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-
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
Related questions
Question
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 the
maximum possible value of the smallest
integer?
(A) 1
(B) 31
(C) 61
(D) 71
(E) 250-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd19722f0-ad57-46c7-85cc-fbe792f30d47%2F87945109-0fbf-4846-aed9-014726b4b3f4%2Feoowv7m_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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-
![Ans:-
Given, the average of five distinct integers
is 65.
let a, b, c, d and e
then
tep2
)
⇒
a+b+c+d+e
5
⇒
a+b+c+d+e= 65x5 = 325
⇒ a+b+c+d+e = 325 .(I)
Let a be the largest integer then a= 75.
From I
from II
are five distinct integers,
Since, b, c, d and e
Let
b= 74,
65
75 +b+c+d+e 325
b+c+d+e=. 325-75 = 250
b+c+d+e = 250
are distinct.
c= 73,d= 72
e = 31
74+73+72 +e= 250
219 += 250
- H
e=250-219
Ans
Hence, e = 31 is the required smallest integer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd19722f0-ad57-46c7-85cc-fbe792f30d47%2F87945109-0fbf-4846-aed9-014726b4b3f4%2Fq4hg7q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Ans:-
Given, the average of five distinct integers
is 65.
let a, b, c, d and e
then
tep2
)
⇒
a+b+c+d+e
5
⇒
a+b+c+d+e= 65x5 = 325
⇒ a+b+c+d+e = 325 .(I)
Let a be the largest integer then a= 75.
From I
from II
are five distinct integers,
Since, b, c, d and e
Let
b= 74,
65
75 +b+c+d+e 325
b+c+d+e=. 325-75 = 250
b+c+d+e = 250
are distinct.
c= 73,d= 72
e = 31
74+73+72 +e= 250
219 += 250
- H
e=250-219
Ans
Hence, e = 31 is the required smallest integer.
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