Aptitude - Percentage - Discussion

Discussion Forum : Percentage - General Questions (Q.No. 15)
15.
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
4.37%
5%
6%
8.75%
Answer: Option
Explanation:

Increase in 10 years = (262500 - 175000) = 87500.

Increase% = 87500 x 100 % = 50%.
175000

Required average = 50 % = 5%.
10

Discussion:
96 comments Page 1 of 10.

Prakash singha said:   7 days ago
Nice, Thanks for explaining.

Prashant said:   3 weeks ago
Anyone, explain me, why we are not using compounding? Please clarify it.
(1)

Yoga said:   1 month ago
Why aren't we using the formula of P (1+R/100) ^N? Please explain to me.

Siddhant said:   1 year ago
262500 = x% of 175000
i.e. 150% of 175000

Thus, a 50% increase from 100%.

50% increase in 10 years, so 5% increase per year.
(10)

Priyanka said:   2 years ago
The population increased in 10 years 262500 - 175000 = 87500. For a decade that is for 10 years
Per year increase 87500/10= 8750.
Avg per cent increase in population per year= 8750/175000*100 = 5 per cent.
(35)

T Muvetha said:   2 years ago
A decade means 10 years.
(15)

S A DEVIKA said:   2 years ago
We are taking 10 years since it has been mentioned in the question that there is an increase in a decade.

So, 1 decade = 10 years.
(10)

Vikash Pal said:   3 years ago
It was mentioned increase was there in a decade means
we have a 50% increase in 10 years.
Now we were asked in 1 year =?
Simply unitary method
10 -> 50
1 -> ?
1*50/10 = 5%.
(21)

Ali said:   4 years ago
Why not directly the formula of compound interest for finding population after nth years is not used?

Anyone, please clarify it.
(15)

Naveen Kumar said:   4 years ago
100-(20+60+10) = 10%.
10% = 30 means 100% = 300.
(2)


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