## CBSE Sample Papers for Class 10 Maths Paper 11

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 11.

## CBSE Sample Papers for Class 10 Maths Paper 11

 Board CBSE Class X Subject Maths Sample Paper Set Paper 10 Category CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 11 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
• There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted.

Section – A

Question 1.
Write whether the rational number $\frac { 7 }{ 75 }$ will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Question 2.
Find the value (s) of k, if the quadratic equation 3x2 – k √3x + 4 = 0 has equal roots.

Question 3.
Find the eleventh term from the last term of the AP: 27, 23, 19,… ,- 65.

How do you round 214.0822 to the nearest tenth? … Hint: We are given a decimal number.

Question 4.
Find the coordinates of the point on y- axis which is nearest to the point (-2, 5).

Question 5.
In given figure, ST || RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of ∆ PST to the area of ∆ PRQ.

Question 6.
If cos A = $\frac { 2 }{ 5 }$ , find the value of 4 + 4 tan2 A.

Section – B

Question 7.
If two positive integers p and q are written as p = a2 b3 and q = a3 b; a, b are prime numbers, then verily: LCM (p, q) × HCF (p, q) = pq

Question 8.
The sum of first n terms of an AP is given by Sn = 2n2 + 3n. Find the sixteenth term of the A.P.

Question 9.
Find the value (s) of k for which the pair of linear equations kx + y = k2 and x + ky = 1 have infinitely many solutions.

Question 10.
If ( 1, $\frac { p }{ 3 }$ ) is the mid- point of the line segment joining the points (2,0) and ( 0, $\frac { 2 }{ 9 }$ ) then show that the line 5x + 3y + 2 = 0 passes through the point (-1 , 3p)

Question 11.
A box contains cards numbered 11 to 123. A card is drawn at random from the box. Find the probability that the number on the drawn card is

1. a square number
2. a multiple of 7.

To convert 10 11/16 to decimal you can use the long division method explained in our article fraction to decimal, which you can find in the header menu.

Question 12.
A box contains 12 balls of which some are red in colour. If 6 more red balls are put in the box and a ball is drawn at random, the probability of drawing a red ball doubles than what it was before. Find the number of red balls in the bag.

Section – C

Question 13.
Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Question 14.
Find all the zeroes of the polynomial 3x4 + 6x3 – 2x2 – 10x – 5 if two of its zeroes are √ $\frac { 5 }{ 3 }$ and – √ $\frac { 5 }{ 3 }$

Question 15.
Seven times a two digit number is equal to four times the number obtained by reversing the order of its digits. Ifthe difference ofthe digits is 3, determine the number.

Question 16.
In what ratio does the x-axis divide the line segment joining the points (-4,-6) and (-1, 7)? Find the co-ordinates of the point of division.
OR
The points A (4, -2), B (7,2), C (0,9) and D (-3,5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.

Question 17.
In given figure ∠1 = ∠2 and ∆NSQ = ∆MTR, then prove that ∆ PTS ~ PRQ

OR
In an equilateral triangle ABC, D is a point on the side BC such that BD = $\frac { 1 }{ 3 }$ BC. Prove that 9 AD2 = 7 AB2

Question 18.
In given figure XY and X’ Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’ Y’ at B. Prove that ∠ AOB = 90°.

Question 19.

Question 20.
In given figure ABPC is a quadrant of a circle ofradius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region

Question 21.
Water in a canal, 6 m wide and 1.5m 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?
OR
A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the remaining solid after the cone is carved out.

Question 22.
Find the mode of the following distribution of marks obtained by the students in an examination :

 Marks obtained 0-20 20-40 40-60 60-80 80-100 Number of students 15 18 21 29 17

Given the mean of the above distribution is 53, using empirical relationship estimate the value of its median.

Question 23.
A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train.
OR
Check whether the equation 5x2 – 6x – 2 = 0 has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.

Question 24.
An AP consists of 3 7 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Question 25.
Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
OR
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Question 26.
Draw a triangle ABC with side BC = 7 cm, ∠B=45°, ∠A= 105°. 1 Then, construct a triangle whose sides are $\frac { 4 }{ 3 }$ times the corresponding sides of ∆ABC.

Question 27.

Question 28.
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Question 29.
Two dairy owners A and B sell flavoured milk filled to capacity in mugs of negligible thickness, which are cylindrical in shape with a raised hemispherical bottom. The mugs are 14 cm high and have diameter of 7 cm as shown in given figure. Both A and B sell flavoured milk at the rate of ₹ 80 per litre. The dairy owner A uses the formula nrh to find the volume of milk in the mug and charges ₹ 43.12 for it. The dairy owner B is of the view that the price of actual quantity of milk should be charged. What according to him should be the price of one mug of milk? Which value is exhibited by the dairy owner B ? [ use π = $\frac { 22 }{ 7 }$ )

Question 30.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency k.

 Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Number of children 3 6 9 13 k 5 4

The following frequency distribution shows the distance (in metres) thrown by 68 students in a Javelin throw competition.

 Distance (in m) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Number of students 4 5 13 20 14 8 4

Draw a less than type O give for the given data and find the median distance thrown using this curve.

Solutions

Solution 1.
Non terminating repeating decimal expansion.

Solution 2.
k = ± 4

Solution 3.
a11 = -25

Solution 4.
(0,5)

Solution 5.
9 : 49

Solution 6.
4 + tan2 A = 4 sec2 A = 4 ( $\frac { 5 }{ 2 }$ ) 2 = 25

Solution 7.
LCM (p, q) = a3 b3
HC,F (p, q) = a2 b
LCM (p, q) × HCF (p, q) = a5 b4 = (a2 b3)
(a3 b) = pq

Solution 8.
Sn = 2n2 + 3n
S1 = 5 = a1
S2 = a1 + a2 = 14 ⇒ a2 = 9
d= a2 – a1 = 4
a16 = a1 + 15d = 5+ 15(4) = 65

Solution 9.

Solution 10.

Solution 11.

1. P (square number) = $\frac { 8 }{ 113 }$
2. P (multiple of 7) = $\frac { 16 }{ 113 }$

Solution 12.
Let number of red balls be = x
∴ P (red ball) = $\frac { x }{ 12 }$
If 6 more red balls are added: The number of red balls = x + 6 P (red ball) = $\frac { x+6 }{ 18 }$
Since, $\frac { x+6 }{ 18 }$ = 2 ( $\frac { x }{ 12 }$ ) ⇒ x = 3
There are 3 red balls in the bag.

Solution 13.
Let n = 3k, 3k + 1 or 3k+ 2.
(i) When n = 3k: n is divisible by 3.
n + 2 = 3k + 2 ⇒ n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3 (k + 1)+ 1 ⇒ n + 4 is not divisible by 3.

(ii) When n = 3k + 1: n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k+ 3 = 3 (k+ 1) ⇒ n + 2 is divisible by 3.
n + 4 = (3k+ l) + 4 = 3k + 5 = 3(k+ 1) + 2 ⇒ n + 4isnotdivisibleby3.

(iii) When n = 3k + 2: n is not divisible by 3.
n + 2 = (3k+2) + 2 = 3k + 4 = 3(k+ 1)+ 1 ⇒ n + 2 is not divisible by 3.
n + 4 = (3k + 2)+ 4 = 3k+ 6 = 3(k + 2) ⇒ n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Solution 14.

Solution 15.
Let the ten’s and the units digit be y and x respectively.
So,the number is 10 y + x.
The number when digits are reversed is 10 x + y.
Now, 7(10y+ x) = 4(10x + y) ⇒ 2y = x …( 1)
Also x – y3 …(ii)
Solving (1) and (2),we get y = 3 and x = 6.
Hence the number is 36.

Solution 16.

Solution 17.
∠SQN = ∠TRM (CPCT as NSQ ≅ MTR)

Since, ∠P + ∠1 + ∠2 = ∠P + ∠PQR + ∠PRQ (Angle sum property)
⇒ ∠1 + ∠2 = ∠PQR + ∠PRQ
⇒ 2 ∠1 = 2 ∠PQR (as ∠1 = ∠2 and ∠PQR= ∠PRQ) ⇒ ∠1 = ∠PQR
Also ∠2 = ∠PRQ and ∠SPT = ∠QPR (common) ∆PTS ~ ∆PRQ (By AAA similarity criterion)

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 22.

Solution 23.

Solution 24.
Let the three middle most terms of the AP be a – d, a, a + d
We have, (a – d) +a + (a + d) = 225
⇒ 3 a = 225 ⇒ a = 75
Now, the AP is a – 18d,…,a – 2d, a – d, a, a + d,a + 2d,…,a + 18d
Sum of last three terms :
(a + 18d) + (a + 17d) + (a + 16d) = 429
⇒ 3a + 51 d = 429 ⇒ (a + 17d) = 143
⇒ 75 + 17d = 143 ⇒ d = 4
Now, first term = a – 18d = 75 – 18(4) = 3
∴ The AP is 3,7,11,…, 147.

Solution 25.
Given : A right triangle ABC right angled at B.
To prove : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof : In ∆ ADB and ∆ ABC

Solution 26.
Draw ∆ABC in which BC = 7 cm, ∠B = 45°, ∠A = 105° and hence ∠C = 30°.
Construction of similar triangle A’ BC’ as shown below.

Solution 27.

Solution 28.

Solution 29.

Solution 30.

We hope the CBSE Sample Papers for Class 10 Maths paper 11 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 11, drop a comment below and we will get back to you at the earliest.

## UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics

Discrete Mathematics Long Answer Type Questions (8 Marks)

Question 1.
How are characters created by Binary Numbers? What are its different codes? Explain with examples. (UP 2011)
Computer Code: Computer codes are used to convert data into binary form to make the computer understand it. Apart from this, they are responsible for error-free signal flow in the computer. Three popular computer codes are:
BCD (Binary Coded Decimal): It is one of the earliest developed (UPBoardSolutions.com) memory codes. In this, every digit is converted into binary form separately:
e.g.,

But four-bit code can handle only 24 = 16 dIfferent characters that are why it is extended to 6-bit code and It can handle 26 = 64 different characters. It is 6-bit code and divided into two parts i.e., zone bit and character code.
Zone bit consists of Z bits and character zone consists of 4 bits.
To understand more look at the table given below:

EBCDIC (Extended Binary Coded Decimal Interchange Code): BCD can convert only 64 characters but we use more than 64 characters in computer to represent data. To overcome this problem, 2 more bits have been added to the zone bit to develop new 8-bit code and, that is why it is known as extended binary coded decimal interchange code. EBCDIC is 8- bit code which can encode 28 = 256 different characters. It is similar to BCD in working but it has 4 bits in bit zone. To understand more table is given below:

ASCII (American Standard Code for Information Interchange): This code is the most popular and widely accepted computer code. It is the standard code for computers, developed by the American National Standards Institute in the year 1963 for (UPBoardSolutions.com) encoding different characters in the computer. It is used by almost every manufacturing company. ASCII codes are of two types:
(a) 7-bit Code: To encode 27 = 128 characters with 3 bits in zone bit and four in character zone.

(b) 8-bit code: To encode 28 = 256 characters with 4 bits in bit zone and 4 bits in character zone.

What is 5/8 as a decimal you ask? Converting the fraction 5/8 into a decimal is very easy.

Question 2.
Define character representation in computers. (UP 2008)
Or
What is meant by “character representation”? Explain one such code in detail. (UP 2009)
Or
What is meant by character coding? Explain one coding methods in detail. (UP 2016)
Or
What is Character Representation? (UP 2018)
Character Representation: Physical devices used to store and process data in computers are two-state devices. A switch, for example, is a two-state device. It can be either ON or OFF. Electronic devices such as transistors used in computers must function reliably when operated as switches. Thus, all data to be stored and processed in computers are transformed or coded as strings of two symbols, one symbol to represent each state.

Coding of characters has been standardized to facilitate the exchange of recorded data between computers. The most popular standard is known as ASCII. Each letter is a unique combination of Binary Digits (BITS). That is, each letter (UPBoardSolutions.com) is a group of charged and uncharged transistors and it is grouped in such a way that a particular combination represents a specific character. A group of 8 BITS which is used to represent a character is called a byte. The length of 1 word is called word length which ranges from 1 byte to 64 bytes.
The internal code representation of string HARSH is:

Question 3.
Explain the Number System.
Number System: Number systems are very important to understand because the design and organization of a computer system depend on it.
Number systems are basically of two types:
1. Non-positional Number System: In this number system, each symbol represents the same value regardless of its position in the number and the symbols are simply added to find out the value of a particular number. Since it is very difficult to perform arithmetical operations with such a number system, positional number systems have been developed.

2. Positional Number System: In a positional number system, there are only a few symbols (UPBoardSolutions.com) called digits, and these symbols represent different values depending on the position they occupy in the number.

The value of each digit in such a number is determined by three considerations :

1. The digit itself
Face Value: The face value of a digit always remains the same regardless of its position in the number, e.g., the face value of 4 in 554, 40567 etc. is 4.
2. The position of the digit in the number.
Place Value: Place value of a digit changes due to change in its position, e.g., place value of 2 in 4210 is 2 hundred, in 32,450 it is 2 thousand, etc.
3. The base of the number system (where the base is defined as the total number of digits available in the number system), e.g., Decimal number system has base 10 since it includes only 10 digits 0, 1, 2, ….., 9, to represent any number.

The various positional systems in use are:

1. Binary number system
2. Octal number system
3. Decimal number system

Convert fraction 6 and 1/2 to decimal. What is 6 1/2 as a decimal? Answer: 6.5.

Question 4.
What are the logical operators? What are their different types? Explain operators making their Truth Table. (UP 2007, 08, 10)
Logical Operators: AND, OR, and NOT are logical operators. Since these operators are operated on logical values 0 and 1, that is why these operators are called logical operators.

AND Operator: An AND operator is represented by the symbol ‘.’. Basically AND operator is used to performing logical multiplication. A, B, and C are three logical variables, where A, B, are input variables and C is the output variable. We can define the AND operator by listing all possible combinations of A and B and the resulting value of C in the operation A.B = C.

It may be noted that since the variables A and B can have only two possible values (0 or 1) so only four (22) combinations of inputs are possible as shown in the following table. The resulting output values for each of the four input combinations (UPBoardSolutions.com) are given in the table. Such a table is known as the truth table. Thus, the table is the truth table for the logical AND operator.

As we can observe from the truth table that in AND operation, the output will be 1 when all inputs are 1 else output will be 0.
OR Operator: An OR operator is represented by the symbol V. Basically an OR operator is used to perform logical addition. As in the previous example, A and B are input variables and C its output variable. We can define the OR operator by listing all possible combinations of A and B and the resulting value of C in the equation A + B = C. The truth table for Logical OR operator is shown in the following Table:

As we can observe from the truth table that in logical addition, the output will be 1 when any one input is 1. It means if all inputs are 0 then the output will be 0.

NOT Operator: The two operators (AND and OR) are binary operators because they operate on two variables. NOT operator denoted by is a Unary operator because it operates on a single variable. NOT operator is also known as complementation operator or inverse operator.
Thus, complement of A is $\bar { A }$. Complement of (A + B) is $\bar { (A+B } )$. If value
of $\bar { A }$ is 0 then value of A is 1 and if value of A is 1 then value of $\bar { A }$ is 0. (UPBoardSolutions.com) The truth table for logical NOT operator is shown in table.

Question 5.
Write about the postulates of Boolean Algebra.
Postulates of Boolean Algebra: Boolean Algebra is an algebraic structure defined on a set of elements B together with two binary operators + and . provided the following postulates are satisfied:
(1) (a) Closure with respect to the operator +
(b) Closure with respect to the operator.

(2) (a) An identity element with respect to +, designated by
0 : X + 0 = 0 + X = X.
(b) An identify element with respect to designated by
1 : X . 1 = 1 . X = X.

(3) (a) Commutative with respect to + : X + Y = Y + X
(b) Commutative with respect to . : X . Y = Y . X

(4) (a) . is distributive over : : X . (Y + Z) = (X . Y) + (X . Z)
(b) + is distributive over . : X + (Y . Z) = (X + Y) . (X + Z)

(5) For every element X ∈ B, there exists an element $\bar { X }$ ∈ B such that:
(a) X × $\bar { X }$ = 1
(b) X . $\bar { X }$ = 0
The postulates listed above are called Huntington (1904) Postulates and need (UPBoardSolutions.com) no proof. They are used to prove the theorems of Boolean Algebra.

Question 6.
What are the different Gates in Boolean Algebra? (UP 2004, 05, 07)
Or
What is ‘Truth Table’? How is it helpful in understanding GATES? (UP 2006)
Or
Explain the working of a NAND Gate. Give its two application. (UP 2009, 10)
Or
How can you show that NAND is a Universal Gate? Explain with diagrams and truth tables. (UP 2011, 19)
Truth Table: A table that shows all the input-output possibilities of a logic circuit is called a truth table.
There are several types of the truth table. AND, OR, NOT, NAND, NOR, XOR, XNOR Gates are described below:
(a) AND Gate: In English language, Input A is ANDed with Input B to get output Y.

The truth table illustrates four ways to express the logical ANDing of A and B.
The AND Gate works on the principle that output will be high when all the inputs (UPBoardSolutions.com) are high otherwise output will be low.

(b) OR Gate: In OR Gate, input A is ORed with input B to get output Y.

The OR Gate works on the principle that, if anyone input is high, the output will be high. Thus, the only case when output will be low is when all inputs are low i.e., 0.

(c) NOT Gate: The NOT Gate is an electronic circuit that generates an output signal which is the reverse of the input signal. A NOT gate is also known as an inverter because it inverts the input.

(d) NAND Gate: A NAND Gate is a complemented AND gate. That is, the output of NAND Gate will be 1 if anyone of the inputs is 0 and will be 0 when all the inputs are 1.

(e) NOR Gate: A NOR Gate is a complemented OR Gate. That is, the output of a NOR Gate will be 1 only when all inputs are 0 and will be 0 if any input represents a 1.

(f) XOR Gate (Exclusive-OR Gate): XOR Gate is a combination of AND, OR, and NOT (UPBoardSolutions.com) Gates. symbol denotes XOR operation. This Gate works on the principle that if an odd number of inputs are 1, the output will be 1 otherwise output will be 0.

As observed from the truth table, the output is 1 when odd numbers of inputs are 1.

(g) XNOR Gate (Exclusive-NOR Gate): Similarly, XNOR gate is also formed with a combination of AND, OR, and NOT gates. symbol denotes XNOR operation. Since this gate is the inverse of XOR gate, the output will be 0 when odd numbers of inputs are 1 otherwise output will be 1.

Question 7.
Explain the basic features of ASCII Code. (UP 2005, 08, 09, 10)
Or
Explain in detail the features of the ASCII character code. (UP 2007)
ASCII: Binary numbers are coded to represent characters in the computer memory. Several codes are used for this purpose. One most commonly used code is the American Standard Code for Information Interchange (ASCII). ASCII has been adopted by several American computer manufacturers as their computer’s internal code. This code is popular in data communications, is used (UPBoardSolutions.com) almost exclusively to represent data internally in microcomputers, and is frequently found in the larger computers produced by some vendors.

ASCII is of two types: ASCII-7 and ASCII-8. ASCII-7 is a 7-bit code that represents 128 (27) different characters.
ASCII-8 is an extended version of ASCII-7. It is an 8-bit code that represents 256 (28) different characters rather than 128.
e.g. (i) A is given ASCII code 65.

Now if we convert 65 into Binary form we get 01000001 → 1 byte
In the same way, every character has its own ASCII value after converting into binary code stored on the computer.

Question 8.
Describe various Binary Arithmetic Operations. (UP 2008, 09, 11)
Or
What is binary arithmetic? Explain with suitable example. (UP 2017)
Four basic arithmetic operations are performed inside a computer using binary numbers. These are addition, subtraction, multiplication, and division. Since binary numbers are made up of 0’s and 1’s, results of arithmetic operations are also in 0’s and 1’s only.

Binary Addition: Binary addition is performed in the same manner as decimal addition. However the binary system has only two digits, the addition table for binary arithmetic is very simple, consisting of only four entries. The complete table for binary addition is as follows:
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

Plus a carry of 1 to next higher column. Carryovers are performed in the same manner as in decimal arithmetic. Since 1 is the largest digit in the binary system, any sum greater than 1 requires that a digit be carried over.
Example:

Binary Subtraction: From the following table, it is clear that the lower digit is subtracted (UPBoardSolutions.com) from the upper digit. If the lower digit is larger than the upper digit, it is necessary to borrow from the column to the left which equals to 2 (10).
0 – 0 = 0
1 – 0 = 1
1 – 1 = 0
0 – 1 = 1
with borrow from the next column. Thus, the only case in which it is necessary to borrow is when 1 is subtracted from 0.
Example

Binary Multiplication: Multiplication in the binary system also follows the same general rules as decimal multiplication. The table for binary multiplication is as follows:
0 × 0 = 0
0 × 1 = 0
1 × 0 = 0
1 × 1 = 1
Example:

Binary Division: Binary division is, again, very simple. As in the decimal system (or in any other system), division by zero is meaningless, here too. Hence, the complete table for the binary division is as follows:
0/1 = 0
1/1 = 1
The division process is performed in a manner similar to the decimal division.
Example:

Discrete Mathematics Short Answer Type Questions (4 Marks)

Question 1.
What is the order of precedence in Boolean Algebra? (UP 2007, 09, 19)
In a Boolean expression, many operators are used. The order in which they are operated is known as precedence. The precedence of Boolean operators is as follows:

1. The expression is scanned from left to right.
2. Expressions enclosed within parentheses are evaluated first.
3. All complement (NOT) operations are performed next.
4. All ‘.’ (AND) operations are performed after that.
5. Finally, all ‘+’ (OR) operations are performed in the end.

Question 2.
Define the Principal of Quality. (UP 2016)
The Huntington Postulates have been listed in two parts: (a) and (b). One part may be obtained from the other if ‘+’ is interchanged ‘+’ with ‘.’ and ‘0’ in interchanged with ‘1’ and vice-versa. This important property of Boolean Algebra (UPBoardSolutions.com) is called Principle of Quality. This principle ensures that, if a theorem is proved using the postulates, then a dual theorem obtained interchanging ‘+’ with ‘.’ and ‘0’ with ‘1’ automatically holds and need not be proved separately.
The table below lists theorems and their corresponding dual theorems.

Question 3.
Write a note on De Morgan’s Theorems to prove it.
Theorem (a): De Morgan’s Theorems: (x + y)’ = x’. y’
Proof: The truth table for proving this theorem is given below:

From the truth table, it is clear that both sides of the theorem are equal. Hence, the theorem is proved.
Theorem (b): (x + y)’ = x’ + y’
Proof: The truth table for proving this theorem is given below:

From the truth table, It is clear that both sides of the theorem are equal. Hence, the theorem is proved.
Theorems 6(a) and 6(b) are very important and useful. They are known as De Morgan’s theorems. They can be extended to n variables as given below:
(X1 + X2 + X3 + ………. + Xn)’ = X1‘ . X2‘ . X3‘ …….. Xn
(X1 . X2 . X3. …… Xn)’ = X1‘ + X’2 + X3‘ + ……….. + Xn

Question 4.
Write AND and OR LAWS of Discrete mathematics.
AND LAWS: AND LAWS are the laws which work on logical multiplication. They are:
1. x . 1 = x
2. x . x’ = 0
3. x . x = x
4. x . 0 = 0
“The tabular representations of truth values of a compound statement based on the truth values of the prime connective ness of statements is called TRUTH TABLE.”
Truth table consists of horizontal lines (rows) and vertical lines (columns). If a compound statement consists of N statements, the number of rows will be 2^N. The number of columns in a truth table depends upon the number of relationships between these statements.

Discrete Mathematics Very Short Answer Type Questions (2 Marks)

Question 1.
Discuss De-Morgan’s Theorem. (UP 2014)
First Theorem: This theorem states that the complement of a sum of the binary variable is equal to the product of the complement of the binary variables.
Second Theorem: The theorem states that the complement of a product of binary (UPBoardSolutions.com) variable is equal to the sum of the complement of the binary variable

Question 2.
What is the full form of ASCII? (UP 2014)
The full form of ASGII is American Standard Code for Information Interchange.

Question 3.
If A = 0 and B = 1, then find the value of y from the following expression:
Y = (A . B)
Y = $\bar { (0.1 } )$ = $\bar { (0 } )$ = 1

Question 4.
Give the name of the Boolean operators. (UP 2014)
AND Operator, Or operator and NOT operator.

Question 5.
Write a full form of EBCDIC. (UP 2017)
Extended Binay Coded Decimal Interchange Code.

Discrete Mathematics Objective Type Questions (1 Marks)

There are four alternative answers for each part of the questions. Select the (UPBoardSolutions.com) correct one and write in your answer book:

Question 1.
Each letter is a unique combination of:
(a) Bits
(b) Bytes
(c) Word length
(d) Binary.
(a) Bits

Question 2.
A group of 8 bits which is used to represent a character is called :
(a) Bits
(b) Bytes
(c) Integer
(d) None of these.
(b) Bytes

Question 3.
The most popular standard is known as:
(a) BCD
(b) ABC
(c) ASC
(d) ASCII
(d) ASCII

Question 4.
In the hexadecimal number system, the base :
(a) 8
(b) 10
(c) 16
(d) None of these.
(c) 16

Question 5.
The binay equivalent of the number (15)10. (UP 2014)
(a) (1101)
(b) (1110)2
(c) (1111)2
(d) (1000)2.
(a) (1101)

Question 6.
Which logic gate has only one input and one output? (UP 2015)
(a) NOT
(b) NOR
(c) OR
(d) AND.
(a) NOT

Question 7.
The value of the binary number (1010)2 would be?
(a) (14)10
(b) (12)10
(c) (10)10
(d) (11)10
(c) (10)10

Question 8.
What is binary equivalent of [31]10. (UP 2017)
(a) 10000
(b) 11111
(c) 100000
(d) 11110.
(b) 11111

Question 9.
Which of the following is a single input logic gate? (UP 2018)
(a) NAND
(b) AND
(c) NOT
(d) NOR.
(c) NOT

## UP Board Solutions for Class 10 Maths Chapter 1 Real Numbers

These Solutions are part of UP Board Solutions for Class 10 Maths. Here we have given UP Board Solutions for Class 10 Maths Chapter 1 Real Numbers.

प्रश्नावली 1.1 (NCERT Page 8)

प्र. 1.
निम्नलिखित संख्याओं का HCF ज्ञात करने के लिए यूक्लिड (UPBoardSolutions.com) विभाजन एल्गोरिथ्म का प्रयोग कीजिए:
(i) 135 और 225
(ii) 196 और 38220
(iii) 867 और 255
हलः

प्र. 2.
दर्शाइए कि कोई भी धनात्मक विषम पूर्णांक 6q +1 या 6q +3 या 6q + 5 के रूप का होता है, जहाँ q कोई पूर्णाक है। [NCERT Exempler]
हलः
मान ‘a’ एक धनात्मक पूर्णाक है। को 6 से विभाजित करने पर (UPBoardSolutions.com) भागफल q और शेष । प्राप्त होता है। .: यूक्लिड प्रमेयिका से,

प्र. 3.
किसी परेड में 616 सदस्यों वाली एक सेना (आर्मी) की टुकड़ी को 32 (UPBoardSolutions.com) सदस्यों वाले एक आर्मी बैंड के पीछे | मार्च करना है। दोनों समूहों को समान संख्या वाले स्तंभों में मार्च करना है। उन स्तंभों की अधिकतम संख्या वाले स्तंभों में मार्च करना है। उन स्तंभों की अधिकतम संख्या क्या है, जिसमें वे मार्च कर सकते हैं?
हलः

प्र. 4.
यूक्लिड विभाजन प्रमेयिका का प्रयोग करके दर्शाइए कि किसी धनात्मक (UPBoardSolutions.com) पूर्णांक का वर्ग, किसी पूर्णांक m के लिए 3m या 3m +1 के रूप का होता है।

हलः
माना x एक धनात्मक पूर्णांक 3q, 3g + 1 या 3g + 2 के रूप में है।

प्र. 5.
यूक्लिड विभाजन प्रमेयिका का प्रयोग करके दर्शाइए कि किसी धनात्मक पूर्णांक का घन 9m, 9m + 1 या 9m +8 के रूप का होता है।
हलः
एक धनात्मक पूर्ण x की कल्पना करें कि यह 3q, (3q + 1) या (3q + 2) (UPBoardSolutions.com) के रूप में है।

प्रश्नावली 1.2 (NCERT Page 13)

प्र. 1.
निम्नलिखित संख्याओं को अभाज्य गुणनखंडों के (UPBoardSolutions.com) गुणनफल के रूप में व्यक्त कीजिए।
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
हुलः

LCM of 15 and 20 is 60.

प्र. 2.
पूर्णाकों के निम्नलिखित युग्मों के HCF और LCM ज्ञात कीजिए तथा इसकी जाँच कीजिए कि वो संख्याओं को गुणनफल = HCF x LCM है।
(i) 28 और 91
(ii) 510 और 92
(iii) 336 और 64
हलः

प्र. 3.
अभाज्य गुणनखण्डन विधि द्वारा निम्नलिखित पूर्णांकों के (UPBoardSolutions.com) HCF और LCM ज्ञात कीजिए।
(i) 12, 15 और 21
(ii) 17, 23 और 29
(iii) 8, 9 और 25
हलः

प्र. 4.
HCF (306, 657) = 9 दिया है। LCM (306, 657) ज्ञात कीजिए।
हलः
HCF (306, 657) अर्थात् 306 और 657 का HCF = 9
चूंकि LCM x HCF = संख्याओं का गुणनफल

प्र. 5.
जाँच कीजिए कि क्या किसी प्राकृत संख्या n के लिए, संख्या 6″ अंक 0 पर समाप्त हो सकती है। [NCERT Exemplar]
हुलः
यहाँ n एक प्राकृत संख्या है और माना 69 अंक 0 पर (UPBoardSolutions.com) समाप्त होती है।

प्र. 6.
व्याख्या कीजिए कि 7 x 11 x 13 + 13 और 7×6 x 5x4x3 x 2 x 1+ 5 भाज्य संख्याएँ क्यों हैं।
हलः

प्र. 7.
किसी खेल के मैदान के चारों ओर एक वृत्ताकार पथ है। इस मैदान का एक चक्कर लगाने में सोनिया को 18 मिनट लगते हैं, जबकि इसी मैदान का एक चक्कर लगाने में रवि को 12 मिनट लगते हैं। मान लीजिए वे दोनों एक ही स्थान (UPBoardSolutions.com) और एक ही समय पर चलना प्रारंभ करके एक ही दिशा में चलते हैं। कितने समय बाद वे पुनः प्रारंभिक स्थान पर मिलेंगे?
हलः
एक चक्कर लगाने में सोनिया का समय = 18 मिनट
एक चक्कर लगाने में रवि का समय = (UPBoardSolutions.com) 12 मिनट
18 और 12 का LCM के समान समय के बाद वे पुनः प्रारंभिक स्थान पर मिलेंगे।

प्रश्नावली 1.3 (NCERT Page 17)

प्र. 1.
सिद्ध कीजिए कि 5 एक अपरिमेय संख्या (UPBoardSolutions.com) है।
हलः

प्र.2.
सिद्ध कीजिए कि 3+24/5 एक अपरिमेय संख्या है।
हलः

प्र. 3.
सिद्ध कीजिए कि निम्नलिखित संख्याएँ अपरिमेय हैं।
(a) $\frac { 1 }{ \sqrt { 2 } }$
(b) $7\sqrt { 5 }$
(c) $6+\sqrt { 2 }$
हलः

परिमेय संख्याओं और उनके दशमलव प्रसारों (UPBoardSolutions.com) का पुनर्भमण

प्रश्नावली 1.4 (NCERT Page 22)

प्र. 1.
बिना लंबी विभाजन प्रक्रिया किए बताइए कि निम्नलिखित परिमेय (UPBoardSolutions.com) संख्याओं के दशमलव प्रसार सांत हैं या असांत आवर्ती हैं।
(i) [latex s=2]\frac { 13 }{ 3125 } [/latex]
(ii) [latex s=2]\frac { 17 }{ 8 } [/latex]
(iii) [latex s=2]\frac { 64 }{ 455 } [/latex]
(iv)  [latex s=2]\frac { 15 }{ 1600 } [/latex]
(v) [latex s=2]\frac { 29 }{ 343 } [/latex]
(vi) [latex s=2]\frac { 23 }{ { 2 }^{ 3 }{ 5 }^{ 2 } } [/latex]
(vii) [latex s=2]\frac { 129 }{ { 2 }^{ 2 }{ 5 }^{ 7 }{ 7 }^{ 5 } } [/latex]
(viii) [latex s=2]\frac { 6 }{ 15 } [/latex]
(ix) [latex s=2]\frac { 35 }{ 50 } [/latex]
(x) [latex s=2]\frac { 77 }{ 210 } [/latex]
हलः
चूंकि किसी भी परिमेय संख्या के हर के अभाज्य गुणनखण्डन में 2n, 5m के (UPBoardSolutions.com) अतिरिक्त गुणनखण्ड नहीं हैं तो इसका दशमलव प्रसार सांत-दशमलव होता है अन्यथा यह असांत-आवर्ती दशमलव प्रसार होता है।

प्र. 2.
ऊपर दिए गए प्रश्न में उन परिमेय संख्याओं के दशमलव (UPBoardSolutions.com) प्रसारों को लिखिए जो सांत हैं।
हलः

प्र. 3.
कुछ वास्तविक संख्याओं के दशमलव प्रसार नीचे दर्शाए गए हैं। प्रत्येक स्थिति के लिए निर्धारित कीजिए कि वह संख्या परिमेय संख्या है या नहीं। यदि यह परिमेय संख्या है और P के रूप की है तो q के अभाज्य गुणनखण्डों के बारे में आप क्या कह सकते हैं?

(i) 43.123456789
(ii) 0.120120012000120000 (UPBoardSolutions.com)
(iii) 43.123456789
हलः
(i) 43.123456789
चूंकि उक्त दशमलव प्रसार सांत है। .:. इसे $\frac { p }{ q }$ के रूप में व्यक्त किया जा सकता है।

Hope given UP Board Solutions for Class 10 Maths Chapter 1 are helpful to complete your homework.

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## UP Board Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

These Solutions are part of UP Board Solutions for Class 10 Maths. Here we have given UP Board Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions.

The prime factors of 50 are the prime numbers which divide 50 exactly, without remainder as defined by the Euclidean division.

प्रश्नावली 5.1 (NCERT Page 108)

प्र. 1. निम्नलिखित स्थितयों में से किन स्थितयों में संबद्ध संख्याओं की सूची A.P है और क्यों?
(i) प्रत्येक किलों मीटर के बाद टैक्सी का किराया, जबकि प्रथम किलो मीटर के लिए किराया 15 रु है और प्रत्येक अतिरिक्त किलो मीटर के लिए किराया 8 रु है|
(ii) किसी बेलन (cylinder) में उपस्थित हवा की मात्रा, जबकि वायु (UPBoardSolutions.com) निकालने वाला पम्प प्रत्येक बार बेलन की हवा का $\frac { 1 }{ 4 }$ भाग बाहर निकाल देता है|
(iii) प्रत्येक मीटर की खुदाई के बाद, एक कुआं खोदने में आई लागत, जबकि प्रथम मीटर खुदाई की लागत 150 रु है और बाद में प्रत्येक खुदाई की लागत 50 रुo बढ़ती जाती है|
(iv) खाते में प्रत्येक वर्ष का मिश्रधन, जबकि 10000 रुo की राशि 8 % वार्षिक की दर से चक्रवृद्धि ब्याज पर जमा की जाती है|

प्र. 2. दी हुई A.P के प्रथम चार पद लिखिए, जबकि प्रथम (UPBoardSolutions.com) पद a और सार्व अंतर d निम्नलिखित हैं :
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, (UPBoardSolutions.com) d = -3
(iv) a = -1, d = $\frac { 1 }{ 2 }$
(v) a = -1.25, d = -0.25

The prime factorization of 38 is 2 × 19.

प्र. 3. निम्नलिखित में से कौन-कौन A.P हैं? यदि कोई (UPBoardSolutions.com) A.P है, तो इसका सार्व अंतर ज्ञात कीजिए और इनके तीन पद लिखिए |
(i) 3, 1, -1, -3, ……
(ii) -5, -1, 3, 7, ……
(iii) $\frac { 1 }{ 3 }$ , $\frac { 5 }{ 3 }$ , $\frac { 9 }{ 3 }$ , $\frac { 13 }{ 3 }$
(iv) 0.6, 1.7, 2.8, 3.9

प्र. 4. निम्नलिखित में से कौन-कौन A.P हैं? यदि कोई A.P है, तो इसका (UPBoardSolutions.com) सार्व अंतर ज्ञात कीजिए और इनके तीन पद लिखिए |

प्रश्नावली 5.2 (NCERT Page 116)

प्र. 1. निम्नलिखित सारणी में, रिक्त स्थानों को भरिए, जहाँ A.P (UPBoardSolutions.com) का प्रथम पद a, सार्व अंतर d और n वाँ पद an है:

प्र. 2. निम्नलिखित में सही उत्तर चुनिए और उसका औचित्य दीजिए:
(i) A.P: 10, 7, 4, …………………. का 30 वाँ पद है: (UPBoardSolutions.com)
(A) 97
(B) 77
(C) -77
(D) -87

प्र. 3. निम्नलिखित समांतर श्रेढ़ी में, रिक्त खानों (boxes) के पदों को ज्ञात कीजिए|

प्र. 4. A.P. : 3, 8, 13, 18, . . . का कौन सा पद 78 है ?

प्र. 5. निम्नलिखित समांतर श्रेढियों में से (UPBoardSolutions.com) प्रत्येक श्रेढ़ी में कितने पद हैं ?
(i) 7, 13, 19, …………….. , 205
(ii) 18, 15$\frac { 1 }{ 2 }$ , 13, ………, -47

प्र. 6. क्या A.P., 11, 8, 5, 2 ……. का एक पद -150 है ? क्यों ?

प्र. 7. उस A.P का 31वाँ पद ज्ञात कीजिए, जिसका 11 वाँ पद 38 है और 16वाँ पद 73 है।

प्र. 8. एक A.P में 50 पद हैं, जिसका तीसरा पद 12 है और (UPBoardSolutions.com) अंतिम पद 106 है। इसका 29वाँ पद ज्ञात कीजिए।

प्र. 9. यदि किसी A.P के तीसरे और नौवें पद क्रमशः 4 और -8 हैं, तो इसका कौन-सा पद शून्य होगा?

प्र. 10. किसी A.P का 17 वाँ पद उसके 10वें पद से 7 अधिक है। इसका (UPBoardSolutions.com) सार्व अंतर ज्ञात कीजिए।

प्र. 11. A.P. : 3, 15, 27, 39, ……… का कौन-सा पद उसके 54वें पद से 132 अधिक होगा?

प्र. 12. दो समांतर श्रेढि़यों का सार्व अंतर समान है। यदि इनके 100 वें पदों का अंतर 100 है, तो इनके 1000वें पदों का अंतर क्या होगा?

इसलिए, 1000वें पदों का अंतर भी 100 है |

प्र. 13. तीन अंकों वाली कितनी संख्याएँ 7 से विभाज्य हैं?

प्र. 14. 10 और 250 के बीच में 4 के कितने (UPBoardSolutions.com) गुणज हैं?

प्र. 15. n के किस मान के लिए, दोनों समांतर श्रेढि़यों 63, 65, 67, ………. और 3, 10, 17, ……… के n वें पद बराबर होंगे?

प्र. 16. वह A.P ज्ञात कीजिए जिसका तीसरा पद (UPBoardSolutions.com) 16 है और 7वाँ पद 5वें पद से 12 अधिक है।

प्र. 17. A.P. : 3, 8, 13, …, 253 में अंतिम पद से 20वाँ पद ज्ञात कीजिए।

प्र. 18. किसी A.P. के चौथे और 8वें पदों का योग 24 (UPBoardSolutions.com) है तथा छठे और 10वें पदों का योग 44 है। इस A.P. के प्रथम तीन पद ज्ञात कीजिए।

प्र. 19. सुब्बा राव ने 1995 में 5000 के मासिक वेतन पद कार्य आरंभ किया और प्रत्येक वर्ष 200 की वेतन वृद्धि प्राप्त की। किस वर्ष में उसका वेतन 7000 हो गया?

प्र. 20. रामकली ने किसी वर्ष के प्रथम सप्ताह में 5 की बचत की और (UPBoardSolutions.com) फिर अपनी साप्ताहिक बचत 1.75 बढ़ाती गई। यदि n वें सप्ताह में उसकी साप्ताहिक बचत 20.75 हो जाती है, तो n ज्ञात कीजिए।

प्रश्नावली 5.3 (NCERT Page 124)

प्र. 1. निम्नलिखित समांतर श्रेढियों का योग ज्ञात कीजिए :
(i) 2, 7, 12, . . ., 10 पदों तक
(ii) -37, -33, -29, . . ., 12 पदों तक (UPBoardSolutions.com)
(iii) 0.6, 1.7, 2.8, . . ., 100 पदों तक
(iv) $\frac { 1 }{ 15 }$ , $\frac { 1 }{ 12 }$ , $\frac { 1 }{ 10 }$ , ……., 11

प्र. 2. नीचे दिए हुए योग्फालों को ज्ञात (UPBoardSolutions.com) कीजिये:
(i) 7 + 10$\frac { 1 }{ 2 }$ + 14 + ….. + 10
(ii) 34 + 32 + 30 + . . . + 10
(iii) -5 + (-8) + (-11) + . . . + (-230)

प्र. 3. एक A.P. में,
(i) a = 5, d = 3 और an = 50 दिया है। n और Sn ज्ञात कीजिए।
(ii) a = 7 और a13 = 35 दिया है। d और S13 ज्ञात कीजिए।
(iii) a12 = 37 और d = 3 दिया है। a और (UPBoardSolutions.com) S12 ज्ञात कीजिए।
(iv) a3 = 15 और S10 = 125 दिया है। d और a10 ज्ञात कीजिए।
(v) d = 5 और S9 = 75 दिया है। a और a9 ज्ञात कीजिए।
(vi) a = 2, d = 8 और Sn = 90 दिया है। n और an ज्ञात कीजिए।
(vii) a = 8, an = 62 और Sn = 210 दिया है। n और d ज्ञात कीजिए।
(viii) an = 4, d = 2 और Sn = -14 दिया है। n और a ज्ञात कीजिए।
(ix) a = 3, n = 8 और S = 192 दिया है। d ज्ञात कीजिए।
(x) l = 28, S = 144 और कुल 9 पद हैं। a ज्ञात कीजिए।

प्र. 4. 636 योग प्राप्त करने के (UPBoardSolutions.com) लिए, A.P. : 9, 17, 25 ……….. के कितने पद लेने चाहिए ?

प्र. 5. किसी A.P. का प्रथम पद 5, अंतिम पद 45 और योग 400 है। पदों की संख्या और सार्व अंतर ज्ञात कीजिए।

प्र. 6. किसी A.P. के प्रथम और अंतिम पद क्रमशः 17 और 350 हैं। (UPBoardSolutions.com) यदि सार्व अंतर 9 है, तो इसमें कितने पद हैं और इनका योग क्या है?

प्र. 7. उस A.P. के प्रथम 22 पदों का योग ज्ञात कीजिए, जिसमें d = 7 है और 22 वाँ पद 149 है।

प्र. 8. उस A.P. के प्रथम 51 पदों का योग ज्ञात कीजिए, (UPBoardSolutions.com) जिसके दूसरे और तीसरे पद क्रमशः 14 और 18 हैं।

प्र. 9. यदि किसी A.P. के प्रथम 7 पदों का योग 49 है और प्रथम 17 पदों का योग 289 है, तो इसके प्रथम n पदों का योग ज्ञात कीजिए।

प्र. 10. दर्शाइए कि a1, a2, . . ., an, . . . से एक A.P. बनती है, (UPBoardSolutions.com) यदि an नीचे दिए अनुसार परिभाषित है:
(i) an = 3 + 4n
(ii) an = 9 – 5n
साथ ही, प्रत्येक स्थिति में, प्रथम 15 पदों का योग ज्ञात कीजिए।

प्र. 11. यदि किसी A.P. के प्रथम n पदों का योग 4n – n2 है, तो इसका प्रथम (UPBoardSolutions.com) पद (अर्थात् S1 ) क्या है? प्रथम दो पदों का योग क्या है? दूसरा पद क्या है? इसी प्रकार, तीसरे, 10वें और n वें पद ज्ञात कीजिए।

प्र. 12. ऐसे प्रथम 40 धन पूर्णांकों का योग ज्ञात कीजिए जो 6 से विभाज्य (UPBoardSolutions.com) हैं।

प्र. 13. 8 के प्रथम 15 गुणजों का योग ज्ञात कीजिए।

प्र. 14. 0 और 50 के बीच की विषम संख्याओं का योग ज्ञात (UPBoardSolutions.com) कीजिए।

प्र. 15. निर्माण कार्य से सम्बन्धी किसी ठेके में, एक निश्चित तिथि के बाद कार्य को विलंब से पूरा करने के लिए, जुर्माना लगाने का प्रावधन इस प्रकार है: पहले दिन के लिए 200 रु, दूसरे दिन के लिए 250 रु, तीसरे दिन के लिए 300 रु इत्यादि, (UPBoardSolutions.com) अर्थात् प्रत्येक उतरोत्तर दिन का जुर्माना अपने से ठीक पहले दिन के जुर्माने से 50 रु अधिक है। एक ठेकेदार को जुर्माने के रूप में कितनी राशि अदा करनी पड़ेगी, यदि वह इस कार्य में 30 दिन का विलंब कर देता है ?

प्र. 16. किसी स्कूल के विद्यार्थियों को उनके समग्र शैक्षिक प्रदर्शन के लिए 7 नकद (UPBoardSolutions.com) पुरस्कार देने के लिए 700 रु की राशि रखी गई है। यदि प्रत्येक पुरस्कार अपने से ठीक पहले पुरस्कार से 20 रु कम है, तो प्रत्येक पुरस्कार का मान ज्ञात कीजिए।

प्र. 17. एक स्कूल के विद्यार्थियों ने वायु प्रदुषण कम करने के लिए स्कूल के अन्दर और बाहर पेड़ लगाने के बारे में सोंचा । यह निर्णय लिया गया कि प्रत्येक कक्षा का प्रत्येक अनुभाग अपनी कक्षा की संख्या के बराबर पेड़ लगाएगा । उदाहरणार्थ, (UPBoardSolutions.com) कक्षा I का एक अनुभाग एक पेड़ लगाएगा, कक्षा II का एक अनुभाग 2 पेड़ लगाएगा, कक्षा III का एक अनुभाग 3 पेड़ लगाएगा, इत्यादि और ऐसा ही कक्षा XII तक के लिए चलता रहेगा । प्रत्येक कक्षा के तीन अनुभाग हैं । इस विद्यालय के विद्यार्थियों द्वारा लगाए गए कुल पेड़ों की संख्या कितनी होगी ?

प्र. 18. केंद्र A से प्रारंभ करते हुए, बारी-बारी से केन्द्रों A और B को लेते हुए, त्रिज्याओं 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm …. वाले उत्तरोत्तर अर्धवृतों को खींचकर एक सर्पिल (spiral) बनाया गया है, जैसा कि आकृति में दर्शाया गया है| (UPBoardSolutions.com) तेरह क्रमागत अर्धवृतों से बने इस सर्पिल की कुल लंबाई क्या है ? (π = $\frac { 22 }{ 7 }$)

प्र. 19. 200 लट्ठों (logs) को ढेरी के रूप में इस प्रकार रखा जाता है: सबसे नीचे वाली पंक्ति में 20 लट्ठे, उससे अगली पंक्ति में 19 लट्ठे, उससे अगली पंक्ति में 18 लट्ठे, इत्यादि (देखिए आकृति )। ये 200 लठ्ठे कितनी पंक्तियों में रखे गए हैं तथा (UPBoardSolutions.com) सबसे ऊपरी पंक्ति में कितने लट्ठे हैं?

प्र. 20. एक आलू दौड़ (potato race) में, प्रारंभिक स्थान पर एक बाल्टी रखी हुई है, जो पहले आलू से 5m की दूरी पर है, तथा अन्य आलुओं को एक सीधी रेखा में परस्पर 3m की दूरियों पर रखा गया है। इस रेखा पर 10 आलू रखे गए हैं (UPBoardSolutions.com) (देखिए आकृति)।

प्रत्येक प्रतियोगी बाल्टी से चलना प्रारंभ करती है, निकटतम आलू को उठाती है, उसे लेकर वापस आकर दौड़कर बाल्टी में डालती है, दूसरा आलू उठाने के लिए वापस दौड़ती है, उसे उठाकर वापस बाल्टी में डालती है, और वह ऐसा तब तक करती रहती है, (UPBoardSolutions.com) जब तक सभी आलू बाल्टी में न आ जाएँ। इसमें प्रतियोगी को कुल कितनी दूरी दौड़नी पड़ेगी?

प्रश्नावली 5.4 (NCERT Page 127)

प्र. 1. A.P : 121, 117, 113,…., का कौन-सा पद सबसे पहला ऋणात्मक पद होगा ?
[संकेत : a< 0 के लिए n ज्ञात कीजिए|]

प्र. 2. किसी A.P. के तीसरे और सातवें पदों का योग 6 है और उनका गुणनफल 8 है| इस A.P. के प्रथम 16 पदों का योग ज्ञात कीजिए|

प्र. 3. एक सीढ़ी के क्रमागत डंडे परस्पर 25 cm की दुरी पर हैं| (देखिए आकृति 5.7)| डंडों की लंबाई एक समान रूप से घटती जाती है तथा सबसे निचले डंडे की लंबाई 45 cm है और सबसे ऊपर वाले डंडे की लंबाई 25 cm है | यदि ऊपरी और (UPBoardSolutions.com) निचले डंडे के बीच की दुरी 2$\frac { 1 }{ 2 }$ m है, तो डंडों को बनाने के लिए लकड़ी की कितनी लंबाई की आवश्यकता होगी ? [संकेत : डंडों की संख्या = $\frac { 250 }{ 25 }$ ÷ 1 हैं|]

प्र. 4. एक पंक्ति के मकानों को क्रमागत रूप से संख्या 1 से 49 तक अंकित किया गया है| दर्शाइए कि x का एक ऐसा मान है x से अंकित मकान से पहले के मकानों की संख्याओं का योग उसके बाद वाले मकानों की संख्याओं के योग के (UPBoardSolutions.com) बराबर है| x का मान ज्ञात कीजिए|
[संकेत : Sx-1 = S49 – Sx है| ]

प्र. 5. एक फुटबॉल के मैदान में एक छोटा चबूतरा है जिसमें 15 सीढीयाँ बनी हुई हैं| इन सीढीयों में से प्रत्येक की लंबाई 50m है वह ठोस कंक्रीट ( concrete) की बनी है प्रत्येक सीढ़ी में $\frac { 1 }{ 4 }$ m की चौड़ाई है और $\frac { 1 }{ 2 }$ m का फैलाव (चौड़ाई) है| (देखिए आकृति 5.8 )| इस चबूतरे को बनाने में लगी कंक्रीट का कुल आयतन परिकलित कीजिए| (UPBoardSolutions.com) [संकेत : पहली सीढ़ी को बनाने में लगी कंक्रीट का आयतन = $\frac { 1 }{ 4 }$ x $\frac { 1 }{ 2 }$ x 50 m3 है|]

Hope given UP Board Solutions for Class 10 Maths Chapter 5 are helpful to complete your homework.

If you have any doubts, please comment below. UP Board Solutions try to provide online tutoring for you.

## UP Board Solutions for Class 10 Computer Science Chapter 1 Computer and Communication

Computer and Communication Long Answer Type Questions (8 Marks)

Question 1.
Write in brief about the evolution of computer. (U. P. 2010, 17)
A computer is a high-speed electronic device that accepts data and instructions from the user, then processes the data accordingly to produce information as output. It is capable of performing arithmetic and logical operations on data. It also stores and executes set of instructions. Data is entered in the computer through some input devices like keyboard, mouse, etc. It is then processed by C.P.U. and the result is displayed through an output device like a monitor.

The invention of the computer has affected many areas of our life. In its early time, it was a very costly and rare machine, limited to scientific laboratories and research centres. It was difficult to work on (UPBoardSolutions.com) and they were very bulky. Slowly, the technology improved and the size of the computer reduced as well as its working became easy.

Computers have completely altered the structure of a business, a large volume of accounting and record-keeping, data can be manipulated, organized, stored, retrieved and used for scientific purposes.

Nowadays, a computer is used in homes, offices, shops and almost every-where. It is used to do simple as well as the most difficult calculations. Every business, no matter big or small, is based on computers. Similarly, an organization without computer is hard to find. Computer has changed the world.

Evolution of Computer: The first mechanical calculator developed by Blaise Pascal acted as a model for modem computers. Since then many machines have been developed which have to lead the way to modem microcomputers. A series of a scientific breakthrough by many scientists have contributed to produce this electronic machine called the computer.
1. Abacus: Movable beads on a wooden frame constituted the first (UPBoardSolutions.com) known calculating device. The abacus was used by the ancient Greeks and Romans.

2. Pascaline: The gear-driven machine capable of addition, subtraction and multiplication, considered as a first mechanical calculator, was invented by French mathematician Blaise Pascal in the year 1642.

3. Jacquard’s Loom: In 1801, a Frenchman named Joseph Jacquard perfected a loom that was controlled by the holes in cardboard punched cards. This machine gave an idea about storage.

4. Difference Engine: In 1822, Charles Babbage invented his first machine (Difference Engine). He designed it to calculate logarithm tables. A series of levers were used to enter the data and a device similar to the typewriter was used to print the output.

5. Analytical Engine: In 1833, it was developed by Charles Babbage to perform addition, subtraction, multiplication and division through the use of the stored program. First programmer lady Ada Augusta Byron Lovelace helped him in its developing.

6. Atanasoff-Berry Computer: This electronic machine was developed by Dr John Atanasoff in 1939 for certain mathematical equations. It was called Atanasoff-Berry Computer or ABC, after its inventor’s name and his assistant Clifford Berry. It used 45 vacuum tubes for internal logic and capacitors for storage.

7. Mark 1: In 1944, Dr. Howard Aiken developed a machine called an (UPBoardSolutions.com) Automatic Sequence Controlled Calculator which was later named as Mark-I. It was the first electromechanical computer.

Age Difference Calculator helps to determine the age gap between two persons easily.

8. ENIAC: First fully electronic computer named ENIAC (Electronic Numerical Integrator and Calculator) was developed by Prosper Eckert and J.W. Mauchly in 1945. It used high-speed vacuum tube switching devices.

9. EDVAC: In 1946, Dr John Von Neumann used the principle of storing in 0 and 1 (Binary Digits) in place of earlier technologies and developed EDVAC with the new concept of ‘stored program’. ED VAC (Electronic Discrete Variable Automatic Computer).

10. UNIVAC: Universal Automatic Computer was developed by Eckert and Mauchly in the year 1951. It was the first commercial computer used by Electronic Corporation. Its memory was MDL (Mercury Delay Line).

11. PDP Series: The computer of this series was developed by DEC (Digital Equipment Corporation).

• PDP-1 → 1961 → 8 bit → 4 KB
• PDP-8 → 1965 → 16 bit → 16 KB
• PDP-11 → 1970 → 16 bit → 32 KB

12. Micro Computers: Intel is the No. 1 company in the microprocessor. The microprocessors developed by this company were the best for microcomputers, few of them are as follows:

• 8080 → 1974 → 8 bits
• 8085 → 1978 → 8 bits
• 8086 → 1980 → 16 bits
• 80286 → 1982 → 32 bits
• 80386 → 1985 → 32 bits
• 80486 → 1986 → 32 bits

13. Pentium Series: After 80486, Intel developed Pentium processors which are used almost in every computer nowadays.

• Pentium → 1993
• Pentium I → 1995
• Pentium II → 1997
• Pentium Mobile → 1998
• Pentium III → 1999
• Pentium IV → 2000
• Pentium Centrino → 2004

Question 2.
What are the different components of the Computer?
The main components of the computer system are as follows:

• Computer Hardware.
• Computer Software.
• Computer User.

Hardware:
Computers are made up of many electronic and electro-mechanical devices. The devices together are referred to as computer hardware.
Hardware is the physical components of the computer which (UPBoardSolutions.com) are tangible and visible to the user. Hardware is mainly categorised into two parts:

1. C.P.U. (Central Processing Unit),
2. Peripherals.

Input/Output System:
The computer will be of no use if it is not communicating with the external world. For a computer, it is must to have a system to receive information from the outside world and must be able to communicate results to the external world. Thus, a computer consists of an Input/Output (I/O) System.

Input Unit: Input unit is a link between user and computer. It takes the input from the user and converts it into a form understandable by the computer. It consists of input devices attached to the computer which feeds the data and instructions into the computer. Examples of Input Devices :

1. Keyboard
2. Mouse
3. Scanner
4. Light Pen
6. Magnetic Ink Character Reader (MICR), etc.

1. Keyboard: The Keyboard is one of the most common input devices for the computer. The layout of the keyboard is like that of the traditional QWERTY type-writer.

Different type of keys of a keyboard are as follows:

• Alphabetical keys (A to Z, a to z)
• Numeric keys (0 to 9)
• Function keys (FI to FI2)
• Special keys (@, !, +, -, <, =, etc.)
• Cursor control keys (space bar, →,  ↓, ↑, etc.)

2. Mouse: A mouse is a pointing device used while display based package. It is a push-button control device that eliminates the need to type computer commands. Instructions are given by the user to the computer by (UPBoardSolutions.com) pointing an arrow on the screen to a picture or word and then pushing the button on the mouse. The user moves the arrow on the screen by sliding the mouse across the desktop. It is used for selection, dragging, drawing, playing games, etc.

3. MICR: Today banks are issuing special types of cheques. These cheques have the cheque number printed by special magnetic ink. The cheque clearance is done using computers. There is a special device called a MICR. It directly reads the cheque number, bank code, branch code etc. and converts it in machine-readable form.

4. OCR: Optical Character Reader is an information processing device that converts typed or handwritten data in a form that the computer understands. Its speed ranges from 50 to 3000 characters per second.

5. Scanner: Scanner is used to convert photographs and other images from documents and drawing into electronic form. The scanner is used to scan the images and store them on the disk, as a file.

6. Light Pen: A light pen is an electronic device in the form of a photodiode, that allows the operator to identify a particular point or character displayed on the screen, and can be used alone or in conjunction with (UPBoardSolutions.com) a keyboard to add, rearrange or delete information, modify images displayed on the screen.

Output Unit:
It is a communication link between the computer and the user. The output unit consists of the output devices attached to the computer. The output devices help in the communication of data and information from machine to man.
The output devices help the computer to communicate with the user by converting the electric signals to human understandable signals. Examples of output devices:

• VDU (Visual Display Unit)
• Printer
• Plotter.

The output on the VDU is called soft copy output whereas the output on paper, which is produced on printers and plotters is referred to as hard copy output.

1. Monitor/VDU: Monitor is a device to interact with the computer. The messages from the computer are displayed on the screen of the monitor. It is the most common output device using a Cathode Ray Tube (CRT) to produce images. Images are formed by a collection of spots, each known as a pixel.
Classification of Monitors: Monitors can be classified based on:
(A) Resolution: The number of pixels that make up the screen
The examples are CGA (Colour Graphic Adapter), EGA (Enhanced (UPBoardSolutions.com) Graphic Adapter), VGA (Video Graphic Array Adapter), SVGA (Super Video Graphic Array Adapter), CGA having the least resolution and SVGA the highest resolution.

(B) Colour Facilities:
Monochrome. Single colour display on a black background.
Colour monitors. Which can display multicolour outputs by combining red, green and blue in varying intensities?

2. Printer: A printer is a device that can record information permanently on paper in the form of printed copies.
Printers are mainly divided into two parts:
(i) Impact,
(ii) Non-Impact

(i) Impact Printer: These printers print characters by striking the character against the inked ribbon, which leaves an impression on the paper. Impact printers are noisy and can be used to create carbon copies.

(a) Dot-Matrix Printer: This is one of the most popular printers used for personal computing systems. These printers are cheaper as compared to other technologies. Dot Matrix Printer uses impact technology and a print head containing banks of wires moving at high speeds against inked ribbon and paper. These printers form characters by building them up by dots. The character matrix (array of wires) are of 7’5, 7’7, 9’7, 9’9.
The speeds range from 40 cps (characters per seconds) to about 1000 cps.

(b) Line Printer: The line printer prints one line at a time. The speed of line printers is measured in lines per minute. The printing speed of these printers varies from 300-3000 lines per minute. The width of the line is 15 (UPBoardSolutions.com) inches and the line has 96 to 160 characters. Different types of line printers are Drum printers, Chain printers and Band printers.

(c) Daisy Wheel: Daisy wheel printer contains a disk of metal or plastic and it has 96 characters on its petals. This disk is capable of rotating. A hammer hits the petal to print the character. Speed of the Daisy wheel printer is 20-100 characters per second. These printers cannot be used for longer duration and also these are very slow in functioning. The main feature of the Daisy wheel printer is that its printing is of superior quality.

(ii) Non-Impact Printer: Non-impact printers print characters without directly striking on the paper. Non-impact printers cannot create carbon copies. They are, however, very quiet.

(a) Laser Printer: This is a high quality, high speed, high volume technology which works in non-impact fashion on plain paper or pre-printed forms.
Printing is achieved by deflecting laser beam on to the photosensitive surface of a drum. The latent image attracts the toner (A special kind to ink) to the image areas. The toner is then electrostatically transferred to the paper and fixed into the permanent image. The speed of printing can range from 10 pages per minute to about 200 pages per minute. This technology is relatively expensive but is becoming very popular because of the quality, speed and noiseless operations.

(b) Ink-Jet Printer: Ink-Jet printers print by spraying a cont-rolled stream of tiny ink droplets (from a fine nozzle) accurately on the paper forming either dot matrix or solid characters. These are non-impact printers.
The typical speed ranges from 50 cps to above 300 cps. This technology has (UPBoardSolutions.com) been used well for the production of colour printing and elaborate graphics.

3. Plotter: Plotter is an output device that is used to produce graphical outputs on the paper. It uses pens, either of a single colour or multi-colour to draw pictures. Engineering designs can be printed with good precision by making use of plotters.
There are two types of plotters:

• Drum Plotter,
• Flatbed Plotter.

Applications of Plotters:

• Map Drawing.
• Architectural Drafting.

System Unit:
The system unit is a box-like case that houses the electronic components of the computer that are used to process data. The system unit is made of metal or plastic and is designed to protect the electronic components from damage. (UPBoardSolutions.com) The electronic components and most storage devices reside inside the system unit and the other devices such as a keyboard, mouse, monitor are located outside the system unit.

MotherBoard:
It is the main printed circuit board in an electronic device, which contains sockets that accept additional boards. In a personal computer, the motherboard contains the bus, CPU, memory sockets, keyboard controller and supporting chips.

Chips that control the video display, serial and parallel ports, mouse and disk drives may or may not be present on the motherboard. If not, they are independent controllers that are plugged into an expansion slot on the motherboard.
Central Processing Unit (CPU)
CPU is the brain of a computer system. It consists of:

• Control Unit (CU),
• Arithmetic Logical Unit (ALU),
• Memory Unit (MU).

The parts of a CPU are connected by an electronic component referred to as a bus, which acts as an electronic highway between them. In order to temporarily store data and instructions, the CPU has special-purpose storage devices (UPBoardSolutions.com) called as registers. The CPU of a modem computer is fully electronic. It is made up of millions of electronic components etched on to a number of silicon chips. These chips are all assembled on a printed circuit board called the Mother Board.

Functions of CPU: The main functions of the CPU are the following:

• To store data and instructions.
• To control the sequence of operations.
• To carry out the processing.

The CPU of a personal computer is mainly recognized by the microprocessor chip that it uses.
Microprocessor Chips: A microprocessor is assembled from thousands of tiny transistors, resistors and other electronic components. A typical microprocessor is fabricated on a single tiny chip of silicon and is combined with other elements. These other elements provide input and output connections, storage and control to form a complete microprocessor onboard.

In the 70’s, Intel engineers built the first microprocessor chip for a Japanese manufacturer of desk calculators. Subsequently, 8-bit data processing capabilities were developed in chips. In 1974, a personal-sized system was developed and called ‘ALTAIR 8800′ which used the Intel 8-bit microprocessor.

In the 80’s, the personal computer systems started being based on a 16-bit microprocessor. The most popular amongst these is IBM PC.

All 16-bit personal computers are also built around a few popular microprocessors. The Intel 8088 is found in IBM PC. Another popular microprocessor used in systems such as Apple, Cromemco and many other vendors (UPBoardSolutions.com) is the Motorola’s 68000.
Intel 80286, 80386 and 80486 are 32-bit microprocessor chips.
Intel’s Pentium series is the latest 32-bit and 64-bit microprocessor chips.
The microprocessor is the brain of a computer system. It is the place where actual processing is done. It consists of :

1. Control Unit
2. ALU or Arithmetic Logic Unit
3. Registers.

1. C.U. (Control Unit): It controls the movement of data and program instructions into, and out of the C.P.U. and to control the operations of A.L.U., C.U., fetches the instructions from the main memory into registers, decodes it and sends control signals to other components of the computer so that the appropriate actions are carried out. This is known as fetch executes cycle.

2. ALU (Arithmetic Logic Unit): The ALU performs all the arithmetic and logical functions, that is, it adds, subtracts, multiplies, divides and does comparisons. It also performs logical operations (AND, OR and NOT). These operations provide the facility of decision making through the computer. The result of a logical operation is either TRUE or FALSE.
A.L.U. only operates on data that is in the internal C.P.U. memory, also known as registers. C.P.U. memory, also known as registers. Registers are very fast, temporary storage whose functions are to receive and hold data. They are not accessible to the programmers.

3. Registers: A register is a special temporary location within the CPU. Registers very quickly accept, store and transfer data and instructions that are being used immediately. To execute an instruction, the control unit of the CPU retrieves it from main memory and places it into a register.

The number and types of registers in a CPU vary according to the CPU design. Their size (capacity) and number can affect the processing power of a computer system. In general, the “larger” is the register (the more bits it (UPBoardSolutions.com) can carry at once), the greater is the processing power.

BUS: The term ‘bus’ refers to an electrical pathway through which bits are transmitted between the various computer components. Depending on the design of a system, several types of buses may be present.
The control bus is the pathway for all timing and controlling functions sent by the control unit to the other units of the system.
The address bus is the pathway used to locate the storage position in memory where the next instruction is to be executed or the next piece of data will be found.
Data bus is the pathway where the actual data transfer takes place.

Question 3.
What do you mean by memory? Explain in brief.
Memory: A computer system has storage areas, referred to as memory. The memory can receive, hold and deliver data when instructed to do so. Data that are being processed are held in Primary Memory, which is capable of sending and receiving the data at very high speeds. Secondary memory stores data not currently being used and operates more slowly, but it is capable of storing large volumes of data.

Primary Memory: Primary Memory consists of semi-conductor memory chips and is used to store data and programs currently in use. Each storage element of memory is directly (randomly) accessible and can be examined and modified without affecting other cells. Main memory can be volatile or non-volatile. Primary memory is classified into two groups:
(A) RAM (Random Access Memory),

RAM is read/write memory. It is a volatile memory. It stores the information as long as power is switched on and the information is lost when the power supply is switched off. RAM is of two types:

• DRAM: It needs constant refreshing in order for the stored data to be maintained.
• SRAM: It does not need refreshing. It is faster and expensive than DRAM.

Features of RAM: The main features of RAM are as follows:

• Data that needs to be processed and the instructions which are used for processing are held in the RAM.
• Each element of RAM is a memory location in which data can be stored. Each location has a unique address. Using this address, data can be directly retrieved or stored.
• Since RAM must hold the data to be processed and the instructions for (UPBoardSolutions.com) processing, its size or capacity is one of the measures of the power of the computer.

Functions of RAM: The principal function of the main memory is to act as a buffer between the CPU and the rest of the computer system components.
Main memory is used for the following purposes:

• Storage of a copy of the main software program that controls the general operation of the computer. This copy is loaded into main memory when the computer is turned on. and it stays there as long as the computer is on.
• The temporary storage of a copy of application programs and instructions to be retrieved by the CPU for interpretation and executed.
• The temporary storage of data that has been input from the keyboard or another input device until instructions call for the data to be transferred into the CPU for processing.

Read-Only Memory (ROM): ROMs are the memories on which it is not possible to write the data when they are on-line to the computer. They can only be read. The ROMs can be used in storing programs provided by the manufacturer of computer for basic operations. ROMs are non-volatile in nature and need not be loaded in a secondary storage device.

All the different pair combinations from the factors of 70 above are the Factor Pairs of 70.

ROMs can be written only at the time of manufacture. It is necessary, and also convenient, to have instructions stored in ROM. For example, if you are using a microcomputer with floppy disk drives, the more instructions in ROM. the fever diskettes you may have to handle.

The process of manufacturing ROM chips and recording data on them was more expensive than the process of producing RAM chips. As a result, manufacturers tended to record in ROM only those instructions that were crucial to the operation of the computer.

PROM, EPROM, EEPROM: In addition to ROM, three additional categories of non-volatile memory are used in some computer systems namely PROM, EPROM and EEPROM.

PROM: Programmable Read-Only Memory is a non-volatile memory which allows the user to program the chip with a PROM writer. The chip can be programmed once, thereafter it cannot be altered. Therefore, PROMs are (UPBoardSolutions.com) more flexible than ROMs.
EPROM stands for Erasable Programmable Read-Only Memory. EPROM chips were developed as an improvement over PROM chips.

The EPROMs can be written electrically. It requires the erasure of whole storage cells by exposing the chip to ultraviolet light, thus brings them to the same initial state. This erasure is a time-consuming process. Once all the cells have been brought to the same initial state, then the EROM can be written electrically.

EEPROM stands for Electrically Erasable Programmable Read-Only Memory. EEPROMs are becoming increasingly popular as they do not require prior erasure of previous contents. It avoids the inconvenience of having to take chips out of the computer to change data and instructions. Instead, changes can be made electrically under software control. These chips are being used in point-of-sale terminals to record price-related data for products. The only disadvantage of EEPROM chips is that they cost more than regular ROM chips.

Secondary Memory: Secondary memory is also known as permanent memory as we can store data on it for future use. It is both read/write memory and consists of different storage devices known as secondary storage devices as follows:

(A) Magnetic tape: Magnetic tape allows large amount of data to be stored economically passes the devices write head, data is recorded by magnetising the iron oxide in different directions. While reading the iron oxide causes a current in the read head.

(B) Floppy disk: It is a plastic film coated with iron oxide metal and protected by a plastic jacket as cover that has an opening which allows the read/write head to retrieve or store data. The most common floppy disk now in use is the 3.5″ disk also known as micro floppy. Earlier floppies were of 5.25″ and 8″.

(C) Hard disk: It comprises at least one rigid disk protected by a strong and airtight casing. It contains disks, read/write head, access arms, servomotor and the electronic circuit board to control the disk (UPBoardSolutions.com) operations. This is the most common secondary storage device having the highest storage capacity (2GB, 4GB, 10GB, 40GB….). It is fast and speeds of less than 10 microseconds are achievable.

(D) Optical disk: These are plastic disks handles more data in comparison to floppy disks. In Optical disk a light source is used to catch data patterns on the disks. Normally two laser lights are used; a weak to read data and stronger to write by burning the surface of the disk. Different types of optical disks are CD-R, CD-R/W, DVD.

Free and easy to use online median calculator. ➤ Calculates the median value of a set of numbers

Question 4.
Describe Software in brief.
Software: Software is one of the primary elements of a computer system. It is a set of computer programs, procedures and associated documentation related to the effective operation of a computer system.
The software’s are classified into the following categories:

1. System Software,
2. Utility Software,
3. Application Software.

1. System Software: System software is programs that directly interact with the hardware. For example, when a file is to be saved on a disk, the system software sends the required instructions to perform this task. It provides the environment to write application programs.
The system software is written by computer professionals called System Programmers.

The system software is of two types:

• Translators
• Operating System.

Translators: Instructions given to a computer in any language, have to be translated into machine code for the computer systems to execute the instructions. This work of translation is done by a Translator.
Translators can be classified as:

• Interpreters,
• Compilers,
• Assemblers.

(a) Interpreters: Interpreters are the translators which are used to convert programming language into machine language for the purpose of execution line by line. It translates one statement at a time and executes it.

(b) Compilers: The compiler translates the whole program code (known as source code) and prints a list of errors which have to be corrected as a whole. Once the program is error-free, the executable code is generated.

(c) Assemblers: This software translates a program written in assembly language to machine code.

(ii) Operating System: An Operating System is a set of routine programs that are used to manage the operations of the computer.
The Operating System isolates the hardware from the user. The user (UPBoardSolutions.com) communicates with the Operating System, supplies application programs and the data that are in a language and format acceptable to the Operating System and receives output.

Some of the popular operating systems are:
DOS, WINDOWS 95, WINDOWS 98, WINDOWS NT, WINDOWS Millennium, XP, OS/2, LINUX, UNIX.

2. Utility Software: A utility program is a type of system software that performs a specific task, usually related to managing a computer, its devices or its programs. Most operating systems include several utility programs.
Some of the utilities are described below:

(i) File Compression Utility: A file compression utility reduces or compresses the size of a file. A compressed file takes up less storage space on a hard disk or floppy disk, which frees up room on the disk and improves system performance.
When a compressed file is received, it must be uncompressed or unzipped to restore it to its original form. Two popular stand-alone file compression utilities are PKZIP and WinZip.

Disk Scanner: A Disk Scanner is a utility that detects and corrects both physical and logical problems on a hard disk or floppy disk. It also searches and removes unwanted files.
Windows 98/XP includes two disk scanner utilities:

• Scan Disk which detects and corrects problems.
• Disk Cleanup searches for and removes unnecessary files such as temporary files.

(ii) Disk Defragmenter: A disk defragmenter is a utility that reorganizes the files and unused space on a computer’s hard disk so that data can be accessed more quickly and programs can run faster.
When a computer stores data on a disk, it places the data in the first (UPBoardSolutions.com) available sector on the disk. Disk defragmenter reorganises the file and unused space. Windows includes a disk defragmenter called Disk Defragmenter.

(iii) Uninstaller: An uninstaller is a utility that removes an application, as well as any associated entries in the system files. When you install an application, the operating system records the information that it uses to run the software in the system files.
If you attempt to remove the application from your computer by deleting the files and folders associated with that program without running the uninstaller, the system file entries remain.

(iv) Anti-virus: An anti-virus program is a utility that prevents v detects and removes viruses from a computer’s memory or storage devices. A virus is a program that copies itself into other programs and spreads through multiple computers. Viruses are designed to damage a computer intentionally by destroying or corrupting their data.

(v) Back-Up: This is used to back-up files on your hard disk. Files can be back-up to a floppy disk, a tape drive or another computer on the network. If the original files are damaged or lost, it can be restored from the backup.

(vi) Search Engine: It is used to search files or data from the disk quickly. It works randomly to give search results to the user as quickly as possible.

3. Application Software: Application software consists of programs designed to perform specific tasks for users.
Application software also called a software application, can be used for the following purposes:

• To assist with graphics and multimedia projects.
• To support household activities, for personal business, or for education.
• To facilitate communications.

Word Processor: Word Processor transforms a screen into ‘sheets of paper’ to be written on, and provides quick and accurate ways to create and revise business documents.
These packages allow:
Text to be written in document form.
Edit any part of a document.
Adjust the format of a document, such as margins, spacing, page numbers.
Save the entire document on a disk and retrieve it later.
Example:
Wordstar, MS-Word, Word Perfect, Page Maker.

Spreadsheets: A spreadsheet is a table consisting of rows and columns, and provides business professionals with a quick and accurate means of performing mathematical calculations involved in answering a question.
It performs various mathematical functions as well as logical and conditional operations in a quick and accurate manner.
Examples: Lotus 1-2-3, Quattro Pro, Excel, Multiplan.

Database Management System: Database Management System (UPBoardSolutions.com) (DBMS) is a set of programs that manipulates a database by appending, deleting and modifying records. It is used to generate the queries, forms, reports etc. on the basis of some condition by using different records stored in it.
The two basic types of data management software are

• File management systems, for example, Dbase, Foxbase, FoxPro etc.
• Relational Database Management System, for example, Ingress, Oracle, Unify etc.

Graphics/Presentation Packages: This software is used to create slide transition, animation, etc. by using its advanced features. Most of the graphics or presentations are being prepared using this software.
PowerPoint, Corel Draw, Macromedia, Director.

Desk Top Publishing (DTP) Software: It is a software system that is used to produce attractive page layouts complete with pictures and text printed in a variety of styles, which is ideal for use in newspaper and magazine publishing companies.
Examples: Page Maker, Word, Corel Draw.

Question 5.
What are the different means of communication available in the modern age? Give their comparison also. (U. P. 2006, 12, 15, 19)
Communication Media: The medium (or media) is the matter or substance that carries the voice or data transmission. It can be copper (wire or coaxial cable), glass (fibre optical cable), or wave (microwave). A circuit (channel or line) is nothing more than the path over which data moves.
Communication media is of two types:
(A) Guided / Wired Media: In this type of media, wires are used.
It is further classified into three categories:
(1) Twisted Pair Cable:

Made of copper, coated with insulating material and continuously twisted throughout its entire length.
Twisted cable helps minimize the effects of noise or electromagnetic interference.
Relatively inexpensive and easy to install.
Low immunity of noise, narrow bandwidth.
Data rates up to 20 Mbps in LAN; 19.2 Kbps for long-distance (UPBoardSolutions.com) communications over the telephone network.
Usage: Voice communication, Data communication e.g., telephone.

2. Coaxial Cable:
Consists of wire surrounded by the insulating layer, shielding layer and an outer jacket.

Quarter-inch or more in diameter, therefore, less flexible than twisted pair.
Less susceptible to noise but more expensive than twisted pair.
Wider bandwidth, more difficult to install/tap, more costly than twisted pair.
Data rate up to 150 Mbps.
Usage:
Data Communication Video Transmission Voice Communication
Used extensively in LAN and relatively short distance (10 miles). For longer distance, repeaters may be necessary.

3. Fibre Optic Cable:

• One or more glass or plastic fibres are woven together to form the core of the cable. This core is surrounded by a glass or plastic layer called the cladding, which in turn is covered with plastic or other material for protection.
• The light source used is either LASER or light-emitting diodes (LED) whereas the detector is a photodiode.
• Low error rate, very high noise immunity, immunity to electrical and magnetic noise.
• Reduced size and weight (than copper or coaxial).
• The high cost of installation with special equipment.
• Very expensive, but maybe economical for high-volume application.
• High data rate-over 2 Gbps possible.

Usage:

• Voice Communication
• Video Communication
• Data Communication
• Networks using fibre-optic are called “Fibre Distributed Data Interface” (FDDI) often ring-based.

(B) Unguided/Wireless Media: In this type of media, wires are not used, means no physical connection.

Question 6.
Why are computer and communication used together? What are its advantages? Give their application. (U. P. 2004, 16, 17)
Communication: Communication is a process in which data is (UPBoardSolutions.com) transferred from one place to another. The three components of communication are:

• Sender: where the information is sent.
• Receiver: to whom the information is sent.
• Medium: through which the information is sent.

A computer is an electronic machine which can perform a variety of tasks like:

• Text formatting
• Producing sound
• Creating graphics
• Show clippings, etc.

All the above tasks are needed in the process of communication and that is why a computer is one of the most popular devices used for communication. In computer, communication information is sent electronically.

Advantages of Computer Communication: There are the following advantages of computer communication:

• A user-friendly environment.
• Information can be sent in multiple ways.
• Information can be transferred in pictures, sound, text, etc. from the same place.
• Secured means of communication.
• It saves time as well as money.

Application of Computer Communication: There are the following applications of computer communication:

• E-mail: Electronic mail is one of the most popular applications of communication.
• Video-conferencing: People from far off places can organize meetings very easily.
• Chatting: Different persons from different places can exchange their thoughts at the same time.
• Information Distributor: Information available on different computers can be used.

Question 7.
Draw the basic model for primary communication. Explain the functions of each block. (U. P. 2007, 10)
or
What is a basic model of communication? Explain its various parts. (U. P. 2008)
or
Give a simple block-view of a primary communication system. Explain the working of each block in detail. (U. P. 2011)
Communication: Transferring the data from one place to another is known as “communication.” For this, computers are playing a very important role in today’s life. We can connect one computer to others in any (UPBoardSolutions.com) part of the world very easily and computers are used to do fast and accurate communication.

Every communication system has the following important components:

1. Sender: It is an electronic device which is responsible for sending information e.g., Telephone, Computer, Mobile, etc.

2. Data Communication Device: It is a device which accepts the data from source sending device and converts it into such a form (analogue signal) that can be transmitted over communication channels e.g., computers with the modem.

3. Communication Medium: In order to transfer the information from one place to another “carriers are needed”, and in a computer network these carriers are generally of two types:

Guided Communication Medium: In this type of medium, wires are used to transfer the data and it is also known as wired media. It consists of:

• Twisted Pair Cable,
• Co-axial Cable,
• Fibre Optics Cable.

Unguided Communication Medium: This is wireless medium, hence, no wires are used. It is categorised into:

• Microwave Waves,
• Satellite Communication.

4. Receiver: It is used to accept signals of the communication system and convert them in an understandable form. Generally, the receiver is a computer with a modem.
Examples:

• TCP/IP (Transmission Control Protocol/Internet Protocol): TCP breaks up the data to be sent into packets. It guarantees that any data sent to the destination computer reaches it. IP is a set of conventions used to send packets from one host to another. It is responsible for routing the packets to a desired destination IP address.
• X.12: This protocol is used to establish the connection between companies to exchange important papers.
• X. 25: This protocol establishes the interface for common (UPBoardSolutions.com) data network.
• HTTP: HyperText Transfer Protocol works on the internet to send and receive files from different locations.

Computer and Communication Short Answer Type Questions (4 Marks)

Question 1.
Define TCP/IP.
The Internet is a packet-switching network, data is transmitted by converting it into packets. This work is done by the set of rules or standard designs to enable a computer to connect to one another and to exchange information known as PROTOCOL.

• TCP/IP is the only protocol used to send data all around the Internet.
• TCP/IP is made up of two components—TCP and IP.

TCP (Transmission Control Protocol): TCP breaks up the data to be sent into packets. It guarantees that any data sent to the destination computer reaches it.

IP (Internet Protocol): IP is a set of conventions used for routing packets from one host to another. It is responsible for routing the packets to a desired destination IP address.

Question 2.
Define different types of Verbal Communication. (U. P. 2008)
Verbal Communication. When any data or information is transferred verbally from one place to another, it is known as verbal communication, such as the communication between two persons on the telephone.
Different types of Verbal Communications are:

• Telephone: With the help of telephone, we can converse with others from a distance. The communication is verbal in this case and the medium is wire.
• Intercom: It is just like telephone technology, but it is limited to a building or campus for the purpose of conversation between the two.
• Mobile: It is the modem technology of the telephone system. In this, we can also converse with each other, but the medium in this is magnetic wave through the air.
• Chatting through Computer: In this technology, we can exchange our views and thoughts verbally by using a microphone. The medium in this technology is the telephone.

Question 3.
Explain the ISO reference model. (U. P. 2016, 19)
An interconnected protocol for the computer to computer communication as recommended by the International Standards Organisation (ISO) is gaining wide acceptance. It is an approach based on defining a (UPBoardSolutions.com) number of distinct layers each addressing itself to one aspect of linking. This is known as the ISO model for Open System Interconnection. It is a seven-layer architecture and defines a separate set of protocols for each layer.

Question 4.
Define satellite Communication.

1. Line of sight required between satellite and earth stations.
2. 12 to 24 transponders per satellite. These transponders receive, amplify, change frequency and transmit.
3. Geosynchronous orbit (22,300 miles).
4. Low security—anyone with satellite dish and right frequency can tune in.
6. Data rates of up to 50 Mbps.
7. Microwave signal at 6 GHz is beamed to it from a transmitter on the earth.

It is amplified and retransmitted to the earth at 4 GHz by a system called a transponder mounted on the satellite to avoid interference.

Question 5.
Explain Data Communication. (U. P. 2017)
Communication refers to the electronic transmission of any type of electronic information encompasses telephone communication, the transmission of television signals, data communication of all forms, electronic mail, facsimile transmission, and so on.

Data communication is the movement of encoded information from one point to another by means of an electrical or optical transmission system, such systems often are called data communication networks.

Computer and Communication Very Short Answer Type Questions (2 Marks)

Question 1.
Give names of various type of communication media. (U. P. 2014)
Communication media are of two types:

1. Guided or Wired media:
• Twisted pair cable
• Coaxial cable
• Optical Fibre Cable.
2. Unguided or Wireless media:
• Microwave
• Satellite.

Question 2.
How many layers are there in the ISO model? (U. P. 2016)
There are seven layers in the ISO model.

Question 3.
Which layer is responsible for file transfer?
The application layer is responsible for file transfer.

Question 4.
Which communication is used to link metropolitan cities?
Microwave communication is used to link metropolitan cities.

Question 5.
Discuss Wireless Communication Media. (U. P. 2014)
In this type of media, wires are not used, means not physical connection.

Question 6.
At how much distance are the repeaters used in microwave transmission?
Repeaters are used at a distance of 30-35 km (UPBoardSolutions.com) in microwave transmission.

Question 7.
What is RS-232-C?
RS-232-C is a kind of protocol which is used to link a digital device to the modem.

Question 8.
Explain the full form of E-mail.
The full form of E-mail is Electronic Mail.

Computer and Communication Objective Type Questions (1 Mark)

There are four alternative answers for each part of the questions. Select the correct one and write in your answer book:

Question 1.
Which communication media has a data rate of up to 20 Mbps?
(a) Coaxial cable
(b) Twisted cable
(c) Fibre optic
(d) Microwave.
(b) Twisted cable

Question 2.
Which is not a protocol? (U. P. 2014)
(a) TCP/IP
(b) HTTP
(c) X-25
(d) ISP.
(d) ISP

Question 3.
Which is not a physical communication channel?
(a) Twisted pair
(b) Coaxial pair
(c) Fibre optic
(d) Microwave
(d) Microwave

Question 4.
In which media, transponders are used? (U. P. 2012, 14)
(a) Satellite
(b) Microwave
(c) Fibre optic
(d) Twisted cable.
(a) Satellite

Question 5.
Transponders are used in which communication media? (U. P. 2014)
(a) Fibre optic
(b) Microwave
(c) Satellite
(d) Co-axial cable.
(d) Co-axial cable

Question 6.
Which of the following is not Hardware? (U. P. 2015)
(a) CPU
(b) RAM
(c) Windows
(d) MODEM.
(c) Windows

Question 7.
In which transmission medium the data transmits in the form of light waves? (U. P. 2017)
(a) Copper wire
(b) Coaxial Cables
(c) Telephone lines
(d) Optical Fibre Cable.
(b) Coaxial Cables

Question 8.
Which of the following is not an input device? (U. P. 2018)
(a) Key-board
(b) Printer
(c) Mouse
(d) Joy-stick
(b) Printer

Question 9.
Which one of the following is an O.S.? (U. P. 2018, 19)
(a) MODEM
(b) Ink-jet
(c) Pen drive
(d) DOS.
(d) DOS.

Question 10.
Which of the following is not a hardware device? (U. P. 2018)
(a) Memory
(b) Cache
(c) Excel
(d) Processor.
(c) Excel

Question 11.
What type of communication happens in Computer Communication? (U. P. 2019)
(a) Simplex
(b) Full duplex
(c) Half-duplex
(d) Quarter.