For a set of five distinct integers, the median, as well as the mean

Question

For a set of five distinct integers, the median, as well as the mean, is 24. If the range of the set is 16, what is the greatest possible value of the smallest integer in the set?

A. 17
B. 18
C. 19
D. 20
E. 22

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chloreton · Answer

lowest value = x
highest value = 16+x (Since range => H - L = 16)
Median is 24

we need to maximize x, which means we need to minimize other values. We will arrange the numbers as x, x+1, 24, 25, 16+x
We know avg = 24 ; therefore sum = 5*24=120

Now we can perform the calculation as x+(x+1)+24+25+(16+x) = 120
we get x =  18 .

Option B

Adit_ · Answer

Take 2nd value right of mean to be +1 in order to maximize smallest value.
Why?
So that the negative deviation to mean is very small such that right side need not be compensated by much and since its all unique the lowest value we can provide to right side of mean is +1
So now we have 3rd value = 24, 4th value = 25.
Now simply pick values from options 17 then last value is 33, diff of 9 with mean 9+1 = 10 and 7+6 = 13 still on the left, so it can be inceased further
18 is thus the answer as:
6+5 deviation on the left is perfectly = 1+10 on the right side.

Answer: Option B

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This approach is the SD approach I have taken as i find it very intuitive and logical and helps save lot of time in complex problems where dealing with equations would be annoying.

GraemeGmatPanda · Answer

We define variables for the five distinct integers in increasing order.
 s < a < m < b < t 
 	ext{where }s=	ext{smallest},\ m=	ext{median},\ t=	ext{largest}

We use the median property Median and translate the relevant part of the question "the median, as well as the mean, is 24" into a value for the middle element.
 m = 24

We apply the average formula Average (Arithmetic Mean) to translate "the mean ... is 24" into a sum for all five integers.
 \frac{s + a + m + b + t}{5} = 24 
 s + a + m + b + t = 120

We use the range property Range to translate "the range of the set is 16" into a relationship between largest and smallest.
 t - s = 16

We minimize the sum of the two unknowns  a  and  b  (subject to order constraints) to maximize  s , then solve the resulting equations.
 t = s + 16 
 s + a + 24 + b + (s+16) = 120 
 2s + a + b + 40 = 120 
 2s + a + b = 80 
 a + b = 80 - 2s

ext{Order constraints: }a \ge s+1,\ a \le 23,\ b \ge 25,\ b \le s+15 
 	ext{Minimal allowed }(a,b) = (s+1,25)	ext{ so }a + b = s + 26

s + 26 = 80 - 2s 
 3s = 54 
 s = 18 
 	ext{Then }a = 19,\ b = 25,\ t = 34 
 18 + 19 + 24 + 25 + 34 = 120	ext{ (works)}

ext{Therefore the greatest possible smallest integer is }18

Answer B

Hope this helps! 
ʕ•ᴥ•ʔ

egmat · Answer

Let's call the  5  integers in order: a, b, c, d, e.

What we know: 
- Median =  24 , so the middle value c =  24 
- Mean =  24 , so a + b + c + d + e =  120 
- Range =  16 , so e = a +  16

Plug in c =  24  and e = a +  16 :
a + b +  24  + d + (a +  16 ) =  120 
 2 a + b + d =  80 
So: b + d =  80  -  2 a

Key Insight:  To maximize a, we need to minimize b + d.

Now here's the key — we have constraints because the integers must be distinct and in order:
- b must be greater than a, so the smallest b can be is a +  1 
- d must be greater than  24 , so the smallest d can be is  25 
- d must be less than e = a +  16

Setting b and d to their minimums: 
(a +  1 ) +  25  =  80  -  2 a
a +  26  =  80  -  2 a
 3 a =  54 
a =  18

Verification:  the set is { 18 ,  19 ,  24 ,  25 ,  34 }
- All  5  are distinct integers ✓
- Median (middle value) =  24  ✓
- Sum =  120 , so mean =  120 / 5  =  24  ✓
- Range =  34  -  18  =  16  ✓

Answer: B (18)

General principle:  When asked to maximize the smallest value, you need to minimize everything else that's flexible. Here, the  2 nd and  4 th values were the flexible pieces, so we pushed them as small as the constraints allowed.

ExpertsGlobal5 · Answer

Video explanation:

https://www.youtube.com/watch?v=2WUrEQTIh1I

For a set of five distinct integers, the median, as well as the mean

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For a set of five distinct integers, the median, as well as the mean

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