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Two fractions are inserted between 1/4 and 1/2 so that the

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harithakishore
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Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   23 Aug 2010, 15:03
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Two fractions are inserted between 1/4 and 1/2 so that the difference between any two successive fractions is the same. Find the sum of the four fractions.

A. 18/12
B. 16/12
C. 5/2
D. 9/19
E. 24/32

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My ans is 16/12
but OA is 18/12...
Can someone please explain the OA.....
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A
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Bunuel
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Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   23 Aug 2010, 15:23
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harithakishore wrote:
Two fractions are inserted between 1/4 and 1/2 so that the diffrence between any two successive fractions is the same. Find the sum of the four fractions.
A.18/12
B.16/12
C.5/2
D.9/19
E.24/32


My ans is 16/12
but OA is 18/12...
Can someone please explain the OA.....


As the difference of any two successive fractions is the same then we have evenly spaced set (arithmetic progression). Sum of the terms of AP is \(\frac{first \ term + last \ term}{2}*(number \ of \ terms)\) --> \(sum=\frac{\frac{1}{4}+\frac{1}{2}}{2}*4=(\frac{1}{4}+\frac{1}{2})*2=\frac{3}{2}\).

Answer: A.
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Re: fractions.... [#permalink] New post   05 May 2013, 20:14
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Hi Bunuel...

Can we solve it like in following procedure...

Let a/b and c/d two fraction..so the sequence 1/2,a/b,c/d,1/4

As given- 1/2-a/b=c/d-1/4=>a/b+c/d=3/4
Question ask us to find 1/2+a/b+c/d+1/4=>1/2+3/4+1/4=3/2 ans....

Rgds
Prasannajeet
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Bunuel
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Re: fractions.... [#permalink] New post   05 May 2013, 23:02
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prasannajeet wrote:
Hi Bunuel...

Can we solve it like in following procedure...

Let a/b and c/d two fraction..so the sequence 1/2,a/b,c/d,1/4

As given- 1/2-a/b=c/d-1/4=>a/b+c/d=3/4
Question ask us to find 1/2+a/b+c/d+1/4=>1/2+3/4+1/4=3/2 ans....

Rgds
Prasannajeet


Yes, that's correct.
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Re: fractions.... [#permalink] New post   05 May 2013, 23:02
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Let the fractions inserted b/w \(\frac{1}{4} & \frac{1}{2}\) be \(\frac{1}{x} and \frac{1}{y}\)We can easily calculate the value of \(\frac{1}{x}\) by:-
\(2(\frac{1}{x}-\frac{1}{4})=\frac{1}{2}-\frac{1}{x}\)Solving we get, \(\frac{1}{x}=\frac{1}{3}\)Also, \(\frac{2}{y}=\frac{1}{3}+\frac{1}{2}\)
Solving we get \(\frac{1}{y}=\frac{5}{12}\)Hence, the 4 fractions are \(\frac{1}{4}, \frac{1}{3}, \frac{5}{12} and \frac{1}{2}\)
The required sum is thus:-
\(sum=\frac{1}{4}+ \frac{1}{3}+ \frac{5}{12}+ \frac{1}{2}=\frac{18}{12}\)
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   25 May 2013, 05:45
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Think of this as 4 points on the number line...

(1/4).....(X).....(Y).....(1/2)

OR

(3/12)....(X).....(Y).....(6/12)

You can easily see that X & Y should be 4/12 and 5/12 respectively. Add and you'll get the sum as 3/2. :)
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   26 Jun 2014, 02:46
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Let the 2 fractions be a & b

\(\frac{1}{2}\) ........... a ............... b ................ \(\frac{1}{4}\)

\(\frac{1}{2} - a = a - b\)

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

Solving the above, we get

\(a = \frac{5}{12}\)

\(b = \frac{1}{3}\)

Addition

\(= \frac{1}{2} + \frac{5}{12} + \frac{1}{3} + \frac{1}{4}\)

\(= \frac{18}{12}\)

Answer = A
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   03 Aug 2020, 05:10
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Write 1/4 as 3/12 and 1/2 as 6/12

therefore,

3/12, 4/12, 5/12, 6/12

Sum 18/12
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   03 Aug 2020, 10:18
Let the two fractions be X and Y

1/4.... X.....Y....1/2

Since we know that the difference between all of them is same

X-1/4 = 1/2-Y

X+Y = 1/2+1/4

X+Y = 3/4....... (1)

And

Y-X = X - 1/4

Y-X-X = -1/4

Y-2X = -1/4

2X - Y = 1/4........ (2)

Slove equations (1) and (2)

X+Y = 3/4.....(1)
2X-Y = 1/4.... (2)

This gives X=1/3 and Y=5/12

So add (1/4)+(1/3)+(5/12)+(1/2) = 18/12.

Hence (A).

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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   16 Aug 2020, 02:02
doesn't usually the gmat present the correct answer choice as a simplified fraction??? i was able to attain 3/2 but with the rush i was not able to identify 18/12 as 3/2 since i was searching 3/2...... it made me guess an answer choice.... i hope to not make the same mistake on the exam
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   21 Mar 2023, 18:24
The midpoint of 1/4 and 1/2 is 3/8 (1/4+1/2 divided by 2). Let the Two fractions be a and b, since they have to be the same distance from the midpoint, thus 1/4 ---a--3/8-----b---1/3.

3/8 - a = b+3/8
a+b = 3/8 + 3/8 = 6/8

sum of 4 fractions

6/8+1/4+1/2= 3/2

option A. (18/12 simplified is 3/2)
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink] New post   12 Aug 2023, 10:18
The question clearly states that - The difference between any two successive fractions is the same.

What does that mean?

It simply means that these numbers are evenly spaced - hence, these numbers are in arithmetic progression

Sum of the terms of an AP = (first term+last term)/2∗(number of terms)

3/2 = 18/12 (Don't get nervous if the options are not in the same form)

I hope it helps! :)
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Re: Two fractions are inserted between 1/4 and 1/2 so that the [#permalink]
12 Aug 2023, 10:18
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