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Bunuel
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A person saves Rs. 200 more each year than in the previous year. If 29 Jun 2021, 03:45
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85% (hard)

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52% (02:33) correct 48% (02:16) wrong based on 162 sessions

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A person saves Rs. 200 more each year than in the previous year. If he started with Rs. 400 in the first year, how many years would he take to save Rs. 18,000 (excluding interest)?

(A) 10 years
(B) 12 years
(C) 15 years
(D) 18 years
(E) None of these
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B
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A person saves Rs. 200 more each year than in the previous year. If 30 Jun 2021, 19:37
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Bunuel
A person saves Rs. 200 more each year than in the previous year. If he started with Rs. 400 in the first year, how many years would he take to save Rs. 18,000 (excluding interest)?

(A) 10 years
(B) 12 years
(C) 15 years
(D) 18 years
(E) None of these
Show more

This is a problem of Arithmetic progression:

First term = a = Rs. 400, common difference = d = Rs. 200, Total sum = S = Rs. 18,000, Number of years = n =?

\(S = \frac{n}{2} * [2a + (n - 1) * d]\)
=> \(18000 = \frac{n}{2} * [800 + 200n - 200]\)
=> \(18000 = n * [300 - 100n]\)
=> \(n^2 - 3n + 180 = 0\)
Solving above equation, we get n = 15 or n = -12
As n is years, it cannot be negative.
=> n = 15 years.

So, correct answer is option C.
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A person saves Rs. 200 more each year than in the previous year. If 03 Jul 2021, 01:11
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MarmikUpadhyay
Bunuel
A person saves Rs. 200 more each year than in the previous year. If he started with Rs. 400 in the first year, how many years would he take to save Rs. 18,000 (excluding interest)?

(A) 10 years
(B) 12 years
(C) 15 years
(D) 18 years
(E) None of these

This is a problem of Arithmetic progression:

First term = a = Rs. 400, common difference = d = Rs. 200, Total sum = S = Rs. 18,000, Number of years = n =?

\(S = \frac{n}{2} * [2a + (n - 1) * d]\)
=> \(18000 = \frac{n}{2} * [800 + 200n - 200]\)
=> \(18000 = n * [300 - 100n]\)
=> \(n^2 - 3n + 180 = 0\)
Solving above equation, we get n = 15 or n = -12
As n is years, it cannot be negative.
=> n = 15 years.

So, correct answer is option C.
Show more

the equation will be
=> \(n^2 + 3n - 180 = 0\)
Solving above equation We get n=-15 or n =12
So n=12years (years cannot be negative)

So, correct answer is option B.[/quote]
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A person saves Rs. 200 more each year than in the previous year. If Updated on: 13 Jul 2024, 06:20
Originally posted by Eilah on 13 Jul 2024, 06:18.
Last edited by Eilah on 13 Jul 2024, 06:20, edited 1 time in total.
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MarmikUpadhyay

Bunuel
A person saves Rs. 200 more each year than in the previous year. If he started with Rs. 400 in the first year, how many years would he take to save Rs. 18,000 (excluding interest)?

(A) 10 years
(B) 12 years
(C) 15 years
(D) 18 years
(E) None of these
This is a problem of Arithmetic progression:

First term = a = Rs. 400, common difference = d = Rs. 200, Total sum = S = Rs. 18,000, Number of years = n =?

\(S = \frac{n}{2} * [2a + (n - 1) * d]\)
=> \(18000 = \frac{n}{2} * [800 + 200n - 200]\)
=> \(18000 = n * [300 - 100n]\)
=> \(n^2 - 3n + 180 = 0\)
Solving above equation, we get n = 15 or n = -12
As n is years, it cannot be negative.
=> n = 15 years.

So, correct answer is option C.
Show more
­The correct answer is B, not C.

Here is the series: 400, 600, 800, 1000, ...
or: 400 + 200*(n-1), with "n" being the number of total years.

The sum of this series would be as follows:

=> 400n + 200 * n(n-1)/2 = 18000
=> 4n + n(n-1) = 180
=> n^2 + 3n = 180

Therefore, n is either -15 or 12.
Since n should be positive, answer B is correct.­
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