Bunuel wrote:
On a number line, \(a < b < c < d\). The distance from \(a\) to \(b\) is \(\frac{1}{4}\) of the distance from \(b\) to \(d\). The distance from \(a\) to \(c\) is \(3\) times the distance from \(c\) to \(d\). What is the value of \(\frac{(b-a)}{(c-a)}\)?
A. \(\frac{3}{20}\)
B \(\frac{1}{5}\)
C. \(\frac{4}{15}\)
D. \(\frac{7}{20}\)
E. \(\frac{3}{5}\)
Distance is always positive but here we do have relation in terms of value of variables for all.
a<b<c<d.
The distance from \(a\) to \(b\) is \(\frac{1}{4}\) of the distance from \(b\) to \(d\)
\(|a-b|=\frac{1}{4}*|b-d|\).
We know b-a > 0 and d-b >0
So, \(b-a=\frac{1}{4}*(d-b)…….4(b-a)=d-b\)….(I)
The distance from \(a\) to \(c\) is \(3\) times the distance from \(c\) to \(d\)
Similarly, \(c-a=3(d-c)\)….(II)
We are looking for value of \(\frac{b-a}{c-a}\)
So we have two equations.
As we do not require d, let us remove d by multiplying I by 3 and subtracting both.
3(4(b-a))-(c-a)=3(d-b)-(3(d-c)
12(b-a)-(c-a)=3d-3b-3d+3c=3(c-b)
As c-b=c-a-(b-a), add 3(b-a) on both sides to convert c-b to c-a
12(b-a)-(c-a)+3(b-a)=3(c-b)+3(b-a)
15(b-a)=3(c-a)+c-a
\(\frac{b-a}{c-a}=4/15\)
An easier method would be to visualise it on number linea……X…….b…….Y………c……..Z……..d
X, Y and Z are the distances.
So, we are looking for the value of \(\frac{X}{X+Y}\)
This means a relation between X and Y is sufficient to answer. The distance from \(a\) to \(b\) is \(\frac{1}{4}\) of the distance from \(b\) to \(d\)
\(X=\frac{1}{4}*(Y+Z)\).
\(4X=Y+Z\)….(I)
The distance from \(a\) to \(c\) is \(3\) times the distance from \(c\) to \(d\)
Similarly, \(X+Y=3Z\)….(II)
As we do not require Z, let us remove Z by multiplying I by 3 and subtracting both.
4X=Y+Z => 12X=3Y+3Z
X+Y=3Z
12X-X-Y=3Y+3Z-3Z………11X-Y=3Y…….11X=4Y……..\(Y=\frac{11X}{4}\)
Now, \(\frac{X}{X+Y}\)= \(\frac{X}{X+(11X/4)}\)
\(\frac{X}{\frac{15X}{4}}=\frac{4}{15}\)
Note you could also start with taking value of X as something and then moving ahead.
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