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A 2-dight number is divisible by both 2 and 5. Half this number is less than 6 times the tens digit of the number by 4. Find this 2-digit number.
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方程式一問 中1
2007-10-29 12:30 am
回答 (2)
2007-10-29 1:31 am
✔ 最佳答案
Let the 2-digit number be 10a + b (列式)
(6 * a) - [0.5 * (10a + b)] = 4
6a - 5a - 0.5b = 4
a = 0.5b + 4 ……..(1)
Also given the 2-digit number is divisible by 2 and 5
That is the number is divisible by 10 (because 2 * 5 = 10, which is the l.c.m.)
Therefore. b = 0
Put b = 0 into (1), we get
a = 0.5 * 0 + 4
a = 4
10a + b = 10 * 4 + 0 = 40 (ans.)
Ans. The 2-digit number is 40.
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2007-10-30 3:13 am
Let the 2-digit number be 10a + b
設 2-位數字是 10a + b
[a 是十位數; b是個 位數; 例子: 12 = (10 乘1 + 2) = 12]
(列式)
(6 * a) - [0.5 * (10a + b)] = 4
[6 乘 十位數 - 一半 乘2-位數字] = 4
6a - 5a - 0.5b = 4
a = 0.5b + 4 ……..(1)
Also given the 2-digit number is divisible by 2 and 5
2-位數字是可以被 2 , 5 除到;
That is the number is divisible by 10 (because 2 * 5 = 10, which is the l.c.m.)
2-位數字是可以被 10 除到 (2乘5 = 10 是最小公倍數);
Therefore. b = 0
Put b = 0 into (1), we get
a = 0.5 * 0 + 4
a = 4
10a + b = 10 * 4 + 0 = 40 (ans.)
Ans. The 2-digit number is 40.
設 2-位數字是 10a + b
[a 是十位數; b是個 位數; 例子: 12 = (10 乘1 + 2) = 12]
(列式)
(6 * a) - [0.5 * (10a + b)] = 4
[6 乘 十位數 - 一半 乘2-位數字] = 4
6a - 5a - 0.5b = 4
a = 0.5b + 4 ……..(1)
Also given the 2-digit number is divisible by 2 and 5
2-位數字是可以被 2 , 5 除到;
That is the number is divisible by 10 (because 2 * 5 = 10, which is the l.c.m.)
2-位數字是可以被 10 除到 (2乘5 = 10 是最小公倍數);
Therefore. b = 0
Put b = 0 into (1), we get
a = 0.5 * 0 + 4
a = 4
10a + b = 10 * 4 + 0 = 40 (ans.)
Ans. The 2-digit number is 40.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071028000051KK03091