REAL NUMBERS TOPIC-1. Euclid s Division Lemma and Fundamental Theorem of Arithmetic WORKSHEET-1 WORKSHEET-2 P-1. qqq SECTION. Solutions.

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1 CHPTER SECTION REL NUMERS TOPIC- Euclid s Division Lemma and Fundamental Theorem of rithmetic WORKSHEET The given number ends in. Hence it is a multiple of. Therefore it is a composite number. [CSE Marking Scheme, 0]. k k. HCF of k. k. k. is k. [CSE Marking Scheme, 0] 90. and HCF 8 LCM 0 [CSE Marking Scheme, 0] and HCF of 0 and 8 HCF of and HCF of and 80 HCF of 80, and is. [CSE Marking Scheme, 0]. LCM of 8, and is. ) ( Remainder Required number 9,99,9.. Prime factors of : HCF No, because HCF LCM Product of three numbers.. WORKSHEET- y 9 8 [CSE Marking Scheme, 0].. 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. and. 0. Given HCF (0,,) 8 LCM (0,,)? Let 8 a 0 b, 8 We know that a b LCM (a, b) HCF (a, b) 0, LCM (a, b) 8 0, LCM (a, b) 8 LCM (0,,),8 S OLUT I ONS,0 P-

2 Let n be any positive integer. y 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 +. [CSE Marking Scheme, 0]. Fundamental theorem of arithmetic : Every composite number can be epressed as the product of powers of primes and this factorization in unique LCM 980 [CSE Marking Scheme, 0] WORKSHEET-. The smallest prime number is and the smallest composite number is. Hence, required HCF (, ) 9, 009. a, 00 b Since and, 00, so c or d or Composite number, ( 0 + ) (909 + ) (90) a composite number [QProduct of more than two factors] [CSE Marking Scheme, 0]. y using Euclid s Division Lemma, we have gain we apply Euclid s Division Lemma of divisor,8 and remainder 8 +, HCF (990, 8) Now, using Euclid s Divison Lemma on and, we have + gain, applying Euclid s Division Lemma on divisor and remainder + 0 Now, HCF (, ) HCF (, ) Hence, HCF of 990, 8 and is.. If n ends with 0, then it must have as a factor. ut we know that only prime factor of n are and. n ( )n n n From the fundamental theorem of arithmetic, we know that the prime factorization of every composite numbers is unique. n can never end with 0. [CSE Marking Scheme, 0] WORKSHEET-. HCF of and P- 9 M TH EM T I C S - X

3 . ( ) + ( + ) (9 + ) 9 and ( ) + ( + ) (8) which is a composite number (more than one prime factor).. ) 0 ( )( 0)( 90 )0(0 0 0 Hence HCF (,,,0) [CSE Marking Scheme, 0],00 ( ) The smallest natural number is. [CSE Marking Scheme, 0]. To find LCM (9,, ) 9 LCM (9,, ) 80 minutes.. The bells will toll together after 80 minutes. HCF (, ) LCM (, ) 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. LHS RHS Hence our answer is correct.. Fundamental theorem of rithmetic : Every composite number can be epressed as a product of primes and decomposition is unique, apart from the order in which the prime factors occur. HCF LCM 0 LCM 0 Let s calculate. not an integer HCF Since LCM is always a multiple of HCF, hence, two numbers cannot have HCF and LCM as and 0 respectively. [CSE Marking Scheme, 0] WORKSHEET-. We know that a b HCF (a, b) LCM (a, b),800 LCM (a, b), 800 LCM (a, b) 0. Let a be any positive integer a q + r, where 0 r < a q, q +, q +, q +, q +, q + a q, q +, q+ [CSE Marking Scheme, 0]. y Euclid s division algorithm, a bq + r Take Since 0 r <, S OLUT I ONS b r 0,,, ny positive odd integer is of the form (q + ) or (q + ).. No. y division algorithm ut a is an even integer 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. ut q +, q + are odd, as they are not divisible by. does not divide. LCM of two numbers should be eactly divisible by their HCF. Two numbers cannot have their HCF as and LCM as.. y Euclid s division algorithm, P-

4 and 8 HCF (0, 9) 9 0 LCM (0, 9) HCF (0, 9) LCM (0, 9) 0 90 Product of two numbers HCF LCM Product of two numbers. We have HCF m m m and Hence, Now, LCM 8 TOPIC- Irrational Numbers, Terminating and NonTerminating Recurring Decimals WORKSHEET The decimal epansion of a rational number p terminates, if the denominator of rational no., q when p and q are co-primes and q can be epressed as mn. where m and n are non-negative integers. e.g [CSE Marking Scheme, 0]. Since Thus smallest rational number is 00 ut, we know that rational. Let , If possible, let a b be a rational number., where a and b are integers and co-primes Squaring both sides, we have a b a b a is divisible by a is divisible by. we can write a c for some c (integer) (c) b is an irrational number (i) 0.8 Thus, our assumption is wrong. Hence, is an irrational number Subtracting, [CSE Marking Scheme, 0]. Let be a rational number, which can be put a in the form, where b 0 ; a and b are co-prime. b a b a b...(i) 9c b b c b is divisible by b is divisible by P-...(ii) M TH EM T I C S - X

5 From equation (i) and (ii), we have p q is rational number is p q q p q q is a factor of a and b which is contradicting the fact that a and b are co-primes. Thus, our assumption that wrong. Hence, is an irrational number. (ii) let us assume to contrary that + is a rational number. + p, q 0 and p, q Integer q + p q p q and q both are integers, hence is a rational number. ut this contradicts the fact that is irrational number. Hence + is an irrational number. WORKSHEET-. rational number is either terminating or nonterminating repeating. n irrational number is non-terminating and nonrepeating. [CSE Marking Scheme, 0] 8. 0 primes. Hence our assumption that, number, is false. So. Let is an irrational number. a, b (a, b are co-prime integers and b 0) Decimal epansion, a 0 00 is a rational be a rational number. Denominator 00. b a a is divisible by a is divisible by...(i) We can write a c for some integer c a 9c b 9c b c b is divisible by b is divisible by...(ii) From equations (i) and (ii), we get as a factor of a and b, which is contradicting the fact that a and b are co b a b Squaring, divides a 0. [CSE Marking Scheme, 0]. Let be a rational number. a. b S OLUT I ONS a b 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 are integers and co-primes) On squaring both the sides, divides a. a and b have as a common factor. ut this contradicts that a, b have no common factor other than. Our assumption is wrong. P-

6 Hence, is irrational. Let be rational a, where a and b are b integers, b 0 b is rational but a is not rational. a Our assumption is wrong. b b a is irrational. WORKSHEET-8 HOTS & Value ased Questions. Let Now lies between the perfect squares of () and () Prime numbers less than are,,,,,,, 9,. Since is not divisible by any of the above numbers. So, is a prime number.. The required number is the LCM of,,,,,,, 8, 9, 0 LCM 0. Let the number of columns be. is the largest number, which should divide both 0 and HCF of 0 and 9 is 8 Hence, 8 columns are required.. Using Euclid s algorithm, the HCF (0, ) (i) 0 HCF (0, ) ( 0 )...(ii)...(iii) [From (ii)] [From (i)] ( + ) 0 + ( ), y lso, 0 + ( ) () + ( ), y Hence, and y are not unique.. (i) Required number of minutes is the LCM of 8 and. 8 and LCM of 8 and Hence, Ravish and Priya will meet again at the starting point after minutes. (ii) LCM of numbers. (iii) Healthy competition is necessary for personal development and progress. P- M TH EM T I C S - X

7 CHPTER SECTION POLYNOMILS TOPIC- Zeroes of a Polynomial and Relationship between Zeroes and Coefficients of Quadratic Polynomials WORKSHEET-9. Factors of + + : Coefficient of Sum of the roots. Coefficient of ( + ) +( + ) 0 b a + b - a b a. Product of roots (zeroes) a α. α ( ) ( ) ( ) ( ), Zeroes are,. Given, a and are the zeroes of + + a. α a p () p + q but p( ) p + q 0 p + q 80 p q 80...(ii) and p( ) ( ) + ( ) + ( ) + p( ) + q but p( ) p + q 0 p + q p q...(iii) On solving eq. (ii) and eq. (iii) by elimination method, we get p q 80 p q + + p Constant term Coefficient of a α a a Let...(i) If p() is eactly divisible by + +, then and are zeroes of p() [from eq (i)] We know that, Product of the zeroes p( ) ( ) + ( ) + ( ) + p( ) + q a a [CSE Marking Scheme, 0]. Q f () ( ) i.e. f () 0 0 or Hence, zeroes are 0 &. p (), p() p + q. Let ( + ) ( + ) 0 On putting the value of p in eq. (i), ( ) q 80 0 q 80 q 0 80 q 0 q 0 Hence, p, q 0 S OLUT I ONS P-

8 WORKSHEET-0. The number of zeroes of p() is.. Sum of zeroes, Product of zeroes 9 Quadratic polynomial is + 9 lso ( ) ( ) 0, Hence zeroes are,.. ccording to the question, Sum of zeroes Sum of zeroes, (a + b) Product of zeroes, (ab) 0 0 Coeff. of c Hence a and c a 8 a + a 8a -. Given, polynomial, f() a + c Let the zeroes of f () are a and b, then according to the question Coeff. of 0 c and ( + ) a+b b a So, quadratic polynomial (Sum of zeroes) + Product of zeroes Now ab Constant term Coeff. of. Let a and b be the zeroes of the polynomial, then as per the question ( + ) and and Product of zeroes a a a a k + a k + k + k + 9 k k + a WORKSHEET-. Let. p() p() ( ) ( ) 0 ( ) ( ) The zeros are,, Hence, zeroes of the polynomial are P-8,. ) + ( Remainder Hence must be added [CSE Marking Scheme, 0] M TH EM T I C S - X

9 p() + + ( + ) ( + ) ( ) ( + ) m and n. Let So, zeroes are, Now, We know that - - Product of roots pq Constant term Coefficient of f () + Coefficient of Sum of roots p + q Coefficient of. m n n m (p + q) p + q + pq p + q (p + q) pq [CSE Marking Scheme, 0] p(y) y y + a + b -. ab and Now α+β / + αβ α β / and α β αβ The required polynomial is y y + [y y + ]. WORKSHEET-. p() a + b + c be the zeroes of p(), then Let a and α c α a So, required condition is, c a. Since, is a zero of the polynomial p() k + k, then p ( ) 0 k ( ) ( ) + k 0 k++k 0 k + 0 k Hence, k Product of zeroes, a and ab c a ab ab a + b ab.. Given, p() and a and b are its zeroes. Sum of zeroes a + b If a and b are the zeroes of +. then b a+b a a+b a+b S OLUT I ONS ( ) Coefficient of Coefficient of - Product of zeroes ab + 0. Constant term Coefficient of For the new polynomial, α+β Sum of zeroes + αβ α β P-9

10 Product of zeroes α β αβ Required quadratic polynomial (Sum of zeroes) + Product of zeroes b -8 8 a a b c a α+β αβ α β Sum of zeroes Product of the zeroes ( + ) ( + ) a+b α β αβ Now for making a polynomial.. From the given polynomial we will find the value, the sum of the zeroes, and the multiple of the zeroes. p() (a + b) + ab The polynomial is : p() ( ) WORKSHEET f() 8 Let other zero be k, then - Sum of zeroes + k k 8 Sum of zeroes. and ab a then b. Given, f() Let one zero be a The other Sum of zeroes Sum of zeroes a+b a a + b 0 Constant term Product of zeroes Coefficient of b k 9 a, a + ( a) 0 Coefficient of Sum of zeroes Coefficient of. k Coefficient of Coefficient of Relation between zeroes and coeff. of polynomial is verified. p () ( ), p () 0. Constant term ( ) 0 zeroes are 0 and Sum of zeroes k k 0 k 0. f() + Coefficient of Product of zeroes - ccording to the question, Sum of zeroes 9 0, - are zeroes of polynomial +. Coefficient of Coefficient of 9 f and Product of zeroes 0 Coefficient of Coefficient of Constant term Coefficient of [CSE Marking Scheme, 0] f + - P-0 M TH EM T I C S - X

11 TOPIC- Problems on Polynomials WORKSHEET- + Remainder. + + g() ( ) + g() + ) + ( Hence g(). [CSE Marking Scheme, 0]. Thus, and + + ) Quotient + Remainder + [CSE Marking Scheme, 0]. + ) ( + + Hence ( + ) must be added [CSE Marking Scheme, 0] + + ) Quotient + ; Remainder + p() g() q() + r() ( + ) ( + ) + ( + ) Verified. and are zeroes. ( + ) ( + ) ) ( The third zero is. [CSE Marking Scheme, 0] WORKSHEET-. g() + +, f() ) S OLUT I ONS Remainder, r() r() 0, g() is not a factor of p(). [CSE Marking Scheme, 0] P-

12 . + ) Remainder So should be added. [CSE Marking Scheme, 0] +. For eact division, remainder is zero, then + ) a + b a (a + ) + b (a + ) + b 0 a + 0, b 0 a, b. + (8 k). + k) k + + ( k) k + + (8 k) ( k) + 0 (8 k) ( k) + (8k k) + (k 9) + (0 8k + k) Given, remainder + a 0 k 9 k and (a + ) + b a 0 8k + k [CSE Marking Scheme, 0] WORKSHEET ) ( a + b + On comparing, we get a and b. [CSE Marking Scheme, 0] + +. ) P- + Remainder 0 is not a factor of the given polynomial.. p() p() () () () () is a factor of p(). p( ) y long division, + + ) M TH EM T I C S - X

13 p( ) ( + ) ( + ), are the other zeroes of p() ll the zeroes of p() are,,. WORKSHEET Hence should be added to to make it completely divisible by ) Thus, Quotient and Remainder +.. s and are the zeroes of + ) Quotient, Remainder 0 [CSE Marking Scheme, 0]. On dividing by +, remainder should be zero.. + ) ( , So ( ) and ( + ) are the factors of + 8 Now, ( ) ( + ) Dividing + 8 by 8 8) + 8 ( ( 8) ( + ) ( 8) ( + ) ( ) ( + )() ( + ) 9 + Hence other two zeroes are 0 and -. WORKSHEET-8 HOTS & Value ased Questions. If + is a factor then ( ) is a root. So f ( ) ( ) + 0 ( ) + ( ) + ( ) + k 0 (when it is a root the polynomial should be equal to zero when value is substituted) k 0 + k k 0 k ( 9) S OLUT I ONS. s ± ( are the zeroes of p(), so ± are the factors of p(). ) { ( )}{ ( )} {( ) }{( ) + } ( ) ( ) Now, + + Dividing p() by + P-

14 +) { ( p() + )} { ( )} ( ) s, ( + )( ) Hence, other two zeroes of p() are and.. Dividend (Divisor Quotient) + Remainder Hence, we get 9 g() ( ) + ( + 8) g() ( ) 9 0 g() ( ) g() 9 0 ( ) + + ) Therefore, g() + + P- M TH EM T I C S - X

15 CHPTER SECTION PIR OF LINER EQUTIONS IN TWO VRILES TOPIC- Graphical Solution of Linear Equations in Two Variables Consistency/Inconsistency WORKSHEET-9 a, a. Here a, b, c and a, b, c Clearly, a b c a b c b - b - Hence lines are coincident. [CSE Marking Scheme, 0]. The given equations can be re-written as : + y 0 y 0 On comparing with a + by + c 0, we have a, b, c a, b, c i.e. a b, - a b Thus, i.e. -8 c c and 8 9 Thus, a b c a b c i.e. Hence, the pair of linear equations is inconsistent.. Let amounts contributed by two sections X- and X- be ` and ` y. + y,00...(i) y 00...(ii) - 00 a b a b Hence, the pair of linear equations is consistent.. The given equations can be re-written as y 8 0 y 9 0 On comparing with a + by + c 0, we have a, b, c 8 a, b, c 9 (00, 800) ,00 Point of intersection (00, 800) Hence 00 and y 800. [CSE Marking Scheme, 0] WORKSHEET-0. Since, S OLUT I ONS am bl a b c l m n So, a + by c and l + my n has no solution. P-

16 . The given equations can be re-written as : + y 0 9 0y 0 On comparing with a + by + c 0, we have a, b, c a 9, b 0, c a 8 a - b b -0 Hence, the pair of linear equations is consistent.. The pair of linear equations is : y 8 y 8 0 (i) and y y 0 (ii) On comparing with a + by + c 0, we have a, b, c 8 a, b, c a b c,, b c a a b c a b c s - Thus, a b a b i.e., the lines are parallel having no solution. So, the pair of linear equations is inconsistent. WORKSHEET-. ad bc a b c d Hence, the pair of given linear equations has unique solution.. i.e., a b a b y 0 y + y Hence, the pair of linear eqations is consistent. [CSE Marking Scheme, 0]. Yes, it is consistent. We have, for the equation + y 9 0 a, b and c 9 and for the equation, + y 8 0 a, b and c 8 a Here a b b y Graph of two equations is : 0 y (, ) + y ' and It is clear that a b c a b c (, ) (, ) c -9-8 c y (, 0) (, 0) 0 Marks Hence, system is consistent and dependent.. Given linear equation is + y 9 (i) Intersecting line is y 0 (ii) Coincident line is + 8y 8. [CSE Marking Scheme, 0] P- + y - y. y' Lines meet -ais at (, 0) and (, 0) respectively. [CSE Marking Scheme, 0] M TH EM T I C S - X

17 WORKSHEET-. Given equations are : + py () and + y () The condition of unique solution, a b a b p. Pair of linear equations k y 0 and y 9 0 a b c Condition for infinite solutions : a b c k k y y (, 0) ' y 0 (, ) (, ) (, ) (, 0) y (, ) 0 y' k + k+ k + y a b c For infinitely many solutions a b c. y k k. For equation, + y 0 a, b, c For equation, (k + ) + y (k + )0 a k +, b, c (k + ) y + y 0 - y p p or Hence, (i) These lines intersect each other at point (, ). Hence and y. (ii) The vertices of triangular region are (, ), (, 0) and (, 0). 8 (iii) rea of rea sq unit. TOPIC- lgebraic Methods to Solve Pair of Linear Equations and Equations Reducible to Linear Equations WORKSHEET- and + y 0 y 0 y +. S OLUT I ONS y 0 y [CSE Marking Scheme, 0]. Let age of father and son be and y respectively. + y 0...(i) y...(ii) y solving eqns. (i) and (ii), we get 0 and y 0 ges are 0 years and 0 years. [CSE Marking Scheme, 0] P-

18 lternative Method : Let age of father and son be and y years respectively. + y 0...(i) y...(ii) Substitute the value of from eqn. (ii) in, (i) y + y 0 y 0 y 0 yrs. From (i), yrs. ge of father is 0 years and ge of son is 0 years.. Substitute X and y Y X + Y X 9Y Multiply eqn. (i) by, X + 9Y dding eqn. (iii) and eqn. (ii), X 9Y X + 9Y X 0X X 0. Let cost of chair ` and cost of table ` y ccording to the question, +y 00...(i) +y 0...(ii) Multiplying eqn. (i) by and eqn. (ii) by, 8 +y 00...(iii) + y 0...(iv) eqn. (iv) eqn. (iii) 00 0 Substituting the value of in (i), y 00 Cost of chair and table ` 0, ` 00 respectively. [CSE Marking Scheme, 0] Now...(i)...(ii)...(iii) Y X Y y9 y Thus, y 9 is the solution. WORKSHEET-. and + y 0...(i) + y 0...(ii) From eqn. (ii). + y 0 y Value of y put in eqn. (i) + y 0 + ( ) Substitute the value of in eq (ii) to get value of y, + y 0 () + y 0 +y 0 y 0 y Hence, values of and y are and respectively. y P-8 y - [CSE Marking Scheme, 0] lternative Method : Given : + y 0 y 0 0 y cross-multiplication method, y (i)...(ii) y y - and and y -. y + 0 y + 0 Multiplying eqn. (i) by, we get ( y + ) 0 0 y + 0 Subtracting eqn. (ii) from eqn. (i), we get...(i)...(ii) M TH EM T I C S - X

19 0 y + 0 y + 0 ( ) (+) ( ) Substituting the value of in eqn. (i), y + ( ) y + y+ y y y Hence, and y WORKSHEET-. y elimination method : Given, + y 0 (i) and y (ii) On multiplying eqn. (i) by and eqn. (ii) by and adding, ( + y) + ( y) y + y 0 + On substituting in eqn. (i), + y 0 y y Hence, and y y substitution method : Given, + y 0 (i) and y (ii) From eqn. (ii), y y (iii) On substituting y from eqn. (iii) in eqn. (i), + ( ) 0 From eqn. (iii), y y Hence, and y. y elimination method : Given, y (i) and 9 y + (ii) On multiplying eqn. (i) by and eqn. (ii) by, 9 y (iii) 9 y (iv) On subtracting eqn. (iv) from eqn. (iii), (9 y) (9 y) y + y y y On substituting y - in eqn. (i), Hence 9 and y y substitution method : Given, y and 9 y + From eqn. (ii), y y (i) (ii) (iii) On substituting y from eqn. (iii) in eqn. (i), On substituting 9 in eqn. (iii), y y Hence, 9 and y S OLUT I ONS P-9

20 WORKSHEET-. Given, a pair of linear equations is : s t s t+ s t and +...(i)...(ii) On substituting s t +, from eqn. (i) in eqn. (ii), we get t+ t + (t + ) + t t + t 0 t From eqn., (i), s +9 Hence, s 9, t. Let the actual speed of the train be km/hr and actual time taken be y hr. Distance Speed Time Q y km ccording to the given condition, y ( + 0)(y ) y y + 0y 0 0y y [divide by ] (i) and y ( 0)(y + ) y y + 0y 0 0y 0 0 (ii) On multiplying eqn. (i) by and subtracting eqn. (ii) from eqn. (i), ( y + 0) ( 0y 0) 0 y 0 y On substituting y in eqn. (i), Hence, the distance covered by the train 0 00 km.. Let the incomes of two persons be and. lso the ependitures of two persons be 9y and y. and 9y 00...(i) y 00...(ii) Multiplying eqn. (i) by and eqn. (ii) by 9 and subtracting, y,000...(iii) y, (iv) On subtracting, , Substituting this value of in eqn. (i), 00 9y 00 9y 00 00,800 y, Their monthly incomes are 00 ` 00 and 00 ` 00.[CSE Marking Scheme, 0] WORKSHEET-. Given, a pair of linear equations is : y or y...(i) and 9 y 9...(ii) On substituting y from eqn. (i) in eqn. (ii), 9 ( ) 9 i.e., 9 9 It is a true statement. Hence, eqn. (i) and (ii) have infinitely many solutions.. Let the two numbers be and y ( > y) We are given that, y (i) and y (ii) On substituting from eqn. (ii) in eqn. (i), we get y y On substituting y in eqn. (ii), we get 9 9 Hence, the two numbers are 9 and.. Let the speed of the car I from km/hr. Speed of the car II from y km/hr. Same dirrection : Distance covered by car I 0 + (distance covered by car II) 0 + y y 0 y 0...(i) y y P-0 M TH EM T I C S - X

21 Opposite direction : Distance covered by car I + distance covered by car II 0 km + y 0...(ii) dding eqns. (i) and (ii), 0 80 Substituting 80 in eqn. (i), y 0 Speed of the car I from 80 km/hr and speed of the car II from 0 km/hr. [CSE Marking Scheme, 0]. Let the fraction be. y lso, - y+ y + y - y+ + 8 y 9 y Subtracting eqn. (ii) from eqn. (i), y y Substituting this value of y in eqn. (i), 8 Hence, 8, y Fraction 8...(i)...(ii) WORKSHEET-8. Given, a pair of linear equations is : y....(i) and y....(ii) From eqn. (i), + y [ Q Multiply by 0] y - y...(iii) On substituting y from eqn. (iii) in eqn. (ii), ( - ) ( ) + ( ) On substituting in eqn. (iii), we get - 9 y i.e., y Hence,, y. Let the supplementary angles be and y ( > y ) Now, + y 80 (i) and y 8 (ii) From eqn. (ii), y 8 (iii) On substituting y from eqn. (iii) in eqn. (i), On substituting 99 in eqn. (iii), y y 8 Hence, the angles are 99 and 8. Let the number of red balls be and white balls be y. ccording to the question, y or y 0...(i) and ( + y) y or y...(ii) Multiplying eqn. (i) by and eqn. (ii) by and then subtracting, we get 9y 0 8y + y y 0 Substituting in eqn. (i), 8 8, y Hence, number of red balls 8 and number of white balls [CSE Marking Scheme, 0] S OLUT I ONS P-

22 WORKSHEET-9. Let length and breadth y Then according to the first condition, ( )(y + ) y 9 y...(i) ccording to the second condition, ( +)(y +) y + + y...(ii) Multiplying eqn. (i) by and eqn. (ii) by and then adding, 9 y y 0 9 Substituting this value of in eqn. (i), () y y y 9 Hence, perimeter ( + y) ( + 9) units.. ( y) y...(i) ( + y) y...(ii) Divide eqns. (i) and (ii) by y, and - y...(iii) + y...(iv) a and b, y Let then equations (iii) and (iv) become a b...(v) a + b...(vi) Multiplying eqn. (v) by and then adding with eqn. (vi), 0a 0 a Substituting this value of a in eqn. (v), b a y Now y b and. WORKSHEET-0. Let the sum of the ages of the children be and the age of the father be y years. y i.e. y 0...(i) and 0 + y + 0 y 0...(ii) Subtracting (ii) from (i), 0 From (i), y 0 0 y 0 Hence, the age of the father 0 years. [CSE Marking Scheme, 0]. Let the number of students in the class be and the number of rows be y. The number of students in each row y ccording to the given condition, + (y ) y + y y y + 0 y (i) and - (y + ) y + y 0 y On putting and y y (ii) u in eqn. (i) and (ii), y u y + 0 (iii) u y 0 (iv) On subtracting eqn. (iii) from eqn. (iv), u 9 0 u 9 On substituting u 9 in eqn. (iii), 9 y + 0 P- M TH EM T I C S - X

23 y y Now, u 9 (v) 9 y [from eqn. (v)] The number of students in the class. WORKSHEET-. Let the ages of ni and iju be yr. and y yr, respectively. ccording to the given condition, y ± (i) lso, age of ni s father Dharam years y nd age of iju s sister years ccording to the given condition, y 0 y 0 Case I : When y On subtracting eqn. (iii) from eqn. (ii), 9 years On putting 9 in eqn. (iii), 9 y y years Case II : When y On subtracting eqn. (iv) from eqn. (ii), 0 + (ii) (iii) (iv) years On putting in eqn. (iv), we get y y years Hence, age of ni is 9 year and age of iju is years or age of ni is years and age of iju is years.. Let the amount of their respective capitals be ` and ` y. ccording to the given condition, + 00 (y 00) y 00 (i) and ( 0) y + 0 y 0 (ii) On multiplying eqn. (ii) by and subtracting from eqn. (), y 00 + y On substituting 0 in eqn. (i), 0 y 00 y 0 y 0 Hence, the amount of their respective capitals are ` 0 and ` 0. S OLUT I ONS P-

24 CHPTER SECTION QUDRTIC EQUTIONS TOPIC- Solution of Quadratic Equations WORKSHEET ( ± Positive root [CSE Marking Scheme, 0] or + 0 or Roots of equation are and. [CSE Marking Scheme, 0]. + 0 ( ) + 0 ( ) 0, ( + ) ( ) ( ) ( + ) ( ) ( + ) 0, [CSE Marking Scheme, 0] Let ) 0 [CSE Marking Scheme, 0] + ( + ) ( + ) ( + )( ). )( or y ( - ) y + y 0 (y + 8) (y ) 0 y, 8 y y 8 [CSE Marking Scheme, 0] WORKSHEET-. Putting - in + k 0 - P- + k - 0 k 0 k - k M TH EM T I C S - X

25 -9 [CSE Marking Scheme, 0] k., ( + ) ( + ) 0 ( + )( ) 0,. [CSE Marking Scheme, 0, 0] - ( +) + 0 [CSE Marking Scheme, 0] ( + ) ( + ) ( + ) + ( ) + ( + 9) ( + ) ( + )( + ) ( + )( + ) ( + + ) ( - ) ± ( - ) - ()( -8 ) ± 8 ± ± + or ( - ) - ( - ) 0 ( - ) ( - ) 0, [CSE Marking Scheme, 0] WORKSHEET ( ( )( +) 0 - ( + )( ) 0 [CSE Marking Scheme, 0, 0], ) ( ) + ( ) 0 ( )( + ) 0, [CSE Marking Scheme, 0, 0] S OLUT I ONS ( ) +( ) 0 ) +( ) 0 ( + )( ) (., [CSE Marking Scheme, 0]. ( ) ( + ) ( ) ( ) + or or + 0 ( ) ( ) 0, [CSE Marking Scheme, 0] or P-

26 . a( b)( c) + b( a)( c) c( a)( b) (a + b c) + ( ab ac ab bc + ac + bc) 0 (a + b c) + ( ab + ac + bc) 0 ac + bc - ab a + b - c [CSE Marking Scheme, 0] WORKSHEET- lternate Method : (9 b + b) a 0 ( b) (a) 0 ( b + a) ( b a) 0, a, C (a b). Here, a ± ( - a ) - ( a - b ) a ± a - a + b 8 8 a ± b a ± b 8 and a ± b +. a a+b a a+b a a + b a a + b. Here, {( b ± ( -b ) a - b )} 9 b ± b + a - b 8 b ± a 8 a + b b - a, ( + )( ) ( + ) ( + ) 0 b ± a 8 a 9, b b, c (a b ) a - b a + b, - a -( a + b ), a a+b [CSE Marking Scheme, 0] [CSE Marking Scheme, 0] lternative Method : ( a + a) b 0 ( a) (b) 0 ( a + b) ( a b) 0 a a a+b a + b a + b a a + b a - b 8 b - a b + a, ( + )( ) 0, [CSE Marking Scheme, 0] P- M TH EM T I C S - X

27 WORKSHEET- ( ) 0 0 or. Let time taken by pipe be minutes. and time taken by pipe be + minutes. In one minute pipe will fill, tank. + b + b a 0 ( + b) a 0 ( + b + a) ( + b a) 0 - (a + b), a-b In one minute pipe will fill [CSE Marking Scheme, 0] + oth pipes + will fill + tank in one minute +. a (b a) 0 Given equation can be written as a + a b 0 or ( a) (b) 0 ( a + b) ( a b) 0 a b, a + b [CSE Marking Scheme, 0] +. ( - )( - ) ( - )( - ) Then according to the question ( + ) 00 00( + ) 9( + ) ( - )( - )( - ) - ( - )( - )( - ) ( - ) ( - )( - )( - ) ( - )( - ), tank + 9( 0) + ( 0) 0 ( 0)(9 + ) 0 0, - 9 rejecting negative value, 0 minutes ( ) ( ) + 0 and + minutes Hence pipe will fill the tank in 0 minutes and pipe will fill in minutes. WORKSHEET- The roots are. Here, a, b p, c (p q ) The roots are given by the quadratic formula p ± p - p + q p ± q 8 S OLUT I ONS 8 [CSE Marking Scheme, 0] - b ± b - ac a p ± p ( p q ) p+q p-q,.. Let the side of the smaller square be y and the side of the longer square by, then. y y y+ ccording to the question, + y 8 (y + ) + y 8 y + y + 8 y + y 0 P-

28 y + y 0 (y + 8)(y ) 0 y 8, Rejecting y 8, as side can not be negative y and 8 [CSE Marking Scheme, 0, 0]. a( a) + b( b) [ (a + b) + ab] a a + b b (a + b) + ab (a + b) + (a + b) 0 (a + b) (a + b) + (a + b) 0 [ (a + b)] [ (a + b)] 0 lternative Method : We know that can be written as - - [ - ] a+b [CSE Marking Scheme, 0] ( - ) + ( - ) 0 ( - )( + ) ( - )( + ) 0 a + b,. or - -. Let the speed while going be km/h 0 0 Therefore ( + 0)( 0) 0 0 Speed while going 0 km/h and speed while returning 0 km/h [CSE Marking Scheme, 0], + [CSE Marking Scheme, 0] WORKSHEET-8. Given, + ( ) or [ ] + [ ] 0 ( )( + ) 0, Squaring both sides ( ) ( ) ( )( ) [CSE Marking Scheme, 0]. Substituting in a + + b 0 a+ +b 9 a + + 9b a + 9b 9a + b 9a + b Solving (i) and (ii), a P and b...(i) ( ) or ( + ) - ( + ) 0 ( + ) ( - ) 0...(ii) + ( ) -, + [CSE Marking Scheme, 0] M TH EM T I C S - X

29 . Let the fraction be +, [ + ( + ) ] 8 ( + ) 0 Fraction ( rejected) + [CSE Marking Scheme, 0] WORKSHEET-9 Required quadratic equation, 0 0 [CSE Marking Scheme, 0]. On completing the square, ( ) 8 0 ( ) 0 ( ) ( ) ( ). + (a + a ) 0 - ± + ( a + a - ) ± ± +, [Q [CSE Marking Scheme, 0]. Let units digit and tens digit of the two digit number be and y Number is 0y + ccording to question, 0y + (y + ) 0y + y + 0y y y y...(i) lso, 0y + y...(ii) 0y + y (y)y y y y y 0 y (y ) 0 y 0 or y Rejecting y 0 as the number can not be zero. Required number is. - ± ( a + ) a - - a -, a, (a + ) [CSE Marking Scheme, 0] +. c cm a C Here b a + b + c 0, c a + b Using Pythagorus theorem a + b Using identity (a + b) a + b + ab + ab. Let the number of wickets taken by Zahir be. The number of wickets taken by Harbhajan ccording to question, ( ) 0 0 -b ± b - ac ] a rea of ab 00 C ab 0 cm. [CSE Marking Scheme, 0] S OLUT I ONS P-9

30 TOPIC- Discriminant and Nature of Roots WORKSHEET-0 i.e.,. (k + ) (k + ) + 0 has equal roots D0 b ac (k + ) (k + ) k+ k0 [CSE Marking Scheme, 0]. is the root of + k + 0 () + k + 0 k + 0 k 8 Put k 8 in + k + q q 0 For equal roots ( 8) ()q 0 q 0 q q [CSE Marking Scheme, 0]. Given, is a root of the equation, + p 8 0 Putting in + p p 8 0 p Given, p + k 0 has equal roots. + + k D () ()(k) k k k k( ) has equal roots b ac k k Here, a k, b k, c 0 Given, roots are equal, D b ac 0 Q ( k) k 0 0 0k 0k 0 0k(k ) 0 k(k ) 0 k 0 k. For equal roots of + p + mn 0, p mn 0 p mn...(i) For equal roots of (m + n) + (m + n + p) 0 (m + n) (m + n + p) 0 m + n + mn m n (mn) 0(From (i)) If roots of + p + mn 0 are equal then those of a(m + n) + (m + n + p) 0 are also equal. [CSE Marking Scheme, 0] WORKSHEET-. a 9, b k, c k Since roots of the equation are equal b ac 0 ( k) ( 9 k) 0 9k k 0 k k 0 k (k ) 0 k 0 or k Since k 0 is not possible for the equation. Since the roots are equal, then D 0 D b ac 0 ( k) (k)() 0 k k 0 k(k ) 0 k 0, ut k 0, as coefficient of can not be zero k [CSE Marking Scheme, 0] k. P-0 M TH EM T I C S - X

31 . b ac ( ) ()() b ac 0 ( 8) k 0 k k...(ii) From (i) and (ii), we get k For k, given equations will have equal solution. [CSE Marking Scheme, 0] Roots are real and equal. Roots are b b -, - a a or,. (i) For + k + 0 to have real roots k 0 [CSE Marking Scheme, 0]. For equation + k + b ac k k k and for equation 8 + k ± 0...(i)...(i) (ii) For 8 + k 0 to have real roots k 0...(ii) For (i) and (ii) to hold simultaneously k [CSE Marking Scheme, 0] WORKSHEET-. s the equation has equal roots i.e., D b ac 0 p p 8 p p b ac (0) ( )( ) 00 s b ac < 0 So, the equation has no real roots. [CSE Marking Scheme, 0] (k + ) (k ) + 0 k(k ) 0 For equal roots D b ac 0 i.e., k 8k + k 0 D b ac 0 a, b 8, c k. ( 8) ()(k) 0 k 0 k k (k ) (k + ) 0. Since equation has equal roots, D 0 i.e., b ac 0 {(k )} (k )(k ) 0 (k k + 9) (k )(k ) 0 k k + 9 k + k + 0k 0 k k + 0 k k k + 0 k(k ) (k ) 0 (k )(k ) 0 k, [CSE Marking Scheme, 0]. Since the given equation has real roots, Here, a k +, b (k ), c k 0, [CSE Marking Scheme, 0] Here,. Given, k + (k ) + 0 i.e., ± a, b 0, c. k k 0 D 0 k S OLUT I ONS P-

32 WORKSHEET-. (i) Let the cost price of the toy be `, then gain % Gain ` 00 HOTS & Value ased Questions ( - 9 ) + ( k - ). 0 ` 00 Q Dist ( - ) + ( y - y ) ( ) + k k + 00 k k k k 0 0 k 0k +k 0 0 k(k 0) + (k 0) 0 (k 0)(k + ) 0 k 0, [CSE Marking Scheme, 0, 0]. O + P Let P be the location of the pole such that its distance from gate, metres. P + is diameter P 90 and m + ( + ) () or ± , 8 m, + m [CSE Marking Scheme, 0] ut S.P. C.P. + Gain + 00 S.P. ` ( + 0) 0( + 0) ( 0)( + 0) 0 or , [Q 0 is not possible] Hence the cost price of the toy is ` 0. (ii) Quadratic equation. (iii) Genuine profit.. Let the usual speed of plane be km/h ( + 000) ( 0) 0 0 Speed of plane 0 km/h. Values depicted here are : (i) Helping the needy. (ii) Quick help to the injured. [CSE Marking Scheme, 0] WORKSHEET- Since number of books cannot be negative, HOTS & Value ased Questions. Let the number of books bought ( + 0)( 0) 0 P- 0 or 0 0 Number of books bought 0. [CSE Marking Scheme, 0]. Let the number of books be and cost of each book be ` y. y 80 lso, ( + )(y ) 80 M TH EM T I C S - X

33 y + y ( + 0) ( + 0) 0 ( )( + 0) 0 (as it cannot be negative) No. of books bought + [CSE Marking Scheme, 0]. Since speed can t be negative, therefore 8 is not possible. Speed of passenger train km/h and Speed of fast train 8 km/h. [CSE Marking Scheme, 0]. (i) Suppose rjun had arrows. Number of arrows used to cut arrows of heeshm Number of arrows used to kill the rath driver Number of other arrows used Remaining arrows So, km Let speed of passenger train be km/h speed of superfast train ( + ) km/h 9 y question, Tpassenger ( + ) ( + 8) ( + 8) ( )( + 8) ( + ) ( + ) or Put y, then y 0 + 8y y 8y 0 0 y 0y + y 0 0 y(y 0) + (y 0) 0 (y 0) (y + ) 0 y 0 or y y 0, [Q y cannot negative] y 00. Hence the number of arrows which rjun had is 00. (ii) Quadratic equations. (iii) ravery, fight for truth. S OLUT I ONS P-

34 CHPTER SECTION RITHMETIC PROGRESSION TOPIC- To Find nth Term of the rithmetic Progression WORKSHEET-. a, d, an 0 an a + (n )d 0 + (n )( ) 0 n + n n + n. Given, t9 i.e. and 8 + (n ) 8 a + 8d a + 8d a d d. n 8 is not a term of given.p. d a+8 a a a + d a + d a + d d d a... (ii) a t + + t. Let the three numbers in.p. be a d, a, a + d. a+8 a0 a + 9d a a + d... (i) from (ii), [CSE Marking Scheme, 0]. n n 9 From, equation (ii) equation (i), we get an a + (n )d, let an 8. Given, a + d 9 [CSE Marking Scheme, 0]. Here, a, d t. Here, a, d Q l 9 9 n n. Hence 0 is not a term of the given.p. a + (n )d + (n )( ) n th term. n an a + (n ) d a a. + P- M TH EM T I C S - X

35 lso, ( d) + + ( + d) 88 8d + d d d + d + d 88 d d d ± The numbers are,,, or,,. [CSE Marking Scheme, 0] WORKSHEET-. Here given sequence is an.p. From (i), a + d 0, 8, 8,... From (ii),,,... Where, a,d a0 a + (0 - ) +9 From (iii), d 00 d 0 a 8, n a a + d a 8 + ( ) 8 8. Hence, is the th term from the end. [CSE Marking Scheme, 0, 0]. Let the nth term be zero. an 0 a + (n )d (n )( ) 0 0 n + 0 n n. Therefore, the first negative term is nd term.. Let the first term of.p. be a and common difference be d. a9 a a + 8d (a + d)...(i) a a + a + d (a + d) + [CSE Marking Scheme, 0] a 9 a + d 9...(i) a8 a (a + d) (a + d) d d...(ii) Substituting (ii) in (i), we get a + () 9 a So,.P. is,, 9,,... [CSE Marking Scheme, 0]. Here given, i.e., 8,..., 0,. a +d 0. Let us write.p. in reverse order then...(iv) Subtracting (iv) from (iii), 0...(iii) a + d a + 0d + a + d, n 0 an a + (n ) d a + 8d a + d...(ii). a,, b, and c are in.p. Let the common difference be d a+d a + d...(i)...(ii) From (i) and (ii), we get a, d 8 b a + d b + 8 b + b c a + d c a, b, c [CSE Marking Scheme, 0] S OLUT I ONS P-

36 WORKSHEET-. Here, a. Given, a8 0 a + d 0 a d, a + d 8 d - Net term + + a8 a + d a8 d + d 0d a8 a + d a8 0d 0d a8 a8 a8. d + d 0d 0 [CSE Marking Scheme, 0] Hence Proved. [CSE Marking Scheme, 0]. Common difference, d. Let the four numbers be a d, a d, a + d, a + d a d + a d + a + d + a + d a a Hence numbers are d, d, + d, + d Now, according to question, - ( - 9- ) - ( - d )( + d ) ( - d )( + d ) The given series is not in.p. as common difference does not eist. [CSE Marking Scheme, 0]. Here, a, a + d 8 d d d ().. Let the st term be a and common difference be d. ccording to the question, a a 9 d 9 d d 9 9 d ±,, +, + 8,,, 0. a a + (p )d, b a + (q )d, c a + (r )d a(q r) [a + (p )d][q r] a + d a + d b(r p) [a + (q )d][r p] a0 a + 9d and c[p q] [a + (r )d][p q] a(q r) + b(r p) + c(p q) a [q r + r p + p 9d + 9d 8d Q ( ) 9 9d The numbers are a + d (a + d) a 9d + ( - ) (9 9d) (9 d) Net term 9 - d 9-9d a a + 0d q] + d [p(q r) q + r + (q ) (r p) + (r ) (p q)] 9d + 0d 9d Hence Proved. a 0 + d[pq pr + qr pq + pr qr + ( q + r r + p p + q)] 0 [CSE Marking Scheme, 0] [CSE Marking Scheme, 0] a0 a P- M TH EM T I C S - X

37 WORKSHEET-8. If k +, k +, k are in.p. then (k ) (k+ ) (k + ) (k + ) k k k + k k k + k k + or k a, d - - ( - ). Here, nth term a +(n )d th term + ( ) + 0 [CSE Marking Scheme, 0]. Given, a (a + d) a + 0d a + d (a + d) 8a8 8(a + d) 8a + d 0 0 a + d 0 a 0. [CSE Marking Scheme, 0]. Let the first term be a and common difference be d. a +d 0...(i) a + d + a + 0d a + 8d...(ii) Solving equations (i) and (ii), common difference d [CSE Marking Scheme, 0]. Sn n + n Sn (n ) + (n ) (n + n) + n n + n + n n n an Sn Sn n + n (n n ) n +.P. is 8,, 0,... a a + d 8 + () 9. [CSE Marking Scheme, 0]. [Topper nswer, 0] S OLUT I ONS P-

38 WORKSHEET-9 TOPIC- Sum of nth Terms of an rithmetic Progression. Here, a, n 0, d Q S S0 n [ a + ( n - )d ] 0 [ + (0 - )]. [0 + 9 ] [0 + ]. Here, a, d, n Q S lso, n [ a + ( n - )d ] [ + 8] 0 and. Here a, d, Sn 0 n [0 + (n ) ( )] 0 n [CSE Marking Scheme, 0]. Here a 8, d, Sn 0 n Therefore [ + (n ) ( )] 0 n 9 [CSE Marking Scheme, 0] n + n Sn an Sn Sn a S S ( ) + ( ) ( ) + ( ) {( ) + ( )} S Now, [CSE Marking Scheme, 0]. S S [ + ( - )] [ + ] ( 9 + ) 80 [CSE Marking Scheme, 0] S n. S upto n terms S upto n terms n( n + ) S S + S n [ + (n )] n [n] n n [ + (n )] n( n - ) n( n + ) n( n - ) + n[n + + n - ] n[ n] n S Hence Proved.. a 8, d / years, Sn 8 n Sn [ a + ( n - )d ] 8 n ( 8 ) + ( n - ) n + n n + n n (n )(n + ) 0 n or n n (n cannot be negative) ge of the eldest participant a + d years [CSE Marking Scheme, 0] P-8 M TH EM T I C S - X

39 WORKSHEET-0. Here, a 0, d 0, n n Q S [ a + ( n - )d ] S. [ 0 + ( - )( - )] S 8[0 + ( )] 8[0 0] 8 ( 0) 0 [CSE Marking Scheme, 0] S + S. Here, a, d, n S. Here a, d, Sn 0 n [ + (n )( )] 0 n 9 [CSE Marking Scheme, 0] or a + d a + d...(i) S0 (a + 9d) or a + 9d...(ii) Solving (i) and (ii), wet get a, d Here.P.,,,... [CSE Marking Scheme, 0] n [0 + n + n ] n [0 + 8(n )] n [0 + (n )]. Let a be the first term and the common difference be d. S [a + d] a + d S8 [a + d] 8a + 8d S [a + d] a + d (S8 S) [(8a + 8d) (a + d)] [a + d] a + d [a + d] S [CSE Marking Scheme, 0] [ + ] [ + ] [] 8 90 [CSE Marking Scheme, 0] S + S ( a + d ) + ( a + d ) S [CSE Marking Scheme, 0]. n [0 + (n )] n [ a + ( n - )d ] S [ + ( - ) ()] Q n [0 + (n )] n S [0 + (n )] S. Let total time be n minutes Total distance convered by thief (00n) metres Total distance covered by policeman (n ) terms n - 00n [00 + (n )0] n n 8 0 (n )(n + ) 0 n Policeman took minutes to catch the thief. [CSE Marking Scheme, 0] WORKSHEET-. Let the sum of n terms of.p. Sn. Given, Sn n + n th Now, n term of.p. Sn Sn an (n + n) [ (n ) S OLUT I ONS + (n )] n + n [ n n + + n ] n + n n n + P-9

40 an n + th term a + 9 [CSE Marking Scheme, 0]. Let the sum of k terms of.p. is Sn k k Now kth term of.p. Sn Sn ak (k k) [ (k ) (k )] k k [k k + k + ] k k k + k k first term a [CSE Marking Scheme, 0]. Here, a, a and d n Q Sn [ a + ( n - ) d ] S [ ( -) + ( - ) ( - )] [8 + ( - )] ( 8 - ) a, d S0 0 [ + ( 0 - )( - )] 0[ + 9 ( )] 0( 8) 0 ( ) 0. Let a, be the first terms and d, D be the common difference of two.p. s Then according to question, n [ a + ( n - )d ] Sn n + n S 'n n + [ + ( n - )D] a + ( n - )d n + + ( n - )D n + n - )d n + n - n + +( )D a+( [ + ( - ) ( - )] Solving (i) and (ii), we get [CSE Marking Scheme, 0] 8 [ 0] 8 ( ) 9 [CSE Marking Scheme, 0]. Here n, tn n Taking n,,,... t t t Given.P. is,,,.... Here, a, d n Now, Sn [ a + ( n - )d ]. Here, a a8 Let first term be a and common difference be d. a d....(i) a 8 a + d 8...(ii) Putting n - m n m a + ( m - )d ( m - ) + ( m - ) + + ( m - )D am m - m 8m [CSE Marking Scheme, 0] WORKSHEET-. Here, an n Taking n, n, P-0 tn t 0 n Sn ( a + l ) S [ + ( - )] [ -] ( ) 9 [CSE Marking Scheme, 0] M TH EM T I C S - X

41 Tn Sn Sn Tn Sn Sn Sn Sn + Sn Sn Sn Sn + Sn (Sn Sn ) (Sn Sn ) Tn Tn d. [CSE Marking Scheme, 0]. Let the sum of first n terms of.p. Sn Given, Sn n n th Now, n term of.p. Sn Sn an (n n) [(n ) (n )] n n [n (n + n)] n n (n n + n) n n n + + n n + an (n ) nth term an (n ).. a, l (given) + (n )d (n )d 0...(i) Given, Sn 00 n 00 (+ ) n n from (i), (n )d 0 d d [CSE Marking Scheme, 0]. Let the first term be a and the common difference be d. a a + d...(i) S [a + d] a + d...(ii) Solving (i) and (ii), we get a 9 and d 0 [ 9 + (0 - )( - )] Thus, S0 [8 8] 0 [CSE Marking Scheme, 0]. Let the first 0 terms or lso, S0 0 ( a + 9d ) 00 0 ( a + 9d ) 00 0 [a + 9d] a + 9d 0 0 ( a + 9d ) S0...(i) 00 0 [a + 9d] or a + 9d 80...(ii) from (i) and (ii), we get a and d 0 S0 [ + (0 - )( )] [ + 9 ] [ + 8] 0 00 [CSE Marking Scheme, 0] WORKSHEET- HOTS & Value based nswers. The sequence goes like this, 0, 0, 0, Since they have a common difference of 0, they form an.p. a 0, an 990, d 0 an a+ (n ) d (n ) (n ) (n ) 0 n 88 n 89 These are 89 terms between 0 and 999 divisible by and.. Let.P. is 0,, 9,..., 99 which is divisible by. Here, a 0, d 0, Tn 99, then Tn a + (n )d (n ) S OLUT I ONS 889 (n ) 889 n n + 8. Therefore, the first negative term is nd term.. Two digit numbers which are divisible by are,, 8, It forms an.p. a, d, an 98 an a + (n )d 98 + (n ) 98 n 8 + n n 9 n [CSE Marking Scheme, 0] P-

42 . Here a 0, d, an 99 an a + (n )d (n ) n 9. [CSE Marking Scheme, 0]. Here, a, l, d l-a - n + + d + + There are multiples of that lie between and.. Here, required money is ` 00 a saving in st week ` 00 d difference in weekly saving ` 0.P. formed by saving, ccording to the question, Sequence is 00, 0, 0,... upto terms n Q Sn [a + (n )d] S [ 00 + ( ) 0] [00 + 0] [00 + 0] 0 0 She will be able to send her daughter to school after weeks. Value : small saving can fulfill your big desires or anyone else s. [CSE Marking Scheme, 0] WORKSHEET- HOTS & Value based nswers. Here an n + d a n an d (n + ) [(n ) + ] n + n + n d n Since common difference depends upon n and not constant, n + can not be the nth term of an.p. [CSE Marking Scheme, 0] S 9 a + d S 89 a + d Solving equations to get a l and d n Hence Sn [ + (n )] n. [CSE Marking Scheme, 0]. Let the three digits be a d, a, a + d. Sum a d + a + a + d a given the three digits are d,, + d. Original number 00( d) ( + d) 99d Revered number 00( + d) ( d) + 99d ccording to question, ( 99d) ( + 99d) 9 98d 9 9 d -98 ( ),, + ( ) i,e., 8, and Original number is The list of -digit odd positive numbers are, This is an.p. with a, d We have, tn a + (n )d So, 99 + (n ) n n Sn [ a +( n - )d ]. The three digits are Sn [()+()] [ + 88].. (i) Since, each section of class plants the same number of trees as the class number and there are three sections of each class. Total number of trees planted by the students [ ] [ +( - ) ] [( + )] 8 Students planted trees. (ii) rithmetic Progression. (iii) Our duty towards earth is to environment. save our P- M TH EM T I C S - X

43 CHPTER SECTION LINES (IN TWO DIMESIONS) TOPIC- Distance etween Two Points and Section Formula WORKSHEET- P.. Let P(, y) is equidistant from (, ) and (, ) P (, y) P P (( + ) + ( y - ) ) (( ) + ( y - ) ) (, ) : y y y + 0 y + y y + y y 9 0 is the required relation.. (, ) (, ) 9a (, y) pplying section formula for co-ordinate, + ( ) + y + ( ) + Hence b -...(ii) From (i), a + a + a 8 a + 9a 9 + y ( ) + ( ) + 9 ( ) + ( -) +...(i) (9a ) a + y (, y) is (, ) b 9a y + 0 or ( 8 a ) + ( a + ) + b Similarly for y co-ordinate, From (ii), + 8 or. y section formula C C Given that, [CSE Marking Scheme, 0] [CSE Marking Scheme, 0] : C P, y 9 (, ) y y + 9 P : P : a [CSE Marking Scheme, 0] S OLUT I ONS P-

44 WORKSHEET-. The point on the y-ais is (0, ) Distance between (, ) and (0, ) P(, y) (, ) + 0 units.. Let the point P be (y, y) P P P P ( ) + (y ) ( + ) + (y ) + + y y + PQ PR ( y ) + ( y + ). P(, y), (, ), (, ) [CSE Marking Scheme, 0] + [CSE Marking Scheme, 0] (0 - ) + ( - ) d + + ; y y y + ( y + ) + ( y ) y y y y + 8y [CSE Marking Scheme, 0] y Solving to get y 8 Hence coordinates of point P are (, 8).. Let the point on y-ais be (0, y) and P : P K Hence proved. [CSE Marking Scheme, 0] k Therefore 0 gives k k +. Let P(, y) and Q(, y) are two points which divide in three equal parts. y section formula Hence required ratio is : ( ) y ( - ) + ( ) ( - ) + ( -), P(, y) + + Hence point on y- ais is 0, ( - ), (0, ) [CSE Marking Scheme, 0] ( - ) + ( ) ( - ) + ( -), Q(, y) + +. P (, y) (, ) : P ( -), (, ) P : P : (, ) [CSE Marking Scheme, 0] WORKSHEET-. (, ) P (, k) (, ) Let the ratio in which P divides be n : Considering co-ordinate for section formula ( - )n + ( -) n+ P- Ratio n n n n n n or : Now, Considering y co-ordinate ( ) + ( - ) k + (n + ) n M TH EM T I C S - X

45 k - k E (, y) PQ PR. (, ). F (, y) [ - ( a + b )] + [ y - ( b - a )] C (, ) (9, 0) [ - ( a - b )] + [ y - ( b + a )] Co-ordinates of point E P (, y) , (8, ) Co-ordinates of point F Q (a + b, b a) + +, R (a b, a + b) (, ) Squaring, we get Length of EF [ (a + b)] + [y (b a)] [ (a b)] + [y (a + b)] [ (a + b)] [ a + b] (8 ) + ( - ) ( ) + () units (y a b) (y b + a) Length of C ( a b + a + b) ( a b + a b) (y a b + y b + a)(y a b y + b a) ( a) ( b) (y b) ( a) ( a) b (y b) a b ay....(i) (9 - ) + (0 - ) (8 ) + ( ) 0 units...(ii) From equation (i) and (ii), we get EF C. Hence proved. Hence proved. [CSE Marking Scheme, 0] [CSE Marking Scheme, 0] WORKSHEET-8. (, ). Given (, 0), (, ) and C(, ) ( ) + (0 ) 9 + C ( + ) + ( ) C ( ) + ( 0) + 9 C C Triangle is isosceles lso, + 0 (, ) + C C C (, t) Since pythagoras theorem is verified, therefore triangle is a right angled triangle. S OLUT I ONS P-

46 ( ) + ( ) 9 + C ( ) + (t + ) + (t + ) C ( + ) + ( t) 9 + ( t) Since C is a right angled triangle C + C 9 + ( t) + + (t + ) 9 + t + t + t + t + t + t 8t 8 t [CSE Marking Scheme, 0]. Co-ordinates of point R (, 0) QR 8 units Let the co-ordinates of point P be (0, y) Since PQ QR ( 0) + (0 y) + y y P (0,y) Q (,0) R (,0) y y ± Coordinates of P are (0, ) or (0, ) [CSE Marking Scheme, 0] WORKSHEET-9. Here,. Here C is a right angle triangle, (, ) Q Q Q Q - Q (p, ) Q : Q :. Points are (, ) and (0, ) (0 + ) + ( - + ) Distance + 0 units. Let P divide in the ratio k : P (/, /) k: (, ), k + / 8k + k + k + k Required ratio : [CSE Marking Scheme, 0] C (, ) + C C (p ) + ( ) + ( p) + ( ) ( ) + ( ) (p ) + ( ) + ( p) + 0 () + ( ) p 8p p p 9 + p p + 8 p p + 0 p p (p ) (p ) 0 p or p, p [CSE Marking Scheme, 0]. Let point be (0, y) + (y + ) () + (y ) + y + 9 y + 8y y Point (0, ) [CSE Marking Scheme, 0] P- M TH EM T I C S - X

47 TOPIC- rea of Triangle WORKSHEET-0. rea of triangle [(y y) + (y y) + (y y)] [0(0 ) + ( 0) + 0(0 0)] [ ] sq. units [( y) + ( ) (y ) + ( )] 0 [0 y y + + ( )] 0 [ 8y + ] 0 + y 0 Hence proved. [CSE Marking Scheme, 0] [CSE Marking Scheme, 0]. rea of the triangle formed by the given points (0, ), (, ) and C(, ) 0( ) + ( ) + ( ) 0 + () () + () ( ) (0) 0 The given points are collinear. [CSE Marking Scheme, 0]. (, y), (, y), C(, y) (, ) (, p) (, ) Since the points are collinear rea 0 [(y y) + (y y) + (y y)] 0 [(p + ) + ( ) + ( p)] 0 [p p] 0 [ p] 0 p 0, p 0. [CSE Marking Scheme, 0]. Since the points are collinear The area of triangle 0 rea of triangle [ (y y) + (y y) + (y y)] S OLUT I ONS. rea of the triangle t (t + t) + (t + ) (t t + ) + (t + ) (t t ) [t + t + t ] sq. units. which is independent of t. [CSE Marking Scheme, 0]. Using mid-point formula, co-ordinaes of P are 0 + 0,, Now as P lies on line joining C and D. So, C, D, P are collinear. Hence, rea of 0...(i) i.e., [(y y) + (y y) + (y y)] 0 Given points are C( 9, ) P(, ) D(, y), y, y, y So, from (i) rea. of [ 9( y) + ( ) (y + ) + ( ) ( )] [ 8 + 9y y + ] 0 y 8 0 y So, the co-ordinates of D are now (, ) Now, finding length of CP and PD using distance formula, P-

48 CP PD 0 required ratio CP : PD WORKSHEET-. Since the points are collinear, then rea of triangle 0 [(y y) + (y y) + (y y)] 0 + y + ( ) Let the points be (, ), (, ) and C(, ) pplying distance formula ( + ) + ( ) + 9 Similarly C ( ) + ( ) C 0 C y + ( ). y C (, ) Now the co-ordinates are (, ), (, ), C(, ) rea of triangle [(y y) + (y y) + (y y)] [( ) + ( + ) ( )] [ ] lso, sq. units C ( ) + ( ) 9 + C and C Clearly + C C + 0 Hence the triangle is right angled, rea of C ase Height (, ) + 0 [( + ) + ( ) ( ) + ( + )] 0 y. [CSE Marking Scheme. 0] (, ) (, ) sq unit. D (, ) ar ( D) (, ) C (, ) (8) + ( ) + ( ) sq. units E (, ) F(0, ) (, y) Let E be the mid point of C (, y) ar ( CD) ( ) + ( 8) (0) 8 sq. units ar (Quad CD) 80 sq. units [CSE Marking Scheme, 0] P-8 M TH EM T I C S - X

49 WORKSHEET-. Formed by the given points (, ), (, 0) and C(, 0) rea of triangle [(y y) + (y y) + (y y)] [(0 0) + ( ) (0 ) + ( 0)]. s, units [ +] + (p ) (p p)] 0 [(p + )( p + ) + (p ) ( ) + (p ) (p )] 0. rea of C 0 rea of C [(y y) + (y y) + (y y) 0 [( ) ( y) (y ) 0 + y y y + 0 [CSE Marking Scheme, 0]. If three points are collinear, then area of 0 rea [(y y) + (y y) + (y y)] Here, and b b y given data, a b i.e., a a (a, b) (, ). If the points are collinear, then rea of triangle 0 [(p + ) [p (p )] + (p )[(p ) (p )] (,, ) (, a, ) (y, y, y) (, b, ) rea [ b ( b)] 0 b + [ p + p p + p + + p p p + ] 0 [ p + ] 0 p p. rea of triangle [(y y) + (y y) + (y y)] [(k + ) ( + ) k( k)] [Q (a b) ] 8 k k + k k + k 0 (k ) (k 9) 0 9 k, k [CSE Marking Scheme, 0] WORKSHEET-. The points are collinear, then area of triangle 0 [(y y) + (y y) + (y y)] 0 [0( y) + (y 0) + (0 ) ] 0 [y ] 0 S OLUT I ONS y 0 [CSE Marking Scheme, 0]. Given points are (8, ), (, k) and (k, ) are collinear. rea of triangle formed 0 [8( k + ) + (, ) + k( + k)] 0 k k + 0 (k ) (k ) 0 k, [CSE Marking Scheme, 0] P-9