Question

一道GMAT数学题……

Sevenpiecesofropehaveanaverage(arithmetic

mean) length of 68 centimeters and a median length
of 84 centimeters. Ifthe length of the longest piece of
rope is 14 centimeters more than 4 times the length
of the shortest piece of rope, what is the maximum
possible length, in centimeters, of the longest piece
of rope?
(A) 82
(B) 118
(C) 120
(D) 1 3 4
(E) 152
求大神解释……！！

Answer 1

我们假设 7 根绳子的长度按从小到大排列分别是 a,b,c,d,e,f,g. 则最短的为 a,d=84,g=4a+14,我们要求 4a+14 最大,则要求 a 最大。所以 b=c=a,在这样的情况下 a 才尽可能大。对于 e,f 的长度,在总长度一定的情况下,要让 4a+14 尽可能大,则 e, f 必须尽可能小,最小的可能是 e=f=84,在这样的情况下可以保证 a 最大。综上,我们 得到 7 根绳子的长度分别为a,a,a,84,84,84,4a+14,再通过 7 个数的平均数是 68 可以解出 a 的值 30,最终 4a+14=134。选 D

Answer 2

Seven pieces of rope have an average(arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

mean (u) = 68
median = (m) = 84
longest - 4(shortest) = 14
x1+x2+..+x7 = 7(68)

x7-4(x1)=14
x4=84
To find max x7

Solution
without loss of generality, assume x1<=x2<=x3<=..<=x7
x7-4(x1)=14
x1= (x7-14)/4
max x7 occurs when
x1, x2=x1,x3=x1, x4, x5=x4,x6=x4, x7

x1+x2+...+x7=7(68)
3(x7-14)/4+ 3(84) + x7 = 7(68)
3x1-42+1008+4x7=1904
x7=134
Ans:D