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Sajjad1994
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Hi bhagyajeet.sahu,

As the question says, if practice and jameson both increase at constant rate in 3 yrs time they both would be same.

So practice rate of increase be x courses/yr then after 3 yrs= 2000+ 3*x
similarly for jameson let rate of increase be y courses/yr then after 3 yrs= 3200+3*y

after 3 yrs both are same so 2000 + 3x = 32000 + 3y => x - y = 400

Final x and y which satisfy the criteria from the given options, they are x = 800 and y = 400.

Hope this clarifies!
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Official Explanation

If Practice Tech increases its course offerings by 800 courses per year, in three years it will offer 4,400 courses per year. If Jameson Vocational increases course offerings by 400 courses per year, in three years it will offer 4,400 courses per year as well. After the three-year mark, Practice Tech will offer more courses per year than Jameson Vocational.
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Assuming Practice Tech currently increasing at rate of x and Jameson Vocational currently increasing at rate of y.
and both of these rates remain constant through out.

Given - in three years both schools will offer the same number of courses for the first time.
Therefore, 
2000 + 3x = 3200 + 3y
3x - 3y = 1200
x - y = 400
x = y + 400
Put values for y and check for x if its there table bingo thats our answer.
y = 250, 400, 540, 600, 760, 800
So,
x = 650, 800, 940, 1000, 1160, 1200.
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­Cant we use the formula 2000[1+x/100]^3 = 3200[1+y/100]^3 ?. if yes then how to arrive at the solution.
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shreyoshee
­Cant we use the formula 2000[1+x/100]^3 = 3200[1+y/100]^3 ?. if yes then how to arrive at the solution.
­No its not that way please refer to the solution posted above.
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I simply used the options to solve this. Since the difference between the 2 course is 1200. We need to select the option for which the difference is 1200 for 3 years.
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