[MATH] GWD30-3

xiaowanwan

Q3:

At a certain food stand, the price of each apple is ￠ 40 and the price of each orange is ￠ 60.  Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is ￠ 56.  How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is ￠ 52?

 

A.  1

B.  2

C.  3

D.  4

E.  5

我查過了 好想沒有人問過這一題

請問 答案為什麼是E 不是B呢?

karen_gao

假设原来10个当中有orange的个数x，则有：40*（10-x）+60*x=56*10，x=8，10个当中有apple的个数10-8=2；

假设y个orange需要放回以保证剩下的水果的平均价格为52，则有：40*2+60*（8-y）=52*（10-y），得y=5，也就是答案E

Radia_ting

这道题好恶心,要不是你提供正确答案,我也错了

主要是题目没仔细看,它并没有说总数还是10,而是拿出来后使得剩下的平均值是52

(560-60n)/(10-n)=52,答案就是5了

documentwei

式子求不出來 5 呀~

但是我试了 (560-60n)/(n)=52, 答案才是5耶~

560-60n=52n

560=112n

n=5

请问这样对吗?

micklefuer

let the number of oranges being put back be x.

(560-60x)/(10-x) = 52

solve x = 5