## An integral fraction between

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### An integral fraction between

by chieftang » Mon Dec 26, 2011 10:04 pm
Which of the following fractions is greater than 7/10 and less than 5/7?

(A) 9/13
(B) 11/15
(C) 11/16
(D) 12/17
(E) None of the above

Show/describe how you'd solve this one during the GMAT without a calculator. A quick trick for this one to come.

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by 6983manish » Mon Dec 26, 2011 11:47 pm
I tried this problem with the basic method and was able to solve it in 1.5 minutes.

Firstly I simplify the limits ::
<ANS> should be greater than 7/10 == 0.7 and
less than 5/7 == 0.71*
No we need to rule out the options which are straight not fitting in this limit.

A. 9/13 == 0.6* ...Rule out this one straight !
B. 11/15 == 0.7* ...Keep it for the moment !
C. 11/16 == 0.6* ...Rule out this one straight !
D. 12/17 == 0.7* ... Keep this one as well !!

No we have to choose between B and D... here we have to do some mental calculations and find out the second decimal place value, which in case of D is 0.70
and in case of B its 0.73.

Since D lies within the limit , we can mark it as the answer.

For these type of questions, I feel we need to be ready with some sort of mental calculations if we are not able to keep all the answer choices to common denominator fractions , which is again a call for some mental calculations. As far as I have seen, mathematical tables till 20 should suffice for even the complex number GMAT problems.

I would appreciate, if someone can please respond with any shortcut for this kind.

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by limestone » Mon Dec 26, 2011 11:57 pm
IMO: D

First, I would say that I solved this question in more than 5 minutes, but mostly in finding a method that can apply in the same issue.

Commonly, when comparing fractions, we find the smallest common denominator, then we compare the numerators.

Ex: compare 1/2 and 1/3

Smallest common denominator: 6

1/2 = 3/6
1/3 = 2/6

2<3, thus 2/6 < 3/6
or 1/3 < 1/2

But how about comparing: 35/37 with 68/71

Finding the smallest common denominator for these two fractions is weird!

Here is the tip: use the subordinate fractions of them

35/37 + 2/37 = 1
53/56 + 3/56 = 1

Now comparing 2/37 with 3/56
Find the smallest common numerator first: 6

2/37 = 6/(37*3) = 6/111
3/56 = 6/(56*2) = 6/112

as 112 > 111, thus 6/111 > 6/112, or 2/37 > 3/56 . You can check with 2/3, 2/4, 2/5 and see 2/3 > 2/4 > 2/5 using traditional method.

Now we know that:
2/37 > 3/56
then 1 - 2/37 > 1 - 3/56
or 35/37 > 53/56

The rule is:

if A > B, then

S(A) < S(B)

if A < B, then

S(A) > S(B)

with A, B is positive fractions less than 1. S(A) and S(B) is subordinate fractions that add up to A & B will create a sum of 1.

Note: I have not tested this with negative fractions.

Now apply this rule to the question:

7/10 has subordinate fractions of 3/10
5/7 has subordinate fractions of 2/7

A. 9/13 ---> 4/13
B. 11/15 ---> 4/15
C. 11/16 ---> 5/16
D. 12/17 ---> 5/17

Solve A:
Smallest common of nominators: 2,3,4 is 12

4/13 = 12/39
3/10 = 12/40
2/7 = 12/42

12/39 > 12/40 > 12/42 => 9/13 < 7/10 < 5/7 (Keep the order, change the smaller or bigger signals only)
Eliminate

Note: in the real test, we should stop at the underlined part to save time. If S(C) is not between S(A) and S(B), how can C be between A and B? I just eleborate for the sake of understand easier.

Solve B:

4/15 = 12/45

12/40 > 12/42 > 12/45 Stop here and eliminate this answer choice (reason said in answer A).

Or if you want to find out more detail: 12/40 > 12/42 > 12/45 => 7/10 < 5/7 < 11/15 Eliminate

Solve C:

Smallest common of 2,3,5 is 30
5/16 = 30/96
3/10 = 30/100
2/7 = 30/105

30/96 > 30/100 > 30/105. Stop here and eliminate this answer choice (reason said in answer A).

Solve D:

5/17 = 30 / 102 (ultilize the answer in C to save time, 16*6 = 96, thus 17*6 = 96+6 = 102)

30/100 > 30/102 > 30/105. Bingo! 30/102 is between 30/100 and 30/105
or 5/17 is between 3/10 and 2/7
or 12/17 is between 7/10 and 5/7
D is the correct answer choice!

When facing this problem against, use the rule. Do not think about "why" again as it cost a lot of time.

Hope it helps.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.

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by user123321 » Tue Dec 27, 2011 5:24 am
chieftang wrote:Which of the following fractions is greater than 7/10 and less than 5/7?

(A) 9/13
(B) 11/15
(C) 11/16
(D) 12/17
(E) None of the above

Show/describe how you'd solve this one during the GMAT without a calculator. A quick trick for this one to come.
simplest way is just calculate numerators when doing the operations...
take option 1) 9/13 - 7/10 = 90 - 91 = -ve so this is wrong

option 2) 11/15 - 7/10 = 110 - 105 = +ve can be true
5/7 - 11/15 = 75 - 77 = -ve so this is wrong

option 3) 11/16 - 7/10 = 110 - 112 = -ve so this is wrong

option 4) 12/17 - 7/10 = 120-119 = +ve can be true
5/7 - 12/17 = 85 - 84 = +ve so this is true.

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by sam2304 » Tue Dec 27, 2011 6:07 am
Even i went with user123321's method. Cross multiply two fractions and the side with larger value is greatest among the two.

say 2/3 and 4/5 => 10 and 12 so 4/5 > 2/3

IMO D. Same explanation as provided by user123321.
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by pemdas » Tue Dec 27, 2011 6:51 am
when i saw this question it reminded me of my favorite game - sudoku picking numbers

7/10 and 5/7 is within interval (49;50), so the difference is tiny

This came to mind during fist minute: look up numbers less than 5/7 and decide about answer choices
(A) 9/13 < 5/7, 63 < 65 possible
(B) 11/15 < 5/7, 77 < 75 No
(C) 11/16 < 5/7, 77 < 80 possible
(D) 12/17 < 5/7, 84 < 85 possible (very close! Like 49 and 50 interval)
(E) None of the above, ?

three candidates 9/13 and 11/16. Need to compare with 7/10
7/10 < 9/13, 91 < 90 No
7/10 < 11/16, 102 < 101 No
7/10 <12/17, 119 < 120 Yes

d
chieftang wrote:Which of the following fractions is greater than 7/10 and less than 5/7?

(A) 9/13
(B) 11/15
(C) 11/16
(D) 12/17
(E) None of the above

Show/describe how you'd solve this one during the GMAT without a calculator. A quick trick for this one to come.
Success doesn't come overnight!

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by chieftang » Tue Dec 27, 2011 7:45 am
And now a quick trick for this one...

Many of you are on to it, by proving via cross multiplication. But let me show you how to solve this one in 2 seconds.

The answer, D, is called a mediant fraction.

The mediant fraction of a/b and c/d is (a+c)/(b+d). (Not to be confused with addition of fractions!)

So in this case:

(7+5)/(10+7) = 12/17. Answer D. One can further verify by exhaustion that the mediant fraction is the only solution with denominator less than or equal to the sum of the denominators of the two fractions!!

Now how cool is that? Feel free to thank me. :mrgreen:
Last edited by chieftang on Tue Dec 27, 2011 10:00 am, edited 1 time in total.

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by pemdas » Tue Dec 27, 2011 9:13 am
very useful method, thanks chieftang

actually, this method would be extremely important to generate the next two numbers suitable for this question as answers
(12+7)/(17+10)=19/27
(12+5)/(17+7)=17/24

we could even check whether 12/7 lies in between two new answers -> (19+17)/(27+24)=36/51 is absolutely identical to 12/17
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by user123321 » Tue Dec 27, 2011 10:48 am
chieftang wrote: Feel free to thank me. :mrgreen:
you got it user123321
Just started my preparation Want to do it right the first time.

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by limestone » Tue Dec 27, 2011 5:16 pm
Nice tip Chieftang,
It does help shortern the time solving this. I try to make a thanks on my iPad but it did not allow. Thus I must reach my PC to thank you for this method.

Hope you keep posting useful maths questions.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.

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