City A' s population is $96000$ and it is decreasing by $800$ per year and city B' s population is $68000$ which is increasing by $1200$ per year. After how many years population of both the city becomes equal? If I have to solve this question in less than 1 minute what I have to do?
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1If your numbers are accurate, then the population of the cities will never become equal. 'A' starts with fewer people and decreases, while 'B' starts with more and increases. – Deepak Aug 22 '18 at 02:04
3 Answers
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The simple approach is to note that they start $28000$ apart and the difference decreases by $2000$ per year. You just have to divide $\frac {28}2=14$ to get the answer.
Ross Millikan
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Thanks I needed the most quick solution thanks for helping me . – Aayush Mukharji Aug 22 '18 at 02:33
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Hint $$96000-800x=y$$$$68000+1200x=y$$
Now solve for $$x=?,y=?$$
In x years, they will both have y people
Key Flex
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OK, OP has clarified, your answer can be reverted back to its original form. – Deepak Aug 22 '18 at 02:12
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@AayushMukharji Ok, I edited the answer. If you find the value of $x$ and $y$ then that will be your answer – Key Flex Aug 22 '18 at 02:13
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Thanks for it. With your guidance I solved it in 14 years both the cities will be at the population of 84,800. – Aayush Mukharji Aug 22 '18 at 02:30
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Create two equations for the population per year for City A and City B. Use a variable like $n$ for the year and then set these two equation equal to each other (in other words same population) and solve for $n$.
Andrew Allen
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