the answer is not 31 because it is asking for minimum not maximum so what is the real answer? How did you get it? The average of five distinct integers is 65.If the largest integer is 75, what is themaximum possible value of the smallestinteger?(A) 1(B) 31(C) 61(D) 71(E) 250- Ans:-Given, the average of five distinct integersis 65.let a, b, c, d and ethentep2)⇒a+b+c+d+e5⇒a+b+c+d+e= 65x5 = 325⇒ a+b+c+d+e = 325 .(I)Let a be the largest integer then a= 75.From Ifrom IIare five distinct integers,Since, b, c, d and eLetb= 74,6575 +b+c+d+e 325b+c+d+e=. 325-75 = 250b+c+d+e = 250are distinct.c= 73,d= 72e = 3174+73+72 +e= 250219 += 250- He=250-219AnsHence, e = 31 is the required smallest integer.

Answered: The average of five distinct integers…

The average of five distinct integers is 65. If the largest integer is 75, what is the maximum possible value of the smallest integer? (A) 1 (B) 31 (C) 61 (D) 71 (E) 250-

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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