Bhagya

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.3

Question 1.
Prove that \(\sqrt{5}\) is irrational.
Solution:
Let us assume, to the contrary, that \(\sqrt{5}\) is rational.
∴ \(\sqrt{5}=\frac{a}{b}\)
∴ b × \(\sqrt{5}\) = a
By Squaring on both sides,
5b² = a² …………. (i)
∴ 5 divides a².
5 divides a.
∴ We can write a = 5c.
Substituting the value of ‘a’ in eqn. (i),
5b² = (5c)² = 25c²
b² = 5c²
It means 5 divides b².
∴ 5 divides b.
∴ ‘a’ and ‘b’ have at least 5 as a common factor.
But this contradicts the fact that a’ and ‘b’ are prime numbers.
∴ \(\sqrt{5}\) is an irrational number.

Question 2.
Prove that 3 +2\(\sqrt{5}\) is irrational.
Solution:
Let 3 + 2\(\sqrt{5}\) is rational.

⇒ From (1), \(\sqrt{5}\) is rational
But this contradicts the fact that \(\sqrt{5}\) is irrational.
∴ Our supposition is wrong.
Hence, 3 + 2\(\sqrt{5}\) is irrational.

Question 3.
Prove that the following are irrationals.
(i) \(\frac{1}{\sqrt{2}}\)
(ii) 7\(\sqrt{5}\)
(iii) 6 + \(\sqrt{2}\)
Solution:
(i) We have

From (1), \(\sqrt{2}\) is rational number which contradicts the fact that \(\sqrt{2}\) is irrational.
∴ Our assumption is wrong.
Thus, \(\frac{1}{\sqrt{2}}\) is irrational.

(ii) Let \(7 \sqrt{5}\) is rational.
∴ We can find two co-prime integers a and b such that \(7 \sqrt{5}=\frac{a}{b}\), where b ≠ 0

This contradicts the fact that \(\sqrt{5}\) is irrational.
∴ Out assumption is wrong.
Thus, we conclude that 7\(\sqrt{5}\) is irrational.

(iii) Let 6 + \(\sqrt{2}\) is rational.
∴ We can find two co-prime integers a and b

= Rational [ ∵ a and b are integers]
From (1), \(\sqrt{2}\) is a rational number,
which contradicts the fact that \(\sqrt{2}\) is an irrational number.
∴ Our supposition is wrong.
⇒ 6 + \(\sqrt{2}\) is an irrational number.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.1

Question 1.
The graphs of y = p(x) are given in the below figures, for some polynomials p(x). Find the number of zeroes of p{x), in each case.

Solution:
(i) The given graph is parallel to x-axis, it does not intersect the x-axis.
It has no zero.
(ii) The given graph intersects the x-axis at one point only.
∴ It has one zero.
(iii) The given graph intersects the x-axis at three points.
∴ It has three zeroes.
(iv) The given graph intersects the x-axis at two points.
∴ It has two zeroes.
(v) The given graph intersects the x-axis at four points.
∴ It has four zeroes.
(vi) The given graph meets the x-axis at three points.
∴ It has three zeroes.

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.6 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 3.6

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:


Solution:
(i) we have


Equations (1) and (2) becomes,
2u + 3v = 2 ⇒ 2u + 3v – 2 = 0 … (3)
4u – 9v = -1 ⇒ 4u – 9v + 1 = 0 … (4)

(iii)

The equations (1) and (2) becomes,
4p + 3y = 14 ⇒ 4y + 3y -14 = 0 …. (3)
3p – 4y = 23 ⇒ 3p – 4y – 23 = 0 …. (4)
Here, a₁ = 4, b₁= 3, c₁ = —14, a₂ = 3, b₂ = – 4, c₂ = -23
Solving (3) and (4) by cross multiplication method, we get

(iv)

Equations (1) and (2) can be expressed
5u + v = 2 ⇒ 5u + v – 2 = 0 …. (3)
6u – 3v = 1 ⇒ 6u – 3v – 1 = 0 …. (4)

Thus, the required solution is x = 4, y = 5

(v)

∴ Equation (3)and (4)can be expressed as:
7q – 2p – 5 = 0 … (5)
8q + 7p -15 = 0 … (6)

Thus, the required solution is x = 1, y = 1

(vi) We have, 6x + 3y = 6xy …. (1)
2x + 4y = 5xy … (2)
From (1) , we get

2q + 4p = 5 ….(6)
Multiply (6) by 3, we get
6q +12p = 15 … (7)
Subtracting (5) from (7) , we get
6q + 12p – 6q – 3p = 15 – 6
9p = 9 ⇒ p = 1
Now, Substituting, p = 1 in (5) , we get
6q + 3(1) = 6

∴ Equations (1) and (2) can be expressed as :
10p + 2q = 4 = ⇒ 10p + 2q – 4 = 0 …(3)
15p – 5q = -2 ⇒ 15p – 5q + 2 = 0 … (4)

⇒ 2x = 6 ⇒ x = 3
Now, substituting the value of x in equation (5) , we get
3 + y = 5 ⇒ y = 2
Thus, the required solution is x = 3, y = 2

(viii)

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let the speed of Ritu in still water = x km/hr
and the speed of the current = y km/hr
∴ Downstream speed = (x + y) km/hr
Upstream speed = (x – y)km/hr

Substituting the value of x in equation (1), we get
6 + y = 10 ⇒ y = 10 – 6 = 4
Thus, speed of rowing in still water = 6 km/hr, speed of current = 4 km/hr

(ii) Let the time taken to finish the task by one woman alone = x days
and the time taken to finish the task by one man alone = y days


∴ 1 man can finish the work in 36 days and 1 woman can finish the work in 18 days,

(iii) Let the speed of the train = x km/hr
and the speed of the bus = y km/hr

Case I: Total journey = 300 km
Distance travelled by train = 60 km
Distance travelled by bus = (300 – 60)km
= 240 km
∵ Total time taken = 4 hours

Case II: Distance travelled by train = 100 km
Distance travelled by bus = (300 – 100)km
= 200 km
Total time = 4 hrs 10 mins

Thus, speed of the train = 60 km/hr
and speed of the bus = 80 km/hr

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.2

Question 1.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
(i) Using factor tree method, we have

∴ 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7

(ii) Using factor tree method, we have

∴ 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

(iii) Using factor tree method, we have

3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

(iv) Using factor tree method, we have

∴ 5005 = 5 × 7 × 11 × 13

(v) Using factor tree method, we have

∴ 7429 = 17 × 19 × 23

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) Using factor tree method, we have

⇒ 26 = 2 × 13 and 91 = 7 × 13
LCM of 26 and 91 = 2 × 7 × 13 = 182
HCF of 26 and 91 = 13 Now,
LCM × HCF = 182 × 13 = 2366 and 26 × 91 = 2366
i. e., LCM × HCF = Product of two numbers,

(ii) Using the factor tree method, we have

⇒ 510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
∴ LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
HCF of 510 and 92 = 2 LCM × HCF
= 23460 × 2 = 46920 and 510 × 92 = 46920
i. e., LCM × HCF = Product of two numbers.

(iii) Using the factor tree method, we have

336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
LCM of 336 and 54
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
HCF of 336 and 54 = 2 × 3 = 6
LCM × HCF = 3024 × 6 = 18144
Also 336 × 54 = 18144
Thus, LCM × HCF = Product of two numbers.

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12,15 and 21
(ii) 17,23 and 29
(iii) 8,9 and 25
Solution:
(i) We have

⇒ 12 = 2 × 2 × 3, 15 = 3 × 5 and 21 = 3 × 7
∴ HCF of 12, 15 and 21 is 3
LCM of 12,15 and 21 is 2 × 2 × 3 × 5 × 7 i.e., 420

(ii) We have
17 = 1 × 17, 23 = 1 × 23, 29 = 1 × 29
⇒ HCF of 17, 23 and 29 is 1 LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339

(iii) We have

⇒ 8 = 2 × 2 × 2, 9 = 3 × 3 and 25 = 5 × 5
∴ HCF of 8, 9 and 25 is 1
LCM of 8, 9 and 25 is 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
H.C.F × L.C.M = a × b
(a, b) × (a, b) = a × b
9 × x = 306 × 657
x = \(\frac{306 \times 657}{9}\)
∴ x = 22338
∴ L.C.M of (306, 657) = 22338

Question 5.
Check whether 6″ can end with the digit 0 for any natural number n.
Solution:
Here, n is a natural number and let 6ⁿ ends with digit 0.
∴ 6ⁿ is divisible by 5.
But the prime factors of 6 are 2 and 3. i.e.,
6 = 2 × 3
⇒ 6ⁿ = (2 × 3)ⁿ
i. e., In the prime factorisation of 6ⁿ, there is no factor 5.
So, by the fundamental theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order in which the prime factorisation occurs.
∴ Our assumption that 6ⁿ ends with digit 0, is wrong.
Thus, there does not exist any natural number n for which 6ⁿ ends with zero.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
i) 7 × 11 × 13 + 13 = 13(77 + 1)
= 13 × 78
∴ It is a composite number
[It is having more than two factors]

ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
=5[7 × 6 × 4 × 3 × 2 × 1 + 1]
= 5 (1008 + 1)
= 5 × 1009 = 5045
∴ It is a composite number.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken by Sonia to drive one round of the field = 18 minutes
Time taken by Ravi to drive one round of the field = 12 minutes
The LCM of 18 and 12 gives the exact number of minutes after which they meet again at the starting point.
We have

⇒ 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
∴ LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
Thus, they will meet again at the starting point after 36 minutes.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
We have the cumulative frequency distribution as follows:

Now, we plot the points corresponding to the ordered pair (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on a graph paper and join them by a free hand to get a smooth curve as shown below:

The curve so obtained is called the less than ogive.

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
Here, the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals.
We plot the points (ordered pairs) (38, 0), (40, 3), (42, 5), (44, 9), (46,14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them by a free hand to get a smooth curve.
The curve so obtained is the less than type ogive.

∵ N = 35
∴ \(\frac{N}{2}=\frac{35}{2}\) = 17.5
The point 17.5 is on y-axis.
From this point (i.e., from 17.5) we draw a line parallel to the x-axis which cuts the curve at P. From this point P, draw a perpendicular to the x-axis, meeting the x-axis at Question The point Q represents the median of the data which is 47.5.
Verification:
To verify the result, let us make the following table in order to find median using the formula :

Thus, the median = 46.5 kg is approximately

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village

Change the distribution to a more than type distribution, and draw its ogive.
Solution:
For more than type distribution, we have

Now, we plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) and join the points with a free hand to get a smooth curve.
The curve so obtained is the ‘more than type ogive’.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:

Question 3.
Evaluate:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
(ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
∵ sin 63° = sin (90° – 27°) = cos 27°
⇒ sin² 63° = cos² 27°
cos² 73° = cos² (90° – 17°) = sin² 17°
∴ \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}=\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}=1\)
[ ∵ cos² A + sin² A = 1]

(ii) sin 25° cos 65° + cos 25° sin 65°
∵ sin 25° = sin (90° – 65°) = cos 65° [ ∵ sin (90° – A) = cos A]
cos 25° = cos (90° – 65°) = sin 65° [ ∵ cos (90° – A) = sin A]
∴ sin 25° cos 65° + cos 25° sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= (cos 65°)² + (sin 65°)²
= cos² 65° + sin² 65° = 1 [∵ cos² A + sin² A = 1]

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec²A – 9 tan²A =
(A) 1
(B) 9
(C) 8
(D) 0

(ii) (1 +tanθ + secθ) (1 + cotθ – cosec0) =
(A) 0
(B) 1
(C) 2
(D) – 1

(iii) (sec A + tan A) (1 – sinA) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\)
(A) sec² A
(B) -1
(C) cot² A
(D) tan² A
Solution:
(i) (B): Since, 9 sec² A – 9 tan² A
= 9 (sec² A – tan² A) = 9 (1) = 9 [∵ sec² A – tan² A = 1]

(ii) (C): Here,

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.


Solution:

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.

Solution:
Median:
Let us prepare a cumulative frequency table:

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
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-135}{20}\)
Now, we have the following table:

∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σf_iu_i] = 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

We observe that the three measures are approximately equal.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Solution:
Here, we have N = 60
Now, cumulative frequency table is:

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

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.

Solution:
The given table is cumulative frequency distribution. We write the frequency distribution as given below :

∵ 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

Thus, the median age = 35.76 years.

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:

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:

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:

Find the median life time of a lamp.
Solution:
To compute the median, let us write the cumulative frequency distribution as given below:

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.

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:

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:

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

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

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.

Solution:
We have cumulative frequency table as follows:

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

Thus, the required median weight of the students = 56.67 kg.

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.5 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 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0
(ii) 2x + y = 5; 3x + 2y = 8
(iii) 3x – 5y = 20; 6x -10y = 40
(iv) x – 3y – 7 = 0; 3x – 3y – 15 = 0
Solution:
(i) For x – 3y – 3 = 0, 3x – 9y – 2 = 0
∴ a₁ = 1, b₁ = – 3, C₁ = – 3, a₂ = 3, b₂ = – 9, C₂ = -2

∴ The given system has no solution.

(ii) 2x + y – 5 = 0, 3x + 2y – 8 = 0
∴ a₁ = 2, b₁ = 1, c₁ = -5, a₂ = 3, b₂ = 2, c₂ = -8

∴ The given system has a unique solution.
Now, using cross multiplication method, we have

(iii) For 3x – 5y – 20 = 0, 6x – 10y – 40 = 0

∴ The given system of linear equation has infinitely many solutions

(iv) For x – 3y – 7 = 0, 3x – 3y – 15 = 0

∴ The given statement has unique solution. Now, using cross multiplication method, we have

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) We have, 2x + 3y = 7 and
(a – b) x + (a + b) y = (3a + b – 2)
⇒ 2x + 3y – 7 = 0
and (a – b) x + (a + b) y -(3a + b – 2) = 0
a₁ = 2, b₁ = 3, c₁ = – 7, a₂ = (a- b) , b₂ = (a + b) , c₂ = -(3a + b – 2)
For infinite number of solutions,

From the first two equations, we get

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9; 3x + 2y = 4
Solution:
Method-1 [Substitution method]:
8x + 5y = 9 … (1)
3x + 2y = 4 … (2)

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist)by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
(i) Let the fixed charges = ₹ x
and charges of food per day = ₹ y
For student A : Number of days = 20
∴ Cost of food for 20 days = ₹ 20y
According to the problem,
x + 20y = 1000
⇒ x + 20y – 1000 = 0 …. (1)
For student B : Number of days = 26
Cost of food for 26 days = ₹ 26y
According to the problem,
x + 26y = 1180
⇒ x + 26y – 1180 = 0 … (2)
Solving these by cross multiplication, we get

Thus x = 400 and y = 30
∴ Fixed charges = ₹ 400
and cost of food per day = ₹ 30

(ii) Let the numerator = x
and the denominator = y

From equations (1) and (2) , we have

(iii) Let the number of correct answers = ₹ x
and the number of wrong answers = ₹ y
Case-I: Marks for all correct answers
= (3 × x) = 3x
Mark for all wrong answers = (1 × y) = y
∴ According to the condition :
3x – y = 40 ⇒ 3x – y – 40 = 0
Case-II: Mark for all correct answers
= (4 × x)=
Marks for all wrong answers = (2 × y)
= 2y
∴ According to the condition :
4x – 2y = 50
⇒ 2x – y = 25 ⇒ 2x – y – 25 = 0 … (2)
From (1)and (2) , we have a₁ = 3, b₁ = -1, c₁ = -40, a₂ = 2, b₂ = -1, c₂ = -25

Now, total number of questions = [Number of correct answers] + [Number of wrong answers]
= 15 + 5 = 20
Thus, required number of questions = 20.

(iv) Let the speed of car-I be x km/hr.
and the speed of car-II be y km/hr.
Case-I:

Distance travelled by car-I = AC
∵ AC = time × speed = 5 × x km, AC = 5x
Distance travelled by car-II, BC = 5y
Since AB = AC – BC,
100 = 5x – 5y
⇒ 5x – 5y – 100 = 0
⇒ x – y – 20 = 0 …. (1)


Thus, speed of car-I = 60 km/hr
Speed of car-II = 40 km/hr

(v) Let the length of the rectangle = x units
and the breadth of the rectangle = y units
∴ Area of rectangle = x × y = xy

Condition-II:
(Length + 3) × (Breadth + 2) = Area + 67
⇒ (x + 3) (y + 2)= xy + 67 ⇒ 2x + 3y + 6 = 67
⇒ 2x + 3y – 61 = 0….(2)
Now, using cross multiplication method in (1)and (2) , where

Thus, length of the rectangle = 17 units and breadth of the rectangle = 9 units.

MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements

MP Board Class 10th Science Chapter 5 Intext Questions

Intext Questions Page No. 81

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.
Answer:
In Newland’s Octaves, the properties of lithium and sodium were found to be the same. This arrangement is also found in Dobereiner triads.

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:
Dobereiner could identify only three ‘triads’ from the elements known at that time. Hence this system of classification into triads was not found to be useful.

Question 3.
What were the limitations of Newlands’ Law of Octaves?
Answer:

The Limitations of Newlands’ Law of Oclaves is as follows:

1. It was found that the Law of Octaves was applicable only upto calcium, as after calcium every eighth element did not possess properties similar to that of the first.
2. It was assumed by Newlands’ that only 56 elements existed in nature and no more elements would be discovered in the future. But later on, several new elements were discovered, whose properties did not fit into the Law of Octaves.
3. In order to fit elements into his Table, Newlands adjusted two elements in the same slot, but also put some unlike elements under the same note.

Intext Questions Page No. 85

Question 1.
Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements: K, C, Al, Si, Ba.
Answer:

1. K is in I group. Its oxide is K₂O
2. C, is in IV group, its oxide is CO₂
3. Al, is in III group, its oxide is Al₂O₃
4. Si, is IV group, its oxide is SiO₂
5. Ba, is in II group, its oxide is BaO

Question 2.
Besides gallium, which other elements have since been discovered that were left by Mendeleev in his Periodic Table? (any two)
Answer:
Scandium and germanium.

Question 3.
What were the criteria used by Mendeleev in creating his Periodic Table?
Answer:
Mendeleev used the relationship between the atomic masses of the elements and their physical and chemical properties. Among chemical properties, he examined the compound formed by elements with oxygen and hydrogen. He found that if the 63 elements known at that time were arranged in the increasing order of their atomic masses, the properties of elements and also formulae of their oxides and hydrides gradually changed from element to element and at a certain interval they suddenly started almost repealing relationship was expressed by Mendeleev’s periodic law. i,e the properties of examinants are the periodic functions of their atomic masses.

Question 4.
Why do you think the noble gases are placed in a separate group?
Answer:
All noble gases are inert elements. Their properties are different from other elements and are the least reactive. Therefore, the noble gases are placed in a separate group.

Intext Questions Page No. 90

Question 1.
How could the Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:

1. In the Modern periodic Table atomic number of an elements is a more fundamental property than its atomic mass.
2. The anomalous position of hydrogen can be discussed after we see what are the basis on which the position of an elements in the Modern Periodic Table depends.
3. The elements present in any one group have the same number of valence electrons.
4. Atoms of different elements with the same number of occupied shells are placed in the same period.
5. In the Modern Periodic Table, a zig-zag line separated metals from non-metals.

Question 2.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer:
Calcium (Ca) and Strontium (Sr) is expected to show chemical reactions similar to magnesium (Mg). This is because the number of valence electrons (2) is the same in all of these three elements and since chemical properties are due to valence electrons, they show the same chemical reactions.

Question 3.
Name:

1. Three elements that have a single electron in their outermost shells.
2. Two elements that have two electrons in their outermost shells.
3. Three elements with filled outermost shells.

Answer:

1. Lithium (Li), Sodium (Na), and Potassium (K) has a single electron in their outermost shells.
2. Magnesium (Mg) and Calcium (Ca) have two electrons in their outermost shells.
3. Neon (Ne), Argon (Ar), and Xenon (Xe) have filled outermost shells.

Question 4.
Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
Answer:

Lithium, sodium and potassium – These three elements have one electron in their outermost orbit.

b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
Both helium (He) and neon (Ne) have filled outermost shells. Helium has a duplet in its K shells, while neon has an octet in its L shells.

Question 5.
In the Modern Periodic Table, which are the metals among the first ten elements?
Answer:
In the modem periodic table, Lithium and Beryllium are the metals among the first 10 elements.

Question 6.
By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic?
Ga Ge As Se Be
Answer:
Since, ‘Be’ lies to the extreme left-hand side of the periodic table, ‘Be’ is the most metallic among the given elements.

MP Board Class 10th Science Chapter 5 Ncert Textbook Exercises

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of the periodic table.
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl₂, which is solid with a high melting point. X would most likely be in the same group of the Periodic table as:
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3.
Which element has:
(a) two shells, both of which are completely filled with electrons?
(b) Electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:
(a) Neon
(b) Magnesium
(c) Silicon
(d) Boron
(e) Carbon

Question 4.
(a) What property do all elements in the same column of the Periodic table as Boron have in common?
(b) What property do all elements in the same column of the Periodic table as Fluorine have in common?
Answer:
(a) Valency equal to 3.
(b) Valency equal to 1.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)
N(7) F(9) P(15) Ar(18)
Answer:
(a) The atomic number of this element is 17.
(b) It would be chemically similar to F(9) with configuration as 2, 7.

Question 6.
The position of three elements A, B and C in the Periodic table is shown below:

(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
(a) A is a non-metal.
(b) C is less reactive than A because reactivity decreases down the group in halogens.
(c) C should be smaller in size than B as moving across a period, the nuclear charge increases and therefore, electrons come closer to the nucleus.
(d) A will form an anion as it will accept an electron to complete its octet.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer:
Nitrogen (7) : 2, 5
Phosphorus (15) : 2, 8, 5
Nitrogen is more electronegative because Metallic character decreases across a period and increased down a group.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modern Periodic table?
Answer:
In the modern periodic table, atoms with similar electronic configurations are placed in the same column. In a group, the number of valence electrons remains the same. Elements across a period show an increase in the number of valence electrons.

Question 9.
In the Modern Periodic table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21, and 38. Which of these have physical and chemical properties resembling calcium?
Answer:
The element with atomic number 12 has the same chemical properties as that of calcium. This is because both of them have same number of valence electrons (2).

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s Periodic table and the Modern Periodic table.
Answer:

Mendeleev s Periodic table:

1. 1. Elements are arranged in the increasing order of their atomic mass.
   2. This table has 8 groups and 6 periods. And each group is subdivided as an A and B.
   3. In this table, Hydrogen has no position.
   4. No position for isotopes, because in Mendeleev period these are not discovered.

Modern Periodic table:

1. 1. Elements are arranged in the increasing order of their atomic number.
   2. It has 18 groups and 7 periods.
   3. Inert gases are placed in separate groups.
   4. In this table, a zigzag line separates Metals from Non-metals.

(or)

Mendeleev’s Periodic table vs Modern Periodic table:

1. Elements are arranged in the increasing order of their atomic masses, while in Modern Periodic table elements are arranged in the increasing order of their atomic numbers.
2. There are a total of 7 groups (columns) and 6 periods (rows) while in Mendeleev’s’ Periodic Table, there are a total of 18 groups (columns) and 7 periods (rows).
3. Elements having similar properties were placed directly under one another, while in Mendeleev’s’ Periodic Table elements having the same number of valence electrons are present in the same group.
4. In Mendeleev’s Periodic Table the position of hydrogen could not be explained, while in Modern Periodic table hydrogen is placed above alkali metals.
5. No distinguishing positions for metals and non-metals in Mendeleev’s Periodic Table while in Modern Periodic Table metals are present at the left-hand side of the periodic table whereas nonmetals are present at the right-hand side.

MP Board Class 10th Science Chapter 5 Additional Questions

MP Board Class 10th Science Chapter 5 Multiple Choice Questions

Question 1.
Which of the following statements is correct about the trends when going down in a group of the periodic table?
(a) Elements become less electropositive in nature.
(b) Element oxides become more acidic.
(c) Valence electrons increases.
(d) Elements lose their electrons more easily.
Answer:
(d) Elements lose their electrons more easily.

Question 2.
Element A forms a chloride with the formula ACl₃, which is a stable compound. A would most likely be the same group of the Periodic Table as –
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(c) Al

Question 3.
Which of the following are coin metals?
(a) Ne, Ca, Na
(b) H₂, N₂, O₂
(c) Li, Na, K
(d) Cu, Au, Ag
Answer:
(d) Cu, Au, Ag

Question 4.
Who gave the triad arrangement of elements?
(a) Mendeleev
(b) Newlands
(c) Dalton
(d) Dobereiner
Answer:
(d) Dobereiner

Question 5.
Newlands periodic table is based on the
(a) Atomic weight
(b) Atomic number
(c) Atomic radius
(d) Atomic volume
Answer:
(a) Atomic weight

Question 6.
Which of the following is not gas in normal atmospheric condition?
(a) Helium (He)
(b) Argon (Ar)
(c) Bromine (Br)
(d) Chlorine (Cl)
Answer:
(c) Bromine (Br)

Question 7.
While moving left to right across a period, the atomic radii –
(a) Remains the same
(b) Approaches zero
(c) Decreases
(d) Increases first then decreases
Answer:
(c) Decreases

Question 8.
Which element is a metalloid?
(a) Carbon
(b) Nitrogen
(c) Oxygen
(d) Silicon
Answer:
(d) Silicon

Question 9.
Moseley’s periodic table is based on
(a) Atomic mass
(b) Mass number
(c) Atomic number
(d) Atomic volume
Answer:
(c) Atomic number

Question 10.
Which of the following is a group of highly electronegative elements?
(a) Cl, Br, I
(b) S, Se, Te
(c) Na, K, Rb
(d) Ca, Sr, Ba
Answer:
(a) Cl, Br, I

Question 11.
Which of the following elements is a non-metal?
(a) Aluminium
(b) Chlorine
(c) Sodium
(d) Silicon
Answer:
(b) Chlorine

Question 12.
As we move down in a group in Modern Periodic Table, the size of elements generally
(a) increases
(b) decreases
(c) remain the same
(d) first, increase then decrease
Answer:
(a) increases

Question 13.
As we move from top to bottom in a group in Modern Periodic Table, the electronegativity of elements
(a) Increases
(b) Decreases
(c) No change
(d) Not certain
Answer:
(b) Decreases

Question 14.
Which group of elements is considered most electropositive?
(a) Group 1
(b) Group 2
(c) Group 17
(d) Group 18
Answer:
(a) Group 1

Question 15.
Group 1 elements are also called as:
(a) Alkali metals
(b) Alkaline earth metals
(c) Halogens
(d) Noble gases
Answer:
(a) Alkali metals

Question 16.
Group 17 elements are also called as:
(a) Alkali Metals
(b) Alkaline Earth Metals
(c) Halogens
(d) Noble Gases
Answer:
(c) Halogens

Question 17.
How many elements were known when Mendeleev started his work?
(a) 100
(b) 215
(c) 65
(d) 80
Answer:
(c) 65

Question 18.
Why Mendeleev left spaces in his Periodic Table?
(a) A mistake
(b) For future elements
(c) For Isotopes
(d) For Isobars
Answer:
(b) For future elements

Question 19.
Why Lanthanoids and Actinoids are placed below in the Periodic Table?
(a) A mistake
(b) Better representation and view
(c) They were found very recently
(d) All of the above
Answer:
(c) They were found very recently

Question 20.
A period may have elements with –
(a) Variable atomic sizes
(b) Variable atomic number
(c) Variable valency
(d) All of the above
Answer:
(d) All of the above

Question 21.
Element A belongs to group 15. The formula of its hydride will:
(a) AH
(b) AH₂
(c) AH₃
(d) A₃H
Answer:
(c) AH₃

Question 22.
An electropositive element, A with 2 valence electron will form which type of oxide?
(a) AO
(b) A₂O
(c) AO₂
(d) AO₃
Answer:
(a) AO

Question 23.
Most electronegative element in our Periodic Table:
(a) Iron
(b) Nitrogen
(c) Carbon
(d) Flourine
Answer:
(d) Flourine

Question 24.
Which of the following elements where not among metals/elements named to fill the gap of Mendeleev’s Periodic Tablespaces?
(a) Cobalt
(b) Scandium
(c) Gallium
(d) Germanium
Answer:
(a) Cobalt

Question 25.
In a group, all elements have similar ………..
(a) Electronic configuration
(b) Valence electron
(c) Electronegativity
(d) All of these
Answer:
(b) Valence electron

Question 26.
Which among the following is a noble gas?
(a) C
(b) N
(c) O
(d) Ne
Answer:
(d) Ne

Question 27.
Which of the following elements has electronic configuration E = 2, 6?
(a) C
(b) N
(c) O
(d) Ne
Answer:
(c) O

Question 28.
Which of the following elements is a metalloid?
(a) B
(b) Al
(c) S
(d) P
Answer:
(a) B

Question 29.
Which of the following is not a halogen?
(a) Br
(b) I
(c) Te
(d) At
Answer:
(c) Te

Question 30.
Which element has a total of two shells, with four valence electrons?
(a) C
(b) N
(c) Br
(d) Co
Answer:
(a) C
(ii) Match column A’s description with column B’s Particulars.

Answers:

1. → 12
2. → 3
3. → 2
4. → 4
5. → 5
6. → 13
7. → 11
8. → 14
9. → 10
10. → 10
11. → 6
12. → 1
13. → 7
14. → 8

MP Board Class 10th Science Chapter 5 Very Short Answer Type Questions

Question 1.
What would be the maximum number of electrons present in the outermost shell of atoms in the first period of Periodic Table?
Answer:
Two

Question 2.
What is Solder?
Answer:
It is an alloy of lead (Pb) and tin (Sn).

Question 3.
What is anode mud?
Answer:
During electrolytic refining, the soluble impurities go into the solution, Whereas, the insoluble impurities settle down at the bottom of anode and are known as anode mud.

Question 4.
Which metal is used with iron oxide to join railway tracks or cracked machine parts?
Answer:
Aluminium.

Question 5.
Give the thermit reaction.
Answer:
Fe₂O₃(s) + 2Al(s) → 2 Fe(l) + Al₂O₃(s) + Heat

Question 6.
Roasting is used for the extraction of which ore?
Answer:
Sulphide ore.

Question 7.
Name the metal lowest inactivity series (relative reactivities of metals).
Answer:
Au or Gold.

Question 8.
Which gas is evolved when a metal reacts with nitric acid?
Answer:
Hydrogen gas.

Question 9.
Name any two metal that does not react with water at all.
Answer:
Lead, copper, gold, silver. (any two)

Question 10.
Complete the following reaction: Metal oxide + water →.
Answer:
Metal hydroxide.

Question 11.
Which material is used to coat electric wires in homes?
Answer:
PVC or Polyvinylchloride.

Question 12.
Name any two metals that are poor conductors of heat.
Answer:
Lead and mercury.

MP Board Class 10th Science Chapter 5 Short Answer Type Questions

Question 1.
What are the limitations of the Modern Periodic Table?
Answer:
The limitations of the Modern Periodic Table:
Position of hydrogen still dicey. It is not fixed till now. Position of lanthanides and actinides has not been given inside the main body of the Periodic Table. It does not reflect the exact distribution of electrons of some of the transition and inner transition elements.

Question 2.
Two elements X and Y have atomic numbers 12 and 16 respectively. Write the electronic configuration for these elements. To which period of the Modern Periodic Table do these two elements belong? What type of bond will be formed between them and why?
Answer:
Electronic configuration of X (Z= 12): 2, 8,2
Electronic configuration of Y (Z = 16): 2, 8,6
Both these elements belong to the third period. An ionic bond is formed between X and Y due to the transfer of two electrons from X to Y.

Question 3.
The present classification of elements is based on which fundamental property of elements?
Answer:
Atomic number.

Question 4.
Li, Na and K are the elements of a Dobereiner’s Triad. If the atomic mass of Li is 7 and that of K is 39, what would be the atomic mass of Na?
Answer:
According to of Dobereiner’s law of triads, the atomic mass of the middle element, in this case, Na, should be the arithmetic mean of Li and K. Thus, Arithmetic mean of Li and K = (7 + 39)/2 = 23.

Question 5.
Define Dobereiner’s law of triads.
Answer:
It states, “When elements are placed in order of the ascending order of atomic masses, groups of three elements having similar properties are obtained. The atomic mass of the middle element of the triad is equal to the mean of the atomic masses of the other two elements of the triad.”

Question 6.
Why did Dobereiner’s system of classification fail?
Answer:
The major drawback of Döbereiner’s classification was that it was valid only for a few groups of elements known during that time. He was able to identify three triads only. Also, more accurate measurements of atomic masses showed that the mid element of the triad did not really have the mean value of the sum of the other two elements of the triad. For elements of very low mass or very high mass, the law did not hold good. For example, Fluorine (F), Chlorine (Cl), Bromine (Br). The atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br.

Question 7.
Explain the position of metalloids in the Modern Periodic Table.
Answer:
In the Modern Periodic Table, a zig-zag line separates metals from non-metals. The borderline elements – boron, silicon, germanium, arsenic, antimony, tellurium and polonium – are intermediate in properties and are called metalloids or semi-metals.

Question 8.
Why silicon is classified as metalloid?
Answer:
Silicon is classified as a semi-metal or metalloid because it exhibits some properties of both metals and non-metals.

Question 9.
State Newlands law of octaves.
Answer;
Elements are arranged in increasing order of their atomic masses such that the properties of the eighth element are the repetition of the properties of the first element (similar to eighth note in an octave of music).

Question 10.
X and Y are the two elements having similar properties which obey Newlands law of octaves. How many elements are there in between X and Y?
Answer:
The law states there are eight elements in an octave (row). A number of elements between X and Y are six.

Question 11.
What are the drawbacks of Newlands law of octaves?
Answer:
Following are the major drawbacks:

1. Worked well with lighter elements (upto calcium. After those elements in the eighth column did not possess properties similar to elements in the first column.
2. Newland assumed only 56 elements existed so far. Later, new elements were discovered which did not fit into octaves table.
3. Newland adjusted few elements in the same slot through their properties were quite different, e.g., Cobalt and nickel are in the same slot and these are placed in the same column as fluorine, chlorine and bromine which have very different properties than these elements. Iron, which resembles cobalt and nickel in properties, has been placed far away from these elements.

MP Board Class 10th Science Chapter 5 Long Answer Type Questions

Question 1.
What are the salient features of the Modern Periodic Table?
Answer:
In a period of the Periodic Table, the number of valence electrons increases as the atomic number increases. As a result, elements change from metal to metalloid to nonmetal to a noble gas. Atomic size is a periodic property. As atomic number increases in a period, the atomic radius decreases. As atomic number increases in a group, atomic radius increases.

Positive ions have smaller atomic radii than the neutral atoms from which they derive. Negative ions have larger atomic radii than their neutral atoms. Positive ions in the same group increase in size down the group. In a group, each element has the same number of valence electrons. As a result, the elements in a group show similar chemical behaviour.

Metallic character decreases from left to right in a period because of the increase in the effective nuclear charge. Non-metallic character increase from left to right in a period because of the increase in effective nuclear charge. Non-metallic character decreases down the group because of increase in the size of the atom.

Question 2.
What periodic trends do we observe in terms of atomic radii or atomic sizes in Modern Periodic table?
Answer:
Following two trends are observed:
1. Within each column (group), atomic radius tends to increase from top to bottom. This trend results primarily from the increase in the number of the outer electrons. As we go down a column, the outer electrons have a greater probability of being farther from the nucleus, causing the atom to increase in size.

2. Within each row (period), the atomic radius tends to decrease from left to right. The major factor influencing this trend is the increase in the nuclear charge as we move across a row. The increasing effective nuclear charge steadily draws the valence electrons closer to the nucleus, causing the atomic radius to decrease.

Question 3.
An element A with atomic number 19 combines separately with NO₃^–and (SO₄)₂^–,(PO₄)₃^–radicals:
(a) Give the electronic configuration of element A.
(b) Write the formulae of the three compounds so formed.
(c) To which group of the periodic table does the element ‘R’ belong?
(d) Does it form covalent or ionic compound? Why?
Answer:
(a) Electronic configuration of A: 2,8, 8, 1.
(b) Compounds formed are A(NO₃), A₂(SO₄) and K₃(PO₄).
(c) A has one valence electron and hence, it belongs to the first group.
(d) It forms the ionic compound.

Question 4.
Describe types of periods, blocks and trends of periodic properties along periods associated with Modern Periodic Table.
Answer:
Periods:
First period (Atomic number 1 and 2): This is the shortest period. It contains only two elements (hydrogen and helium).

Second Period: (Atomic number 3 to 10): It contains eight elements (lithium to neon).

Third period (Atomic number 11 to 18): It contains eight elements (sodium to argon).

Fourth period (Atomic number 19 to 36): Row contains eighteen elements (potassium to krypton). i.e., 8 normal elements and 10 transition elements.

5th period (Atomic number 37 to 54): Contains 18 elements (rubidium to xenon) includes 8 normal elements and 10 transition elements.

Sixth period (Atomic number 55 to 86): The longest period. It contains 32 elements (caesium to radon) has 8 normal elements, 10 transition elements and 14 inner transition elements (lanthanides).

7th period (Atomic number 87 to 118): As like the sixth period, this period also can accommodate 32 elements. Till now 26 elements have been authenticated by IUPAC.

Blocks in Periodic Table:
The periodic table includes “blocks” defined in terms of which type of orbital is being filled via the Aufbau principle. This gives us the s-block, p-block, d-block, and f-block.

Blocks:
The s-, p-, d-, and f-blocks contain elements with outer electrons in the same type of orbital. Another key link between electron arrangement and position in the periodic table is that elements in any one main group have the same number of electrons in their highest energy level. The number of elements discovered so far is 118. The last element authenticated by IUPAC is Cn112 (Copernicium).

Properties of Periods: As you proceed to the left in a period or as you proceed down within a group:

1. The metallic strength increases (Non-Metallic Strength decreases).
2. The atomic radius increases.
3. The ionization potential decreases.
4. The electron affinity decreases.
5. The electronegativity decreases.

MP Board Class 10th Science Chapter 5 Textbook Activities

Class 10 Science Activity 5.1 Page No. 84

1. Looking at its resemblance to alkali metals and the halogen family, try to assign hydrogen a correct position in Mendeleev’s Period Table.
2. To which group and period should hydrogen be assigned?

Observations:

1. No position can be fixed for hydrogen in the Mendeleev’s Periodic Table.
2. Properties of hydrogen fit with alkali metal as it combines with halogens, oxygen and sulphur to form compounds.
3. Properties of hydrogen also fit or are similar to halogen as it exists in the form of diatomic molecules and combines with metals and non-metals forming covalent compounds.

Class 10 Science Activity 5.2 Page No. 85

1. Consider the isotopes of chlorine, Cl-35 and CI-37.
2. Would you place them in different slots because their atomic masses are different?
3. Or would you place them in the same position because their chemical properties are the same?

Observations:
Two isotopes of chlorine are Cl-35 and Cl-37. Both isotopes have the same chemical properties and hence, both isotopes should be placed in the same position.

Class 10 Science Activity 5.3 Page No. 85

1. How were the positions of cobalt and nickel resolved in the Modern Periodic Table?
2. How were the positions of isotopes of various elements decided in the Modern Periodic Table?
3. Is it possible to have an element with atomic number 1, 5 placed between hydrogen and helium?
4. Where do you think should hydrogen be placed in the Modern Periodic Table?

Observations:
The position of Cobalt and Nickel were decided by placing them in the increasing order of atomic number in the Modern Periodic Table. Since isotopes are elements with the similar atomic number they are placed in the same position as its basic elements in the modern periodic table.

Class 10 Science Activity 5.4 Page No. 87

1. Look at group 1 of the Modern Periodic Table, and name the elements present in it. Write down the electronic configuration of the first three elements of group 1.
2. What similarity do you find in their electronic configurations?
3. How many valence electrons are present in these three elements?

Observations:
Elements present in Group 1 are:

Electronic configuration of the first three elements of Group I are as below:

Class 10 Science Activity 5.6 Page No. 87

1. If you look at the Modern Periodic Table, you will find that the elements Li, Be, B, C, N, O, F, and Ne are present in the second period. Write down their electronic configurations.
2. Do these elements also contain the same number of valence electrons?
3. Do they contain the same number of shells?

Observations:
No, these elements contain variable valence electrons as they belong to different groups:

Class 10 Science Activity 5.6 Page No. 88

1. How do you calculate the valency of an element from its electronic configuration?
2. What is the valency of magnesium with atomic number 12 and sulphur with atomic number 16?
3. Similarly, find out the valencies of the first twenty elements.
4. How does the valency vary in a period on going from left to right?
5. How does the valency vary in going down a group?

Observations:

1. Valency of an element can be calculated by the numbers of valence electron present.
2. Valency of Magnesium: 2
3. Valency of Sulphur: 2 Variation of valency while moving left to right in a period.
   1 → 2 → 3 → 4 → 3 → 2 → 1 → 0
4. Variation of valency while going down in a group does not change.

Class 10 Science Activity 5.7 Page No. 88

1. Atomic radii of the elements of the second period are given below:
   
2. Arrange them in decreasing order of their atomic radii.
3. Are the elements now arranged in the pattern of a period in the Periodic Table?
4. Which elements have the largest and the smallest atoms?
5. How does the atomic radius change as you go from left to right in a period?

Observations:
Decreasing order of atomic radii of following elements:

1. O < N < C < B < Be < Li
2. No in pattern.
3. Oxygen is smallest as per given data while Li is largest.
4. Atomic radius reduces while moving right in a group a, nuclear charge increase.

Class 10 Science Activity 5.8 Page No. 89

1. Study the variation in the atomic radii of first group elements given below and arrange them in increasing order.
   
2. Name the elements which have the smallest and the largest atoms.
3. How does the atomic size vary as you go down a group?

1. Sodium (Na) has the smallest atom and calcium (Ca) has the largest atom.
2. Atomic size increases as we go down a group.

Class 10 Science Activity 5.9 Page No. 89

1. Examine elements of the third period and classify them as metals and non-metals.
2. On which side of the Periodic Table do you find the metals?
3. On which side of the Periodic Table do you find the non-metals?

Observations:
Elements of the third period are:

Class 10 Science Activity 5.10 Page No. 89

1. How do you think the tendency to lose electrons change in a group?
2. How will this tendency change in a period?

Observations:
Metallic property reduces while moving right in a period.

Class 10 Science Activity 5.11 Page No. 90

1. How would the tendency to gain electrons change as you go from left to right across a period? How
2. would the tendency to gain electrons change as you go down a group?

Observations:

1. The electrons increases as we go left to right in a period up to 17th group. It decreases in the 18th
2. group. The tendency of gaining the electrons decreases as we go down a group.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.5 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) The sides are : 7 an, 24 cm, 25 cm
Here, (7 cm)² = 49 cm²
(24 cm)² = 576 cm²
(25 cm)² = 625 cm²
∵ (49 + 576)cm² = 625 cm²
∴ It is a right triangle.
Hypotenuse = 25 cm.

(ii) The sides are: 3 cm, 8 cm, 6 cm
Here, (3 cm)² = 9 cm²
(8 cm)² = 64 cm²
(6 cm)² = 36 cm²
∵ (9 + 36) ≠ 64 cm²
∴ It is not a right triangle.

(iii) The sides are : 50 cm, 80 cm, 100 cm
Here, (50 cm)² = 2500 cm²
(80 cm)² = 6400 cm²
(100 cm)² = 10000 cm²
∵ (2500 + 6400) cm² ≠ 10000 cm²
∴ It is not a right triangle.

(iv) The sides are : 13 cm, 12 cm, 5 cm
Here, (13 cm)² = 169 cm²
(12 cm)² = 144 cm²
(5 cm)² = 25 cm²
∵ (144 + 25)cm² = 169 cm²
∴ It is a right triangle.
Hypotenuse = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR.
Solution:
In ∆QMP and ∆QPR,
∠QMP = ∠QPR [Each 90°]
∠Q = ∠Q [Common]
⇒ ∆QMP ~ ∆QPR …. (1) [AA similarity]
Again, in ∆PMR and ∆QPR,
∠PMR = ∠QPR [Each = 90°]
∠R = ∠R [Common]
⇒ ∆PMR ~ ∆QPR …… (2) [AA similarity]
From (1) and (2), we have

Question 3.
In the figure, ABD is a triangle, right angled at A and AC ⊥ BD.

Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
We have, right ∆ABC such that ∠C = 90° and AC = BC.

∴ By Pythagoras theorem, we have AB² = AC² + BC² = AC² + AC² = 2AC²
[∵ BC = AC (given)]
Thus, AB² = 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.
Solution:
We have, an isosceles AABC such that BC = AC.

Also, AB² = 2AC²
∴ AB² = AC² + AC²
But AC =BC
∴ AB² = AC² + BC²
∴ Using the converse of Pythagoras theorem, ∠ACB = 90°
f.e., ∆ABC is a right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
In equilateral triangle, altitude bisects the base.
⇒ AD = DB

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Let us have a rhombus ABCD.

∵ Diagonal of a rhombus bisect each other at right angles.
∴ OA = OC and OB = OD
Also, ∠AOB = ∠BOC [Each = 90°]
And ∠COD = ∠DOA [Each = 90°]
In right ∆AOB, we have,
AB² = OA² + OB² …… (1)
[Using Pythagoras theorem]
Similarly, in right ∆BOC,
BC² = OB² + OC² …… (2)
In right ∆COD,
CD² = OC² + OD² …… (3)
In right ∆AOD,
DA² = OD² + OA² ……. (4)
Adding (1), (2), (3) and (4)
AB² + BC² + CD² + DA²
= [OA² + OB²] + [OB² + OC²] + [OC² + OD²] + [OD² + OA²]
= 2OA² + 2OB² + 2 OC² + 2OD² = 2[OA² + OB² + OC² + OD²]
= 2[OA² + OB² + OA² + OB²]

Thus, sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question 8.
In the figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²

Solution:
We have a point in the interior of a ∆ABC such that
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
(i) Let us join OA, OB and OC.

In right ∆OAF, by Pythagoras theorem
OA² = OF² + AF² …(1)
Similarly, from right triangle ODB and OEC, we have
OB² = BD² + OD², …(2)
and OC² = CE² + OE² …(3)
Adding (1), (2) and (3), we get OA² + OB² + OC²
= (AF² + OF²) + (BD² + OD²) + (CE² + OE²)
⇒ OA² + OB² + OC²
= AF² + BD² + CE² + (OF² + OD² + OE²)
⇒ OA² + OB² + OC² – (OD² + OE² + OF²)
= AF² + BD² +CE²
⇒ OA² + OB² + OC² – OD² – OE² – OF²
= AF² + BD² + CE²

(ii) In right triangle OBD and triangle OCD, by Pythagoras theorem:
OB² = OD² + BD² and OC² = OD² + CD²
⇒ OB² – OC² = OD² + BD² – OD² – CD²
⇒ OB² – OC² = BD² – CD² ….. (1)
Similarly, we have
OC² – OA² = CE² – AE² …… (2)
and OA² – OB² = AF² – BF² ….. (3)
Adding (1), (2) and (3), we get (OB² – OC²) + (OC² – OA²) + (OA² – OB²) = (BD² – CD²) + (CE² – AE²) + (AF² – BF²)
⇒ 0 = BD² + CE² + AF² – (CD² + AE² + BF²)
⇒ BD² + CE² + AF² = CD² + AE² + BF²
or AF² + BD² + CE² = AE² + BF² + CD²

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Let PQ be the ladder and PR be the wall

⇒ PQ = 10 m, PR = 8 m
Now, in the right ∆PQR, PQ² = PR² + QR²
⇒ 10² = 8² + QR²
[using Pythagoras theorem]
⇒ QR² = 10² – 8² = (10 + 8)(10 – 8)
= 18 × 2 = 36
QR = \(\sqrt{36}\) = 6m
Thus, the distance of the foot of the ladder from the base to the wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is the wire and BC is the vertical pole. The point A is the stake.

Now, in the right AABC, using Pythagoras Theorem, we have

Thus, the stake is required to be taken at \(6 \sqrt{7}\)m from the base of the pole to make the wire taut.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \(1 \frac{1}{2}\) hours?
Solution:
Let the point A represent the airport. Plane-I fly towards North,
∴ Distance of the plane-I from the airport after \(1 \frac{1}{2}\) hours = speed × time

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops?
Solution:
Let the two poles AB and CD are such that the distance between their feet AC = 12m.
∵ Height of pole-1, AB = 11 m

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Solution:
We have a right ∆ABC such that ∠C = 90°.
Also, D and E are points on CA and CB respectively.
rain We have a right ∆ABC such that ∠C = 90°.

Let us join AE and BD.
In right ∆ACB, using Pythagoras theorem
AB² = AC² + BC² …… (1)
In right ∆DCE, using Pythagoras theorem,
DE² = CD² + CE² …… (2)
Adding (1) and (2), we get
AB² + DE² = [AC² + BC²] + [CD² + CE²]
= AC² + BC² + CD² + CE² = [AC² + CE²] + [BC² + CD²] …. (3)
In right ∆ACE,
AC² + CE² = AE² …… (4)
In right ∆BCD,
BC² + CD² = BD² …….. (5)
From (3), (4) and (5), we have AB² + DE² = AE² + BD² or AE² + BD² = AB² + DE²

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB² = 2AC² + BC².

Solution:

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD² = 7AB².
Solution:
We have an equilateral ∆ABC; in which D is a point on BC such that BD = \(\frac{1}{3}\) BC.

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
We have an equilateral ∆ABC, in which AD ⊥ BC.
Since, an altitude in an equilateral A, bisects the corresponding side.

∴ D is the mid-point

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = \(6 \sqrt{3}\) cm, AC = 12 cm and BC = 6 cm. The angle B is
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
(C): We have, AB = \(6 \sqrt{3}\) cm, AC = 12 cm, and BC = 6 cm