से-कम तीन वाक्यों में लिखिए

(क)
व्यक्ति को जीवन में सफल होने के लिए क्या आवश्यक है?
उत्तर
व्यक्ति को जीवन में सफल होने के लिए द्रढ़ निश्चय और कढ़ी मेहनत की तो आश्यकता होती ही है, किंतु सफलता में निश्चयता लाने के लिए उत्साह की जरूरत होती है।

(ख)
कवि के अनुसार बुरी जिंदगी कौन-सी है?
उत्तर
जीवन के प्रत्येक कार्य में फिसड्डी बने रहना तथा हमेशा दूसरों के दया-धर्म पर जिंदा रहना, कवि ने ऐसी जिंदगी को सबसे बुरी बना है।

(ग)
शर्मिदगी से बचने के लिए हमें क्या करना चाहिए?
उत्तर
शर्मिंदगी से बचने के लिए हमें आत्म-निर्भर तथा मेहनती बनना चाहिए। हमें कदापि दूसरों के टुकड़ों पर नहीं पलना चाहिए। यदि हमें पूर्ण विकास करना है तो निरंतर आगे बढ़ते रहना चाहिए।

(घ)
हम ‘बड़े शेर’ कब थे और कैसे?
उत्तर
प्राचीन काल में भारत प्रत्येक क्षेत्र जैसे साहित्य, व्यापार, चिकित्सा तथा कला आदि में सर्वश्रेष्ठ था। इसीलिए हम कहते है कि हम कभी शेर थे। किंतु हमें यह सोचकर हमेशा खुश नहीं रहना चाहिए बल्कि अपने वर्तमान को सुधारने के लिए आगे बढ़ते रहना चाहिए।

(ङ)
उत्साह कविता का मूल-भाव क्या है? संक्षेप में लिखिए।
उत्तर
कवि कविता के माध्यम से हम सबको संदेश देते हैं कि हमें अपने आलस्य को त्याग कर तथा दूसरों पर निर्भरता छोड़ कर स्वयं की मेहनत और कर्म पर विश्वास करना चाहिए। कभी हम सोने की चिढ़िया कहलाते थे, यह सोचकर हमेशा खुश नहीं होते रहना ‘चाहिए बल्कि हमें वर्तमान की मेहनत और लान पर भरौसा करना चाहिए।

भाषा की बात

Class 7 Hindi Chapter 8 Question Answer प्रश्न 4.
निम्नलिखित शब्दों का शुद्ध उच्चारण कीजिए
शर्मिंदगी, जिंदगी, यात्रा, जमाना
उत्तर
छात्र स्वयं करें।

Class 7 Hindi Chapter 8 प्रश्न 5.
निम्नलिखित शब्दों में शुद्ध वर्तनी वाले शब्दों को गोला लगाइए
उत्तर
फिसड्डी, फसड्डी, फिसीड्डी खियाल, ख्याल, खयाल सर्मिन्दगी, शर्मिंदगी, शर्मंदगी

Path 8 Hindi Class 7 प्रश्न 6.
निम्नलिखित शब्दों के तत्सम रूप लिखिए
महूरत, दूध, जात्रा, भाग
उत्तर
शब्द – तत्सम रुप
महूरत – मुहूर्त
दुध – दुग्ध
जात्रा – यात्रा
भाग – भाग्य

Class 7 Hindi Lesson 8 Question Answer प्रश्न 7.
निम्नलिखित मुहावरों का अर्थ लिखकर वाक्यों में प्रयोग कीजिए
फिसड्डी रहना, किसी का दिया दूध पीना, मुहूर्त सोचना, बड़े शेर होना, ख्यालों में बना।
उत्तर

उत्साह कविता का परिचय

1. प्रस्तुत कविता में कवि ने देशवासियों को सर्वदा बढते रहने के लिए कहा है। जीवन में चाहे कितनी ही कठिनाइयाँ उत्पन्न हों, हमें रूकना नहीं चाहिए। हमें किसी की दया और सहारे पर नहीं बल्कि अपने आत्मविश्वास और मेहनत पर भरोसा करना चाहिए। समय हमारे साथ है, हमें आगे बढ़ने के लिए किसी का इंतजार नहीं करना चाहिए। यह भी सोचना व्यर्थ है कि सैकड़ों वर्ष पहले हम श्रेष्ठ थे, “अब क्या है?” यह महत्त्वपूर्ण है, हम आज की कमियों को पूरा करने के लिए बढ़ते चले।

उत्साह संदर्भ-प्रसंग सहित व्याख्या

1. चलते चलो और चलते चलो लहरों में,
लपटों में पलते चलो।

फिसड्डी रहे और जीते रहे,
किसी का दिया दूध पीते रहे,
तो इससे बुरी जिंदगी कौन-सी?
बड़ी और शर्मिंदगी कौन-सी?

ये आराम छोड़ों, उबलते चलो
लहरों में, लपटों में चलते चलो।

शब्दार्थ-लहरों में = जीवन का उतार चढ़ाव में। लपटों में = जीवन में आने वाले कष्टों की ज्वालाओं में। फिसड्डी = पीछे रहने वाले। किसी का दिया दूध पीते रहे = दी हुई सहायता का कर्ज से पलते रहे। शर्मिंदगी = उत्साह या जोशपूर्वक चलते चलो।

संदर्भ- प्रस्तुत काव्य पंक्तियाँ हमारी पाठ्य-पुस्तक ‘सुगम भारती’ (हिंदी सामान्य) भाग-7 के पाठ-8 ‘उत्साह’ से ली गई हैं। इसके रचयिता भवानीप्रसाद मिश्र हैं।

प्रसंग-इन पंक्तियों में कवि ने शर्मिंदगी की जिंदगी को छोड़ आगे बढ़ने का उत्साह दिया है।

व्याख्या-कवि कहता है कि हमें कठिन से कठिन परिस्थितियों में भी आगे बढ़ते रहना चाहिए। सैकड़ों सालों से हम रूढ़िवादी और फिसड्डी बने रहे हैं। हमने कभी भी आत्मनिर्भरता प्राप्त नहीं की। यह हमारे लिए बड़े शर्म की बात है। हमे सारे विश्व के साथ आगे बढ़ना चाहिए। अब हमें आलस्य को त्याग कर अपने अंदर एक उत्साह पैदा करना चाहिए। और उस उत्साह में हम बढ़ते चले जाएँ।

विशेष

• सरल भाषा का प्रयोग है।
• आम बोलचाल के शब्दों का उपयोग है।

2. भला यात्रा का महूरत भी है,
कहीं बैठे रहने की सूरत भी है,
निकलता चला जा रहा है समय,
महूरत न सोचो मचलते चलो।
लहरों में, लपटों में पलते चलो।

कभी तुम शेर थे, ठीक है,
उसी ख्याल में डूबना लीक है,
जमाना कहाँ से कहाँ जा चुका,
जरा भाग अपना बदलते चलो।
लहरों में, लपटों में पलते चलो।

शब्दार्थ- महूरत = शुभ-समय। बैठ रहने की सूरत = रुककर बैठे रह जाने की स्थिति या काम बंद कर देने की दशा। मचलते चलो = जिद करके अपनी धुन में बढ़े चलो। बड़े शेर = बहुत शक्तिशाली। ख्याल में डूबना = विचारों में खो जाना। लीक = परंपरा । भाग = भाग्य।

संदर्भ-पूर्ववत्।

प्रसंग-इसमें कवि ने कहा है कि इतिहास की सफलता पर ही खुश न होकर वर्तमान में भी आगे बढ़ना चाहिए।

व्याख्या-कवि के अनुसार समय कभी किसी के लिए रूकता नहीं है जो उसके साथ चलते है वहीं कामयाब रहते हैं। रूकने का कारण मत ढूंढों, बल्कि आगे बढ़ते चलो। किसी जमाने में हम सर्वश्रेष्ठ थे, किंतु उस बीती सफलता को सोच-सोच कर खुश नहीं होना चाहिए। जीवन वर्तमान संघर्ष पर विद्यमान रहता है। सारा संसार विकांस कर रहा है और हम दर्शक बने मात्र देख रहे हैं। अब हमें अपनी किस्मत बदलनी होगी और आगे बढ़ता चला जाऊँगा।

विशेष – वर्तमान सफलता पर जोर दिया गया है।

MP Board Class 7th Maths Solutions Chapter 3 Data Handling Ex 3.3

MP Board Class 7 Maths Solutions Chapter 3 Question 1.
Use the given bar graph to answer the following questions.
(a) Which is the most popular pet?
(b) How many students have dog as a pet?

Solution:
(a) Since, the bar representing number of students for cats is the tallest, so cat is the most popular pet.
(b) The number of students having dog as a pet is 8.

Class 7 Maths Chapter 3 Exercise 3.3 Solutions Question 2.
Read the given bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:

(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
Solution:
(i) In 1989, 175 books were sold. In 1990, 475 books were sold. In 1992, 225 books were sold.
(ii) From the graph, it can be concluded that in the year 1990 about 475 books were sold and in the year 1992 about 225 books were sold.
(iii) From the graph, it can be concluded that in the year 1989 and 1992, the number of books sold were less than 250.
(iv) From the graph, it can be concluded that the number of books sold in the year 1989 is about 1 and \(\frac{3}{4}\) th part of 1 cm.
We know that the scale is taken as 1 cm = 100 books.
∴ Number of books sold in 1989

Therefore, about 175 books were sold in the year 1989.

Class 7th Maths Chapter 3 Exercise 3.3 Question 2 Question 3.
Number of children in six different classes are given below. Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of class eighth.
Solution:

(a) We will choose a scale as 1 unit = 10 children because we can represent a more
clear difference between the number of students of class 7^th and that of class 9^th by this scale.
(b) (i) Since, the bar representing the number of children for class fifth is the tallest. So, there are maximum number of children in class fifth. Similarly, the bar representing the number of children for class tenth is the smallest. So, there are minimum number of children in class tenth.
(ii) The number of students in class sixth is 120 and the number of students in class eighth is 100. Therefore, the ratio of students of class sixth to the students 5.
of class eighth \(=\frac{120}{100}=\frac{6}{5}\) i.e., 6 : 5

Class 7th Maths Chapter 3 Exercise 3.3 Question 4 Question 4.
The performance of a student in 1^st Term and 2^nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:

(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Solution:
A double bar graph for the given data is as follows.

(i) There was a maximum increase in the marks obtained in Maths. Therefore, the child has improved his performance the most in Maths.
(ii) From the graph, it can be concluded that the improvement was the least in
S. Science.
(iii) From the graph, it can be observed that the performance in Hindi has gone down.

Class 7 Maths Exercise 3.3 Question 5 Question 5.
Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Solution:
(i) A double bar graph for the given data is as follows:

The double bar graph represents the number of people who like watching and participating in different sports.
(ii) From the bar graph, it can be observed that the bar representing the number of people who like watching and participating in cricket is the tallest among all the bars. Hence, cricket is the most popular sport.
(iii) The bars representing watching sport are longer than the bars representing participating in sport. Hence, watching different types of sports is more preferred than participating in the sports.

Class 7 Maths Chapter 3 Exercise 3.3 Solution Question 6.
Take the data giving the minimum and the maximum temperature of various cities given in the table. Plot a double bar graph using the data and answer the following:

(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
Solution:
A double bar graph for the given data is constructed as follows.

(i) From the graph, it can be concluded that Jammu has the largest difference in its minimum and maximum temperature on 20.6.2006.
(ii) From the graph, it can be concluded that Jammu is the hottest city and Bangalore is the coldest city.
(iii) Bangalore and Jaipur, Bangalore and Ahmedabad.
For Bangalore, the maximum temperature was 28°C, while minimum temperature of both cities, Ahmedabad and Jaipur, was 29°C.
(iv) From the graph, it can be concluded that the city which has least difference between its minimum and maximum temperature is Mumbai.

MP Board Class 7th Maths Solutions

MP Board Class 9th Special English Composition: Visual Stimulus

1. Harish wants to make a speech at the assembly titled Patriotism and the Young on Independence Day. He notices the chart given below in the paper and decides to use it as the basis for his speech. Using the chart and your own ideas write out the speech in about 150-200 words. [10 marks]

Answer:
In the present day it has become imperative to make the young people of India realize the importance of being patriotic for acquiring political strength which was. always keenly felt by the leaders of India’s national movement.

While the call for throwing out the British struck a responsive chord in every Indian heart, the significance of the campaign in favour of patriotism has seldom seen fully appreciated. We should take pride in being Indians and value the unity in diversity that our culture possesses. When a foreign leader praises our country or an Indian wins an international award, are the times when our pride for our country should soar high. A wave of nationalist fervor should sweep our countrymen on hearing the National Anthem, while seeing the National Flag being hoisted and while seeing the Republic Day parade. The plant of patriotism which sprouted when India gained her independence, was the culmination of more than a century of sacrifices offered by numerous soldiers at the altar of national freedom. At p.cz-mt, when we hear or read about the sacrifice of these soldiers, we should try to keep alive in our hearts the sacred memory of those great martyrs who made tremendous sacrifices for the attainment of freedom. From their memories should we derive our inspiration of patriotism?

Just as it is the moral duty of everyone to maintain and preserve good health, it is the bounden duty of our countrymen to build up their patriotism so that they can exist in this violence-ridden world with self-respect and proudly proclaim, ‘We are Indians’.

2. ABC Bank, a private bank is celebrating its 10th anniversary in Jaipur. A director of the company, is requested to give a short speech on the bank’s service to the town. Study the information given, and together with your own ideas, write a speech stressing the importance of this public service and make some practical suggestions too. Write your speech in about 150-175 words. [10 marks]

Answer:
As a director of ABC Bank (Ltd.), I feel extremely privileged in reading out the achievements and services rendered by our bank during the last ten years in the historical and beautiful city of Jaipur.

We came to Jaipur with a dream ten years ago to build up a nationalised bank to cater to the demands of the people. “Customer satisfaction” has been our motto ever since. Today our bank is regarded as one of the most prestigious banks of the city having 15 local branches all over Jaipur. Our bank has felicitated the growth and development of many small-scale industries by providing almost 60% of the total loan to them. We have also offered modest loans of 7% and 9% to housing and medical aid respectively. In our move towards advancement we have not ignored even our rural sector and have provided farmers with 24% of the loans I feel our greatest achievement during these ten years has been our attitude of ‘service before self which is imbibed most naturally by each and every person associated with our bank. Our attitude has indeed been rather successful as is evident from the increasing number of people who have their accounts in the various branches of our bank. To further enhance and strengthen our services, it is our proposal to fully computerise all the branches of our bank in the near future and to further propagate the idea that for us each of our customers is special in his own right.

With this 1 would like to conclude my speech by congratulating all the employees who have contributed greatly towards the development of our bank.

3. A survey was conducted to find out how teenagers spend then free time. The following trends were observed. Using the data given in the pie-chart given below together with your own ideas write an article for your school magazine on the topic “How teenagers spend their free time?” Write your answers in about 150 words giving your own reasons and conclusions for the trends.

Answer:

“How Teenagers Spend Their Free Time?”
By : XYZ ‘

With the changing times the priorities of all teenagers have also changed The revolutionisation of modern times has made the teenagers of today’s aware of more and more options to spend their free time.

Recently a survey was conducted by the ‘Weekly “Magazine to determine ‘ “how teenagers spend their free time?” It clearly indicated that about 50% of the girls and 40% of the boys spend most of their time watching television programs. These programs include movies, songs and serials or the sports channels on television. Though there are also about 30% boys who spend their leisure time in playing indoor and outdoor games, only 5% girls spend their time in playing these games. Reading comics and General Knowledge Books still holds good with almost 25% girls but barely 10% boys spend their time on reading books. Talking to friends comes next on the teenagers’ priority list. 15% girls and about 10% boys spend time in chatting with their peers. With so much to occupy their friends in their free time, pursuing a hobby has taken a complete back seat with the teenagers. Hardly 5% girls and 10% boys collect materials for their hobbies. Thus, we note a steady downward trend in the percentage of teenagers who now pursue a constructive recreational activity as they are more prone to indulge in options that are easily accessible to them.

4. Gauri sees the-rowing scene as he looks out of his bedroom window. He is alarmed at the way his city has changed in the past five years. He writes an article for his city newspaper expressing his alarm and painting the present position cautioning people against environmental pollution. Write the article in not more than 150-200 words. [10 marks]

Answer:
In this vast universe, the earth is so far the only planet to be endowed with an environment that can support life-form with which we are familiar. What sustains life on earth is a thin cover of air and water encasing the earth and known as the biosphere. Without that our planet would have been just another lifeless, desolate form spinning around in the musky depths of space.

But nature has been suffering due to the thoughtlessness of mankind. The air we breathe, the water we drink, the two main ingredients that sustain life, have been polluted. The air that we inhale is polluted due to the smoke emitted by vehicles and chimneys of industries, indiscriminate felling of trees and over-population. Water becomes contaminated due to a variety of pollutants. Water-borne diseases have always posed a threat to mankind. The use of loudspeakers on various functions and occasions causes noise-pollution. The atmospheric garbage in a permanent or chronic state is fouling our environment at a fantastic pace. It is a serious concern today for environmental scientists and enlightened statesmen all over the world regarding man’s misuse and abuse of nature and his deliberate fouling of the atmosphere under the guise of modernisation, automation and scientific or technological progress. If we. do not undertake ways and means to check environmental pollution, it will gradually lead to our doom, for one day may come when nature might be in no mood to stand any more abuses; and excesses at the hands of humanity.

5. You Anand/Anandi. Write a speech to be delivered at the morning assembly in your school about the increase in violence and social unrest in your area, interpreting the data given below. Also give suggestions to curb them. Write your answer in about 150-200 words. [10 marks]

Answer:
It is indeed unfortunate that there is a visible increase in the cases of violence and social unrest in our area in the past few years. Some anti-social elements are working very hard and have as good as succeeded i in disturbing the peace and harmony of our area.

No longer do the people of our area feel a sense of security while walking on the roads and there is constant fear in the minds of the people j that some unnecessary violence may crop up anywhere. A data study shows a constant and steady rise in the number of cases of violence during the last six years. In the year 1992-93, the number of cases of violence was ten. and these cases have gone up by almost four times during the last 5-6 years. If effective and concrete steps are not taken up by the police and local authorities to curb the cases of violence, our area will surely end up as being declared as one of the most troubled areas of the city l where it will become almost impossible for the common man to even exist peacefully.

The anti-social elements, who are bent upon disturbing the peace of our area, should be dealt with very strictly so that just thinking about the severity of the penalty for causing social unrest should make them feel scared constantly. Each and every one of us should also make an effort to voice our protest against this increase in social unrest more loudly and emphatically and ensure, that our complaints are seriously dealt with by the concerned authorities so. that our lives can be more secure and peaceful.

So, the need of the time is to collectively stand up against the increase in these violent activities so that we can make our area a better place to live in without having to fear constantly about the next hour.

6. You are Varun/Aruna. Write a speech to be delivered at the morning assembly of your school about the increase in road accidents in your city, using the following data. Also give suggestions to curb them. Write your answer in about 150-200 words. [10 Marks]

Answer:
We at our school have decided to observe the Road Safety Week by making students aware about the hazards and dangers they face while walking on the road. It is extremely important to keep in mind and follow the traffic rules while one is on the road considering that there has been a constant and steady rise in the number of road accidents in our city.

It is indeed unfortunate that due to the carelessness of pedestrians- and drivers alike, there has been an increase in the. road accidents in our city. The number of road accidents has gone up by more than 100% in the past four years. A data study shows that while there were 50 cases of road accidents in x the year 1995, the number went up to 110 in the year 1996 and after a gradual yet steady rise, in the year 1999 the number of road accidents has gone up to as many as 205. Let us not forget that life is precious and a valuable and precious thing as life should not be wasted by losing it due to sheer carelessness. We must always remember that traffic rules are made keeping in mind the safety of the people and so it becomes our duty to observe these rules. The result of rash and negligent driving is always dangerous so one should always practice patience and restraint while driving. Likewise while talking on the road, pedestrians should be alert and vigilant so as to avoid any untoward incident. Only if each and every one of us makes a combined and collective ;
effort can we possibly lessen road accidents and make our city a safe place to walk in.

7. Based on the cartoon given below and your own ideas, write a letter to the editor on-the need to educate people about s keeping the environment clean. Write your letter in not more than 750 words. [10 marks]

Answer:
75 M.G. Road
New Delhi
The Editor
The Daily News, New Delhi
3 March, 20xx

Dear sir,
It is indeed unfortunate that people have adopted a most callous attitude about keeping the environment clean.

The air, that we breathe in, the water that we drink and our whole surroundings have become polluted. It is indeed high time that we stopped nature’s suffering due to the atrocities at the hands of human beings. If each and every one of us, in our own little way, help in keeping the environment clean, we will be leading more cleaner and healthier lives. This awareness about cleansing the environment will also enable us to hand over to our future generations an environment where living will be easy and hygienic.

Unless we do not realise the need to keep the environment clean soon, it may be too late to repair the damage done.

Yours sincerely
XYZ

8. You come across the following advertisement for graduate students. Write a letter to the Editor of a local newspaper expressing your views on the lack of job opportunities and counseling to the youth after class X.
Write the letter in your answer sheet in not more than 150 words. [10 marks]

Answer:
The Editor
Jaago Bharat
New Delhi 10th March, 20xx

Sir,
I had the opportunity of going through the advertisement on page 3 of your 5th March 20xx issue regarding expert counseling to students who want to study in USA. I thought I must write to you in the hope that through the medium of your esteemed newspaper 1 am able to bring to light a fact that is troubling me most.

It is most astonishing that we Indians are more concerned to impart counseling for foreign universities for graduates whereas our own Xth standard students lack not only the ]ob opportunities but also the avail¬ability of a counseling centre for them after they pass out their Matric Examination. Our priority at present should be to try and help the students of Xth standard to choose the proper career and field in the most formative years of their life. Times have greatly changed and life has become much more complex. There is an infinite variety of jobs and vocations and hence a Xth standard student is confused as to what his next step should be. So, it is the duty of our society to assess the native intelligence of the child, develop and groom it and decide where it should be placed.

To successfully harness this concept may take time. So, let us seriously divert our attention on to the accomplishment of this target.

Yours sincerely,
XYZ

9. Sunil Surana is an active member of the Youth Wing of the SPCA (Society for the Prevention of Cruelty to Animals). He saw the following scene on one of the roads. He decided to write a letter to the editor of a newspaper, protesting against cruelty to animals. Using the information in the picture, to-gether with your own ideas, write the letter in your answer sheet in not more than 175-200 words. [10 marks]

Answer:
13, Bairn Road
Puri (Orissa)
The Editor,
The Bugle, Bhubaneshwar
2nd March, 20xx ;

Dear Sir,
I am an, active member of ike Youth Wing of the SPCA and am writing to you to protest against the cruelty to animals after I witnessed the most overburdened bullock-cart on the road, the terrible weight of the burden almost making it impossible for the poor bull to even move.

This is an example of just one of the innumerable cruelties towards animals ! by man. !

Man has become so self-centered that to suit his own selfish means he does not mind inflicting any kind of unspeakable cruelty on animals as long as his own purpose is duly served. This cruelty may include over burdening animals rather mercilessly to carry loads from one place to another, using the cane to make the animals perform to amuse the children, caging them, selling them or even slaughtering them. It is hard to believe that man, who himself is a member of the living species, can treat another member of the same living species (animals) so terribly so as to even deny them the very right to live and survive. It is high time that we realized that it is highly improper and wrong on,our part that we are being most cruel and inhumane in our treatment to animals and find effective means and ways of remedying our folly. Even if each one of us decides to take one step in a positive direction it will be a most healthy sign.

I sincerely hope that through your esteemed newspaper 1 am able to put forth to the readers the need for a change in our attitude of taking animals for granted to serve us in each and every way that we think is possible.

Yours faithfully
Sunil Surana

10. Harish has to make a speech at the assembly on The Qualities Required for Success. He notices the chart given below in the paper and decides to use it as the basis for his speech. Using the chart and your own ideas write out the speech in about 150-200 words. [10 marks]

Answer:
The present-day generation is very conscious and highly aware about the most important things or qualities in life and the qualities required to attain success. Knowledge, by far, is regarded as the most important thing in life by the young people. Knowledge leads to excellence of the mind; it facilitates the creation of a critical, creative outlook, seeking the happiness of all and perfection where attainable. Love is regarded as another important virtue which enables one to win over people and value human relationships. The money factor is also important because it enables us to live a life of comfort and be free from daily worries such as how to make both ends meet. Thirst far fame and hunger far power are also considered as essential ingredients in modern life without which it becomes almost difficult to fulfill our aims and ambitions.

If one is fortunate enough to follow and abide by those virtues which are considered important in our present life, one must not ignore the most essential factors that ensure success. It is an old saying that ‘hard work never goes waste’. So, one should never give up on hard work. True talent and sincerity to do work also help us to be successful. Relying solely on one’s luck is regarded as a sign of foolish behavior because luck also favors only those, who think and work positively. Good contacts also help us to be successful but gone are the days when one needed only contacts to be successful. The most encouraging factor in the present day is that if .one has the talent and the inclination to do hard work, it is certain enough that slowly but surely success will definitely come one’s way and hindrances such as lack of money or contacts can very easily be overcome. The right approach to be successful is ‘To do as much as you can and then pray to God to fill in the blank.

MP Board Class 9th English Solutions

MP Board Class 11th Maths Solutions Chapter 6 सम्मिश्र संख्याएँ और द्विघातीय समीकरण Ex 6.2

निम्नलिखित असमिकाओं को आलेखन विधि से द्विविमीय तल में निरूपित कीजिए। (प्रश्न 1 से 10 तक)
प्रश्न 1.
x + y < 5.
हल:
समीकरण x + y = 5 को लीजिए। यह एक सरल रेखा है जो बिन्दु (5, 0), (0, 5) से होकर गुजरती है।
x = 0, y = 0 असमिका x + y < 5 में रखने पर,
अर्थात 0 + 0 < 5 या 0 < 5
⇒ मूल बिन्दु x + y < 5 के क्षेत्र में है।
छायाकिंत क्षेत्र x + y < 5 को निरूपित करता है जो इसका हल है।

प्रश्न 2.
2x +y ≥ 6.
हल:
2x + y ≥ 6
समीकरण 2x + y = 6 को लीजिए, यह रेखा (3, 0) और (0, 6) से गुजरती है।
x = 0, y = 0 को 2x + y ≥ 6 में रखें तो 0 ≥ 6, जो सत्य नहीं है।
∴ मूल बिन्दु 2x + 2 6 के क्षेत्र में नहीं हैं।
2x + y ≥ 6 का क्षेत्र छायांकित किया गया है।

प्रश्न 3.
3x + 4y ≤ 12.
हल:
दी गई असमिका 3x + 4y ≤ 12 सरल रेखा 3x + 4y = 12 बिन्दु (4, 0), (0, 3) से होकर जाती है।
असमिका 3x + 4y ≤ 12 में (0, 0) रखने पर,
0 + 0 ≤ 12 अर्थात 0 ≤ 12 जो सत्य है
∴ मूल बिन्दु 3x + 4y ≤ 12 के क्षेत्र में आता है।
इसका आलेख साथ वाली आकृति में दिखा गया है।

प्रश्न 4.
y+8 ≥ 2x.
हल:
दी हुई रैखिक असमिका y + 8 ≥ 2x सरल रेखा 2x – y = 8 बिन्दु (4,0). और (0, – 8) से होकर जाती है।
असमिका
y + 8 ≥ 2x,
x = 0, y = 0 रखने पर
0 + 8 ≥ 0 अर्थात 8 ≥ 0 जो सत्य है।
∴ मूल बिन्दु y + 8 ≥ 2x के क्षेत्र में आता है। इसका आलेख साथ दी हुई आकृति में बनाया गया है।

प्रश्न 5.
x – y ≤ 2.
हल:
दी हुई असमिका x – y ≤ 2
सरल रेखा x – y = 2 बिन्दु (2, 0), (0, – 2) से होकर जाती है।
x = 0, y = 0 असमिका x – y ≤ 2 में रखने पर 0 ≤ 2 जो सत्य है।
∴ मूल बिन्दु x – y ≤ 2 के क्षेत्र में है।
असमिका x – y ≤ 2 का आलेख साथ वाली आकृति में बनाया गया है।

प्रश्न 6.
2x – 3y > 6.
हल:
दी हुई रैखिक असमिका 2x – 3y > 6
सरल रेखा 2x – 3y = 6, (3, 0) और (0, – 2) से होकर जाती है।
असमिका 2x – 3y > 6 में x = 0, y = 0 रखने पर 0 > 6 जो सत्य नहीं है।
∴ मूल बिन्दु (0, 0) दी हुई असमिका में नहीं आता है।
∴ इसका आलेख दी हुई आकृति में दर्शाया गया है।

प्रश्न 7.
– 3x + 2y ≥ – 6.
हल:
दी हुई रैखिक असमिका – 3x + 2y ≥ – 6 या 3x – 2y ≤ 6
सरल रेखा – 3x + 2y = – 6 बिन्दु (2, 0) और (0, – 3) से होकर जाती है।
– 3x + 2y ≥ – 6 में x = 0, y = 0 रखने पर 0 ≥ – 6, जो सत्य है।
∴ मूल बिन्दु (0, 0), 3x + 2y ≥ – 6 असमिका के क्षेत्र में है।
∴ इसका आलेख दी हुई आकृति में दर्शाया गया है।

प्रश्न 8.
3y – 5x < 30.
हल:
दी हुई असमिका 3y – 5x < 30
सरल रेखा 3y – 5x = 30, बिन्दु (-6, 0) और (0, 10) से होकर जाती है।
असमिका 3y – 5x < 30 में x = 0, y = 0 रखने पर
∴ 0 < 30 सत्य है।
मूल बिन्दु (0, 0), 3y – 5x < 30 के क्षेत्र में है। इसका आलेख दी गई आकृति में दर्शाया गया है।

प्रश्न 9.
y < – 2
हल:
दी हुई रैखिक असमिका y < – 2 सरल रेखा y = – 2 बिन्दु (2, – 2) और (- 2, – 2) से होकर जाती है।
y <- 2 में y = 0 रखने पर 0 < – 2, यह सत्य नहीं है।
∴ मूल बिन्दु (0, 0), y < – 2 में नहीं।
दी हुई आकृति में छायांकित क्षेत्र से दर्शाया गया है।

प्रश्न 10.
x > – 3
हल:
दी हुई रैखिक असमिका x > – 3
सरल रेखा x = – 3 बिन्दु (- 3, 2), (- 3, – 2) से होकर जाती है।
x > – 3 में x = 0 रखने पर,
0 > – 3, यह सत्य है।
∴ मूल बिन्दु (0, 0), x > – 3 में है। दी हुई आकृति में x > – 3 छायांकित क्षेत्र से दर्शाया गया है।

MP Board Class 11th Maths Solutions