Find four consecutive integers such that 8 times the sum of the first and the third is 40 greater than 10 times the fourth.

解

F in d f o u r co n sec u t i v e in t e g ers s u c h t ha t 8 t im es t h e s u m o f t h e f i rs t an d t h e t hi r d i s 40 g re a t er t han 10 t im es t h e f o u r t h .

The numbers : - 33 \frac{2}{3}, - 32 \frac{2}{3}, - 31 \frac{2}{3}, - 30 \frac{2}{3}

解答ステップ

Represent the numbers with

x

First number : x

Consecutive number 1 : x + 1

Consecutive number 2 : x + 2

Consecutive number 3 : x + 3

A.T.Q.:

8 (x + \frac{1}{3}) = 40 + 10 (x + 3)

解く

8 (x + \frac{1}{3}) = 40 + 10 (x + 3) : x = - \frac{101}{3}

x + 1 = - \frac{101}{3} + 1 = - \frac{98}{3}

x + 2 = - \frac{101}{3} + 2 = - \frac{95}{3}

x + 3 = - \frac{101}{3} + 3 = - \frac{92}{3}

仮分数を帯分数に変換する:

\frac{101}{3} = 33 \frac{2}{3}

仮分数を帯分数に変換する:

\frac{98}{3} = 32 \frac{2}{3}

仮分数を帯分数に変換する:

\frac{95}{3} = 31 \frac{2}{3}

仮分数を帯分数に変換する:

\frac{92}{3} = 30 \frac{2}{3}

The numbers : - 33 \frac{2}{3}, - 32 \frac{2}{3}, - 31 \frac{2}{3}, - 30 \frac{2}{3}