Four vessels are kept on a table in a manner that their quantity keeps on becoming 3 times when measured from left towards right. 25% of the first vessel, 30% of the second vessel, 40% of the third vessel and 20% of the fourth vessel are empty. If all the four vessels are emptied into the fifth vessel, then the total quantity of the fifth vessel is what percent of total quantity of the four vessels?
Answers
Answer:
I'm going to assume that by "quantity" you mean to say that the "volume" of each vessel is three times that of the one immediately to its left. If that is the case, then if the first vessel has a capacity equal to x, the second has a capacity of 3x, the third 9x, and the fourth 27x -- total 40x. Given the proportions stated above that each vessel is empty, the total volume of stuff in the four vessels is:
.75x + .7(3x) + .6(9x) + .8(27x) = 29.85x
Since the combined capacity of the four vessels is 40x, the percent of total capacity that is being used is 29.85 divided by 40 or 74.625%,
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Answer:
74.625%
Given:
Four vessels are kept on a table in a manner that their quantity keeps on becoming 3 times when measured from left towards right. 25% of the first vessel, 30% of the second vessel, 40% of the third vessel and 20% of the fourth vessel are empty. If all the four vessels are emptied into the fifth vessel.
To find:
Total quantity of the fifth vessel is what percent of total quantity of the four vessels?
Solution:
A B C D
1 3 9 27
.25 0.9 3.6 5.4
= .25 + .9 + 3.6 + 5.4 = 10.15
Total capacity = 1 + 3 + 9 + 27 = 40
Filled = 40 - 10.15 = 29.85
% filled = 29.85 / 40 × 100
= 149.25 / 2 = 74.625
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