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There are 1600 jelly beans divided between two jars, X and Y. If the
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06 Feb 2015, 09:03
2
9
00:00
A
B
C
D
E
Difficulty:
25% (medium)
Question Stats:
74% (02:04) correct 26% (02:17) wrong based on 186 sessions
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There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?
Re: There are 1600 jelly beans divided between two jars, X and Y. If the
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06 Feb 2015, 13:43
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Hi All,
This question can certainly be solved with Algebra; with 2 variables and 2 unique equations, it's just a matter of translating the text into equations and doing "system" math.
The answers are numbers though, and there is a logical pattern in the prompt that you can use to quickly TEST THE ANSWERS.
We're told that there are a total of 1600 marbles in two jars. Jar X has 100 less than THREE TIMES the number of marbles in Jar Y. We're asked for the number of marbles in Jar X.
100 marbles is a relatively small amount, compared to the 1600 total marbles. If we ignore the "100" and do a rough estimation, Jar X would be about 1200 and Jar Y would be about 400 (which matches the information about THREE TIMES the number of marbles). Since we DO have to factor in the 100 marbles though, the number of marbles in X has to be LESS than 1200. With this deduction, the answer has to be C or D.
Let's TEST C (since it looks like the "nicer" number to deal with).
IF.... X = 1150 marbles X + 100 = 1250 1250/3 = 416.6666 marbles 1150 + 416.66666 is NOT 1600 total marbles. Eliminate C.
Re: There are 1600 jelly beans divided between two jars, X and Y. If the
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06 Nov 2018, 20:08
pacifist85 wrote:
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?
A. 375
B. 950
C. 1150
D. 1175
E. 1350
Let Jelly beans in jar \(X = x\) and \(Y = y\)
\(x + y = 1600\) ----- (\(i\))
Given jar \(X\) has \(100\) fewer jelly beans than three times the number of beans in jar \(Y\).
Therefore; \(x + 100 = 3y\)
\(x = 3y - 100\)
Substituting value of \(x\) in equation (\(i\)), we get;
\(3y - 100 + y = 1600\)
\(4y = 1600 + 100 = 1700\)
\(y = \frac{1700}{4} = 425\)
Substituting value of \(y\) in equation (\(i\)), we get;
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74% (02:04) correct 
