I have the following maths questions, please help to solve the probelm. thks thks a lot!
1) Ann had $100, she used part of her money to buy 7 ceramic cats, in which 3 of them are more expensive than the rest by $3. If she had $7 left, how much doe the two types of ceramic cat cost?
2) The cost of tea A is $30 per kg, the cost of tea Bis $20 per kg. If "w" kg of tea A and (w +1)kg of tea B are mixed together, and the cost of the mixed tea is $24 per kg. Find "w".
Form 2 maths question
2007-08-25 9:57 pm
回答 (2)
2007-08-25 10:49 pm
✔ 最佳答案
1)Let $y be the cost of the cheaper ceramic cat,
and $(y+3) be the cost of the more expensive one.
100 - [3(y+3) + 4y] = 7
100 - (3y + 9 + 4y) = 7
100 - 9 -7y = 7
91 - 7y = 7
91 - 7 = 7y
84 = 7y
7y = 84
y = 84 / 7
y = 12
Thus, the cheaper ceramic cat costs $12 ,
then, the more expensive one costs $(12+3) = $15.
~Check~
100 - ( 3*15 + 4*12 )
= 100 - ( 45 + 48 )
= 100 - 93
= 7
So, the answer is correct.
2)
-------------------------------------------
w kg of TEA A = $30(w)
(w+1) kg of TEA B = $20(w+1)
-------------------------------------------
30(w) + 20(w+1) = 24
30w + 20w + 20 = 24
50w = 24-20
50w = 4
w = 4 / 50
w = 0.08
Thus, w is 0.08kg.
~Check~
30(0.08) + 20(0.08+1)
= 2.4 + 21.6
= 24
So,it is a right answer.
Hope that it can help you!=]]
參考: yamie
2007-08-25 10:06 pm
1) Let x be the cost of type 1 ceramic cat
Therefore, x+3 will be the cost of another type of ceramic cat
3(x+3) + 4x + 7 = 100
3x + 9 + 4x + 7 = 100
7x + 16 = 100
7x= 84
x= 12
12+3 = 15
Therefore, the two types of ceramic cat are $12 and $15.
2) 30w + (w+1)20 = 24
30w + 20w + 21 = 24
50w = 3
w = 0.06
The value of w is 0.06.
Therefore, x+3 will be the cost of another type of ceramic cat
3(x+3) + 4x + 7 = 100
3x + 9 + 4x + 7 = 100
7x + 16 = 100
7x= 84
x= 12
12+3 = 15
Therefore, the two types of ceramic cat are $12 and $15.
2) 30w + (w+1)20 = 24
30w + 20w + 21 = 24
50w = 3
w = 0.06
The value of w is 0.06.
參考: myself
收錄日期: 2021-04-13 13:07:14
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