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Chapter No Chapter Name Page No , 01 Real Numbers 1. 02 Polynomials 24, 03 Pair of Linear Equations in Two Variables 62. 04 Quadratic Equations 136, 05 Arithmetic Progressions 185. 06 Triangles 240, 07 Coordinate Geometry 315, 08 Introduction to Trigonometry 363. 09 Some Applications of Trigonometry 427, 10 Circles 458.

11 Constructions 490, 12 Areas Related to Circles 536. 13 Surface Areas and Volumes 594, 14 Statistics 643. 15 Probability 754, Board Question Paper March 2017. marked questions are not from examination point of view . The pattern of Board Question Paper will be revised from the academic year 2017 2018 However the. Board Question Paper of March 2017 has been included to give a fair idea to the students about the kind. of questions asked in the Board Examination , 01 Real Numbers Chapter 01 Real Numbers. Step III Continue the process till the remainder is. Euclid s Division Lemma, zero When the remainder is zero the.

divisor at that stage is the required HCF , Given positive integers a and b there exist unique. For the above algorithm HCF a b HCF b r , integers q and r satisfying. a bq r 0 r b Example , Use Euclid s division algorithm to find the HCF of. where a is dividend b is divisor q is quotient, 1467 and 453 . and r is the remainder , Note q and r can also be zero .

Step I Apply Euclid s division lemma to 1467 and 453 . Examples 1467 453 3 108, Consider the following pair of integers 3 q. i 29 8 b 453 1467 a, Here a 29 and b 8 1359, By using Euclid s Division lemma 108 r. a bq r 0 r b, i e 29 8 3 5 0 5 8 Step II Since r 0. apply Euclid s division lemma to 453 and 108 , 3 quotient q 453 108 4 21. divisor b 8 29 dividend a 4 q, 24 b 108 453 a, 5 remainder r .

ii 77 7, Step III Again r 0, Here a 77 and b 7, apply Euclid s division lemma to 108 and 21 . By using Euclid s Division lemma , 108 21 5 3 5, a bq r 0 r b. i e 77 7 11 0 0 0 7 21 108, 11 quotient q 3, divisor b 7 77 dividend a . 77 Step IV Apply Euclid s division lemma to 21 and 3 . 0 remainder r 21 3 7 0, iii 9 12 3 21, Here a 9 and b 12 0. By using Euclid s Division lemma , Since r 0, a bq r 0 r b HCF 1467 453 3.

i e 9 12 0 9 0 9 12 3 HCF 21 3 HCF 108 21 , 0 quotient q HCF 453 108 HCF 1467 453 . divisor b 12 9 dividend a , Things to Remember, 9 remainder r Euclid s division algorithm can be extended. for all integers except zero i e b 0 , Euclid s Division Algorithm . Euclid s division algorithm is a technique to compute. the Highest Common Factor HCF of two given NCERT Exercise 1 1. positive integers , 1 Use Euclid s division algorithm to find the. Euclid s Division Algorithm to find HCF of two HCF of . positive integers a and b a b i 135 and 225 ii 196 and 38220. Step I By Euclid s division lemma find whole iii 867 and 255. numbers q and r Solution , where a bq r 0 r b i Since 225 135 we apply the division lemma.

Step II If r 0 the HCF is b If r 0 apply the to 225 and 135 to get. division lemma to b and r 225 135 1 90, Class X Mathematics. Since the remainder 90 0 we apply the Solution , division lemma to 135 and 90 to get HCF 616 32 will give the maximum number. 135 90 1 45 of columns in which they can march , We consider the new divisor 90 and the new Let us use Euclid s algorithm to find the. remainder 45 and apply the division lemma to get HCF . 90 45 2 0 616 32 19 8, As the remainder is zero we stop 32 8 4 0. Since the divisor at this stage is 45 the HCF the HCF of 616 and 32 is 8 . of 135 and 225 is 45 the maximum number of columns in which. ii Since 38220 196 we apply the division they can march is 8 . lemma to 38220 and 196 to get, 38220 196 195 0 4 Use Euclid s division lemma to show that.

As the remainder is zero we stop the square of any positive integer is either. Since the divisor at this stage is 196 the HCF of the form 3m or 3m 1 for some integer. of 196 and 38220 is 196 m , iii Since 867 255 we apply the division lemma Hint Let x be any positive integer then it is. to 867 and 255 to get of the form 3q 3q 1 or 3q 2 Now square. each of these and show that they can be, 867 255 3 102. rewritten in the form 3m or 3m 1 , Since the remainder 102 0 we apply the. CBSE 2015 , division lemma to 255 and 102 to get. 255 102 2 51 Let x be any positive integer and b 3 . We consider the new divisor 102 and the new Then by Euclid s division lemma x 3q r. remainder 51 and apply the division lemma to, for some integer q 0 and.

r 0 1 2 because 0 r 3, 102 51 2 0, x 3q or 3q 1 or 3q 2. As the remainder is zero we stop , When x 3q , Since the divisor at this stage is 51 the HCF. x 2 3q 2 9q2, of 867 and 255 is 51 , 2 Show that any positive odd integer is of the 3m where m is a integer. form 6q 1 or 6q 3 or 6q 5 where q is When x 3q 1, some integer CBSE 2014 x 2 3q 1 2 9q2 6q 1. Solution 3 3q2 2q 1, Let a be any positive integer and b 6 3m 1 where m is a integer.

Then by Euclid s algorithm a 6q r for When x 3q 2 . some integer q 0 and r 0 1 2 3 4 5 because x 2 3q 2 2 9q2 12q 4. 0 r 6 3 3q2 4q 1 1, a 6q or 6q 1 or 6q 2 or 6q 3 or 6q 4 3m 1 where m is a integer. or 6q 5 the square of any positive integer is either of. Here a cannot be 6q or 6q 2 or 6q 4 as the form 3m or 3m 1 for some integer m . they are divisible by 2 , 6q 1 5 Use Euclid s division lemma to show that. 6 is divisible by 2 but 1 is not divisible by 2 the cube of any positive integer is of the. 6q 3 form 9m 9m 1 or 9m 8 , 6 is divisible by 2 but 3 is not divisible by 2 Solution . 6q 5 Let a be any positive integer and b 3 , 6 is divisible by 2 but 5 is not divisible by 2 Then by Euclid s division lemma a 3q r. Since 6q 1 6q 3 6q 5 are not divisible for some integer q 0 and. by 2 they are odd numbers r 0 1 2 because 0 r 3, Therefore any odd integer is of the form a 3q or 3q 1 or 3q 2.

6q 1 or 6q 3 or 6q 5 When a 3q , a3 3q 3 27q3, 3 An army contingent of 616 members is to 9 3q3 . march behind an army band of 32 members 9 m where m is a integer. in a parade The two groups are to march When a 3q 1 . in the same number of columns What is the a3 3q 1 3 27q3 27q2 9q 1. maximum number of columns in which 9 3q3 3q2 q 1, they can march 9m 1 where m is a integer. Chapter 01 Real Numbers, When a 3q 2 the maximum capacity of a container which. a3 3q 2 3 27q3 54q2 36q 8 can measure the petrol of each tanker in exact. 9 3q3 6q2 4q 8 number of times is 170 litres , 9m 8 where m is a integer. the cube of any positive integer is of the form 5 A sweetseller has 420 kaju barfis and 130. 9m or 9m 1 or 9m 8 badam barfis She wants to stack them in. such a way that each stack has the same, Problems based on Exercise 1 1 number and they take up the least area of.

the tray What is the maximum number of, 1 Using Euclid s division algorithm find the barfis that can be placed in each stack for. HCF of 240 and 228 CBSE 2012 this purpose , Solution Solution . By Euclid s division algorithm HCF 420 130 will give the maximum. 240 228 1 12 number of barfis that can be placed in each. 228 12 19 0 stack , HCF 240 228 12 By Euclid s division algorithm . 420 130 3 30, 2 Find the HCF by Euclid s division 130 30 4 10. algorithm of the numbers 92690 7378 and 30 10 3 0. 7161 CBSE 2013 HCF 420 130 10, the sweetseller can make stacks of 10 for both.

By Euclid s division algorithm , kinds of barfi . 92690 7378 12 4154, 7378 4154 1 3224 6 The length breadth and height of a room. 4154 3224 1 930 are 8m 25 cm 6m 75 cm and 4 m 50 cm. 3224 930 3 434 respectively Find the length of the longest. 930 434 2 62 rod that can measure the three dimensions. 434 62 7 0 of the room exactly CBSE 2012 , HCF 92690 7378 62 Solution . 7161 62 115 31 Since 1m 100 cm, 62 31 2 0 8 m 25 cm 825 cm. HCF 7161 62 31 6 m 75 cm 675 cm, HCF 92690 7378 7161 31 4 m 50 cm 450 cm.

HCF 825 675 450 will give the length of the, 3 Using Euclid s division algorithm find longest rod . whether the pair of numbers 231 396 are 825 675 1 150. coprime or not 675 150 4 75, Solution 150 75 2 0, By Euclid s division algorithm . HCF 825 675 75, 396 231 1 165, 450 75 6 0, 231 165 1 66. HCF 450 75 75, 165 66 2 33, 66 33 2 0 HCF 825 675 450 75. HCF 231 396 33 the length of the longest rod is 75 cm . the numbers are not coprime , NCERT Exemplar, 4 Two tankers contain 850 litres and.

680 litres of petrol Find the maximum 1 Write whether every positive integer can be. capacity of a container which can measure of the form 4q 2 where q is an integer . the petrol of each tanker in exact number of Justify your answer . times CBSE 2012 Solution , Solution No every positive integer cannot be only of. HCF 850 680 will give the maximum the form 4q 2 , capacity of container Justification . 850 680 1 170 Let a be any positive integer Then by Euclid s. 680 170 4 0 division lemma we have, HCF 850 680 170 a bq r where 0 r b. Class X Mathematics, Putting b 4 we get The numbers are of the form 3q 3q 1 and. a 4q r where 0 r 4 3q 2 , Hence a positive integer can be of the form 3q 2 9q2 3 3q2 .

4q 4q 1 4q 2 and 4q 3 3m where m is a integer , 3q 1 2 9q2 6q 1 3 3q2 2q 1. 2 The product of two consecutive positive 3m 1 , integers is divisible by 2 Is this statement where m is a integer . true or false Give reasons 3q 2 2 9q2 12q 4 , Solution which cannot be expressed in the. True form 3m 2 , Justification Square of any positive integer cannot be. Let a a 1 be two consecutive positive expressed in the form 3m 2 . By Euclid s division lemma we have 5 A positive integer is of the form 3q 1 q. a bq r where 0 r b being a natural number Can you write its. For b 2 we have square in any form other than 3m 1 i e . a 2q r where 0 r 2 i 3m or 3m 2 for some integer m Justify. Putting r 0 in i we get your answer , a 2q which is divisible by 2 Solution .

a 1 2q 1 which is not divisible by 2 No , Putting r 1 in i we get Justification . a 2q 1 which is not divisible by 2 Consider the positive integer 3q 1 where q. a 1 2q 2 which is divisible by 2 is a natural number . Thus for 0 r 2 one out of every two 3q 1 2 9q2 6q 1. consecutive integers is divisible by 2 3 3q2 2q 1. The product of two consecutive positive 3m 1 where m is an integer . integers is divisible by 2 Thus 3q 1 2 cannot be expressed in any. other form apart from 3m 1 , 3 The product of three consecutive positive. integers is divisible by 6 Is this statement 6 The numbers 525 and 3000 are both. true or false Justify your answer divisible only by 3 5 15 25 and 75 What is. Solution HCF 525 3000 Justify your answer , True Solution . Justification HCF 525 3000 75, At least one out of every three consecutive Justification . positive integers is divisible by 2 3 5 15 25 and 75 are the only common. The product of three consecutive positive factors of 525 and 3000 . integers is divisible by 2 75 is highest among the common factors . At least one out of every three consecutive HCF 525 3000 75. positive integers is divisible by 3 , The product of three consecutive positive 7 Show that the square of any positive integer.

integers is divisible by 3 is either of the form 4q or 4q 1 for some. Since the product of three consecutive positive integer q . integers is divisible by 2 and 3 Solution , It is divisible by 6 also Let a be the positive integer and b 4 . Then by Euclid s algorithm a 4m r for, 4 Write whether the square of any positive some integer m 0 and r 0 1 2 3 because. integer can be of the form 3m 2 where m 0 r 4 , is a natural number Justify your answer a 4m or 4m 1 or 4m 2 or 4m 3 . Solution 4m 2 16m2 4 4m2 , No 4q where q is some integer . Justification 4m 1 2 16m2 8m 1, Let a be any positive integer Then by Euclid s 4 4m2 2m 1.

division lemma we have 4q 1 where q is some integer . a bq r where 0 r b 4m 2 2 16m2 16m 4, For b 3 we have 4 4m2 4m 1 . a 3q r where 0 r 3 i 4q where q is some integer , Chapter 01 Real Numbers. 4m 3 2 16m2 24m 9 10 Show that the square of any positive integer. 4 4m2 6m 2 1 cannot be of the form 6m 2 or 6m 5 for. 4q 1 where q is some integer any integer m , The square of any positive integer is either of Solution . the form 4q or 4q 1 where q is some integer Let a be the positive integer and b 6 . Then by Euclid s algorithm a 6q r for, 8 Show that cube of any positive integer is of some integer q 0 and r 0 1 2 3 4 5. the form 4m 4m 1 or 4m 3 for some because 0 r 5 , integer m a 6q or 6q 1 or 6q 2 or 6q 3 or 6q 4.

Solution or 6q 5 , Let a be the positive integer and b 4 6q 2 36q2 6 6q2 . Then by Euclid s algorithm a 4q r for 6m where m is any integer . some integer q 0 and r 0 1 2 3 because 6q 1 2 36q2 12q 1. 0 r 4 6 6q2 2q 1, 6m 1 where m is any integer , a 4q or 4q 1 or 4q 2 or 4q 3 . 6q 2 2 36q2 24q 4, 4q 3 64q3 4 16q3 , 6 6q2 4q 4. 4m where m is some integer , 6m 4 where m is any integer . 4q 1 3 64q3 48q2 12q 1, 6q 3 2 36q2 36q 9, 4 16q3 12q2 3 1 6 6q2 6q 1 3.

4m 1 where m is some integer 6m 3 where m is any integer . 4q 2 3 64q3 96q2 48q 8 6q 4 2 36q2 48q 16, 4 16q3 24q2 12q 2 6 6q2 7q 2 4. CBSE 2012 Solution HCF 850 680 will give the maximum capacity of container 850 680 1 170 680 170 4 0 HCF 850 680 170 the maximum capacity of a container which can measure the petrol of each tanker in exact number of times is 170 litres 5 A sweetseller has 420 kaju barfis and 130 badam barfis She wants to stack them in