mangamma wrote:
Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, then what is the value of a + b + c + d?
A) -1
B) −(10/3)
C) 12
D) 6
E) -2
You do not have to solve this problem all the way.
(1) Express three of the variables in terms of one other.
Use \(a\) or \(d\) for the sake of visual simplicity; both are on one end of the first string of equalitites.
\(d+4=c+3\)
\(c=(d+1)\)
\(d+4=b+2\)
\(b=(d+2)\)
\(d+4=a+1\)
\(a=(d+3)\)
(2) Solve for \(d\) using the final equality.
Substitute for \(a, b,\) and \(c\)
\((d+4)=a+b+c+d+5\)
\(d+4=d+3+d+2+d+1+d+5\)
\(d+4=4d+11\)
\(-7=3d\)
\(3d=-7\)
\(d=-\frac{7}{3}\)
Stop. The answer must be a fraction with a denominator of 3.
If we add an integer to a fraction, the answer is a fraction with the same denominator.
Only one option has a denominator of 3.
The answer is (B).Why stop?
\(a, b,\) and \(c\) all have values that equal
\(d\)(a fraction) + an integer
Test: If we add \(1\) to \(-\frac{7}{3}\), then \(1\) must be changed to \(\frac{3}{3}\)
\(c=(d+1)\)
\(c=(-\frac{7}{3})+(\frac{3}{3})\)
\(c=-\frac{4}{3}\)
\(c\) = another fraction with a denominator of 3.
\(a\) and \(b\) will also be fractions with a denominator of 3.
-- The sum of all four variables will be a fraction with a denominator of 3.
-- Only one option has a denominator of 3.
The answer is (B).(3) if you are not sure, do all of the arithmetic
Find each value
\(a=(d+3)\)
\(a=(-\frac{7}{3})+(\frac{9}{3})\)
\(a=\frac{2}{3}\)
\(b=(d+2)\)
\(b=(-\frac{7}{3})+(\frac{6}{3})\)
\(b=-\frac{1}{3}\)
From above, \(c=-\frac{4}{3}\)
\(a+b+c+d=\)
\(\frac{2}{3}+((-\frac{1}{3})+(-\frac{4}{3})+(-\frac{7}{3}))=\)
\(\frac{2}{3}+(-\frac{12}{3})=(\frac{2}{3}-\frac{12}{3})=-\frac{10}{3}\)
Answer B
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