小六數學問題_很趕

(1) 設該數為7A4B能被55整除,即可被5及11整除
B可以是0或5;
(7+4)-(A+B)必為11的倍數
=>11-(A+B) 為11的倍數
=>A+B為11的倍數
若B=0,則A=0該數為7040
若B=5,則A=6該數為7645
(2) (A+B+C+D)/4 = 38 => A+B+C+D = 38*4 = 152
(A+B)/2=60 => A+B=120
因此C+D=152-120=32
C和D的平均數=(C+D)/2 = 32/2 = 16
(3) 健身徑一旁街燈 = 7500/15 +1 = 501盞
兩旁街燈 = 501*2 = 1002 盞
(4) 餘下面積=15.75平方米, 即整個正方面積>15.75,所以正方形邊長應大若4米
若正方形邊長＝4米,即餘下長方形濶度＝3米,面積=3*4 = 12平方米太少
若正方形邊長＝5米,即餘下長方形濶度＝4米,面積=4*5 = 20平方米太大
若正方形邊長＝4.5米,即餘下長方形濶度＝3.5米,面積=3.5*4.5=15.75平方米剛好
劃出長方形土地的面= 4.5*1 = 4.5 平方米
中學解法：設邊長為x米
x(x – 1)= 15.75
x^2 – x – 15.75 = 0
(x – 4.5)(x + 3.5) = 0
x = 4.5 或x=-3.5(不合)

1. 可能7040,7645
2. (38x4)-(60x2)÷2
3. (7500÷15+1)x2 加1因為計埋頭果一盞
4. 4.5平方米

1) answer is 7040 as 5 or 0 can only divided by 5;


2) A + B + C + D = 4 x 38..............(1)
A + B = 2 x 60...............................(2)

(1) - (2) we have, C+D = 152 - 120, therefore C+D = 32 and average of
C and D is equal to 32/2 =16;



3) 7500 / 15 = 500
so that there will be 500 + 1 = 501 since the street lamps shall be installed at starting point and ended at finish point;


4) this is quite a quadratic problem

Let Y be the unknown length (m) of the rectangular and width is given as 1 m
15.75 = Y (Y-1) (sq.m) and the sqaure is equal to Y x Y (sq.m)
hence, Y^2 -Y-15.75 = 0


resolve the quadratic we have Y = 4.5 or Y = -3.5 (rejected as no negative result).

Therefore, the area of the rectangular is equal to 4.5 x 1 = 4.5 (sq. m).






2010-01-19 23:26:37 補充：
Question (2) is 501 x 2 = 1002 nos. of street lamps as the question is requested to install in both sides of the road.

1.7040
2.38x2-60=16
3.7500/15+1=501

1.一個四位數7 [ ]4 [ ]能比55整除,求出有可能的四位數。
答 7040
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