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A

If you have 20 different positive numbers and you want one number to be as big as possible, then you want the other numbers to be as small as possible. Make the first 19 numbers the numbers 1-19 (19 different positive numbers, I presume they have to be integers, otherwise the right answer is not a choice). The sum of the numbers 1-19 is 190. Then you know (190+x)/20 = 20 where x is your biggest number. Multiply both sides by 20 to get 190 + x = 400. Then subtract 190 from both sides to get x = 210.

2006-10-05 07:08:11 · answer #1 · answered by Cara B 4 · 0 0

a)210

(The average of 20 different positive numbers is 20 .
For different positive no.:
Let x be the greatest no., then
(1+2+3+4+5+6+7+8+9+10+11+12+13+14+16+17+18+19+x)/20=20
(1+2+3+4+5+6+7+8+9+10+11+12+13+14+16+17+18+19+x)
=20*20
=400,
so,190+x=400
x=400-190
x=210)

2006-10-05 07:35:07 · answer #2 · answered by Anonymous · 0 1

the average of 20 positive numbers = 20
hence their sum = 20*20 = 400

we need to assume that the numbers are integers here....it is an important assumption... else we could consider 0.1, 0.2, 0.3 or even 0.000001 etc as a positive number....certainly this is not what we are given .....

to choose the 'greatest' number out of these 20, we must make the other 19 numbers as small as possible... the only such numbers are 1,2,3....19

the sum of 1+2+...+19 = 19*20/2 = 190

so we can have the 20th number = 400 - 190 = 210

2006-10-05 07:35:40 · answer #3 · answered by m s 3 · 0 0

actually the greatest possible number would be 229.
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+229=400... 400/20=20

2006-10-05 07:14:50 · answer #4 · answered by jackal 1 · 0 1

20=400/20 is the average
the lowest the 1st 19 can be is 1+2+3+...18+19=190

the largest possible integer is 400-190=210

2006-10-05 07:32:27 · answer #5 · answered by yupchagee 7 · 0 0

The way you put it, there's no greatest number. All of the numbers are less than 400 = 20 x 20, but it's possible to get average 20 making one of the numbers as close to 400 as you want and making the others as close to 0 as required. For example, for every 0 < eps < 20, you can take a_1 = 400 - eps and set a_2, a_3,.... a_20 in such a way that they are pairwise distinct and a_2 +a_3 ....+ a_20 = eps. There are infinitely many ways to do it, For example, a_2 = eps/190, a_3 = 2eps/190, a_4 = 3eps/190,....a_20 = 19eps/190

2006-10-05 07:17:27 · answer #6 · answered by Steiner 7 · 0 1

a, if you are using positive integers.

The other numbers would be 1-19, which total 190.

However, if you really are just using positive numbers, the other numbers could be .00000001 through .000000019, and the 20th number could be 399.9999981.

2006-10-05 07:08:13 · answer #7 · answered by Anonymous · 0 0

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