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31 Jul 2021, 05:52
Kudos
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C
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Difficulty:
85%
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Question Stats:
57% (02:38) correct
43%
(02:49)
wrong
based on 341
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Five integers in a set are written ascending order. The median of this set is 17, and the average of the smallest and largest integers is 16. When the smallest and largest numbers are removed from the set, the average of the new smallest and largest integers is 15. What can be the minimum value of the largest of the original five integers?
a. 17
b. 18
c. 19
d. 20
e. 21
Posted from my mobile device
a. 17
b. 18
c. 19
d. 20
e. 21
Posted from my mobile device
31 Jul 2021, 08:10
Kudos
Bookmarks
Here we have N1, N2, N3, N4, N5 where the median N3 = 17.
We are given that (N1+N5)/2 = 16 and (N2+N4)/2 = 30. Simplify these to be N1+N5=32 and N2 + N4 = 30.
The difference in the values of N1+N5 and N2+N4 is 2. This means that from the lower and upper bounds, the highest value number and/or the lowest value number has to be at least 2 from the median (N4 cannot be valued below 17 and N2 cannot be above 17 - that would break the rule). We are asked what can be the minimum value of the largest of original five inetegers AKA what is the value of N5? I think we found the answer in the prior sentence (N5 is at least 2 higher than N3), but let's test cases to be sure.
N5 + N1 = 32
19 + N1 = 32 => N1 = 13
Now we must address N2 + N4 = 30.
Our bounds are now 13 ≤ N2 ≤ 17 and 17 ≤ N4 ≤ 19 but we have to make the overall value drop by 2. We obviously cannot have N2 < N1 so we must have N2 ≥ N1 and manipulate the upper bound.
13+N4=30 => N4 = 17.
If we were to increase N2 to 14, then N4 would have to equal 16 but we cannot have N4 cross the value of the median so this does not work.
As such you can see 19 is the correct answer for the lowest value for N5. IMO answer is C.
We are given that (N1+N5)/2 = 16 and (N2+N4)/2 = 30. Simplify these to be N1+N5=32 and N2 + N4 = 30.
The difference in the values of N1+N5 and N2+N4 is 2. This means that from the lower and upper bounds, the highest value number and/or the lowest value number has to be at least 2 from the median (N4 cannot be valued below 17 and N2 cannot be above 17 - that would break the rule). We are asked what can be the minimum value of the largest of original five inetegers AKA what is the value of N5? I think we found the answer in the prior sentence (N5 is at least 2 higher than N3), but let's test cases to be sure.
N5 + N1 = 32
19 + N1 = 32 => N1 = 13
Now we must address N2 + N4 = 30.
Our bounds are now 13 ≤ N2 ≤ 17 and 17 ≤ N4 ≤ 19 but we have to make the overall value drop by 2. We obviously cannot have N2 < N1 so we must have N2 ≥ N1 and manipulate the upper bound.
13+N4=30 => N4 = 17.
If we were to increase N2 to 14, then N4 would have to equal 16 but we cannot have N4 cross the value of the median so this does not work.
As such you can see 19 is the correct answer for the lowest value for N5. IMO answer is C.
04 Aug 2021, 06:57
Kudos
Bookmarks
My approach:
work backwards
a. 3 out of the 5 numbers in the set are 17 ---> x y 17 17 17
17 + x = 32, x = 15
17 + y = 30, y = 13
y has to be bigger than x, therefore a is the wrong option
b. set of numbers: x y 17 z 18, where z could be either 17 or 18
18 + x = 32, x = 14
z + y = 30, y = 13 (if z = 17) or y = 12 (if z = 18)
y has to be bigger than x, therefore b is the wrong option
c. set of numbers: x y 17 z 19, where z could be either 17, 18 or 19
19 + x = 32, x = 13
z + y = 30, if z = 17 --> y = 13
set of numbers: 13 13 17 17 19 --> c is the correct option
No need to go further.
work backwards
a. 3 out of the 5 numbers in the set are 17 ---> x y 17 17 17
17 + x = 32, x = 15
17 + y = 30, y = 13
y has to be bigger than x, therefore a is the wrong option
b. set of numbers: x y 17 z 18, where z could be either 17 or 18
18 + x = 32, x = 14
z + y = 30, y = 13 (if z = 17) or y = 12 (if z = 18)
y has to be bigger than x, therefore b is the wrong option
c. set of numbers: x y 17 z 19, where z could be either 17, 18 or 19
19 + x = 32, x = 13
z + y = 30, if z = 17 --> y = 13
set of numbers: 13 13 17 17 19 --> c is the correct option
No need to go further.
