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2015-12-18T01:03:20+05:30

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Three things to do :

1. 
Finding the respective improvements of each student - A and B (I prefer to do it in percentage)

2. Do comparison between the improvement rates of each student.

3. The one with higher improvement rate showed more improvement out of the two.


1. FINDING IMPROVEMENT RATES

A's Improvement:

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(Fraction)

IMPROVEMENT:  \frac{14}{12}

After cancellation, in the simplest form \frac{7}{6} or 1 \frac{1}{6}
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(Percentage)

IMPROVEMENT:  \frac{14}{12}*100%

Simplification 

 \frac{14}{12}*100%

 \frac{7}{6}*100%

7*50% /3

=  \frac{350}{3} %

116.666666666...% it really goes on forever and it is sufficient to just use the number up to 2 or 3 decimal places. 
---------------------------------------------------


So, A's overall improvement is   \frac{7}{6} or 1 \frac{1}{6} or 116.66%


B's Improvement: 

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(Fraction)

IMPROVEMENT:  \frac{12}{10}

After cancellation, in the simplest form -  \frac{6}{5} or 1 \frac{1}{5}

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(Percentage)

IMPROVEMENT:  \frac{12}{10}*100%

Simplification 

 \frac{12}{10}*100%

 \frac{6}{5}*100%

6*20%

= 120%
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(Decimal)

Fraction=120%= \frac{120}{100}

Simplification =  \frac{12}{10} =1.2
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So, B's overall improvement is   \frac{6}{5} or 1 \frac{1}{5} or 120% or 1.2




2. COMPARISON

(Fraction) 

A's improvement:  \frac{7}{6}

B's improvement: 
 \frac{6}{5}

 \frac{6}{5}\ \textgreater \  \frac{7}{6}

(Percentage)

A's improvement: 116.66%

B's improvement: 
120%

120116.66%



3. RESULT

So, after comparison, we can conclude that B student has shown more improvement than A student.
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