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Question 1205435: A box contains 15 identical reams of paper. The total mass of the box with 1/5 of all the reams of papers is 33.6 kg lighter than the box with al the reams of paper. The mass of the empty box is 2/7 of a ream of paper. What is the mass of the empty box?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
p = mass of one ream of paper in kg
(2/7)p = mass of the empty box in kg
A = mass of empty box and 3 reams of paper
A = (2/7)p + 3p
A = (2/7 + 3)p
A = (2/7 + 21/7)p
A = (23/7)p
Note that 3 reams is 1/5 of the total 15 reams.
B = mass of empty box and 15 reams of paper
B = (2/7)p + 15p
B = (2/7 + 15)p
B = (2/7 + 105/7)p
B = (107/7)p
Mass A is 33.6 kg lighter compared to mass B.
A = B - 33.6
(23/7)p = (107/7)p - 33.6
7*(23/7)p = 7*( (107/7)p - 33.6 )
23p = 107p - 235.2
23p-107p = -235.2
-84p = -235.2
p = -235.2/(-84)
p = 2.8
Each ream of paper has a mass of 2.8 kg.
(2/7)*p = (2/7)*2.8 = 0.8 kg is the mass of the empty box.
Notes:
2.8 = 14/5
0.8 = 4/5
Answer by ikleyn(52855)  (Show Source):
You can put this solution on YOUR website! .
A box contains 15 identical reams of paper.
The total mass of the box with 1/5 of all the reams of papers
is 33.6 kg lighter than the box with al the reams of paper.
The mass of the empty box is 2/7 of a ream of paper.
What is the mass of the empty box?
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Usually, it is assumed, that after reading such problem,
the student will solve it MENTALLY in a few lines.
Indeed, compare these two statements
A box contains 15 identical reams of paper.
The total mass of the box with 1/5 of all the reams of papers
is 33.6 kg lighter than the box with al the reams of paper.
From this reading, it should be clear that 12 = 15 - 3 reams of paper weight 33.6 kg,
and hence, one ream of paper weights 33.6/12 = 2.8 kg.
One simple step further, and we answer the question: the mass of the empty box is  = 2*0.4 = 0.8 kg.
ANSWER. The mass of the empty box is 0.8 kilograms.
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It is the most detailed solution, which should accompany this problem.
I am 179% sure that the problem is expected to be solved this way, in a few lines and a few words.
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