The average (arithmetic mean) of five positive integers a, b, c, d, e

Question

Competition Mode Question

The average (arithmetic mean) of five positive integers a, b, c, d, e is 22 and a < b < c < d < e. If e is 40, what is the least possible value of the median of the five integers?

A. 10
B. 11
C. 12
D. 13
E. 14

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firas92 · Accepted Answer

\frac{a+b+c+d+e}{5}=22

a+b+c+d+40=110

a+b+c+d=70

Since  a<b<c<d<e , we know that  c  is the median. Let's plug in the answer choices,

A. 10

If  c=10, b=9  and  a=8 ,

d=70-10-9-8=43

This is not possible since  d<e  and so  d  must be less than  40

B. 11

If  c=11, b=10  and  a=9 ,

d=70-11-10-9=40

Again this is not possible since  d  must be less than  40

So the answer must be the next smallest value which is  12

Answer is  (C)

Staphyk · Answer

The average (arithmetic mean) of five  positive integers  a, b, c, d, e is 22 and  a < b < c < d < e .If  e is 40 , what is the  least possible value  of the  median  of the five integers?
Median =c?

a+b+c+d+e= 110(5•22)
a+b+c+d+40=110
Now to minimize c we have to maximize d and a<b<c<d<e
Where d<40, dmax =37
10+11+12+37+40= 110 , c=12

A. 10
B. 11
 C. 12 
D. 13
E. 14

Hit that C

Posted from my mobile device

rajatchopra1994 · Answer

Explanation:
average a, b, c, d, e is 22 
a+b+c+d+e = 22 x 5 = 110
e=40
a+b+c+d = 70

as we need least possible value of the median, 
Put, d=39 ( d < e), b=c-1, a = c-2
a+b+c+d = 70
(c-2)+(c-1)+c+39=70
3c = 34
c = 11.33

So, least possible value of the median c = 12

IMO-C

freedom128 · Answer

a < b < c < d < 40, where a,b,c,d are all positive integers.
a + b + c + d + 40 = 22*5 ---> a + b + c + d = 70.

QUESTION: What is least possible value of c (median)?

To get least possible value of c (median), we need to assume d_max=40-1=39. Thus, a + b + c + 39 = 70 --> a + b + c = 31 (minimum).

Let's try the option that satisfies a + b + c = 31 (a < b < c)
A. 10 --> a + b + c = 8 + 9 + 10 = 27 <31 (No!)
B. 11 --> a + b + c = 9 + 10 + 11 = 30 <31 (No!)
C. 12 --> a + b + c = 10 + 11 + 12 = 33 >31 (YES!), because value of d can be lower than d_max=39

FINAL ANSWER IS (C)

debjit1990 · Answer

Ans: B

am=22
total=5*22=110
so, x+(x+1)+(x+2)+(x+y)+40=110(median c=x+2)
4x+4+y=70
4x+y=66
we have to make y as much as possible, but x+y<40, as highest number is 40
going by the answers-->
a)10,so, x+2=10,x=8
4x+y=66
32+y=66,so,y=34...x+y(4th term)=34+8=42>40 not possible

b)11,so, x+2=11,x=9
4x+y=66
36+y=66,so,y=30...x+y(4th term)=30+8=38<40 possible

Archit3110 · Answer

sum of a, b, c, d, e= 100
and a+b+c+d = 70
since a<b<c<d<e
e=40 
solve using plug in 
at c= 12 ; a=10 , b=11 we get d= 37 
hence sufficient
IMO C ;12

The average (arithmetic mean) of five positive integers a, b, c, d, e is 22 and a < b < c < d < e. If e is 40, what is the least possible value of the median of the five integers?

A. 10
B. 11
C. 12
D. 13
E. 14

sukoon9334 · Answer

This question requires very direct calculation using logic.

If mean of a,b,c,d,e is 22, then the a+b+c+d+e= 110
Given: e=40 which implies a+b+c+d=70

Now,
Since a<b<c<d<e, 'c' is the median. 
For the median to be the lowest possible, we'll have to distribute the remaining 70 in a way that a,b,c are the lowest. (since a<b<c)
For a,b,c to be the lowest, d has to be the highest possible i.e. d=39 (d<e=40)

Now, a+b+c=31. By Brute Force method, we find that the closest we ca go to a sum total of 31 without breaking the rule a<b<c is 9,10,11 BUT this sums up to only 30!
So, we can easily add just 1 and get the series 9,10,12 i.e. a=9, b=10, c=12. 
Therefore the lowest possible median is 12.

exc4libur · Answer

avg=22, n=5, sum=110
a < b < c < d < e = m-2 < m-1 < m < x < 40
3m-3+x+40=110
3m+x=73 (we must maximize x to find the smallest m)
m=(73-x)/3=integer (x<40)
73- =34≠divby3
73- =35≠divby3
73- =36=divby3
m=36/3=12 (10<11<12< <40)

Ans (C)

eakabuah · Answer

a=110 3c+76>=110 3c>=34 c>=11.33 Since c is an integer, the minimum integer that satisfy the criteria above is 12 Hence c=12 The answer is C. Posted from my mobile device

Win2020 · Answer

The average (arithmetic mean) of five positive integers a, b, c, d, e is 22 and a < b < c < d < e. If e is 40, what is the least possible value of the median of the five integers?

A. 10
B. 11
C. 12
D. 13
E. 14

Soln

Given - a,b,c,d,e== Positive Integers & a < b < c < d < e
 
 a+b+c+d+e/5=22
 so a+b+c+d+e=110
 
we are given e=40 so now we know a+b+c+d+40=110

Sum of remaining four integers (a+b+c+d)=70 , also we already know e is largest Integer among all 5 , so remaining four integers have to be less than 40 and Positive.

To find least value of median ( median in odd number of terms is the middle value) .So Median here is c,
Question has asked us to find least possible value of c.

So far we know sum of a,b,c,d=70 ,

We want least value of c, so lets maximize the d , condition is d<e so as e=40 , d can max be 39
 
So now we are left with a+b+c= 70-39=31

sum of remaining 3 integers has to be 31, now think of ways of distributing 31 among a,b,c such that a<b<c

Take clue from ans options Ans choice A : if c=10, b=9, a=8 Not equal to 31 Eliminate A
 Ans choice B: if c=11, b=10, a=9 Not equal to 31 Eliminate B
  Ans choice C: if c=12, b=10, a=9 Yes now its sum =31 CORRECT 
 Ans choice D: if c=13, b =10 a=7 also equals 31 but we are asked least value of c
 Ans choice E: if c=14, b=10, a=6 also equals 31 but we are asked least value of c

Ans is B

Please hit kudos ,if it helped you

CareerGeek · Answer

Median =  c 
Given, average =  22  &  e = 40 
-->  \frac{(a + b + c + d + e)}{5} = 22 
-->  a + b + c + d + 40 = 110 
-->  a + b + c + d = 70

Note that, " c " is least when  a, b  &  d  are highest.
So, maximum possible value of  d = e - 1 = 40 - 1 = 39 
-->  a + b + c = 70 - 39 = 31 
--> The only possible set of  (a, b, c)  in which " c " is least =  (9, 10, 12) 
--> Least possible value of  c = 12

Option C

lacktutor · Answer

a+b+c+d+e =22*5= 110 e =40 a d = 39 a+b+c = 110–40–39= 31 The optimal option —9+10 +12 =31 —> c = 12 The answer is C Posted from my mobile device

GMATSheep · Answer

(a + b + c + d + 40)/5 = 22

a + b + c + d = 70 , We want to minimize c (the median) so let d = 39 as that is its max value. Now a + b + c = 31

If c is 10 then a + b = 21, which is impossible because both a and b must be smaller than c. If c is 11 then a + b = 20, which is not possible because both a & b cannot be 10. If c is 12 then a + b = 19, which is possible if for example b is 11 and a is 8 or b is 10 and a is 9.

ANS: C

Jawad001 · Answer

a + b + c + d +e /5 =22
a + b + c + d +e =110
a + b + c + d +40 = 110 e = 40
a + b + c + d = 110-40 = 70

Now, a < b < c < d < e, Here ,the value of median should be = the value of c
For being the value of median least, the value of a, b and d should be maximum as much as possible.
d =39, and a + b + c =31, that is c = 12, b = 10, a = 9.
Median, C= 12(C)

fauji · Answer

Approach:

Given: a < b < c < d < 40, Total = 22*5 = 110

To find c or median as smallest possible value:
- We need to have d as largest possible value, d = 39
- To get a+b+c = 31 with a < b < c, only c = 12 possible with a and b be 10 and 11.

IMO Option C!

ostrick5465 · Answer

The median of the five positive integers is c. While c is at the least possible value, d must be at the greatest possible value. because a < b < c < d < e and e=40 => the greatest possible value of d is 39 => the sum of a, b ,c is 22*5 - 40-39 = 31 because a the least possible value of c is greater than the average of a,b,c => c > \frac{31}{3} . if c =11 => b=10, a=9 => 9+10+11 = 30 <31 => eliminate c=11 if c =12 => b=11, a=8 or b=10, a=9 => Choice C : c=12

unraveled · Answer

The average (arithmetic mean) of five positive integers a, b, c, d, e is 22 and a < b < c < d < e. If e is 40, what is the least possible value of the median of the five integers?

A. 10
B. 11
C. 12
D. 13
E. 14

a + b + c + d + e = 22 * 5 = 110

Looking at the options median(c) must be as far from e as possible which is possible if a, b and c are closer to each other i.e. if they are consecutive numbers.

Let a = 9, b = 10 and c = 11 then
d = 110 - (9 + 10 + 11) - 40
 d = 40   which is not possible since d < e

Let a = 10, b = 11 and c = 12 then
d = 110 - (10 + 11 + 12) - 40
 d = 37  which is possible since d < e

No need to check further here. c = 12.

Answer C.

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The average (arithmetic mean) of five positive integers a, b, c, d, e

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The average (arithmetic mean) of five positive integers a, b, c, d, e

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