MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 1
Solution:
Median:
Let us prepare a cumulative frequency table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 2
Now, we have N = 68 ⇒ \(\frac{N}{2}=\frac{68}{2}\) = 34
The cumulative frequency just greater than 34 is 42 and it corresponds to the class 125 – 145.
∴ 125 – 145 is the median class.
∴ l = 125, cf = 22, f= 20 and h = 20
Using the formula,
Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 125 + \(\left[\frac{34-22}{20}\right]\) × 20
= 125 + \(\frac{12}{20}\) × 20 = 125 + 12 = 137 units.
Mean: Let assumed mean, a = 135
∵ Class size, h = 20
∴ ui = \(\frac{x_{i}-a}{h}=\frac{x_{i}-135}{20}\)
Now, we have the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 3
∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σfiui] = 135 + 20 × \(\frac{7}{68}\)
= 135 + 2.05 = 137.05 units.
Mode:
∵ Class 125 – 145 has the highest frequency i.e., 20.
∴ 125 – 145 is the modal class.
We have: h = 20, l = 125 , f1 = 20, f0 = 13, f2 = 14
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 4
We observe that the three measures are approximately equal.

Use this median calculator to quickly find the median of a set of numbers. Just enter your numbers in the box and click on the button that says calculate.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 5
Solution:
Here, we have N = 60
Now, cumulative frequency table is:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 6
Since, median = 28.5 (Given)
∴ Median class is 20 – 30 and l = 20, f = 20, cf = 5 + x, N = 60
∴ l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20}\right]\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ 57 = 40 + 25 – x
⇒ x = 40 + 25 – 57 = 8
Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 45 – 8 = 7.
Thus x = 8, y = 7

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

The percentage difference calculator is here to help you compare two numbers.

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 7
Solution:
The given table is cumulative frequency distribution. We write the frequency distribution as given below :
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 8
∵ The cumulative frequency just greater than 50 is 78.
∴ The median class is 35 – 40.
Now, \(\frac{N}{2}\) = 50, l = 35, cf = 45, f = 33 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 9
Thus, the median age = 35.76 years.

Also you can check the age difference between your loved ones, friends using the Age difference Calculator.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 10
Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5,
126.5 – 135.5 ………… 171.5 – 180.5.]
Solution:
After changing the given table as continuous classes we prepare the cumulative frequency table as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 11
The cumulative frequency just above 20 is 29 and it corresponds to the class 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
We have: \(\frac{N}{2}\) = 20, l = 144.5, f= 12, cf = 17 and h = 9
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 144.5 + \(\left[\frac{20-17}{12}\right]\) × 9
= 144.5 + \(\frac{3}{12}\) × 9 = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25 = 146.75
Median length of leaves = 146.75 mm.

Question 5.
The following table gives the distribution of the life time of 400 neon lamps:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 12
Find the median life time of a lamp.
Solution:
To compute the median, let us write the cumulative frequency distribution as given below:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 13
Since, the cumulative frequency just greater than 200 is 216.
∴ The median class is 3000-3500 and so l = 3000, cf= 130, f = 86, h = 500
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\frac{70}{86}\) × 500 = 3000 + \(\frac{35000}{86}\)
= 3000 + 406.98 = 3406.98
Thus, median life time of a lamp = 3406.98 hours.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 14
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution:
Median: The cumulative frequency distribution table is as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 15
Since, the cumulative frequency just greater than 50 is 76.
∴ The class 7-10 is the median class.
We have, \(\frac{N}{2}\) = 50 , f = 7, cf = 36, f = 40 and h = 3
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 16
Mode:
Since the class 7 – 10 has the maximum frequency i.e., 40.
∴ The modal class is 7 – 10.
So, we have l = 7,h = 3, f1 = 40, f0 = 30, f2 = 16
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 17
Thus, the required median = 8.05, mean = 8.32 and mode = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 18
Solution:
We have cumulative frequency table as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 19
The cumulative frequency just greater than 15 is 19, which corresponds to the class 55 – 60.
So, median class is 55-60 and we have \(\frac{N}{2}\) = 15,
l = 55, f = 6, cf = 13 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 20
Thus, the required median weight of the students = 56.67 kg.

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

In this article, we will share MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Question 1.
Using appropriate properties, find.
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 1
Solution:
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 2
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 3

Question 2.
Write the additive inverse of each of the following
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 4
Solution:
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 5
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 6

(iv) We have given \(\frac{2}{-9}\)
Multiplying numerator and denominator by -1, we get \(\frac{2}{-9}\)
The additive inverse of \(\frac{2}{-9}\) is \(\frac{2}{9}\)
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 7

(v) We have \(\frac{19}{-6}\)
Multiplying numerator and denominator by -1, we get \(\frac{-19}{6}\)
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 8

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Question 3.
Verify that -(-x) = x for
(i) x = \(\frac{11}{15}\)
(ii) x = \(-\frac{13}{17}\)
Solution:
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 9

Question 4.
Find the multiplicative inverse of the following.
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 10
Solution:
(i) We have given, -13
The multiplicative inverse of -13 is \(\left(\frac{-1}{13}\right)\)
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 11

(ii) We have given, \(\frac{-13}{19}\)
The multiplicative inverse of is \(\frac{-13}{19}\) is \(\frac{-19}{13}\)
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 12

(iii) We have given, \(\frac{1}{5}\)
The multiplicative inverse of \(\frac{1}{5}\) is 5.
∵ \(\frac{1}{5} \times 5=1\)

(iv) We have given, \(\frac{-5}{8} \times \frac{-3}{7}\)
The multiplicative inverse of
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 13

(v) We have given, \(-1 \times \frac{-2}{5}\)
The multiplicative inverse of
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 14

(vi) We have given, -1.
The multiplicative inverse of -1 is -1.
∵ (-1) × (-1) = 1.

Question 5.
Name the property under multiplication used in each of the following.
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 15
Solution:
(i) We have given, \(\frac{-4}{5} \times 1=1 \times \frac{-4}{5}=\frac{-4}{5}\)
i.e., 1 is the multiplicative identity. Thus, it is a identity property under multiplication
(ii) We have given, \(\frac{-13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\), which shows the commutativity.
Thus, it is a commutative property under multiplication.
(iii) We have given \(\frac{-19}{29} \times \frac{29}{-19}=1\), which shows that \(\frac{29}{-19}\) is a multiplicative inverse of \(\left(\frac{-19}{29}\right)\)
Thus, it is a inverse property under multiplication.

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Question 6.
Multiply \(\frac{6}{13}\) by the reciprocal of \(\frac{-7}{16}\).
Solution:
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 16

Question 7.
Tell what property allows you to compute \(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right) \text { as }\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}\)
Solution:
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 17
Above property is associativity.
[ ∵ a × (b × c) = (a × b) × c]

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Question 8.
Is \(\frac{8}{9}\) the multiplicative inverse of \(-1 \frac{1}{8}\) ? Why or why not?
Solution:
We have given a fraction \(\frac{8}{9}\) and \(-1 \frac{1}{8}=\frac{-9}{8}\)
No, \(\frac{-9}{8}\) is not a multiplicative inverse of \(\frac{8}{9}\) because \(\frac{8}{9} \times\left(\frac{-9}{8}\right)=-1 \neq 1\)

Question 9.
Is 0.3 the multiplicative inverse of \(3 \frac{1}{3}\) ? Why or why not?
Solution:
MP Board Class 8th Maths Chapter 1 Rational Numbers Ex 1.1 18

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Factoring simplifying rational expressions calculator.

Question 10.
Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution:
(i) 0 is the rational number, which does not have a reciprocal.
(ii) 1 and (-1) are the rational numbers, that are equal to their reciprocals.
(iii) 0 is the rational number that is equal to its negative.

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.1

Question 11.
Fill in the blanks.
(i) Zero has …… reciprocal.
(ii) The numbers …… and ……. are their own reciprocals.
(iii) The reciprocals of -5 is ……
(iv) Reciprocal of \(\frac{1}{x}\), where x ≠ 0 is ……
(v) The product of two rational numbers is always a ……
(vi) The reciprocal of positive rational number is …….
Solution:
(i) Zero has no reciprocal.
(ii) The numbers 1 and -1 are their own reciprocals.
(iii) The reciprocal of -5 is \(\frac{-1}{5}\) .
(iv) Reciprocal of \(\frac{1}{x}\), where x ≠ 0 is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of positive rational number is positive.

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4

Question 1.
Find the square root of each of the following numbers by Division method.
(i) 2304
(ii) 4489
(iii) 3481
(iv) 529
(v) 3249
(vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900.
Solution:
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 1
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 2
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 3
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 4
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 5
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 6
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 7

Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625
Solution:
(i) Since number of digits in 64 is 2 (= n), which is even.
Then its square root will have \(\frac{n}{2}\) digits.
Number of digits = \(\frac{2}{2}\) = 1.

(ii) Since the number of digits in 144 is 3 (= n), which is odd.
Then its square root will have \(\left(\frac{n+1}{2}\right)\) digits.
Number of digits \(=\frac{3+1}{2}=\frac{4}{2}=2\).

(iii) Since number of digits in 4489 is 4(= n), which is even.
Then its square root will have \(\frac{n}{2}\) digits.
Number of digits = \(\frac{4}{2}\) = 2

(iv) Since the number of digits in 27225 is 5( = n), which is odd.
Then its square root will have digits \(\left(\frac{n+1}{2}\right)\)
Number of digits \(=\frac{5+1}{2}=\frac{6}{2}=3\)

(v) Since number of digits in 390625 is 6 (= n), which is even.
Then its square root will have \(\left(\frac{n}{2}\right)\) digits.
Number of digits = \(\frac{6}{2}\) = 3

Online calculator that adds and subtracts roots calculator, describes all solution steps.

MP Board Solutions

Question 3.
Find the square root of the following decimal numbers.
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 8
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 9
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 10

MP Board Solutions

Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000.
Solution:
(i) 402
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 11
Thus, if we subtract 2 from 402, we get a perfect square number whose square root is 20.

(ii) 1989
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 12
Thus, if we subtract 53 from 1989, we get a perfect square number whose square root is 44.

(iii) 3250
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 13
Thus, if we subtract 1 from 3250, we get a perfect square number whose square root is 57.

(iv) 825
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 14
Thus, if we subtract 41 from 825, we get a perfect square number whose square root is 28.

(v) 4000
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 15
Thus, if we subtract 31 from 4000, we get a perfect square number whose square root is 63.

MP Board Solutions

Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution:
(i) 525
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 16
Hence, the number to be added is 529 – 525 = 4 and square root of 529 is 23.

(ii) 1750
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 17
Clearly, 412 = 1681 < 1750
422 = 1764 > 1750.
Hence, the number to be added is 1764 – 1750 = 14 and square root of 1764 is 42.

(iii) 252
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 18
Clearly, 152 = 225 < 252
162 = 256 > 252
∴ The number to be added is 256 – 252 = 4 and square root of 256 is 16.

(iv) 1825
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 19
Clearly, 422 = 1764 < 1825
432 = 1849 >1825
∴ The number should be added is 1849 – 1825 = 24 and square root of 1849 is 43.

(v) 6412
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 20
Clearly, 802 = 6400 < 6412
812 = 6561 > 6412
∴ The number should be added is 6561 – 6412 = 149 and square root of 6561 is 81.

MP Board Solutions

Question 6.
Find the length of the side of a square whose area is 441 m2.
Solution:
Let the length of the side of a square is x m. Area of the square = x2 ⇒ 441 = x2
⇒ x = \(\sqrt{441}\)
⇒ x = 21
Thus, the required length of side of the square is 21 m.

Question 7.
In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC.
(b) If AC = 13 cm, BC = 5 cm, find AB.
Solution:
(a) AB = 6 cm, BC = 8 cm
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 21
By using Pythagoras theorem,
AC2 = AB2 + BC2
⇒ AC2 = (6)2 + (8)2
⇒ AC2 = 36 + 64
⇒ AC2 = 100
⇒ AC = \(\sqrt{100}\)
⇒ AC = 10 cm.

(b) AC = 13 cm, BC = 5 cm
By using Pythagoras theorem,
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 22
AC2 = AB2 + BC2
⇒ (13)2 = AB2 + (5)2
⇒ 169 = AB2 + 25
⇒ 169 – 25 = AB2
⇒ 144 = AB2
⇒ AB = \(\sqrt{144}\)
⇒ AB = 12 cm.

MP Board Solutions

Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution:
Total number of plants = 1000.
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 23
Since, the plants are planted in a garden in such a way that the number of rows and the number of columns remain same.
Clearly, 312 = 961 < 1000
322 = 1024 >1000
∴ 1024 – 1000 = 24.
Thus, gardener needs 24 more plants.

MP Board Solutions

Question 9.
There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
Solution:
Total number of children = 500.
Since the number of rows is equal to the number of colums, in which children have to stand.
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 24
Clearly, 16 children would be left out in this arrangement.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3

Question 1.
What could be the possible ‘ones’ digits of the square root of each of the following numbers?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i) We know that the ‘ones’ place of the square of 1 and 9 is 1.
∴ The possible ‘ones’ digits of the square root of 9801 are 1 and 9.
(ii) We know that the ‘ones’ place of the square of 4 and 6 is 6.
∴ The possible ‘ones’ digits of the square root of 99856 are 4 and 6.
(iii) We know that the ‘ones’ place of the square of 1 and 9 is 1.
∴ The possible ‘ones’ digits of the square root of 998001 are 1 and 9.
(iv) We know that ‘ones’ place of the square of 5 is 5.
∴ The possible ‘ones’ digit of the square root of 657666025 is 5.

MP Board Solutions

Solve a quadratic equation by completing the square calculator mathpapa.

Question 2.
Without doing any calculation, find the numbers which are surely not perfect squares,
(i) 153
(ii) 257
(iii) 408
(iv) 441
Solution:
We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares. So, (i), (ii) and (iii) are surly not perfect squares.
(iv) Since, the number 441 ends with 1. Thus, 441 may or may not be a perfect square.

MP Board Solutions

Question 3.
Find the square root of 100 and 169 by the method of repeated subtraction.
Solution:
First consider 100.
(1) 100 – 1 = 99
(2) 99 – 3 = 96
(3) 96 – 5 = 91
(4) 91 – 7 = 84
(5) 84-9 = 75
(6) 75 – 11 = 64
(7) 64 – 13 = 51
(8) 51 – 15 = 36
(9) 36 – 17 = 19
(10) 19 – 19 = 0.
∴ \(\sqrt{100}\) = 10.
Now, consider 169
(1) 169 – 1 = 168
(2) 168 – 3 = 165
(3) 165 – 5 = 160
(4) 160 – 7 = 153
(5) 153 – 9 = 144
(6) 144 – 11 = 133
(7) 133 – 13 = 120
(8) 120 – 15 = 105
(9) 105-17 = 88
(10) 88 – 19 = 69
(11) 69 – 21 = 48
(12) 48 – 23 = 25
(13) 25 – 25 = 0.
∴ \(\sqrt{169}\) = 13.

MP Board Solutions

Question 4.
Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Solution:
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 1
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 2
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 3
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 4
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 5
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 6
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 7
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 8

Question 5.
For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768
Solution:
(i) We have, 252 = 2 × 2 × 3 × 3 × 7
The smallest whole number is 7 by which 252 should be multiplied so as to get a perfect square.
252 × 7 = 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 252 × 7 = 1764 is a perfect square.
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 9

(ii) We have, 180 = 2 × 2 × 3 × 3 × 5
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 10
The smallest whole number is 5, by which 180 should be multiplied so as to get a perfect square.
180 × 5 = 2 × 2 × 3 × 3 × 5 × 5
So, \(\sqrt{900}\) = 2 × 3 × 5 = 30.

(iii)
We have,
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 11
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
The smallest whole number is 7, by which 1008 should be multiplied so as to get a perfect square.
1008 × 7 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in pair. Therefore,
1008 × 7 = 7056 is a perfect square.
So, \(\sqrt{7056}\) = 2 × 2 × 3 × 7 = 84.

(iv) We have, 2028 = 2 × 2 × 3 × 13 × 13
The smallest whole number is 3 by which 2028 should be multiplied so as to get a perfect square.
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 12
2028 × 3 = 2 × 2 × 3 × 3 × 13 × 13
Now each prime factor is in pair. Therefore, 2028 × 3 = 6084 is a perfect square.
So, \(\sqrt{6084}\) = 2 × 3 × 13 = 78.

(v) We have, 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
The smallest whole number is 2, by which 1458 should be multiplied so as to get a perfect square.
1458 × 2 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 13
Now each prime factor is in pair.
Therefore, 1458 × 2 = 2916 is a perfect square.
So, \(\sqrt{2916}\) = 2 × 3 × 3 × 3 = 54

(vi) We have, 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 14
The smallest whole number is 3, by which 768 should be multiplied so as to get a perfect square.
768 × 3 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Now each prime factor is in pair.
Therefore, 768 × 3 = 2304 is a perfect square.
So, \(\sqrt{2304}\) = 2 × 2 × 2 × 2 × 3 = 48.

MP Board Solutions

Question 6.
For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained,
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
Solution:
(i) We have 252 = 2 × 2 × 3 × 3 × 7
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 15
We find that 252 should be divided by 7, to get a perfect square.
252 ÷ 7 = 36 = 2 × 2 × 3 × 3
Therefore, the required smallest number is 7.

(ii) We find that 2925 = 3 × 3 × 5 × 5 × 13
We find that 2925 should be divided by 13, to get a perfect square.
2925 ÷ 13 = 225 = 3 × 3 × 5 × 5
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 16
Therefore, the required smallest number is 13.
Also, \(\sqrt{225}\) = 3 × 5 = 15.

(iii) We have, 396 = 2 × 2 × 3 × 3 × 11
We find that 396 should be divided by 11, to get a perfect square.
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 17
396 ÷ 11 = 36 = 2 × 2 × 3 × 3
Therefore, the required smallest number is 11.
Also, \(\sqrt{36}\) = 2 × 3 = 6.

(iv) We have, 2645 = 5 × 23 × 23
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 18
We find that 2645 should be divided by 5, to get a perfect square.
2645 ÷ 5 = 529 = 23 × 23
Therefore, the required smallest number is 5.
Also, \(\sqrt{529}\) = 23.

(v) We have, 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
We find that 2800 should be divided by 7, to get a perfect square
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 19
2800 ÷ 7 = 400 = 2 × 2 × 2 × 2 × 5 × 5
Therefore, the required smallest number is 7.
Also, \(\sqrt{400}\) = 2 × 2 × 5 = 20.

(vi) We have,
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 20

Question 7.
The students of Class VIII of a school donated ₹ 2401 in all, for the Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
We have, 2401 = 7 × 7 × 7 × 7
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 21
We find that the number of students in the class is 49.

MP Board Solutions

Question 8.
2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Total number of plants = 2025
The plants are planted in a garden in such a way that each row contains as many plants as the number of rows.
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 22
2025 = 3 × 3 × 3 × 3 × 5 × 5
∴ Number of plants in each row = \(\sqrt{2025}\) = 3 × 3 × 5 = 45
So, number of rows = number of plants.
Thus, the number of rows = 45
and number of plants in each row = 45.

Question 9.
Find the smallest square number that is divisible by each of the numbers 4,9 and 10.
Solution:
The smallest number divisible by each 4, 9 and 10 is their LCM.
The LCM of 4, 9 and 10 is 2 × 2 × 3 × 3 × 5 = 180.
Now, prime factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 23
In order to get a perfect square, each factor of 180 must be paired.
So, we need to make pair of 5.
∴ 180 should be multiplied by 5.
Hence, the required number is 180 × 5 = 900.

MP Board Solutions

Question 10.
Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution:
The smallest number divisible by each 8, 15 and 20, is their LCM.
The LCM of 8, 15 and 20 is 2 × 2 × 2 × 3 × 5 = 120
MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.3 24
Now prime factorisation of 120 is
120 = 2 × 2 × 2 × 3 × 5 ….. (i)
In order to get a perfect square, each factor of 120 must be paired.
Thus we multiply (i) by 2 × 3 × 5 = 30, we get 120 × 30 = 3600.

MP Board Class 8th Maths Solutions

MP Board Class 10th General English Unseen Passages

In this article, we will share MP Board Class 10th English Solutions General Unseen Passages Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th General English Unseen Passages

In this part of your syllabus you will be given three unseen passages of 150 words each. You are required to read the passage carefully and answer the questions given below it. Each passage will carry 5 marks with 1 mark for vocabulary included in it.

 

Discursive Passages

1. Read the following passage and answer the questions given below it.

Man’sourney of life from childhood to old age is very charming and colorful. Youth is the most exciting period of man’s life when it is time to grow and dream. A young man is full of hope, energy and zeal. Nothing is difficult or impossible or dangerous for him. The old people say that youth is not daring but thoughtless. A young man bums the candle at both ends. He commits mistakes and leams only after burning his finger. Sometimes the young men misuse their freedom and thus invite difficulties by their foolish actions. They are full of strength, energy and enthusiasm. They become rebels and are no longer afraid of facing the forces of realities. A young man accepts the challenge of evil difficulties and hardships, to win or lose the game of life is the mission of his career. He loves to lead an adventurous life and has a keen desire to build up a new world of his dream. But the period of youth does not last long. Soon it is followed by old age when he regrets his past mistakes and failures. The weak old man feels helpless, depressed and disappointed. He becomes unfit for any adventure. But some fortunate old people never grow old and continue to feel young and active and make the most of even the last years of their lives. It will not be wrong to say that youth brings honor and old age commands respect.

Questions:
(a) Man’sourney is :
(i) Charming and colorful (ii) exciting (iii) dangerous
(b) What is the opinion of the old people about the young men?
(c) What is the mission of a young man’s career?
(d) Why does the old man feel helpless and disappointed?
(e) Find the antonym of ‘clever’ from the passage.
Answers:
(a) (i) charming and colorful.
(b) The old people say that youth is not daring but thoughtless.
(c) To win or lose the game of life is the mission of his, career.
(d) He feels helpless and disappointed because he becomes unfit for any adventure.
(e) Foolish.

2. Read the following passage and answer the questions given below it.

Whatever may be the cause of their suffering, we have to treat the handicapped with sympathy and understanding. In many instances, physically handicapped children are neglected and left to themselves in their homes. This makes them extremely sad and lonely. Our first duty is to make these children happier and less lonely. Secondly, we have to educate these children and help them to live useful lives. We should secure for them the benefits of education in schools specially intended for them. We have to make them useful citizens by creating suitable opportunities in them to be employed. They will then have a sense of achievement and we can be happy that we have done our duty towards them.

Questions:
(a) The physically handicapped children suffer from:
(i) fever (ii) neglect (iii) cold
(b) Which are the two important duties towards the handicapped children that we should perform?
(c) How can we make them useful citizens?
(d) What will be the outcome of these efforts?
(e) Give the synonym of ‘disabled’ from the passage?
Answers:
(a) (ii) neglect.
(b) Our first duty is to make handicapped children happier and less lonely. Secondly, we should educate them and help them to live useful lives.
(c) We can make them useful citizens by creating suitable opportunities for them to be employed.
(d) The outcome of these efforts will be that they will have a sense of achievement and we can be happy that we have done our duty towards them.
(e) Physically handicapped.

3. Read the following passage and answer the questions given below it.

Reading has a variety of meanings. To some people it means little more than the ability to pronounce aloud the printed word; to others it means it an ability to gain merely a general impression of what they read. Even students daily engaged in the study of books develop a superficial ability to read rapidly, and with apparent understanding, what they it subsequently prove to have understood imperfectly. Ability to read properly, to understand not only the general sense of a given passage but its particular implications, to appreciate so to speak, the light and shade of the passage, the precise meaning of the parts as well as of the whole, what it hints at as well as what it states, to distinguish between what is clearly proved and established and what is merely suggested or put forward as a supposition is still a comparatively rare quality.

MP Board Solutions

Questions:
(a) has a variety of meanings.
(i) Reading (ii) Studying (iii) Understanding.
(b) What does it mean to other people?
(c) How do students develop?
(d) What is a comparatively rare quality?
(e) precise’ means
(i) exact (ii) short (iii) incorrect.
Answers:
(a) (i) Reading.
(b) Others think that it is an ability to gain merely a general impression of what they read.
(c) They develop a superficial ability to read quickly and with apparent understanding, what they subsequently prove to have understood imperfectly.
(d) To distinguish between what is clearly proved and established and what is merely suggested or put forward as a supposition is still a comparatively rare quality.
(e) (i) exact.

4. Read the following passage aid answer the questions given below it.

A cheerful person is always more disposed to be happy than miserable. He tends to look at the bright side of things and thus often derives pleasure from circumstances which would ordinarily sadden a person. “^ cheerful beggar is happier than a low-spirited millionaire. As a source of happiness neither wealth nor fame nor beauty nor power nor even health can be compared even for a moment with cheerful temperament. A great advantage of cheerfulness is that it enables man to do his work more efficiently and prevents him from being easily exhausted. The laborer who whistles over his work goes homeless tired and can work harder than another who deeply thinks over real or imaginary troubles. ’

Questions:
(a) A cheerful person is always :
(i) miserable (ii) happy (iii) beautiful
(b) What is the tendency of a cheerful person?
(c) What cannot be compared with cheerful moment?
(d) What is the real advantage of cheerfulness?
(e) ‘Exhausted’ means :
(i) tired (ii) vigor (iii) low-spirited.
Answers:
(a) (ii) happy.
(b) A cheerful person tends to look at the bright side of things.
(c) Neither wealth nor fame nor beauty nor power nor even health can be compared with a cheerful moment.
(d) The real advantage of cheerfulness is that it enables man to do his work more efficiently and prevents him from being easily exhausted.
(e) (i) tired.

5. Read the following passage and answer the questions given below it.

Discipline means obedience to the established rules of conduct. Certain rules have been laid in every society to control and regulate the life and activities of its members so that the society as a whole may progress in harmony and peace. If any of these rules is broken, there is trouble and society suffers. In fact, discipline is the very basis of progress in every sphere, public or private. A man without discipline is like an engine without a brake. A society that has no rules or whose members do not conform to its rules soon falls into pieces. In game too, discipline is necessary. Every player has to obey his captain and carry out his commands whether he likes them or not. In army, discipline is more necessary. An army without discipline is no better than a lawless mob. In the same way a school or a college cannot run if the boys do not observe the rules and regulations of the institution. Teaching is impossible if the boys do not keep discipline. Discipline cultivates a spirit of respect for elders and superiors, teaches gentlemanly behavior in society and meek submission to any punishment that may be inflicted due to indiscipline. It is the duty of every student to observe them if they want to build their character and prosper in life.

Questions:
(a) When rules are broken suffers :
(i) life (ii) society (iii) man
(b) Why have certain rules of conduct been laid down by the society?
(c) What is a man without discipline?
(d) What does discipline cultivate among the students?
(e) The synonym of’ crowd’ is :
(i) society (ii) mob (iii) members.
Answers:
(a) (ii) society.
(b) Certain rules of conduct have been laid down in every society to control and regulate the life and activities of its members so that society as a whole may progress in harmony and peace.
(c) A man without discipline is an engine without a brake.
(d) Discipline cultivates a spirit of respect for elders and superiors, teaches gentlemanly behavior in society and meek submission to any punishment that may be inflicted due to indiscipline.
(e) (ii) mob.

MP Board Solutions

6. Read the following passage and answer the questions given below it.

Home is the first and the most important school of character. It is here that every human being receives his moral training, or his worst, for it is here! that he imbibes those principles of conduct which endure throughout manhood, and cease only with life.

It is a common saying that, “Manners make the man” and there is a second that, “Mind makes the man”, but truer than either is a third that “Home makes the man”. For the home, training includes not only manners and mind, but also character. It is mainly in the home that the heart is opened, the habits are formed, the intellect is awakened, and the character is molded for good or for evil.

From that source be it pure or impure, issue the principles and maxims of society. Law itself is but the reflection of homes. The finest bits of opinions sown in the minds of children in private life afterward issue forth to the world and become its public opinion, for nations are grown out of nurseries. Those who hold the leading-strings of children may even exercise greater power than those who wield the reins of Government.

Questions:
(a) Which is the most important school of character:
(i) Office (ii) Home (iii) Factory
(b) What does hole training include?
(c) How is public opinion formed at home?
(d) What is done at home with the man?
(e) Find a word from the passage which means ‘to bear’.
(i) intellect (ii) endure (iii) imbibe.
Answers:
(a) (ii) Home.
(b) Home training includes not only the manners and mind, but also character.
(c) The small opinions sown in the minds of children in private life afterward issue forth to the world and become its public opinion.
(d) At home heart is opened, habits are formed, the intellect is awakened and the character of man is molded for good or for evil.
(e) (ii) endure.

literary Passages

1. Read the following poem and answer the questions given below it.

Life ! I know not what thou art,
But know that, thou and I must part.
And when or how, or where we met
I own to me’s secret yet.
Life ! We’ve been long together
Through pleasant and through cloudy weather
Tis hard to part when friends are dear
Perhaps it will cost a sigh, a tear;
Then steal away, give little warning,
Choose thine own time :
Say not good night; but in some better clime,
Bid me good morning.

Questions:
(a) Whom is the poet addressing to :
(i) Friends (ii) Weather (iii) Life
(b) When is it hard to part?
(c) What is a secret to the poet?
(d) How were life and poet associated together?
(e) A word from the poem that means the same as ‘ to leave’ is :
(i) part (ii) steal (iii) clime (iv) none of these.
Answers:
(a) (iii) Life.
(b) It is hard to part from life when the friends and dear ones are close.
(c) It is a secret to the poet that when, how and where he met with life.
(d) Poet and life were associated together in all good and bad times.
(e) (ii) part.

2. Read the following poem and answer the questions given below it.

We have no wings, we cannot soar
But we have feet to scale and climb,
By slow degrees, by more and more,
The cloudy summits of our time.
The heights by great men reached and kept,
Were not attained by sudden flight
But they, while their companions slept,
Were toiling upward in the night.

MP Board Solutions

Questions:
(a) What do we not have :
(i) feet (ii) hands (iii) wings
(b) How did great men reach heights and kept them?
(c) What do we have if not wings?
(d) What lesson do you get from the poem?
(e) Which word in the poem means ‘to fly’?
(i) climb (ii) flighty (iii) soar (iv) scale
Answers:
(a) (iii) wings.
(b) Great men reached heights by working hard even when their friends were sleeping.
(c) We have feet to climb upward if not wings to fly.
(d) We learn from the poem that if one works hard continuously then he reaches great heights.
(e) (iii) soar.

3. Read the following passage and answer the questions given below it.

A poor, villager once saved the life of a wealthy goldsmith by attacking a robber who was about to kill him. When the villager had knocked the robber down with his lathi and bound his hands and feet, the goldsmith said to the villager, “I have no money with me. So I shall give you my watch.” He did so and went on his way. The man was greatly pleased with his watch and spent hours in listening to its ticking and watching the second-hand go round. Next day the watch stopped as the man did not know how to wind it. He was very sad and said, “Alas! It is dead.” Thinking its dead body might be of value, he took the watch to a Mahajan who gave him fifty rupees for it, as it was well worth for two hundred. As he was leaving the room, the villager who was at heart an honest man, turned back and said, “Here take your money. It is dead and I have cheated you.” But the Mahajan only laughed and told him to keep the money and go.

Questions:
(a) The goldsmith presented to the villager:
(i) a watch (ii) money (iii) a pen
(b) Why did the watch stop?
(c) What did the villager do with the dead watch?
(d) What did the villager admit in front of Mahajan?
(e) The synonym of ‘dead’ in the passage is :
(i) stopped (ii) lifeless (iii) immovable
Answers:
(a) (i) a watch.
(b) The watch stopped because the villager did I not know how to wind it.
(c) The villager sold the dead watch to a Mahajan for fifty rupees.
(d) The villager admitted in front of the Mahajan that the watch was dead and he had cheated him.
(e) (i) stopped.

4. Read the following passage and answer the questions given below it.

A deer who was very thirsty went to a pool to s quench his thirst. At the time of drinking water, he saw himself in the clear water “How handsome lam! i thought he. The horns on my head are branching like trees. My coat is smooth and glossy. My eyes sparkle like stars. Only my legs are so long and thin that I am ashamed of them.” just then he heard the sound; of the hunter’s foot steps. He dashed away through the forest. His long, thin legs bearing him swiftly on. ; The forest grew thicker and at last he could not run on account of his branching horns. So he was caught by the hunter. :

“How foolish I have been! ” cried the dying deer.
“Oh ! my splendid horns are the cause of my death.”

Question :
(a) The cause of deer’s death was :
(i) his legs (ii) his coat (iii) his horns
(b) What was he ashamed of?
(c) Why could he not escape?
(d) What did he think at the time of his death?
(e) Which word in the passage means ‘bright’?
(i) splendid (ii) glossy (iii) sparkle
Answers:
(a) (iii) his horns.
(b) He was ashamed of his long and thin legs.
(c) He could not escape because of his horns which got stuck in the dense forest.
(d) At the time of his death, he thought of his foolishness that the horns for whose beauty he was proud were the cause of his death.
(e) (ii) glossy.

5. Read the following passage and answer the questions given below it.

How delightful to Sita, Ram and Lakshman were the years of their forest exile. Wherever they went, they were welcomed by the companies of hermits and admitted to the forest ways of life. Thus they were quickly established in huts made of leaves and carpeted with the sacred grass, like other ascetics. Quickly they had also arranged their articles of worship, and gathered together their small stores of necessities and without any loss of time, Sita fell into the habit of cooking for her husband and brother like any peasant-woman and serving them with her own fair hands. Now and‘then it would happen, during their first years in the forest, that they came across some great saint, who would recognize Ram at the glance as the Lord himself.

Questions.:
(a) The years of forest exile of Ram, Laxman and Sita were :
(i) delightful (ii) gloomy (iii) frustrating
(b) What were the huts made of?
(c) How did Sita help Ram and Laxman?
(d) What did the saint recognize?
(e) Which word in the passage means ‘articles of need’?
(i) necessities (ii) ascetics (iii) none of these.
Answers :
(a) (i) delightful.
(b) The huts were made of leaves and carpeted with the sacred grass.
(c) Sita helped Ram and Laxman by cooking and serving them food.
(d) The saint recognized Ram as the Lord himself.
(e) (i) necessaries.

MP Board Solutions

6. Read the following passage and answer the questions given below it.

Mahmud of Ghazni had conquered so many countries that he could not rule over them properly. In one of these countries, robbers attacked a caravan of merchants and killed many of them and stole then- goods. The mother of one of the merchants walked a long way to Ghazni and made a complaint to the sultan. “My good woman,” said Mahmud, “how can I keep order in that distant land? It is hundreds of miles from Ghazni. I cannot put down robbers nor keep the road safe so far away.” “Why then, ” replied the old woman, “do you take countries which you cannot rule? For the bad rule of every country of which you are the king, God will call you to account, when you die.”

Questions:
(a) Where did the woman go to :
(i) Ghazni (ii) Agra (iii) Delhi
(b) What had the robbers done?
(c) Whiat did the old woman question him?
(d) Why did she say that he would have to give an account to God?
(e) Find the antonym of ‘near’ from the passage.
Answers:
(a) (i) Ghazni.
(b) The robbers had attacked a caravan of merchants and killed many of them and stolen their goods.
(c) The old woman questioned him that why had he taken over countries which he could not rule.
(d) She said so because of his bad rule of every country of which he was the king.
(e) distant.

Factual Passages

1. Read the following passage and answer the questions given below it.

Kashmir is economically a backward state. It has been ruled despotically for several centuries. Progress therefore has been extremely slow. Arts and crafts of Kashmir could not flourish as our handicrafts did not find any market due to difficulties of transport. The peasant in Kashmir still clings to the age-old methods of farming.

As agriculture occupies the most important position in the plan, special attention was given to promote it. More and more facilities were offered to farmers to improve the conditions of agriculture. The launching of the National Extension Service Scheme is the most important step in this direction. This aims at the social, economic and educational uplift of the poor people of Kashmir. The people engaged in the service are enthusiastic workers. The Government is determined to see the prompt accomplishment of responsibilities of reconstructing the countryside through the basic principle of Extension.

Questions:
(a) Kashmir is an economical state :
(i) forward (ii) developed (iii) backward
(b) Why could art and craft not flourish in Kashmir?
(c) What was done to improve agriculture?
(d) What does the National Extension Service Scheme aim at?
(e) Find out the synonyms of ‘sticks’ from the passage.
Answers:
(a) (iii) backward.
(b) Art and craft did not flourish in Kashmir because the handicrafts did not find any market due to difficulties of transport.
(c) More and more facilities were given to the farmers to improve agriculture.
(d) National Extension Service Scheme aims at the social, economic and educational uplift of the poor people in Kashmir.
(e) clings.

2. Read the following passage and answer the questions given below it.

Indira Gandhi was the first woman Prime Minister of India. She was born in Allahabad on November 19, 1917. She was the only child of Pt.awahar Lai Nehru and Kamla Nehru. Indira was a lovely child so her parents and grandparents called her Indira Priyadarshini. They lived in a big house. Its name was Anand Bhawan.

Little Indira had many dolls. Some of them were foreign. She loved to play with them. She dressed them like brides and bridegrooms or Raja and Rani or Satyagrahis and Policemen. At the age of four, Indira went to Gandhiji’s I Ashram at Sabarmati. There she slept on the floor, and ate simple food.

CBSE Class 9 English Beehive Book Poem 6 No Men are Foreign MCQ Questions with Answers from Beehive Book.

Questions:
(a) Gandhiji’s ashram was at:
(i) Allahabad (ii) Lucknow (iii) Sabarmati
(b) When and where was she born?
(c) What is Anand Bhawan?
(d) How did she dress her dolls?
(e) Which word in the passage means ‘newly married man’?
Answers:
(a) (iii) Sabarmati.
(b) She was born at Allahabad on November 19, 1917.
(c) Anand Bhawan is the name of the house I where Indira Gandhi and her parents lived,
(d) She dressed her dolls as brides and bridegrooms or Raja and Rani or Satyagrahis and policemen.
(e) bridegroom.

3. Read the following passage and answer the questions given below it.

A hockey team from Delhi went to Sri Lanka last month. They came to Chennai by plane. They left Delhi at 7 o’clock in the morning and reached Chennai at 12 o’clock. They stayed in a hotel for the day. They visited the museum and went in for shopping.

The next morning they went by bus to the railway station. They went in two buses. They traveled from Chennai to Dhanushkoti by ship. They went by train to Talaimannar in Sri Lanka. From Talaimannar they went to Colombo by taxi The team stayed in Sri Lanka for ten days.

MP Board Solutions

Questions:
(a) The team stayed in Sri Lanka for:
(i) ten days (ii) one month (iii) three months.
(b) When and where did the hockey team go?
(c) What did the team do in Chennai?
(d) How did the team travel from Chennai to Colombo?
(e) Find a word from the passage which means a place where antique things are kept’.
Answers:
(a) (i) ten days.
(b) Hockey team went from Delhi to Sri Lanka last month.
(c) The team visited the museum and went in for shopping in Chennai.
(d) The team traveled from Chennai to Dhanushkoti by ship. They went by train to Talaimannar in Sri Lanka. From Talaimannar they went to Colombo by taxi.
(e) museum.

4. Read the following passage and answer the questions given below it.

A peacock is a beautiful bird. It is the national bird of India. It feeds on plants and animals. Seeds, fruits, bulbs, roots, grass and leaves are its staple food. It also eats white ants, insects, spiders and worms. It devours lizards and frogs. It is an enemy of all kinds of snakes. It drinks water an hour after sunrise and returns to roost after a heavy meal at dusk. In the monsoon when there is rainfall, it is ready to dance. Its breeding season starts with the rainy season. It dances to see the clouds in the sky. It is considered as an auspicious bird.

Questions:
(a) Which is our National Bird :
(i) Sparrow (ii) Crow (iii) Peacock
(b) What does it feed on?
(c) When does it dance?
(d) When is its breeding season?
(e) Which word in the passage means ‘swallows’?
Answers:
(a) (iii) Peacock.
(b) It feeds on plants and animals. Seeds, fruits, bulbs, roots, grass and leaves are its staple food. It also eats white ants, insects, spiders and worms.
(c) It dances in the monsoon when there is rainfall.
(d) Its breeding season starts with the rainy season.
(e) devours.

5. Read the following passage and answer the questions given below it.

Pandit Jawaharlal Nehru was our first Prime Minister. His father was Pandit Motilal Nehru. He was an advocate and he was a rich man. He lived , in Allahabad. The name of his house is ‘Anand Bhawan’. It is a big house and there is beautiful . garden around it.

Pandit Jawaharlal Nehru studied in India and England. He played cricket too. He was a good writer and a good speaker. He was brave and kind.
Nehru loved children and roses very much. Children lovingly called him ‘Chacha Nehru’. His birthday, 14th November, is celebrated as Children’s
Day. He was a great freedom-fighter. He will always be remembered.

Questions:
(a) Pt. Motilal Nehru was :
(i) a businessman (ii) an advocate (iii) an industrialist
(b) Mention the qualities of Jawaharlal Nehru.
(c) Where did Jawaharlal Nehru study?
(d) When is Children’s Day celebrated?
(e) Find the synonym of ‘author’ from the passage.
Answers:
(a) (ii) an advocate.
(b) Jawaharlal Nehru could play cricket. He was a good writer and speaker. He was brave and kind.
(c) He studied in India and England.
(d) Children’s day is celebrated on 14th November on Jawaharlal Nehru’s birthday.
(e) Writer.

6. Read the following passage and answer the questions given below it.

Kalpana Chawla, the astronaut who died on board the space shuttle Columbia, had been sponsoring two students from her school (Tagore Bal Niketan in Kamal) each year since 1997 for. International Space School Camp in Houston.

Manpreet Kaur and Namita Along visited NASA in August 2002. “We spent an excellent day with KC she even cooked for us and made us feel totally at home”, said Namita. These two and other youngsters who benefitted from this program, say, KC, as she was popularly known, might have been the first Indian-born woman in space but was entirely unaffected by her success.

Gaurav Goel of the 1999 batch and now an engineering student at Ambala said, “Not only was KC down to earth, she still retained the Indian in herself despite living in US for so many years”.

Even as a student, Kalpana had looked out for other students. She used to pay the fees for two of her college mates who could not afford it. “It is possible that to this day they are not aware that Kalpana used to pay their tuition fees,” says Sovina Sood, Kalpana’s sunior in the Punjab Engineering College.

Sovina who now teaches Civil Engineering said that sometimes Kalpana would hand over the money to her “but ask me not to disclose it to anybody”.

MP Board Solutions

Questions:
(a) Kalpana Chawla’s school was at:
(i) Ambala (ii) Kamal (iii) Houston
(b) Who visited NASA in August 2002?
(c) What did Kalpana Chawla do for her college mates when she was a student?
(d) Give a word from a passage that means the same as ‘getting advantage’.
(e) Sponsoring means :
(i) to provide support (ii) contributing (iii) participating.
Answers:
(a) (ii) Kamal.
(b) Manpreet Kaur and Namita Along visited NASA in August 2002.
(c) Kalpana Chawla used to pay the fees for two of her college mates when she was a student.
(d) benefitted.
(e) (ii) contributing.

MP Board Class 9th General English Unseen Passages

MCQ Questions for Class 9 English Moments Chapter 1 The Lost Child MCQ with Answers

MP Board Class 9th General English Unseen Passages

Factual Passages

Q. 1. Read the following paragraph and answer the questions.

The first computer was built in the early nineteenth century by Charles Babbage. He called it a ‘difference engine’: It could carry out long calculations and print the results. Later he invented a better machine called an ‘analytical engine’. The computer that you use has been developed by several scientists working independently or as a team. It is being improved everyday.

The computer can work out calculations and provide answers to complicated problems in a flash. A man would take days, or even months, to work the same thing out. But it cannot do things on its own. All the information it needs to do something is fed into it. This information is called data. Then it must be given clear and precise instructions about what to do with this data. These instructions have to be in a language that the computer understands. The information and the instructions make up a computer program. If a computer gives an incorrect answer, something must be wrong with the programme and not with the machine.

The computer also has a memory. It can store away information and use it later. It can, therefore, be compared to the human brain. But it can solve problems much faster and more accurately than the human brain. It can be made to play games like chess and can translate words from one language to another. (If you like, it can correct your spelling mistakes in what you write.) But it cannot take decisions or think up new deas. That is where it cannot compete with the human brain.

MP Board Solutions

Questions
(1) What two things could the first computer developed by Charles Babbage do?
(2) What is a computer program?
(3) In what way is the computer better than a human brain?
(4) Give a word from the passage that means same as ‘correctly’.
(5) Find a word from the passage that means ‘without anybody’s help’.
Answers
(1) It could carry out long calculations and print the results.
(2) A computer program consists of information and instructions.
(3) A computer is better than a human brain because it solves problems much faster and more accurately than human brain.
(4) accurately.
(5) independently.

Q. 2. Read the given passage and answer the following questions.

Kalpana Chawla, the astronaut who died on aboard the space shuttle Columbia, had been sponsoring two students from her school (Tagore Bal Niketan in Kamal) each year since 1997 for the International Space School Camp in Houston.

Manpreet Kaur and Namita Alung visited NASA in August 2002. “We spent an excellent day with KC; she even cooked for us and made us feel totally at home,” said Namita. These two and other youngsters who benefited from this programme say KC, as she was popularly known might have been the first Indian-born woman in space but was entirely unaffected by her success.

Gaurav Goel of the 1999 batch and now an engineering student at Ambala said, “Not only was KC down to earth, she still retained the Indian in herself despite living in US for so many years.”

Even as a student, Kalpana had looked out for other students. She used to pay the fees for two of her college mates who could not afford it. “It is possible that to this day they are not aware that Kalpana used to pay their tuition fees,” says Sovina Sood, Kalpana’s junior in the Punjab Engineering College.

Sovina who now teaches Civil Engineering said that sometimes Kalpana would hand over the money to her “but-ask me not to disclose it to anybody.”

Questions
(1) What did Kalpana Chawla do since 1997?
(2) Who visited NASA in August 2002?
(3) What did Kalpana Chawla do for her college mates when she was a student?
(4) Give a word from the passage that means same as ‘getting advantage’.
(5) Sponsoring means—
(a) to provide support, (b) contributing, (c) participating.
Answers
(1) Kalpana Chawla had been sponsoring two students from her school since 1997.
(2) Sanpreet Kaur and Namita Along visited NASA in August 2002.
(3) Kalpana Chawla used to pay the fees for two of her college mates.
(4) benefited.
(5) (b) contributing.

Q. 3. Read the following passage and answer the questions below it.

Ramcharit Manas is the holy book of Hindus.

It was written by Tulsidas. It is the story of Ram, the son of Dashrath, the king of Ayodhya. Ram is described as an ideal character. He is an ideal son, ideal husband, ideal brother and ideal ruler. He left Ayodhya to obey his father and lived in forest. When Ravana took away Sita, he attacked Lanka. Ravana was killed in the war and Sita was released. They returned to Ayodhya after 14 years of exile. Then Ram became the king of Ayodhya.

MP Board Solutions

Questions
(1) Who wrote Ramcharit Manas?
(2) What does Ramcharit Manas describe?
(3) How is Ram described in the book?
(4) Find a-word from the passage that means same as ‘to set free’.
(5) Exile means—
(a) banishment, (b) movement, (c) to expel.
Answers
(1) Ramcharit Manas was written by Tulsidas.
(2) Ramcharit Manas describes the life of Lord Ram.
(3) Ram is described as an ideal character in the book.
(4) released.
(5) (a) banishment.

Q. 4. Read the following paragraph and answer the questions givep below it.

India is facing many dangerous problems. The main problems among them are unemployment, pollution, population and illiteracy. Due to these harmful problems our country cannot progress well. All of us and specially our government should make effective efforts to solve them. The role of students, teachers and of the whole society has great importance in this field. We should change our tendency, policy of education and should end these problems. Our state governments are trying their best to solve such huge problems. Every change needs enough time. Questions

(1) What are the main problems of India?
(2) Who can play the main role in this field?
(3) Give a suitable heading to the above passage.
(4) Give a word from the passage that means same as ‘producing a desired result’.
(5) Tendency means—
(a) habit, (b) behavior, (c) incline.
Answers
(1) The main problems of India are unemployment, pollution, population and illiteracy.
(2) Students, teachers and whole society can play the main role in this field.
(3) ‘Problems of India’.
(4) effective.
(5) (a) habit.

Q. 5. Read the following passage and answer the questions given below it.

Smoking is the single largest preventable cause of death worldwide. It is killing about 2,200 people in India everyday that means one every forty seconds. It is alarming that in spite of this, smoking is increasing among the youth of India. According to World Bank Study, India, Indonesia and China are the only countries in the world where incidence of smoking is going up. It is increasing not only in cities but also in towns and villages. Higher educated groups are also taking to smoking more readily.

Questions
(1) How many people in India die in a day due to smoking?
(2) In which countries is smoking going up?
(3) Who are taking up smoking more readily?
(4) Give a word from the passage that means the same as ‘reason’.
(a) cause (b) work (c) begin (d) habit.
(5) Find a word from the passage that means frightening or worrisome.
(a) beating (b) ringing (c) alarming (d) singing.
Answers
(1) Smoking is killing about 2,200 people in India everyday.
(2) Smoking is going up in India, Indonesia and China.
(3) Higher educated groups are taking to smoking more readily.
(4) (a) cause.
(5) (c) alarming.

Literary Passages

Q. 1. Read the following stanzas and answer the questions given below.

Life has loveliness to sell,
All beautiful and splendid things.
Blue waves whitened on a cliff,
Soaring fire that sways and sings,
And children’s faces looking up,
Holding wonder like a cup.
Life has loveliness to sell,
Music like a curve of gold,
Scent of fine trees in the rain,
Eyes that love you, arms that hold,
And for your spirit’s still delight,
Holy thoughts that star the night.

MP Board Solutions

Questions
(1) What does life have to sell?
(2) What do soaring fire do?
(3) With what music is compared?
(4) Find out a word from the poem that means same as ‘magnificent’.
(a) wonder (b) splendid (c) delight (d) spirit.
(5) Find a word from the passage that means ‘steep rock’.
(a) a small hill (b) island (c) cliff (d) curve.
Answers
(1) Life has. loveliness to sell.
(2) Soaring fire sways and sings.
(3) Music is compared with a curve of gold.
(4) (a) splendid
(5) (c) cliff.

Q. 2. Read the following paragraph and answer the questions given below.

A few days later, Premchand resigned his job of Inspector of Schools after having worked in the department for twenty years. He was a free man after all. Now he could write novels and stories about his country and its people. In his books he dealt with the lives of the peasants and workers. He revealed the greed and meanness of the money-lenders, landlords and priests. He attacked the social evils like dowry and early marriage. He held society responsible for the sins of women.

Questions
(1) Which job did Premchand resign?
(2) What kind of man was Premchand after his resignation?
(3) What did he reveal?
(4) Find out a word from the passage that means same as—‘Farmers’.
(5) Find out a word from the passage that means ‘one who lends money’.
Answers
(1) Premchand resigned the job of Inspector of Schools.
(2) Premchand was a free man.
(3) He revealed the greed and meanness of money lenders.
(4) Peasants.
(5) Money lenders.

Q. 3. Read the following poem and answer the questions given below it.

Unfolding Bud
One is amazed
By a water-lily bud
Unfolding
With each passing day
Taking on a richer colour
And new dimensions
One is not amazed
At a first glance,
By a poem,
Which is as tight closed
As a tiny bud.
Yet on is surprised
To see a poem
Gradually unfolding,
Revealing its rich feelings
As one reads it
Again
And over again.

MP Board Solutions

Questions
(1) What two things amaze the poet?
(2) With what does a water-lily bud arrest one’s attention?
(3) What does a poem reveal?
(4) Find out a word from the passage that means same as : ‘leaf or a flower at the beginning of its growth’:
(a) bud (ii) colour (iii) thorn (iv) stem.
(5) ‘Gradually’ means :
(a) slowly, step by step (b) rapidly (c) surprisingly (d) frighteningly.
Answers
(1) An unfolding bud and an unfolding poem amaze the poet.
(2) A water-lily bud arrests one’s attention by taking on a richer colour.
(3) A poem reveals its rich feelings.
(4) (a) bud.
(5) (a) slowly, step by step.

Q. 4. Read the following passage and answer the questions given below it.

Long, long ago there lived an old woman who was very rich. She had a number of maids. They were made to work very hard to do their fixed duties at any given time.

Since there were np clocks to tell time, the old woman kept a cock. When it started crowing, everyone got up. The poor maids had to leave their beds as soon as the cock crew early in the morning every day. And the maids did not like it.’ But what could they possibly do?

The cock was their enemy, they thought. “The cock is a nuisance,” said one maid to another, “Why don’t we get rid of it ? ”

“How ?” asked her friends. “If we wring its neck and kill it, there will be no one to disturb us in our sleep. We could get up late every morning and then our mistress would not be able to say anything to us. ”

“A good idea”, cried the others. One of the maids, therefore, quietly caught hold of the bird and killed it. As a result, the next morning they were able to sleep late. The old woman slept late too, as there was no cock to wake her up. She woke up at last to discover how late it was, and that the cock was dead. She was, of course, angry, not only because she had lost her bird but also because her routine for the day had been upset.

She thought of a plan. From that day onwards she began to keep a watch herself at nights, making up for her loss of sleep by resting during the day. And as she kept watch, she went at odd hours of the night to wake up her maids so that they did not get their usual hours of sleep.

The maids found that they were worse off now.

“This is very bad”, said they. “Now the old woman wakes us up at all odd hours. We thought we had solved our problem by killing the cock. But we have not.”

You see, they were helpless now. They had to obey the old woman and work according to her set routine.

Questions
(1) Why did the maids kill the cock ?
(2) Why was the old woman angry ?
(3) Why were the maids worse off now ?
(4) Find a word from the passage that means ‘one that causes difficulty’.
(5) Wring means—
(a) to cut (b) to squeeze tightly (c) to kill.
Answers
(1) The maids killed the cock because it was their enemy, they thought.
(2) The old woman was angry not only because she had lost her bird but also because her routine for the day had been upset.
(3) The maids were worse off now because they did not get their usual hours of sleep.
(4) nuisance.
(5) (b) to squeeze tightly.

Discursive Passages

Q. 1. Read the following passage carefully and answer the questions given below it.

Home is the first and the most important school of character. It is a common saying that “Manners make the man”, and there is a second that “Mind makes the man”, but truer than either is a third that “Home makes the man”. For the home training includes not only manners and mind, but also character. It is mainly in the home that the heart is opened, the habits are formed, the intellect is awakened, and the character is moulded for good or for evil.

Questions
(1) Which is the first and most important school of character?
(2) What does home training include?
(3) Which is the common saying?
(4) Give a word from the passage that means same as ‘moral strength’.
(5) Find a word from the passage that means ‘reasoning power of the mind’.
Answers
(1) Home is the first and most important school of character.
(2) Home training includes not only manner and mind, but also character.
(3) The common saying is that “Manners make the man”.
(4) Character.
(5) Intellect.

Q. 2. Read the following passage carefully and answer the questions given below it.

There is no doubt that a common language used throughout the world does much to bring countries closer to each other. Though it is becoming increasingly easy to move from place to place, our inability to communicate with one another, gives rise to numerous misunderstandings and makes real contact between people of different “nationalities impossible. Many attempts have been made to overcome the problem and they have all failed. The fear of foreign influence and domination rules out the universal acceptance of any one of the existing major languages.

MP Board Solutions

Questions
(1) What will help bring the countries of the world close to each other?
(2) What gives rise to numerous problems?
(3) What is the major problem in accepting one of the existing major languages as a common language?
(4) Find out a word from the passage that means the same as ‘surely’:
(a) no doubt (b) of course (c) easily (d) really.
(5) ‘So many’ means (a) contact (c) fear
Answers
(1) A common language will help the countries of the world to communicate with one another and create goodwill. Thus, the countries will come closer to one another.
(2) Our inability to communicate with each other gives rise to many problems.
(3) The fear of foreign influence and domination rules out the universal acceptance of any one of the existing major languages.
(4) (a) no doubt.
(5) (b) numerous.

Q. 3. Read the following passage carefully and answer the questions given below it.

Proper food is the basic requirement of health. Health does not mean absence of disease; it rather means the presence of energy and vitality. Most of the food we eat daily does not contain essential nutrients needed for preservation of health and prevention of disease. There are various kinds of food. Protective foods are essential for prevention of disease and are needed both by the healthy and sick. They are rich in proteins, vitamins and minerals. Energy giving food are rich in carbohydrates.

Questions
(1) What is the real meaning of health?
(2) Name two essential elements of protective food?
(3) What do energy giving foods contain?
(4) Find out a word from the passage that means same as ‘need’.
(a) requirement (b) prevention (c) disease (d) energy.
(5) ‘Power to live or grow’ means :
(a) variety (b) vitality (c) health (d) protein.
Answers
(1) Real meaning of health means the presence of energy and vitality.
(2) Vitamins and proteins.
(3) Energy giving foods are rich in carbohy-drates.
(4) (a) requirement.
(5) (b) vitality.

Q. 4. Read the following passage carefully and answer the questions given below it.

Discipline must be enforced early in life. Discipline at home makes for the future greatness of a child. It forms his character and makes him a fit citizen. The child who is allowed to have his own way becomes wayward. The child who is allowed all sorts of exercises like running in sun, exposing himself to cold, eating unwholesome things will fall ill very frequently. A child whose habits have been disciplined and who has been taught to rise early, attend to his lessons properly, take physical exercise at the proper time and avoid things that are injurious will grow up to be a useful member of the society. The spoilt child who has been allowed to run his own course will show vices contained in younger days. His parents would wish that he had not been at all.

MP Board Solutions

Questions
(1) When should discipline be enforced?
(2) What does discipline do for a child?
(3) If a child is allowed to do what he likes, what will happen to him?
(4) Pick out from the passage a word which means the same as ‘often’.
(5) Wayward means—
(a) spoilt, (b) self willed, (c) destructive.
Answers
(1) Discipline should be enforced in early life.
(2) Discipline forms a child’s character and makes him a fit citizen.
(3) If a child is allowed to do what he likes he would become wayward and spoilt. He would show vices contained in younger days.
(4) frequently.
(5) (b) self willed.

MP Board Class 9th English Solutions

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

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.3 Pdf, 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.

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)
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 1
From (1), we have s = (3 + t) … (2)
Substituting this value of s in (2), we get
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 2
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),
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 3
Substituting the value of y in (2), we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 4
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 5
Substituting the value of x in (1), we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 6

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

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 13

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°
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 7
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)
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 8
Substituting this value of y in (1) , we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 9

(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

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

(v) Let the numerator = x
and the denominator = y
∴ Fraction = \(\frac{x}{y}\)
According to the given condition,
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 10
⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y + 4 = 0 … (1)
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 11
Substituting this value of x in (1) , we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 12

(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 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 1.
What is the sum of any two
(a) Odd numbers?
(b) Even numbers?
Solution:
(a) The sum of any two odd numbers is an even number.
As like, 1 + 3 = 4, 3 + 5 = 8

(b) The sum of any two even numbers is an even number.
As like, 2 + 4 = 6, 6 + 8 = 14

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

The factor pair of 96 or any number as a set of two factors, which, when multiplied together, give a particular product.

Question 2.
State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Solution:
(a) False
Since, sum of two odd numbers is even and sum of one odd number and one even number is always odd.
(b) True
Since, sum of two odd numbers is even and sum of two even numbers is always even.
(c) True
(d) False
If an even number is divided by 2, then the quotient is either odd or even.
(e) False
Since, prime number 2 is even.
(f) False
Factors of prime numbers are 1 and the number itself.
(g) False
Sum of two prime numbers is either even or odd.
(h) True
(i) False
Since, even number 2 is prime i.e., not composite.
(j) True

Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution:
Pairs of prime numbers having same digits upto 100 are 17 and 71; 37 and 73; 79 and 97

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 4.
Write down separately the prime and composite numbers less than 20.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

GCF Calculator is a free online tool that displays the Greatest Common Factor of two numbers given the inputs with detailed steps on how to approach.

Question 5.
What is the greatest prime number between 1 and 10?
Solution:
The greatest prime number between 1 and 10 is 7.

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 6.
Express the following as the sum of two odd primes.
(a) 44
(b) 36
(c) 24
(d) 18
Solution:
(a) 44 = 3 + 41
(b) 36 = 5 + 31
(c) 24 = 7 + 17
(d) 18 = 7 + 11

Question 7.
Give, three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes].
Solution:
Three pairs of prime numbers whose difference is 2 are 3 and 5; 5 and 7; 11 and 13.

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 8.
Which of the following numbers are prime?
(a) 23
(b) 51
(c) 37
(d) 26
Solution:
23 and 37 are prime numbers and 51 and 26 are composite numbers.
Thus, numbers in option (a) and (c) are prime.

Question 9.
Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
Seven consecutive composite numbers less than 100 are 90, 91, 92, 93, 94, 95, 96

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 10.
Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61
Solution:
(a) 21 = 3 + 7 + 11
(b) 31 = 3 + 11 + 17
(c) 53 = 13 + 17 + 23
(d) 61 = 13 + 19 + 29

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2

Question 11.
Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
(Hint :3 + 7 = 10)
Solution:
Since, 2 + 3 = 5; 7 + 13 = 20; 3 + 17 = 20; 2 + 13 = 15; 5 + 5 = 10 and 5, 10, 15, 20 all are
divisible by 5.
So, five pairs of prime numbers less than 20 whose sum is divisible by 5 are 2, 3; 2, 13; 3, 17; 7, 13; 5, 5.

Question 12.
Fill in the blanks:
(a) A number which has only two factors is called a ___.
(b) A number which has more than two factors is called a ___.
(c) 1 is neither ___ nor ___.
(d) The smallest prime number is ___.
(e) The smallest composite number is ___.
(f) The smallest even number is ___.
Solution:
(a) Prime number
(b) Composite number
(c) Prime number, composite number
(d) 2
(e) 4
(f) 2

MP Board Class 6th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2

MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2

Question 1.
Represent these numbers on the number line.
(i) \(\frac{7}{4}\)
(ii) \(\frac{-5}{6}\)
Solution:
(i) We have to represent \(\frac{7}{4}\) on the number line. \(\frac{7}{4}\) can be written as \(1 \frac{3}{4} \cdot 1 \frac{3}{4}\) lies between 1 and 2.
Step-1: Draw a number line and mark O on it to represents ‘0’ (zero)
Step-2 : Take a point A to represent
1 and B to represent 2.
Step-3 : Divide the distance of A and B in four equal parts A A1, A1A2, A2A3, A3B.
Step-4 : Count from 1 and reach to the third point A3 and A3 is the required point on number line.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-1

(ii) \(\frac{-5}{6}\) is lies between 0 and -1.
Step-1: Draw a number line and mark O on it to represent ‘0’ (zero).
Step-2 : Take a point A to represent -1.
Step-3 : Divide the distance of A and O in six equal parts AO5, O5O4, O4O3, O3O2, O2O1, O1O.
Step-4 : Count from 0 and reach to the fifth point O5.
O5 is the required point.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-2

MP Board Solutions

Free Least Common Denominator calculator – Find the LCD of two or more numbers step-by-step.

Question 2.
Represent \(\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}\) on the number line.
Solution:
We have to mark \(\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}\) on the same number line.
Since \(\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}\) all are less than 0 but greater than -1.
∴ All these lie between 0 and -1.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-3
Thus, A, B and C are the required points.

Question 3.
Write five rational numbers which are smaller than 2.
Solution:
Five numbers less than 2 lies on the left of 2 on the number line.
∴ Five rational numbers are \(0, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\)

MP Board Solutions

Question 4.
Find ten rational numbers between \(\frac{-2}{5}\) and \(\frac{1}{2}\).
Solution:
We have given, two rational numbers \(\frac{-2}{5}\) and \(\frac{1}{2}\).
First we make the same denominator of both rational numbers.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-4
Now, we have to find 10 rational numbers between \(\frac{-4}{10}\) and \(\frac{5}{10}\). so we have to multiply the numerator and denominator by a number such that difference between numerators is atleast 10.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-5

Question 5.
Find five rational numbers between
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-6
Solution:
First we make the same denominator of both rational numbers
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-7
Since, we have to find five rational numbers between \(\frac{10}{15}\) and \(\frac{12}{15}\) so we multiply the numerator and denominator by a number such that difference between the numerators is atleast 5.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-8
∴ The five rational numbers between \(\frac{1}{4}\) and \(\frac{2}{4}\) are \(\frac{41}{60}, \frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}\)

(ii) First we make the same denominator of both rational numbers.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-9
Since, we have to find five rational numbers between and \(-\frac{9}{6}\) and \(\frac{10}{6}\), so we do not need to multiply the numerator and denominator of \(-\frac{9}{6}\) and \(\frac{10}{6}\) by any number, because we can see that the difference between the numerators is 19 > 5.
∴ Five rational numbers between \(\frac{-3}{2}\) and \(\frac{5}{3}\) are \(\frac{-8}{6}, \frac{-7}{6}, \frac{0}{6}, \frac{1}{6}, \frac{2}{6}\)

(iii) First we make the same denominator of both rational numbers.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-10
Since we have to find five rational numbers between \(\frac{1}{4}\) and \(\frac{2}{4}\), so we multiply the numerator and denominator by a number such that difference between the numerators is atleast 5.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-11
∴ Five rational numbers between \(\frac{1}{4}\) and \(\frac{2}{4}\) are \(\frac{9}{32}, \frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}\).

Question 6.
Write five rational numbers greater than -2.
Solution:
Five rational numbers greater than -2 lies on the right side of -2 on number line.
∴ Any five rationals on the right of -2 are \(\frac{-3}{2},-1, \frac{-1}{2}, 0, \frac{1}{2}\).

MP Board Solutions

Question 7.
Find ten rational numbers between \(\frac{3}{5}\) and \(\frac{3}{4}\)
Solution:
Make the common denominator.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-12
Since we have to find ten rational numbers between \(\frac{3}{5}\) and \(\frac{3}{4}\) so, we multiply the numerator and denominator by a number such that difference between the numerators is atleast 10.
MP Board Class 8th Maths Solutions Chapter 1 Rational Numbers Ex 1.2 img-13

MP Board Class 8th Maths Solutions

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

Question 1.
Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) HCF of 135 and 225
Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get
225 = (135 × 1) + 90, since 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and
90, we get 135 = (90 × 1) + 45
But 45 ≠ 0
∴ Applying Euclid’s lemma to 90 and 45, we get 90 = (45 × 2) + 0
Here, r = 0, so our procedure stops. Since, the divisor at the last step is 45,
∴ HCF of 225 and 135 is 45.

(ii) HCF of 196 and 38220
We start dividing the larger number 38220 by 196, we get
38220 = (196 × 195) + 0
Here, r = 0
∴ HCF of 38220 and 196 is 196.

(iii) HCF of 867 and 255 Here, 867 > 255
∴ Applying Euclid’s Lemma to 867 and 255, we get
867 = (255 × 3) + 102, 102 ≠ 0
∴ Applying Euclid’s Lemma to 255 and 102, we get
255 = (102 × 2) + 51, 51 ≠ 0
∴ Applying Euclid’s Lemma to 102 and 51, we get
102 = (51 × 2) + 0, r = 0
∴ HCF of 867 and 255 is 51.

Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, let q be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we get a = 6q + r
where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,
a = 6q + 0 = 6q or a = 6q + 1
or a = 6q + 2 or a = 6q + 3
or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ‘a’.
[∵ 6q = 2(3q) = 2m1 6q + 2 = 2(3q + 1) = 2m2,
6q + 4 = 2(3 q + 2) = 2m3]
But ‘a’ being an odd integer, we have :
a = 6q + 1, or a = 6q + 3, or a = 6q + 5

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
Total number of members = 616
∴ The total number of members are to march behind an army band of 32 members is HCF of 616 and 32.
i. e., HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns.
∴ Applying Euclid’s lemma to 616 and 32, we get
616 = (32 × 19) + 8, since, 8 ≠ 0
Again, applying Euclid’s lemma to 32 and 8, we get
32 = (8 × 4) + 0, r = 0
∴ HCF of 616 and 32 is 8
Hence, the required number of maximum columns = 8.

Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3g, 3q + 1 or 3g + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]
Solution:
Let us consider an arbitrary positive integer as ‘x’ such that it is of the form
3q, (3q + 1) or (3q + 2)
For x = 3q, we have x2 = (3q)2
⇒ x2 = 9q2 = 3(3q2) = 3m ………. (1)
Putting 3q2 = m, where m is an integer.
For x = 3q + 1,
x2 = (3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1 = 3m + 1 ………… (2)
Putting 3q2 + 2q = m, where m is an integer.
For x = 3q + 2,
x2 = (3q + 2)2
= 9q2 + 12q + 4 = (9q2 + 12q + 3) + 1
= 3(3q2 + 4q + 1) + 1 = 3m + 1 ……….. (3)
Putting 3q2 + 4q +1 = m, where m is an integer.
From (1), (2) and (3),
x2 = 3m or 3m + 1
Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q +1) or (3q + 2).
For x = 3q
x3 = (3q)3 = 27q3 = 9(3q3) = 9m ……… (1)
Putting 3q3 = m, where m is an integer.
For x = 3q + 1
x3 = (3 q + 1)3 = 27q3 + 27q2 + 9q + 1
= 9(3q3 + 3q2 + q) + 1 = 9m + 1 ………… (2)
Putting 3q3 + 3q2 + q = m, where m is an integer.
For x = 3q + 2,
x3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8
= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8 ……………. (3)
Putting 3q3 + 6q2 + 4q = m, where m is an integer.
From (1), (2) and (3), we have
x3 = 9m, (9m + 1) or (9m + 8)
Thus, cube of any positive integer can be in the form 9m, (9m + 1) or (9m + 8) for some integer m.

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