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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]

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20 Apr 2012, 23:54
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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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21 Apr 2012, 00:13
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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. Answer: B. P.S. Please always post answer choices for PS problems and do not reword or shorten the questions. _________________ Manager Status: I will not stop until i realise my goal which is my dream too Joined: 25 Feb 2010 Posts: 235 Schools: Johnson '15 Followers: 2 Kudos [?]: 50 [1] , given: 16 Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] Show Tags 21 Apr 2012, 05:59 1 This post received 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 hence Answer is B: 8 _________________ Regards, Harsha Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat Satyameva Jayate - Truth alone triumphs Intern Joined: 04 Jan 2012 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 2 Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] Show Tags 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 Current Student Joined: 30 Mar 2012 Posts: 141 Location: United States (TX) Concentration: Strategy, Finance GMAT 1: 670 Q45 V38 GPA: 2.64 WE: Supply Chain Management (Consumer Electronics) Followers: 3 Kudos [?]: 11 [0], given: 14 Re: Need help with exponents problem (bothering me) [#permalink] Show Tags 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]

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21 Apr 2012, 12:55
Is that really a sub level 600 question ?!
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21 Apr 2012, 13:02
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]

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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. Answer: B. 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? Thanks in advance Math Expert Joined: 02 Sep 2009 Posts: 36596 Followers: 7093 Kudos [?]: 93422 [0], given: 10563 Re: Need help with exponents problem (bothering me) [#permalink] Show Tags 25 Sep 2012, 05:52 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]

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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]

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06 Jan 2013, 08:40
bunuel,
do we really need to know ap gp & hp for gmat?
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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]

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19 Jul 2014, 01:56
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]

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29 Aug 2014, 13:09
Sachin9 wrote:
bunuel,
do we really need to know ap gp & hp for gmat?

No, there are other ways:

My way:
Mouth - Payment - Sum
1st - 100 - 100 (4^0*100)
2nd - 300 - 400 (4^1*100)
3rd - 1200 - 1600 (4^2*100)
4th - 4800 - 6200 (4^3*100)
...
So we can say:
Sum = 4^(mouth-1)*100
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27 Sep 2015, 01:41
Hello from the GMAT Club BumpBot!

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Re: Joan took out a mortgage from hel local bank. Each monthly   [#permalink] 27 Sep 2015, 01:41
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