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# Joan took out a mortgage from hel local bank. Each monthly

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Joan took out a mortgage from hel local bank. Each monthly [#permalink]  20 Apr 2012, 23:54
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Question Stats:

68% (03:04) correct 32% (01:50) wrong based on 113 sessions
Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is$328000, how many months will it take Joan to pay back her mortgage?

A. 6
B. 8
C. 10
D. 11
E. 13

[Reveal] Spoiler:
The answer, according to Veritas is "8." I put 6, which seems to be incorrect. Really bothers me but if you guys can help... kudos!
[Reveal] Spoiler: OA

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Last edited by Bunuel on 21 Apr 2012, 00:08, edited 1 time in total.
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Re: Need help with exponents problem (bothering me) [#permalink]  21 Apr 2012, 00:13
2
KUDOS
Expert's post
jxatrillion wrote:
Joan took out a mortgage from her bank. Each monthly payment she makes must triple the amount of the previous month's payment. If her first payment is $100, and the total that she must pay back is$328,000, how many months will it take Joan to pay her mortgage?

[Reveal] Spoiler:
The answer, according to Veritas is "8." I put 6, which seems to be incorrect. Really bothers me but if you guys can help... kudos!

Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is$328000, how many months will it take Joan to pay back her mortgage?
A. 6
B. 8
C. 10
D. 11
E. 13

Since her first payment is $100 and all subsequent payments must be three times as many as the payment for the previous month, then her monthly payments will be:$100;
$300;$900;
$2,700; ... As you can see we have a geometric progression with the first term of 100 and the common ratio of 3. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, (where b is the first term, n # of terms and r is a common ratio \neq{1}). So, \frac{100*(3^{n}-1)}{3-1}=$328,000 --> 3^n-1=6,560 --> 3^n=6,561. Now, the unit's digit of 3 in a positive integer power repeats in pattern of 4: {3, 9, 7, 1}. Since the last digit of 3^n is 1 then n can be 4 (3^4=81), 8, 12, ... only 8 is present among options.

P.S. Please always post answer choices for PS problems and do not reword or shorten the questions.
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  21 Apr 2012, 05:59
1
KUDOS
jxatrillion wrote:
Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is$328000, how many months will it take Joan to pay back her mortgage?

A. 6
B. 8
C. 10
D. 11
E. 13

[Reveal] Spoiler:
The answer, according to Veritas is "8." I put 6, which seems to be incorrect. Really bothers me but if you guys can help... kudos!

followed long manual method of calculating the amount
1st month = 100
2: 300
3: 900
4: 2700
5: 8100
6: 24300
7: 72900
8: 217800
9: more than 6L

hence if we add 8 to 1 months amount it should be more than 328000

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Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  21 Apr 2012, 08:06
An alternate way of solving this is as follows - Brunei plz confirm it.

Its geometric progression defined as Cn = 100 x (3 ^ (n-1) ) (100 times 3 raised to power of n-1)

100+300+900+....+100 x 3 ^ (n-1) = 3280 x 100

3^0 + 3 + 3^2 +....+3^(n-2) + 3^(n-1) = 3280 ..... Equation A
3 + 3^2 +....+3^(n-2) + 3^(n-1) = 3280 - 1 = 3279
3(1 + 3 +.....+3^(n-2)) = 3279
1 + 3 +.....+3^(n-2) = 1093

Putting above value in equation A

1093+3^(n-1) = 3280
3(n-1) = 2187 = 3 x 729 = 3 x 81 x 9 = 3^7

n-1 = 7 so n =8
Manager
Joined: 30 Mar 2012
Posts: 121
Location: United States (TX)
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Kudos [?]: 6 [0], given: 13

Re: Need help with exponents problem (bothering me) [#permalink]  21 Apr 2012, 11:06
Bunuel wrote:
jxatrillion wrote:
Joan took out a mortgage from her bank. Each monthly payment she makes must triple the amount of the previous month's payment. If her first payment is $100, and the total that she must pay back is$328,000, how many months will it take Joan to pay her mortgage?

[Reveal] Spoiler:
The answer, according to Veritas is "8." I put 6, which seems to be incorrect. Really bothers me but if you guys can help... kudos!

Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is$328000, how many months will it take Joan to pay back her mortgage?
A. 6
B. 8
C. 10
D. 11
E. 13

Since her first payment is $100 and all subsequent payments must be three times as many as the payment for the previous month, then her monthly payments will be:$100;
$300;$900;
$2,700; ... As you can see we have a geometric progression with the first term of 100 and the common ration of 3. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, (where b is the first term, n # of terms and r is a common ratio \neq{1}). So, \frac{100*(3^{n}-1)}{3-1}=$328,000 --> 3^n-1=6,560 --> 3^n=6,561. Now, the unit's digit of 3 in positive integer power repeats in pattern of 4: {3, 9, 7, 1}. Since the last digit of 3^n is 1 then n can be 4 (3^4=81), 8, 12, ... only 8 is present among options.

P.S. Please always post answer choices for PS problems and do not reword or shorten the questions.

I apologize. I wasn't able to copy and paste from the online application so I must've missed something from typing it. Thank you for your and everyone's answers though. I really appreciate it.

I'm not sure how I got 6 now. Was just really bugging me how I didn't see 8 last night. Thanks again, team.
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  21 Apr 2012, 12:55
Is that really a sub level 600 question ?!
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  21 Apr 2012, 13:02
Expert's post
Alexmsi wrote:
Is that really a sub level 600 question ?!

No, it's not. I'd say it's around 650 level question.
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Re: Need help with exponents problem (bothering me) [#permalink]  25 Sep 2012, 02:29
Bunuel wrote:
jxatrillion wrote:
Joan took out a mortgage from her bank. Each monthly payment she makes must triple the amount of the previous month's payment. If her first payment is $100, and the total that she must pay back is$328,000, how many months will it take Joan to pay her mortgage?

[Reveal] Spoiler:
The answer, according to Veritas is "8." I put 6, which seems to be incorrect. Really bothers me but if you guys can help... kudos!

Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is$328000, how many months will it take Joan to pay back her mortgage?
A. 6
B. 8
C. 10
D. 11
E. 13

Since her first payment is $100 and all subsequent payments must be three times as many as the payment for the previous month, then her monthly payments will be:$100;
$300;$900;
$2,700; ... As you can see we have a geometric progression with the first term of 100 and the common ratio of 3. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, (where b is the first term, n # of terms and r is a common ratio \neq{1}). So, \frac{100*(3^{n}-1)}{3-1}=$328,000 --> 3^n-1=6,560 --> 3^n=6,561. Now, the unit's digit of 3 in a positive integer power repeats in pattern of 4: {3, 9, 7, 1}. Since the last digit of 3^n is 1 then n can be 4 (3^4=81), 8, 12, ... only 8 is present among options.

P.S. Please always post answer choices for PS problems and do not reword or shorten the questions.

Hi Bunuel,
I solved this question in 10 minutes by logic,
but how do you get geometric progression formula, could you please give me any explainations?
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Posts: 19030
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Re: Need help with exponents problem (bothering me) [#permalink]  25 Sep 2012, 05:52
Expert's post
mario1987 wrote:
Bunuel wrote:
jxatrillion wrote:
Joan took out a mortgage from her bank. Each monthly payment she makes must triple the amount of the previous month's payment. If her first payment is $100, and the total that she must pay back is$328,000, how many months will it take Joan to pay her mortgage?

[Reveal] Spoiler:
The answer, according to Veritas is "8." I put 6, which seems to be incorrect. Really bothers me but if you guys can help... kudos!

Joan took out a mortgage from hel local bank. Each monthly mortgage payment she makes must be triple the amount of the previous month's payment. If her first payment is $100, and the total amount she must pay back is$328000, how many months will it take Joan to pay back her mortgage?
A. 6
B. 8
C. 10
D. 11
E. 13

Since her first payment is $100 and all subsequent payments must be three times as many as the payment for the previous month, then her monthly payments will be:$100;
$300;$900;
$2,700; ... As you can see we have a geometric progression with the first term of 100 and the common ratio of 3. The sum of the first n terms of geometric progression is given by: sum=\frac{b*(r^{n}-1)}{r-1}, (where b is the first term, n # of terms and r is a common ratio \neq{1}). So, \frac{100*(3^{n}-1)}{3-1}=$328,000 --> 3^n-1=6,560 --> 3^n=6,561. Now, the unit's digit of 3 in a positive integer power repeats in pattern of 4: {3, 9, 7, 1}. Since the last digit of 3^n is 1 then n can be 4 (3^4=81), 8, 12, ... only 8 is present among options.

P.S. Please always post answer choices for PS problems and do not reword or shorten the questions.

Hi Bunuel,
I solved this question in 10 minutes by logic,
but how do you get geometric progression formula, could you please give me any explainations?

Check here: sequences-progressions-101891.html It might help.
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  27 Sep 2012, 22:41
Joan starts off with 100 \$ .. which is to be tripled every month

Her monthly payments look like this :

100 , 300 , 900 , 2700 ......... Upto 328000

This can be re written as :

100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 ...... 100 x 3280

So we have 1 , 3 , 9 , 27 ..... 32800 in GP

We know that a =1 , and r = 3 ( its easy to figure it out by looking at the question , but regardless of it being mentioned in the question we can still compute the value of r using the formula Tn = a3^n-1 ...)

Therefore to find the Sum of n terms of a GP we use this formula :

Sn = a (1-r^n) / 1 -r

Using this and plugging in the information we get ...

3280 = 1 - 3^n / 1-3 ; 1-3^n / -2

Cross multiplying we get

3280 x -2 = 1- 3^n

- 6560 = 1 - 3^n

- 6561 = - 3 ^n

6561 = 3 ^n (negatives cancel out)

6561 can also be re written as 3 ^ 8

Therefore ; 3 ^8 = 3 ^n

Thus n = 8 (B)
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  06 Jan 2013, 08:40
bunuel,
do we really need to know ap gp & hp for gmat?
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  07 Jan 2013, 01:35
I came to this 100 * 3^t = 328000 --> 3^t = 32800, is this correct and how do I calculate t from this? Can somebody please help. Thanks in advance
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]  19 Jul 2014, 01:56
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Re: Joan took out a mortgage from hel local bank. Each monthly   [#permalink] 19 Jul 2014, 01:56
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