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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.
4m where m is some integer 6m 4 where m is any integer . 4q 3 3 64q3 144q2 108q 27 6q 5 2 36q2 60q 25, 4 16q3 36q2 27q 6 3 6 6q2 10q 4 1. 4m 3 where m is some integer 6m 1 where m is any integer . The cube of any positive integer is of the form The square of any positive integer is of the. 4m 4m 1 or 4m 3 for some integer m form 6m 6m 1 6m 3 6m 4 and cannot. be of the form 6m 2 or 6m 5 for any, 9 Show that the square of any positive integer integer m . cannot be of the form 5q 2 or 5q 3 for, any integer q 11 Show that the square of any odd integer is. Solution of the form 4q 1 for some integer q , Let a be the positive integer and b 5 Solution . Then by Euclid s algorithm a 5m r for Let a be any odd integer and b 4 . some integer m 0 and r 0 1 2 3 4 Then by Euclid s algorithm a 4m r for. because 0 r 5 some integer m 0 and r 0 1 2 3 because. a 5m or 5m 1 or 5m 2 or 5m 3 or 0 r 4 , 5m 4 a 4m or 4m 1 or 4m 2 or 4m 3.
5m 2 25m2 5 5m2 a 4m 1 or 4m 3, 5q where q is any integer Here a cannot be 4m or 4m 2 as they are. 5m 1 2 25m2 10m 1 divisible by 2 , 5 5m2 2m 1 4m 1 2 16m2 8m 1. 5q 1 where q is any integer 4 4m2 2m 1, 5m 2 25m2 20m 4. 4q 1 where q is some integer , 5 5m2 4m 4 4m 3 2 16m2 24m 9. 5q 4 where q is any integer 4 4m2 6m 2 1, 5m 3 2 25m2 30m 9 4q 1 where q is some integer .
5 5m2 6m 1 4 The square of any odd integer is of the form. 5q 4 where q is any integer 4q 1 for some integer q . 5m 4 2 25m2 40m 16, 5 5m2 8m 3 1 12 If n is an odd integer then show that n2 1. 5q 1 where q is any integer is divisible by 8 , The square of any positive integer is of the Solution . form 5q 5q 1 5q 4 and cannot be of the Any odd integer n is of the form 4m 1 or. form 5q 2 or 5q 3 for any integer q 4m 3 , Class X Mathematics. n2 1 4m 1 2 1 HCF 3125 1250 625, 16m2 8m HCF 1250 9375 15625 625. 8 2m2 m the largest number is 625 , which is divisible by 8 .
Also n2 1 4m 3 2 1 16 Show that the cube of a positive integer of. 16m2 24m 8 the form 6q r q is an integer and r 0 1 . 8 2m2 3m 1 2 3 4 5 is also of the form 6m r , which is divisible by 8 Solution . n2 1 is divisible by 8 for any odd integer n 6q r is a positive integer where q is an. integer and r 0 1 2 3 4 5, 13 Prove that if x and y are both odd positive Then the positive integers are of the form 6q . integers then x2 y2 is even but not 6q 1 6q 2 6q 3 6q 4 and 6q 5 . divisible by 4 Taking cube of each term we have ,Solution 6q 3 216 q3 6 36q3 0. Since x and y are odd positive integers we 6m 0 where m is an integer. have 6q 1 3 216q3 108q2 18q 1, x 2m 1 and y 2n 1 6 36q3 18q2 3q 1. x2 y2 2m 1 2 2n 1 2 6m 1 where m is an integer, 4m2 4m 1 4n2 4n 1 6q 2 3 216q3 216q2 72q 8.
4 m2 n2 4 m n 2 6 36q3 36q2 12q 1 2, 2 2, x y is an even number but not divisible by 4 6m 2 where m is an integer. 6q 3 3 216q3 324q2 162q 27, 14 Use Euclid s division algorithm to find HCF 6 36q3 54q2 27q 4 3. of 441 567 693 6m 3 where m is an integer,Solution 6q 4 3 216q3 432q2 288q 64. By Euclid s division algorithm 6 36q3 72q2 48q 10 4. 6m 4 where m is an integer, 693 567 1 126, 6q 5 3 216q3 540q2 450q 125. 567 126 4 63, 6 36q3 90q2 75q 20 5, 126 63 2 0 6m 5 where m is an integer.
HCF 693 567 63 the cube of a positive integer of the form. 441 63 7 0 6q r q is an integer and r 0 1 2 3 4 5 is. HCF 441 63 63 also of the form 6m r , HCF 693 567 441 63. 17 Prove that one and only one out of n n 2, 15 Using Euclid s division algorithm find the and n 4 is divisible by 3 where n is any. largest number that divides 1251 9377 and positive integer . 15628 leaving remainders 1 2 and 3 Solution , respectively By Euclid s division lemma we have. Solution a bq r 0 r b, Since 1 2 and 3 are the remainders of 1251 For a n and b 3 we have. 9377 and 15628 respectively n 3q r i , 1251 1 1250 is exactly divisible by the where q is an integer.
required number and 0 r 3 i e r 0 1 2 , 9377 2 9375 is exactly divisible by the Putting r 0 in i we get. required number n 3q, 15628 3 15625 is exactly divisible by the n is divisible by 3 . required number n 2 3q 2, required number HCF of 1250 9375 and n 2 is not divisible by 3 . 15625 n 4 3q 4, By Euclid s division algorithm n 4 is not divisible by 3 . 15625 9375 1 6250 Putting r 1 in i we get, 9375 6250 1 3125 n 3q 1.
6250 3125 2 0 n is not divisible by 3 , HCF 15625 9375 3125 n 2 3q 3 3 q 1 . 3125 1250 2 625 n 2 is divisible by 3 , 1250 625 2 0 n 4 3q 5. Chapter 01 Real Numbers, n 4 is not divisible by 3 The product n3 n is divisible by 2 . Putting r 2 in i we get Since one out of every three consecutive. n 3q 2 integers is divisible by 3 , n is not divisible by 3 The product n3 n is divisible by 3 . n 2 3q 4 Any number which is divisible by 2 and 3 is. n 2 is not divisible by 3 also divisible by 6 , n 4 3q 6 3 q 2 The product n3 n is divisible by 6 .
n 4 is divisible by 3, Thus for each value of r such that 0 r 3 20 Show that one and only one out of n n 4 . only one out of n n 2 and n 4 is divisible n 8 n 12 and n 16 is divisible by 5 . by 3 where n is any positive integer , 18 Prove that one of any three consecutive By Euclid s division lemma we have. positive integers must be divisible by 3 a bq r 0 r b. Solution For a n and b 5 we have, Let the three consecutive positive integers be n 5q r i . n n 1 and n 2 where n is any integer Where q is an integer. By Euclid s division lemma we have and 0 r 5 i e r 0 1 2 3 4 . a bq r 0 r b Putting r 0 in i we get, For a n and b 3 we have n 5q. n 3q r i n is divisible by 5 , Where q is an integer n 4 5q 4.
and 0 r 3 i e r 0 1 2 n 4 is not divisible by 5 , Putting r 0 in i we get n 8 5q 8. n 3q n 8 is not divisible by 5 , n is divisible by 3 n 12 5q 12. n 1 3q 1 n 12 is not divisible by 5 , n 1 is not divisible by 3 n 16 5q 16. n 2 3q 2 n 16 is not divisible by 5 , n 2 is not divisible by 3 Putting r 1 in i we get. Putting r 1 in i we get n 5q 1, n 3q 1 n is not divisible by 5 .
n is not divisible by 3 n 4 5q 5 5 q 1 , n 1 3q 2 n 4 is divisible by 5 . n 1 is not divisible by 3 n 8 5q 9, n 2 3q 3 3 q 1 n 8 is not divisible by 5 . n 2 is divisible by 3 n 12 5q 13, Putting r 2 in i we get n 12 is not divisible by 5 . n 3q 2 n 16 5q 17, n is not divisible by 3 n 16 is not divisible by 5 . n 1 3q 3 3 q 1 Putting r 2 in i we get, n 1 is divisible by 3 n 5q 2.
n 2 3q 4 n is not divisible by 5 , n 2 is not divisible by 3 n 4 5q 6. Thus for each value of r such that 0 r 3 n 4 is not divisible by 5 . only one out of n n 1 and n 2 is divisible n 8 5q 10 5 q 2 . by 3 n 8 is divisible by 5 , n 12 5q 14, 19 For any positive integer n prove that n3 n n 12 is not divisible by 5 . is divisible by 6 CBSE 2012 n 16 5q 18,Solution n 16 is not divisible by 5 . n3 n n n2 1 n n 1 n 1 Putting r 3 in i we get, n3 n is product of three consecutive positive n 5q 3. integers where n is any positive integer n is not divisible by 5 . Since one out of every two consecutive n 4 5q 7, integers is divisible by 2 n 4 is not divisible by 5 .
Class X Mathematics, n 8 5q 11 11 An army contingent of 104 members is to. n 8 is not divisible by 5 march behind an army band of 96 members in. n 12 5q 15 5 q 3 a parade The two groups are to march in the. n 12 is divisible by 5 same number of columns What is the. n 16 5q 19 maximum number of columns in which they. can march CBSE 2011 , n 16 is not divisible by 5 . Putting r 4 in i we get 12 144 cartons of Coke cans and 90 cartons of. n 5q 4 Pepsi cans are to be stacked in a canteen If. each stack is of the same height and if it, n is not divisible by 5 . contain cartons of the same drink what would, n 4 5q 8. be the greatest number of cartons each stack, n 4 is not divisible by 5 would have CBSE 2011 .
n 8 5q 12, 13 Pens are sold in pack of 8 and notepads are. n 8 is not divisible by 5 , sold in pack of 12 Find the least number of. n 12 5q 16 pack of each type that one should buy so that. n 12 is not divisible by 5 there are equal number of pen and notepads . n 16 5q 20 5 q 4 CBSE 2014 , n 16 is divisible by 5 . Thus for each value of r such that 0 r 5, only one out of n n 4 n 8 n 12 and. n 16 is divisible by 5 1 23 2 5 3 5, 4 Yes 5 64 6 3.
Practice problems based on Exercise 1 1 11 8 12 18 13 2 3. Multiple Choice Questions,1 Find the HCF of 1656 and 4025 by Euclid s. division algorithm CBSE 2013 1 Euclid s division lemma states that for two. 2 Use Euclid s division algorithm to find HCF positive integers a and b there exist unique. of 65 and 175 CBSE 2015 integers q and r such that a bq r where r. must satisfy NCERT Exemplar , 3 Find the HCF of 1620 1725 and 255 by A 1 r b B 0 r b. Euclid s division algorithm CBSE 2012 C 0 r b D 0 r b. 2 For some integer m every even integer is of, 4 Using Euclid s division algorithm find the form NCERT Exemplar . whether the pair of numbers 847 2160 are A m B m 1. coprime or not CBSE 2012 C 2m D 2m 1, 5 Find the largest number that divides 2053 and 3 For some integer q every odd integer is of the. 967 and leaves remainders 5 and 7 form NCERT Exemplar . respectively A q B q 1, C 2q D 2q 1,6 Find HCF of 81 and 237 and express it as a.
linear combination of 81 and 237 i e HCF of 4 n2 1 is divisible by 8 if n is. 81 237 81x 237y for some x and y NCERT Exemplar , CBSE 2012 A an integer. B a natural number, 7 Show that the square of an odd positive C an odd integer. integer is of the form 8m 1 for some whole D an even integer. 5 The largest number which divides 70 and 125 , 8 Show that the square of any positive integer is leaving remainders 5 and 8 respectively is. of the form 4m or 4m 1 where m is any NCERT Exemplar . integer CBSE 2012 A 13 B 65, C 875 D 1750,9 Show that any positive odd integer is of the. form 4q 1 or 4q 3 where q is some 6 For any positive integer a and 3 there exist. integer CBSE 2012 unique integers q and r such that a 3q r . where r must satisfy CBSE 2012 , 10 Prove that n2 n is divisible by 2 for every A 0 r 3 B 1 r 3.
positive integer n CBSE 2012 C 0 r 3 D 0 r 3, Chapter 01 Real Numbers. 30 2 3 5,The Fundamental Theorem of Arithmetic, 144 2 2 2 2 3 3 24 32. Every composite number can be expressed HCF 2 3 6 and. factorised as a product of primes and this LCM 24 32 5 720. factorisation is unique apart from the order in which Things to Remember. the prime factors occur , This fundamental theorem of arithmetic help us to HCF of two numbers is always a factor of. find HCF and LCM of the numbers their LCM , This method is also called the prime factorisation LCM is always a multiple of HCF . method HCF p q r LCM p q r p q r , Example where p q r are positive integers However .
Consider a composite number 2352 the following results hold good for three. 2352 numbers p q and r , p q r HCF p q r , LCM p q r . 2 1176 HCF p q HCF q r HCF p r , p q r LCM p q r . HCF p q r , 2 588 LCM p q LCM q r LCM p r , NCERT Exercise 1 2. 2 294, 1 Express each number as a product of its, 2 147 prime factors . i 140 ii 156, 3 49 iii 3825 iv 5005, Prime factorization of 2352 is.
2352 2 2 2 2 3 7 7, 2 70, 24 3 72,HCF and LCM 2 35. HCF Product of the smallest power of each, common prime factor in the numbers 5 7. LCM Product of the greatest power of each prime, factor in the numbers 140 2 2 5 7 22 5 7. Relation between HCF and LCM of any two ii 156,positive integers . For any two positive integers a and b 2 78,HCF a b LCM a b a b.
Example 2 39, Find HCF and LCM of the following pairs of integers . i 60 and 72 3 13, ii 12 30 and 144, 156 2 2 3 13 22 3 13. i The prime factorization of 60 and 72 gives , 60 2 2 3 5 22 3 5 iii 3825. 72 2 2 2 3 3 23 32, HCF 22 3 12 and 3 1275, LCM 23 32 51 360. 60 72 60 72 3 425, OR LCM , HCF 12, LCM 360 5 85,ii The prime factorization of 12 30 and 144 5 17.
12 2 2 3 22 3 3825 3 3 5 5 17 32 52 17, Class X Mathematics. iv 5005 Solution , i 12 2 2 3 22 3, 5 1001 15 3 5, 21 3 7. 7 143 LCM 12 15 21 22 3 5 7 420, HCF 12 15 21 3, 11 13 ii 17 1 17. 23 1 23, 5005 5 7 11 13, 29 1 29,v LCM 17 23 29 1 17 29 23 11339. HCF 17 23 29 1, 17 437 iii 8 2 2 2 23, 9 3 3 32, 19 23 25 5 5 52.
LCM 8 9 25 23 32 52 1800, 7429 17 19 23, HCF 8 9 25 1. 2 Find the LCM and HCF of the following 4 Given that HCF 306 657 9 find LCM. pairs of integers and verify that 306 657 , LCM HCF product of the two numbers Solution . i 26 and 91 LCM HCF Product of two numbers, ii 510 and 92 CBSE 2011 LCM 9 306 657. iii 336 and 54 306 657, LCM 22338,Solution 9, i 26 2 13 5 Check whether 6n can end with the digit 0. 91 7 13 for any natural number n , LCM 26 91 2 7 13 182 Solution .
HCF 26 91 13 If the number 6n for any natural number n ends. Verification with the digit zero then it is divisible by 5 That. LCM HCF 182 13 2366 is the prime factorization of 6n contains the. Product of two numbers 26 91 2366 prime 5 This is not possible because. LCM HCF Product of two numbers prime factorisation of 6n 2 3 n 2n 3n . so the only primes in the factorisation of 6n are. ii 510 2 3 5 17 2 and 3 and the uniqueness of the fundamental. 92 2 2 23 22 23 theorem of arithmetic guarantees that there are. LCM 510 92 22 3 5 17 23 23460 no other primes in the factorization of 6n . HCF 510 92 2 So there is no natural number n for which 6n. Verification ends with the digit zero , LCM HCF 23460 2 46920. 6 Explain why 7 11 13 13 and, Product of two numbers 510 92 46920. 7 6 5 4 3 2 1 5 are composite, LCM HCF Product of two numbers. iii 336 2 2 2 2 3 7 24 3 7 Solution , 54 2 3 3 3 2 33 7 11 13 13 7 11 1 13. LCM 336 54 24 33 7 3024 77 1 13 78 13, HCF 336 54 2 3 6 2 3 13 13.
Verification 2 3 132, LCM HCF 3024 6 18144 Since 7 11 13 13 can be expressed as a. Product of two numbers 336 54 18144 product of primes it is a composite number . LCM HCF Product of two numbers 7 6 5 4 3 2 1 5, 7 6 4 3 2 1 1 5. 3 Find the LCM and HCF of the following 1008 1 5 1009 5. integers by applying the prime factorization 5 1009. method Since 7 6 5 4 3 2 1 5 can be, i 12 15 and 21 ii 17 23 and 29 expressed as a product of primes it is a. iii 8 9 and 25 composite number , Chapter 01 Real Numbers. 7 There is a circular path around a sports field Solution . Sonia takes 18 minutes to drive one round of Every composite number can be expressed. the field while Ravi takes 12 minutes for the factorised as a product of primes and this. same Suppose they both start at the same factorisation is unique apart from the order in. point and at the same time and go in the which the prime factors occur . same direction After how many minutes will Yes , they meet again at the starting point Justification .
Solution HCF of two numbers is always a factor of, Sonia and Ravi start at the same time from the their LCM . same point and go in the same direction and at But here 18 is a factor of 378 . the same time so to find the time when they Two numbers can have 18 as their HCF and. will meet again at the starting point we have 380 as their LCM . to find the LCM of 12 and 18 6 Find the missing numbers a b c and d in. 12 2 2 3 22 3 the given factor tree , 18 2 3 3 2 32. LCM 12 18 22 32 36, Sonia and Ravi will meet again at the starting 2 9009. point after 36 minutes , a 3003,Problems based on Exercise 1 2. 3 1001,1 Given that HCF 306 1314 18 Find, LCM 306 1314 CBSE 2013 b 143.
LCM HCF Product of two numbers c d, LCM 18 306 1314 CBSE 2012 . 306 1314 Solution , LCM 22338 9009, 18 9009 a 3003 a 3. 2 Find the HCF and LCM of 90 and 144 by 1001, the method of prime factorization 1001 b 143 b 7. CBSE 2012 , 143 c d, Since 143 11 13, 90 2 3 3 5 2 32 5. c 11 d 13 or c 13 d 11, 144 2 2 2 2 3 3 24 32, HCF 90 144 2 32 18 7 Complete the following factor tree and find.
LCM 90 144 24 32 5 720 the composite number x ,3 Find the LCM of 96 and 360 by using x. fundamental theorem of arithmetic , CBSE 2012 2 3381. 96 2 2 2 2 2 3 25 3 3 , 360 2 2 2 3 3 5 23 32 5, 7 161. LCM 96 360 25 32 5 1440, 4 Determine the values of p and q so that the 7 CBSE 2012 . prime factorization of 2520 is expressible as Solution . 23 3p q 7 CBSE 2014 6762 2 3381 6762, Since 2520 2 2 2 3 3 5 7 2 3381.
23 32 5 7, p 2 and q 5 3 1127 7 161 1127,5 State Fundamental theorem of arithmetic 7 161. Is it possible for the HCF and LCM of two, numbers to be 18 and 378 respectively 7 23 161 7 23. Justify your answer CBSE 2014 x 6762, Class X Mathematics. 8 Explain why 7 13 11 11 and NCERT Exemplar, 7 6 5 4 3 2 1 3 are composite. numbers CBSE 2012 1 Explain why 3 5 7 7 is a composite. Solution number , 7 13 11 11 7 13 1 11 Solution , 91 1 11 92 11 3 5 7 7 3 5 1 7.
2 2 23 11 15 1 7, 2 2 11 23 16 7 7 16, Since 7 13 11 11 can be expressed as a Since 3 5 7 7 can be expressed as a. product of primes it is a composite number product of primes it is a composite number . 7 6 5 4 3 2 1 3, 7 6 5 4 2 1 1 3 2 Can two numbers have 18 as their HCF and. 1680 1 3 1681 3 380 as their LCM Give reasons , 41 41 3 3 41 41 Solution . Since 7 6 5 4 3 2 1 3 can be No , expressed as a product of primes it is a Justification . composite number HCF of two numbers is always a factor of. their LCM , 9 Three bells toll at intervals of 9 12 15 minutes.
But here 18 is not a factor of 380 , respectively If they start tolling together after. Two numbers cannot have 18 as their HCF, what time they next toll together CBSE 2011 . and 380 as their LCM , 9 3 3 32 3 Show that 12n cannot end with the digit 0 or. 12 2 2 3 22 3 5 for any natural number n , 15 3 5 Solution . LCM 9 12 15 22 32 5 180 If the number 12n for any natural number n . the bells will toll together after 180 minutes ends with the digit 0 or 5 then it is divisible by. 10 The HCF of 65 and 117 is expressible in the 5 That is the prime factorization of 12n contains. form 65m 117 Find the value of m Also the prime 5 This is not possible because prime. find the LCM of 65 and 117 using prime factorisation of 12n 22 3 n 22n 3n so. factorization method CBSE 2011 the only primes in the factorisation of 12n are. Solution 2 and 3 and the uniqueness of the fundamental. 117 65 1 52 theorem of arithmetic guarantees that there are. 65 52 1 13 no other primes in the factorization of 12n . 65 13 5 0 So there is no natural number n for which 12n. HCF 65 117 13, ends with the digit zero , Since HCF 65 117 65m 117.
13 65m 117 4 On morning walk three persons step off. 65m 117 13 130 together and their steps measure 40 cm . 130 42 cm and 45 cm respectively What is the, m 2 minimum distance each should walk so that. each can cover the same distance in, Now 65 5 13, complete steps . and 117 3 3 13 32 13, LCM 65 117 32 5 13 585 Since the three persons start walking together . 11 If two positive integers x and y are The minimum distance covered by each of. expressible in terms of primes as x p2q3 them in complete steps LCM of the. and y p3q what can you say about their measures of their steps. LCM and HCF Is LCM a multiple of LCM 40 42 45 , HCF Explain CBSE 2014 40 2 2 2 5 23 5. Solution 42 2 3 7, x p2q3 and y p3q 45 3 3 5 32 5, HCF x y p2 q LCM 40 42 45 23 32 5 7.
LCM x y p3 q3 2520, LCM x y p q2 HCF x y Each person should walk a minimum distance. LCM is a multiple of HCF of 2520 cm in complete steps . Chapter 01 Real Numbers, Practice problems based on Exercise 1 2 Multiple Choice Questions. 1 Find the HCF of 960 and 432 1 If the HCF of 65 and 117 is expressible in the. form 65m 117 then the value of m is,2 Find the HCF and LCM of 404 and 96 and. NCERT Exemplar , verify HCF LCM Product of the two given. A 4 B 2, numbers CBSE 2012 , C 1 D 3,3 Express 2120 as the product of its prime.
2 If two positive integers a and b are written as. factors CBSE 2012 , a x3y2 and b xy3 x y are prime numbers . 4 Find the HCF and LCM of 60 120 and 288 then HCF a b is NCERT Exemplar . CBSE 2012 A xy B xy2, C x y D x2y2,5 The HCF and LCM of two numbers are 9 and. 360 respectively If one number is 45 write 3 If two positive integers p and q can be. the other number expressed as p ab2 and q a3b a b being. prime numbers then LCM p q is, 6 The HCF of 45 and 105 is 15 Write their NCERT Exemplar . LCM CBSE 2010 A ab B a2b2, 7 The LCM of 2 numbers is 14 times their HCF C a3b2 D a3b3. The sum of LCM and HCF is 600 If one 4 The least number that is divisible by all the. number is 280 then find the other number numbers from 1 to 10 both inclusive is. CBSE 2012 NCERT Exemplar , 8 Find the HCF of the numbers 520 and 468 by A 10 B 100.
prime factorization method CBSE 2014 C 504 D 2520. 9 Find LCM of the numbers given below 5 Given that LCM 91 26 182 then HCF. m 2m 3m 4m and 5m where m is any 91 26 is CBSE 2011 . positive integer CBSE 2014 A 13 B 26, C 7 D 9,10 Show that the number 4 when n is a natural. number cannot end with the digit zero 6 If the HCF of 55 and 99 is expressible in the. CBSE 2012 form 55m 99 then the value of m is, CBSE 2011 . 11 Show that 7n cannot end with the digit zero A 4 B 2. for any natural number n CBSE 2012 C 1 D 3, 12 Can two numbers have 15 as their HCF and 7 The values of x and y in the given figure are. 175 as their LCM Give reasons , CBSE 2012 y 3,13 State Fundamental theorem of arithmetic 7. Is it possible that HCF and LCM of two, numbers be 24 and 540 respectively Justify CBSE 2012 .
your answer CBSE 2015 A x 10 y 14 B x 21 y 84, C x 21 y 25 D x 10 y 40. 14 Explain why the number 7 5 3 2 3 is, not a prime number CBSE 2015 8 LCM of 23 32 and 22 33 is. CBSE 2012 ,Answers A 23 B 33, C 23 33 D 22 32,2 HCF 4 LCM 9696 Revisiting Irrational Numbers. 3 23 5 53, Rational Number ,4 HCF 12 LCM 1440, A number r is called a rational number if it can be. 5 72 6 315, 7 80 8 52 written in the form where p and q are integers and.


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