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# Fractions : Faster calculation

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Intern
Joined: 30 Jul 2010
Posts: 9

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01 Nov 2014, 07:23
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1.Fractions - Comparing two fractions

Simple way to compare two fractions

$$\frac{45}{77}$$ and $$\frac{63}{52}$$

Cross multiple ==> $$45*52 < 63*77$$. Hence $$\frac{45}{77} < \frac{63}{52}$$

2. Fractions - finding highest/least value among multiple fractions with consistent pattern.

LEGEND
N= Numerator
D=Denominator
a= increase in numerator
b= increase in denominator

1. If N and D increased by constants value as the sequence of fractions progresses and if increase in numerator greater than or equal to increase in denominator then the last fraction is greatest among all given fractions.

EX: Which of the following fractions is greatest?
$$\frac{19}{24}$$, $$\frac{28}{27}$$, $$\frac{10}{21}$$,$$\frac{1}{18}$$

Solution:
Re-writing the list to apply the above formula. $$\frac{1}{18}$$, $$\frac{10}{21}$$, $$\frac{19}{24}$$,$$\frac{28}{27}$$
After clear observation we can find that the in above fractions both N and D are increased by constant values. i.e, N is incremented by 9 and D is incremented by 3. Clearly $$9>3$$

Hence 28/27 is the greatest value among the given fractions. It's pretty straight forward and can deduce the solution in seconds with clear observation.

Genralizing the formula:

$$\frac{x}{y}$$, $$\frac{x+a}{y+b}$$, $$\frac{x+2a}{y+2b}$$,$$\frac{x+3a}{y+3b}$$...$$\frac{x+na}{y+nb}$$
Then $$\frac{x+na}{y+nb}$$ is greatest among all given fractions.

1. Both numerator and denominator increase in constant values.(Numerator by a, denominator by b)
2. ($$a>=b$$)

what if $$a<b$$

Rule 2:

If a<b,
Then compare $$\frac{a}{b}$$ to first fraction of the list i.e$$\frac{x}{y}$$

1. If $$\frac{a}{b} > \frac{x}{y}$$

Then the last fraction is greatest . i.e $$\frac{x+na}{y+nb}$$

2. If $$\frac{a}{b} < \frac{x}{y}$$
Then the last fraction is least among all . i.e $$\frac{x+na}{y+nb}$$

3. If $$\frac{a}{b} = \frac{x}{y}$$
Then all fractions are equal.

EX: Rule 2. Type#1. Which of the following fractions is greatest?
$$\frac{4}{39}$$, $$\frac{2}{25}$$, $$\frac{3}{32}$$,$$\frac{1}{18}$$

Solution:
Re-writing the list to apply the above formula. $$\frac{1}{18}$$, $$\frac{2}{25}$$, $$\frac{3}{32}$$,$$\frac{4}{39}$$
After clear observation we can find that the in above fractions both N and D are increased by constant values. i.e, N is incremented by 1 and D is incremented by 7.

1. N increased by 1 and D increase by 7 ($$1<7$$) i.e $$a<b$$
2. compare $$a<b$$ with first fraction $$1/18$$ . This will give us $$1/7 >1/18$$

Hence the last fraction is greatest.

EX: Rule 2. Type#2. Which of the following fractions is least?
$$\frac{105}{401}$$, $$\frac{100}{301}$$, $$\frac{95}{201}$$,$$\frac{90}{101}$$

Re-writing the list to apply the above formula. $$\frac{90}{101}$$, $$\frac{95}{201}$$, $$\frac{100}{301}$$,$$\frac{105}{401}$$

a=5 and b=100
compare $$\frac{a}{b}$$ i.e, $$\frac{1}{20}$$ with first fraction i.e, $$\frac{90}{101}$$. Clearly $$\frac{1}{20}<\frac{90}{101}$$
Hence the last fraction in the sequence is the least value. i.e, $$\frac{105}{401}$$ is least among all.
Manager
Joined: 23 Oct 2014
Posts: 85
Concentration: Marketing
Re: Fractions : Faster calculation  [#permalink]

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05 Nov 2014, 15:40
Thank you. This was enlightening.
Intern
Joined: 14 Feb 2019
Posts: 2
Re: Fractions : Faster calculation  [#permalink]

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01 Mar 2019, 00:37
great. It is very healpful
Re: Fractions : Faster calculation   [#permalink] 01 Mar 2019, 00:37
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