SECTION CHPTER B Real Numbers TOPIC- Euclid s Division Lemma Sol.. a and b are two positive integers such that the least prime factor of a is and the least prime factor of b is. Then least prime factor of (a + b) is. Sol.. y 6 x 9 8 [CBSE Marking Scheme, 0] Sol.. HCF of k k. k. isk k. k. [CBSE Marking Scheme, 0] Sol.. 00 ( ) The smallest natural number is. [CBSE Marking Scheme, 0] Sol.. Let n be any positive integer. By Eucild s division lemma, n q + r, 0 r < n q, q +, q +, q + or q +, where q ω q is a whole number now n (q) q (q ) m n (q + ) q + 0q + m + n (q + ) q + 0q + m + Similarly n (q + ) m + and n (q + ) m + Thus square of any positive integer cannot be of the form m + or m +. [CBSE Marking Scheme, 0] Sol. 6. If 8 n ends with 0, then it must have as a factor. But we know that only prime factor of 8 n is. 8 n ( ) n n n n From the fundamental theorem of arithmetic, we know that the prime factorisation of every composite number is unique. 8 n can never ends with 0. WORKSHEET- Sol. 7. Since 6 > 6, so we state with 6 as dividend and 6 as divisor, we have 6 6 7 + Now, is the remainder, which is not zero, so we again apply Euclid s Division lgorithm to 6 and, we have 6 9 + 0 Now, the remainder has become zero and is the divisor. Hence, the HCF of 6 and 6 is LCM 6 8 6 LCM (6, 6) 8 6 9 0 HCF LCM Product of the two number 0 6 6 96 96 Hence Verified. Sol. 8. Since, 7 > 8 On applying Euclid s division algorithm, we get 7 8 + 7 (i) 8 7 + 6 (ii) 7 6 + (iii) 6 + 0 (iv) Hence, H.C.F. (8, 7). In order to write in the form of 8x + 7y, we move backwards as follows : 7 6 [From (iii)] 7 (8 7 ) [Replace 6 From (ii)] 7 (8 7 ) 7 8 + 7 7 + 7 8 7 ( + ) 8 7 8 (7 8 ) 8 [Replace 7 From (i)] 7 8 8 7 8 (6 + ) 7 8 8 8 ( 8) + 7 () 8x + 7y \ Hence x 8 and y Note that the values of x and y are not unique. P-
Sol.. The smallest prime number is and the smallest composite number is. Hence, required HCF (, ) Sol.. If the number n, for any n, were to end with the digit zero, then it would be divisible by. That is, the prime factorization of n would contain the prime. This is not possible because n () n ; so the only prime in the factorisation of n is. So, the uniqueness of the Fundamental Theorem of rithmetic guarantees that there are no other primes in the factorisation of n. So, there is no natural number n for which n ends with the digit zero. [CBSE Marking Scheme, 0] Sol.. 0 6 6 8 7 8 7 7 7 7 x 0 Sol.. 0 8 + and 8 9 + 0 HCF of 0 and 8. Sol.. The greatest number of cartons is the HCF of and 90 90 HCF 8 The greatest number of cartons 8. Sol.. We know that a b HCF (a, b) LCM (a, b) 800 LCM LCM (a, b) 800 0 WORKSHEET- Sol. 6. If 6 n ends with 0, then it must have as a factor. But we know that only prime factor of 6 n are and. 6 n ( ) n n n From the fundamental theorem of arithmetic, we know that the prime factorisation of every composite numbers is Unique. 6 n can never end with 0 [CBSE Marking Scheme, 0] Sol. 7. n n n(n ) n(n + )(n ) (n ) n(n + ) product of three consecutive positive integers. Now, we have to show that the product of three consecutive positive integers is divisible by 6. We know that any positive integer a is of the form q, q + or q + for some integer q. Let a, a +, a + be any three consecutive integers. Case I : If a q. a(a + )(a + ) q(q + )(q + ) q (r) 6qr, which is divisible by 6. ( Product of two consecutive integers (q + ) and (q + ) is an even integer, say r) Case II : If a q +. a(a + )(a + ) (q + )(q + )(q + ) (r) ()(q + ) 6r(q + ), which is divisible by 6. Case III : If a q +. a(a + )(a + ) (q + )(q + )(q + ) multiple of 6 for every q 6r (say), which is divisible by 6. Hence, the product of three consecutive integers is divisible by 6. [CBSE Marking Scheme, 0] WORKSHEET- Sol.. 0 + ( 0 + ) (909 + ) (90) a composite number [ Product of more than two factors] [CBSE Marking Scheme, 0] P- M T H E M T I C S - X T E R M -
Sol.. 66) 0 ( 7)66( 6 0)7( 690 )0(0 0 0 Hence HCF (66, 0) [CBSE Marking Scheme, 0] Sol.. Given HCF (06, ) 8 LCM (06, )? Let a 06 b We know that a b LCM (a, b) HCF (a, b) 06 LCM (a, b) 8 06 LCM (a, b) 8 \ LCM (06, ) 8 Sol.. To find LCM (9,, ). 9 LCM (9,, ) 80 minutes The bells will toll together after 80 minutes. Sol. 6. By Euclid s division algorithm, for two positive integers a and b, we have a bq + r, 0 r < b Let b 6, r 0,,,,, Sol.. HCF of and 9 Sol.. 90 HCF 8 LCM 70. [CBSE Marking Scheme, 0] Sol.. (7 ) + (7 + ) (9 + ) 9. and (7 6 ) + (7 6 + ) (68). which is a composite number (more than one prime factor). So a 6q, 6q +, 6q +, 6q +, 6q +, 6q + Clearly, a 6q, 6q +, 6q + are even, as they are divisible by. But 6q +, 6q +, 6q + are odd, as they are not divisible by. ny positive odd integer is of the form 6q +, 6q + or 6q +. Sol. 7. Fundamental theorem of rithmetic : Every composite number can be expressed as a product of primes and decomposition is unique, apart from the order in which the prime factors occur. HCF LCM 0 Let s calculate LCM HCF 0 Since LCM is always a multiple of HCF,. not an integer hence, two numbers cannot have HCF and LCM as and 0 respectively. [CBSE Marking Scheme, 0] Sol. 8. ny positive integer is of the form q or q +, for some integer q. When n q n n q(q ) m, where m q(q ) which is divisible by. When n q + n n (q + )(q + ) q(q + ) m, when m q(q + ) which is divisible by. Hence, n n is divisible by for every positive integer n. WORKSHEET- Sol.. By Euclid s division algorithm a bq + r Take b Since 0 r <, r 0,,, So, a q, q +, q +, q + Clearly, a q, q + are even, as they are divisible by. Therefore cannot be q, q + as a is odd But q +, q + are odd, as they are not divisible by. ny positive odd integer is of the form (q + ) or (q + ). Sol.. We have 7 6 + 6 + + 0 Hence, HCF 6m 7 6m 7 + 0 m 0 6 Now, 6 P-
7 LCM 8 Sol. 6. 6 6 HCF (6, 6) LCM (6, 6) 6 9 We can check HCF and LCM are correct or wrong by using. Formula HCF (a, b) LCM (a, b) Product of the numbers a b Product of the HCF and LCM should be equal to product of the numbers. 6 6 76 76 LHS RHS Hence our answer is correct. Sol. 7. Let a q + r, 0 r < a q, q +, q + or q + Case I : a (q) 6q (q ) m, m q Case II : a (q + ) 6q + 8q + (q + q) + m +, where m q + q Case III : a (q + ) 6q + 6q + (q + q + ) m, where m q + q + Case IV : a (q + ) 6q + q + 9 6q + q + 8 + (q + 6q + ) + m +, where m q + 6q + From cases I, II, III and IV, we conclude that the square of any +ve integer is of the form m or m +. TOPIC- Irrational Numbers, Terminatting and Non-Terminating Recurring Decimals Sol.. Sol.. Let 6 8 7 0.7 0 be a rational number, which can be put in the form a, where b 0 ; a and b are co-prime. b 6 a b (Integer) (Integer) 6 a b 6 rational But, we know that 6 is an irrational number. Thus, our assumption is wrong. Hence, 6 is an irrational number. Sol.. + 0 + 0 7 + 00 000.7 + 0..87. WORKSHEET- Sol.. Let p be a prime number and if possible, let p be rational m Let p, where m and n are integers having no n common factor other than and n 0. Then, p m n Squaring on both sides, we get ( ) p m n p m n pn m...(i) p divides m and p divides m. [ p divides pn ] [ p is prime and p divides m p divides m] Let m pq for some integer q. On putting m pq in eq. (i), we get pn p q n pq p divides n [ p divides pq ] and p divides n. [ p is prime and p divides n p divides n] P- M T H E M T I C S - X T E R M -
Thus p is a common factor of m and n but this contradicts the fact that m and n have no common factor other than. The contradiction arises by assuming that p is rational. n + n+ P Q Q P n + n+...(i) Sol.. Let Hence, p is irrational. be a rational number. a b. (a and b are integers and co-primes) On squaring both the sides, a b b a a is divisible by a is divisible by...() We can write a c for some integer c a 9c b 9c b c b is divisible by b is divisible by...() From () and (), we get as a factor of a and b, which is contradicting the fact that a and b are coprimes. Hence our assumption that is a rational number, is false. So is an irrational number. Sol. 6. Let us assume that there is a positive integer n for which n + n+ is rational and equal to P Q, where P and Q are positive integers (Q 0). Sol.. 6 0 6 0 6 6 6 6 (0) 0.088 will terminate after decimal placed. Sol.. 0 \ will terminate after decimal places. n+ n Q P pply [(i) + (ii)], we get n n+ ( n + n+)( n n+) n n+ n n+ ( n ) ( n+ ) n+ P Q + Q P n+ pply [(i) (ii)], n P P + Q PQ + Q PQ P + Q PQ...(ii)...(iii)...(iv) From (iii) and (iv) we can say n+ and n both are rational because P and Q both are rational. But it is possible only when n + and n both are perfect squares. But they differ by and two perfect squares never differ by. So both n + and n cannot be perfect squares, hence there is no positive integer n for which n + n+ is rational. WORKSHEET-6 Sol.. Denominator 00 Decimal expansion, 7 00 7 0 Sol.. Let be a rational number. a b, 0. [CBSE Marking Scheme, 0] (a, b are co-prime integers and b 0.) a b a b is a factor of a is a factor of a Let a c, (c is some integer) a c b c b c is a factor of b P-
is a factor of b is a common factor of a, b. But this contradicts the fact that a, b are co-primes. is irrational. Let be rational a a a is rational, so is. But is not rational contradiction is irrational. Sol.. Let + is a rational number. + p q, q 0 + p q p q p q q q q is irrational and is rational q But rational number cannot be equal to an irrational number. + is an irrational number. FORMTIVE SSESSMENT Note : Students should do this activity themselves. Sol. 6. Let be a rational number. a b, (a, b are co-prime integers and b 0) a b Squaring, a b divides a divides a So we can write a c for some integer c, substitute for a, b c, b c This means divides b, so divides b. a and b have as a common factor. But this contradicts that a, b have no common factor other than. Our assumption is wrong. Hence, Let \ integers, b 0 is irrational. be rational b a b a a, where a and b are b b a is rational but is not rational. Our assumption is wrong. is irrational. WORKSHEET-7 P-6 M T H E M T I C S - X T E R M -
SECTION CHPTER B Polynomials TOPIC- Zeroes of a Polynomial and Relationship between zeroes and co-efficient of quadratic polynomial. Sol.. Sum of the roots Coefficient of x Coefficient of x a + b b a b a WORKSHEET-8 Sol.. If a and b are the zeroes of x x +, then a + b b a a + b Sol.. The number of zeroes of p(x) is. and ab c a Sol.. a + b b a ab ( 6) 6 c k a b k a (6) k 0 6 k 0 k a + b (a + b) a b 0 k Sol.. x x + 0 [CBSE Marking Scheme, 0] If a and b are the zeroes of x x +. then a + b b a a + b ( ) a + b and ab c a ab ab \ a + b ab. New quadratic polynomial whose zeroes are a and b is : x (Sum of the roots) x + Product of the roots x (a + b)x + a b x (a + b)x + 9 ab x x + 9 x 9 x + 9 (x 9x + 9) Hence, required quadratic polynomial is (x 9x + 9). Sol. 6. Given, a and a are the zeroes of x + x + a We know that, Product of the zeroes Constant term Coefficient of x a a a a \ a Sol. 7. Factors of x + 7x + : x + 7x + 0 x + x + x + 0 x(x + ) +(x + ) 0 P-7
(x + ) (x + ) 0 x,...(i) Let p (x) x + 7x + 7x + px + q If p(x) is exactly divisible by x + 7x +, then x and x are zeroes of p(x) (from eq (i)) p(x) x + 7x + 7x + px + q p( ) ( ) + 7( ) + 7( ) + p( ) + q but p( ) 0 \ 0 6 8 + p + q 0 p + q 80 p q 80...(ii) p( ) ( ) + 7( ) + 7( ) + p( ) + q but p( ) 0 \ 0 8 89 + 6 p + q 0 p + q Sol. Let, p(x) x x + 9 x 6x 6x + 9 p(x) x (x ) (x ) 0 (x ) (x ) The zeros are, x, Hence, zeroes of the polynomial are,. Sol.. p(x) x kx + 6 Coeff. of x Sum of the zeroes Coeff. of x ( k ) k 9 Sol.. Since α and b are the zeroes of the polynomial, then Coeff. of x α + β Coeff. of x α + β...() Given α β 9...() From () and (), α, b αβ Constant term Coefficient of x, αβ k () ( ) k k 0 Sol.. f (x) x x x (x ) i.e. f (x) 0 x 0 or x Hence, zeroes are 0 &. p q...(iii) On solving eq. (ii) and eq. (iii) by elimination method, we get p q 80 p q + + p On putting the value of p in eq. (i), ( ) q 80 0 q 80 q 0 80 q 60 q 60 Hence, p, q 60. zeroes are 0 and WORKSHEET-9 Coefficient of x Sum of zeroes Coefficient of x Constant term Product of zeroes 0 Coefficient of x [CBSE Marking Scheme, 0] Sol. 6. Let p(x) x + 8x x + 0x x x (x + ) (x + ) (x + ) (x ) Hence, zeroes of the quadratic polynomial x + 8x are and. Verification : Sum of zeroes + 8 Product of zeroes ( ) Coeff. of x gain sum of zeroes Coeff. of x Constant term Product of zeroes Coeff. of x Thus relationship is verified. 8 Sol. 7. Since α and β are the zeroes of polynomial x + x +. Hence, α + β Sol.. p (x) x x x(x ), p (x) 0 x (x ) 0 and αβ Now for the new polynomial, P-8 M T H E M T I C S - X T E R M -
Sum of the zeroes a b + +a +b Sum of zeroes ( a+b ab ) + ( +a b ab) ( +a )( +b) ab +a+b+ab + Sol.. f(x) x 7x 8 Let other zero be k, then 7 Sum of zeroes + k 7 k 8 Sol.. From the graph it is clear that the curve y f (x) cuts the x-axis at two places between and. \ Required number of zeroes. coeff. of x Sol.. Sum of zeroes coeff. of x a + b a a + b 0 constant term Product of zeroes coeff. of x ab b a then b Sol.. Let p (x) x 8x + Zeroes are, x 6x x + x (x ) (x ) ( x ) (x ) x, Sol.. p(y) 6y 7y + 7 7 α + β 6 6 and αβ 6 Now, + a b a+b ab 7 6 7 6 Product of zeroes a b ( a)( b) +a +b ( +a )( +b) a b + ab ( a + b ) + ab +a+b+ab + ( a+b ) +ab 6 + + Product of zeroes + Hence, Required polynomial x (Sum of zeroes)x + Product of zeroes x x +. [CBSE Marking Scheme, 0] WORKSHEET-0 and a b ab The required polynomial is y 7 y + [y 7y + 6]. Sol. 6. Given polynomial, f(x) ax x + c Let the zeroes of f (x) are α and β, then according to the question Sum of zeroes, (α + β) Product of zeroes, (αβ) 0 Now, Coeff. of x α + β Coeff. of x a + 0 a a Constant term and αβ Coeff. of x 0 c c Hence, a and c. Sol. 7. Since α & β are the zeroes of the cubic polynomial x + x + then, α + β α β b a Sum of zeroes + + + a b ab + b + ab + a ab a + b + ab ab P-9
( a+b) ab ( ) 6 Product of zeroes + b + a a b a b ab + + + b a ab a + b + ab ab ( a+b) ab ( ) 6 But required polynomial is : x (Sum of the zeroes)x + Product of the zeroes 6 6 x x+ 6 6 or x x+ (x 6x + 6) Sol.. If is the zero of the polynomial x + kx, then Let p(x) x + kx p() 0 (given) () + k() 0 k 0 k Sol.. If is the zero of the polynomial x + kx, then p() 0 () + k () 0 8 + k 0 k + 0 k + 0 k Now, p(x) x x x 6x + x x(x ) + (x ) (x + ) (x ) p(x) 0 x + 0 or x 0 Hence, other zero is. Sol.. ccording to the question, Sum of zeroes 8 Product of zeroes 6 So, quadratic polynomial x (Sum of zeroes) x + Product of zeroes x 8 x + 6 6 (6x x + ) (6x x + ) 6. Sol.. Let p (x) x + x x + x x x (x + ) (x + ) WORKSHEET- (x ) (x + ) So, zeroes are : m and n Now, m n + n m +. Sol.. Given, p(x) x x 7 and a and b are its zeroes. Coeff. of x Sum of zeroes α + β Coeff. of x Constant term Product of zeroes αβ Coeff. of x For the new polynomial, Sum of zeroes + a b 7 7 a+b ab 7 7 Product of zeroes a b ab 7 7 Required quadratic polynomial x (Sum of zeroes) x + Product of zeroes x x + 7 7 7 (7x + x ) (7x + x ) 7. P-0 M T H E M T I C S - X T E R M -
Sol. 6. Let α and β be the zeroes of the polynomial, then as per the question β 7α 8 α + 7α 8α α Sol. 7. ccording to the question, α and β are zeroes of p(x) 6x x + k So, Sum of zeroes α + β...(i) 6 6 k Product of zeroes α β 6 and α 7α k + 7α 7 k + k + 7 9 k + 7 k k. TOPIC- Problems on Polynomials α β (Given)...(ii) 6 dding equations (i) and (ii), we get α α Put the value of a in equation (ii), we get β 6 β 6 6 β αβ k 6 \ k. Sol.. x + x x + ) x + 8x + 8x + 7 x x + x + + 0x + 6x + 7 + 0x x + + + x + Thus, Quotient x + and Remainder x +. [CBSE Marking Scheme, 0] Sol.. x + x x + ) 6x + x x + 6x x + x + 6x 6x + 6x x + + x + Quotient x + ; Remainder x + p(x) g(x) q(x) + r(x) WORKSHEET- (x x + ) (x + ) + ( x + ) 6x x + x + 6x x + x + 6x + x x +. Verified. Sol.. x + x x + x ) 8x + x x + ax + b 8x + 6x x + 8x + x + ax 8x + 6x x + x + (a + )x + b x x + + + (a + 7)x + b For exact division, remainder is zero, then (a + 7) x + b 0 a + 7 0, b 0 a 7, b. P-
Sol.. p(x) 8x + x x + 7x 8 g(x) rate of the each box x + x q(x) number of boxes r(x) amount of money he donated to child. By using long division for Division x + x x + x ) 8x + x x + 7x 8 8x + 6 x x + 8x + x + 7x 8x + 6x x + x + x 8 x x + + + x 0 Find q(x) and r(x) q(x) x + x number of Boxes r(x) x 0 Values : kindness, helpfulness, care, well- wisher of society and promoting education. [CBSE Marking Scheme, 0] Sol.. g(x) x + x +, f(x) x + x 7x + x + x x + x + x + ) x + x 7x x + x + 9x + x x 0x + x x x x + + + x + 6x + x + 6x + Remainder, r(x) r(x) 0, g(x) is not a factor of p (x). [CBSE Marking Scheme, 0] Sol.. x + x + x (x + 0) g(x) + (6x ) x + x x+0 g(x) x+0 x x + x + 0) x + x x + 0 x + 0x 6x x 6x 0x + + 9x + 0 9x + 0 0 Hence, g(x) x x +. Sol.. p(x) x + x x + x g(x) rate of the each box x + x q(x) number of boxes r(x) amount of money he donated to child. By using long division for Division We need to find q(x) and r(x) On dividing p(x) by g(x). x + x ) x + x x + x (x 6x + x + 8x x + 6x + 0x + x 6x x + 8x + + x 7x x + x 66 + 6x + 6 number of boxes q(x) x 6x + mount of money he donated to child r(x) 6x + 6 Values : kindness, helpfulness, care, well-wisher of society and promoting education. WORKSHEET- Sol.. x + x + x ) x + x + x x x x + x + x x x x + x x 6 + + Remainder 0 x is not a factor of the given polynomial. Sol.. To get share of each member, we have to divide x +8x + 6x + 9 by x+. x + ) x + 8x + 6x + 9 (x + 7x + 9 x + x 7x + 6x 7x + 7x 9x + 9 9x + 9 0 Hence, share to each member is x + 7x + 9 Mr. Kulkarni has taken such initiative because. () To control pollution () To save fossil fuels. P- M T H E M T I C S - X T E R M -
Sol.. x x x ) x x x + x x + x x x + 0x + x + x + + 0 Sol.. x x x + ) x 6x + x + 8 x x + x + x + 9x + 8 x + 9x 6 + + Remainder So should be added. [CBSE Marking Scheme, 0] Sol.. Dividend x x x + x + Divisor g(x) Quotient x x Remainder x We know that Dividend Divisor Quotient + Remainder Dividend Remainder Divisor Quotient x x x + x+ x x x x x x + x+ x x x x ) x x x + x + (x x x x + + x + x + x + x + + 0 Hence, g(x) x Now, x x x x + x x (x ) + (x ) (x + ) (x ) ll the zeroes are,,. [CBSE Marking Scheme, 0, 0] WORKSHEET- Sol.. x x + (8 k) x x + k) x 6x + 6x x + 0 x x + kx + x + (6 k)x x + 0 x + 8x kx + + (8 k)x ( k)x + 0 (8 k)x (6 k)x + (8k k ) + (k 9)x + (0 8k + k ) Given, remainder x + a k 9 k 0 and a 0 8k + k 0 0 +. [CBSE Marking Scheme, 0] Sol.. g(x) (x + ) (x ) x is a factor of the given polynomial x x + x ) x x + x + 0x 8 x x + x + x + 0x x + 0x + x 8 x 8 + 0 x x + (x ) (x ) Other zeroes are and. P-
Sol.. x + x x + x )8x + x x + 8x 8x + 6x x + 8x + x + 8x 8x + 6x x + Sol.. x + x + ) 6x + 8x + 7x + x 7 (x + 6x + 8x + x x + x + 7 x + 0x + x + \ ax + b x + On comparing, we get a and b. [CBSE Marking Scheme, 0] Sol.. x x x ) x 6x + 6x + k x x + x + 6x + k x + 9x + x + k x + 9 + k 9 On division, Quotient x x Remainder k 9 Remainder 0 k 9 [CBSE Marking Scheme, 0] Sol.. On dividing x + x 9x + by x +, remainder should be zero. x + ) x + x 9x + (x x x + x x 9x + x 8 x + + x + x + + Hence should be added to x + x 9x + to make it completely divisible by x +. Sol.. p(x) x x x 6 p() () () () 6 (7) 9 9 6 0 x is a factor of p(x) px ( ) x x x 6 x x x + x x x + + + x Subtract x (or) add x + for exact division. By long division, x + x + x ) x x x 6 x 6x + x x x x + x 6 x 6 0 0 px ( ) x x + x WORKSHEET- x + x + x + (x + ) (x + ), are the other zeroes of p(x) ll the zeroes of p(x) are,,. Sol.. s x and are the zeroes of x + 6x + x x 0 So (x ) and (x + ) are the factors of x + 6x + x x 0 Now, (x ) (x + ) 0 x + x x 0 0 x + x 0 0 Dividing x + 6x + x x 0 by x + x 0 x + x + x + x 0) x + 6x + x x 0 x + x 0 x + x + x x 0 x + 9x 0x + x + 6x 0 x + 6x 0 + 0 x + 6x + x x 0 (x + x 0) (x + x + ) (x ) (x + ) (x + ) (x + ) Hence other two zeroes are and. P- M T H E M T I C S - X T E R M -
Sol.. x x ) x x + 6 x x + x + 6 x + 6 + 0 Quotient x, Remainder 0 [CBSE Marking Scheme, 0] Sol.. x x + ) x x + 6 x x + x + 6 Thus, Quotient x and Remainder x + 6. Sol.. Since x divides f(x) completely (x ) is a factor of f (x) (x ) is a factor of f (x) + x x is a factor of f (x) and are zeroes of f (x) x + x x ) x + x x x + 0 x x + x 6x x x x + 6x + 0 6x + 0 + (x + x ) is a factor of p(x) (x + x x ) is a factor of p(x) (x + ) (x ) is a factor of p(x) and are zeroes of p(x) ll the zeroes of p(x) are,, and. Sol.. s x ± are the zeroes of p(x), so x ( ± ) are the factors of p(x). FORMTIVE SSESSMENT Note : Students should do this activity themselves. Now, { x ( + )}{ x ( ) } {( x ) }{( x ) + } ( x ) ( ) WORKSHEET-6 x x + Dividing p(x) by x x + x x x x +) x 6x 6x + 8x x x + x + x 7x + 8x x + 8x x + + x + 0 x x + 0 x + + p(x) { x ( + ) } x ( ) ( x x ) 0 { } s, x x (x + )(x 7) Hence, other two zeroes of p(x) are and 7. Sol.. s x and are the zeroes of x + x 7x 8x, So (x ) and (x + ) are the factors of x + x 7x 8x Now, (x ) (x + ) 0 x 8 0 Dividing x + x 7x 8x by x 8 x 8) x + x 7x 8x (x + x x 7x + x 8x x 8x + 0 x + x 7x 8x (x 8) (x + x) (x 8) x (x + ) Hence other two zeroes are 0 and (x ) (x + ) (x) (x + ). WORKSHEET-7 P-
SECTION CHPTER B Pair of Linear Equations in Two Variables TOPIC- Graphical Solution of Linear Equations in Two Variables and Their Possibilities of Solution/Inconsistency Sol.. The pair of equations y 0 and y has no solution. Sol.. (i) The given equations can be re-written as : x + y 0 x y 7 0 On comparing with ax + by + c 0, we have a, b, c a, b, c 7 a a, b b Thus, a b i.e. a b Hence, the pair of linear equations is consistent. (ii) The given equations can be re-written as x y 8 0 x 6y 9 0 On comparing with ax + by + c 0, we have a, b, c 8 a, b 6, c 9 a a, and Thus, b b c c 6 8 9 8 9 8 9 a i.e. b a b c c Hence, the pair of linear equations is inconsistent. (iii) The given equations can be re-written as : x + y 7 0 WORKSHEET-8 9x 0y 0 On comparing with ax + by + c 0, we have Thus, a, b, c 7 a 9, b 0, c a a 8 6 b b 0 6 6 6 a i.e. b a b Hence, the pair of linear equations is consistent. (iv) The given equations can be re-written as : x y 0 0x + 6y + 0 On comparing with ax + by + c 0, we have a, b, c a 0, b 6, c a a 0 and Thus, b b c c 6 a i.e. b a b c c Hence, the pair of linear equations is consistent. (v) The given equations can be re-written as : x + y 8 0 P-6 M T H E M T I C S - X T E R M -
x + y 0 On comparing with ax + by + c 0, we have a a a, b, c 8 a, b, c Sol.. (i) The pair of linear equations is : x + y x + y 0 (i) and x + y 0 x + y 0 0 (ii) On comparing with ax + by + c 0, we have a, b, c a, b, c 0 and a Since b c a b c a a c c, b b 0 So, the pair of linear equation is coincident having many solutions. Thus, the equation is consistent. Now, we need to solve it graphically. We have x + y y x x 0 y 0 Points B and x + y 0 y x 0 y 0 Points C D E C(0, ) x + y O x + y D(, ) x + y 0 B E(, 0) 0 x On plotting these points, we observe that the lines are coincident having infinite solutions. b b c 8 and c Thus, a i.e. b a b c c Hence, the pair of linear equations is consistent. (ii) The pair of linear equations is : x y 8 WORKSHEET-9 x y 8 0 (i) and x y 6 x y 6 0 (ii) On comparing with ax + by + c 0, we have a, b, c 8 a, b, c 6 a a, b b, c c a b c s a b c the lines are parallel having no solution. So, the pair of linear equations is inconsistent. (iii) The pair of linear equations are : x + y 6 0 (i) and x y 0 (ii) On comparing with ax + by + c 0, we have a, b, c 6 a, b, c a a, b b c 6 c a a, b b c, c a b c Since, a b c Therefore, the pair of linear equations is consistent. Now, we need to solve it graphically We have, x + y 6 0 y 6 x x 0 y 6 0 and, x y 0 y x P-7
x' y x x 0 y 0 y 7 6 0 (0, 6) (, 0) 6 7 (, 0) x + y 6 0 (0, ) y' (, ) x y 0 (iv) x y 0 x y 0 The pair of linear equation is x y 0 (i) and x y 0 (ii) On comparing with ax + by + c 0, we have a, b, c a, b, c a a, b b c c a a, b b c c Since a b c a b c Therefore, the pair of linear equations is inconsistent. Sol.. x y 7 x y 7 0 a, b, c 7 x + y + 0 a, b, c x X' Thus a a b b a a b b Hence, given pair of linear equation has a unique solution. We have x y 7 0 y x 7 x 0 y 7 and x + y + 0 y x x y 0 Y 9 8 7 6 (, ) (, ) (, 0) 7 6 0 6 7 8 9 0 X 6 7 8 9 (0, 7) Y' (, ) \ x and y is the solution. x y 7 x + y WORKSHEET-0 Sol.. Since, am bl a b c l m n So, ax + by c and lx + my n has no solution. Sol.. Yes, it is consistent. We have, for the equation x + y 9 0 a, b and c 9 and for the equation, x + 6y 8 0 a, b 6 and c 8 Here and a a b b 6 c c 9 8 a It is clear that b c a b c Hence, system is consistent and dependent. P-8 M T H E M T I C S - X T E R M -
Sol.. Given equations are : x + py + 8 0...() x + y + 0...() Sol.. x + y y x The condition of unique solution, a a b b Hence, p or p p. Sol.. Let the length of the garden be x m and its width be y m. Then, Perimeter of rectangular garden (Length + Width) (x + y) Therefore, Half perimeter (x + y) But it is given as 6 m. (x + y) 6 (i) lso, x y + i.e., x y (ii) For finding the solution of eqs. (i) and (ii) graphically, we form the following table. For x + y 6 X' 6 0 8 6 x 0 y 6 For x y x 0 6 0 y 6 6 Draw the graphs by joining the points (0, 6) and (, ) and points (0, 6) and (6, ). The two lines intersect at point (0, 6) as shown in the graph. Y x y 0 Scale unit sq m (0, 6) (0, 6) (6, ) 6 8 0 6 8 0 X (, ) x + y 6 x' x y 0 x y y x + x y 0 Graph of two equatio ns is : x + y (, 0) 0 y (, ) y' (, ) (, ) x y (, 0) Marks Lines meet x axis at (, 0) and (, 0) respectively. [CBSE Marking Scheme, 0] x Y' Hence, length is 0 m and width is 6 m. Sol.. ad bc a c b d. Hence, the pair of given linear equations has unique solution. WORKSHEET- Sol.. Pair of linear equations kx y 0 6x y 9 0 Condition for infinite solutions : a a b b c c P-9
k 6 9 k. Sol.. Given, linear equation is x + y 8 0 (i) (i) For intersecting lines, we know that a a b b One of the possible line may be taken as x + y 9 0 a (ii) For parallel lines, b a b c c One of the possible line parallel to eq. (i) may be taken as 6x + 9y + 7 0 (iii) For coincident lines, a a b b c c One of the possible line coincident to eq. (i) may be taken as x + 6y 6 0 s (, ) is common therefore solution is (, ). Solution (, ) Sol.. x y + 0 y x + x 7 y 6 0 and x + y 0 Sol.. The equation of one line is x + y. We know that if two lines a x + b y + c 0 and a x + b y + c 0 are parallel, then a b c a b c c a a b c b 9 Hence one of the possible, second parallel line is x + 9y. Sol.. For equation, x + y 0 a, b, c For equation, (k + ) x + 6y (k + )0 a k +, b 6, c (k + ) For infinitely many solutions a a b c b c k + 6 k + k + 6 6 k k. x' y x 8 7 6 (, 0) 0 6 7 8 9 x 7 y 0 9 y 7 6 6 7 8 9 0 y' (, ) (, ) (7, 6) (7, 0) x y + 0 (, ) x x + y 0 (i) These lines intersecting each other at point (, ). Hence x and y. (ii) The vertices of triangular region are (, ), (, 0) and (7, 0). (iii) rea of 8 rea sq unit. Sol.. The condition for no solution, a b a b c c WORKSHEET- k k When k, we get k 6 k i.e., k ± 6 k 6, so k 6. ( For k 6, a b c a b c ) Sol.. (m ) x + y 0...() On comparing with the eqn. a x + b y +c 0 a m, b, c x + (n )y 0...() On comparing with the eqn. a x + b y + c 0 a, b (n ), c For a pair of linear equations to have infinite number of solutions. a b a b c c P-0 M T H E M T I C S - X T E R M -
m n (m ) and (n ) 6 m 7, n Sol.. Let the present age of father be x years and the age of daughter be y years. 7 years ago father s age (x 7) years 7 years ago daughter s age (y 7) years ccording to the question, (x 7) 7(y 7) x 7y...(i) fter years father s age (x + ) years fter years daughter s age (y + ) years ccording to the condition, x + (y + ) x y 6...(ii) To find the equivalent geometric representation we find some points on the line representing each question, these solutions are given below in the table. From eq. (i) x 7y x 0 7 y x + 7 6 7 8 Y x y 6 x 6 8 y x 6 0 Points D E F G 0 G (0,6) (7,7) (,8) (,) C 0 B (,) (8,) x y 6 (6,0) E F X' O D 0 0 0 0 X x 7 y Y' Plot the points (0, 6), B (7, 7) and C (, 8) and join them to get a straight line BC. Similarly, plot the points D (6, 0), E (, ) and F (8, ) and join them to get a straight line DEF. These two lines intersect at point G (, ). Thus we conclude that the present ages of ftab and his daughter are years and years respectively. Points B C G TOPIC- lgebraic Methods to Solve Pair of Linear Equations and Equations Reducible to Linear Equations Sol.. x + y 7 0...(i) x + y 6 0...(ii) From eq. (ii). x + y 6 0 y 6 x Value of y put in eq. (i) x + y 7 0 x + (6 x) 7 0 x + 8x 7 0 x 0 x \ x Substitute the value of x in eq (ii) to get value of y, x + y 6 0 () + y 6 0 + y 6 0 WORKSHEET- y 0 y Hence, values of x and y are and respectively. Sol.. Let age of father and son be x and y respectively. x + y 0...() x y...() x 0 y 0 ges are 0 years and 0 years. [CBSE Marking Scheme, 0] lternative Method Let age of father and son be x and y years respectively. x + y 0...() x y...() Substitute the value of x from () in () \ y + y 0 P-
Sol.. y 0 y 0 yrs. from () x + 0 0 x 0 yrs. \ ge of father is 0 years. and ge of son is 0 years. x 80 8 x and y y + 00 60 [CBSE Marking Scheme, 0] lternative method Given : x + y 0...() x y 0 0...() By cross-multiplication method, x y 0 0 x 80 8 x 8 y + 00 60 6 and 96 y 6 x and y Sol.. (i) Given, a pair of linear equations is : x + y...(i) and x y or y x...(ii) On substituting the value of y from eq. (ii) in eq. (i), we get x + x x 8 x 9 On substituting x 9 in eq (ii), we get y 9 i.e., y Hence, x 9 and y (ii) Given, a pair of linear equation is : s t s t +...(i) and s t + 6...(ii) On substituting s t +, from eq. (i) in eq. (ii), we get t+ t + 6 (t + ) + t 6 t + 6 6 t 0 t 6 Sol.. Let cost of chair ` x and cost of table ` y ccording to the question, x +y 00...() x +y 70...() Multiplying () by and () by 8x +6y 00...() x + 6y 0...() () () 7x 00 x 0 Substituting the value of x in (), y 00 Cost of chair and table ` 0, ` 00 respectively [CBSE Marking Scheme, 0] Sol.. Substitute x X and y Y X + Y...() X 9Y...() () 6X + 9Y 6...() dding () & (), X 9Y 6X + 9Y 6 + + + 0X 0X X 0 x Now Y X Y y 9 y Thus x, y 9 is the solution. WORKSHEET- From eq., (i), s 6 + 9 Hence, s 9, t 6 (iii) Given, a pair of linear equations is : x y or y x...(i) and 9x y 9...(ii) On substituting y from eq. (i) in eq. (ii), 9x (x ) 9 i.e., 9 9 It is a true statement. Hence, eq. (i) and (ii) have infinitely many solutions. (iv) Given, a pair of linear equations is : 0. x + 0. y....(i) and 0. x + 0. y....(ii) From eq. (i), x + y [ Multiply by 0] i.e., y x i.e., y x...(iii) On substituting y from eq. (iii) in eq. (ii), ( x) x + 0 0 0 P- M T H E M T I C S - X T E R M -
x + ( x) x + ( x) x + 6 0x 69 x 69 6 x On substituting x in eq. (iii), we get y 9 i.e., y Hence, x, y (v) Given, a pair of linear equations is : x+ y 0 (i) x 8y 0 or y x 8 (ii) On substituting y from eq. (ii) in eq. (i), x x + 8 0 x x + 0 8 x 8 + x 0 x + x 0 7x 0 x 0 On substituting x 0 in eq. (ii), y 0 8 0 i.e., y 0 Hence, x 0, y 0 (vi) Given, a pair of linear equations is : x y (i) and From eq. (ii), x y + 6 y x x 6 6 ( ) y x 6 (ii) Sol.. x y + x+ y 6 x + y 6...(i) and x y ( ) y x (iii) On substituting y from eq. (iii) in eq. (i), we get x ( x) 8 x ( x) 9 x 8 9 ( x) 8 [ Multiply by 8] 7x 0 ( x) 6 7x 0 + 0x 6 7x + 0x 0 6 7x 9 x On substituting x in eq. (iii), y 9 i.e., y Hence, x, y Sol.. Let fixed charge be ` x and cost of food per day be ` y x + 0y 000 x + y 00 x 000, and y 00 Fixed charge and cost of food per day are ` 000 and ` 00. [CBSE Marking Scheme, 0] lternative method Let fixed charge be ` x and cost of food per day be ` y x + 0y 000...() x + y 00...() Substract () from () : x+ y 00 x+ 0y 000 y 00 y 00 from (), x + 0 (00) 000 x 000 \ x 000 and y 00 Fixed charge and cost of food per day are ` 000 and ` 00. WORKSHEET- x y x y 9...(ii) On subtracting eq. (ii) from eq. (i), x + y 6 x y 9 + y P-
\ y y put in eq (i) x + ( ) 6 x 6 x 6 x 6 \ x Hence required value of x and y are and respectively. Sol.. x y + 0...() x y + 0...() Multiplying eq. () by, we get (x y + ) 0 0x y + 0 Subtracting eq () from eq (), we get 0x y + 0 x y + 0 ( ) (+) ( ) 7x + 0 7x x Substituting the value of x in eq. () x y + 0 ( ) y + 0 y + 0 y 0 y \ y Hence, x and y Sol.. (i) By elimination method : Given, x + y (i) and x y (ii) On multiplying eq. (i) by and eq. (ii) by and adding, (x + y) + (x y) + x + y + x y + x 9 9 x On substituting x 9 9 + y in eq. (i), y 9 y y 6 9 Hence, x 9 and y 6 By substitution method : Given, x + y (i) and x y (ii) From eq. (i), y x (iii) On substituting y from eq. (iii) in eq. (ii), x ( x) x + x x 9 x 9 On substituting x 9 in eq. (iii), we get y 9 y 6 Hence x 9 and y 6 (ii) By elimination method : Given, x + y 0 (i) and x y (ii) On multiplying eq. (i) by and eq. (ii) by and adding, (x + y) + (x y) 0 + x + y + x y 0 + 7x x On substituting x in eq. (i), + y 0 y y Hence, x and y By substitution method : Given, x + y 0 (i) and x y (ii) From eq. (ii), y x y x (iii) On substituting y x from eq. (iii) in eq. (i), x + (x ) 0 7x x From eq. (iii), y i.e., y Hence, x and y (iii) By elimination method : Given, x y (i) and 9x y + 7 (ii) On multiplying eq. (i) by and eq. (ii) by, 9x y (iii) 9x y 7 (iv) On subtracting eq. (iv) from eq. (iii), (9x y) (9x y) 7 y + y y y P- M T H E M T I C S - X T E R M -
On substituting y x 9x + 9x 7 9 x in eq. (i), Hence x 9 and y By substitution method : Given, x y (i) and 9x y + 7 (ii) From eq. (ii), y 9x 7 y 9 x 7 (iii) On substituting y from eq. (iii) in eq. (i), Sol.. Given, a pair of linear equations is : x + y (i) and x y (ii) From eq. (ii), y x + x + y (iii) On substituting y from eq. (iii) in eq. (i), we get x + x + x + (x + ) x + x + 6 7x 6 7x x From eq. (iii), + y i.e., y On substituting x and y in the equation y mx +, m ( ) + m + m m Sol.. (i) Let the fraction be x y ccording to the given conditions x + x and y y + x + y and x y + x y (i) x y (ii) On subtracting eq. (i) from eq. (ii), 9 x 7 x 6x x + 8 9x 7 9 x On substituting x 9 in eq. (iii), 9 9 7 y 0 6 y 8 9 Hence, x 9 and y WORKSHEET-6 (x y) (x y) + x On substituting x in eq. (i), y y Hence, the fraction is. (ii) Let the present age of Nuri x (in years) Present age of Sonu y (in years) ccording to the given conditions, year ago, x (y ) i.e., x y 0 (i) 0 yr later, x + 0 (y + 0) i.e., x y 0 (ii) On subtracting eq. (i) from eq. (ii), (x y) (x y) 0 + 0 y + y 0 y 0 On substituting y 0 in eq. (ii), x y + 0 x 0 + 0 x 0 Present age of Nuri is 0 years and present age of Sonu is 0 years. (iii) Let x be the digit at unit s place and y be the digit at ten s place of the value of the number. ccording to the given condition x + y 9 (i) Value of the number x + 0y When we reverse the order of the digits, the value of the new number y + 0x We are given that, 9 (x + 0y) (y + 0x) 9x + 90y y + 0x 88y x P-
x 8y x 8y 0 (ii) Subtract (ii) from (i), x + y 9 x 8y 0 + 9y 9 y From (i) x + 9 x 8 Hence, the number is : 0x + y 8 0 + 8 (iv) Let number of ` 0 notes x Number of ` 00 notes y ccording to the given condition, x + y (i) and 0 x + 00 y 000 x + y 0 ( divide by 0) (ii) On subtracting eq. (i) from eq. (ii), (x + y) (x + y) 0 y On substituting y in eq. (i), x + x 0 Sol.. (i) Let the two numbers be x and y (x > y) We are given that, x y 6 (i) and x y (ii) On substituting x from eq. (ii) in eq. (i), we get y y 6 y 6 y On substituting y in eq. (ii), we get x 9 i.e., x 9 Hence, the two numbers are 9 and. (ii) Let the supplementary angles be x and y (x > y ) Now, x + y 80 (i) and x y 8 (ii) From eq. (ii), y x 8 (iii) On substituting y from eq. (iii) in eq. (i), x + x 8 80 x 98 x 99 On substituting x 99 in eq. (iii), y 99 8 8 i.e., y 8 Hence, the angles are 99 and 8 (iii) Let, Cost of one bat ` x Cost of one ball ` y Cost of 7 bats and 6 balls ` 800 i.e., 7x + 6y 800 (i) Cost of bats and balls ` 70 i.e., x + y 70 (ii) From eq. (ii), y 70 x Hence number of ` 0 notes is 0 and number of ` 00 notes is. (v) Let the fixed charges for the first three days be ` x Let the additional change per day be ` y ccording to the given conditions, Sarita paid for 7 days ` 7 i.e., x + y 7 i.e., x + y 7 (i) [ ` y are to be paid for extra days] In the case of Susy, x + y...(ii) [ ` y are to be paid for extra days] On substracting eq. (ii) from eq. (i), y 7 y 6 y On substituting y in eq. (i), x + 7 x 7 x Hence, fixed charge for first three days ` dditional charge per extra day `. WORKSHEET-7 i.e., y 70 x on substituting y from eq. (iii) in eq. (i), 7x + 6 70 x 800 On multiplying by, x + 6 (70 x) 800 x + 000 8x 9000 8x 9000 000 7x 800 x 00 On substituting x 00 in eq. (iii), 70 00 y 70 00 0...(iii) i.e., y 0 Hence, cost of one bat ` 00 and cost of one ball ` 0 (iv) Let fixed charge be ` x and charge per km be ` y. We are given that, x + 0y 0 (i) x + y (ii) From eq. (i) x 0 0y (iii) On substituting x from eq. (iii) in eq. (ii) 0 0y + y y 0 0 i.e., y 0 On substituting y 0 in eq. (iii), x 0 0 0 P-6 M T H E M T I C S - X T E R M -
0 00 i.e., x Hence, fixed charges ` Rate per km ` 0 mount to be paid for travelling km ` + ` 0 ` + ` 0 ` (v) Let x y integers. Given, be the fraction, where x and y are positive x + 9 y + and x + y + 6 (x +) 9 (y + ) and 6 (x + ) (y + ) x + 9y + 8 and 6x + 8 y + x 9y + 0 and 6x y + 0 The required equations are : x 9y + 0 (i) and 6x y + 0 (ii) From eq. (ii), y 6x + y 6 x + (iii) On substituting y from eq. (iii) in eq. (i), 6x + x 9 + 0 Sol.. From fig., x + y (i) and x y 6 (ii) dding (i) and (ii), we get x 8 x 9 Substitute the value of x in equation (i), we get 9 + y y 9 Hence, x 9 and y. Sol.. (i) Given, a pair of linear equations is : x y 0 (i) and x 9y 0 (ii) On comparing eq. (i) and (ii) with ax + by + c 0 a, b, c and a, b 9, c a b, c, a b 9 c a b a b c c Hence, no solution exists. x 9 (6x + ) + 0 0 x x 7 + 0 0 x 7 On substituting x 7 in eq. (iii), y 6 7 + i.e. y 9 Hence, the fraction x y is 7. 9 (vi) Let x (in years) be the present age of Jacob s son and y (in years) be the present age of Jacob. years hence, it has relation : (y + ) (x + ) y + x + x y + 0 0 (i) years ago, it has relation (y ) 7(x ) 7x y 0 0 (ii) From Eq. (i), y x + 0 (iii) On substituting in eq. (ii), we get 7x (x + 0) 0 0 x 0 0 x 0 On substituting x 0 in eq. (iii), y 0 + 0 y 0 Hence, the present age of Jacob and his son is 0 years and 0 years respectively. WORKSHEET-8 (ii) Given, a pair of linear equations is : x + y x + y 0 (i) and x + y 8 x + y 8 0 On comparing eq. (i) and (ii) with ax + by + c 0, a, b, c and a, b, c 8 Now, a a b b a b a b Hence, we have a unique solution. By cross multiplication method, x y 8 8 x y z b c c a a b b c c a a b P-7
x {()( 8) ()( )} x y ( 8 + 0) ( + 6) y {( )() ( 8)()} { ()() ()() } x y x y and ( ) x and (iii) Given, a pair of linear equations is : x y 0 x y 0 0 (i) and 6x 0y 0 6x 0y 0 0 (ii) On comparing eq. (i) and (ii) with ax + by + c 0, a, b, c 0 and a 6, b 0, c 0 Now, 0 6 0 0 a b c a b c Hence, they have infinitely many solutions. (iv) Given, a pair of linear equations is : x y 7 0 (i) and x y 0 (ii) On comparing eq. (i) and (ii) with ax + by + c 0, a, b, c 7 and a, b, c Sol.. Given, a pair of linear equation is : 8x + y 9 8x + y 9 0 (i) and x + y x + y 0 (ii) On comparing eq. (i) and (ii) with ax + by + c 0, a 8, b, c 9 and a, b, c By cross-multiplication method, x y 9 9 8 8 x b c b c y c a a b c a a b a b and b a a a b b Hence, we have a unique solution. By cross-multiplication, x 7 x b c b c x {( )( ) ( )( 7)} x ( ) y 7 y z c a a b c a a b y ( 7)() ( )() {( )} { ()( ) ()( ) } y ( + ) ( + 9) x y 6 6 x 6 and y 6 6 x 6 6 6 x and y WORKSHEET-9 x y {()( ) ()( 9)} {( 9)() ( )(8)} { 8 } x y x and y x and y By substitution method : from (ii), x y x y...(iii) Substitute the value of x in (iii) in (i), P-8 M T H E M T I C S - X T E R M -
y \ 8 + y 9 6y + y 7 y 7 y \ from (iii), x () 0 x \ x and y. Sol.. (i) Let fixed part of monthly hostel charges ` y Cost of food for one day ` x In the case of student, 0x + y 000 (i) In the case of student B, 6x + y 80 (ii) On subtracting eq. (i) from eq. (ii), 6x 0x 80 000 x 0 On substituting x 0 in eq. (i), y 000 0 0 000 600 00 Hence, monthly fixed charges ` 00 Cost of food per day ` 0 (ii) Let the fraction be x y. ccording to the given conditions, x y and x y + 8 We get, x y (i) and x y 8 (ii) On subtracting eq. (i) from eq. (ii), (x y) (x y) 8 x On putting x in eq. (i), y y Hence, the fraction is. (iii) Let the number of right answers be x. and the number of wrong answers be y. Total number of equations x + y In first case, marks awarded for x right answers x. Marks lost for y wrong answers y y Sol.. 99x + 0y 99 () 0x + 99y 0 () dding eq. () and (), 00x + 00y 000 x + y () x y 0 (i) In second case, marks awarded for x right answers x. Marks lost for y wrong answers y ccording to the given condition, x y 0 (ii) From eq. (i), y x 0 (iii) On substituting the value of y from eq. (iii) in eq. (ii), x (x 0) 0 x 6x + 80 0 x 0 x On substituting the value of x in eq. (iii), y 0 Total number of questions x + y + 0 (iv) Let speed of car I x km/hr and speed of car II y km/hr Car I starts from point and car II starts from point B First Case : Car I Car II? 00 km B C Two cars meet at C after hr. C Distance travelled by car I in hr x km BC Distance travelled by car II in hr y km We know that, C BC B x y 00 [ B 00 km] x y 0 [ divide by ] (i) Second case : Car I Car II x 00 km B? C c B Two cars meet at C after one hour x + y 00 (ii) On adding eq. (i) and (ii), x 0 x 60 On substituting x 60 in eq. (ii), 60 + y 00 y 0 Hence, the speed of the two cars are respectively 60 km/hr and 0 km/hr. y WORKSHEET-0 Subtracting eq. () from eq. () x + y x y () dding equations () and () x 6 P-9
x Substitute the value of x in eq. (), y Sol.. Let the man can finish the work in x days and the boy can finish the same work in y days. Work done by one man in one day x Now, work done by one boy in one day y ccording to the question, 7 + x y and Let + x y x a + 7b a + b a and b, then y Multiply eq. () by and subtract from it eq. () a + b b Put b a + b 0b 60 y...()...()...()...() 6 y 60 days. 60 in equation (), a + 7 60 a a So 7 60 x x days. Sol.. Since BC DE and BE CD with BC CD, BCDE is a rectangle. Opposite sides are equal. i.e., BE CD, x + y...() Sol.. x y y x (i) and DE BC x y Since perimeter of BCDE is. B + BC + CD + DE + E + x y + x + y + x y + 6 + x y x y dding () and (), we get x 0...() x On substituting the value of x in (), we get y 0 x and y 0. Sol.. Let the amount of their respective capitals be ` x and ` y. ccording to the given condition, x + 00 (y 00) x y 00 (i) and 6(x 0) y + 0 6x y 70 (ii) On multiplying eq. (ii) by and subtracting from eq. (), x y 0 x x 00 0 x 0 x ` 0 On substituting x 0 in eq. (i), 0 y 00 y 0 y ` 70 Hence, the amount of their respective capitals are ` 0 and ` 70. Sol.. Let the incomes of two persons be x and 7x. lso the expenditures of two persons be 9y and y. x 9y 00...() and 7x y 00...() Multiplying eq. () by and eq. () by 9 and subtracting, x y 000...() 6x y 600...() + On subtracting, 8x 600 600 x 8 00 Substituting this value of x in eq. (), 00 9y 00 9y 00 00 800 y 800 00 9 Their monthly incomes are 00 ` 00 and 7 00 ` 00. [CBSE Marking Scheme, 0] WORKSHEET- x + y (ii) Substituting the value of y from eq. (i) in eq. (ii), P-0 M T H E M T I C S - X T E R M -
x + 6x 6 7x \ x From eq. (), y x and y. x+ y Sol.. + 9 Sol.. (x + ) + (y ) x + y...() x y+ + 8 (x ) + (y + ) 8 x + y 7... () Multiply eq. () by (), 9x + 6y 9 Multiply eq. () by, x + 6y 9 On subtracting x 6 Substitute the value of x in eq. (), () + y 7 y 7 6 y 7 x 6 Hence x, y 7. Let p and q x y Then given equations become 6p q...() and p + q...() Multiply eq. () by and add in eq. (), p + q 6 p 7 p 7 Putting this value of p in equation (), 6 q q q FORMTIVE SSESSMENT Note : Students should do this activity themselves. q Now, x p x x and y q y y Hence, x and, y. Sol.. Let the fraction be x, then according to the y question, x + 9 y + x 9y...() and x + y + 6 6x y...() Now from () and (), x 9y + 0 and 6x y + 0 Solving these equations by cross-multiplication method, we get x ( 9)() ( )() x 7 + 0 x 7 y ()(6) ()() ()( ) (6)( 9) y + y 9 Hence, x 7, y 9 Fraction 7 9 WORKSHEET- P-
SECTION CHPTER B Triangles Sol.. In PO and QBO PO QBO B 90 PO QOB O OB (Given) (Vertically Opposite ngle) P QB (by ) WORKSHEET- Sol.. (a) The given 's are not similar. (b) In PQR, R 80 ( + 78 ) 7 In LMN, N 80 ( + 7 ) 78 PQR MNL (By similarly criterion) [CBSE Marking Scheme, 0] Sol.. 6. QB F E QB. 6 QB cm. Sol.. BC PQR (Given) B PQ BC QR C PR z 8 6 y z 8 6 z 8 6 and 8 6 y and y 6 8 z and y y + z + Sol.. B O + OB () + () (Pythagoras th.) B cm B + C () + () () BC CB 90 (Converse of Pythagoras th.) Sol.. X XB, Y, YC 9 (Given) X XB Y and YC 9 X XB Y YC Hence XY is not parallel to BC. B D C In BC, Given that F, E and D are the mid-points of B, C and BC respectively. Hence FE BC, DE B and DF C by midpoint theorem. If DE B then DE BF and FE BC then, FE BD FEDB is a parallelogram in which DF is diagonal and we know that diagonal of Parallelogram divides its into equal reas. Hence ar ( BDF) ar ( DEF)...() Similarly ar ( CDE) ar ( DEF)...() or ( FE) ar ( DEF)...() or ( DEF) ar ( DEF)...() On adding eqns. (), (), () and (), ar ( BDF) + ar ( CDE) + ar ( FE) + ar ( DEF) ar ( DEF) ar ( BC) ar ( DEF) ar ( DEF) ar BC ( ) WORKSHEET- Sol.. Let BC be an equilateral triangle of side cm and D is altitude which is also a perpendicular bisector of side BC. BC Hence BD cm B cm D B BD P- M T H E M T I C S - X T E R M -
( ) ( ) 76 D cm The length of the altitude is cm. Sol.. D BD CD D BD CD D DDC BD by SS (Q D 90 ) DBD CD; DC DB (Corresponding angles of similar of triangles) BD+ CD + DC + DB 80 BD + DC 80 BD + DC 90 90 Sol.. D CDE CB (alternate angles) CDE CB (By similarity rule) ar ( CB) CD CD Now ar ( DCE) D + DC C ( ) ar ( CB) CD ar ( DCE) ( ) CD 9 [CBSE Marking Scheme, 0] Sol.. In DDE and D BC, (Common) DE BC (given) \ DE BC ( similarity) D DE B BC D DE E + BE BC 7.6 7. +. DE DE 8. 7.6 8...6 cm. B E D CD In CDE and CB C C (common) Sol.. In BC, DE B CD C CE CB CD CD + D x + x+ + x x x + x + C (Given) (By Thales s Theorem) CE CE + BE x + x x x (x + ) (x ) x (x + ) x x + 9x x + x 8x x 8x x x x WORKSHEET- Sol.. BC DEF (Given) ar( BC) B ar DEF DE ( ) 00 96 B (7) 00 96 B 9 B 9 00 96 B 900 96 B B cm P-
Sol.. F Sol.. G D B E C D C B BCE CF (equilateral 's are equiangular) hence they are similar by similar criterion DBC is a right triangle. \ By Pythagoras theorem, C B + BC C BC (\ B BC in a square) C BC now ar CB C ar DCE BC E ( BC ) BC ar ( CF) ar ( BEC) ar ( BEC) ar ( CF) [CBSE Marking Scheme, 0] Sol.. No, ngle included should be same. Sol.. Given : D. cm ; BD cm and B D + BD. +.0. cm. Proof : In triangle DE and BC, is common DE BC DE BC (Corresponding angles) DE BC, ( similarity) ar( DE) D ( ). ar( BC) B (.) ar( DE) ar( BC) ar( DE) 9 ar( DE) ar(trapezium BCED) 8 9 Sol.. ccording to the question, QPR 90 QR QP + PR PR PKR 90 PK 6 00 0 cm 0 8 00 6 6 6 cm. F Draw CG FD (Construction) Given BED BDE BE BD EC...() (given that E is mid-point of BC) In BCG, DE GC BD DG BE EC (from ()) BD DG EC BE (using ()) In DF, CG FD G C GD CF (By BPT) Sol.. OB COD ar ar G GD + C CF + D GD F CF ( COD) ( OB) Sol.. Given : BC PQR To Prove : ar ar D ( BC) ( PQR) F CF D BE (using ()) WORKSHEET-6 CD B CD ( ) O CD 9 C ( similarity) B [CBSE Marking Scheme, 0] B PQ BC QR C PR Construction : Draw D BC and PE QR Proof : BC PQR P- M T H E M T I C S - X T E R M -
B D C Q B PQ BC QR C PR (Corresponding sides of similar triangles)...(i) B Q In DB and PEQ B Q DB PEQ [each 90 ] DB PEQ ( similarity) D PE B PQ (Corresponding sides of similar triangles)...(ii) From eq. (i) and eq. (ii), Sol.. In the triangles OB and DOC, O Since OC BO OD and B cm O OC BO OD P E R (Given) OB DOC (Vertically opposite angles) OB ~ DOC (SS similarity) O OC BO B OD DC ( In similar triangles corresponding sides are proportional) DC B DC 6 cm Sol.. In BC and DEF, B DE and B DF B DE ; C DF ; P P BC EF D Now ar ar ( BC) ( PQR) ( BC) ( PQR) B PQ BC QR C PR BC D QR PE BC QR D PE D PE...(iii) ar BC ar QR...(iv) [from eq. (iii)] From eq (iii) and eq (iv) ar( BC) B ar( PQR ) PQ B DE BC QR C PR WORKSHEET-7 C DF BC EF BC DEF (SSS Criterian) ar( BC) B ar( DEF) () 9. DE Sol.. ccording to the question, O 0 cm, BO cm, PB 8 cm In QO and BPO, OP BOP (Vertically opposite angles) B 90 QO BPO ( criterion) Q QO O BP PO BO Sol.. Q 8 0 Q 0 cm. D E B C E F B DE BC C P-
D DB E EC x + x x + 6 x + (By BPT ) (x + ) (x + ) x (x + 6) x + x + 6x + 0 x + 6x x + 0 6x 6x x 0 x 0 x Sol.. (ii) CE BC BC C BC CE C...(ii) On adding eq. (i) and (ii) B + BC E C + CE C B + BC C (E + CE) B + BC C C B + BC C Hence proved. D C E O B Given : B BC Construction : BE C To Prove : B + BC C Proof : In EB and BC (Common) E B (each 90) EB BC (by similarity) E B B C B E C...(i) Now, In CEB and CB C C (Common) E B (each 90 ) CEB CB (by similarity) Sol.. Since G is the mid-point of PQ, PG GQ PG GQ ccording to the question, GH QR PG GQ PH HR (BPT) PH HR PH HR. Hence, H is the mid-point of PR. C Given : BCD is a rhombus O OC C and BO OD BD C BD To Prove : B C + BD Proof : OB 90 (Diagonal of rhombus bisect each other at right angle) B O + OB B B C + BD B Sol.. ccording to the question, Given, C BD + B C + BD Hence Proved. WORKSHEET-8 ST QR PS PQ PT (By BPT) PR PS PQ and PR 8 cm P-6 M T H E M T I C S - X T E R M -
PT 8 PT 8 Sol.. To prove : B + C D + BC B E D C 6.8 cm. Draw E BC In DBE B E + BE (Pythagoras theorem) B D DE + (BD DE) D DE + BD + DE BD DE B D + BD BD DE...() In DEC C E + EC C (D ED ) + (ED + DC) D ED + ED + DC +ED DC C D + CD + ED CD C D + DC + DC DE...() dding eqns. () & (), B + C (D + BD ) ( BD DC) D + BC (as BD BC) or B + C (D + BD ) Hence proved. Sol.. So, PQ MN KP PM KQ QN KP PM KQ KN KQ KQ 0. KQ 0. KQ KQ 7 KQ 0. (By BPT) Sol.. B D C Given : P B C, Q R Prove : let P x In BC + B + C 80 x + B + B 80 ( B C given) B 80 x 80 x B...(i) Now, In PQR P + Q + R 80 x + Q + Q 80 ( Q R given) Q 80 x 80 x Q...(ii) In BC and PQR, P [Given] B Q (from eq. (i) and (ii)) BC PQR ( similarity) ar( BC) D ar( PQR) PE Q 6 D PE D PE D PE P E R WORKSHEET-9 KQ 0. 7.8 cm. Sol.. Draw C intersecting EF at G. B D E G F C P-7
In CB, GF B G CG BF FC In DC, EG DC E ED From equations () and (), E ED G CG (By BPT)...() (By BPT)...() BF FC Sol.. In CB, B (Given) C CB (By isosceles triangle property) But, D BE (Given)...() C D CB BE Dividing equation () by (), CD D CE BE CD CE...() By converse of BPT, DE B. N Sol.. W,000 O S,00 Sol.. Given B DE and BC 8 cm BC DEF So, B BC DE EF B 8 B EF EF 8 6 cm. Sol.. From given figures, PQ ZY. 8. ; E Distance covered by first aeroplane due North after two hours 00,000 km. Distance covered by second aeroplane due East after two hours 60,00 km. Distance between two aeroplane after hours Sol.. Given NE ON + OE (,000) + (,00) 0,00,000 + 6,90,000 6,90,000 60. km. B + D BC + CD D BC + CD B D h + d x In rt CD D C + DC (h + d x) (x + h) + d (h + d x) (x + h) d (h + d x x h) (h + d x + x + h) d (a b ) (a b) (a + b)) (d x) (h + d) d hd + d hx xd d PR ZX QR YX PQ ZY hd hx + xd (h + d) x hd x h+ d [CBSE Marking Scheme, 0] WORKSHEET-0 6 ; 7 PR ZX QR YX PQR ZYX (SSS) X R X 80 (60 + 70 ) 0 Sol.. Proof : In OB and COD, B CD D C O B P-8 M T H E M T I C S - X T E R M -