## Knowledge in Class XII Real Numbers - Ncert Solutions for Class 10 Maths CBSE

Chapter 1 - Real Numbers Excercise Ex. 1.1Solution 1(i) 135 and 225&nbsp;Step 1:&nbsp;&nbsp;Since 225 &gt; 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r&nbsp;&nbsp;such that 225 = 135q+r, 0&nbsp;&nbsp;r&lt;135&nbsp;On dividing 225 by 135 we get quotient as 1 and remainder as 90&nbsp;i.e 225 = 135 x 1 + 90&nbsp;Step 2:&nbsp;Remainder r which is 90&nbsp;&nbsp;0, we apply Euclid's division lemma to b =135 and r = 90 to find whole numbers q and r such that&nbsp;135 = 90 x q + r,&nbsp;0&nbsp;&nbsp;r&lt;90On dividing 135 by 90 we get quotient as 1 and remainder as 45i.e 135 = 90 x 1 + 45Step 3:&nbsp;&nbsp;Again remainder r = 45&nbsp;&nbsp;0 so we apply Euclid's division lemma to b =90 and r = 45 to find q and r&nbsp;&nbsp;such that&nbsp;90 = 90 x q + r,&nbsp;&nbsp;0&nbsp;&nbsp;r&lt;45On dividing 90 by 45 we get quotient as 2 and remainder as 0&nbsp;i.e 90 = 2 x 45 + 0Step 4:&nbsp;Since the remainder is zero, the divisor at this stage will be HCF of (135, 225).&nbsp;Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.&nbsp;(ii)&nbsp;&nbsp;&nbsp;196 and 38220Step 1:&nbsp;&nbsp;Since 38220 &gt; 196, apply Euclid's division lemmato a =38220 and b=196&nbsp;to find&nbsp;whole numbers q and r&nbsp;&nbsp;such that&nbsp;38220 = 196 q + r, 0&nbsp;&nbsp;r &lt; 196On dividing 38220 we get quotient as 195 and remainder r as 0&nbsp;i.e 38220 = 196 x 195 + 0Since the remainder is zero, divisor at this stage will be HCF&nbsp;Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.NOTE: HCF( a,b) = a if a is a factor of&nbsp;b. Here, 196 is a factor of&nbsp;38220 so HCF is 196.&nbsp;(iii)&nbsp;&nbsp;&nbsp;867 and 255&nbsp;Step 1:&nbsp;Since 867 &gt; 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r&nbsp;&nbsp;such that 867 = 255q + r, 0&nbsp;&nbsp;r&lt;255On dividing 867 by 255 we get quotient as 3 and remainder as 102i.e 867 = 255 x 3 + 102&nbsp;Step 2:&nbsp;Since remainder 102&nbsp;&nbsp;0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that&nbsp;255 = 102q + r where 0&nbsp;&nbsp;r&lt;102On dividing 255 by 102 we get quotient as 2 and remainder as 51&nbsp;i.e 255 = 102 x 2 + 51Step 3:&nbsp;Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51&nbsp;to find whole numbers q and r such that&nbsp;102 = 51 q + r where 0&nbsp;r &lt; 51&nbsp;&nbsp;On dividing 102 by 51 quotient is 2 and remainder is 0i.e 102 = 51 x 2 + 0Since the remainder is zero, the divisor at this stage is the HCFSince the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51.&nbsp;Concept Insight: To crack such problem remember to apply the Euclid's division Lemma which states that "Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0&nbsp;&nbsp;r &lt; b" in the correct order.&nbsp;Here, a &gt; b.&nbsp;Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives&nbsp;a number which divides a and b exactly.&nbsp;i.e&nbsp;&nbsp;&nbsp;HCF(a,b) =HCF(b,r)Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps. Information about the SECOND HAND books at reasonable rates

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