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M is the mean of 5,x, and 7
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01 Feb 2024, 06:14
01 Feb 2024, 06:14
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67% (01:09) correct
32% (01:56) wrong based on 34 sessions
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M is the mean of 5,x, and 7
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(M\) |
\(\frac{5+x+M+7}{4}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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M is the mean of 5,x, and 7
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01 Feb 2024, 15:05
01 Feb 2024, 15:05
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The mean of a set of terms = sum of terms / number of terms
Since M is the mean of 5, x, and 7, \(\frac{5+x+7}{3}= M\)
If you multiply both sides by 3 to get rid of the denominator, \(5+x+7 = 3M\)
We see that (5+x+7) is part of the numerator of Quantity B. Replace it with 3M.
\(\frac{5+x+M+7}{4} = \frac{3M + M}{4} = \frac{4M}{4} = M\)
Thus, Quantity B is also M.
Since M is the mean of 5, x, and 7, \(\frac{5+x+7}{3}= M\)
If you multiply both sides by 3 to get rid of the denominator, \(5+x+7 = 3M\)
We see that (5+x+7) is part of the numerator of Quantity B. Replace it with 3M.
\(\frac{5+x+M+7}{4} = \frac{3M + M}{4} = \frac{4M}{4} = M\)
Thus, Quantity B is also M.
M is the mean of 5,x, and 7
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25 Apr 2026, 04:20
25 Apr 2026, 04:20
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The solution by
So after getting rid of the denominator \(\frac{5+x+7}{3}=M\) --->\(5+x+7=3M\) we can think of a number that makes \(12+x\) divisible by 3 we get a nice integer for M. Let's pick \(x=3\) and we get \(M=5\).
Then just substitute the numbers in the quantity B and you get \(\frac{5+x+7+M}{4}=\frac{20}{4}=5\); the two quantities are equal.
Moreover this can also be answered by reasoning: since the second quantity its the average of all the previous numbers plus their average the result must still be the average.
Answer: C
Quote:
nurirachel
is for sure the quickest, but when I se variables in the answers I always think about working backwards substituting variables with numbers.So after getting rid of the denominator \(\frac{5+x+7}{3}=M\) --->\(5+x+7=3M\) we can think of a number that makes \(12+x\) divisible by 3 we get a nice integer for M. Let's pick \(x=3\) and we get \(M=5\).
Then just substitute the numbers in the quantity B and you get \(\frac{5+x+7+M}{4}=\frac{20}{4}=5\); the two quantities are equal.
Moreover this can also be answered by reasoning: since the second quantity its the average of all the previous numbers plus their average the result must still be the average.
Answer: C
