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Anonymous users2024-02-05You can assume that you bring one yuan, then 1 15 yuan is the ** of the language book, and 1 24 yuan is the ** of the math book.
10 language books, 10 * 1 15 = 10 15 = 2 3 yuan 1-2 3 = 1 3 yuan Then the money to buy 10 math books is 1 3 yuan, 1 3 * 1 24 (1 3 divided by 1 24) = 8 books. The answer is: 8 copies. Thank you.
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Anonymous users2024-02-04The smallest common multiple of 15 and 24 is: 120 and then suppose the money he brings is 120, then 120 15 = 8 (yuan) 120 15 = 8 (yuan) 120 24 = 5 (yuan) 10 language books 8 * 10 = 80 (yuan) the rest of the money is 120-80 = 40 (yuan) 40 5 = 8 (yuan).
A: The rest of the money is all for math books, and you can also buy 8 math books.
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Anonymous users2024-02-03Solution: Set up a language book for x yuan, a mathematics book for y yuan, and you can also buy z mathematics books.
15x=24y （1）
15x+24y=2(10x+zy) (2) Organize (2) as: 5x=(24-2z)y, x:y=(24-2z): 5 from (1) to get x:y=24:15
24-2z):5=24:15
z=8 so you can buy 8 math books.
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Anonymous users2024-02-028 copies. Because, the money of 15 language books = the money of 24 math books.
So, the money of 5 language books = the money of 8 math books.
10 language books have been bought, and 5 language books can be bought, i.e. 8 math books.
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Anonymous users2024-02-01Stupid! Won't ask the teacher It's too stupid.
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Anonymous users2024-01-31Suppose the money he brings with him is x yuan, then the ** of a language book is x 15, and the ** of a math book is x 24
Suppose he can also buy y math books, then y = (x - x 15 10) (x 24) = 8
So you can also buy 8 math books.
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Anonymous users2024-01-3015x=24y then x:y=8:5, if the unit price of Chinese books is 8 and the unit price of math books is 5, then this person has a total of 15*8=120 yuan.
The price of ten language books is 10 * 8 = 80, 120-80 = 40, the remaining 40 yuan, and the unit price of math books is 5 yuan, so the rest of the money can also buy 40 5 = 8 copies.
A: You can also buy 8 math books.
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Anonymous users2024-01-29It's too easy, do it yourself.
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