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Ans D 37

Sum is 30 x 7 = 210

since median is 27 that means a4 is 27. we want to find the minimum value of a7

we will put in the max values of a1 2 and 3 which will be 24,25,26 since all value are different.

24+25+26+27+a7-2 + a7-1 +a7 =210

a7-2 + a7-1 +a7=108

3a7=111

a7= 37



Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


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Mean. = 30, Sum of 7 integers = 210

Now median is 27 = a4 value
to minimize the highest value, we have to maximize the lowest values
then, and each is different than a3 = 26, a2 = 25, and a1 = 24
Sum of a1 to a4 =. 24+25+26+27 = 102
then a5+a6+a7 = 210-102 = 108
Let's say all are equal, then 3a = 108, a = 36

to create a5<a6<a7, remove 1 from a5 and add to a7, 35<36<37 = lowest possible maximum value ( D )
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27 is median and all marks are different. To get minimum high score, other marks must be the highest possible values. So,

_ _ _ 27 _ _ _ . 24,25,26 are maximum values possible before median 27. 7*30=210. So , 210-24-25-26-27=108. Minimum highest value can be obtained if all the values after median are near by values. 108/3=36. As they are given varying marks, 36,36,36 can be transformed to 35,36,37. Hence 37.
Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


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for the GMAT Club Olympics Competition

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Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


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for the GMAT Club Olympics Competition

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Let the seven students who took the exams be A,B,C,D,E,F and G.

The average marks scored by all 7 = 30.

Total marks by all seven students = 7*30 =210

Moreover, given that the median is 27. Let the students marks be arranged from lowest to highest.

And their ranking be same as A,B,C,D,E,F and G. With A be the lowest and G being the maximum.

The marks of Students A,B,C should be below median ( as all values are distinct and like between 0 and 100).

we need to get the minimum value of Max G.

so, we maximise the values of A,B, and C as 24,25,26. Hence, the sum of A+B+C+D = 102

Remaining sum = 210 -102 = 108.

E+F+G = 108

Case 1: G= 34, then E+F = 108-34 = 74. E,F should be less than 34 and distinct, so 33+32 = 65 < 74. Hence eliminated.

Case 2: G= 35, then E+F = 108-35 = 73. E,F should be less than 35 and distinct, so 33+34 = 67 < 73. Hence eliminated.

Case 3: G= 36, then E+F = 108-36 = 72. E,F should be less than 36 and distinct, so 35+34 = 69 < 72. Hence eliminated.

Case 4: G= 37, then E+F = 108-37 = 71. E,F should be less than 37 and distinct, so 35+36 = 71 = 71. Hence correct.

The minimum value for max G = 37

option D
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Let's say the scores of the 7 students in ascending order are: A,B,C,D,E,F, and G

We are given the median is 27, which is the middle value; so D = 27
Also, we are given that mean is 30, so sum of all the scores is 210

A+B+C+D+E+F+G = 30*(7) = 210 [equation 1]

We are asked to find the minimum value of G.


To minimize the value of G, we have to maximize the value of A,B, and C
Since we are told that all scores are different integers, we cannot assume A,B, and C to be equal to the median.

So, we take:
A = D-3 = 24
B = D-2 = 25
C = D-1 = 26

This ensures that all these values are different integers, and that they are the max possible.

Plugging this in equation 1, we get,

E+F+G = 108,

Now to find the minimum value of G; E and F would have to be their greatest values, and that is only possible when they are

E = G-2
F = G-1

Plugging this back in the equation we get,

3G - 3 = 108
3G = 111
G = 37


Answer D. 37
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Let the scores be s1,s2,s3,s4,s5,s6 and s7.
avg=30 that means
s1+s2+s3+s4+s5+s6+s7=210
and s4= 27 (As s4 is the median)

To minimize s7, we need to maximize the other scores, s1,s2,s3,s5,s6, while respecting the conditions.
As s1,s2,s3 will be less than 27. Lets put them 24,25,26 respectively.

Therefore,
24+25+26+27+s5+s6+s7=210
s5+s6+s7=108

Now, To minimize s7, we need s5 and s6 to be as large as possible.
Then s6 must be at most s7−1. And s5 must be at most s7−2.

s7-2+s7-1+s7=108
3s7=108+3
s7=37

Option D) 37
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the average of 7 scores was 30, so the total points scored = 210

we're told that 27 = the median. Because this is an odd numbered set, we know that 27 is the middle number --> i.e., there are 3 numbers above and 3 below.

we are also told that each student received a different integer score, and we are asked to determine the minimum possible score for the highest score

it's important to recognize that the answer choices represent the information sought after by the question stem; therefore, we can PLUG IN THE ANSWER choices
--> since the question asks for the MINIMUM possible value for the highest score, we might save time by starting with 34 and seeing if it works:

__ __ __ 27 __ __ 34 = 210:
27+34= 61

b/c each of the scores have to be DIFFERENT Integers, let's just make the next two highest scores one and two less than 34 --> i.e. 33, and 32 = 65 + 61 = 126

210- 126= 84, which means the remaining 3 scores would each have to be greater than 28 on average, which means one of the scores has to be greater than the median of 27, which is not mathematically permitted if 27 is in fact the median of an odd numbered set.

therefore, 34 will not work; let's try 36:

__ __ __ 27 __ __ 36= 210
27+36= 63

make the next numbers 35 and 34 = 69 + 63= 132
210-110=100-22= 78, making the remaining scores each a different integer, we can subtract 26, then 25, and we are left with 27,
which will not work because it is the same as the median.


try 37:
__ __ __ 27 __ __ 37= 210
37+27= 64 + 36+ 35 = 64 + 71 = 135
210-135= 75- 26 = 49-25= 24

Bingo, 37 works: 24 + 25 + 26 + 27 + 35 +36 + 37 = 210 Answer: D

Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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Let the 7 integers be:

\(a<b<c<d<e<f<g\)

\(d\) is the median, and \(d=27\)

As per the question,
\(a+b+c+27+e+f+g = 30*7 = 210\) ..... (1)

To derive the minimum value of \(g\), we make all other values as high as possible.
So, \(a=24\), \(b=25\), \(c=26\)

Now, \(e= g-2\), and \(f=g-1\)

Using these in the eqn (1)
\(\\
24+25+26+27+(g-2)+(g-1)+g = 210\\
3g-3=108\\
g=37\\
\)

Answer D
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To minimize the highest score, we need to maximize every other score.
The mean is \(30\). Therefore, the sum of the scores is \(30*7 = 210\).
The median of these \(7\) scores is \(27\). So the \(4th\) value is \(27\).
The scores look something like this -> \(x_1, x_2, x_3, 27, x_5, x_6, x_7\)

Every score is different, so the maximum values for \(x_3\) is \(1\) less than \(x_4\). The maximum value for \(x_2\) is \(1\) less than \(x_3\) and the max value for \(x_1\) is \(1\) less than \(x_2\). We know \(x_4\) is \(27\).

Now the scores look like this -> \(24, 25, 26, 27, x_5, x_6, x_7\)

Max value for \(x_6\) is \(1\) less than \(x_7\) and the max value for \(x_5\) is \(2\) less than \(x_7\)

Scores -> \(24, 25, 26, 27, (x_7-2), (x_7-1), x_7\)

Their sum needs to be \(210\)

\(24+25+26+27+(x_7-2)+(x_7-1)+x_7 = 210\)
\(102 + 3x_7 -3 = 210\)
\(3x_7 = 111\)
\(x_7 = 37\)

For this question, it is very important to remember that the scores are all distinct. If you make the mistake of maximizing the first three scores by keeping them equal to the median and the second last and third last scores by keeping them equal to the highest score, you will get the question wrong (I made this mistake lol. Didn't read the question properly)
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Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

In this case, i use a very straight forward approach.
That said, we take the total value, which is the average * the number of entities: 30 * 7 = 210
and we look at the median. The only fixed value we have: 27.

No we look at the left side. To get to the smallest value possible, we neet to maximize the left side (as well as the right one).

The first iteration is:
24, 25, 26 and 27.
This is the max possible for the left and in sum: 103.

Now we need to fill the right space with 210-103 = 107.
107/3 = 35(2/3) - Would it be an even number like 35, we could just take 34,35 and 36 and it would be the ideal minimum.
As this is not possible, we would hit to high, if we would take 35,36 and 37 and too low, if 34,35 and 36. Therefore. we have to take one away in the row of: 35,36 and 37 -> 34,36 and 37, and here we go. Our ideal smallest max.

Therefore, D) is the correct answer.
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Question analysis: If n = 7 and avg = 30, the sum of all scores (S1 to S7) = 7x30=210. We must find the minimum possible value for the highest score (S7) and this will happen when all other values are the maximum possible.

Finding the answer: If med = 27 (S4) and we need to maximize all scores but the highest, scores from S1 to S3 will be 24, 25 and 26, resulting in a subtotal of 102. Therefore, the lack to be filled with 3 highests scores is of 108 (36x3). As all scores must be differents, this result in scores from S5 to S7 being 35, 36 and 37, respectively. So minimum highest score is 37 -> D

-------------------
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38
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We can solve this question with visibility method
we have 7 students 3+1+3 = to see median and median is 27.
Avg is 30 so total is 7*30 = 210.
Now to minimize highest score which will be 7th in position maximize all left part of 27 and that can be 27 as we have median 27. now adding all 4 we got 108 and 210 is total from avg, we are left with 102.

To minimize highest we can equally distribute this 102 in 3 students right of median because this way we can minimize max value. 102 /3 = 34
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Topic(s)- Statistics, Max/Min
Strategy- Max/Min
Variable(s)- Values = "x"; Smallest Value -> Largest Value = {x1, x2, x3, x4, x5, x6, x7}
Rephrase the Question: What is (x7)min?

1. To minimize x7, maximize all other values
{(x1)max, (x2)max, (x3)max, (x4)max, (x5)max, (x6)max, (x7)min}
i. For an odd # of values: median = middle number
(x4)max = 27
ii. Determine the values to the left of the median
(x3)max = (x4)max -1 = 26
(x2)max = (x3)max -1 = 25
(x1)max = (x2)max - 1 = 24
iii. To maximize the values to the right of the median, we want these to be as close to (x7) as possible
(x7)min = (x7)min
(x6)max = (x7)min -1
(x5)max = (x6)max -1 = (x7)min - 2

2. Use the average formula to find (x7)min
Average = (sum of numbers)/(number of numbers)
30 = (24 + 25 + 26 + 27 + ((x7)min-2) + ((x7)min -1) + (x7)min)) / (7)
3(x7)min = 111 -> (x7)min = 37

Answer: D
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D. 37
You are looking for the minimu of the highest score, so you need the rest of the numbers to be as high as possible. Therefor you take the highest 3 numbers below the median for the first 4: 24, 25, 26, 27
now you need the highest fifth and sixth number, so the seventh can be the lowest, while average stays at 30.
Hence, you need 3 consecutive numbers, such that their summed differences with 30 is equal to 30-27, 30-26, etc.
6+5+4+3=18
n-30 + (n+1)-30 + (n+2)-30 = 18
3n = 105
n = 35
heighest = 35+2 = 37
Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 


Let us assume the scores of the students as s1,s2,s3,s4,s5,s6,s7. We know that s4=27. and avg =30. We also know that s1<s2<s3<s4<s5<s6<s7. As they are different scores

To get the min value s7, All the other scores has to take the max possible values.

Max(s1) = 24, Max(s2) = 25, Max(s3) = 26 , s4 = 27. Now we already know that the average is 30. Each of the students on the left is short by 6,5,4,3. (18 marks).
We will compensate this with the right side. s6= 36, s5=35, s7=37.

Hence the min s7 is 37. IMO D
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Avg score \(=30\)

Total score of 7 students \(=30*7=210\)

Median \(=27\) and each score is a different integer

Let's say the scores are \(x,y,z,27,a,b,c\)

\(x+y+z+a+b+c=210-27=183\)

To minimize \(a,b,c\), \(x,y,z\) needs to be maximized. Maximum \(x,y,z\) must be \(24,25,26\)

\(a+b+c=108\). Average of \(abc=36\). \(a=b=c=36\) is not possible, since each needs to be a different integer.

The possibility with minimum values for \(a,b,c\) are \(35,36,37\)

\(c=37\)

Answer: D
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For AM to be 30, we have a total aggregate of 7 x 30 = 210.

If median is 27, the minimum for the 3 values below / equal to 27 can be taken as 27.

Reducing 27 x 4 from 210 gives us 102.

102 / 3 = 34, the minimum possible value for the highest score.



Bunuel
A class of seven students took a quiz, and each received a different integer score between 0 and 100. If the average (arithmetic mean) score was 30 and the median was 27, what is the minimum possible value for the highest score?

A. 34
B. 35
C. 36
D. 37
E. 38


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

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