Page 1 Question 1. Find the value of 32 48 812 + + Solution: Since 32 16 2 =× = 42; 48 16 3 4 3 =×= 8 = 42 × = 22and 12 4 3 2 3 =× = ? 32 48 812 + + = 42 4 3 22 2 3 + + = () () 42 3 2 22 3 + = + Question 2. If 2 = 1.4142 then find the value â€“ 21 21 + . Solution: Rationalising the denominator of 2â€“1 21 + , we get 2â€“1 21 + = 2â€“1 2â€“1 21 2â€“1 × + = () () () 2 2 2 2â€“1 2â€“1 = () 2 2â€“1 2â€“1 = () 2 2â€“1 ? 2â€“1 21 + = () 2 2â€“1 or 2â€“1 21 + = 2â€“1 ........ (1) ä 2 = 1.4142 â 2â€“1 21 + = 2â€“1 = 1.4142 â€“ 1 = 0.4142 Question 3. Find the value of 'a' in â€“ â€“ 35 19 a5 32 5 11 = + Solution: L.H.S. = 3â€“ 5 32 5 + = 3â€“ 5 32 5 + × 3â€“ 2 5 3â€“2 5 Rationalising the denominator = () () 2 2 9â€“6 5 â€“3 5 10 3â€“25 + 22 using (a b) (a â€“ b) a â€“ b += = 19 â€“ 9 5 19 â€“ 9 5 9 â€“ 20 â€“11 = = 919 5â€“ 11 11 R.H.S. = 19 5â€“ 11 a since, L.H.S. = R.H.S. Page 2 Question 1. Find the value of 32 48 812 + + Solution: Since 32 16 2 =× = 42; 48 16 3 4 3 =×= 8 = 42 × = 22and 12 4 3 2 3 =× = ? 32 48 812 + + = 42 4 3 22 2 3 + + = () () 42 3 2 22 3 + = + Question 2. If 2 = 1.4142 then find the value â€“ 21 21 + . Solution: Rationalising the denominator of 2â€“1 21 + , we get 2â€“1 21 + = 2â€“1 2â€“1 21 2â€“1 × + = () () () 2 2 2 2â€“1 2â€“1 = () 2 2â€“1 2â€“1 = () 2 2â€“1 ? 2â€“1 21 + = () 2 2â€“1 or 2â€“1 21 + = 2â€“1 ........ (1) ä 2 = 1.4142 â 2â€“1 21 + = 2â€“1 = 1.4142 â€“ 1 = 0.4142 Question 3. Find the value of 'a' in â€“ â€“ 35 19 a5 32 5 11 = + Solution: L.H.S. = 3â€“ 5 32 5 + = 3â€“ 5 32 5 + × 3â€“ 2 5 3â€“2 5 Rationalising the denominator = () () 2 2 9â€“6 5 â€“3 5 10 3â€“25 + 22 using (a b) (a â€“ b) a â€“ b += = 19 â€“ 9 5 19 â€“ 9 5 9 â€“ 20 â€“11 = = 919 5â€“ 11 11 R.H.S. = 19 5â€“ 11 a since, L.H.S. = R.H.S. i.e. 919 5â€“ 11 11 = 19 5â€“ 11 a ? 9 11 = a Question 4. Find the value of 'a' and 'b' 75 7â€“5 â€“ 7â€“ 5 7 5 + + = 7 a5b 11 + Solution: L.H.S. = 75 7â€“5 â€“ 7â€“ 5 7 5 + + = ()() ()() 22 75 â€“7â€“5 7â€“ 5 7 5 + + = 49 514 5â€“49â€“514 5 49 â€“ 5 ++ + = 47 5 75 44 11 ?? ?? = = 7 05 11 + R.H.S. = a + 7 5b 11 Since, L.H.S. = R.H.S ? 0 + 7 5 11 = a + 7 5b 11 ? a = 0 and b = 1 Question 5. If a = 35 2 + , then find the value of + 2 2 1 a a . Solution: a = 35 2 + ? 1 a = 2 35 + Now 2 2 1 + a a = 2 1 â€“2 ?? + ?? ?? a a ? 2 2 1 + a a = 2 35 2 â€“2 23 5 ?? + + ?? + ?? = () () 2 2 2 35 2 â€“2 23 5 ?? ++ ?? ?? + ?? ?? = 2 95 2 3 5 4 â€“2 23 2 5 ?? ++ × × + ?? ×+ ?? = 2 18 6 5 â€“2 62 5 ?? + ?? + ?? = () () 2 63 5 â€“2 23 5 ?? + ?? ?? + ?? = 2 6 â€“2 2 ?? ?? ?? = () 2 3â€“2 =9 â€“ 2 = 7 So, 2 2 1 + a a = 7Read More

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