Month: October 2024

Jharkhand Board JAC Class 10 Science Important Questions Chapter 5 Periodic Classification of Elements Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 5 Periodic Classification of Elements

Additional Questions and Answers

Question 1.
Answer the following questions in short:
(1) From the following elements :
20^Ca, 3^Li, 11^Na, 10^Ne
(a) Select the element which has two shells, both of which are completely filled with electrons.
(b) Select two elements of the same group.
Answer:
(a) Element in which two shells are completely filled with electrons is 10^Ne (2, 8).
(b) Two elements of the same group are 3^Li(2, 1) and 11^Na (2, 8, 1).

(2) Answer the following questions [for an element having atomic number 17] :
(a) Name the element.
(b) In which period will you find this element?
(c) To which group of the periodic table does this element belong?
(d) State the electronic configuration of the element.
Answer:
(a) Chlorine
(b) Third period
(c) Group 17
(d) Electronic configuration : 2, 8, 7

(3) An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide.
(a) What is the position of elements X and Y in the periodic table?
(b) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed.
Answer:
Element X is a non-metal while element Y is a metal.
Molecular formula of dihalide is YX₂.
(a)

Position	Element X	Element Y
In a group	17	2
In a period	3	4

(b) Basic oxide : YO
Nature of the bond : Ionic

(4) Two elements M and N belong to the same period of the modern periodic table and are in group I and group II respectively. Compare their following properties:
(a) Atomic size
(b) Metallic character
(c) Valency of oxides
(d) Molecular formula of their chlorides
Answer:
(a) Element ‘M’ has larger atomic size than that of ‘N’.
(b) Element ‘M’ possess more metallic? character than that of ‘N’.
(c) Group I : Valency : 1
Group II : Valency : 2
(d) MCl, MCl₂

(5) A part of the periodic table has been shown below :

Answer the following questions on the basis of position of elements in the above table:
(a) Which element is a noble gas? Give reason.
(b) Which element is most electronegative? Give reason.
(c) Write the electronic configuration of B and E.
Answer:
(a Element G is a noble gas, because it is present in group 18 and has zero valency.
(b) Element E is the most electronegative element due to its small size and greater tendency to gain electrons.
(c) Electronic configuration of ‘B’ = 2,8, 1 Electronic configuration of ‘E’ = 2, 7

(6) The positions of elements A, B, C, D, E, F, G and H in their respective groups are as follows :

Group	1	2	13	14	15	16	17	18
Element	A	B	C	D	E	F	G	H

Answer the following questions:
(a) Which elements have the largest and smallest atomic size?
(b) Which elements have the valency 3 and 0 respectively?
Answer:
(a) Largest atomic size : A
Smallest atomic size : G
(b) Valency 3 : Element C
Valency O : Element H

(7) Consider the part of periodic table given below and answer the following questions :

(i) State the most reactive metal.
(ii) How many shells does element d have?
(iii) Identify the element having valency 2.
(iv) Write the number of electrons in the valence shell of j.
(v) Out of h and i, which element is more non-metallic in nature?
(vi) Out of e and h, which element possess large atomic size?
Answer:
(i) d
(ii) 4
(iii) e and g
(iv) 2
(v) h
(vi) e

(8)

Using the above table, answer the following questions:
(i) Which element will form only covalent compounds?
(ii) Which element is a metal with valency 3?
(iii) Which element is a non-metal with valency 3?
(iv) Out of D and E, which one has a bigger atomic size?
(v) Write common name for the family of elements, C and F.
Answer:
(i) E
(ii) D
(iii) B
(iv) D
(v) Inert gases or noble gases

Question 2.
Distinguish between the following :
(1) Elements of a group and Elements of a period
Answer:

Elements of a group	Elements of a period
1. The atomic number of elements increases on moving down the group.	1. The atomic number of elements increases on moving from left to right along a period.
2. All the elements in a group have same number of valence electrons.	2. The number of valence electrons of the elements in a period increases.
3. The chemical reactivity of elements in a group are same or identical.	3. The chemical reactivity of elements in a period are not identical.
4. The atomic radius and metallic character increase on moving down the group.	4. The atomic radius and metallic character decrease on moving from left to right along a period.

(2) Mendeleev’s periodic table and the Modern periodic table
Answer:

Mendeleev’s periodic table	Modern periodic table
1. Mendeleev’s periodic table consists of seven periods and eight groups.	1. Modern periodic table consists of seven periods and eighteen groups.
2. Transition elements are not separated in the Mendeleev’s periodic table.	2. Transition elements are placed in a separate groups in the modern periodic table.
3. In Mendeleev’s periodic table, elements are arranged in increasing order of their atomic masses.	3. In the modern periodic table, elements are arranged in increasing order of their atomic numbers.
4. Period number and group number of an element cannot be predicted.	4. Period number and group number of an element can be determined easily.
5. Mendeleev’s periodic table has descripancies and limitations.	5. Modern periodic table is almost errorless.
6. Periodicity in the properties of elements cannot be explained.	6. Periodicity in the properties of elements can be explained.

(3) Metallic elements and Non-metallic elements
Answer:

Metallic elements	Non-metallic elements
1. They are electropositive elements.	1. They are electronegative elements.
2. Metals have tendency of losing the electrons during the bond formation process.	2. Non-metals have tendency of gaining the electrons during the bond formation process.
3. Oxides of metals are basic.	3. Oxides of non-metals are acidic.
4. Metals possess 1, 2 or 3 electrons in their respective valence shells.	4. Non-metals possess 5, 6 or 7 electrons in their respective valence shells.

Question 3.
Give scientific reasons for the following statements:
(1) Dobereiner’s triads could not arrange all the elements known at that time.
Answer:
In Dobereiner’s triads, three elements were arranged in the order of increasing atomic masses of elements in which the atomic mass of the middle element is the average of the atomic masses of the other two elements.

Dobereiner could identify triads from the elements known at that time was only a coincidence. Thus, this classification of elements into triads could not be applied to all the elements known at that time.

(2) Newlands’ law of octaves could not classify all the elements known at that time.
Answer:
Newlands’ law of octaves was applicable upto calcium. After calcium, every eighth element s did not possess properties similar to that of the first element.

• Newlands’ assumed that only 56 elements existed s in nature.
• Newlands’ law of octaves was found correct only to lighter elements.
• Thus, all the elements could not be classified s by the Newlands’ law of octaves.

(3) No fixed position can be assigned to hydrogen in the periodic table.
Answer:
Electronic configuration of hydrogen resembles with alkali metals.

• Like alkali metals, hydrogen combines with halogens, oxygen and sulphur to form compounds having similar formulae.
• Hydrogen also resembles halogens as it exists in the form of diatomic molecules.
• Hydrogen combines with metals and non-metals to form ionic and covalent bonds respectively.
• Thus, hydrogen cannot be assigned fixed position in the periodic table.

(4) The atomic size decreases in a period on moving from left to right.
Answer:
As moving from left to right in a period, the atomic number of elements increases by 1.

• As atomic number increases, nuclear charge also increases.
• Due to this increased nuclear charge, the electrons are attracted strongly towards the nucleus and hence the atomic size decreases.

(5) On moving down in a group, the atomic radii of elements increases gradually.
Answer:
On moving down in a group, the atomic number of element increases.

• A new shell of electrons is added with increase in atomic numbers of elements.
• Thus, the distance between the valence shell and nucleus increases and atomic size increases down the group inspite of the increase in nuclear charge.

Objective Questions and Answers

Question 1.
Answer the following questions in one word:

1. How many elements are known till date?
2. How many amongst the known elements are naturally occurring?
3. In which two types the elements were classified in early attempts?
4. By which name the law proposed by Dobereiner is known?
5. How many elements were classified by Dobereiner’s triads?
6. Cl, X and I are the elements of Dobereiner’s triad, then what would be the element X?
7. Name the elements Newlands started and ended the classification of elements.
8. With what Newlands’ law of octaves was compared?
9. Name the element having similar property with lithium in Newlands’ law of octaves.
10. Which element possess similar property with boron in Newlands’ law of octaves?
11. Name two elements which were placed in the same slot in Newlands’ octaves.
12. How many elements were known when Mendeleev started the classification of elements?
13. Name two elements which formed compound on which Mendeleev had concentrated.
14. What are the vertical columns and the horizontal rows called in Mendeleev’s periodic table?
15. Which Sanskrit prefix was used by Mendeleev in the naming of elements which were not discovered at that time?
16. Which element was not placed in Mendeleev’s periodic table properly?
17. Which scientist proposed tire ‘periodic law’ for the modern periodic table?
18. Write the number of elements present in first, third and fourth period in the modern periodic table.
19. State the number of elements present in sixth period of the modern periodic table.
20. How many groups are present in the modern periodic table?
21. How many elements exist as gases in the modern periodic table?
22. By what names the elements of group I are known in the modern periodic table?
23. By which formula the maximum number of electrons that can be accommodated in a shell determines?
24. Write the valency of an element having atomic number 13.
25. Name the elements in the modern periodic table having lowest and highest atomic radii.
26. State the radius of a hydrogen atom.

Answer:

1. 118
2. 98
3. Metals and non-metals
4. Law of Triads
5. 9
6. Br
7. Hydrogen and thorium
8. Indian musical notes
9. Sodium
10. Aluminium
11. Cobalt and nickel
12. 63
13. Hydrogen and oxygen
14. Groups and periods
15. Eka
16. Hydrogen
17. Henry Moseley
18. 2, 8, 18
19. 32
20. 18
21. 11
22. Alkali metals
23. 2n²
24. 3
25. He and Fr
26. 39

Question 2.
Define :
(1) Isotopes
Answer:
Atoms of the same element having same atomic number but different atomic masses are known as isotopes.

(2) Periodic properties
Answer:
Properties of the elements which are periodic function of their electronic configuration and are repeated after definite interval of atomic numbers.
OR
Properties which show the regular gradation in the same group (moving from top to bottom) or along a period (moving from left to right) are known as periodic properties.

(3) Valency
Answer:
The valency is the combining capacity of an atom of an element to acquire noble gas configuration.

(4) Atomic radius
Answer:
The distance between the centre of the nucleus and the outermost shell of an isolated atom is called atomic radius.

(5) Metalloids (Semi-metal)
Answer:
Elements which possess intermediate properties of both metals and non-metals are called metalloids or semi-metals.

Question 3.
Fill in the blanks :

1. Lithium, sodium and …………………. are the members of Dobereiner’s triad.
2. Newlands’ law of octaves is applicable for …………………. elements.
3. According to Newlands, …………………. elements occur in nature.
4. Mendeleev named scandium as ………………….
5. The element known as eka-silicon is ………………….
6. The position of cobalt in Mendeleev’s periodic table is prior to ………………….
7. If the valency of an element is 2, then it lies in group ………………….
8. The electronic configuration of an element is 2, 8, 3, then it is an element of …………………. period.
9. In the modern periodic table, noble gases are placed in group ………………….
10. Modern periodic table consists of …………………. periods and …………………. groups.
11. Position of …………………. in the periodic table is controversial.
12. The valency of noble gases is ………………….
13. On moving down in any group, the metallic character of the elements ………………….
14. Oxides of non-metallic elements are …………………. in nature.
15. Oxides of …………………. elements are basic in nature.

Answer:

1. potassium
2. lighter
3. 56
4. eka-boron
5. germanium
6. nickel
7. 2
8. third
9. 18
10. 7, 18
11. hydrogen
12. zero
13. increases
14. acidic
15. metallic

Question 4.
State whether the following statements are true or false:

1. At present, naturally occurring elements are 98.
2. The atomic masses of elements form a Dobereiner’s triad are 14 u, 31 u and 74.9 u respectively.
3. Calcium, strontium and barium form a Dobereiner’s triad.
4. According to the Newlands’ law of octaves, every eighth element had properties similar to that of the first element.
5. Sodium is the eighth element after lithium.
6. Oxygen is the eighth element after sulphur.
7. Phosphorus is tire eighth element after nitrogen.
8. Dobereiner’s triads are observed in Newlands’ octaves.
9. Mendeleev’s periodic law was based on atomic number of element.
10. Molecular formula of oxide of barium is BaO.
11. Mendeleev named gallium for eka-silicon.
12. An element having atomic number 3.5 can be placed between Be and B.
13. There are three valence electrons present in the elements of group I.
14. Electrons are filled in K, L and M shells in the elements of the third period.
15. Each period starts with the filling of electrons in a new shell.

Answer:

1. True
2. False
3. True
4. True
5. True
6. False
7. False
8. True
9. False
10. True
11. False
12. False
13. False
14. True
15. True

Question 5.
Match the following properly :
(1)

Column I	Column II
1. Dobereiner	a. Law of octaves
2. Newlands	b. Periodic law
3. Mendeleev	c. Modern periodic table
4. Henry Moseley	d. Law of triads

Answer:
(1-d), (2-a), (3- b), (4 – c).

(2)

Column I	Column II
1. Li, Na, K	a. Metalloids
2. S, E Cl	b. Noble gases
3. B, Si, Ge	c. Non-metallic elements
4. He, Ne, Ar	d. Metallic elements

Answer:
(1-d), (2-c), (3-a), (4-b).

Question 6.
Draw the following graphs :
1. Draw the graph of Atomic radii → Atomic numbers for the elements of second period.
Answer:

2. Draw the graph of Atomic radii → Atomic numbers for the alkali metal elements.
Answer:

Question 7.
Which one of the following depict the correct representation of atomic radius of an atom?

Answer:
(ii) and (iii)

Question 8.
(A) Choose the correct option from those given below each question:
1. For which of the following element, Mendeleev didn’t left gap in his periodic table?
A. Gallium
B. Beryllium
C. Germanium
D. Scandium
Answer:
B. Beryllium

2. Newlands’ law of octaves was found to be applicable upto …………………
A. nickel
B. cobalt
C. phosphorus
D. calcium
Answer:
D. calcium

3. According to Mendeleev’s periodic law, the elements were arranged in the periodic table in the order of…
A. increasing atomic number.
B. decreasing atomic number.
C. increasing atomic masses.
D. decreasing atomic masses.
Answer:
C. increasing atomic masses.

4. The elements Si, B and Ge are …………………
A. metallic elements
B. non-metals
C. metalloids
D. metal, non-metal and metalloid respectively
Answer:
C. metalloids

5. In Mendeleev’s periodic table, gaps were left for the elements to be discovered later. Which of the following elements found a place in the periodic table later?
A. Be
B. Ge
C. Si
D. Se
Answer:
B. Ge

6. The three imaginary elements X, Y and Z represent a Dobereiner’s triad. If the atomic mass of element X is 14 u and that of element Y is 46 u, then the atomic mass of element Z will be …………………
A. 28
B. 60
C. 78
D. 72
[Hint: According to the law of Dobereiner’s triad,
Atomic mass of Y

Answer:
C. 78

7. The atomic numbers of four elements R Q, R and S are 6, 8, 14 and 16 respectively. Out of these, the element known as metalloid is ……….
A. P
B. Q
C. R
D. S
[Hint: Elements in which generally four? electrons are present in outermost shell are called metalloids.]
Answer:
C. R

8. Which of the following statements is correct with regard to the classification of elements?
A. Elements in modern periodic table are arranged on the basis of increasing atomic masses.
B. Elements in modern periodic table are arranged on the basis of decreasing atomic numbers.
C. In modern periodic table, the element nickel of lower atomic mass is kept before the element cobalt of higher atomic mass.
D. In modern periodic table, the isotopes of chlorine having different atomic masses are kept in the same group.
Answer:
D. In modern periodic table, the isotopes of chlorine having different atomic masses are kept in the same group.

9. Which of the following statements about the modern periodic table is correct?
A. It has 18 horizontal rows known as periods.
B. It has 8 vertical columns known as periods.
C. It has 18 vertical columns known as groups.
D. It has 7 horizontal rows known as groups.
Answer:
B. It has 8 vertical columns known as periods.

10. An element X forms an oxide X₂O₃. In which : group of Mendeleev’s periodic table is this element placed?
A. Group II
B. Group III
C. Group V
D. Group VIII
Answer:
B. Group III

11. Who proposed the ‘Modern periodic law’ for the modern periodic table?
A. Dobereiner
B. Newlands
C. Henry Moseley
D. Mendeleev
Answer:
C. Henry Moseley

12. Which fundamental particle forms the real basis for the modern classification of elements?
A. Proton
B. Electron
C. Neutron
D. Nucleon
Answer:
A. Proton

13. Which of the following is not correct about the trends when going from left to right across the periods of the periodic table?
A. The elements become more non-metallic in nature.
B. The number of valence electrons increases.
C. The atoms lose their electrons easily.
D. The oxides become more acidic.
Answer:
C. The atoms lose their electrons easily.

14. The electronic configuration of the atom of an element X is 2, 8, 4. In modern periodic table, the element X is placed in
A. Group 2
B. Group 14
C. Group 4
D. Group 8
Answer:
B. Group 14

15. The atomic number of an element is 20. In modern periodic table, this element is placed in ……………..
A. 2nd period
B. 3rd period
C. 1st period
D. 4th period
Answer:
D. 4th period

16. The elements A, B, C, D and E have atomic numbers of 2, 3, 7, 10 and 18 respectively. The elements which belong to the same period of the periodic table are
A. A, B, C
B. B, C, D
C. A, D, E
D. B. D, E
Answer:
B. B, C, D

17. The elements A, B, C, D and E have atomic numbers 9, 11, 17, 12 and 13 respectively. The pair of elements which belong to the same group of the periodic table is ……………..
A. A and B
B. B and D
C. D and E
D. A and C
Answer:
D. A and C

18. Which of the following element would lose an electron easily?
A. Mg
B. Na
C. K
D. Ca
Answer:
C. K

19. Which of the following element will gain an electron easily?
A. Na
B. F
C. Mg
D. Al
Answer:
B. F

20. Where would you place the element with electronic configuration 2, 8 in the modern periodic table?
A. Group 8
B. Group 2
C. Group 18
D. Group 10
Answer:
C. Group 18

21. An element which is an essential constituent of all organic compounds belong to ……………..
A. Group 4
B. Group 10
C. Group 16
D. Group 14
Answer:
D. Group 14

22. Which of the following is the valence shell for the elements of second period of the modern periodic table?
A. M-shell
B. K-shell
C. N-shell
D. L-shell
Answer:
D. L-shell

23. The element which has the maximum number of valence electrons is ……………..
A. 15^P
B. 11^Na
C. 14^Si
D. 13^Al
Answer:
A. 15^P

24. The correct increasing order of the atomic radii of the elements oxygen, fluorine and nitrogen is ……………..
A. O, F, N
B. N, F, O
C. O, N, F
D. F, O, N
Answer:
D. F, O, N

25. Which one of the following does not increase while moving down the group of the periodic table?
A. Atomic radius
B. Metallic character
C. Valence electrons
D. Basicity of oxides
Answer:
C. Valence electrons

(B) Choose more than one correct options from those given below each question:
1. Mention the drawbacks of Mendeleev’s periodic table.
A. Position of hydrogen
B. Position of isotopes
C. Arrangement of Noble gases
D. Arrangement of more than one elements in the same slot.
Answer:
A, B, C, D

2. Which of the following increase while moving down the group 17 elements?
A. Atomic radius
B. Valence electrons
C. Metallic character
D. Acidity of oxides
Answer:
A, C

3. Which of the following statements are correct for modern periodic table?
A. New elements can be easily arranged.
B. Predictions of properties of the elements become easy.
C. The elements have been divided into metals and non-metals by the thick zig-zag line running diagonally across the periodic table.
D. Atomic volume of elements decreases on moving down in the group.
Answer:
A, B, C

4. The electronic configuration of the atom of an element is 2, 8, 7. In this reference, which of the following statements are correct?
A. This element belong to group 17.
B. This element has a tendency to gain one electron.
C. This element belongs to fourth period.
D. The element fluorine is placed above this element.
Answer:
A, B, D

(C) In each of the following questions, a statement of Assertion A is given, followed by a Reason R. Read the statements carefully and choose the correct option as under:
A. Assertion A and reason R are true, and reason R is the correct explanation of assertion A.
B. Assertion A and reason R are true, but reason R is not the correct explanation of assertion A.
C. Assertion A is corret but reason R is false.
D. Assertion A is false but reason R is true.
1. Assertion A: The position of hydrogen in modern periodic table is a matter of controversy.
Reason R : The properties of hydrogen resembles with the properties of alkali metals and halogens.
Answer:
A. Assertion A and reason R are true, and reason R is the correct explanation of assertion A.

2. Assertion A: In the modern periodic table, metallic character of elements increases on moving down the group.
Reason R: In the modern periodic table, the elements are arranged in order of their increasing atomic masses.
Answer:
C. Assertion A is corret but reason R is false.

3. Assertion A : Newlands arranged the then known elements in order of musical notes.
Reason R: According to Newlands, proton is responsible for the arrangement of elements in the periodic table.
Answer:
C. Assertion A is corret but reason R is false.

4. Assertion A : Chemical reactivity of the element of group 18 is very less.
Reason R : Their outermost shell is completely filled with electrons.
Answer:
A. Assertion A and reason R are true, and reason R is the correct explanation of assertion A.

Value Based Questions With Answers

Question 1.
Six elements of periodic table A, B, C, D, E and F have atomic numbers of 2, 12, 20, S 18, 4 and 10 respectively (where A, B, C,? D, E and F are not the chemical symbols of s these elements). Based on this information, answer the following questions :
(1) Which of these elements belong to the same groups of the periodic table? Why?
(2) Which of these elements belong to the same periods of the periodic table? Why?
(3) Which of these elements are (i) metals and (ii) non-metals?
(4) Which of these elements are chemically (i) reactive and (ii) unreactive?
(5) What values are indicated in a student in answering the above questions?
Answer:
First of all, write the electronic configurations? of all the given six elements as follows :

Element	Atomic number	Electronic configuration
		K	L	M	N
A	2	2
B	12	2	8	2
C	20	2	8	8	2
D	18	2	2	8
E	4	2	8
F	10	2	8

(1) (i) Elements B, C and E belong to the s group 2 of the periodic table because all of them have an equal number of valence electrons equal to 2.

(ii) Elements A, D and F also belong to the group 18 because they are all inert gases having completely filled outermost electron shells with 8 valence electrons.

(2) (i) Element B (2, 8, 2) and element D (2, 8, 8) belong to the third period because they both have 3 electron shells – K, L and M each.

(ii) Element E (2, 2) and element F (2, 8) belong to the second period because they both have 2 electron shells – K and L each.

(3) (i) Elements B, C and E belonging to group 2 are metals.
(ii) Elements A, D and F belonging to group 18 are non-metals (noble gases).

(4) (i) Elements B, C and E of group 2 are chemically reactive elements.
(ii) Elements A, D and F of group 18 are chemically unreactive elements.

(5) Answering these questions by a student are indicating awareness towards modern periodic classification of the elements and ability to apply his knowledge in solving the related questions.

Question 2.
In his periodic table, Mendeleev arranged all the then known 63 elements in the order of increasing atomic masses in horizontal rows but in such a way that elements having similar properties came directly under one another in the same vertical column. In the classification of the then known elements, Mendeleev was guided mainly by two factors.

In order to make sure that the elements having similar properties fall in the same vertical column, Mendeleev left some gaps in his periodic table. Though leaving gaps in the periodic table was considered to be a drawback of his classification of elements at that time but Mendeleev was firm on his decision.
Answer the following questions:
(1) What are the horizontal rows of Mendeleev’s periodic table known as? How many horizontal rows of elements were there in Mendeleev’s periodic table?

(2) What are the vertical columns of Mendeleev’s periodic table known as? How many vertical columns were there? in Mendeleev’s periodic table?

(3) What were the similar properties used by Mendeleev to classify the then known elements into vertical columns?

(4) What were the two main guiding factors for Mendeleev in the classification of the then known elements?

(5) For what purpose were some gaps left by Mendeleev in his periodic table? Does the modern periodic table also have the 5 gaps left by Mendeleev?

(6) What values were displayed by Mendeleev in presenting his classification of elements?
Answer:
(1) The horizontal rows of elements in the Mendeleev’s periodic table are called periods. There were seven periods in s Mendeleev’s periodic table.

(2) The vertical columns of Mendeleev’s periodic table are called groups. There were eight groups in Mendeleev’s periodic s table.

(3) The similar properties used by Mendeleev to classify the elements into vertical columns were the similar formulae of S their oxides and hydrides.

(4) The two main guiding factors for Mendeleev in the classification of the then known elements were

• increasing t order of atomic masses
• grouping together the elements having similar properties.

(5) These gaps were left for the elements which had not been discovered at that time. He thought that these elements would be discovered later in future. The modern periodic table does not have any gaps.

(6) The various values displayed by Mendeleev were-

• self-confidence
• courage
• foresight
• prophecy.

Question 3.
There are three elements X, Y and Z having atomic numbers of 6, 16 and 19 respectively. Based on this information, Het has been asked to answer the following questions :
(1) In which group of the periodic table would you expect to find (i) element X (ii) element Y and (iii) element Z?
(2) Which two elements will form ionic bonds? Why?
(3) What will be the formula of ionic compound formed?
(4) Which two elements will form covalent bonds? Why?
(5) What will be the formula of covalent compound formed?
(6) What values are displayed by Het in answering the above questions?
Answer:
In order to answer these questions, Het first wrote down the electronic configurations of the elements X, Y and Z by using their atomic numbers, as follows :

Element	Atomic number	Electronic configuration
		K	L	M	N
X	6	2	4
Y	16	2	8	6
Z	19	2	8	8	1

(1) (i) Element X has 4 valence electrons in its atom, so it will be placed in group 4 + 10 = 14 of the modern periodic table.
(ii) Element Y has 6 valence electrons in its atom, so it will be placed in group 6 + 10 = 16 of the modern periodic table.
(iii) Element Z has 1 valence electron in its atom, so it will be placed in group 1 of the modern periodic table.

(2) An ionic bond is formed between a metal element and a non-metal element. Here, the elements Z and Y will combine to form ionic bond.

(3) The element Z has 1 valence electron, so its valency is +1 (after losing this electron). The element Y need two electrons to complete its octet. So, element Y will gain two electrons and form Y^2- ion. Here, two Z⁺ ions will combine with one Y^-2 ion, so that the formula of ionic compound formed will be Z₂Y.

(4) A covalent bond is formed between two non-metallic elements. Here, element X of group 14 is a non-metal and element Y of group 16 is also a non-metal. Therefore, the elements X and Y will combine together to form covalent bonds.

(5) The element X has 4 valence electrons, so its valency is 4. The element Y has 6 valence electrons, so its valency is 2. So, one atom of element X will combine with two atoms of element Y to form a covalent compound having the formula XY₂.

(6) The values displayed by Het in answering these questions are (1) knowledge of modern classification of elements, and (2) understanding of periodic table.

Practical Skill Based Questions With Answers

Question 1.
Complete the following cross-word puzzle :

Across:
(1) An element with atomic number 12.
(3) Metal used in making cans and member of group 14.
(4) A lustrous non-metal which has 7 electrons ; in its outermost shell.

Down:
(2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene.
(5) The first element of second period.
(6) An element which is used in making fluorescent bulbs and is second member of group 18 in the modern periodic table.
(7) A radioactive element which is the last member of halogen family.
(8) Metal which is an important constituent of steel and forms rust when exposed to moist air.
(9) The first metalloid in modern periodic table whose fibres are used in making bullet-proof vests.
Answer:

Question 2.
In this ladder, symbols of elements are jumbled up.
(a) Rearrange these symbols of elements in the increasing order of their atomic numbers in the periodic table.
(b) Arrange them in the order of their group also.

Answer:
(a) H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, E S, Cl, Ar, K, Ca.
(b)

Group	Elements
1	H, Li, Na, K
2	Be, Mg, Ca
13	B, Al
14	C, Si
15	N, P
16	O, S
17	F, Cl
18	He, Ne, Ar

Memory Map

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.3

Take π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
R = 4.2, r = 6
Volume of the cylinder = Volume of the melted sphere
πr²h = \(\frac{4}{3}\)πR³
3πr²h = 4πR³
h = \(\frac{4 \pi \mathrm{R}^3}{3 \pi \mathrm{r}^2}\)
= \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6}\)
= 4 × 1.4 × 0.7 × 0.7
= 5.6 × 0.49
= 2.744 cm.
Height of the cylinder = 2.74 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
r₁ = 6 cm, r₂ = 8 cm, r₃ = 10 cm.
Let the radius of the resulting sphere be r cm.
Its volume = \(\frac{4}{3}\)πR³
Volume of the resulting sphere = Volume of the sphere of radius r₁ + Volume of the sphere of radius r₂ + Volume of the sphere of radius r₃

Radius of the resulting sphere = 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.
Solution:
Volume of the earth got by digging the well = πr²h
= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14\)
= 99 cm.
R = 1.5 + 4
r = 1.5

Area of the circular embankment around the well = πR² – πr²
= π[(5.5)² – (1.5)²]
= π(5.5 + 1.5) (5.5 – 1.5)
= π(7) (4).

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
Volume of ice cream in the cylindrical container =

Alternative Method:
Number of cones will be = \(\frac{\text { Volume of cylinder }}{\text { Volume of cone }}\)
For the cylinder part,
Radius = \(\frac{12}{2}\) = 6 cm, Height = 15 cm
∴ Volume of cylinder = π × r² × h = 540π
For the cone part,
Radius of conical part = \(\frac{6}{2}\) = 3 cm, Height = 12 cm
Radius of hemispherical part = \(\frac{6}{2}\) = 3 cm
Now,
Volume of cone = Volume of conical part + Volume of hemispherical part
\(\left(\frac{1}{3}\right)\) × π × r² × h + \(\left(\frac{2}{3}\right)\) × π × r³
= 36π + 18π
= 54π.
∴ Number of cones = \(\frac{540 \pi}{54 \pi}\) = 10

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:

Volume of sand in the bucket = πr²h
= π × 18 × 18 × 32 cc.
Volume of the conical heap of sand = \(\frac{1}{3}\)πr²h
\(\frac{1}{3}\)πr²h = π × 18 × 18 × 32
πr²h = 3 × π × 18 × 18 × 32
πr² × 24 = 3 × π × 18 × 18 × 32
r² = \(\frac{3 \times \pi \times 18 \times 18 \times 32}{\pi \times 24}\)
r² = 18² × 2²
r = 18 × 2 = \(\frac{72}{2}\) = 36 cm.
l² = h² + r²
= (24)² + (36)²
= (24 × 24) + (36 × 36)
= (2 × 12 × 12 × 2) + (3 × 12 × 12 × 3)
= 12² (4 + 9)
= 12² × 13
l = 12\(\sqrt{13}\) cm.

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Quantity of water flowing in the canal in 1 hour = lbh
l = 10 km, b = 6 m, h = 1.5 m
10 × 1000 × 6 × 1.5 cubic metres.
The area this water can irrigate in 1 hour if 8 cm of standing water is needed
= \(\frac{10 \times 1000 \times 6 \times 1.5}{0.08}\)
8 cm = 0.08 mm
Area needed to irrigate in 30 mins. or \(\frac{1}{2}\) hr.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
d = 20 cm, r = 10 cm = \(\frac{10}{100}\) = 0.1 mt, h = 3 km = 3 × 1000 m.
Volume of water that comes out of the pipe of diameter 20 cm
= πr²h
= π × 0.1 × 0.1 × 3000
= π × 30 cm.
In 1 hr. (60 mins.), volume of water that flows into the tank = π × 30
In 1 minute = \(\frac{\pi \times 30}{60}=\frac{\pi}{2}\) cm.
Volume of the cylindrical tank = πr²h = π × 5 × 5 × 2. d = 10, r = 5
\(\frac{\pi}{2}\) cubic metres of water will be filled in 1 minute
π × 5 × 5 × 2……. \(\frac{\pi \times 5 \times 5 \times 2 \times 2}{\pi}\) = 100 mins.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A, (ii) sin C, cos C.
Solution :

(i) sin A, cos A
AC² = AB² + BC²
= 24² + 7² = 576 + 49
= 625
∴ AC = 25

Question 2.
In the figure, find tan P – cot R.
Solution :

QR² = PR² – PQ² = 13² – 12²
= 169 – 144 = 25.
∴ QR = 5.
tan P = \(\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{5}{12}\)
cot R = \(\frac{\mathrm{QR}}{\mathrm{QP}}=\frac{5}{12}\)
∴ tan P – cot R = \(\frac{5}{12}=\frac{5}{12}\) = 0

Question 3.
If sin A = \(\frac{3}{4}\), calculate cos A and tan A.
Solution:

Given, sin A = \(\frac{3}{4}\) ⇒ \(\frac{P}{H}\) = \(\frac{3}{4}\)
Let P = 3k and H = 4k.
In right angled ΔABC,
P² + B² = H² (By Pythagoras theorem)
⇒ (3k)² + B² = (4k)²
⇒ 9k² + B² = 16k²
⇒ B² = 16k² – 9k² = 7k²
∴ B = +k\(\sqrt{7}\) (∵ Base cannot be -ve)

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Given, 15 cot A = 8 ⇒ cot A = \(\frac{8}{15}\)
\(\frac{\mathrm{B}}{\mathrm{P}}=\frac{8}{15}\)
B = 8k and P = 15 k
In right angled ΔABC, H² = B² + P² (By Pythagoras theorem)
= (8k)² + (15k)²
= 64k² + 225k²
= 289k²
∴ H = 17k (∵ Side cannot be -ve)
Now, sin A = \(\frac{\mathrm{P}}{\mathrm{H}}=\frac{15 \mathrm{k}}{17 \mathrm{k}}=\frac{15}{17}\)
sec A = \(\frac{\mathrm{H}}{\mathrm{B}}=\frac{17 \mathrm{k}}{8 \mathrm{k}}=\frac{17}{8}\)

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios.
Solution :

sec θ = \(\frac{1}{cos θ}\)
sec θ . cos θ = 1
\(\frac{13}{12} \times \frac{12}{13}\) = 1
AB² = AC² – BC²
= 13² – 12²
= 169 – 144
= 25
AB = \(\sqrt{25}\) = 5

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution :

cos A = \(\frac{AC}{AB}\)
cos B = \(\frac{BC}{AB}\)
cos A = cos B
\(\frac{AC}{AB}=\frac{BC}{AB}\)
∴ AC = BC
∴ \(\hat{A}\) = \(\hat{B}\) (∵ Angles opposite to equal sides are also equal)

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate: (i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\) (ii) cot² θ
Solution :

AC² = AB² + BC²
= 7² + 8²
= 49 + 64 = 113
AC = \(\sqrt{113}\)

(ii) cot² θ = (cot θ)²
= (\(\frac{7}{8}\))² = \(\frac{49}{64}\)
Given cot θ = \(\frac{7}{8}\)

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos² A – sin² A or not.
Solution :

3 cot A = 4
cot A = \(\frac{4}{3}\)
AC² = AB² + BC²
= 4² + 3²
= 16 + 9
= 25
AC = \(\sqrt{25}\) = 5

Question 9.
In triangle ABC, right-angled at B, if tan A = find the value of :
(i) sin A cos C + cos A sin C,
(ii) cos A cos C – sin A sin C.
Solution :
(i) sin A cos C + cos A sin C

(ii) cos A . cos C – sin A . sin C
\(\frac{AB}{AC} \cdot \frac{BC}{AC}\) – \(\frac{BC}{AC} \cdot \frac{AB}{AC}\)

Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
PR + QR = 25
QR = 25 – PR
PQ² + QR² = PR²
(5)² + (25 – PR)² = PR²
25 + (625 – 50PR + PR²) = PR²
650 – 50PR = PR² – PR² = 0
– 50PR = – 650
PR = \(\frac{-650}{-50}\) = 13.
PR = 13, QR = 25 – PR = 25 – 13 = 12, PQ = 5.

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac{12}{5}\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution :
(i) False, because the value of tan A increases from 0 to ∞. Also, tan 45° = 1.
(ii) True, because the value of sec A increases from 1 to ∞.
(iii) False, cos A is the abbreviation used for the cosine of angle A.
(iv) False, because cot A is one symbol. We cannot separate cot and A.
(v) False, because the value of sin θ always lies between 0 and 1. Here, sin θ = \(\frac{4}{3}\) which is greater than 1. So, it is not possible.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 3 दो चरों वाले रखिक समीकरण का युग्म Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 3 दो चरों वाले रखिक समीकरण का युग्म

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
एक कक्षा में विद्यार्थियों को पंक्तियों में खड़ा किया जाता है। यदि एक पंक्ति में एक विद्यार्थी को अतिरिक्त खड़ा किया जाए तो पंक्तियों की संख्या 2 कम हो जाती है और यदि एक पंक्ति में एक विद्यार्थी कम खड़ा किया जाय तो पंक्तियों की संख्या 3 बढ़ जाती है। कक्षा कुल विद्यार्थियों की संख्या ज्ञात कीजिए।
हल :
माना कि मूल पंक्तियों की संख्या x है तथा मूलतः प्रत्येक पंक्ति में y विद्यार्थी खड़े किये जाते हैं।
विद्यार्थियों की कुल संख्या = x × y = xy
अब यदि एक पंक्ति में i विद्यार्थी अतिरिक्त खड़ा किया जाये तो एक पंक्ति में विद्यार्थियों की संख्या (y + 1) हो जायेगी तथा दिये गये प्रतिबन्ध से पंक्तियों की संख्या (x – 2) हो जायेगी।
अतः विद्यार्थियों की कुल संख्या
= (y + 1) (x – 2)
∴ (y + 1) (x – 2) = x × y
⇒ xy – 2y + x – 2 = xy
⇒ x – 2y = 2 ……..(1)
पुन: एक पंक्ति में एक विद्यार्थी कम खड़ा किया जाता है तो एक पंक्ति में विद्यार्थियों की संख्या (y – 1) होगी तथा प्रश्न में दिये गये प्रतिबन्ध के अनुसार पंक्तियों की संख्या (x + 3) हो जायेगी।
अतः विद्यार्थियों की कुल संख्या
= (y – 1) (x + 3)
∴ (y – 1) (x + 3) = xy
⇒ xy + 3y – x – 3 = xy
⇒ 3y – x – 3 = 0
⇒ x – 3y = – 3 ……..(2)
समीकरण (1) में से (2) को घटाने पर,

समीकरण (1) में y का मान रखने पर,
x – 2 × 5 = 2
⇒ x = 2 + 10
∴ x = 12
कक्षा में विद्यार्थियों की कुल संख्या = (xy)
= 12 × 5 = 60.

प्रश्न 2.
k का मान ज्ञात कीजिए ताकि निम्न समीकरण युग्म का कोई हल नहीं हो
(3k + 1)x + 3y – 2 = 0
(k² + 1)x + (k – 2)y – 5 = 0
हल :
दिए गए समीकरण युग्म का कोई हल नहीं होने के लिए प्रतिबन्ध

k = – 1 के लिए निम्न वक्तव्य सही है-
\(\frac{3}{k-2}\) ≠ \(\frac{2}{5}\)
अतः दिए गए समीकरण का कोई हल नहीं होगा। यदि k = – 1 है।

प्रश्न 3.
अशोक ने एक टेस्ट में 65 अंक अर्जित किए, जब उसे प्रत्येक सही उत्तर पर 5 अंक मिले तथा प्रत्येक गलत उत्तर पर 2 अंक की कटौती की गई। यदि उसे सही उत्तर पर 3 अंक मिलते तथा प्रत्येक गलत उत्तर पर 1 अंक कटता, तो अशोक 40 अंक अर्जित करता। इस समस्या को बीजगणितीय रूप में व्यक्त कर ग्राफ विधि से हल कीजिए। टेस्ट में कुल कितने प्रश्न थे ?
हल :
माना अशोक द्वारा सही हल किये गये प्रश्नों की संख्या = x
गलत हल किये गये प्रश्नों की संख्या = y
बीज गणितीय निरूपण – पहली शर्त के अनुसार,
⇒ 5x – 2y = 65 …….(1)
दूसरी शर्त के अनुसार,
3x – y = 40 ……………(2)
ज्यामितीय निरुपण – समी. (1) से
5x – 2y = 65
x = \(\frac{65+2 y}{5}\)
y = – 5 रखने पर, x = \(\frac{65-10}{5}=\frac{55}{5}\) = 11
y = 0 रखने पर, x = \(\frac{65-0}{5}=\frac{65}{5}\) = 13
y = 5 रखने पर, x = \(\frac{65+10}{5}=\frac{75}{5}\) = 15
सारणी – I

x	11	13	15
y	-5	0	5

समी. (2) से
3x – y = 40
x = \(\frac{40+y}{3}\)
y = – 7 रखने पर x = \(\frac{40-7}{3}=\frac{33}{3}\) = 11
y = 2 रखने पर x = \(\frac{40+2}{3}=\frac{42}{3}\) = 14
y = 5 रखने पर y = \(\frac{40+5}{3}=\frac{45}{3}\) = 15

सारणी – II

x	11	14	15
y	-7	2	5

आलेख से स्पष्ट है कि दिए गए समीकरण युग्म की दो सरल रेखाऐं बिन्दु (10, 5) पर काटती है।
x = 10, y = 5
टेस्ट में प्रश्नों की संख्या
= 10 + 5 = 15.

प्रश्न 4.
x तथा y के लिए हल कीजिए-
27x + 31y = 85.
31x + 27y = 89.
हल :
दी गयी समीकरणं है-
27x + 31y = 85 ………….(1)
31x + 27y = 89 ………….(2)
समीकरण (1) तथा (2) को जोड़ने पर,

x का मान समीकरण (3) में रखने पर
2 + y = 3
⇒ y = 3 – 2 = 1
अतः x = 2 तथा y = 1 समीकरणों के अद्वितीय हल हैं।

प्रश्न 5.
एक तालाब को दो पाइपों द्वारा भरने में 12 घंटे लगते हैं। यदि बड़े व्यास वाले पाइप को 4 घन्टे तथा छोटे व्यास वाले पाइप को 9 घंटे प्रयोग किया जाता है तो तालाब का आधा भाग भरा जाता है। बताइये कितने समय में तालाब प्रत्येक पाइप द्वारा अलग-अलग भरा जायेगा।
हल :
माना बड़े व्यास वाला पाइप तालाब को x घंटे में तथा छोटे व्यास वाला पाइप y घंटे में भरता है।
∴ बड़े व्यास वाले पाइप द्वारा 1 घंटे में भस गया भाग = \(\frac{1}{x}\)
बड़े व्यास वाले पाइप द्वारा 4 घंटे भरा गया भाग = \(\frac{4}{x}\)
छोटे व्यास वाले पाइप द्वारा 1 घंटे में भरा गया भाग = \(\frac{1}{y}\)
छोटे व्यास वाले पाइप द्वारा 9 घंटे में भरा गया भाग = \(\frac{9}{y}\)
प्रश्नानुसार, \(\frac{4}{x}=\frac{9}{y}\) = \(\frac{1}{2}\) ……….(1)
और यदि दोनों पाइपों से तालाब 12 घंटे में भरा जाता है।
∴ \(\frac{12}{x}=\frac{12}{y}\) = 1 ……….(2)
माना कि \(\frac{1}{x}\) = a तथा \(\frac{1}{y}\) = b यह मान समीकरण (1) तथा (2) में रखने पर
4a + 9b = \(\frac{1}{2}\) ……….(3)
12a + 12b = 1 ……….(4)
समीकरण (3) को 3 से गुणा करके इसमें से समीकरण (4) को घटाने पर

अतः बड़े व्यास वाला पाइप 20 घन्टे तथा छोटे व्यास वाला पाइप 30 घंटे लेगा।

प्रश्न 6.
7 रबड़ और 5 पेन्सिलों का कुल मूल्य ₹ 58 है, जबकि 5 रबड़ और 6 पेन्सिलों का कुल मूल्य ₹ 56 है। इस समस्या को बीजगणितीय रूप में व्यक्त कर ग्राफ विधि से हल कीजिए ।
हल :
माना 1 रबड़ का मूल्य ₹ x तथा एक पेन्सिल का मूल्य ₹ y है।
प्रश्नानुसार,
∴ 7x + 5y = 58
तथा 5 रबड़ और 6 पेन्सिलों का कुल मूल्य = ₹ 56
⇒ 7x + 5y = 56
बीजगणितीय निरूपण
7x + 5y = 58 ……(1)
5x + 6y = 56 …….(2)
ज्यामितीय निरूपण:
समीकरण (1) से,
7x + 5y = 58
⇒ 5y = 58 – 7x
⇒ y = \(\frac{58-7 x}{5}\)
x = 4 रखने पर, y = \(\frac{58-7 \times 4}{5}=\frac{58-28}{5}\) = 6
x = – 1 रखने पर y = \(\frac{58-7 \times(-1)}{5}=\frac{58+7}{5}\) = 13
सारणी – I

x	11	13
y	-5	0

समीकरण (2) से
5x + 6y = 56
⇒ 6у = 56 – 5x
⇒ y = \(\frac{56-5 x}{6}\)
x = – 2 रखने पर, y = \(\frac{56-5 \times(-2)}{6}=\frac{56+10}{6}\)
y = 11
x = 4 रखने पर y = \(\frac{56-5 \times 4}{6}=\frac{56-20}{6}\) = 6

सारणी – II

x	11	13
y	-5	0

सारणी I और सारणी II से प्राप्त x और y के मानों का आलेखन करने पर हमें निम्न आलेख प्राप्त होता है।

आलेख से स्पष्ट है कि दोनों रैखिक समीकरणों से प्राप्त सरल रेखाएँ बिन्दु P(4, 6) पर प्रतिच्छेदित होती हैं।
∴ x = 4 तथा y = 6
अतः एक रबड़ का मूल्य = ₹ 4 तथा एक पेन्सिल का मूल्य = ₹ 6

प्रश्न 7.
अभ्यास पुस्तिका और 3 पेन्सिलों का कुल मूल्य 17 रूपए है, जबकि 3 अभ्यास पुस्तिका और 4 पेन्सिलों का कुल मूल्य 24 रूपए है। इस समस्या को बीजगणितीय रूप में व्यक्त कर ग्राफ विधि से हल कीजिए।
हल :
माना एक अभ्यास पुस्तिका का मूल्य ₹ x तथा एक पेन्सिल का मूल्य ₹ y है।
प्रश्नानुसार,
2 अभ्यास पुस्तिका और 3 पेन्सिलों का मूल्य = ₹ 17
⇒ 2x + 3y = 17
3 अभ्यास पुस्तिका और 4 पेन्सिलों का मूल्य 3x + 4y = ₹ 24
⇒ 3x + 4y = 24
बीजगणितीय निरूपण :
2x + 3y = 17 …….(1)
3x + 4y = 24 …….(2)
ज्यामितीय निरूपण :
समीकरण (1) से,
2x + 3y = 17
⇒ 3y = 17 – 2x
⇒ y = \(\frac{17-2 x}{3}\)
x = 1 रखने पर, y = \(\frac{17-2 \times 1}{3}=\frac{15}{3}\) = 5
x = 4 रखने पर, y = \(\frac{17-2 \times 4}{3}=\frac{9}{3}\) = 3
x = 7 रखने पर, y = \(\frac{17-2 \times 57}{3}=\frac{3}{3}\) = 1
सारणी – I

x	11	13	15
y	-5	0	5

समीकरण (2) से,
3x + 4y = 24
⇒ 4y = 24 – 3x
⇒ y = \(\frac{24-3 x}{4}\)
x = 0 रखने पर y = \(\frac{24-3 \times 0}{4}=\frac{24}{4}\) = 6
x = 4 रखने पर y = \(\frac{24-3 \times 4}{4}=\frac{12}{4}\) = 3
x = 8 रखने पर y = \(\frac{24-3 \times 8}{4}=\frac{0}{4}\) = 0

सारणी – II

x	11	13	15
y	-5	0	5

सारणी I और सारणी II से प्राप्त x और y के मानों का आलेखन करने पर हमें निम्न आलेख प्राप्त होता है :

आलेख से स्पष्ट है कि दोनों रैखिक समीकरणों से प्राप्त सरल रेखाएँ बिन्दु (4, 3) पर प्रतिच्छेद करती हैं।
∴ x = 4 तथा y = 3
अतः एक पुस्तिका का मूल्य = ₹ 4
तथा एक पेन्सिल का मूल्य = ₹ 3

प्रश्न 8.
एक रेलगाड़ी एक निश्चित दूरी एक समान चाल से तय करती है। यदि यह रेलगाड़ी 6 किमी / घंटा तेज चाल से चलती तो यह निश्चित समय से 4 घंटे कम लेती तथा यदि 6 किमी / घंटा धीमी चाल से चलती तो यह निश्चित समय से 6 घंटे अधिक लेती। यात्रा की निश्चित दूरी ज्ञात कीजिए।
हल :
माना रेलगाड़ी की चाल x किमी / घंटा तथा लिया गया समय y घंटे हैं।
∴ रेलगाड़ी द्वारा तय की गई दूरी = चाल × समय = x × y = xy किमी
तेज चलने पर रेलगाड़ी की चाल = (x + 6) किमी / घंटा
रेलगाड़ी द्वारा लिया गया समय = (y – 4) घंटे
अब, दूरी = (x + 6) (y – 4)
⇒ xy = xy – 4x + 6y – 24
⇒ 4x – 6y = – 24 ………..(1)
धीमी गति से चलने पर रेलगाड़ी की चाल = (x – 6) किमी / घंटा
समय = (y + 6) घंटे
अब, दूरी = चाल × समय
⇒ xy = (x – 6) × (y + 6)
⇒ xy = xy + 6x – 6y – 36
⇒ – 6x + 6y = – 36 ………..(2)
समीकरण (1) और समीकरण (2) को जोड़ने पर

समीकरण (1) में x = 30 रखने पर
4 × 30 – 6y = 24
⇒ – 6y = – 24 – 120
⇒ y = \(\frac{-144}{-6}\) = 24
अत: रेलगाड़ी की चाल = 30 किमी / घंटा
रेलगाड़ी द्वारा लिया गया समय = 24 घंटे
तथा रेलगाड़ी द्वारा तय की गई दूरी = 30 × 24
= 720 किमी

प्रश्न 9.
दो संपूरक कोणों में से बड़े कोण का मान छोटे कोण के मान से 18° अधिक है। दोनों कणों के मान ज्ञात कीजिए ।
हल :
माना दिए गए कोण A तथा B हैं।
दिए है,
∠A = ∠B + 18° ………(i)
∵ ∠A व B संपूरक कोण हैं,
∴ ∠A + ∠B = 180°
[∵ संपूरक कोणों का योग 180° होता है]
समीकरण (i) से,
∠B + 18° + ∠B = 180°
2∠B = 162°
∠B = 81°
∴ अभीष्ठ ∠A = 81° + 18° = 99°
अतः अभीष्ट कोण 99° तथा 81° होगे।

प्रश्न 10.
सुमित की आयु उसके बेटे की आयु की तीन गुनी है। पाँच वर्षा के बाद, उसकी आयु अपने बेटे की आयु की ढ़ाई गुना हो जाएगी। इस समय सुमित की आयु कितने वर्ष है?
हल :
माना सुमित की वर्तमान आयु x वर्ष तथा उसके पुत्र की वर्तमान आयु y वर्ष है।
प्रश्नानुसार, x = 3y ………(i)
पाँच वर्ष बाद, सुमीत की आयु = (x + 5)
पाँच वर्ष बाद, पुत्र की आयु = (y + 5)
प्रश्नानुसार, x + 5 = 2\(\frac{1}{2}\)(y + 5) …………(ii)
समीकरण (i) का मान समीकरण (ii) में रखने पर,
3y + 5 = \(\frac{5}{2}\)(y + 5)
3y + 5 = \(\frac{5y}{2}=\frac{25}{2}\)
3y – \(\frac{5y}{2}\) = \(\frac{25}{2}\) – 5
\(\frac{6 y-5 y}{2}=\frac{25-10}{2}\)
\(\frac{y}{2}=\frac{15}{2}\)
समीकरण (i) से,
y = 15 वर्ष
x = 3 × 15 = 45 वर्ष
अतः सुमित की वर्तमान आयु 45 वर्ष तथा उसके पुत्र की वर्तमान आयु 15 वर्ष है।

प्रश्न 11.
एक पिता की आयु अपने दो बच्चों की आयु के योग के तीन गुने के समान है। 5 वर्ष के पश्चात् उसकी आयु बच्चों की आयु के योग के दुगुने के समान होगी। पिता की वर्तमान आयु ज्ञात कीजिए।
हल :
माना, पिता की वर्तमान आयु x वर्ष है तथा उसके दोनों बच्चों की वर्तमान आयु का योग y वर्ष है।
प्रश्नानुसार,
x = 3y …………….(i)
पाँच वर्ष बाद, पिता की आयु = x + 5
पाँच वर्ष बाद पुत्रों की आयु = (y + 5 + 5)
प्रश्नानुसर,
x + 5 = 2 (y + 5 + 5)
x + 5 = 2 ( y + 10 ) …………(ii)
x = 3y, समीकरण (ii) में रखने पर,
3y + 5 = 2y + 20
y = 15 वर्ष
अब, y = 15 समीकरण (i) में रखने पर,
x = 3 × 15
x = 45 वर्ष
अतः पिता की वर्तमान आयु 45 वर्ष है।

प्रश्न 12.
एक भिन्न \(\frac{1}{3}\) हो जाती है, अब उसके अंश से 2 घटाया जाता है, और \(\frac{1}{2}\) वह हो जाती है, जब हर में से 1 घटाया जाए। वह भिन्न ज्ञात कीजिए।
हल :
मान कि भिन्न \(\frac{x}{y}\) है
प्रश्नानुसार, \(\frac{x-2}{y}\) = \(\frac{1}{3}\) …………(i)
तथा \(\frac{x}{y-1}\) = \(\frac{1}{2}\)
⇒ 2x = y – 1
⇒ 2x + 1 = y …………(ii)
समीकरण (i) में y = 2x + 1, रखने पर,
\(\frac{x-2}{2 x+1}=\frac{1}{3}\)
3x – 6 = 2x + 1
x = 7
अब, समीकरण (ii) में x = 7 रखने पर,
y = 2 × 7 + 1
y = 15
अतः अभीष्ट भिन्न \(\frac{7}{15}\) है।

प्रश्न 13.
5 पेंसिलों तथा 7 पेनों का कुल मूल्य ₹ 250 है जबकि 7 पेंसिलों तथा 5 पेनों का कुल मूल्य ₹ 302 है। एक पेंसिल तथा एक पेन का मूल्य ज्ञात कीजिए।
हल :
माना एक पेंसिल का मूल्य ₹ तथा एक पेन का मूल्य ₹ y है।
प्रश्नानुसार,
5x + 7y = 250 ……..(1)
7x + 5y = 302 ………(2)
समीकरण (1) को 5 से तथा समीकरण (2) को 7 से गुणा करने पर
25x + 35y = 1250 ……..(3)
49x + 35y = 2114 ……..(4)
समीकरण (4) में समीकरण (3) घटाने पर

समीकरण (1) में x = 36 रखने पर,
5 × 36 + 7y = 250
7y = 250 – 180
y = \(\frac{70}{7}\) = 10
अत: एक पेंसिल का मूल्य = ₹ 36
तथा एक पेन का मूल्य = ₹ 10

प्रश्न 14.
निम्नलिखित समीकरण युग्म को व्रज-गुणन विधि से हल कीजिए:
x – 3y – 7 = 0
3x – 5y – 15 = 0
हल :
दिया गया समीकरण युग्म
x – 3y – 7 = 0
3x – 5y – 15 = 0
व्रज गुणन विधि से,

तथा \(\frac{y}{-6}=\frac{1}{4}\) ⇒ y = \(\frac{-6}{4}=\frac{-3}{2}\)
अत: x = \(\frac{5}{2}\) और y = \(\frac{-3}{2}\)

प्रश्न 15.
दो संख्याओं का अन्तर 26 है तथा बड़ी संख्या, छोटी संख्या के तीन गुने से 4 अधिक है। संख्याएँ ज्ञात कीजिए।
हल :
माना दो संख्याएँ x और y हैं। यहाँ x > y है।
प्रश्नानुसार x – y = 26 ………(i)
तथा x = 3y +4 ………..(ii)
समीकरण (ii) x = 3y + 4 समीकरण (i) में रखने पर
3y + 4 – y = 26
⇒ 2y = 26 – 4
⇒ y = \(\frac{22}{2}\) = 11
समीकरण (i) में y = 11 रखने पर,
x – 11 = 26
⇒ x = 26 + 11 = 37
अत: संख्याएँ 37 और 11 हैं।

प्रश्न 16.
x और y के मान ज्ञात कीजिए :
\(\frac{2}{x}+\frac{3}{y}\) = 13
\(\frac{5}{x}-\frac{4}{y}\) = – 2
हल :
माना \(\frac{1}{x}\) = a तथा \(\frac{1}{y}\) = b, तब दिए गए समीकरण निम्न प्रकार से होगे :
2a + 3b = 13 …………(i)
5a – 4b = – 2 …………(ii)
समीकरण (i) को 4 से तथा (ii) को उसे गुणा करके जोड़ने पर,
8a + 12b = 52
15a – 12b = – 6
23a = 46
⇒ a = \(\frac{46}{23}\) = 2
समीकरण (i) में a = 2 रखने पर,
2 × 2 + 3b = 13
⇒ 3b = 13 – 4 = 9
⇒ b = \(\frac{9}{3}\) = 3
अब a = \(\frac{1}{x}\) ⇒ \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)
और b = \(\frac{1}{y}\) ⇒ \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)
अतः x = \(\frac{1}{2}\) तथा y = \(\frac{1}{3}\)

प्रश्न 17.
k के किन मानों (किस मान) के लिए निम्न समीकरणों के युग्म का एक अद्वितीय हल है :
x + 2y = 5 और 3x + ky + 15 = 0
हल :
दिया है, समीकरण युग्मः
x + 2y = 5
और 3x + ky + 15 = 0
दिए गए समीकरणों के युग्म की समीकरणों a₁x + b₁y + c₁ = 0 तथा a₂x + b₂y + c₂ = 0 से तुलना करने पर,
a₁ = 1, b₁ = 2, c = – 5, a₂ = 3, b₂ = k, c₂ = 15
∵ दिए गए युग्म का एक अद्वितीय हल है,
∴ \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)
अर्थात् \(\frac{1}{3}\) ≠ \(\frac{2}{k}\)
k ≠ 6
अत: 6 के अतिरिक्त, k के प्रत्येक मान के लिए दिए हुए समीकरणों के युग्म का एक अद्वितीय हल होगा।

प्रश्न 18.
c का मान कीजिए, यदि समीकरण निकाय cx + 3y + (3 – c) = 0, 12x + cy – c = 0 के अपरिमित रूप से अनेक हल हैं।
हल :
दिए गए समीकरण हैं :
cx + 3y + (3 – c) = 0
तथा 12x + cy – c = 0
दिए गए समीकरण युग्म की समीकरण युग्म a₁x + b₁y + c₁ = 0 तथा a₂x + b₂y + c₂ = 0 से तुलना करने पर,
a₁ = c, b₁ = 3, c₁ = 3 – c
तथा a₂ = 12, b₂ = c, c₂ = – c
रैखिक समीकरणों के युग्म के अपरिमित रूप से अनेक हल होने के लिए शर्त:

प्रश्न 19.
K का मान ज्ञात कीजिए यदि समीकरण निकाय 2x + 3y = 7, (k + 1) x + (2k – 1) y = 4k + 1 के अपरिमित रूप से अनेक हल हैं।
हल :
दिए गए समीकरणों का निकाय है:
2x + 3y = 7
तथा (k + 1) x + (2k – 1)y = 4k + 1
जहाँ, a₁ = 2, b₁ = 3, तथा c₁ = – 7
और a₂ = (k + 1), b₂ = (2k – 1) तथा c₂ = – (4k + 1)
रैखिक समीकरणों के युग्म के अपरिमित रूप में अनेक हल होने के लिए शर्त

⇒ 4k – 2 = 3k + 3
4k – 3k = 3 + 2
k = 5
तथा 12k + 3 = 14k – 7
12k – 14k = – 7 – 3
– 2k = – 10
k = 5
अतः k = 5 है।

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. दो चरों वाले रैखिक समीकरण का आलेख सदैव एक ………………. रेखा को निरूपित करता है।
2. वह समीकरण युग्म जिसका हल अद्वितीय होता है, रैखिक समीकरणों का ………….. युग्म कहलाता है।
3. वह समीकरण युग्म जिसका कोई हल नहीं है, रैखिक समीकरणों का ………………… युग्म कहलाता है।
4. k का वह मान जिसके लिए समीकरण निकाय x + 2y = 3 तथा 5x + ky = 7 का कोई हल नहीं है, है ……………..
5. ………………… विधि में दोनों समीकरणों के दो चरों में से एक चर के गुणांक के समान करके विलुप्त कर दूसरे चर का मान ज्ञात करते हैं।

हल :

1. सरल,
2. संगत,
3. असंगत,
4. 10 या ± \(\frac{14}{3}\)
5. विलोपन

निम्न कथनों में सत्य / असत्य बताइए :

प्रश्न (ख)

1. दो चरों वाले रैखिक समीकरण का आलेख सदैव एक सरल रेखा को निरूपित करता है।
2. x और y के मानों से सम्बद्ध कोई भी युग्म जो दोनों समीकरणों को सन्तुष्ट करता हो, युग्म का शून्यक कहलाता है।
3. प्रतिच्छेदी रेखाओं के अनन्त हल होते हैं।
4. संपाती रेखाओं का केवल एक हल होता है।
5. समांतर रेखाओं का निकाय असंगत होता है।

हल :

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
k का वह मान जिसके लिए समीकरण निकाय x + y – 4 = 0 तथा 2x + ky = 3 का कोई हल नहीं है, है:
(A) – 2
(B) ≠ 2
(C) 3
(D) 2
हल :
दिया है, निकाय का कोई हल नहीं है।

अत: सही विकल्प (D) है।

प्रश्न 2.
K का वह मान जिनके लिए रैखिक समीकरण युग्म kx + y = k² तथा x + ky = 1 के अपरिमित रूप से अनेक हल हैं, है:
(A) ± 1
(B) 1
(C) – 1
(D) 2
हल :
अपरिमित रूप से अनेक हल के लिए शर्त :

अतः सही विकल्प (A) है।

प्रश्न 3.
k का वहमान जिसके लिए रैखिक समीकरण निकाय x + 2y = 3, 5x + ky + 7 = 0 असंगत है, है :
(A) \(\frac{-14}{3}\)
(B) \(\frac{2}{5}\)
(C) 5
(D) 10
हल :
रैखिक समीकरण निकाय के लिए अंसगत होने के लिए शर्त्त :

अत: सही विकल्प (D) है।

प्रश्न 4.
रैखिक समीकरणों y = 0 तथा y = – 6 के युग्म का एक:
(A) अद्वितीय हल है
(B) कोई हल नहीं है
(C) अनेक हल हैं
(D) सिर्फ एक हल (0, 0) है
हल :
सही विकल्प (B) है।

प्रश्न 5.
रैखिक समीकरणों \(\frac{3 x}{2}+\frac{5 y}{3}\) = 7 तथा 9x + 10y = 10 का युग्म :
(A) संगत है
(B) असंगत है
(C) संगत है तथा सिर्फ एक हल है।
(D) संगत है तथा अनेक हल हैं
हल :

अतः सही विकल्प (B) है।

प्रश्न 6.
k का मान जिसके लिए समीकरण 3x – y + 8 = 0 तथा 6x + ky = – 16 संपाती रेखाओं को व्यक्त करें, है:
(A) –\(\frac{1}{2}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) – 2
हल :
∵ संपाती रेखाओं के लिए शर्त :

अत: सही विकल्प (D) है।

प्रश्न 7.
यदि रैखिक समीकरणों का कोई युग्म संगत हो, तो इसके आलेख की रेखाएँ होंगी :
(A) समांतर
(B) सदैव संपाती
(C) प्रतिच्छेदी या संपाती
(D) सदैव प्रतिच्छेदी
हल :
सही विकल्प (C) है।

प्रश्न 8.
समीकरण x = a और y = b का युग्म आलेखीय रूप से वे रेखाएँ निरूपित करता है, जो :
(A) समांतर हैं-
(B) (b, a) पर प्रतिच्छेद करती हैं।
(C) संपाती हैं
(D) (a, b) पर प्रतिच्छेद करती हैं
हल :
सही विकल्प (D) हैं।

प्रश्न 9.
आश्रित रैखिक समीकरणों के युग्म का एक समीकरण – 5x + 7 = 2 है। दूसरा समीकरण हो सकता है –
(A) 10x + 14y + 4 = 0
(B) – 10x – 14y + 4 = 0
(C) – 10x + 14y + 4 = 0
(D) 10x – 14y = – 4
हल :
सही विकल्प (D) है।

प्रश्न 10.
मेरी आयु पुत्र की आयु की तिगुनी है। 13 वर्ष बाद मेरी आयु पुत्र की आयु की दुगुनी रह जाएगी। मेरी और मेरे पुत्र की आयु बताइए-
(A) 39 वर्ष, 13 वर्ष
(B) 45 वर्ष, 15 वर्ष
(C) 30 वर्ष, 10 वर्ष
(D) 36 वर्ष, 12 वर्ष ।
हल :
मान लीजिए मेरी आयु (वर्षो में) x और मेरे पुत्र की आयु (वर्षो में) y है ।
प्रश्नानुसार, x = 3y ⇒ x – 3y = 0 … (i)
और x + 13 = 2(y + 13)
अर्थात् x – 2y = 26 – 13
⇒ x – 2y – 13 = 0
वज्रगुणन द्वारा समी. (i) व (ii) को हल करने पर

अर्थात् मेरी आयु 39 वर्ष और मेरे पुत्र की आयु 13 वर्ष है।
अत: सही विकल्प (A) है।

प्रश्न 11.
दो अंकों वाली संख्या के अंकों का योगफल 7 है। अंकों का क्रम उलट देने पर प्राप्त संख्या पहली संख्या से 9 अधिक है। वह संख्या ज्ञात कीजिए :
(A) 43
(B) 34
(C) 52
(D) 25
हल :
मान लीजिए दी हुई संख्या में दहाई अंक x और इकाई अंक y है। तब
दी हुई संख्या = 10x + y
अंकों का क्रम उलटने पर संख्या = 10y + x
प्रश्नानुसार, x + y = 7 ⇒ x + y – 7 = 0 ………….(i)
और (10x + y) + 9 = 10y + x
और 9x – 9y + 9 = 0
या x – y + 1 = 0 ……(ii)
समी. (i) व (ii) को वज्रगुणन द्वारा हल करने पर,

अतः अभीष्ट संख्या = 10x + y = 10 × 3 + 4 = 30 + 4 = 34 है।
सही विकल्प (B) है।

प्रश्न 12.
एक लड़के की आयु अभी अपनी माता की आयु की एक तिहाई है। यदि माता की वर्तमान आयु x वर्ष है तो 12 वर्ष बाद लड़के की आयु होगी :
(A) \(\frac{x}{3}\) + 12
(B) \(\frac{x+12}{3}\)
(C) x + 4
(D) \(\frac{x}{3}\) – 12
हल :
यहाँ माता की वर्तमान आयु = x वर्ष
माना पुत्र की वर्तमान आयु = y वर्ष
लेकिन प्रश्नानुसार, y = x × \(\frac{1}{3}\)
y = \(\frac{x}{3}\)
पुत्र की 12 वर्ष पश्चात् आयु = (y + 12) वर्ष
y = \(\frac{x}{3}\) रखने पर
अतः 12 वर्ष बाद लड़के की आयु = \(\frac{x}{3}\) + 12
सही विकल्प (A) है।

प्रश्न 13.
C के किस मान के लिए समीकरण युग्म Cx – y = 2 तथा 6x – 2y = 3 के अनन्त हल हैं:
(A) 3
(B) – 3
(C) – 12
(D) कोई मान नहीं
हल :
समीकरण युग्म को निम्न प्रकार लिखा जा सकता है :
Cx – y – 2 = 0 ……………(i)
तथा 6x – 2y – 3 = 0 ……………(ii)
समीकरण युग्म के अनन्त हल के लिए प्रतिबन्ध

अत: विकल्प (D) सही है।

प्रश्न 14.
k के किस मान के लिए समीकरण युग्म 4x – 3y = 9, 2x + ky = 11 का कोई हल नहीं है-
(A) \(\frac{9}{11}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{-3}{2}\)
(D) \(\frac{-2}{3}\)
हल :
दिया गया समीकरण युग्म है :
4x – 3y – 9 = 0, 2x + ky – 11 = 0
कोई हल न होने के लिए प्रतिबन्ध

⇒ k = \(\frac{-3}{2}\) या k ≠ \(\frac{-11}{3}\)
अतः k = \(\frac{-3}{2}\)
अत: सही विकल्प (C) है।

प्रश्न 15.
यदि समीकरणों 5x + 2y = 16 और 3x + \(\frac{6}{5}\)y = 2 का हल होगा :
(A) संगत
(B) असंगत
(C) दोनों (A) और (B)
(D) इनमें से कोई नहीं।
हल :
दिए गए समीकरण युग्म की निम्न प्रकार से भी लिख सकते हैं-
5x + 2y – 16 = 0 ……….(i)
15x + 6y – 10 = 0 ……….(ii)

अतः दिए गए समीकरण युग्म का कोई हल नहीं है। दिया गया समीकरण युग्म असंगत है ।
सही विकल्प (B) है।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution :
(i) We know that, cosec²A = 1 + cot² A ⇒ cosec A = \(\sqrt{1+\cot ^2 \mathrm{~A}}\)

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution :
(i) sin A sin² A + cos² A = 1
sin² A = 1 – cos² A

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 (90 – θ) = cos θ
sin (90 – 27°) = cos 27°
sin 63° = cos 27°
sin² 63 = cos² 27°

cos (90 – θ) = sin θ
cos (90° – 73°) = sin 73°
cos (17°) = sin 73°
cos² 17° = sin² 73°

(ii) sin 25° cos 65° + cos 25° sin 65°
sin (90° – θ) = cos θ
sin (90° – 25°) = cos 25°
sin (90° – 65°) = cos 65°
sin 25° = cos 65°

cos (90° – θ) = sin θ
cos (90° – 65°) = sin 65°
cos 25° = sin 65°

sin 25° = cos 65°
cos 65° . cos 65° + sin 65° . sin 65°
∴ cos² 65 + sin² 65 = 1

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A=
(A) 1
(B) 9
(C) 8
(D) 0
Solution :
9 sec² A – 9 tan² A
9(1 + tan² A) – 9 tan² A
9 + 9 tan² A – 9 tan² A = 9.
∴ 9 sec² A – 9 tan² A = 9.

(ii) (1 + tan θ + sec θ) (1 + cot 0 – cosec θ) =
(A) 0
(B) 1
(C) 2
(D) – 1
Solution :

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

(iv) \(\frac{1+\tan ^2 A}{1+\cot ^2 A}\) =
(A) sec² A
(B) – 1
(C) cot² A
(D) tan² A
Solution :

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

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

LHS = sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + (1 + cot² A) + 2 sin A . \(\frac{1}{sin A}\) + (1 + tan² A) + 2 cos A . \(\frac{1}{cos A}\) (∵ 1 + cot² A = cosec² A and sec² A = 1 + tan² A)
= 1 + 1 + cot² A + 2 + 1 + tan² A + 2 (∵ sin² A + cos² A = 1)
= 7 + tan² A + cot² A = RHS.

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
Solution :
[Hint: Simplify LHS and RHS separately].
LHS = (cosec A – sin A) (sec A – cos A)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

In each of the following, give the justification of the construction also:

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution :

Steps of construction:
1. Draw the line segment AB = 7.6 cm.
2. At A, below AB, make an angle BAx = 30°.
3. At B, above AB, make an angle ABy = 30°.
B\(\hat{A}\)x = A\(\hat{B}\)y = 30°
These are alternate angles. ∴ Ax || By.
4. With a convenient radius cut off five equal parts
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ in Ax.
5. With the same radius cut off eight equal parts.
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈.
6. Join A₅B₈. Let it cut AB at C.
AC : CB = 5 : 8.

In ΔACA₅ and CBB₈
1. A\(\hat{C}\)A₅ = B\(\hat{C}\)B₈ (V.O.A.)
2. C\(\hat{A}\)A₅ = A\(\hat{B}\)B₈ (Alternate angles)
3. C\(\hat{A}\)₅A = B\(\hat{B}\)₈C (Alternate angles)
Δs are equiangular.
∴\(\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{\mathrm{CA}_5}{\mathrm{AB}_8}=\frac{\mathrm{CA}}{\mathrm{BC}}\)
∴ \(\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{5}{8}\)

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution :

Steps of construction:

1. Construct ΔABC given AB = 4 cm, BC = 5 cm, AC = cm.
2. At B, make an acute angle CBx.
3. Divide Bx into three equal parts with a convenient radius.
4. Join B₃C.
5. From B₂ draw a parallel to B₃C.
6. Let it cut BC at C’.
7. At C’ make angle A’C’B = ACB.

Join A’C’.
∴ ΔABC ||| A’BC’.

In ΔABC and ΔA’BC’,
1.A\(\hat{B}\)C = A’\(\hat{B}\)C’ (Common angle)
2. A\(\hat{C}\)B = A’\(\hat{C’}\)B (Corresponding angles)
3. B\(\hat{A}\)C = B\(\hat{A’}\)C’ (Remaining angles)
Δs are equiangular.
∴ \(\frac{\mathrm{AC}}{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC} \mathrm{C}^{\prime}}\) = \(\frac{3}{2}\) ∴ [latex]\frac{BC’}{BC}=\frac{2}{3}[/latex]

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution :
Steps of construction:

1. Construct a triangle ABC given BC= 7 cm, AB = 5 cm, AC = 6 cm.
2. At B, below BC, make an acute angle CBx.
3. With a convenient radius cut off seven equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
4. Join B₅C.
5. From B₁ draw a parallel to B₅C to cut BC produced at C’.
6. At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle.
In ΔABC and ΔA’BC’

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution :

Steps of construction:

1. Draw a line segment BC = 8 cm.
2. Draw its perpendicular bisector.
3. Cut off DA = 4 cm. (altitude given)
4. Join AB and AC. ABC is the required triangle.
5. At B, below BC, draw an acute angle CBx.
6. With a convenient radius cut off three equal parts BB₁ = B₁B₂ = B₂B₃.
7. Join B₂C.
8. At B₃ draw a parallel to B₂C to meet BC extended at C’.
9. At C’ draw a parallel to CA to meet BA produced at A’.
10. Join A’C’. A’BC’ is the required triangle similar to ΔABC.

In ΔA’BC’ and ABC
1. A’\(\hat{B}\)C’ = A\(\hat{B}\)C (Common angle)
2. B\(\hat{A}\)‘C’ = B\(\hat{A}\)C (Corresponding angles)
3. A’\(\hat{C}\)‘B = A\(\hat{C}\)B
Δs are equiangular.
∴ \(\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{\mathrm{BC}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}\)
\(\frac{BC’}{BC}\) = \(\frac{3}{2}\)
BC = 8 cm, BC’ = 8 × \(\frac{3}{2}\) = 12 cm.
BA = CA = 5.6 cm ∴ A’B = A’C’ = 5.6 × \(\frac{3}{2}\) = 8.4 cm.

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution :

Steps of construction:

1. Construct ΔABC given BC = 6 cm, AB = 5 cm, A\(\hat{B}\)C = 60°.
2. At B, below BC, make an acute angle CBx.
3. In Bx cut off four equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Join B₄C.
5. From B₃ draw a parallel to B₄C to meet BC at C’.
6. At C’ draw a parallel to CA to meet CA at A’.

A’BC’ is the required triangle similar to ΔABC.
In Δs A’BC’ and ABC

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ΔABC.
Solution :

\(\hat{A}\) + \(\hat{B}\) + \(\hat{C}\) = 180°
\(\hat{A}\) + \(\hat{B}\) = 150° (105° + 45°)
∴ \(\hat{C}\) = 30°
Steps of construction:
1. In ΔABC, BC = 7 cm.
\(\hat{B}\) = 45°, \(\hat{C}\) = 105°
∴ \(\hat{A}\) = 180° – 45° – 105°
= 180° – 150° = 30°.
Construct ΔABC given BC = 7 cm, \(\hat{B}\) = 45°, \(\hat{C}\) = 30°.
2. At B draw an acute angle CBx.
3. In Bx cut off four equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄ with a convenient radius.
4. Join B₃C.
5. At B₄ draw a parallel to B₃C to meet BC produced at C’.
6. At C’ make angle of 30° equal to A\(\hat{C}\)B. Let it meet BA produced at A’.
A’BC’ is the required triangle similar to ΔABC.
In ΔC’BB₄ CB₃ || C’B₄

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution :

BC’ = \(\frac{5}{3}\) × BC = \(\frac{5}{3}\) × 3 = 5 cm
BA’= \(\frac{5}{3}\) × 4 = \(\frac{20}{3}\) = 6.6 cm
A’C’ = \(\frac{5}{3}\) × 5 = \(\frac{25}{3}\) = 8.3 cm.

Steps of construction:

1. Construct ΔABC given BC= 3 cm, \(\hat{B}\) = 90°, BA = 4 cm.
2. At B, make an acute angle CBx.
3. Cut off five equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ along Bx with a convenient radius.
4. Join B₃C.
5. At B₅ draw a parallel to B₃C to meet BC produced at C’.
6. At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle similar to ΔABC.
BC’ : BC = BB₅ = BB₃
\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_5}{\mathrm{BB}_3}\) = \(\frac{5}{3}\)

In Δs A’BC’ and ABC
1. A’\(\hat{B}\)C’ = A\(\hat{B}\)C = 90°
2. B\(\hat{A}\)‘C’ = B\(\hat{A}\)C (Corresponding angles)
3. B\(\hat{C}\)‘A’ = B\(\hat{C}\)A
Δs are equiangular.
∴ \(\frac{A’C’}{AC}\) = \(\frac{BC’}{BC}\) = \(\frac{BA’}{BA}\) = \(\frac{5}{3}\)

Students should go through these JAC Class 10 Maths Notes Chapter 15 प्रायिकता will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 10 Maths Notes Chapter 15 प्रायिकता

भूमिका :
कक्षा IX में हमने किसी घटना के प्रायोगिक या अनुभाविक प्रायिकता के बारे में अध्ययन किया था। स्मरण करें जिस प्रायिकता का अनुमापन हम वास्तविक प्रयोगों के परिणामों तथा घटनाओं के घटित होने की पर्याप्त रिकार्डिंग के आधार पर करते हैं, उस प्रायिकता को प्रायोगिक या आनुभाविक प्रायिकता कहते हैं।

प्रायिता-एक सैद्धान्तिक दृष्टिकोण :
उदाहरण के लिए-जब हम एक सिक्का उछालें तो यह समतल पर गिरेगा। यह एक क्रिया या प्रयोग है जिससे दो परिणाम प्राप्त हो सकते हैं-चित (अशोक वाला) तल दिखायी दे अथवा पट (दूसरा तल) दिखायी दे ये दो सम्भव घटनाएँ (Events) हैं।
(i) इस प्रकार हम देखते हैं कि सिक्का उछालने के किसी प्रयोग में कुल सम्भावित घटनाएँ दो हैं चित गिरने की सम्भावना 2 घटनाओं में से 1 है और पट गिरने की सम्भावना भी उतनी ही है। तब गणित की भाषा में हम कहते हैं कि सिक्के के चित गिरने की प्रायिकता \(\frac{1}{2}\) है। सिक्के के पट गिरने की भी प्रायिकता \(\frac{1}{2}\) है।

(ii) किसी चुनाव में A, B, C, D चार उम्मीदवार खड़े हैं, तब सफल प्रत्याशी के चुनाव की कुल 4 सम्भावनाएँ हैं, क्योंकि चारों प्रत्याशियों में से कोई भी सफल हो सकता है, तब प्रत्याशी A की सफलता की सम्भावना 4 में से 1 है।
हम गणित की भाषा में कह सकते हैं कि प्रत्याशी A की सफलता की प्रायिकता \(\frac{1}{4}\) है; शेष प्रत्याशियों B, C व D की सफलता की भी प्रायिकता उतनी ही है। यहाँ हम कह सकते हैं कि किसी प्रत्याशी की असफलता की प्रायिकता है क्योंकि एक प्रत्याशी के सफल होने की दशा में शेष 3 के असफल होने की घटना अवश्य घटित होगी।

(iii) ताश की गड्डी में से पत्ता खींचिए। इसके लाल होने की प्रायिकता पर विचार कीजिए। कुल 52 पत्तों की गड्डी में 26 लाल और 26 काले पत्ते होंगे, तब पत्ते के लाल होने की प्रायिकता \(\frac{26}{52}\) या \(\frac{1}{2}\) है। खींचे गए पत्ते के लाल न होने की भी प्रायिकता \(\frac{26}{52}\)या \(\frac{1}{2}\) ही है।
→ प्रायिकता (Probability) : जब किसी घटना के घटित होने की सम्भावना संख्यात्मक रूप में व्यक्त की जाती है तो उसे प्रायिकता कहते हैं।

→ प्रयोग (Experiment) : वह विधि जिसके द्वारा हमें किसी प्रेक्षण का परिणाम प्राप्त होता है प्रयोग कहलाता है।

→ परिणाम (Outcome) : किसी प्रयोग के विशिष्ट निष्कर्ष को परिणाम कहते हैं।

→ यादृच्छया प्रयोग (Random Experiment) : ऐसे प्रयोग जिनके निश्चित परिणाम नहीं हों यादृच्छया प्रयोग कहलाते हैं।

→ अभिप्रयोग (Trial) : किसी भी यादृच्छिक प्रयोग को करने की प्रक्रिया को एक अभिप्रयोग (trial) कहते है।

→ पूरक घटनाएँ (Complimentary events) : यदि ‘E’ किसी घटना के घटित होने की प्रायिकता है तथा ‘E नहीं’ न घटित होने की प्रायिकता है तब E और ‘E नहीं’ पूरक घटनाएँ कहलाती हैं।
∴ P(E) + P(E नहीं) = 1.

→ मिश्रित घटनाएँ (Compound events ) : किसी यादृच्छिक प्रयोग से जुड़ी ऐसी घटनाएँ जो दो या दो से अधिक प्रारम्भिक घटनाओं के मिश्रण से प्राप्त होती हैं, मिश्रित घटनाएँ कहलाती हैं।

→ अनुकूल प्रारंभिक घटना (Favourable elementary event) एक ऐसी प्रारम्भिक घटना जो किसी मिश्रित घटना के अनुकूल हो अनुकूल प्रारम्भिक घटना कहते हैं।

→ पासा (Dice or die) : एक छोटा घन जिसके पृष्ठों पर 1 से 6 तक संख्याएँ लिखी हों पासा कहलाता है। प्रायिकता एक सैद्धान्तिक दृष्टिकोण

→ यादृच्छया उछाल (Random toss) : किसी सिक्के को बिना किसी पक्षपात (bias) या रुकावट के स्वतन्त्रता पूर्वक गिरने दिया जाता है तो उसे यादृच्छया उछाल कहते हैं।

→ समप्रायिक (Equally likely) जब किसी यादृच्छया प्रयोग में किसी घटना E के सभी परिणामों के प्राप्त होने की सम्भावना समान होती है तो यह परिणाम समप्रायिक कहलाते हैं।
जैसे- किसी सिक्के को उछालने में चित या पट आना अथवा पाँसे की उछाल में पासे पर अंकित अंकों में से कोई अंक प्राप्त होने की घटना समप्रायिक घटनाएँ हैं।

→ सैद्धान्तिक प्रायिकता (Theoretical Probability) किसी प्रयोग में यदि एक घटना E घटित होती है और यह कल्पना करने पर प्रयोग में घटना E के सभी परिणाम बिल्कुल एक हों तो.

इस प्रकार से अवकलित प्रायिकता को सैद्धान्तिक प्रायिकता कहते हैं।

→ प्रारम्भिक घटना (Elementary event) : वह घटना है जिसका केवल एक ही परिणाम है। किसी पासे को उछालने पर 5 प्राप्त होना प्रारम्भिक घटना है क्योंकि 5 प्राप्त होने का केवल एक ही परिणाम है।
परन्तु विषम या सम संख्या प्राप्त होना प्रारम्भिक घटना नहीं है क्योंकि विषम संख्या प्राप्त होने की घटना के 1, 3, 5 तीन परिणाम हैं। ऐसे ही सम संख्या या 5 से छोटी संख्या प्राप्त होने की घटना भी प्रारम्भिक घटना नहीं है क्योंकि इनके परिणाम 1 से अधिक हैं।
किसी प्रयोग की सभी प्रारम्भिक घटनाओं की प्राथकिताओं का योग होता है।

→ असम्भव घटना (Impossible event) : ऐसी घटना जिसकी प्रायिकता शून्य होती है अर्थात् उस घटना का घटित होना असम्भव हो तो उस घटना को असम्भव घटना (Impossible Event) कहते हैं।
किसी पासे की उछाल में “6 से बड़ा अंक प्राप्त होने की घटना” असम्भव घटना है।

→ निश्चित घटना (Sure event) यदि किसी घटना की प्रायिकता हो तो उस घटना को निश्चित घटना (Sure event) कहते हैं।
जैसे-किसी पासे की फेंक में 7 से छोटी संख्या प्राप्त होने की घटना के सभी परिणाम 1, 2, 3, 4, 5, 6 सम्भव हैं। यह एक निश्चित घटना है।

→ घटना (E) घटित नहीं हो : इसे E1 या \(\bar{E}\) से व्यक्त करते हैं तथा घटना को “E नहीं” से जाना जाता है। यदि किसी घटना की प्रायिकता सदैव शून्य के बराबर या उससे अधिक तथा के बराबर या उससे कम होती है।
0 ≤ P(E) ≤ 1

ताश के पत्ते (Playing Cards) :
ताश की गड्डी में कुल पत्तों की संख्या = 52
कुल पत्ते 4 समूहों में विभाजित होते हैं-
(1) हुकुम (Spades), (2) पान (Hearts), (3) ईट (Diamonds), (4) चिड़ी (Clubs)

प्रत्येक समूह में तेरह पत्ते होते हैं अर्थात्
हुकुम के 13, पान के 13, ईंट के 13 और चिड़ी के 13 पत्ते होते हैं।
काले रंग के पत्ते : हुकुम 13 व चिड़ी 13 = कुल 26
लाल रंग के पत्ते : पान 13 व ईंट 13 = कुल 26
प्रत्येक समूह में इक्का (ace), बादशाह (king), बेगम (queen) और गुलाम (jack) उच्च क्रम के पत्ते होते हैं। इक्का (ace) पहले क्रम का पत्ता है और बादशाह (king), बेगम (queen) तथा गुलाम (jack) फेस पत्ते (face cards) कहलाते हैं।

ध्यान देने योग्य बिन्दु :

उक्त प्रकार से अवकलित प्रायिकता को सैद्धान्तिक प्रायिकता कहते हैं।
→ किसी प्रयोग की सभी प्रारम्भिक घटनाओं की प्रायिकता का योग 1 है। यह व्यापक रूप में भी सत्य है।
→ व्यापक रूप में किसी घटना E के लिए सत्य है कि P(\(\bar{E}\)) = 1 – P(E)
→ एक निश्चित या निर्धारित घटना की प्रायिकता 1 होती है।
→ एक असम्भव घटना की प्रायिकता 0 होती है।
→ घटना E की प्रायिकता एक ऐसी संख्या है, कि (PE)
0 ≤ P(E) ≤ 1
→ किसी भी घटना E के लिए P(E) + P(\(\bar{E}\)) = 1 होता है।
जहाँ \(\bar{E}\) घटना ‘E नहीं’ को व्यक्त करता है। E और \(\bar{E}\) पूरक घटनाएँ कहलाती हैं।
→ एक असम्भव घटना की प्रायिकता 0 (या 0%) होती है। जैसे-सूर्य के कभी अस्त नहीं होने की प्रायिकता 0 है।

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 15 प्रायिकता Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 15 प्रायिकता

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
किसी प्रयोग की सभी प्रारम्भिक घटनाओं की प्रायिकताओं का योग लिखिए।
हल:
किसी प्रयोग की सभी प्रारम्भिक घटनाओं की प्रायिकताओं का योग 1 होता है।

प्रश्न 2.
एक थैले में 3 लाल और 5 काली गेंदें हैं। इस थैले में से एक गेंद यादृच्छया निकाली जाती है। इसकी प्रायिकता क्या है कि गेंद काली नहीं है ?
हल:
थैले में गेंदों की कुल संख्या = 3 लाल + 5 काली = 8
∴ कुल सम्भव परिणाम = 8
काली गेंद न होने की घटना के अनुकूल परिणामों की संख्या = 3
∴ गेंद काली न होने की प्रायिकता = \(\frac{3}{8}\)

प्रश्न 3.
दो पासों को एक साथ फेंका जाता है। इसकी प्रायिकता क्या है कि दोनों पासों पर आने वाली संख्याओं का योग 7 है।
हल:
जब दो पासों को एक साथ फेंका जाता है तब सम्भावित परिणामों की संख्या = 6 × 6 = 36
दोनों पासों पर आने वाली ऐसी संख्याएँ जिनका योग 7 है :
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2) और (6, 1)
∴ अनुकूल परिणामों की संख्या = 6
दोनों पासों पर आने वाली संख्याओं का योग 7 की प्रायिकता = \(\frac{6}{36}=\frac{1}{6}\)

प्रश्न 4.
52 पत्तों की अच्छी प्रकार से फेंटी गई ताश की एक गड्डी में से यादृच्छया एक पत्ता निकाला गया। प्रायिकता ज्ञात कीजिए कि निकाला गया पत्ता-
(i) लाल रंग का बादशाह है।
(ii) एक बेगम अथवा गुलाम है।
हल:
ताश की गड्डी में 52 पत्ते होते हैं तथा एक पत्ता 52 तरीकों से निकाला जा सकता है।
∴ कुल सम्भावित परिणामों की संख्या = 52
(i) माना कि लाल रंग का बादशाह होने की घटनां R है।
अतः घटना (R) के अनुकूल परिणामों की संख्या = 2
P(R) = \(\frac{2}{52}=\frac{1}{26}\)

(ii) माना कि एक गुलाम अथवा बेगम होने की घटना A है।
∴ घटना (A) के अनुकूल परिणामों की संख्या = 8
∴ P(A) = \(\frac{8}{52}=\frac{2}{13}\)
अत: (I) P(R) = \(\frac{1}{26}\), (ii) P(A) = \(\frac{2}{13}\)

प्रश्न 5.
52 पत्तों की ताश की एक गड्डी में से सभी बादशाह, बेगम तथा इक्के निकाल दिए गए। शेष बचे पत्तों को भली प्रकार फेंटने के पश्चात् उनमें से एक पत्ता निकाला गया। प्रायिकता ज्ञात कीजिए कि निकाला पत्ता-
(i) एक काले रंग का तस्वीर वाला पत्ता है।
(ii) एक लाल रंग का पत्ता है।
हल:
ताश की गड्डी में कुल पत्तों की संख्या = 52
बादशाह, बेगम तथा इक्के के पत्तों की संख्या = 12
बादशाह, बेगम तथा इक्के के पत्तों को निकालने के बाद शेष बचे पत्तों की संख्या 52 – 12 = 40
∴ कुल सम्भव परिणामों की संख्या = 40
(i) माना कि निकाले गए पत्ते की काले रंग के तस्वीर वाले पत्ते होने की घटना ‘A’ है।
40 पत्तों में काले रंग के तस्वीर वाले पत्तों की संख्या = 2
∴ घटना (A) के अनुकूल परिणामों की संख्या = 2
⇒ P(A) = \(\frac{2}{40}=\frac{1}{20}\)

(ii) माना कि निकाले गए पत्ते की लाल रंग का होने की घटना ‘B’ है।
लाल रंग के पत्तों की संख्या = 20
∴ घटना (B) के अनुकूल परिणामों की संख्या = 20
⇒ P(B) = \(\frac{20}{40}=\frac{1}{2}\)
अत: (i) P(A) = \(\frac{1}{20}\)
(ii) P(B) = \(\frac{1}{2}\)

प्रश्न 6.
एक पेटी में 30 डिस्क हैं, जिन पर 1 से 30 तक संख्याएँ अंकित हैं। यदि इस पेटी में से एक डिस्क यादृच्छया निकाली जाती है तो इसकी प्रायिकता ज्ञात कीजिए कि इस डिस्क पर अंकित होगी-
(i) दो अंकों की एक संख्या
(ii) एक पूर्ण वर्ग संख्या
हल:
दिया है : पेटी में कुल डिस्क = 30
(i) 30 डिस्क में से दो अंकों की संख्याएँ हो सकती हैं-
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30
दो अंकों की डिस्क के अनुकूल परिणाम = 21 डिस्क संभावित परिणाम = 30
अतः यादृच्छया दो अंकों की एक संख्या होने की प्रायिकता

(ii) एक पूर्ण वर्ग संख्या = 4, 9, 16, 25
अनुकूल परिणाम = 4
संभावित परिणाम = 30
अतः यादृच्छया एक पूर्ण वर्ग संख्या होने की प्रायिकता = \(\frac{4}{30}=\frac{2}{15}\)

प्रश्न 7.
अच्छी तरह से फेंटी गई एक ताश की गड्डी से एक पत्ता यादृच्छया निकाला गया। प्रायिकता ज्ञात कीजिए कि निकाला गया पत्ता-
(i) हुकुम का पत्ता है या एक इक्का है
(ii) एक काले रंग का बादशाह है
(iii) न तो गुलाम है तथा न ही बादशाह है
(iv) या तो बादशाह हैं या बेगम है
हल:
कुल पत्तों की संख्या = 52
(i) हुकुम और इक्का के पत्तों की संख्या 13 + 3 = 16
हुकुम और इक्का के निकलने की प्रायिकता = \(\frac{16}{52}=\frac{4}{13}\)
(ii) गड्डी में काले रंग का बादशाह = 2
काले रंग के बादशाह की प्रायिकता = \(\frac{2}{52}=\frac{1}{26}\)
(iii) गुलाम और बादशाह के पत्तों की संख्या = 4 + 4 = 8
एक गड्डी में न तो गुलाम है तथा न ही बादशाह के पत्तों की संख्या = 52 – 8 = 44
न गुलाम न बादशाह होने की प्रायिकता = \(\frac{44}{52}=\frac{11}{13}\)
(iv) एक गड्डी में बादशाह और बेगम के पत्तों की संख्या = 4 + 4 = 8
या तो बादशाह या बेगम होने की प्रायिकता = \(\frac{8}{52}=\frac{2}{13}\)

प्रश्न 8.
दो भिन्न पासों को एक साथ उछाला गया दोनों पासों के ऊपरी तलों पर आई संख्याओं का गुणनफल 6 आने की प्रायिकता ज्ञात कीजिए।
हल:
जब दोनों पासों को एक साथ फेंका जाता है तो संभावित परिणामों की संख्या 6 × 6 = 36
दोनों पासों पर संख्याओं का गुणनफल 6 प्राप्त होने की अनुकूल स्थितियाँ = (1, 6), (6, 1), (2, 3), (3, 2)
दोनों पासों का गुणनफल 6 होने के अनुकूल परिणामों की संख्या = 4
अभीष्ट परिणाम = \(\frac{4}{36}=\frac{1}{9}\)

प्रश्न 9.
एक जार में केवल लाल, नीली तथा नारंगी रंग की गेंदें हैं। यादृच्छया एक लाल रंग की गेंद के निकालने की प्रायिकता \(\frac{1}{4}\) है। इसी प्रकार उसी जार से यादृच्छया एक नीली गेंद निकालने की प्रायिकता \(\frac{1}{3}\) है। यदि नारंगी की कुल गेंदें 10 हैं, तो बताइए कि जार में कुल कितनी गेंदें हैं।
हल:
माना लाल, नीली तथा नारंगी गेंद निकालने की प्रायिकता क्रमश: E₁, E₂, E₃ है।
हम जानते हैं,
P(E₁) = \(\frac{1}{4}\)
P(E₂) = \(\frac{1}{3}\)
P(E₃) = 1 – PE₂ – PE₁
P(E₃) = \(1-\frac{1}{3}-\frac{1}{4}\)
P(E₃) = \(\frac{5}{12}\)
हमयह भी जानते हैं कि

कुल गेदों की संख्या = \(\frac{120}{5}\) = 24
अतः जार में गेंदों की कुल संख्या = 24 है।

प्रश्न 10.
एक जार में केवल नीले, काले तथा हरे कंचे हैं। इस जार में से यादृच्छया एक नीले कंचे के निकालने की प्रायिकता \(\frac{1}{5}\) है तथा उसी जार में से एक काले कंधे के यादृच्छयता निकालने की प्रायिकता \(\frac{1}{4}\) है। यदि जार में 11 हरे रंग के कंचे हैं, तो जार में कुल कंचों की संख्या ज्ञात कीजिए।
हल:
दिया है,
जार से नीले कंचे निकालने की प्रायिकता P(x) = \(\frac{1}{5}\)
जार से काले कंचे निकालने की प्रायिकता P(y) = \(\frac{1}{4}\)
जार में हरे कंचों की कुल संख्या = 11
माना, जार से हरे कंचे निकालने की प्रायिकता P(z) है।
अतः P(x) + P(y) + P(z) = 1

प्रश्न 11.
एक पिग्गी बैंक में, ₹ 1 के सौ सिक्के, ₹ 2 के 25 सिक्के, फू 5 के 15 सिक्के और ₹ 10 के दस सिक्के हैं। यदि पिग्गी बैंक को हिलाकर उल्टा करने पर कोई एक सिक्का गिरने के परिणाम समप्रायिक हैं, तो इसकी क्या प्रायिकता है कि वह गिरा हुआ सिक्का :
(i) ₹ 2 का होगा ?
(ii) ₹ 5 का होगा ?
हल:
कुल सम्भव परिणामों की संख्या = 100 + 25 + 15 + 10 = 150
(i) अनुकूल परिणामों की संख्या = 25
∴ ₹ 2 का सिक्का गिरने की प्रायिकता = \(\frac{25}{150}=\frac{1}{6}\)
(ii) अनुकूल परिणामों की संख्या = 15
∴ ₹ 5 का सिक्का गिरने की प्रायिकता = \(\frac{15}{150}=\frac{1}{10}\)

प्रश्न 12.
एक डिब्बे में 7 लाल कंचे, 10 सफेद कंचे और 5 हरे कंचे हैं। इस डिब्बे में से एक कंचा यादृच्छया निकाला जाता है। इसकी प्रायिकता क्या है कि निकाला गया कंचा (i) लाल नहीं है? (ii) सफेद है ? (iii) हरा है?
हल:
कुल सम्भव परिणामों की संख्या = 7 + 10 + 5 = 22
(i) अनुकूल परिणामों की संख्या = 7
लाल कंचे के निकलने की प्रायिकता = \(\frac{7}{22}\)
∴ लाल कंचे के न निकलने की प्रायिकता = \(1-\frac{7}{22}\)
= \(\frac{22-7}{22}=\frac{15}{22}\)
(ii) अनुकूल परिणामों की संख्या = 10
सफेद कंचे के निकलने की प्रायिकता = \(\frac{10}{22}=\frac{5}{11}\)
(iii) अनुकूल परिणामों की संख्या = 5
हरे कंचे के निकलने की प्रायिकता = \(\frac{5}{22}\)

प्रश्न 13.
दो विभिन्न पासों को एक साथ उछाला गया। निम्न के आने की प्रायिकता ज्ञात कीजिए :
(i) एक द्विक आना।
(ii) दोनों पासों पर आई संख्याओं का योग 10 आना।
हल:
दो पासों को एक साथ उछालने पर आने वाले कुल परिणाम = 36
(i) एक द्विक आने के संभावित परिणाम = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
कुल सम्भावित परिणामों की संख्या = 6
P (एक द्विक आना) = संभावित परिणाम / कुल परिणाम
= \(\frac{6}{36}=\frac{1}{6}\)
(ii) संख्याओं का योग 10 आने के संभावित परिणाम = [(4, 6), (5, 5) (6, 4)]
कुल संभावित परिणामों की संख्या = 5
P (योग 10) = संभावित परिणाम / कुल परिणाम
= \(\frac{6}{36}=\frac{1}{12}\)

प्रश्न 14.
1 से 100 के बीच की संख्याओं में से यादृच्छया एक संख्या चुनी गई। प्रायिकता ज्ञात कीजिए कि यह संख्या
(i) 8 से भाज्य है।
(ii) 8 से भाज्य नहीं है।
हल:
1 और 100 के बीच आने वाली कुल संख्या = 98
(i) 8 से भाज्य संख्या
= {(8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
कुल संभावित परिणामों की संख्या = 12
P(8 से भाज्य संख्या) = संभावित परिणाम / कुल परिणाम
= \(\frac{12}{98}=\frac{6}{49}\)

(ii) P(8 से अभाज्य संख्या)
\(P \bar{E}\) = 1 – P(E)
= 1 – \(\frac{6}{49}\) = \(\frac{43}{49}\)

प्रश्न 15.
एक थेले में 5 लाल गेंदे तथा कुछ नीली गेंदे हैं। यदि थैले में से यादृच्छया एक नीली गेंद निकालने की प्रायिकता, एक लाल गेंद के निकलने की प्रायिकता की तीन गुना है, तो थैले में नीली गेंदों की संख्या ज्ञात ‘कीजिए।
हल:
थैले में लाल गेंदों की संख्या = 5
माना थैले में नीली गेंदों की संख्या x है।
∴ थैले में से एक लाल गेंद निकालने की प्रायिकता = \(\frac{5}{x+5}\)
तथा थैले में से एक नीली गेंद निकालने की प्रायिकता = \(\frac{x}{x+5}\)
प्रश्नानुसार, \(\frac{x}{x+5}=3 \times \frac{5}{x+5}\)
⇒ \(\frac{x}{x+5}=\frac{15}{x+5}\)
⇒ x = 15
अतः थैले में नीली गेंदों की संख्या = 15

प्रश्न 16.
किसी यादृच्छया लिए गए वर्ष के नवम्बर मास में 5 रविवार होने की प्रायिकता ज्ञात कीजिए।
हल:
यादृच्छया लिए गए वर्ष के नवम्बर मास में 4 पूर्ण सप्ताह तथा 2 अन्य दिन होते हैं।
ये दो अन्य दिन निम्न में से कोई भी हो सकते हैं:
(सोम, मंगल), (मंगल, बुध), (बुध, गुरु), (गुरु, शुक्र), (शुक्ल, शनि), (शनि, रवि), (रवि, सोम)
पाँच रविवार होने के लिए अनुकूल परिणाम (शनि, रवि) तथा (रवि सोम)
अत: वांछित प्रायिकता = \(\frac{2}{7}\) ₹

प्रश्न 17.
तीन बच्चों वाले एक परिवारों में, कम-से-कम दो लड़के होने की प्रायिकता ज्ञात कीजिए।
हल:
तीन बच्चों वाले परिवारों में बच्चों के होने के परिणाम निम्नलिखित हैं :
{BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG}
कुल सम्भव परिणामों की संख्या = 8
कम-से-कम दो लड़के होने के अनुकूल परिणाम = {BBB, BBG BGB, GBB}
अनुकूल परिणामों की संख्या = 4
∴ कम से कम दो लड़के होने की प्रायिकता = \(\frac{4}{8}=\frac{1}{2}\)

वस्तुनिष्ठ प्रश्न :

(क) रिक्त स्थानों की पूर्ति कीजिए :
1. यदि P(E’) ‘नहीं की’ प्रायिकता 0.95 है, तो P(E) की प्रायिकता …………… होगी।
2. एक सिक्के को एक बार उछाला जाता है। इसे पट नहीं आने की प्रायिकता …………… है।
3. एक निश्चित घटने वाली घटना की प्रायिकता ……………. होती है।
4. अंग्रेजी वर्णमाला में से एक अक्षर यदृच्छया चुने जाने पर उसके व्यंजक होने की प्रायिकता ……………… है।
5. यदि किसी के जीतने की प्रायिकता 0.07 है तो उसके हारने की प्रायिकता …………….. है।
उत्तर:
1. 0.05,
2. \(\frac{1}{2}\)
3. 1
4. \(\frac{21}{26}\)
5. 0.93

(ख) निम्न में सत्य / असत्य बताइए :
1. एक पासा एक बार उछाले जाने पर 3 से छोटी संख्या प्राप्त करने की प्रायिकता \(\frac{1}{3}\) है।
2. 52 पत्तों की एक गड्डी में से एक लाल रंग के बादशाह आने की प्रायिकता \(\frac{1}{26}\) है।
3. यदि कल वर्षा होने की प्रायिकता 0.85 है, तो वर्षा न होने की प्रायिकता 0.58 है।
4. दो पासों को एक साथ फेंकने पर दोनों पासों पर संख्याओं का योग 13 आने की प्रायिकता \(\frac{2}{13}\) है।
5. यदि पासों के एक युग्म को एक बार उछाला गया, तो योगफल 8 आने की प्रायिकता \(\frac{5}{36}\) है।
उत्तर:
1. सत्य,
2. सत्य,
3. असत्य,
4. असत्य,
5. सत्य

बहुविकल्पीय प्रश्न :

प्रश्न 1.
एक थैले में 3 लाल, 5 काली तथा 7 सफेद गेंदे हैं। इस थैले में से एक गेंद को यादृच्छया निकाला जाता है। निकाली गई गेंद काली नहीं है, इसकी प्रायिकता है :
(A) \(\frac{1}{3}\)
(B) \(\frac{9}{15}\)
(C) \(\frac{5}{10}\)
(D) \(\frac{2}{3}\)
हल:
कुल गेंदे, n(S) = 3 + 5 + 7 = 15
अनुकूल परिणामों की संख्या = 5
∴ एक काली गेंद निकलने की प्रायिकता

अत: सही विकल्प (D) है।

प्रश्न 2.
संख्याओं 1, 2, 3, ……. 15 से यादृच्छया 4 का एक गुणज चुने जाने की प्रायिकता है:
(A) \(\frac{4}{15}\)
(B) \(\frac{2}{15}\)
(C) \(\frac{1}{15}\)
(D) \(\frac{1}{5}\)
हल:
कुल परिणाम की संख्या = 5
अनुकूल परिणाम = {4, 8, 12}
अनुकूल परिणामों की संख्या = 3
∴ 4 का गुणज चुने जाने की प्रायिकता = \(\frac{3}{15}=\frac{1}{5}\)
सही विकल्प (D) है।

प्रश्न 3.
दो सिक्के एक साथ उछाले गए। अधिक से अधिक एक चित आने की प्रायिकता है:
(A) \(\frac{1}{4}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{3}{4}\)
हल:
सम्भव परिणाम = {HT, HH, TH, TT}
कुल सम्भव परिणामों की संख्या = 4
अनुकूल परिणाम = {HT, TH}
अनुकूल परिणामों की संख्या = 2
अधिक से अधिक एक चित आने की प्रायिकता = \(\frac{2}{4}=\frac{1}{2}\)
सही विकल्प (B) है।

प्रश्न 4.
अच्छी प्रकार से फेंटी गई 52 पत्तों की ताश की गड्डी में से एक पत्ता यदृच्छया निकाला गया है। एक गुलाम के आने की प्रायिकता क्या है?
(A) \(\frac{3}{26}\)
(B) \(\frac{1}{52}\)
(C) \(\frac{1}{13}\)
(D) \(\frac{3}{52}\)
हल:
कुल सम्भव परिणामों की संख्या = 52
अनूकूल परिणामों की संख्या = 4
∴ गुलाम के आने की प्रायिकता = \(\frac{4}{52}=\frac{1}{13}\)
सही विकल्प (C) है।

प्रश्न 5.
किसी असम्भव घटना के होने की प्रायिकता है:
(A) 1
(B) \(\frac{3}{4}\)
(C) परिभाषित नहीं
(D) 0
हल:
सही विकल्प (D) हैं।

प्रश्न 6.
52 ताशों की एक गड्डी में से एक ताश निकाला जाता है। इसके लाल रंग का मुख कार्ड होने की प्रायिकता है :
(A) \(\frac{3}{26}\)
(B) \(\frac{1}{52}\)
(C) \(\frac{1}{13}\)
(D) \(\frac{3}{52}\)
हल:
ताश की गड्डी में कार्डों की कुल संख्या = 52
∴ कुल सम्भव परिणामों की संख्या = 52
लाल रंग के मुख कार्डों की संख्या = 6
∴ घटना के अनुकूल परिणामों की संख्या = 6
ताश की गड्डी में मुख कार्ड होने की प्रायिकता = \(\frac{6}{52}=\frac{3}{26}\)
अत: सही विकल्प (A) है।

प्रश्न 7.
52 ताशों की एक गड्डी में से एक कार्ड निकाला जाता है। कार्ड का ईंट का इक्का न होना घटना E है। E के अनुकूल परिणामों की संख्या है:
(A) 4
(B) 13
(C) 48
(D) 51
हल:
गड्डी में ताशों की कुल संख्या = 52
ताश की गड्डी में से एक कार्ड निकाला जाता है।
कार्ड का ईंट का इक्का न होना घटना E है।
∴ E के अनुकूल परिणामों की संख्या = 52 – 1 = 51
अत: सही विकल्प (D) है।

प्रश्न 8.
कोई लड़की यह परिकलित करती है कि उसके द्वारा एक लाटरी में प्रथम पुरुस्कार जीतने की प्रायिकता 0.08 है। यदि 6000 टिकट बेचे गए हैं तो उस लड़की ने कितने टिकट खरीदे हैं ?
(A) 40
(B) 240
(C) 480
(D) 750
हल:
बेचे गए टिकटों की संख्या = 6000
∴ कुल सम्भव परिणामों की संख्या = 6000
माना कि लड़की द्वारा खरीदे गए टिकटों की संख्या = x
∴ घटना के अनुकूल परिणामों की संख्या = x
प्रथम पुरुस्कार जीतने की प्रायिकता

⇒ x = 0.08 × 6000
⇒ x = 480
अत: सही विकल्प (C) है।

प्रश्न 9.
ताश की गड्डी में एक काले रंग के गुलाम न होने की प्रायिकता होगी :
(A) \(\frac{1}{26}\)
(B) \(\frac{3}{4}\)
(C) \(\frac{11}{52}\)
(D) \(\frac{25}{26}\)
हल:
ताश की गड्डी में कुल पत्तों की संख्या = 52
अतः कुल सम्भावित परिणामों की संख्या = 52
ताश की गड्डी में काले रंग के गुलाम (हुकम, चिढ़ी) होने के अनुकूल परिणामों की संख्या = 2
ताश की गड्डी में काले रंग के गुलाम होने की प्रायिकता = \(\frac{2}{52}=\frac{1}{26}\)
ताश की गड्डी में काले रंग के गुलाम न होने की प्रायिकता = \(1-\frac{1}{26}=\frac{25}{26}\)
अत: सही विकल्प (D) है।

प्रश्न 10.
एक थैले में कार्ड हैं जिन पर 2, 3, 4, …, 11 संख्याएँ अंकित हैं। थैले में से यादृच्छया एक कार्ड निकाला गया है। निकाले गए कार्ड पर एक अभाज्य संख्या होने की प्रायिकता है:
(A) \(\frac{1}{2}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{3}{10}\)
(D) \(\frac{5}{9}\)
हल: थैले में कार्डों पर अंकित संख्याएँ हैं :
2, 3, 4, ………., 11
थैले में कार्डों की संख्या = 10
कार्डों पर अंकित अभाज्य संख्याएँ हैं :
2, 3, 5, 7, 11
कार्ड पर एक अभाज्य संख्या होने के अनुकूल परिणामों की संख्या = 5
∴ कार्ड पर एक अभाज्य संख्या होने की प्रायिकता = \(\frac{5}{10}=\frac{1}{2}\)
अतः सही विकल्प (A) है।

प्रश्न 11.
एक बक्से में कार्ड हैं जिन पर 6 से 50 तक की संख्याएँ अंकित हैं। बक्से में से यादृच्छया एक कार्ड निकाला गया। इस कार्ड पर अंकित संख्या के एक पूर्ण वर्ग होने की प्रायिकता है:
(A) \(\frac{1}{45}\)
(B) \(\frac{2}{15}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{4}{45}\)
हल:
बक्से में कार्डों पर 6 से 50 तक की संख्याएँ अंकित हैं।
कुल सम्भव परिणामों की संख्या = 45
कार्डों पर अंकित पूर्ण वर्ग संख्याएँ हैं:
9, 16, 25, 36, 49
कार्ड पर पूर्ण वर्ग संख्या होने के अनुकूल परिणामों की संख्या = 5
कार्ड पर पूर्ण वर्ग संख्या होने की प्रायिकता = \(\frac{5}{45}=\frac{1}{9}\)
अतः सही विकल्प (C) है।

प्रश्न 12.
एक पासे को एक बार फेंका जाता है। एक विषम संख्या के आने की प्रायिकता है :
(A) 1
(B) \(\frac{1}{9}\)
(C) \(\frac{4}{6}\)
(D) \(\frac{2}{6}\)
हल:
एक पासे को एक बार फेंका जाता है तो पासे के पृष्ठों पर आने वाली संख्याएँ हैं :
{1, 2, 3, 4, 5, 6}
∴ सम्भव परिणामों की संख्या = 6
तथा अभाज्य संख्याएँ हैं : {1, 3, 5}
अभाज्य संख्या आने के अनुकूल परिणामों की संख्या = 3
अभाज्य संख्या आने की प्रायिकता = \(\frac{3}{6}=\frac{1}{2}\)
अतः सही विकल्प (B) है।

प्रश्न 13.
किसी स्कूल में पाँच सदन A, B, C, D और E हैं। किसी कक्षा में 23 विद्यार्थी हैं, जिनमें से 4 सदन A से, 8 सदन B से, 5 सदन C से 2 सदन D से तथा शेष सदन E से हैं। इनमें से एक विद्यार्थी को कक्षा का मॉनीटर बनाने के लिए चुना जाता है। चुने गए इस विद्यार्थी के सदनों A, B और C से न होने की प्रायिकता है :
(A) \(\frac{4}{23}\)
(B) \(\frac{6}{23}\)
(C) \(\frac{8}{23}\)
(D) \(\frac{17}{23}\)
हल:
कक्षा में विद्यार्थियों की कुल संख्या = 23
सदन A, B तथा C में विद्यार्थियों की संख्या का योग = 4 + 8 + 5 = 17
कुल सम्भव परिणामों की संख्या = 23
घटना के अनुकूल परिणामों की संख्या = 17
विद्यार्थी के सदनों A, B और C से होने की प्रायिकता = \(\frac{17}{23}\)
विद्यार्थी के सदनों A, B और C से न होने की प्रायिकता = \(1-\frac{17}{23}=\frac{6}{23}\)
अतः सही विकल्प (B) है।

प्रश्न 14.
यदि दो पासों को एक साथ फेंका जाता है तो दोनों पाँसों पर एक ही संख्या प्राप्त होने की प्रायिकता होगी :
(A) \(\frac{1}{6}\)
(B) \(\frac{5}{6}\)
(C) \(\frac{1}{36}\)
(D) इनमें से कोई नहीं
हल:
जब दोनों पासों को एक साथ फेंका जाता है तो सम्भावित परिणामों की संख्या = 6 × 6= 36
दोनों पाँसों पर एक जैसी संख्या प्राप्त होने की अनुकूल स्थितियाँ
(1, 1); (2, 2); (3, 3); (4, 4); (5, 5); (6, 6)
दोनों पासों पर एक जैसी संख्या प्राप्त होने के अनुकूल परिणामों की संख्या = 6
अभीष्ट प्रायिकता = \(\frac{6}{36}=\frac{1}{6}\)
अंत: सही विकल्प (A) है।

प्रश्न 15.
इसकी प्रायिकता कि यादृच्छिक रूप से चुने गए एक ऐसे वर्ष में, जो अधिवर्ष (leap year) न हो 53 रविवार हों, निम्नलिखित हैं :
(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{3}{7}\)
(D) \(\frac{5}{7}\)
हल:
एक साधारण वर्ष (Non leap year) में 365 दिन होते हैं।
365 दिन में 52 सप्ताह व 1 दिन होते हैं।
अर्थात् 365 दिन में 52 सप्ताह में 52 रविवार होंगे जो 1 दिन बचा है। वह निम्न में से एक हो सकता है।
रविवार, सोमवार, मंगलवार, बुधवार, गुरूवार, शुक्रवार, शनिवार
∴कुल सम्भव परिणामों की संख्या = 7
रविवार आने की घटना के अनुकूल परिणामों की संख्या = 1
रविवार होने की प्रायिकता = \(\frac{1}{7}\)
अत: सही विकल्प (A) है।

JAC Board Class 9th Social Science Important Questions Geography Chapter 6 Population

I. Objective Type Questions

1. What is the counting of people in a country known as?
(a) census
(b) migration
(c) birth rate
(d) death rate.
Answer:
(a) census

2. Which of the following is calculated as the number of persons per unit area?
(a) population distribution
(b) population density
(c) total population
(d) None of the these.
Answer:
(b) population density

3. The population is generally grouped into which of the following categories?
(a) Aged (Above 59 years)
(b) Children (generally below 15 years)
(c) Working Age (15-59 years)
(d) All of the above.
Answer:
(d) All of the above.

4. From which of the following years, the birth rates started declining, resulting in a gradual decline in the birth of India?
(a) 1976
(b) 1990
(c) 1981
(d) 1988.
Answer:
(c) 1981

5. Which of the following along with basic sanitation amenities is available to only one- third of the rural population?
(a) Food Security
(b) Health
(c) Education
(d) Safe Drinking Water.
Answer:
(d) Safe Drinking Water.

II. Very Short Answer Type Questions

Question 1.
What is census?
Answer:
The census is the process of collection, compilation and publication of information relating to different aspects of people living in a country at a specific point of time.

Question 2.
When was the first census held in India?
Answer:
The first census in India (on a limited scale) was held in 1872. The first complete census was taken in 1881 and subsequently it has been taken every 10 years.

Question 3.
What is the total population of India according to 2011 Census?
Answer:
1,210.6 million.

Question 4.
What is India’s share in world population?
Answer:
About 17.5%.

Question 5.
Name the state having highest population in India.
Answer:
Uttar Pradesh.

Question 6.
Which are the most populated and least populated states in India?
Answer:
The most populated state in India is Uttar Pradesh and least populated is Sikkim.

Question 7.
Almost 50% of India’s population lives in five states. Write their names.
Answer:
Almost 50% of India population lives in the five states of Uttar Pradesh, Maharashtra, Bihar, West Bengal and Andhra Pradesh.

Question 8.
Name the less populated states of India.
Answer:
Rajasthan, Madhya Pradesh, Kashmir, Himachal Pradesh, Assam, Tripura, Naga-land, Meghalaya, Manipur, Mizoram, Sikkim and Arunachal Pradesh.

Question 9.
Name the states of India having high density of population.
Answer:
Uttar Pradesh, Maharashtra, Bihar, West Bengal and Andhra Pradesh.

Question 10.
What is the population density of India according to 2011 Census?
Answer:
It is 382 persons per square kilometre.

Question 11.
What is the major reason for the state of Kerala having a very high population density?
Answer:
Kerala has a very high population density because it has fertile soil and gets abundant rainfall, thus resulting in good prospects for agriculture.

Question 12.
What is called as annual growth rate?
Answer:
The rate or the pace of population increase per annum is called as the annual growth rate.

Question 13.
Mention the factors responsible for the population change.
Answer:

1. Death Rate,
2. Birth Rate,
3. Migration.

Question 15.
What kind of migration does not change the size of the population in a country?
Answer:
Internal migration from one city to another or from rural areas to urban areas within a country does not change the size of the population.

Question 16.
What type of migration leads to changes in the distribution of population within the nation?
Answer:
Internal migration leads to changes in the distribution of population within the nation.

Question 17.
What are the three divisions of age composition?
Answer:

1. Children (0-14 years age group),
2. Working Age (15-59 years), and
3. Aged (Above 59 years).

Question 18.
What is the sex-ratio in India according to 2011 Census?
Answer:
It is 943 females per 100 males.

Question 19.
Which states of India have the highest and the lowest sex ratio?
Answer:
Kerala has the highest sex ratio of 1084 and Haryana has the lowest sex ratio of 877 (as per the 2011 census).

Question 20.
In which state of India the sex ratio is favourable to women?
Answer:
Kerala.

Question 21.
Who is called a literate?
Answer:
A person aged 7 years and above, who can read and write with understanding in any language, is treated as literate. ‘

Question 22.
What is the literacy rate for the country as a whole?
Answer:
It is about 73 per cent.

Question 23.
What is the male and female literary rate and the general literacy level in India as per census 2011?
Answer:
As per census of 2011, the male literacy rate is 80.90% female literacy rate is 64.6% and general literacy rate is 74.04%.

Question 24.
Name the state having the highest literacy rate in India.
Answer:
Kerela.

Question 25.
What is known as occupational structure?
Answer:
The distribution of working population of an economy according to different occupations is known as occupational distribution of population or occupational structure.

Question 26.
What are the three sectors of occupations?
Answer:

1. Primary sector,
2. Secondary sector,
3. Tertiary sector.

Question 27.
What are primary activities?
Answer:
Primary activities include agriculture, animal husbandary, forestry, fishing, mining and quarrying etc.

Question 28.
What are secondary activities?
Answer:
Secondary activities include manufacturing industry, building and construction work etc.

Question 29.
What are tertiary activities?
Answer:
Tertiary activities include transport, communication, commerce, banking, administration and other services.

Question 30.
What is adolescent population?
Answer:
Adolescents are generally grouped in the age-group of 10 to 19 years. It constitutes one-fifth of the total population of India.

Question 31.
When was the comprehensive Family Planning Programme launched?
Answer:
In 1952.

Question 32.
When did the National Population Policy come into effect?
Answer:
The National Population Policy come into effect in the year 2000.

III. Short Answer Type Questions

Question 1.
What are the three major aspects of population study?
Answer:
1. Population size and distribution:
First of all, we have to see how many people are there in India and where they are located. Then we have to see which states are the most populated states and which are sparsely populated states.

2. Population growth and processes of population change: The second major aspect of population study is how the population has grown and how it has changed overtime.

3. Characteristics or qualities of the population: It includes the study of age, sex composition, literacy, occupational structure and health conditions of the people.

Question 2.
How is population density calculated? Where does India stand as compared to other countries with respect to population?
Answer:
The population density is calculated as the number of persons per unit of area. India’s stand in population density with respect to other countries is discussed below: India is one of the most densely-populated countries in the world.

As per census report 2011, the population density of India was 382 persons per sq. km. Due to change in climatic conditions, economic opportunities and other geographical factors, India has very regular distribution of population ranging from 1102 persons per sq. km. in Bi’ ar to only 17 persons per sq. km in Arunachal Pradesh. India is the third most dense country in the world after Bangladesh and Japan.

Question 3.
What is the growth of population?
Answer:
Growth of population refers to the change in the number of inhabitants of a country territory during a specific period of time, say during the last ten years. It can be expressed in two ways:

1. In terms of absolute numbers,
2. In terms of percentage change per year.

Question 4.
Describe the term annual growth rate of population. How is it affected by the birth rate?
Answer:
1. Annual Growth Rate: The rate at which the number of individuals in a population increase in one year as a fraction of the initial population is called annual growth rate of population. Effects of Birth

2. Rate on Annual Growth Rate: The annual growth rate is affected by the birth rate in the following ways.

1. With the increase in birth rate, the annual growth rate generally increases.
2. For a larger population even having a lower birth rate, the annual growth rate keeps on increasing.
3. For example, since 1981, the birth rate declined rapidly; still, 182 million people were added to the total population in 1990s alone. If we calculate annual growth rate based on this data, it becomes very high.

Question 5.
What are the causes of migration in India from rural to urban areas?
Answer:
Migration from rural to urban areas in India has taken place mainly due to:

1. Rising population in rural areas.
2. Poverty and unemployment in rural areas.
3. Lack of demand for labour in agriculture.
4. Increased employment opportunities, better education and living standard in urban areas.
5. Expansion of industrial and service sectors in the urban areas.

Question 6.
Explain the categorisation of the population of a nation on the basis of age composi¬tion.
Answer:
The population of a nation is generally grouped into three categories :

1. Children (below 15 years), who are economically unproductive and need to be provided with food, clothing, education and healthcare.
2. Working age (15 to 59 years), who are economically productive and biologically reproductive. They comprise the working population.
3. The aged or elderly (60 years and above), who can be economically productive though they may have retired. They may be working voluntarily, but they are not available for employment through recruitment.

Question 7.
What is the relationship between age composition and dependency ratio ? Briefly explain.
Answer:
Relationship between age composition and dependency ratio is as follows :

1. Children below 15 years of age are economically unproductive and people aged above 59 years do not get employment through recruitment.
2. The percentage of children and the aged affects the dependency ratio because these groups are not producers but are usually only consumers.

Question 8.
What are the reasons for low literacy rate among women in India?
Answer:

1. In India, women generally look after domestic work and are left with no time to get education, which leads to low literacy rate among them, mostly in rural areas of the country.
2. Lack of awareness and economic backwardness are other reasons for low literacy rate among women.

Question 9.
What do you understand by occupational structure? Briefly describe the occupational structure of India.
Answer:
1. Occupational structure: The distribution of population according to different types of occupations is referred to as the occupational structure.

2. Occupational Structure of India: In India, about 64 percent of the population is engaged only in agriculture. The proportion of population dependent on secondary and tertiary sectors is about 13 and 20 percent respectively. There has been an occupational shift in favour of secondary and tertiary sectors because of growing industrialisation and urbanisation in recent times.

Question 10.
Into how many categories are occupations generally classified?
Answer:
Occupations are generally classified into three categories:

1. Primary occupations: These include agriculture, animal husbandry, forestry, fishing, mining and quarrying etc.
2. Secondary occupations: These include manufacturing industry, building and construction works etc.
3. Tertiary occupations: These include transport, communication, commerce, banking, administration and other services.

III. Long Answer Type Questions

Question 1.
Write an essay on the population distribution in India.
Answer:
Population of our country is not evenly distributed. Some regions have high density of population. The population density of India in 2011 was 382 persons per sq. km. Bihar has the highest density of population about 1,102 per sq. km, whereas Arunachal Pradesh has the lowest density of population, i.e. 17 persons per sq.km.

1. Densely Populated Areas: These are those areas which have population of more than 300 persons per sq. km. The population is dense in these areas due to fertile soil and good rainfall.
   • Areas: Sutlej and Gangetic plain; Malabar coastal plain, Coromandel coast.
   • States: Punjab, Uttar Pradesh, Bihar, West Bengal, Kerala, Tamil Nadu.
2. Medium Density:
   These are those areas which have population about 100-300 persons per sq. km.
   • Areas: Brahmaputra valley, industrial areas, areas around the main ports.
   • States: Gujarat, Maharashtra, Goa, Karnataka, Odisha, Andhra Pradesh and Tamil Nadu.
3. Thinly or Sparsely Populated Areas:
   These are areas which have population less than 100 persons per sq. km. These are the areas of low unreliable and of hilly terrain where there is less levelled land for agriculture.
   • Areas: Great Indian Desert, Hills of north-eastern states.
   • States: Rajasthan, Madhya Pradesh, Kashmir, Himachal Pradesh, Assam, Tripura, Nagaland, Meghalaya, Manipur, Mizoram, Sikkim and Arunachal Pradesh.

Question 2.
What is age composition ? Why does it affect the population’s social and economic structure?
Answer:
The age composition of a population refers to the number of people indifferent age groups in a country. It affects the population’s social and economic structure because:

1. Children (0-14 years) do not contribute to the economy in any way. They require resources for their health, education etc.
2. Adults (15-59 years) contribute to the nation’s economy by earning money. They are the working population as they feed and look after the two age groups.
3. Aged (60 + years) do not contribute to the economy in any way. The depend on their children or their own saving. However, sometimes people belonging to this age group do work, but it is for private agencies as they are not considered for recruitment after that age.

Question 3.
What is sex ratio? Why has sex ratio been unfavourable to females? Explain it.
Answer:
Sex ratio: It is defined as the number of females per 1000 males in the population. This information is an important social indicator to measure the extent of equality between males and females in a society at a given time.
Reasons for unfavourable sex ratio for females:

1. The infant mortality rate in India is high and female infant mortality rate is still higher.
2. Preferential treatment is given to a male child and female children get neglected in most Indian homes.
3. People go though prenatal sex determination test. In case of a girl child, they abort the child.
4. Women generally have lower social political and economic status in the Indian society. We find dowry, murder, opposition to widow remarriage and low nutritional levels in women.
5. Lack of social awareness programmes among females, especially in rural areas.

Question 4.
Define the term ‘literate’. Describe the features of literacy rate in India.
Answer:
Definition of Literate: According to the census of 2001, a person aged 7 years and above, who can read and write with understanding in any language, is termed as literate. Features of Literacy Rate in India

1. The literacy rate is steadily improving in India.
2. As per census 2011, the literacy rate of India is 74.04 per cent.
3. The male literacy rate is 80.9 per cent.
4. The female literacy rate is 64.6 per cent.
5. India has a large gap in literacy rate between male and female population.
6. It also exhibits social inequality between males and females.
7. This gap in literacy rate is further increased by the declining sex ratio. Literacy is an important quality of population. For overall development and economic progress of the country, there should be high literacy rate among both males and females.

Question 5.
What is the health status of people at present? What measures have been taken to improve the health of the people? Why is the health situation still an issue of major concern for India?
Answer:
Health is an important component of population and composition, which affects the process of development. Following efforts of government programmes have registered significant improvement in the health conditions of the people :

1. Death rate has declined from 25 per 1000 population in 1951 to 7.2 per 1000 in 2011.
2. Life expectancy at birth has increased from 36.7 years in 1951 to 67.9 years in 2012.
3. This substantial improvement is the result of many factors, including improvement in public health, prevention of infectious diseases and application of modern medical practices in diagnosis and treatment of ailments.

However, despite considerable achievements, the health situation is still an issue of major concern for India due to the following reasons:
(a) The per capita calorie consumption is much below the recommended levels and malnutrition affects a large percentage of our population.

(b) Safe drinking water and basic sanitation amenities are available to only one- third of the rural population.

(c) These problems need to be tackled through an appropriate population policy.

Question 6.
Explain any six significant characteristics of the adolescent population of India.
Answer:
Six significant characteristics of the adolescent population of India are as follows:

1. Adolescent population is generally categorised in the age group of 10 to 19 years.
2. They constitute one-fifth of the total population of India.
3. They are the most important future resource.
4. Nutritional requirements of adolescents are higher than those of normal children or adults.
5. In India, a large number of adolescent girls suffer from anaemia.
6. The adolescent girls have to be sensitised to the problems they confront.

Question 7.
What is the National Population Policy (NPP 2000) ? Why was NPP 2000 initiated by the government?
Answer:
National Population Policy (NPP 2000) It is a comprehensive family planning programme initiated by the government of India. It provides a reliable and relevant policy framework for improving family welfare services and for measuring and monitoring the delivery of family welfare services and their demographic impact in future.
Reasons for Initiations NPP 2000: It was initiated by the government:

1. To improve healthcare quality and coverage, measuring and monitoring the delivery of family welfare programme.
2. To enable the increasingly literate and aware families to achieve their reproductive goals in the country.
3. To achieve rapid population stabilisation.
4. To promote synergy with the on-going educational, info-technology and socio¬economic transition.
5. To achieve rapid population stabilisation and sustainable development as well as improvement in economic, social and human development in the new millennium.

Question 8.
Write a note on National Population Policy (NPP) 2000 and Adolescents.
Answer:
NPP 2000 identified adolescents as one of the major sections of the population that need greater attention. Besides nutritional requirements, the policy provides greater emphasis on other important needs of adolescents, including protection from unwanted pregnancies and Sexually Transmitted Diseases (STDs) like AIDS.

It called for programmes that aim towards encouraging delayed marriage and child bearing, education of adolescents about the risks of unprotected sex, making contraceptive services accessible and affordable, providing food supplements, and strengthening legal measures to prevent child marriages.

Location and Labelling
1. The state having highest and lowest density population.
2. The state having highest and lowest sex ratio.
3. Largest and smallest state according to area.
Answer:
1. (a) The state having the highest density of population is Bihar.
(b) The state having the lowest density of population is Arunachal Pradesh.

2. (a) The state having highest sex ratio-Kerala.
(b) The state having lowest sex ratio-Haryana.

3. (a) Largest state – Rajasthan.
(b) Smallest state – Goa.

JAC Class 9 Social Science Important Questions

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60°

Solution :
(i) sin 60° cos 30° + sin 30° cos 60°
\(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\) + \(\frac{1}{2} \times \frac{1}{2}\)
\(\frac{3}{4}+\frac{1}{4}\) = 1

(ii) 2 tan² 45° + cos² 30° – sin² 60°
2(tan 45)² + (cos 30)² – (sin 60)²
2(1)² + (\(\frac{\sqrt{3}}{2}\))² – (\(\frac{\sqrt{3}}{2}\))² = 2

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
Solution :

(ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) =
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0
Solution :
\(\frac{1-(-1)^2}{1+(1)^2}=\frac{0}{2}\) = 0

(iii) sin 2A = 2 sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°
Solution :
sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2(0) = 0.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\) =
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution :

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\), 0° < A + B ≤ 90°; A > B, find A and B.
Solution :
tan(A + B) = \(\sqrt{3}\)
tan 60° = \(\sqrt{3}\)
∴ A + B = 60°
tan (A – B) = \(\frac{1}{\sqrt{3}}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ A – B = 30°
∴ \(\hat{\mathbf{A}}\) = 45°, \(\hat{\mathbf{B}}\) = 15°

A + B = 60°
45 + B = 60°
B = 60 – 45 = 15
∴ B = 15°

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution :
i) Let A = 30°, B = 60°.
sin 30° = \(\frac{1}{2}\)
sin 30 + sin 60°
sin (A + B) = sin 90° = 1
sin 60° = \(\frac{\sqrt{3}}{2}\)
\(\frac{1}{2}+\frac{\sqrt{3}}{2}\) = \(\frac{1+\sqrt{3}}{2}\)
sin (A+B) sin A + sin B
False.

iv) [θ = 0] sin 0 = 0
cos 0 = 1

θ = 30° sin 30° = \(\frac{1}{2}\)
cos 30° = \(\frac{\sqrt{3}}{2}\)

θ = 45° sin 45° = \(\frac{1}{\sqrt{2}}\)
cos 45° = \(\frac{1}{\sqrt{2}}\)

θ = 60° sin 60° = \(\frac{\sqrt{3}}{2}\)
cos 60° = \(\frac{1}{2}\)

θ = 90° sin 90° = 1
cos 90° = 0
False, because it is true only for θ = 45°.

(v) cot A = \(\frac{cos A}{sin A}\) . cot 0° = \(\frac{cos 0°}{sin 0°}\) = \(\frac{1}{0}\) = Undefined. True.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a, the nth term of the AP:

Solution:
1. Here, a = 7, d = 3, n = 8 and a_n is to be found.
We have a_n = a + (n – 1)d
a₈ = 7 + (8 – 1) 3 = 7 + 21 = 28

2. Here, a = -18, n = 10, a_n = a₁₀ = 0 and d is to be found.
a_n = a + (n – 1)d
∴ 0 = – 18 + (10 – 1)d
∴ 18 = 9d ∴ d = 2

3. Here, d = -3, n = 18, a_n = a₁₈ = -5 and a is to be found.
an = a + (n – 1)d
∴ -5 = a + (18 – 1)(-3)
∴ -5 = a – 51
∴ a = 51 – 5 ∴ a = 46

4. Here, a = -18.9, d = 2.5, a_n = 3.6 and n is to be found.
a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1)(2.5)
∴ 22.5 = 2.5 (n – 1)
∴ (n – 1) = \(\frac{22.5}{2.5}\)
∴ n – 1 = 9 ∴ n = 10

5. Here, a = 3.5, d = 0, n = 105 and a_n is to be found.
a_n = a + (n – 1) d
∴ a₁₀₅ = 3.5 + (105 – 1) (0)
∴ a₁₀₅ = 3.5

Question 2.
Choose the correct choice in the following and justify:
1. 30th term of the AP: 10, 7, 4, ……, is
(A) 97
(B) 77
(C) -77
(D) -87
2. 11th term of the AP: -3, –\(\frac{1}{2}\), 2, ….. is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
1. For the given AP 10, 7, 4,… a = 10,
d = 7 – 10 = -3 and n = 30
a_n = a + (n – 1)d
∴ a₃₀ = 10 + (30 – 1) (-3)
∴ a₃₀ = 10 – 87
∴ a₃₀ = -77
Thus, the correct choice is (C) -77.

2. For the given AP -3, –\(\frac{1}{2}\), 2, ….., a = -3.
d = –\(\frac{1}{2}\) – (-3) = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) and n = 11.
a_n = a + (n – 1)d
∴ a₁₁ = -3 + (11 – 1)(\(\frac{5}{2}\))
∴ a₁₁ = – 3 + 25
∴ a₁₁ = 22
Thus, the correct choice is (B) 22.

Question 3.
In the following APs, find the missing terms in the boxes:

Solution:
For the given AP, first term = a = 2 and third term = a + 2d = 26.
a = 2 and a + 2d = 26 gives d = 12.
Then, second term = a + d = 2 + 12 = 14
Thus, the missing term in the box is

2. For the given AP,
second term = a + d = 13 …….(1)
fourth term = a + 3d = 3 …….(2)
Solving equations (1) and (2) we get d = -5 and a = 18.
Now, first term = a = 18 and third term = a + 2d
= 18 + 2(-5) = 8.
Thus, the missing terms in the boxes are

Alternative Method:
Let the terms of the given AP be a₁, a₂, a₃, a₄.
Here, a₂ = 13 and a₄ = 3.
Now, a₄ – a₃ = a₃ – a₂ = d
∴ 3 – a₃ = a₃ – 13
∴ 2a₃ = 16
∴ a₃ = 8
Again, a₂ – a₁ = a₃ – a₂
∴ 13 – a₁ = 8 – 13
∴ 13 – a₁ = -5
∴ a₁ = 18
Thus, the missing terms in the boxes are

3. For the given AP
first term a = 5 ……….(1)
fourth term = a + 3d = 9\(\frac{1}{2}\) ……….(2)
From equations (1) and (2), we get a = 5 and d = 1\(\frac{1}{2}\).
Now, second term = a + d = 5 + 1\(\frac{1}{2}\) = 6\(\frac{1}{2}\)
and third term = a + 2d = 5 + 2 (1\(\frac{1}{2}\)) = 8
Thus, the missing terms in the boxes are

4. For the given AP
first term a = -4 ……….(1)
sixth term = a + 5d = 6 ……….(2)
From equations (1) and (2), we get
a = -4 and d = 2
Now, second term = a + d(-4) + 2 = -2,
third term = a + 2d = (-4) + 2 (2) = 0.
fourth term = a + 3d = (-4) + 3(2) = 2 and
fifth term = a + 4d = (-4) + 4(2) = 4.
Thus, the missing terms in the boxes are

5. For the given AP,
second term = a + d = 38 ……….(1)
sixth term = a + 5d = -22 ……….(2)
Solving equations (1) and (2), we get
d = -15 and a = 53.
Now, first term = a = 53,
third term = a + 2d = 53 + 2(-15) = 23,
fourth term = a + 3d = 53 + 3(-15) = 8 and
fifth term = a + 4d = 53 + 4(-15) = -7.
Thus, the missing terms in the boxes are

Question 4.
Which term of the AP: 3, 8, 13, 18, … is 78 ?
Solution:
Suppose nth term of the AP 3, 8, 13, 18, … is 78.
Here, a = 3, d = 8 – 3 = 5, a_n = 78 and n is to be found.
a_n = a + (n – 1)d
∴ 78 = 3 + (n – 1)5
∴ 75 = 5 (n-1)
∴ 15 = n – 1 ∴ n = 16
Thus, the 16th term of the AP 3, 8, 13, 18, …….., is 78.

Question 5.
Find the number of terms in each of the following APs:
1. 7, 13, 19, …….., 205
2. 18, 15\(\frac{1}{2}\), 13, ……., -47
Solution:
1. For the given finite AP 7, 13, 19, ….. 205, a = 7, d = 13 – 76 and last term l = 205.
Let us consider that the last term is the nth term.
a_n = a + (n – 1)d
∴ 205 = 7 + (n – 1)6
∴ 198 = 6 (n – 1)
∴ n – 1 = 33
∴ n = 34
Thus, there are 34 terms in the given finite AP.

2. For the given finite AP 18, 15\(\frac{1}{2}\), 13….. -47, a = 18, d = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\) and last term l = -47.
Let us consider that the last term is the nth term.
a_n = a + (n – 1) d
∴ -47 = 18 + (n-1) (-\(\frac{5}{2}\))
∴ -65 = –\(\frac{5}{2}\)(n – 1)
∴ n – 1 = 26
∴ n = 27
Thus, there are 27 terms in the given finite AP.

Question 6.
Check whether -150 is a term of the AP: 11, 8, 5, 2…
Solution:
If possible, let -150 be the nth term of the AP 11, 8, 5, 2,…
Here, a = 11; d = 8 – 11 = -3 and a_n = -150.
a_n = a + (n – 1)d
∴ -150 = 11 + (n – 1)(-3)
∴ -161 = -3 (n – 1)
∴ n – 1 = \(\frac{161}{3}\)
∴ n = \(\frac{164}{3}\)
∴ But n must be positive integer for an AP.
Hence, -150 cannot be a term of the AP 11, 8, 5, 2…..

Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
For any AP, a_n = a + (n – 1)d.
a₁₁ = a + 10 d
∴ a + 10 d = 38 ………(1)
∴ a₁₆ = a + 15d
∴ a + 15d = 73 ………(2)
Solving equations (1) and (2), we get
d = 7 and a = -32.
Now, 31st term = a₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
Thus, the 31st term of the given AP is 178.
Note: d = \(\frac{a_{16}-a_{11}}{16-11}=\frac{73-38}{5}=\frac{35}{5}=7\) can also be used.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
The given finite AP has 50 terms, hence its last term is a₅₀.
So, a₃ = 12 and a₅₀ = 106.
Now, a_n = a + (n – 1)d.
That gives, a₃ = a + 2d = 12 ………(1)
and a₅₀ = a + 49d = 106 ………(2)
Solving equations (1) and (2), we get
d = 2 and a = 8.
Now, 29th term = a₂₉ = a + 28d
∴ a₂₉ = 8 + 28 (2)
∴ a₂₉ = 64
Thus, the 29th term of the given AP is 64.

Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
For the given AP, a₃ = 4 and a9 = -8
We know, a_n = a + (n – 1)d.
∴ a₃ = a + 2d = 4 ………(1)
and a₉ = a + 8d = -8 ………(2)
Solving equations (1) and (2), we get
d = -2 and a = 8.
Now, let nth term of the AP be 0.
a_n = a + (n – 1)d
∴ 0 = 8 + (n – 1) (-2)
∴ 2(n – 1) = 8
∴ n – 1 = 4
∴ n = 5
Thus, the 5th term of the given AP is zero.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
For the given AP,
a₁₇ + 16d = a + 9d + 7a [∵ a_n = a + (n – 1) d]
∴ 7d = 7
∴ d = 1.
Thus, the common difference of the given AP is 1.

Question 11.
Which term of the AP:3, 15, 27, 39, …….. will be 132 more than its 54th term?
Solution:
For the given AP 3, 15, 27, 39, …….., a = 3
and d = 15 – 3 = 12.
Suppose nth term of the AP is 132 more than its 54th term.
∴ a_n = a₅₄ + 132
∴ a + (n – 1)d = a + 53d + 132
∴ 3 + (n – 1) (12) = 3 + 53(12) + 132
∴ 12(n – 1) = 12 (53 + 11)
∴ 12(n – 1) = 12 × 64
∴ n – 1 = 64
∴ n = 65
Thus, the 65th term of the given AP is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first term of two given APs be a₁ and a₂ respectively and let the same common difference be d.
Also, let a₁ > a₂
Then, 100th term of the first AP = a₁ + 99d
(a_n = a + (n – 1)d)
100th term of the second AP = a₂ + 99d.
The difference between their 100th terms is 100.
∴ (a₁ + 99d) – (a₂ + 99d) = 100 (a₁ > a₂)
∴ a₁ – a₂ = 100 ……(1)
Now, 1000th term of the first AP = a₁ + 999d
1000th term of the second AP = a₂ + 999d.
Then, the difference between their 1000th terms
= (a₁ + 999d) – (a₂ + 999d)
= a₁ – a₂
= 100 (by (1))
Thus, the difference between the 1000th terms of the two APs is 100.

Question 13.
How many three digit numbers are divisible by 7?
Solution:
The list of three digit numbers divisible by 7 is as below:
105, 112, 119, …….., 987, 994.
These numbers form a finite AP with a = 105, d = 112 – 105 = 7 and last term l = 994.
Suppose the last term of the AP is its nth term.
∴ l = a_n
∴ 994 = a + (n – 1)d
∴ 994 = 105 + (n – 1)7
∴ 7(n – 1) = 889
∴ n – 1 = 127
∴ n = 128
Hence, there are 128 terms in the AP.
Hence, 128 three digit numbers are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The multiples of 4 lying between 10 and 250 give rise to following finite AP:
12, 16, 20, ….., 244, 248.
Here, a = 12, d = 16 – 12 = 4 and last term l = 248.
If the last term is the nth term of the AP then l = a_n.
∴ l = a + (n – 1) d
∴ 248 = 12 + (n – 1)4
∴ 236 = 4 (n – 1)
∴ n – 1 = 59
∴ n = 60
Thus, there are 60 terms in the AP.
Thus, 60 multiples of 4 lie between 10 and 250.

Question 15.
For what value of n, are the nth terms of two APs 63, 65, 67, …… and 3, 10, 17,… equal?
Solution:
For the first AP 63, 65, 67, ….., a = 63, d = 65 – 63 = 2.
Then, nth term of the first AP a is given by a_n = a + (n – 1)d = 63 + (n – 1)(2).
For the second AP 3, 10, 17, ……, A = 3, D = 10 – 3 = 7.
Then, nth term of the second AP An is given by
A_n = A + (n – 1) D = 3 + (n – 1) (7).
Now, a_n = A_n
∴ 63 + (n – 1) (2) = 3 + (n – 1)(7)
∴ 63 – 3 = (n – 1)(7 – 2)
∴ 60 = 5(n – 1)
∴ n – 1 = 12
∴ n = 13
Thus, for n = 13, the nth term of two given APs are equal.

Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
For the given AP a₃ = 16 and a₇ = a₅ + 12.
For any AP, a_n = a + (n – 1) d.
∴ a + 2d = 16 and a + 6d = a + 4d + 12
a + 6d = a + 4d + 12 gives 2d = 12,
i.e., d = 6.
Substituting d = 6 in a + 2d = 16, we get a = 4.
Then, the required AP is 4, 4 + 6, 4 + 2 (6), 4 + 3(6), …..
Hence, the required AP is 4, 10, 16, 22, ……

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ….., 253.
Solution:
For the given finite AP 3, 8, 13, …., 253,
a = 3, d = 8 – 3 = 5 and last term l = 253.
Let the last term be its nth term.
∴ l = a_n
∴ l = a + (n – 1)d
∴ 253 = 3 + (n – 1)(5)
∴ 250 = 5 (n – 1)
∴ n – 1 = 50
∴ n = 51
Thus, there are in all 51 terms in the AP.
Now, the 20th term from the last term is (51 – 20 + 1)th term = 32nd term from the beginning.
a₃₂ = a + 31d
∴ a₃₂ = 3 + 31(5)
∴ a₃₂ = 158
Thus, the 20th term from the last term is 158.

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
For any AP, a_n = a + (n – 1) d.
a₄ = a + 3d, a₈ = a + 7d, a₆ = a + 5d and a₁₀ = a + 9d.
Now, a₄ + a₈ = 24 (Given)
∴ (a + 3d) + (a + 7d) = 24
∴ 2a + 10d = 24
∴ a + 5d = 12 ……..(1)
Again, a₆ + a₁₀ = 44 (Given)
∴ (a + 5d) + (a + 9d) = 44
∴ 2a + 14d = 44
∴ a + 7d = 22 …….(2)
Solving equations (1) and (2), we get d = 5 and a = -13.
Then, a₂ = a + d = 13 + (5) = -8 and
a₃ = a + 2d = -13 + 2(5) = -3.
Thus, the first three terms of the AP are -13, 8, 3.

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7,000?
Solution:
Subba Rao’s income in first year = ₹ 5,000
His income in second year = ₹ 5,000 + ₹ 200
= ₹ 5200
His income in third year = ₹ 5200 + ₹ 200
= ₹ 5400
and so on.
These numbers of his income (in rupees) form the AP 5000, 5200, 5400, …….
Here, a = 5000; d = 5200 – 5000 = 200;
a_n = 7000 and n is to be found.
a_n = a + (n – 1)d
∴ 7000 = 5000 + (n – 1)(200)
∴ 2000 = 200(n – 1)
∴ n – 1 = 10
∴ n = 11
Thus, Subba Rao’s income will reach ₹ 7000 in 11th year, i.e., in the year 2005.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
Ramkali’s savings in first week = ₹ 5
her savings in second week = ₹ 5 + ₹ 1.75
= ₹ 6.75,
her savings in third week = ₹ 6.75 + ₹ 1.75
= ₹ 8.50.
and so on.
Thus, the weekly savings (in rupees) of Ramkalt form the AP 5, 6.75, 8.50, …..
Here, a = 5; d = 6.75 – 5 = 1.75; a_n = 20.75 and n is to be found.
a_n = a + (n – 1)d
∴ 20.75 = 5 + (n – 1)(1.75)
∴ 1.75(n – 1) = 15.75
∴ n – 1 = 9
∴ n = 10
Thus, if Ramkali’s weekly savings is ₹ 20.75 in nth week, then n = 10.