Brief Introduction ’The last lesson’ written by Alphonse Daudet narrates about the year 1870 when the Prussian forces under Bismarck attacked and captured France. The French districts of Alsace and Lorraine went into Prussian hands. The new Prussian rulers discontinued the teaching of French in the schools of these two districts The French teachers were asked to leave. Now M. Hamel could no longer stay in his school. Still he gave lesson to his students with utmost devotion and sincerity as ever. One such student of M. Hamel, Franz who dreaded French class and M. Hamel’s iron rod, came to the school that day thinking he would be punished as he had not learnt his lesson on participles. But on reaching school he found Hamel dressed in his fine Sunday clothes and the old people of the village...

Exercise 1.4 Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/23 × 52 (vii) 129/22 × 57 × 75 (viii) 6/15 (ix) 35/50 (x) 77/210 Answer (i) 13/3125 Factorize the denominator we get 3125 =5 × 5 × 5 × 5 × 5 = 55 So denominator is in form of 5m so it is terminating . (ii) 17/8 Factorize the denominator we get 8 =2 × 2 × 2 = 23 So denominator is in form of 2m so it is terminating . (iii) 64/455 Factorize the denominator we get 455 =5 × 7 × 13 There are 7 and 13 also in denominator so denominator is not in form of 2m × 5n . so it is not terminating. (iv) 15/1600 Factorize the denominator we get...

Exercise 1.3 Q1. Prove that √5 is irrational. Answer: Let take √5 as rational number If a and b are two co prime number and b is not equal to 0. We can write √5 = a/b Multiply by b both side we get b√5 = a To remove root, Squaring on both sides, we get 5b2 = a2 … (i) Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also. That means 5 will divide a. So we can write a = 5c Putting value of a in equation (i) we get 5b2 = (5c)2 5b2 = 25c2 Divide by 25 we get b2/5 = c2 Similarly, we get that b will divide by 5 and we have already get that a is divide by 5 but a and b are co prime number. so it contradicts. Hence √5 is not a rational number, it is irrational. Q2. Prove that 3 + 2√5 is irrational. Answer: Let tak...

Exercise 1.2 Q1. Express each number as product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Answer (i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7 (ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13 (iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17 (iv) 5005 = 5 × 7 × 11 × 13 (v) 7429 = 17 × 19 × 23 Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 Answer: (i) 26 = 2 × 13 91 =7 × 13 HCF = 13 LCM =2 × 7 × 13 =182 Product of two numbers 26 × 91 = 2366 Product of HCF and LCM 13 × 182 = 2366 Hence, product of two numbers = product of HCF × LCM (ii) 510 = 2 × 3 × 5 × 17 92 =2 × 2 × 23 HCF = 2 LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460 Product of two numbers 510 × 92 = 46920 Product of HCF a...

Exercise 1.1 Q1. Use Euclid’s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 Answer (i) 225 > 135 we always divide greater number with smaller one. Divide 225 by 135 we get 1 quotient and 90 as remainder so that 225= 135 × 1 + 90 Divide 135 by 90 we get 1 quotient and 45 as remainder so that 135= 90 × 1 + 45 Divide 90 by 45 we get 2 quotient and no remainder so we can write it as 90 = 2 × 45+ 0 As there are no remainder so divisor 45 is our HCF. (ii) 38220 > 196 we always divide greater number with smaller one. Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as 38220 = 196 × 195 + 0 As there is no remainder so divisor 196 is our HCF. (iii) 867 > 255 we always divide ...

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