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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  [#permalink]

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28 Jan 2020, 23:38
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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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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 00:06
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1
$$\frac{a+b+c+d+e}{5}=22$$

$$a+b+c+d+40=110$$

$$a+b+c+d=70$$

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 $$12$$

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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 00:29
1
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

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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 00:58
1
1
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
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The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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Updated on: 30 Jan 2020, 01:24
2
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

Originally posted by chondro48 on 29 Jan 2020, 01:19.
Last edited by chondro48 on 30 Jan 2020, 01:24, edited 1 time in total.
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 01:32
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
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
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 01:55
1
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
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 02:13
2
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.
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 03:10
1
Quote:
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

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-[39]=34≠divby3
73-[38]=35≠divby3
73-[37]=36=divby3
m=36/3=12 (10<11<12<[37]<40)

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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 03:18
1
a<b<c<d<e and a+b+c+d+e=22*5=110
But we know that e=40,
Hence for the least possible median (i.e. c), then d must be 39.
also, let’s assume that b = c-1 and a = c-2
Then c-2+c-1+c+39+40>=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

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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 05:06
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
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 05:51
1
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
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 07:52
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a+b+c+d+e =22*5= 110
e =40
a <b < c< d <e
The least value of c ( median) —?
e = 40 —> d = 39
a+b+c = 110–40–39= 31
The optimal option —9+10 +12 =31
—> c = 12

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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 08:42
1
(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
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 08:49
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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)
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 09:08
1
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!
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 10:00
1
Quote:
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

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<b<c => 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
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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e  [#permalink]

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29 Jan 2020, 10:06
1
1
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.

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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e   [#permalink] 29 Jan 2020, 10:06
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