Karan911 wrote:
VeritasKarishma wrote:
EgmatQuantExpert wrote:
Set S contains 26 distinct natural numbers. When the elements are sorted in ascending order, the first 23 numbers are consecutive, and their average is 24. What can be the value of the highest element in S such that the average of all the elements present in S is 30?
A. 52
B. 78
C. 104
D. 153
E. 155
Avg = 24
Total 23 consecutive numbers. So 24 must be the 12th number (middle).
First 11 numbers must be 13 to 23. Next 11 numbers must be 25 to 35.
Since these are first 23 numbers, next 3 numbers must be greater than 35 so let's say the next two numbers are 36 and 37. Now we need the value of the greatest/last number such that avg is 30.
First 23 numbers are 6*23 = 138 less than 30.
36 and 37 are 6+7 = 13 more than 30.
So last number must be 138 - 13 = 125 more than 30 i.e. it must be 155.
Karan911 - This is how the method of deviations discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/0 ... eviations/works its charm
Hi
VeritasKarishma,
I realised I made a mistake, so let the three numbers be a, b, c when added result in a deviation of +3 from the average, the extra amount they bring in gets divided over all of the numbers and each number gets an extra 3,
So letting the numbers be a,b,c, their deviations would from avg would be (a-24 + b -24 + c-24)/ 26 = 6, [ 6 i deviation from 24 to 30]
hence a + b + c = 156 + 72 = 228, now to minimize 2 out of these 3 , i Let them be 36 and 37, hence so the last number is 228 - (36 + 37) = 155, is that correct?
Also one question, if the numbers drop the average, we should be taking (a-24, b-24, c-24)/ 26 = -6 right?
I checked this on a smaller set using arbitrary integers and got the correct result :
for eg if i have 10,20,30, avg is 20, now if I need avg to be dropped to say 15,
so a-20/ 4 = -5, so a = 0,
is this correct?
Karan911 -
Work only with deviations, not the actual numbers. It will be far easier.
I have 23 numbers with average 24.
I need to add 3 numbers (all greater than 35) to make the average 30.
Now think this way:
If I were to add the three numbers each 24 only, the avg would stay the same i.e. 24. But if I want the avg to go to 30, it means the 3 numbers bring 6 extra for everybody including themselves. So if I add three 30s, I still need another 23*6 = 138 extra to makes up the 6 extra for 23 numbers.
Since the smallest 2 numbers can be 36 and 37, I have already utilised 6+7 = 13 of the 138.
So the third number must be 138 - 13 = 125 more than 30 which gives 155.
Start your though process from the highlighted step. It brings a lot of clarity of the situation. I still do.
Quote:
"if the numbers drop the average..."
for eg if i have 10,20,30, avg is 20, now if I need avg to be dropped to say 15,
so a-20/ 4 = -5, so a = 0,
Think: I have 3 numbers with avg 20. If I add a number at 20, the avg doesn't change. But I need to add the number such that the avg goes down to 15. So the number must reduce 5 from everyone including itself. Hence, if I add the number 15, I still need to reduce 3*5 for the other 3 numbers. So the number I must add becomes 0.
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