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  Q.1: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. Solution: Let x be any positive integer and y = 3. By Euclid’s division algorithm; x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3) Therefore, x = 3q, 3q + 1 and 3q + 2 As per the given question, if we take the square on both the sides, we get; x 2  = (3q) 2  = 9q 2  = 3.3q 2 Let 3q 2  = m Therefore, x 2  = 3m ………………….(1) x 2  = (3q + 1) 2 = (3q) 2  + 1 2  + 2 × 3q × 1 = 9q 2  + 1 + 6q = 3(3q 2  + 2q) + 1 Substituting 3q 2 +2q = m we get, x 2  = 3m + 1 ……………………………. (2) x 2  = (3q + 2) 2 = (3q) 2  + 2 2  + 2 × 3q × 2 = 9q 2  + 4 + 12q = 3(3q 2  + 4q + 1) + 1 Again, substituting 3q 2  + 4q + 1 = m, we get, x 2  = 3m + 1…………………………… (3) Hence, from eq. 1, 2 and 3, we conclude that the square of any positive integer is either of...

Important Questions & Answers For Class 10 Maths Chapter 14 Statistics The important questions o f the statistics chapter for class 10 are given here in both shor t answer type and long answer type. Short Answer Type Questions Q.1.   Find the mean of the 32 numbers, such that if the mean of 10 of them is 15 and the mean of 20 of them is 11. The last two numbers are 10. Solution:  The given mean of 10 numbers = 15 So, Mean of 10 numbers = sum of observations/ no. of observations 15 = sum of observations / 10 Sum of observations of 10 numbers = 150 Similarly, Mean of 20 numbers = sum of observations/ no. of observations 11 = sum of observations / 20 Sum of observations of 20 numbers = 220 Hence, Mean of 32 numbers = (sum 10 numbers + sum of 20 numbers + sum of last two numbers)/ no. of observations Mean of 32 numbers = (150 + 220 + 20 ) / 32 = 390 /32 = 12.188 Q.2. Find the mean of the first 10 natural numbers. Solution:  The first 10 natural numbers are 1, 2, 3, 4, 5, ...