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

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21 Apr 2012, 00:54

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

65% (03:07) correct
35% (02:16) wrong based on 237 sessions

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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?

Re: Need help with exponents problem (bothering me) [#permalink]

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21 Apr 2012, 01:13

2

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Expert's post

2

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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?

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. _________________

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

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21 Apr 2012, 06:59

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?

Re: Need help with exponents problem (bothering me) [#permalink]

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21 Apr 2012, 12:06

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?

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.

Answer: B.

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. _________________

Don't let a low GPA destroy your dreams of a business education.

Re: Need help with exponents problem (bothering me) [#permalink]

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25 Sep 2012, 03:29

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?

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

Re: Need help with exponents problem (bothering me) [#permalink]

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25 Sep 2012, 06:52

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?

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

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

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27 Sep 2012, 23:41

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) _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

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

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19 Jul 2014, 02:56

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

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27 Sep 2015, 02:41

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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