On a number line, a

Question

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}

chetan2u · Answer

Distance is always positive but here we do have relation in terms of value of variables for all. 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 line a……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.

tanghoathien · Answer

You should draw a<b<c<d on the number line

b – a = 1/4 (d - b). So we have: b - a = 1/5 (d – a)

c – a = 3 (d - c). So we have: c – a = 3/4 (d – a)

Then:
(b-a)/(c-a)= 1/5 (d-a): 4/3 (d-a) = 1/5 : 4/3 = 4/15

bv8562 · Answer

Given:

|b-a| = 1/4|d-b| => |d-b| = 4|b-a| , and
|c-a| = 3|d-c|
 
Let the distance between A and B be = x
Let the distance between C and D be = y
|b-a| = x
|d-c| = y
|d-b| = 4|b-a| = 4x
|c-a| = 3|d-c| = 3y

∴ |b-a| + |d-b| = 5x
 |c-a| + |d-c| = 4y

As shown in the picture below:

so, 5x=4y
x/y=4/5

if x=4 and y=5 then,

b-a = x = 4
c-a = 3y = 15

hence, b-a/c-a = 4/15 (C)

Seb2m02 · Answer

Hi everyone, I think I have found an easy way to solve this question
Considering AD = 10 (A=0 and D=10)

(1) We have: AB = 1/4(BD) -> 4AB = BD
Since we are on a line and a<b<c<d -> AB + BD = AD
AB + 4AB = AD
5AB = 10
AB = 2

(2) We have AC = 3CD
Since we are on a line and a<b<c<d -> AC + CD = AD
3CD + CD = AD
4CD = 10
CD = 2.5 (so the point C is at D-C distant from 0, s0 10 - 2.5 = 7.5)
We can define AC from (2)

AC = 3CD
AC = 3x2.5
AC = 7.5

So we have
A=0
B=2
C=7.5
D=10

We can now answer
B-A = 2
C-A = 7.5 (or 15/2)
Consequently, B-A / C-A = 4/15 -> answer C

Tarunw111 · Answer

1:3, total = 4
1:4, total = 5
Make the ratios equal by multiplying 5 in 1st and 4 in the 2nd equation.
The equations become 5:15 and 4:16

equation 1 (modified) shows, (c-a):(d-c)=15:5
equation 2 (modified) shows, (b-a):(d-b)=4:16

Now, take the required ratio, 4/15 is the correct answer.

luisdicampo · Answer

Deconstructing the Question

Let

x=b-a,\quad y=c-b,\quad z=d-c

Then

c-a=x+y

and

d-b=y+z

We translate the given distance relationships into equations.

Step-by-step

The first condition says

b-a=\frac{1}{4}(d-b)

So

x=\frac{1}{4}(y+z)

4x=y+z

The second condition says

c-a=3(d-c)

So

x+y=3z

From   4x=y+z  , we get

y=4x-z

Substitute into   x+y=3z  :

x+(4x-z)=3z

5x-z=3z

5x=4z

z=\frac{5}{4}x

Then

y=4x-\frac{5}{4}x=\frac{11}{4}x

So

c-a=x+y=x+\frac{11}{4}x=\frac{15}{4}x

Therefore

\frac{b-a}{c-a}=\frac{x}{15x/4}=\frac{4}{15}

Answer: C

On a number line, a<b<c<d. The distance from a to b is 1/4 of the

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