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Question: The number of seat increases by 10% every year. If the number of seats in 2001 was 400, what was the...

The number of seat increases by 10% every year. If the number of seats in 2001 was 400, what was the number of seats in 2003?
A) 824
B) 484
C) 500
D) 480

Explanation

Solution

The number of seats in next year will increase by 10% of the previous year, we will go on this trend, and find the series containing the number of seats over consecutive years. There we will see the series is a geometric progression. We will compare it with the general form of a GP, i.e. a,ar,ar2,...a, ar, a{r^2},... and find the first term and common ratio of GP. Number of seats in 2003 will be the third term of the GP, so we will find it by applying the formula for nth{n^{th}} term of GP, i.e.
an=arn1{a_n} = a{r^{n - 1}}.

Complete step by step solution: The number of seats increases by 10% every year,
Let, the number of seats originally be aa
So, number of seats in next year will increase by: 10100a=a10\dfrac{{10}}{{100}}a = \dfrac{a}{{10}}
Thus, number of seats in next year: a+a10=11a10a + \dfrac{a}{{10}} = \dfrac{{11a}}{{10}}
Similarly, in the one more next year, total seats will be 1110×11a10=(1110)2a\dfrac{{11}}{{10}} \times \dfrac{{11a}}{{10}} = {\left( {\dfrac{{11}}{{10}}} \right)^2}a
Thus, we have
Number of seats over years as: a,11a10,(1110)2a,...a,\dfrac{{11a}}{{10}},{\left( {\dfrac{{11}}{{10}}} \right)^2}a,...
General form of a GP is: a,ar,ar2,...a,ar,a{r^2},...
Comparing it with above series,
The above series is a geometric progression with first term aaand common ratio r=1110r = \dfrac{{11}}{{10}}
Now, as given, number of seats in 2001 was 400
So, we can considera=400a = 400
Now, we have to find number of seats in 2003, which will be the third term of the GP,
Let the third term of the GP be a3{a_3}
General term of a GP is:
an=arn1{a_n} = a{r^{n - 1}}
Third term of the GP is (1110)2a{\left( {\dfrac{{11}}{{10}}} \right)^2}awhere a=400a = 400
a3=(1110)2400\Rightarrow{a_3} = {\left( {\dfrac{{11}}{{10}}} \right)^2}400
a3=(121100)400\Rightarrow{a_3} = \left( {\dfrac{{121}}{{100}}} \right)400
a3=121×4\Rightarrow{a_3} = 121 \times 4
a3=484\Rightarrow{a_3} = 484
Thus, the third term of the GP is 484.
Hence, the number of seats in 2003 was 484, so option B is correct.

Note: A different approach to the question could be calculating the number of seats in 2002 with the given relation i.e.
400+10%  of  400400 + 10\%\; \text{of}\;400
400+10100×400=400+40=440\Rightarrow400 + \dfrac{{10}}{{100}} \times 400 = 400 + 40 = 440
Now, for 2003,
Number of seats will be
440+10%  of  440440 + 10\%\; \text{of}\;440
440+10100×440\Rightarrow440 + \dfrac{{10}}{{100}} \times 440
440+44=484\Rightarrow440 + 44 = 484
Hence, the number of seats in 2003 was 484.
This approach will be beneficial and can be used only when the number of steps is less.