數學功課一問

設A 班人有 a 個 , B for b 人 ,如此類推
a + b = 78
b + c = 72
c + d = 90
三條式加起
a + d + 2b + 2c = 240
由於b + c = 72
2b + 2c = 144
所以a + d = 96
1D和1A共有學生96人

1A+1D
= (78-1B) + (90-1C)
= (78-1B) + [90 - (72-1B)]
= 78-1B + (90 - 72 + 1B)
= 78 - 1B + 90 - 72 + 1B
= 78 + 90 - 72
= 96 (人)

Let 1A student be a,
1B student be b,
1C student be c,
1D student be d.
a+b=78 ...1
b+c=72 ...2
c+d=90 ...3

Since there are 3 equcation but there are 4 un-kown nunmbers,
therefore, no ansewer is given to 1D student and 1A student's sum.
Ans. : no solution

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假設1A 學生=a
1B 學生=b
1C 學生=c
1D 學生=d
a+b=78 ...1
b+c=72 ...2
c+d=90 ...3

由於只有三條式但是有四個未知數，
所以1D 和 1A 共有多少學是沒有答案。
答案：沒有解決答案。

設1A班有A個學生, 1B班有B個學生,1C班有C個學生, 1D班有D個學生
即
A + B = 78 ........ (1)
B + C = 72 ........ (2)
C + D = 90 ........ (3)
A + D = ?

(1) + (2) + (3)
(A + B) + (B + C) + (C + D) = 78 + 72 + 90
A + 2B + 2C + D = 240
A + 2(B+C) + D = 240
A + 2(72) + D = 240 <--------- [將(2)代入]
A + D = 240 - 2(72)
A + D = 96

1D和1A共有學生96人

1A+1B =78
1A= 78-1B


1B+1C = 72
1C=72-1B


1C+1D = 90
1D = 90-1C



Then: -

1A+1D
= (78-1B) + (90-1C)
= (78-1B) + [90 - (72-1B)]
= 78-1B + (90 - 72 + 1B)
= 78 - 1B + 90 - 72 + 1B
= 78 + 90 - 72
= 96 (人)

這是一個趣味題

設1A,1B,1C,1D的人為A,B,C,D
A+B=78 --(1)
B+C=72 --(2)

(2)-(1)::
C-A= -6 --(3)
C+D=90 --(4)

(4)-(3)::
D-(-A)=96

1D和1A共有學生96人

2007-05-24 16:30:22 補充：
有個快D的方法 (1A 1B) (1C 1D) - (1B 1C)=78 90-72=96解題:(1A 1B) (1C 1D) - (1B 1C)=1A 1B 1C 1D-1B-1C=1A 1D

設1A班有a人
1B班有b人
1C班有c人
1D班有d人
a+b=78---(1)
b+c=72---(2)
c+d=90----(3)
(1)+(2)+(3)
a+b+b+c+c+d=78+72+90
a+2b+2c+d=240
a+2(b+c)+d=240
a+2x72+d=240
a+144+d=240
a+d=96
1D和1A共有學生96人

1D和1A共有學生80人

90+78-72