Web8 aug. 2024 · Best answer Here , 117 > 65 117 > 65 ∴ ∴ ∴ 117 = 65 × 1 + 52 ∴ 117 = 65 × 1 + 52 65 = 52 × 1 + 13 65 = 52 × 1 + 13 52 = 13 × 4 + 0 52 = 13 × 4 + 0 Since remainder = 0 recent divisor is the H.C.F. ∴ ∴ H.C.F. (117, 65) = 13 Now , 13 = 65 − 52 × 1 13 = 65 - 52 × 1 ⇒ 13 = 65 − (117 − 65 × 1) ⇒ 13 = 65 - ( 117 - 65 × 1) WebHence, find the LCM (A) 1000 (B) 9600 (C) 9640 (D) 9696 21.If the HCF of 210 and 55 is expressible in the form 210 x 5 + 55y, find y (A) 19 (B) 15 (C) -19 (D) -21 22.H.C.F of two integers 26, 91 is 13 what will be its L.C.M.? 23.Determine .875 is terminating or non-terminating. 24.Express 140 in its prime factor. 25.<$ 26.Why is7x11x13+7 a …
If the HCF of 210 and 55 is expressible in the form 210 × 5 - Toppr …
WebIf the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ? Answer: D) -19 Explanation: HCF of 210 and 55 is 5 Now, 210x5 + 55P = 5 => 1050 + 55P = 5 => 55P = -1045 => P = -1045/55 => P = -19. Subject: HCF and LCM - Quantitative Aptitude - Arithmetic Ability Exam Prep: AIEEE , Bank Exams , CAT , GATE WebFind the [HCF×LCM] for the numbers 100 and 190. [2009] 13. The HCF of 45 and 105 is 15. Write their LCM. [2010] 14. Has the rational number 441 ... If the HCF of 210 and 55 is expressible in the form 210×5+55× , then the value of … how thermal imagers work
If the HCF of 210 and 55 is expressible in the for - Sawaal.com
WebOpen in App. Solution. HCF of 210 55 210 = 55* 3 + 45 ..(i) Find HCF of 210 and 55 and express it as a linear . The only common prime factor of 210 and 55 is 5. Hence, the ... Find the HCF of 210 and 55 .If the HCF is expressible in the form of 210m + 55n . … WebIf the HCF of 210 and 55 is expressible in the form 210 × 5 + 55y, find y. 2Pi classes 45.3K subscribers Subscribe 260 views 11 months ago Maths Class 10 (Real Numbers-RD Sharma... Web26 nov. 2024 · 210 = 55 × 3 +45 say it is eq . 1) 55 = 45 ×1 + 10 2) 45 = 4×10 +5 3) now we consider the divisor 10 and the remainder 5 and apply DL to get. 10 = 5×2 +0. we observe here that r =0 and our present divisor which is 5 will be it HcF. A/Q. 5 = 210 ×5 + 55y. metal gate hold back