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Joan took out a mortgage from hel local bank. Each monthly [#permalink]
20 Apr 2012, 23:54
7
This post was BOOKMARKED
00:00
A
B
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E
Difficulty:
55% (hard)
Question Stats:
65% (03:08) correct
35% (02:11) wrong based on 231 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?
Re: Need help with exponents problem (bothering me) [#permalink]
21 Apr 2012, 00:13
2
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
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]
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?
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?
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]
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?
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]
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?
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]
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) _________________
"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]
19 Jul 2014, 01:56
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Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]
27 Sep 2015, 01:41
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).
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