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

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Two fractions are inserted between 1/4 and 1/2 so that the  [#permalink]

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New post 23 Aug 2010, 15:03
2
24
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A
B
C
D
E

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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

My ans is 16/12
but OA is 18/12...
Can someone please explain the OA.....
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Two fractions are inserted between 1/4 and 1/2 so that the  [#permalink]

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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]

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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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Re: fractions....  [#permalink]

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New post 05 May 2013, 23:02
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]

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New post 05 May 2013, 23:02
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]

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New post 25 May 2013, 05:45
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1
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]

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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]

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