There are 3000 books in 3 different piles. Pile #1 has 10 more books than Pile #2. Pile # 2 has twice as many books as Pile #3. How many books are in Pile #3? Solve this without using algebra.
2006-12-08 11:22:52 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3000-10=2990
now piles 1 and 2 have the same number of books and both of them have twice as many as pile 3. So pile 1 and 2 each have 40% of 2990 and pile 3 has 20% of 2990, or 598 books
2006-12-08 11:28:12 · answer #1 · answered by kelsey 7 · 1⤊ 0⤋
Pile#1=1206
Pile#2=1196
Pile#3=598
2006-12-08 20:01:47 · answer #2 · answered by Albania 1 · 0⤊ 0⤋
Pile One= 1206 Pile 2= 1196 Pile 3=598
I really don't know how I answered it. I just equally divided each and then messed around until the answer came out right! I know I didn't use algebra!
2006-12-08 19:41:19 · answer #3 · answered by Anonymous · 0⤊ 0⤋
pile#1 = x
pile#2 = y
pile#3 = z
x = y + 10
y = 2z
x + y + z = 3000
(y + 10) + y + (y/2) = 3000
2(y + 10) + 2y + y = 6000
2y + 20 + 2y + y = 6000
5y = 5980
y = 1196
x = y + 10 = 1196 + 10 = 1206
z = y/2 = 598 --> pile#3
2006-12-08 19:30:13 · answer #4 · answered by Xiangwei Xi 3 · 0⤊ 0⤋
Take out the 10 books; that's 2990 books. Then you would have two doubled piles and a single pile. So the answer should be 1/5 of 2990 or 598.
2006-12-08 19:28:22 · answer #5 · answered by Puzzling 7 · 1⤊ 0⤋
3000 - 10 = 2990
2990 : 5 = 598
Pile 3 = 598
Pile 2 = 598 x 2 = 1196
Pile 1 = 10 + 1196= 1206
.::.
2006-12-08 19:43:24 · answer #6 · answered by aeiou 7 · 0⤊ 0⤋
598
2006-12-08 19:35:58 · answer #7 · answered by michael 2 · 0⤊ 0⤋
It's impossible to do without algebra. And even when i do it with algebra I get the following values
Pile 1 has 742.5 books
Pile 2 has 752.5 books
Pile 3 has 1505 books
Your question is flawed, you can't get half a book (unless you look in a bargain bin)
2006-12-08 19:30:37 · answer #8 · answered by Anonymous · 0⤊ 3⤋
598, impossible without algebra.
2006-12-08 19:28:21 · answer #9 · answered by Wise_Guy_57 4 · 0⤊ 3⤋