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CBSE CLASS X MATHEMATICSCHAPTER 1 REAL NUMBERS RECAP FROM PREVIOUS CLASS/ES1.NUMBER LINE: Representation of fractions on a number line2.Whole Numbers (W): 0, 1, 2, 3, ..3. Natural Numbers (N): 1, 2, 3, ..4. Integers (Z): .. 3, -2, -1, 0, 1, 2, 3, .(Comes from German word, Zahlen, which means To Count.5. Rational Numbers (r): Can be written in the form of p/q, where, q 0, As, anything divided by 0 is not defined. ( These Include N, W & Z). p & q have no common factors other than 1, ie. They are co-prime.Equivalent Rational Numbers or Fractions: Their Standard form is same. Eg. 1/2 = 2/4 = 10/20.Finding a Rational Number between Two Given Rational Numbers, a & b: (a + b)/2, is the required number; Proceeding in the same way, we may find more Rational Numbers between the given two Rational Numbers. There are infinitely many Rational Numbers between the given two Rational Numbers.The other way is to convert the given rational numbers into like fractions ( ie. With the same denominators). 6. Irrational Numbers (s) are the ones which cant be written in the form of p/q, where p & q are integers, and, q0.Pythagoreans were the first ones to discover irrational numbers.7. Real Numbers (R): The Rational and Irrational Numbers together form the Real Numbers. A Real Number is either a Rational or Irrational Number. Every Real Number is represented by a unique point on a number line. And, also, every point on a number line, represents a unique real number. (Shown by German Scientists Cantor & Dedekind.)Thus, the number line is also called Real Number Line.The decimal expansion of a rational number is either terminating or non - terminating recurring (which can be pure recurring or mixed recurring - Explained Later) ; & vice - versa.The decimal expansion of a irrational number is non - terminating non recurring; & vice - versa.Let a > 0, be a real number, and, n be a positive integer, then, = b, if bn= a. The Symbol is called the Radical Sign.8. Laws of Indices: Let a and b be positive real numbers, then,(a, n, m are natural numbers; a is called the base; m & n are the exponents)Now, a0 = 1; so, 1/an = a0 / an = a0-n = a-n.1/2 = 2 / 2; which is half of 2; and Thence 1/2 can be represented on a Number Line.When the denominator of an expression contains a term with a square root (or a number with a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called Rationalising the Denominator.

9. FEW RELEVANT POINTS RELATED TO THE CHAPTER:Note the Following points about the Real Numbers:1.Sum or Difference of a rational and an irrational number is irrational [eg. (3 + 7) or (7 - 3) is irrational.]2.The Product or Quotient of a non - zero rational number with an irrational number is irrational [eg. The Expressions 22 OR 32 are irrational.]3.If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational [eg. 2 - 2 is 0, which is rational; but, 7 - 2 is irrational].Square root of all positive integers which are not perfect squares; cube roots of all integers which are not perfect cubes, and so on, are all irrational numbers.Pure Recurring Decimals are the ones, in which, all the digits after the decimal point are repeated; eg. 0.32323232 = 0.32.Mixed Recurring Decimals are the ones, in which, at least one digit after the decimal point is not repeated; eg. 18.33249494949 = 18.33249Prime Numbers are the numbers, other than 1, whose only factors are1 and the number itself; Eg. 3, 13, 17Composite Numbers are the numbers which have more than 2 common factors; Eg. 12, 15 etc.Co - Prime Numbers are those two Natural Numbers (Not necessary prime numbers), ehich have their Highest Common Factor as One. Eg. (3, 10); (15, 33) etc.1 (One) is a Unique Number; ie. It is neither a Prime or a Composite Number.-CBSE CLASS X CHAPTER ONE: REAL NUMBERS1. A non-zero integer a is said to divide an integer b, if there exists an integer c such that: b = acEuclids Division Lemma: Let there be two positive integers a and b. Then, there exist unique integers q and r such that a = bq + r, 0 r < b. Euclids Division Lemma is a restatement of the long division process, and the integers q & r are called the Quotient and Remeinder.((Find integers q and r for follg. pairs of positive integers a and b: (i) 10, 3 (ii) 4, 19 (iii) 81, 3))The Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes (or the powers of primes) in a unique way (ie. The factorization is unique); apart from the order in which the prime factors occur. Composite Number = Product of Primes.This Theorum is used: (i) to prove the irrationality of many of the numbers, and,(ii) to find out when exactly the decimal expansion of a rational number is terminating, and when it is non-terminating, repeating.

An algorithm is a series of well defined steps which gives a procedure for solving a type of problem.A lemma is a proven statement used for proving another statement.2. Finding HCF & LCM by prime Factorisation Method: We can find HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic. This method is also called the prime factorisation method.HCF = Product of the smallest power of each common prime factor in the numbers.LCM = Product of the greatest power of each prime factor, involved in the numbers.For any two positive integers (a & b), HCF (a, b) X LCM (a, b) = a X b

Let x = p/q be a rational Number (where p q are co-prime):If q = 2n 5m, then p/q is a Terminating Decimal Expansion. If q 2n 5m, then p/q is a Non-Terminating Repeating Decimal Expansion.(n& m are non-negative integers)3. Finding HCF of two Positive Integers using Euclids Division Lemma:Let a & b be two positive integers.Obtain two whole numbers, q1 & r1, such that a = bq1 + r1; 0 r1 < b.If r1 = 0, b is the HCF of a & b.If r1 0, Apply Euclids Division Lemma to b & r1, and Obtain two whole numbers, q2 & r2, such that b = r1q2 + r2; Proceed as such, till the remainder becomes zero. The divisor at this stage is the HCF of a & b. 4. Theorum: Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.Proof: Let P1, P2, P3...Pn be factors of a. Thus, a = P1.P2.P3...Pn ; where, P1, P2, P3...Pn are primes, not necessarily all distinct.Thus, a2 = (P1.P2.P3...Pn ). (P1.P2.P3...Pn ); Thus, a2 = P12.P22.P32...Pn2P divides a2 (i) => P is a Prime factor of a2, from The Fundamental Theorem of Arithmetic.Prime factors of a2 are only (P1.P2.P3...Pn ) (ii) (Uniqueness of factors from Fund. Th. Of Arith.)From (i) & (ii), we have P is a prime factor of (P1.P2.P3...Pn ) => P divides a.

HANDLING INDICES1.Introduction:Why Exponents were Invented:10,000,000,000,000 = Ten trillion, or simply 1013; 7x7x7x7x7x7x7x7 = 78; 6,000,000 = 6 x 1,000,000 = 6 x 106 In am or an , a is the base, and m & n are called the exponents;a0 = 1, ie. Anything raised to the power 0 is One (1); Thus, 30=1, x0=1, 100=1.When no exponent is mentioned on to a number, The Exponent is taken to be as 1, ie. 3 = 31; x = x1; 7 = 71A Negative Index implies the Reciprocal of a number, ie. 3-2 = 1/32 = 1/9; x-4 = 1/x4; 2-3 = 1/23 = 1/8; an = 1/a-n; a-n = 1/an.A root is presented as such: 2 = 21/2 [A second root of 2, or Square root of 2, or simply, under-root of 2]; 33 = 31/3 [A Third root of 3, or Cube root of 3]; nx = x1/n [ nth root of x] 2.Laws of Indices:am. an = am+n [32. 33= 3x3x3x3x3 = 35 = 32+3];am / an = am-n [56 / 54 = (5x5x5x5x5x5) / (5x5x5x5) = 5x5 = 52 = 56-4];(am)n = amn [(73)2 = (7x7x7)2 = 7x7x7 x 7x7x7 = 76 = 73x2 ];(ab)m = am x bm [(2x5)3 = 2x5 x 2x5 x 2x5 = 2x2x2x5x5x5 = 23 x 53 ];(a/b)m = am / bm [ (3/5)2 = 3/5 x 3/5 = (3x3) / (5x5) = 32 / 52 ][ 1/an = a0 / an = a0-n = a-n ]3. Applying Indices to Roots:ab = a b [(ab)1/2 = a1/2 . b1/2 = a . b];a/b = a / b [(a/b)1/2 = a1/2 / b1/2 = a / b]; (a - b) x (a + b) = a - b [ (a - b) x (a + b) = (a)2 - (b)2 = (a1/2)2 - (b1/2)2 = a1/2 x 2 - b1/2 x 2 = a1 - b1 = a - b]

CBSE CLASS X MATHEMATICSCHAPTER 1 REAL NUMBERS - NCERT EXERCISES SOLUTIONSEXERCISE 1.1 (P. 7)Q1. Use Euclids division algorithm to find the HCF of :(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255.A.(i) 225 = 135 X 1 + 90; 135 = 90 X 1 + 45; 90 = 45 X 2 + 0; HCF = 45 (ii)HCF = 196 (iii) 867 = 255 X 3 + 102; 255 = 102 X 2 + 51; 102 = 51 X 2 + 0;HCF (867, 255) = HCF (255, 102) = HCF (102, 51) = 51.Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.A. By Euclids division algorithm, a = bq + r (i), 0 r < b(The value of r will always be less than the value of b; ie. Less than the value of divisor)(Also. The degree of r is always less than the degree of the divisor)Putting b = 6 in (i), we get a = 6q + r [ 0 r < 6; ie. r = 0, 1, 2, 3, 4, 5]If r = 0, a = 6q, 6q is divisible by 6 => 6q is even.(Divisibility Criteria: If a number is divided by 2 & 3, it is divisible by 6!) If r = 1, a = 6q + 1, 6q +1 is not divisible by 2.If r = 2, a = 6q + 2, 6q + 2 is even and is divisible by 2.If r = 3, a = 6q + 3, 6q + 3 is not divisible by 2.If r = 4, a = 6q + 4, 6q + 4 is even and is divisible by 2.If r = 5, a = 6q + 5, 6q + 5 is not divisible by 2.As 6q, 6q + 2, 6q + 4 are even; therefore, 6q + 1, 6q + 3, 6q + 5 are odd.Q3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?A. To find the maximum number of columns, we have to find the HCF of 616 and 32.By Euclids division Lemma, HCF of 616 & 32 is 8.Hence, maximum number of columns is 8.Q4. Use Euclids division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.A. By Euclids division algorithm, a = bq + r, 0 r < b (i);On putting, b = 3 in (i), we get a = 3q + r, 0 r < 3, ie. R = 0, 1, 2.If r = 0, a = 3q=> a2 = 9q2;(ii)If r = 1, a = 3q + 1=> a2 = 9q2 + 6q + 1;(iii)If r = 2, a = 3q + 2=> a2 = 9q2 + 12q + 4;(iv)From (ii), 9q2 is a square of the form 3m, where m = 3q2;From (iii), 9q2 + 6q + 1, ie. 3(3q2 + 2q) + 1 is a square of the form 3m + 1, where m = 3q2 + 2q;From (iv), 9q2 + 12q + 4, ie. 3(3q2 + 4q + 1) + 1 is a square of the form 3m + 1, where m = 3q2 + 4q + 1.Thus, square of any positive integer is either of the form 3m or 3m + 1.Q5. Use Euclids division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.A. Let x be any positive integer. Then, it is of the form 3m, 3m + 1, 3m + 2. now, we have to prove that the cube of each of these can be rewritten in the form 9q, 9q + 1, 9q + 8.Now, (3m)3 = 27m3 = 9(3m3) = 9q, where q = 3m3;(3m + 1)3 = 27m3 + 27m2 + 9m + 1 = 9(3m3 + 3m2 + m) + 1 = 9q + 1, where q = 3m3 + 3m2 + m;(3m + 2)3 = 27m3 + 54m2 + 36m + 8 = 9(3m3 + 3m2 + m) + 8 = 9q + 8, where q = 3m3 + 6m2 + 4m.EXERCISE 1.2 (P. 11) Q1. Express each number as a product of its prime factors:(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429. (Use Division Method)A. (i) 140 = 22 X 5 X 7(ii) 156 = 22 X 3 X 13(iii) 3825 = 32 x 52 x 17(iv) 5005 = 5 X 7 X 11 X 13(v) 7429 = 17 X 19 X 23Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM HCF = product of the two numbers.(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54A.Do this solution yourself.Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25A.(i) 12 = 2X2X3; 15 = 3X5; 21 = 3X7; HCF = 3; LCM = 2x2x3x5x7 = 420(ii) The given three numbers don't have any common factor; They are primes.HCF = 1; LCM = 17X23X29 = 11339.(iii) Do this yourself!Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).A.For any two positive integers a and b, HCF (a, b) LCM (a, b) = The product of the numbers (a b)Q5. Check whether 6n can end with the digit 0 for any natural number n.A.If the number 6n ends with the digit zero; then, it is divisible by 5. Therefore, the prime factorisation of 6n contains the prime number 5. This is not possible, as the only prime in the factorisation of 6n is 2 & 3; and, the uniqueness of the fundamental theorum of arithmetic guarantees that there are no other prime in the factorisation of 6n .So, there is no value of n in natural numbers for which 6n ends with the digit zero. Q6. Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.A. (i) 7 11 13 + 13 = 1001 + 13 = 1014; and, 1014 = 2X3X13X13. Thus, 1014 is the product of prime factors. Hence, it is a composite number.(ii) 7 6 5 4 3 2 1 + 5 = 5040 + 5 = 5045 = 5 X 1009. Thus, it is a product of prime factors 5 and 1009. Hence, it is a composite number.Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?A. They will be again at the starting point after common multiples of 18 & 12.18 = 2x3x3; 12 = 2x2x3; Thus, LCM of 18 & 12 = 36. In 36 mts. Ravi arrives at the starting point after making 3 rounds, as he takes 12 mts. to drive one round & Sonia also arrives at the starting point after making 2 rounds, as he takes 18 mts. to drive one round. Thus, they will meet again at the starting point after 36 minutes.EXERCISE 1.3 (P. 14) Q1. Prove that 5 is irrational.A. Let 5 is a rational number. Then, 5 can be written in the form p/q, where p, q are integers and have no common factor (other than 1), q 0. ie. 5 = p/q; Squaring both sides: 5 = p2 / q2 => p2 = 5q2 (i)=> 5 is a factor of p2, & thus divides p2.This implies that 5 divides p (ii). Now, p = 5m => p2 = 25m2. Putting this in equation (i), we get 25m2 = 5q2 => 5m2 = q2 => 5 divides q2 => 5 divides q (iii).From (ii), 5 divides p, and from (iii) 5 divides q. This means 5 is a common factor of p and q. This contradicts the supposition that there is no common factor of p and q. Hence, 5 is an irrational number. Q2. Prove that 3 + 25 is irrational.A. Let 3 + 25 be a rational number. Now, let 3 + 25 = a/b, where a and b are co-prime & b 0.So, 25 = a/b - 3; Or, 5 = a/(2b) - 3/2.Since, a and b are integers, therefore a/(2b) - 3/2 is a rational number. And, thus 5 is a rational number. But 5 is an irrational number. Thus, our supposition is wrong.Hence, 3 + 25 is an irrational number. Q3. Prove that the following are irrationals :(i) 1/2, (ii) 75,(iii) 6 + 2(From Previous Class: Square root of all the numbers which are not perfect squares; and, Cube root of all the numbers which are not perfect cubes are Irrational Numbers.)A. (i) Let 1/2 be a rational number. Now, let 1/2 = a/b, where a and b are co-prime & b 0.Or, (1 X 2 )/(2 X2 ) = a / b; Or, 2 / 2 = a / b; Or, 2 = 2a/b.((Always try to isolate the Irrational Component))Since a and b are integers, 2a/b is rational and so 2 is rational. But, 2 is irrational.Thus, 1/2 is Irrational.(ii) Let 75 be a rational number. Now, let 75= a/b, where a and b are co-prime & b 0.Or, 5= a / 7b; ((Always try to isolate the Irrational Component))Since a and b are integers, a/7b is rational and so 5 is rational. But, 5 is irrational.Thus, 75is Irrational.(iii) Let 6 + 2 be a rational number. Now, let 6 + 2= a/b, where a and b are co-prime & b 0.Or, a/b - 6 = 2; ((Always try to isolate the Irrational Component))Since a and b are integers, a/b - 6 is rational and so 2 is rational. But, 2 is irrational.Thus, 6 + 2 is Irrational.======================================================================EXERCISE 1.4 (P. 17) Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:i.13 / 3125. 3125 = 20 x 55. Thus, Denominator is of the form 2m X 5n. Thus, 13 / 3125 is a Terminating Decimal.ii.17 / 8. 8 = 23 x 50. Thus, Denominator is of the form 2m X 5n. Thus, 8 is a Terminating Decimal.iii.64 / 455. 455 = 5 X 7 X 13. Thus, Denominator is not of the form 2m X 5n. Thus, 64 / 455 is a Non - Terminating Repeating Decimal.iv.15 / 1600. 1600 = 26 x 52. Thus, Denominator is of the form 2m X 5n. Thus, 15 / 1600 is a Terminating Decimal.v.29 / 343.Denominator is not of the form 2m X 5n. Thus, 64 / 455 is a Non - Terminating Repeating Decimal. (As 343 is a odd number, it is not divisible by 2, and hence its denominators factors cant have powers of 2)vi.23 / 23 52. Denominator is of the form 2m X 5n. Thus, 23 / 23 52 is a Terminating Decimal.vii.129 / 225775. Denominator is not of the form 2m X 5n. Thus, 129 / 225775 is a Non - Terminating Repeating Decimal. viii. 6/15 = 2/5. 5 = 20 x 51. Thus, Denominator is of the form 2m X 5n. Thus, 6 / 15 is a Terminating Decimal.ix.35 / 50 = 7 / 10 = 7 / (21 x 51). Thus, Denominator is of the form 2m X 5n. Thus, 35 / 50 is a Terminating Decimal. x.77 / 210. 210 = 7x3x2x5. Thus, Denominator is not of the form 2m X 5n. Thus, 77 / 210 is a Non - Terminating Repeating Decimal.Q2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.A. (i) 13 / 3125 = 13 / (5x5x5x5x5) = (13x2x2x2x2x2) / (5x2x5x2x5x2x5x2x5x2) = 416/100000 = 0.00416.(ii) 17 / 8 = 17 / (23 x 50) = (17x53) / (23 x 53) = (17x125)/103 = 2125/1000 = 2.125.(iii) Non - Terminating Repeating.(iv) 15 / 1600 = 15 / (26 x 52) = 15 / (24x22x52) = (15x54)/(24x54x22x52) = (15x625)/106 = 9375/1000000 = 0.009375.(v) Non - Terminating Repeating.(vi) 23 / 23 52 = (23x5)/2x5x22x52) = 115/103 = 0.115(vii) Non - Terminating Repeating.(viii) 6/15 = 2/5 = (2x2)/(5x2) = 4/10 = 0.4.(ix) 35/50 = 35/(5x10) = (35x2)/(2x5x10) = 70/100 = 0.7(x) Non - Terminating Repeating.Q3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?(i) 43.123456789 (ii) 0.120120012000120000. . . (iii) 43.123456789A. (i) 43.123456789 is terminating, and hence is a Rational Number. 43.123456789 = 43123456789 = p/q. q = 1000000000 = 109 = 29x59. 1000000000 (ii) 0.120120012000120000. . . is Non - Terminating Non - Repeating; and hence is Irrational. (iii) 43.123456789 is Non - Terminating but Repeating; and hence is Rational.43.123456789 = 4312345646 = p/q. Hence, q = 999999999. 999999999

CBSE CLASS X MATHEMATICSCHAPTER 2 POLYNOMIALSRECAP FROM PREVIOUS CLASS/ES1.A Function p(x) of the form p(x) = a0 + a1x + a2x2 + a3x3 +.+ an xn is called a polynomial, where (i) a0, a1, a2, ..an are real numbers, and, (ii) n is a non - negative (ie. 0 or positive) integer. ((Thus, (2/x + 3), (x2 + 2x + 4/x) are not Polynomials.a0, a1, a2, ..an are called co - efficient of the polynomial. The expression of this form is called Polynomial in one variable. We can also have polynomials in more than one variable, eg. x2 + y2 + xyz is a polynomial in three variables.2. Terms: The polynomial 3x2 + 4x + 5 has three terms.3. Co - efficient: Each term of the polynomial has a co-efficient. In 3x2 + 4x + 5, the co-efficients of x2 , x and x0 are 3, 4, and 5.4. Constant Polynomial: -3, 0, 2, 5.. are examples of Constant Polynomials. The Constant Polynomial 0 is called a Zero Polynomial.A Polynomial written either in the descending or ascending powers of x is called the Standard Form of a Polynomial. A Polynomial can be Monomial, Binomial and so on.5. Degree of a Polynomial: The highest exponent in various terms of one variable is called its degree. Eg. A Polynomial can be Linear, Quadratic, Cubic, Biquadratic, etc. If a0 = a1 = a2 = ..=an = 0, ie. If all the constants are zero, we get a Zero Polynomial, which is denoted by 0. The Degree of a Zero Polynomial is not defined. 6. Zeroes of a Polynomial: The value of a polynomial p(x) at x = a, is p(a), obtained on replacing x by a. In general, a Zero of a Polynomial p(x) is a, such that p(a) = 0. Consider a polynomial p(x) = x - 1; Now, p(1) = 0, ie. 1 is a Zero (or root) of the Polynomial p(x), which gives x=1 Now, consider a constant polynomial 5. It has no zero because replacing x by any number in 5x0, still gives 5. Thus, a non - zero constant polynomial has no zero.We will be dealing with Polynomials in one variable only.Zeroes of a Zero (0)Polynomial: By Convention, every Real Number is a zero of the Zero Polynomial.Number of Zeroes of a polynomial of degree n is n.7. Division of a Polynomial: Say, on dividing polynomial p(x) by another polynomial g(x), we get quotient q(x) & remainder r(x). Dividend = ( Divisor X Quotient) + Remainder; ie. P(x) = g(x) . Q(x) + r(x). The degree of the remainder r(x) is always less than the degree of the divisor g(x), ie. Degree of r(x) < Degree of g(x).If the divisor is linear (ie. One degree), then the degree of remainder will be 0, ie. The remainder will be a constant.2x3 + x2 + x = x(2x2 + x + 1), We say that x and (2x2 + x + 1) are Factors of (2x3 + x2 + x ), and (2x3 + x2 + x ) is a multiple of x as well as (2x2 + x + 1) 8. Remainder Theorum: Let p(x) be any polynomial of degree greater than or equal to one, and, let a be any real number. If p(x) is divided by a linear polynomial (x-a), then the remainder is p(a).======================================================================Remember The Following Points:A zero of a polynomial need not be zero.Every Linear Polynomial has one and only one zero.A polynomial can have more than one zero.Number of Zeroes is the degree of the polynomial.[Division of a polynomial by other polynomial using long division method has to be thoroughly practiced by the student.]Proof of Remainder Theorum: Let p(x) be any polynomial with degree 1. Lets say, when p(x) is divided by (x-a), the quotient is q(x), and, the remainder is r(x); ie. P(x) = (x-a) . q(x) + r(x).Since, the degree of (x-a) is one, and the degree of r(x) is less than the degree of (x-a), the degree of r(x) = 0. This means that r(x) is a constant, say r. Thus, for every value of x, r(x) = r.Therefore, p(x) = (x-a) . q(x) + r. In particular, if x=a, this equation gives us p(a) = (a-a) . q(a) + r = r; which proves the theorem.9. Factor Theorum: If p(x) is a polynomial of degree n 1, and a is any real number, then:(i) (x-a) is a factor of p(x), if p(a) = 0, and,(ii) p(a) = 0, if (x-a) is a factor of p(x). This actually follows from the Remainder Theorum.10. Factorisation of Polynomials:A) Splitting the Middle Term: (ie. Quadratic Equations): To factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ac. Let p & q be those two numbers, ie. p+q = b, and, p.q = c.[If on multiplying a & c, we get + sign, well get the middle term by adding; Sign will be that of the middle term.]

B) By Factor Theorum: Let f(x) be a given polynomial. 1.Proceed as follows: (Factorising equations of Degree 3) Write separately, the constant term of the polynomial. Write down its factors. If the number is big, and has many factors, write down the smallest, first few factors.Let a be the zero of polynomial; ie. (x-a) is factor of polynomial f(x).Divide the given polynomial f(x) by (x-a), and get the Quotient.Factorise the Quotient by splitting the middle term.You would thus obtain the factors of the given polynomial. READ THE FOLLOWING EXAMPLE:Let f(x) = x3 - 3x2 - 9x - 5. The constant term is 5, so, its factors are 5, 1 Now, p(-1) = 0; Thus (x+1) is a factor of p(x). Now, do the long division. = (x2 - 4x - 5) (Remainder is zero);Now, by splitting the middle term, (x2 - 4x - 5) = (x+1) (x - 5).Thus, x3 - 3x2 - 9x - 5 = (x+1)(x+1) (x - 5) = (x+1)2 (x-5)In Regards to this method, remember the following important points:If the sum of coefficients of odd powers of any polynomial p(x) is the same as that the sum of its coefficients of even powers, then, (x+1) is always the factor of p(x).If the sum of the coefficients of given expression p(x) is zero, then, (x-1) is always the factor of p(x).C) Factorisation by Algebric Identities: Read the following identies. Understand them, prove them if you have to, and Remember them.(a + b)2= a2+ 2ab + b2 (a - b)2= a2- 2ab + b2 a2 b2 = (a + b) (a b) (x + a)(x + b) = x2+ (a + b)x + ab (a + b + c)2= a2+ b2+ c2+ 2ab + 2ac + 2bc (a + b)3= a3+ b3 + 3ab(a+b) (a - b)3= a3- b3 - 3ab(a-b) a3+ b3= (a + b) (a2 ab + b2) a3 b3= (a b) (a2+ ab + b2) a3+ b3+ c3- 3abc = (a + b + c)(a2+ b2+ c2- ab - bc - ca); Thus, a3+ b3+ c3= 3abc, if (a + b + c) = 0.11. Least Common Multiple (LCM) of Polynomials = (Highest Power of Common Term) X Remaining Terms. --

CBSE CLASS X CHAPTER TWO: POLYNOMIALS1.A Quadratic Polynomial can have either two distinct zeroes, two equal zeroes or no zeroes; ie. A polynomial of degree 2 has at the most two zeroes.Similarly, a Cubic Polynomial can have at the most 3 zeroes. Consider the following:If k is a zero of p(x) = ax + b; then p(k) = ak + b = 0, ie. k = (-b) / a. Thus, the zero of the linear polynomial [ak + b] is [(-b) / a] = (-)Constant Term / Coefficient of x.

Type of PolynomialGeneral Form No. of Zeroes Relationship between ZeroesLinear ax + b, a0 1 k = = Constant term Co-efficient of xQuadratic ax2 + bx + c; a0 2 Sum of Zeroes ( + )= (Let , be two zeroes) Product of Zeroes (.) = .Cubic Polynomial ax3 + bx2 + cx + d 3 Sum of Zeroes ( + +)(Let , , be three zeroes) = = -Co-efficient of x2 Co-efficient of x3 Product of Zeroes (..) = =

Sum of the product of zeroes taken two at a time (For Cubic Polynomial) = (. + . + ) = = 2. To form a Quadratic Polynomial with its given zeroes:Let , be the zeroes of a Quadratic Polynomial. x = , => x - = 0; x = , => x - = 0. Thus, (x - ).(x - ) is the Quadratic Polynomial, ie. x2 - ( + )x + ; ie. x2 - (Sum of Zeroes) x + Product of Zeroes.3. Division Algorithm for Polynomial: If p(x) & g(x) are any two polynomials with g(x) 0, then we can find polynomials q(x) & r(x) such that p(x) = g(x) X q(x) + r(x); where r(x) = 0, or degree of r(x) is less than the degree of g(x).Dividend = Divisor X Quotient + Remainder. 4. Geometric Meaning of the zeroes of the Polynomial. a.Graph of y = ax + b is a straight line which intersects the x-axis at ( , 0). This x-Coordinate (of the point of intersection of the graph with x-axis) is the zero of the polynomial y = ax + b .

b.Graph of the Quadratic Equation y = ax2 + bx + c, a 0, can have two distinct zeroes (graph cutting the x-axis at two distinct points), one zero (or two equal zeroes), or no zeroes (graph not cutting the x-axis at all (in which case, it would not be possible to factorise the quadratic polynomial).In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards or open downwards depending on whether a > 0 or a < 0. (These curves are called parabolas.)c. A cubic polynomial of the form y = a3 can have at the most 3 zeroes.In general, a polynomial p(x) of degree n has at the most n zeroes. N.B.: Formation of a cubic polynomial: Let , , be three zeroes of the Polynomial. Then, the required cubic polynomial is (x - ) (x - )(x - ).

CBSE CLASS X MATHEMATICSCHAPTER 2 POLYNOMIALS - NCERT EXERCISES SOLUTIONSEXERCISE 2.1 (P. 28)Q1. The graphs of y = p(x) are given in the Fig. below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. A.Self Explanatory.-EXERCISE 2.2 (P. 33)Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.Compare the quadratic equations given with the general quadratic equation, ie. ax2 + bx + c. A.(i) x2 2x 8 = (x-4)(x+2). The zeroes are x = 4, -2Sum of Zeroes = 4 - 2 = 2 = -(-2)/1 = - Coefficient of x = (-b)/a Coefficient of x2Product of zeroes = -8 = Constant Term Coefficient of x2 (ii) 4s2 4s + 1 (iii) 6x2 3 7x (iv) 4u2 + 8u (v) t2 15 (vi) 3x2 x 4Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.A. Let the polynomial be ax2 + bx + c, and its zeroes be & .(i) 1/4, -1; + = 1/4; . = -1The polynomial formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - 1/4 x + (-1) = x2 - x/4 - 1. The other possible polynomials would be k(x2 - x/4 - 1).If k = 4, then the polynomial is 4x2 - x - 4. (ii) 2, 1/3; + = 2; . = 1/3The polynomial formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - 2x + 1/3. The other possible polynomials would be k(x2 - 2x + 1/3).If k = 3, then the polynomial is 3x2 - 32x + 1.(iii) 0, 5; + = 0; . = 5The polynomial formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - 0.x + 5 = x2 + 5. (iv) 1, 1; + = 1; . = 1The polynomial formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - 1.x + 1 = x2 - x + 1.(v) -1/4, 1/4; + = -1/4; . = 1/4The polynomial formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - (-1/4)x + 1/4 = x2 + x/4 + 1/4. The other possible polynomials would be k(x2 + x/4 + 1/4).If k = 4, then the polynomial is 4x2 + x + 1.(vi) 4, 1; + = 4; . = 1The polynomial formed is x2 - (Sum of Zeroes) x + Product of Zeroes = x2 - 4x + 1.--

EXERCISE 2.3 (P. 36) Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :(i) p(x) = x3 3x2 + 5x 3, g(x) = x2 2(x3 3x2 + 5x 3) = (x2 2).(x-3) + (7x - 9). (ii) p(x) = x4 - 3x2 + 4x + 5, g(x) = x2 x + 1; (x4 + 0x3 - 3x2 + 4x + 5) = (x2 x + 1).(x2 + x - 3) + (8). (iii) p(x) = x4 + 0.x2 - 5x + 6, g(x) = 2 - x2 = -x2 + 2(x4 + 0.x2 - 5x + 6) = (-x2 + 2).(-x2 - 2) + (-5x + 10). Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:Remainder is 0, So,(t2 - 3) is a factor of (2t4 + 3t3 - 2t2 - 9t - 12). (ii) x2 + 3x + 1; 3x4 + 5x3 - 7x2 + 2x + 2Remainder is 0, So,(x2 + 3x + 1) is a factor of (3x4 + 5x3 - 7x2 + 2x + 2).(iii) x3 - 3x + 1; x5 - 4x4+ + x2 + 3x + 1As Remainder is not 0, So,(x3 - 3x + 1) is not a factor of (x5 - 4x4+ + x2 + 3x + 1).Q3. Obtain all other zeroes of 3x4 + 6x3 2x2 10x 5, if two of its zeroes are (5/3), and - (5/3)A.Since the two zeroes are , therefore, is a factor of p(x).Now, applying division algorithm to the given polynomial and 3x2 - 5.

Now, x2 + 2x + 1 = (x + 1)2. Zeroes of (x + 1)2 are -1, -1.Hence, all its zeroes are (5/3), (-)(5/3), -1,-1.Q4. On dividing x3 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were (x 2) and (2x + 4), respectively. Find g(x).A. By Division Algorithm, p(x) = q(x).g(x) + r(x)Thus, (x3 3x2 + x + 2) = (x 2).g(x) + (2x + 4);Or, (x 2).g(x) = (x3 3x2 + x + 2) - (2x + 4) = (x3 3x2 + 3x - 2)Thus, g(x) = (x3 3x2 + 3x - 2) (x 2). This gives g(x) as x2 - x + 1Q5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0A. Accordingto the division algorithm, ifp(x) andg(x) are two polynomials withg(x) 0, then we can find polynomialsq(x) andr(x) such thatp(x) =g(x) q(x) +r(x),wherer(x) = 0 or degree ofr(x) < degree ofg(x)Degree of a polynomial is the highest power of thevariablein the polynomial (i) deg p(x) = deg q(x) Degree of quotient will be equal to degree ofdividendwhendivisoris constant ( i.e., when any polynomial isdividedby a constant) Let p(x) = 12x2 + 8x + 24; q(x) = 3x2 + 2x + 6; g(x) = 4; & r(x) =0.Degree ofp(x) andq(x) is the same i.e., 2. Checking fordivision algorithm, p(x) =g(x) q(x) +r(x);(12x2 + 8x + 24) = 4(3x2 + 2x + 6) + 0; Thus, the division algorithm is satisfied.(ii) deg q(x) = deg r(x) (Here, we have to remember that the degree of r(x) is always less than the degree of g(x) )Let us assume the division ofx3+ xbyx2,Here, p(x) =x3+ xg(x) =x2q(x) =xandr(x) =xClearly, thedegreeofq(x) andr(x) is the same i.e., 1.Checking fordivision algorithm,p(x) =g(x) q(x) +r(x)x3+ x= x2 x+xx3+ x = x3+ xThus, the division algorithm is satisfied.(iii) deg r(x) = 0Degree of remainder will be 0 when remainder comes to a constant.Let us assume the division ofx3+1byx2.Here, p(x) =x3+1g(x) =x2q(x) =xandr(x) = 1Clearly, the degree ofr(x) is 0.Checking fordivision algorithm,p(x) =g(x) q(x) +r(x)x3+1 = (x2) x+ 1x3+1= x3+1Thus, the division algorithm is satisfied.

CBSE CLASS X MATHEMATICSCHAPTER 3 LINEAR EQUATIONS IN TWO VARIABLESRECAP FROM PREVIOUS CLASS/ES1. An Equation is a statement of equality of two algebraic expressions involving one or more unknown quantities called the variables.The Equations involving only one variable are called equations in one variable, eg. 3x - 6=0; y3 - 8 =0.An Equation involving only linear polynomial is called a linear equation; eg. 3x+7=0.The value of the variable, which when substituted for the variable (in the equation), makes both sides of the given equation equal, is called a solution, or root of the equation; eg. x=3 is root of the equation 3x+10 = 19.2. Properties of Equation: We can add, subtract, to both sides of the equation by the same number,We can perform multiplication or division to both sides of the equation by the same non-zero number, without changing the equality.3. Linear equation in two variables: An Equation of the form ax + by + c = 0, where a, b, c are real numbers, and a 0, b 0, is called a linear equation in two variables, eg. x - y = 0. A linear equation in two variables has infinitely many solutions . Solutions to a Linear Equation lie on a straight line.Often, the condition, a & b are not both zero is denoted by a2 + b2 0The reason that a degree one polynomial equation ax + by + c = 0 is called a linear equation is that its geometrical representation is a straight line.4. Drawing graph of the equation ax + by + c = 0:Express the equation in the form y= - ax + c ; b Give integral values to x, and find corresponding values of y;Plot the points (a1,b1), (a2,b2), (a3,b3) so obtained on the graph paper;Join the points to get a line which represents the equation ax + by + c = 0.Each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.CBSE CLASS X CHAPTER THREE: LINEAR EQUATIONS IN TWO VARIABLES1. The general form for a pair of linear equations in two variables x and y is a1x + b1y + c1 = 0and a2x + b2y + c2 = 0, where,a1 , b1, c1, a2, b2, c2 are all real numbers, and a12 + b12 0; a22 + b22 0For any two given lines in a plane, only one of the following three possibilities can happen:The two lines will intersect at one point,The two lines will not intersect, i.e., they are parallel, &The two lines will be coincident.(( When solving the linear equation in two variable, when one of the variable is put to zero, the equation reduces to a linear equation in one variable, which is solved easily.))A pair of linear equations in two variables is said to form a system of Simultaneous Linear Equations. A pair of values of x & y satisfying each one of the equation in x & y is called a solution of the system.2. Graphical Method of Solution of a Pair of Linear Equations:A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent.Lines representing a pair of linear equations in two variables:The lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations with unique solution),The lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations with infinite solutions],The lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations with no solution).

Summarizing the above facts:Again, consider the general form for a pair of linear equations in two variables x and y a1x + b1y + c1 = 0and a2x + b2y + c2 = 0, where,a1 , b1, c1, a2, b2, c2 are all real numbers, and a12 + b12 0; a22 + b22 0 (a & b are not both zero)Intersecting lines (Consistent pair of equations with unique solution): a1/a2 b1/b2Coincident lines (Consistent pair of equations with infinite solutions): a1/a2 = b1/b2 = c1/c2 Parallel Lines (Inconsistent pair of equations with no solution): a1/a2 = b1/b2 c1/c2 3. Algebraic solution of a system of linear equations: The graphical method is not convenient in cases when the point representing the solution of the linear equations has non-integral coordinates like ( 3, 27 ), (1.05, 3.8) etc.Three algebraic methods available to solve a pair of linear Equations are:3.1 Substitution Method: Consider the following steps to understand this method of solving a given pair of equations: Step 1 : Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.Step 2 : Substitute this value of y in the other equation. This reduces it to an equation in one variable, i.e., in terms of x, which can now be solved. Sometimes, we may get statements with no variable. If the statement obtained is true (eg. 23 = 23), we can conclude that the pair of linear equations has infinitely many solutions. If the statement obtained is false, then the pair of linear equations is inconsistent (& has no solutions).Step 3 : Substitute the value of x obtained in Step 2 in any of the original equation or the one obtained in Step 1 to obtain the value of the other variable.Remark : We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method.3.2 Elimination Method: This is the method of eliminating (i.e., removing) one variable. The steps involved are:Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If we get an equation in one variable, go to Step 3.If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.Step 3: Solve the equation in one variable (x or y) so obtained to get its value.Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.3.3 Cross - Multiplication Method: Consider the two linear equations in their general form:a1x + b1y + c1 = 0 (i), and, a2x + b2y + c2 = 0 (ii). Follow the following steps:Step 1: Multiply equation (i) by b2, & Equation (ii) by b1. We get:b2a1x + b2b1y + b2c1 = 0 (iii), and, b1a2x + b1b2y + b1c2 = 0 (iv). Step 2: Subtracting (iv) from (iii), we get: x=(b1c2 - b2c1)/(a1b2 - a2b1); where, a1b2 - a2b1 0 (v)Step 3: Substituting this value of x in (i) or (ii), we get,y=(c1a2 - c2a1)/(a1b2 - a2b1) (vi); Now, Here two cases may arise:Case 1. a1b2 - a2b1 0. Thus, a1/a2 b1/b2 => The pair of linear equations has a unique solution.Case 2. a1b2 - a2b1 = 0. If we write a1/a2 = b1/b2 = k, then, a1 = ka2; b1 = kb2.Putting the values of a1 & a2 in (i), k(a2x + b2y) + c1 = 0 (vii)Now, Equations (vii) & (ii) can both be satisfied only if c1 = kc2 => c1/c2 = k.If c1 = kc2, , any solution of Equation (ii) will satisfy the Equation (i), and vice versa. So, if, a1/a2 = b1/b2 = c1/c2 = k, then there are infinitely many solutions to the pair of linear equations given by (i) & (ii).If c1 kc2 , then any solution of Equation (i) will not satisfy Equation (ii) and vice versa. Therefore the pair has no solution. Thus, we have:When a1/a2 b1/b2; we get a unique solution;When a1/a2 = b1/b2 = c1/c2; we get infinitely many solutions;When a1/a2 = b1/b2 c1/c2; we don't get any solutions (Inconsistent pair of equations).==>The solutions given in equations (v) & (vi), may be written as: x = y = 1 (viii) b1c2 - b2c1 c1a2 - c2a1 a1b2 - a2b1Use the following diagram to memorise the above result.:xy1b1c1a1b1b2c2 a2b2 Thus, we follow the following steps, in solving a pair of linear equations by the cross-multiplication method:Step 1 : Write the given equations in the form (i) and (ii),Step 2 : Taking the help of the diagram above, write Equations as given in (viii),Step 3 : Find x and y, provided a1b2 - a2b1 0

CBSE CLASS X MATHEMATICSCHAPTER 3 LINEAR EQUATIONS IN TWO VARIABLES - NCERT EXERCISES SOLUTIONSEXERCISE 3.1 (P. 44)Q1. Aftab tells his daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. (Isnt this interesting?) Represent this situation algebraically and graphically.A. Let thepresentage of Aftab bex, and, present age of his daughter =ySeven years ago: Age of Aftab =x 7, and, Age of his daughter =y 7Three years hence, Age of Aftab =x+ 3 Age of his daughter =y+ 3Q2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.A. Let the cost of a bat be Rsx, and of ball be RsyAccording to the question, Preparing solution table for the above two equations:y = (3900 - 3x)/6; and, y = (1300 - x)/2 Q3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.A. Let the cost of 1 kg of apples be Rsx, & cost of 1 kg ofgrapes= Rsy. Acc. to the question, Preparing solution table for the above two equations: y = 160 - 2x; and, y = (300 - 4x)/2, we get the above graph.EXERCISE 3.2 (P. 49)Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.A. Let the number of girls bexand the number of boys bey.Accordingto the question, x+y= 10 (ie. y = 10 - x); and xy= 4 (ie. y = x - 4).Drawing the solution table for the two equations, we get the required graph:(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.A. Let the cost of 1 pencil be Rsxand the cost of 1 pen be Rsy.According to the question, 5x+ 7y= 50 (ie. y = (50 - 5x)/7); and, 7x+ 5y= 46 (ie. y = (46 - 7x)/5)Drawing the solution table for the two equations, we get the required graph:Q2. On comparing the ratios a1/a2, b1/b2 & c1/c2 , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:(i) 5x 4y + 8 = 0; 7x + 6y 9 = 0Comparing the equations with general form, we have Since, a1/a2 b1/b2; Hence, the lines representing the given pair of equations have a unique solution and the pair of linesintersectsat exactly one point. (ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0Comparing the equations with general form, we have Since, a1/a2 = b1/b2 = c1/c2; Hence, the lines representing the given pair of equations are coincident and there are infinitepossiblesolutionsfor the given pair of equations. (iii) 6x 3y + 10 = 0 2x y + 9 = 0Comparing the equations with general form, we have Since, a1/a2 = b1/b2 c1/c2; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution.

Q3. On comparing the ratios a1/a2, b1/b2 & c1/c2 , find out whether the following pair of linear equations are consistent, or inconsistent.(i) 3x+ 2y= 5; 2x 3y= 7 Comparing the equations with general form, we have Since, a1/a2 b1/b2; Hence, the lines representing the given pair of equations have a unique solution and the pair of linesintersectsat exactly one point. (ii) 2x 3y = 8; 4x 6y = 9Comparing the equations with general form, we have Since, a1/a2 = b1/b2 c1/c2 ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution.(iii) 3/2 x + 5/3 y = 7; ; 9x 10y = 14Comparing the equations with general form, we have Since, a1/a2 b1/b2; Hence, the lines representing the given pair of equations have a unique solution and the pair of linesintersectsat exactly one point. Hence, the pair of linear equations is consistent. (iv) 5x 3y = 11; 10x + 6y = 22Comparing the equations with general form, we have Since, a1/a2 = b1/b2 = c1/c2 ; Hence, the lines representing the given pair of equations are Coincident lines (ie. Consistent pair of equations with infinite solutions).(v) 4/3 x + 2y = 8; 2x + 3y = 12Comparing the equations with general form, we have Since, a1/a2 = b1/b2 = c1/c2 ; Hence, the lines representing the given pair of equations are Coincident lines (ie. Consistent pair of equations with infinite solutions).Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:A. (i)x+y= 5; 2x+ 2y= 10 Comparing the equations with general form, Since, a1/a2 = b1/b2 = c1/c2 ; Hence, the lines representing the given pair of equations are Coincident lines (ie. Consistent pair of equations with infinite solutions).x+y= 5 => y = 5 - x; 2x+ 2y= 10 => y = (10 - 2x)/2Drawing the solution table for the two equations, we get the required graph:It can be seen that the two lines are overlapping each other. (ii) x y = 8, 3x 3y = 16Comparing the equations with general form, we haveSince, a1/a2 = b1/b2 c1/c2 ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution.(iii) 2x + y 6 = 0, 4x 2y 4 = 0Comparing the equations with general form, we haveSince, a1/a2 b1/b2; Hence, the lines representing the given pair of equations have a unique solution and the pair of linesintersectsat exactly one point (ie. They are consistent). 2x + y 6 = 0, ie. y = - 2x + 6; 4x 2y 4 = 0, ie. y = (4x - 4)/2 Drawing the solution table for the two equations, we get the required graph:(iv) 2x 2y 2 = 0, 4x 4y 5 = 0Since, a1/a2 = b1/b2 c1/c2 ; Hence, the given pair of Equations are inconsistent (Parallel Lines) with no solution.(To demonstrate by drawing a graph, that the two lines are indeed parallel.)Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.A. Let the length of the garden bex, & the width be y.According to the question, x= y + 4 (1)y+x= 36 (2) Drawing the solution table for the two equations, we will get the required graph.The point where the two lines will intersect, will give us the required dimensions of the garden.Length = x = 20m; Width = y = 16m.Q6. Given the linear equation 2x + 3y 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident linesA. (i) For intersecting lines: 6x + 12y - 8 = 0. (ii) For parallel lines: 4x+ 6y 8 = 0. (iii) For coincident lines: 6x+ 9y 24 = 0.

Q7. Draw the graphs of the equations x y + 1 = 0 and 3x + 2y 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.A. x y + 1 = 0, ie. y = x + 1; and, 3x + 2y 12 = 0, ie. y = -3x + 12.Drawing the solution table for the two equations, we will get the required graph:From the figure, the three lines are intersecting each other at point (2, 3) andx-axis at (1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (1, 0), and (4, 0).--

EXERCISE 3.3 (P. 53)Q1. Solve the following pair of linear equations by the substitution method. (i) x + y = 14 (1); x - y = 4 (2).From (1), we obtain y = 14 - x(3); Putting this value in eq. (2), x - (14 - x) = 4; ie. x = 9. Putting the value of x in eq. (3), we get y = 5.(ii) s t = 3 (1); s/3 + t/2 = 6 (2).From (1), we obtain s = t + 3(3); Putting this value in eq. (2), (t + 3)/3 + t/2 = 6; ie. t = 6. Putting the value of t in eq. (3), we get s = 9.(iii) 3x y = 3 (1); 9x 3y = 9 (2).From (1), we obtain y= 3x 3 (3 ); Putting this value in eq. (2), 9x - 3(3x 3) = 9; ie. 9 = 9. This is always true. Hence, the given pair of equations has infinite possible solutions One of its possible solutions isx= 0,y= -3.(iv) 0.2x + 0.3y = 1.3 (1); 0.4x + 0.5y = 2.3 (2).From (1), we obtain y = (1.3 - 0.2x)/0.3(3); Putting this value in eq. (2), 0.4x + 0.5(1.3 - 0.2x)/0.3 = 2.3; ie. x = 2. Putting the value of x in eq. (3), we get y = 3.(v) 2 x + 3 y = 0 (1); 3 x - 8 y = 0 (2).From (1), we obtain y = (-2 x)/3 (3); Putting this value in eq. (2), 3 x - 8 ((-2 x)/3) = 0; ie. x = 0. Putting the value of x in eq. (3), we get y = 0.(vi) 3x/2 - 5y/3 = -2 (1); x/3 + y/2 = 13/6 (2).DO IT ON YOUR OWN!!x = 2, y = 3.Q2. Solve 2x + 3y = 11 and 2x 4y = 24 and hence find the value of m for which y = mx + 3.A. 2x + 3y = 11 (1); 2x - 4y = -24 (2).Solving for x & y, using substitution method, we get x = -2; y = 5.y = mx + 3; ie. 5 = m (-2) + 3; or, 2 = -2m; or, m = -1.Q3. Form the pair of linear equations for the following problems and find their solution by substitution method.(i) The difference between two numbers is 26 and one number is three times the other. Find them.A. Let the first number bexand the other number beysuch thaty>x.According to giveninfo., y = 3x (1); y - x = 26 (2)Substitutingthe value ofyfrom equation (1) into equation (2), 3x - x = 26 => x = 13 (3) Putting this value in eq. (1), y = 3 x 13 = 39.(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.A. Let the larger angle bexandsmallerangle bey. As per question. x - y = 180 (1), and as the two are supplementary angles, x + y = 1800 (2)From (1) y = x - 180 (3). Put this value in eq. (2), x + x - 180 = 1800; Thus, x = 990.Putting the value of x in eq. (3), we get y = 990 - 180 = 810.(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.A. Let the cost of a bat and a ball bexandy.According to the given info. 7x + 6y = 3800 (1); and, 3x + 5y = 1750 (2)From (1), y = (3800 - 7x)/6 (3); Putting this value in eq. (2) => 3x + 5 [(3800 - 7x)/6] = 1750, which gives x = 500 (4).Putting this value in eq. (3), we get y = 50.(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?A. Let the fixedchargebe Rsxand per km charge be Rsy.As per the question, x + 10y = 105 - (i); and, x + 15y = 155 - (ii)From (i), we get, x = 105 - 10y - (iii). Putting this in eq. (ii), we get, 105 - 10y + 15y = 155,which gives, y = 10 - (iv) Putting this in eq. (iii), we get, x = 5.Hence, fixed charge = Rs 5, and, Per km charge = Rs 10.Charge for 25 km = x + 25y = Rs 255(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. A. Let the fraction be x/y. As per the given information,(x+2)/(y+2) = 9/11 => 11x - 9y = (-)4 - (i);(x+3)/(y+3) = 5/6 => 6x - 5y = (-)3 - (ii)From (i), x = (-4 + 9y)/11 - (iii). Substituting this in equation (ii), we get, y = 9 - (iv)Substituting y = 9 in equation (iii), we get, x = 7. Thus the fraction is 7/9(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacobs age was seven times that of his son. What are their present ages?A. Let the age of Jacob bexand the age of his son bey.According to the given information, (x + 5) = 3(y + 5) => x - 3y = 10 - (i) (x - 5) = 7(y - 5) => x - 7y = -30 - (ii)From, (i), x = 3y + 10 - (iii). Substituting this in equation (ii), we get, y = 10 - (iv)Substituting y = 10 in equation (iii), we get, x = 40. Hence, the present age of Jacob is 40 years, and the present age of his son is 10 years.--EXERCISE 3.4 (P. 56) Q1. Solve the following pair of linear equations by the elimination method and the substitution method :(i) x + y = 5 and 2x 3y = 4By Elimination Method:x + y = 5 (i); 2x 3y = 4 (ii)Multiplying (i) by 2, we get, 2x + 2y = 10 (iii)Subtracting (ii) from (iii), we get, y = 6/5 (iv). Putting this in eq. (i), we get, x = 19/5By Substitution Method:x + y = 5 (i); 2x 3y = 4 (ii)x + y = 5 (i), gives, x = 5 - y (iii). Putting this in (ii), we get, y = 6/5. Putting y = 6/5 in (iii), we get, x = 19/5(ii) 3x + 4y = 10 and 2x 2y = 2; (iii) 3x 5y 4 = 0 and 9x = 2y + 7; and, (iv) x/2 + 2y/3 = -1 and x - y/3 = 3These three pair of equations can be solved following the method described above.Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?A. Let the fraction be x/y. According to given information,(x + 1) / (y - 1) = 1=> x - y = -2 (i)x / (y + 1) = 1/2 => 2x - y = 1 (ii)Subtracting (i) from (ii), we get x = 3 (iii). Putting this in (i), we get y = 5. Thus, the fraction is 3/5(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?A.Do it yourself! Age of Nuri = 50 years; and, Age of Sonu = 20 years.(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.A. Let the unit digit and tens digits of the number bexandyrespectively. Then, number (written as yx) = 10y+x Number afterreversingthe digits = 10x+yAccording to the given information,x+y= 9 (i); 9(10y+x) = 2(10x+y) => 88y 11x= 0 => x+ 8y=0 (ii)Adding equation (i) and (ii), we get 9y= 9, or, y= 1 (iii)Substituting the value in equation (1), we get, x= 8Hence, the number is 10y+x= 10 1 + 8 = 18.(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.A. Let the number of Rs 50 and Rs 100 notes bexandy.According to the given information, x + y = 25 (i); 50x + 50y = 2000 (ii)Multiplying equation (i) by 50, we get, 50x + 50y = 1250 (iii)Subtracting equation (iii) from equation (ii), we get, 50y = 750 => y = 15.Substituting y = 15, in equation (i), we get, x= 10Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100. (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.A. Let the fixedchargefor first three days and each day charge thereafter be Rsxand Rsy.According to the given information, x + 4y = 27 (i); x + 2y = 21 (ii)Subtracting equation (ii) from equation (i), we get, 2y = 6 => y = 3 (iii)Substituting in equation (i), we get, x + 12 = 27 => x = 15Hence, fixed charge = Rs 15, and, Charge per day = Rs 3 --EXERCISE 3.5 (P. 62)Q1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.(i) x 3y 3 = 0; 3x 9y 2 = 0a1 / a2 = 1/3; b1 / b2 = 1/3; c1 / c2 = 3/2. Therefore, a1 / a2 = b1 / b2 c1 / c2 Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.(ii) 2x + y = 5; 3x + 2y = 8a1 / a2 = 2/3; b1 / b2 = 1/2; c1 / c2 = 5/8. Therefore, a1 / a2 b1 / b2 . Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.By cross-multiplication method, x = y = 1____ b1c2 - b2c1 c1a2 - c2a1 a1b2 - a2b1 This gives, x/2 = y/1 = 1. Therefore, x = 2, and, y = 1(iii) 3x 5y = 20; 6x 10y = 40a1 / a2 = 1/2; b1 / b2 = 1/2; c1 / c2 = 1/2. Therefore, a1 / a2 = b1 / b2 = c1 / c2Therefore, the given sets of lines are coincident to each other and thus, there areinfinitesolutions possible for these equations. (iv) x 3y 7 = 0; 3x 3y 15 = 0a1 / a2 = 1/3; b1 / b2 = 1; c1 / c2 = 7/15. Therefore, a1 / a2 b1 / b2 Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication, x = y = 1____ b1c2 - b2c1 c1a2 - c2a1 a1b2 - a2b1 Thus, x = y = 1____ 45 - 21 -21 - (-15) -3 - (-9)Or, x / 24 = y / (-6) = 1/6; This gives, x = 4 & y = -1Q2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?2x + 3y = 7; (a b) x + (a + b) y = 3a + b 2A. a1 / a2 = 2/(a - b); b1 / b2 = 3/(a + b); c1 / c2 = 7/(3a +b - 2). For infinitely many solutions, a1 / a2 = b1 / b2 = c1 / c22/(a - b) = 7/(3a +b - 2), which gives, a - 9b = -4 (i)2/(a - b) = 3/(a + b), which gives, a - 5b = 0 (ii)Solving equations (i), and (ii), we get, a = 5, and, b = 1; for which the given equations will have infinitely many solutions.Q2. (ii) For which values of k will the following pair of linear equations have no solutions?3x + y = 1; (2k 1) x + (k 1) y = 2k + 1A. a1 / a2 = 3/(2k - 1); b1 / b2 = 1/(k - 1); c1 / c2 = 1/(2k + 1). For no solution, a1 / a2 = b1 / b2 c1 / c2 ; which gives k = 2.Hence, for k = 2, the given equation will have no solution.Q3. Solve the following pair of linear equations by the substitution and cross-multiplication methods :8x + 5y = 9; 3x + 2y = 4A. Do It Yourself!Q4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.A. Letxbe the fixed charge of the food andybe the charge for food per day.According to the giveninformation, x + 20y = 1000 (i); x + 26y = 1180 (ii)Subtracting equation (i) from equation (ii), we get, 6y = 180 => y = 30Substituting this value in equation (i), we get, x = 400Hence, fixed charge = Rs 400; and charge per day = Rs 30 Q4.(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.A. Let thefractionbex/y.According to the given info., (x - 1) / y = 1/3 => 3x - y = 3 (i); x / (y + 8) = 1/4 => 4x - y = 8 (ii)Subtracting equation (i) from equation (ii), we get, x = 5. Puttingthis (i), we get, y = 12Hence, the fraction is5/12.Q4.(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?A. Let the number of right answers and wrong answers bexandy.According to the given information, 3x - y = 40 (i); 4x - 2y = 50 => 2x - y = 25 (ii)Subtracting equation (ii) from equation (i), we get, x = 15 (iii). Puttingthis in (ii), y = 5Therefore, number of right answers = 15; and, number of wrong answers = 5.Total number ofquestions= 20Q4.(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?A. Let the speed of 1stcar and 2ndcar beukm/h andvkm/h. Relative speed of both cars while they are travelling in same direction = (u - v) km/h. (u > v)Relative speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u + v) km/hAccording to the given information, 5(u - v) = 100 => u - v = 20 (i); and, 1(u + v) = 100 (ii)Adding both the equations, we get, u = 60 km/h (iii)Substituting this value in equation (2), we get, v = 40 km/hHence, speed of one car = 60 km/h and speed of other car = 40 km/h.Q4.(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.A. (v) Let length and breadth of rectangle bexunit andyunit. Area =xy. Acc. to the question,(x - 5)(y + 3) = xy - 9 => 3x - 5y - 6 = 0 (i); (x + 3)(y + 2) = xy + 67 => 2x + 3y - 61 = 0 (ii)By cross-multiplication method, we get, x = y = 1____ ; b1c2 - b2c1 c1a2 - c2a1 a1b2 - a2b1 Or, x = y = 1___; 305 - (-18) -12 - (-183) 9 - (-10) Or, x / 323 = y / 171 = 1 / 19 => x = 17, and, y = 9Hence, the length and breadth of the rectangle are 17 units and 9 unitsrespectively.

EXERCISE 3.6 (P. 67) Q1. Solve the following pairs of equations by reducing them to a pair of linear equations:Q1. (i) 1 + 1 = 2; 1 + 1 = 13/6; 2x 3y 3x 2y Let 1/x = p, & 1/y = q; The equations thus become,p/2 + q/3 = 2 => 3p + 2q - 12 = 0 (i)p/3 + q/2 = 13/6 => 2p + 3q - 13 = 0 (ii)Using cross multiplication method, we get, p = 2, and, q = 3; which gives, x = 1/2, and, y = 1/3.Q1. (ii) 2/(x) + 3/(y) = 2; /(x) - 9/(y) = -1Putting 1/(x) = p, and, 1/(y) = q. The equations thus become,2p + 3q = 2 (i); 4p - 9q = -1 (ii)Solving, we get, p (=1/(x)) = 1/2; q (=1/(y)) = 1/3Solving for x, and y, we get, x = 4; y = 9Q1. (iii) 4/x + 3y = 14; 3/x - 4y = 23Putting 1/x = p, the equations become,4p + 3y = 14 (i); 3p - 4y = 23 (ii)Solving, we get, p (=1/x) = 5, => x = 1/5; y = -2Q1. (iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 2) = 1Putting 1/(x - 1) = p, and, 1/(y - 2) = q. The equations thus become,5p + q = 2 (i); 6p - 3q = 1 (ii)Solving, we get, p (=1/(x - 1) ) = 1/3; which gives, x = 4Also, we get, q (=1/(y - 2)) = 1/3; which gives, y = 5Therefore, x = 4, and, y = 5Q1. (v) (7x - 2y)/xy = 5 => 7/y - 2/x = 5 (i); (8x + 7y)/xy = 15 => 8/y + 7/x = 15 (ii)Putting 1/x = p, and, 1/y = q. The equations thus become,-2p + 7q = 5 (iii); 7p + 8q = 15 (iv)Solving, we get, p (=1/x) = 1; which gives, x = 1; Also, we get, q (=1/y) = 1; which gives, y = 1Therefore, x = 1, and, y = 1Q1. (vi) 6x + 3y = 6xy => 6/y + 3/x = 6 (i); 2x + 4y = 5xy => 2/y + 4/x = 5 (ii)Putting 1/x = p, and, 1/y = q, the equations thus become, 3p + 6q - 6 = 0 (iii); 4p + 2q - 5 = 0 (iv)Solving, using cross multiplication method, we get, p (=1/x) = 0 => x = 1; q (=1/y) = 1/2 => y = 2Q1. (vii) 10/(x + y) + 2/(x - y) = 4 (i); 15/(x + y) - 5/(x - y) = -2 (ii)Putting 1/(x+y) = p, and, 1/(x-y) = q, the equations become, 10p + 2q = 4(iii); 15p - 5q = -2 (iv)Solving, using cross multiplication method, we get, p (=1/(x+y)) = 1/5 => x + y = 5 (v); and, q (= 1/(x - y) = 1 => x - y = 1 (vi)Solving equations (v) and (vi), we get, x = 3, and y = 2Q1. (viii) 1/(3x + y) + 1/(3x - y) = 3/4 (i); 1/[2(3x + y)] - 1/[2(3x - y) = -1/8 (ii)Putting 1/(3x+y) = p, and, 1/(3x-y) = q, the equations become, p + q = 3/4(iii); p/2 - q/2 = -1/8 => p - q = -1/4 (iv)Solving, we get, p [= 1/(3x+y) ] = 1/4 => 3x + y = 4 (v), and, q (= 1/(3x - y) = 1/2 => 3x - y = 2 (vi)Solving equations (v) and (vi), we get, x = 1, and y = 1Q2. Formulate the following problems as a pair of equations, and hence find their solutions:(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.A. Let the speed of Ritu in still water be x km/h, and the speed of stream beykm/h. Speed of Ritu whilerowing: Upstream =(x - y)km/h; Downstream =(x + y)km/hAccording toquestion: 2(x + y) = 20 => x + y = 10 (i); 2(x - y) = 4 => x - y = 2 (ii)solving the equations, we get, x = 6, and, y = 4.Hence,Ritusspeed in still water is 6 km/h and the speed of thecurrentis 4 km/h. (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.A. Let the number of days taken by a woman and a man, to complete the work bexandy.Therefore, work done by a woman in 1 day = 1/x, and,Work done by a man in 1 day =1/yAccording to the question: 2/x + 5/y = 1/4 (i); and, 3/x + 6/y = 1/3 (ii)Putting1/x = p, and 1/y = q in these equations, we get, 2p + 5q = 1/4 => 8p + 20q = 1 (iii), and, 3p + 6q = 1/3 => 9p + 18q = 1 (iv)Solving equations (iii) & (iv), by cross-multiplication, we get, p (= 1/x) = 1/18 => x = 18; and, q (= 1/y) = 1/36 => y = 36 Hence, number of days taken by a woman to finish the work = 18, and, Number of days taken by a man to finish the work= 36(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.A. Let the speed of train and bus beukm/h andvkm/h. According to the giveninformation,60/u + 240/v = 4 (i); [4Hrs10Mts = 25/6Hrs]; 100/u + 200/v = 25/6 (ii).Putting 1/u = p, and1/v = q, in these equatns, we get,60p + 240q = 4 (iii), 600p + 1200q = 25 (iv)Solving equations (iii) and (iv), we get, p = 1/60; q = 1/80Hence, Speed of train = u = 60km/h, and, Speed of bus = 80km/h.

CBSE CLASS X MATHEMATICSCHAPTER 4 QUADRATIC EQUATIONS1. Introduction: Quadratic polynomial is of the form ax2 + bx + c, a 0. When we equate this polynomial to zero, we get a quadratic equation.2. Quadratic Equations: A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, a 0. For example, 2x2 + x 300 = 0 is a quadratic equation.In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax2 + bx + c = 0, a 0 is called the standard form of a quadratic equation.Q. Check whether the following are quadratic equations:(i) (x 2)2 + 1 = 2x 3;(ii) x(x + 1) + 8 = (x + 2) (x 2);(iii) x (2x + 3) = x2 + 1;(iv) (x + 2)3 = x3 43. Solution of a Quadratic Equation by Factorisation:A real number is called a root of the quadratic equation ax2 + bx + c = 0, a 0 if a2 + b + c = 0. We also say that x = is a solution of the quadratic equation, or that satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. Any quadratic equation can have at-most two roots.We can find the roots of a quadratic equation by factorising it into two linear factors (by splitting the middle term) and equating each factor to zero.Ex. Find the roots of the quadratic equation 6x2 x 2 = 0.A. 6x2 x 2 = (3x 2)(2x + 1)The roots of 6x2 x 2 = 0 are the values of x for which (3x 2)(2x + 1) = 0.Therefore, 3x 2 = 0 or 2x + 1 = 0, ie. x = 2/3; or, x = -1/2We verify the roots, by checking that 2/3 & -1/2 satisfy 6x2 x 2 = 0 4. Solution of a Quadratic Equation by Completing the SquareHerein, the term containing x is brought completely inside a square, and thence, the roots are found easily by taking the square roots. We can convert any quadratic equation to the form (x + a)2 b2 = 0 and then we can easily find its roots. The process is:x2 + 4x = (x + 4/2)2 - 4ie. Make the term containing x2, a perfect square (by dividing or multiplying appropriate number); Half the co-efficient of x, and bring it in the term containing perfect square.Few Examples: (i) Solve the equation 3x2 - 5x + 2 = 0.Dividing both sides by 3; x2 - 5/3x + 2/3 = 0 => (x - 5/6)2 + 2/3 - 25/36 = 0Solving we get, (x - 5/6)2 = (1/6)2 => x = 1, or x = 2/3; which are the roots of the given equation.(ii) Solve the equation 2x2 - 5x + 3 = 0 => x2 - 5/2 x + 3/2 = 0 => (x - 5/4)2 + 3/2 - 25/16 = 0;Or, (x - 5/4)2 = 1/16 => x - 5/4 = 1/4 => x = 3/2 or x = 1

(iii) Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the square.A. 4x2 + 3x + 5 = 0 => x2 + 3/4x + 5/4 = 0 => (x + 3/8)2 - 9/64 + 5/4 = 0 => (x + 3/8)2 = (-71)/64 < 0.But, (x + 3/8)2 cant be negative for any real value of x (as ve x ve = +ve). So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.Generalising the above method of completing the squares:Consider the quadratic equation ax2 + bx + c = 0 (a 0). Dividing throughout by a, we get x2 + b/a x + c/a = 0=> (x + b/2a)2 - (b/2a)2 + c/a =0=> (x + b )2 - (b2 - 4ac) => 2a 4a2=> (x + b )2 = b2 - 4ac (i) 2a 4a2If b2 4ac 0, then by taking the square roots in (i), we get (x + b/2a) = ((b2 - 4ac) / 2aThus, x = -b ((b2 - 4ac) [which are the two roots of quadratic equation ax2 + bx + c = 0 (a 0) 2a if (b2 - 4ac) 0. This is also known as the Quadratic Formula.]Solve the following equations by quadratic formula (find the roots if they exist):(i) x2 + 2x 143 = 0 [ x = 11, or x = -13 ](ii) x2 - 3x - 4 = 0 [ x = 4, or x = -1 ](iii) x2 + 4x + 5 = 0 [ b2 4ac < 0 ](iv) 2x2 22 x + 1 = 0 [ x = 1/2, or, x = 1/2 ]5. Nature of Roots:The roots of the equation ax2 + bx + c = 0 are given by x = - b (b2 - 4ac). 2aIf b2 - 4ac > 0, we get two distinct real roots:x = - b + (b2 - 4ac), and, x = - b - (b2 - 4ac). 2a 2aIf b2 - 4ac = 0, then x = -b/2a 0, ie. x = -b/2a or b/2a.So, the roots of the equation ax2 + bx + c = 0 are both b/2a. Thus, the quadratic equation ax2 + bx + c = 0 has two equal real roots in this case.If b2 4ac < 0, then there is no real number whose square is b2 4ac. Therefore, there are no real roots for the given quadratic equation in this case.Since b2 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 4ac is called the discriminant of this quadratic equation.So, a quadratic equation ax2 + bx + c = 0 has(i) two distinct real roots, if b2 4ac > 0,(ii) two equal real roots, if b2 4ac = 0, (iii) no real roots, if b2 4ac < 0.Eg. Find the Discriminant of the equation 3x2 - 2x + 1/3 = 0, and hence find the nature of its roots. Find them if the are real.Ans. a = 3, b = -2, c = 1/3. Therefore, Discriminant b2 4ac = 0Hence, the given quadratic equation has two equal real roots.The roots are b/2a, -b/2a; ie. 1/3, 1/3

CBSE CLASS X MATHEMATICSCHAPTER 4 QUADRATIC EQUATIONS- NCERT EXERCISES SOLUTIONSEXERCISE 4.1 (P. 73):Q1. Check whether the following are quadratic equations :(i) (x + 1)2 = 2(x 3) => x2 + 7 = 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic equation.(ii) x2 2x = (2) (3 x) => x2 4x + 6 = 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic equation.(iii) (x 2)(x + 1) = (x 1)(x + 3) => 3x - 1 = 0It is not of the form ax2 + bx + c = 0; and hence is not a quadratic equation.(iv) (x 3)(2x +1) = x(x + 5) => x2 10x - 3 = 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic equation.(v) (2x 1)(x 3) = (x + 5)(x 1) => x2 11x + 8 = 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic equation.(vi) x2 + 3x + 1 = (x 2)2 => 7x - 3 = 0.It is not of the form ax2 + bx + c = 0; and hence is not a quadratic equation.(vii) (x + 2)3 = 2x (x2 1) => x3 - 6x2 - 14x - 8 = 0.It is not of the form ax2 + bx + c = 0; and hence is not a quadratic equation.(viii) x3 4x2 x + 1 = (x 2)3 => 2x2 - 13x + 9 = 0.It is of the form ax2 + bx + c = 0; and hence is a quadratic equation.Q2. Represent the following situations in the form of quadratic equations :(Assume what is to be found out in the given question)(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.A. Let thebreadthof the plot bexm => lengthof the plot is (2x+ 1) m.Area of a rectangle =Length Breadth; 528 =x(2x+ 1) => 2x2 + x - 528 = 0(ii) The product of two consecutive positive integers is 306. We need to find the integers.A. Let the consecutive integers bexandx+ 1. It is given that their product is 306. x(x + 1) = 306 => x2 + x - 306 = 0.(iii) Rohans mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohans present age.A. Let Rohans age bex. Hence, his mothers age =x+ 26.Three years hence: Rohans age =x+ 3; Mothers age =x+ 26 + 3 =x+ 29It is given that the product of their ages after 3 years is 360. (x + 3)(x + 29) = 360 => x2 + 32x - 273 = 0.(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.A. Let the speed of train bexkm/h. Time taken to travel 480 km = 480/x hrs.If speed = (x - 8) km/h; time taken to cover 480 kms is (480/x + 3) hrs.Now, distance = speed X time => 480 = (x - 8) X (480/x + 3)=> x2 - 8x + 1280 = 0.EXERCISE 4.2 (P. 76):Q1. Find the roots of the following quadratic equations by factorisation: (ie. By factorizing the middle term)(i) x2 3x 10 = 0; or, (x - 5)(x + 2) = 0.Roots of thisequationare thevaluesfor which (x - 5)(x + 2) = 0 => x - 5 = 0; or x + 2 = 0.Thus, i.e.,x= 5 orx= 2

(ii) 2x2 + x 6 = 0; or, (x + 2)(2x - 3) = 0. Roots of this equation are the values for which (x + 2)(2x - 3) = 0; => (x + 2) = 0; or, (2x - 3) = 0i.e.,x= 2 orx= 3/2(iii) 2 x2 + 7x + 52 = 0; or, (2x + 5)(x + 2) = 0.Roots of this equation are the values for which (2x + 5)(x + 2) = 0; => (2x + 5)= 0; or, (x + 2) = 0; i.e.,x= 5/2 orx= -2(iv) 2x2 - x + 1/8 = 0; or, 1/8(16x2 - 8x + 1) = 1/8 (4x - 1)2 = 0Roots of this equation are the values for which (4x - 1)2 = 0; => (4x - 1)= 0; or, (4x - 1) = 0; i.e.,x= 1/4, or x = 1/4(v) 100x2 20x + 1 = 0; or, (10x - 1)2 = 0.Roots of this equation are the values for which (10x - 1)2 = 0; => (10x - 1)= 0; or, (10x - 1) = 0; i.e.,x= 1/10, or x = 1/10Q2. Solve the problems given in Example 1.Example 1:(i) Required representation of the problem: x2 - 45x + 324 = 0. (Solve by splitting the middle term)(ii) Required representation of the problem: x2 - 55x + 750 = 0. (Solve by splitting the middle term)Q3. Find two numbers whose sum is 27 and product is 182.A. Let the first number bex; then, thesecondnumber is 27 x.As per question,x(27 x) = 182 => x2 - 27x + 182 = 0 => (x - 13)(x - 14) = 0.Thus, either (x - 13) = 0; or, (x - 14) = 0; ie. x = 13, or x = 14.If x = 13, the other number is 14, and vice versa. Thus, the numbers are 13 and 14.Q4. Find two consecutive positive integers, sum of whose squares is 365.A. Let theconsecutivepositiveintegersbexandx+ 1. Thus, x2 + (x + 1)2 = 365 => (x + 14)(x - 13) = 0; Thus, x = -14; or, x = 13.Since the integers are positive, x = 13. Thus, the two consecutive positive integers are 13 and 14.Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.A. Let the base of the righttrianglebexcm. Then, its altitude = (x 7) cm.By Pythagoras Theorum, x2 + (x - 7)2 = 132 => (x - 12)(x + 5) = 0Thus, x = 12, or, x = -5. As sides cant be negative, x = 12.Thus, the base of the triangle is 12cm, and the altitude is 5cm.Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.A. Let the number of articles produced bex.Therefore, cost of production of each article = Rs (2x+ 3)It is given that the total production is Rs 90. Thus, x(2x+ 3) = 90 => (2x + 15)(x - 6) = 0Thus, x = -15/2, or, x = 6.As the number of articles produced can only be a positive integer, therefore,xcan only be 6.Hence, number of articles produced = 6, and, Cost of each article = 2 6 + 3 = Rs 15

EXERCISE 4.3 (P. 87):Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:(i) 2x2 7x + 3 = 0 => x2 - 7/2x + 3/2 = 0 => (x - 7/4)2 - 49/16 + 3/2 = 0 => (x - 7/4)2 = 49/16 - 3/2 => (x - 7/4) = 5/4 => x = 3, or x = 1/2(ii) 2x2 + x 4 = 0 => x2 + x/2 - 2 = 0 => (x + 1/4)2 - 1/16 - 2 = 0=> (x + 1/4)2 = 1/16 + 2 => (x + 1/4)2 = 33/16. Solving, x = (33 - 1)/4 or ( - 33 - 1)/4(iii) 4x2 + 43 x + 3 = 0; (x = -3/2; x = -3/2)(iv) 2x2 + x + 4 = 0; (No real roots, as the square of a number cannot be negative)Q2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.A. Compare the equation with ax2 + bx + c = 0. Get the value of a, b, and c.Find the value of (b2 - 4ac), and proceed. Part (iv) is being solved:(iv) 2x2 + x + 4 = 0. Using quadratic formula, x = - b (b2 - 4ac), we get, 2ax = -1 (-31). But, the square of a number cant be negative. Thus, there are no real roots for the 4 given equation.Q3. Find the roots of the following equations:(i) x - 1/x = 3, x 0. Use quadratic formula, x = (3 13) / 2Form an equation, factorise it by splitting the middle term.We get x = 1 or x = 2Q4. The sum of the reciprocals of Rehmans ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.A. Let x be Rehmans present age. As per question, 1/(x - 3) + 1/(x + 5) = 1/3.Forming a quadratic equation, and factorizing it by splitting the middle term, we get,x = 7, or x = -3. But, as age cant be negative, x = 7.Q5. In a class test, the sum of Shefalis marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.A. Let the marks in Maths bex. Then, the marks in English will be 30 x. Acc. to the question, (x + 2)(30 - x - 3) = 210. Factorizing we get, (x - 12)(x - 13) = 0 => x = 12, or x = 13.If the marks in Maths are 12, then marks in English will be 30 12 = 18If the marks in Maths are 13, then marks in English will be 30 13 = 17 ==>Q6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.A. Let the shorter side of the rectangle bexm.Then, larger side of the rectangle = (x+ 30) m. Diagonal is given to be 60 more than the shorter side.

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.Hence, length of the larger side will be (90 + 30) m = 120 m.

Q7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.A. Let the larger and smaller number bexandyrespectively.

According to the given question, However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will also be negative which is not possible.Therefore, the larger number is 18. Thus, y2 = 144 => y = 12.Thus, the numbers are 18 & 12; or, 18 & -12. Q8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.A. Let the speed of the train bexkm/hr. Time taken to cover 360 km = 360/x hrsAccording to the given question,

However, speed cannot be negative. Therefore, the speed of train is 40 km/h.Q9. Two water taps together can fill a tank in9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.A. Let the time taken by the smaller pipe to fill the tank bexhr.Time taken by the larger pipe = (x 10) hrPart of tank filled by smaller pipe in 1 hour = 1/xPart of tank filled by larger pipe in 1 hour = 1 / (x - 10)It is given that the tank can be filled in9 3/8 = 75/8hours by both the pipes together. Therefore, Part of tank filled by both the pipes together in 1 hour = 8/75Thus, 1/x + 1/(x - 10) = 8/75 => (x - 25)(8x - 30) = 0 => x = 25, or, x = 30/8Time taken by the smaller pipe cannot be30/8 = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 10 =15 hours respectively. Q10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.A. Let the average speed of passenger train bexkm/h.Average speed of express train = (x+ 11) km/hIt is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Now, Speed cannot be negative.Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

Q11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.A. Let the sides of the two squares bexm andym. Therefore, their perimeter will be 4x and 4yrespectively and their areas will bex2andy2respectively.It is given that, 4x 4y= 24 => xy= 6 => x=y+ 6

Thus, y = -18; y = 12 However, side of a square cannot be negative.Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m ---------------------------------------------------------------------------------------------------------------------------------- EXERCISE 4.4 (P. 91):Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:A. For a quadratic equationax2+bx+c= 0,discriminantisb2 4ac.(A) Ifb2 4ac> 0 twodistinctreal roots(B) Ifb2 4ac= 0 two equal real roots(C) Ifb2 4ac< 0 no real roots (i) 2x23x+ 5 = 0Comparing this equation withax2+bx+c= 0, we obtain, a= 2,b= 3,c= 5Discriminant =b2 4ac= ( 3)2 4 (2) (5) = 9 40 = 31Asb2 4ac< 0. Therefore, no real root ispossiblefor the given equation. (ii) 3x2 - 43 x + 4 = 0Comparing this equation withax2+bx+c= 0, we obtain, a= 3,b= - 43,c= 4Discriminant =b2 4ac= (- 43)2 4 (3) (4) = 48 48 = 0 Asb2 4ac= 0. Therefore, real roots exist for the given equation and they are equal to each other.The roots are: -b/2a, and b/2a. [ x = -b/2a = -(- 43)/(2x3) = 2/3]; that is 2/3 & 2/3(iii) 2x2 - 6x + 3 = 0Comparing this equation withax2+bx+c= 0, we obtain, a= 2,b= 6,c= 3Discriminant =b2 4ac= ( 6)2 4 (2) (3) = 36 24 = 12Asb2 4ac> 0, therefore, distinct real roots exist for this equation asfollows:Therefore, the roots are, (3 + 3)/2, or (3 - 3)/2Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.A. If an equationax2+bx+c= 0 has two equal roots, its discriminant (b2 4ac) will be 0.(i) 2x2+kx+ 3 = 0Comparing this equation withax2+bx+ c = 0, we obtain, a= 2,b=k,c= 3Discriminant =b2 4ac=(k)2 4(2) (3) = k2 24. For equal roots, Discriminant = 0 => k2 24 = 0=> k2= 24 => k = 24 = 26(ii) kx (x 2) + 6 = 0 => kx2 - 2kx + 6 + 0Comparing this equation withax2+bx+c= 0, we obtain, a=k,b= 2k,c= 6Discriminant =b2 4ac= ( 2k)2 4 (k) (6) = = 4k2 24kFor equal roots, b2 4ac= 0 => 4k2 24k= 0 => 4k(k 6) = 0=> Either 4k= 0 ork= 6 = 0 => k= 0 ork= 6However, ifk= 0, then the equation will not have the terms x2 and x.Therefore, if this equation has two equal roots,kshould be 6 only. Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800m2? If so, find its length and breadth.A. Let the breadth of mango grove bex. Thus, length of mango grove = 2x.Area of mango grove = (2x)(x) = 2x2.Given, 2x2 = 800 => x2 - 400 = 0.Comparing this withax2+bx+c= 0, we obtain, a= 1,b= 0,c= 400Discriminant =b2 4ac= (0)2 4 (1) ( 400) = 1600Here,b2 4ac> 0 => The equation will have real roots. And hence, the desired rectangular mango grove can be designed.x = 20. However, length cannot be negative.Therefore, breadth of mango grove = 20 mLength of mango grove = 2 20 = 40 mQ4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.A. Let the age of onefriendbexyears => Age of the other friend = (20 x) years.Four years ago, age of 1stfriend = (x 4) yrs; &, age of 2ndfriend = (20 x 4)= = (16 x) yrsGiven that, (x 4) (16 x) = 48 => 16x- x2 - 64 + 4x= 48 => x2+ 20x 112 = 0=> x2 20x+ 112 = 0. Comparing this equation withax2+bx+c= 0, we obtain,a= 1,b= 20,c= 112Discriminant =b2 4ac= ( 20)2 4 (1) (112) = 400 448 = 48Asb2 4ac< 0, no real root is possible for this equation and hence, this situation is not possible.Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.A. Let the length and breadth of the park bexandy.Perimeter = 2(x+y) = 80 => x+y= 40 => y = 40 - xArea =xy=x(40 x) => Area = 40xx2 As per the question, 40xx2= 400 => x2 - 40x + 400 = 0Comparing this equation withax2+bx+c= 0, we obtain, a= 1,b= 40,c= 400Discriminate =b2 4ac= ( 40)24 (1) (400) = 1600 1600 = 0Asb2 4ac= 0, therefore, this equation has equal real roots. And hence, this situation is possible.Root of this equation are: x = -b/2a = - (-40)/2(1) = 20.Therefore, length of park,x= 20 m, and, breadth of park,y= 40 x= 40 20 = 20CBSE CLASS X MATHEMATICSCHAPTER 5 ARITHMETIC PROGRESSIONS1. Introduction: Some of the patterns observed in everyday life are such that, the succeeding terms are obtained by adding a fixed number, or by multiplying with a fixed number, or we find that they are squares of consecutive numbers, and so on.Here-in, we shall learn about a pattern in which succeeding terms are obtained by adding a fixed number to the preceding terms. And, also, to find their nth terms and the sum of n consecutive terms, and use this knowledge in solving some daily life problems.2. Arithmetic Progressions: Consider the following lists of numbers :1, 2, 3, 4, . . .100, 70, 40, 10, . . .Each of the numbers in the list is called a term. In each of the above two lists, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression (AP). So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.This fixed number is called the common difference of the AP, which can be positive, negative or zero.Let us denote the first term of an AP by a1, second term by a2, . . ., nth term by an and the common difference by d. Then the AP becomes a1, a2, a3, . . ., an.So, a2 a1 = a3 a2 = . . . = an an - 1 = d.a, a + d, a + 2d, a + 3d, . . . represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.A finite AP has finite number of terms (The last term is known). An infinite Arithmetic Progression does not have a last term.3. nth term of an A.P.: Let a