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Joan took out a mortgage from hel local bank. Each monthly
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Updated on: 21 Apr 2012, 00:08
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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 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!
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Originally posted by Two64 on 20 Apr 2012, 23:54.
Last edited by Bunuel on 21 Apr 2012, 00:08, edited 1 time in total.
Edited the question and added answer choices




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Re: Need help with exponents problem (bothering me)
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21 Apr 2012, 00:13
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: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{100*(3^{n}1)}{31}=$328,000\) > \(3^n1=6,560\) > \(3^n=6,561\). 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.
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Re: Joan took out a mortgage from hel local bank. Each monthly
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21 Apr 2012, 05:59
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 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
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Re: Joan took out a mortgage from hel local bank. Each monthly
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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 ^ (n1) ) (100 times 3 raised to power of n1)
100+300+900+....+100 x 3 ^ (n1) = 3280 x 100
3^0 + 3 + 3^2 +....+3^(n2) + 3^(n1) = 3280 ..... Equation A 3 + 3^2 +....+3^(n2) + 3^(n1) = 3280  1 = 3279 3(1 + 3 +.....+3^(n2)) = 3279 1 + 3 +.....+3^(n2) = 1093
Putting above value in equation A
1093+3^(n1) = 3280 3(n1) = 2187 = 3 x 729 = 3 x 81 x 9 = 3^7
n1 = 7 so n =8



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Re: Need help with exponents problem (bothering me)
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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: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{100*(3^{n}1)}{31}=$328,000\) > \(3^n1=6,560\) > \(3^n=6,561\). 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.
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Re: Joan took out a mortgage from hel local bank. Each monthly
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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
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Re: Need help with exponents problem (bothering me)
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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?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: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{100*(3^{n}1)}{31}=$328,000\) > \(3^n1=6,560\) > \(3^n=6,561\). 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



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Re: Need help with exponents problem (bothering me)
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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?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: \(sum=\frac{b*(r^{n}1)}{r1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)). So, \(\frac{100*(3^{n}1)}{31}=$328,000\) > \(3^n1=6,560\) > \(3^n=6,561\). 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 Check here: sequencesprogressions101891.html It might help.
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Re: Joan took out a mortgage from hel local bank. Each monthly
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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^n1 ...) Therefore to find the Sum of n terms of a GP we use this formula : Sn = a (1r^n) / 1 rUsing this and plugging in the information we get ... 3280 = 1  3^n / 13 ; 13^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
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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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Re: Joan took out a mortgage from hel local bank. Each monthly
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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
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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^(mouth1)*100



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Re: Joan took out a mortgage from hel local bank. Each monthly
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31 Jan 2018, 04:30
Two64 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 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! Check out this post on Geometric progressions: https://www.veritasprep.com/blog/2012/0 ... gressions/
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Joan took out a mortgage from hel local bank. Each monthly
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31 Jan 2018, 22:02
Two64 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 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! Let \(x = 100\). As per the question, \(328,000 = x + 3x + 9x + ....................\).. Its an GP Series. Sum of GP Series = \([a(r^n  1)/r1\). So, \(328,000 = 100 (3^n  1)/2\) > \(3^n = 6561.\) Now, the logic here is we dont have to factorize 6561... All cyclicity of 3 is 4 for numbers ending with 1. From, the option only B satisfies.
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Re: Joan took out a mortgage from hel local bank. Each monthly
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02 Feb 2018, 11:04
Two64 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 We are given that her first mortgage payment is $100. Since her successive monthly payments have to be triple the previous month’s payment, her second would be $300, and the third would be $900 and so on. Therefore, we have 100 + 300 + 900 + … + P = 328,000 where P is the final monthly payment and with the total payments being $328,000. Notice that the left hand side of the equation is a geometric series, with a_1 = 100, a_2 = 3(100) , a_3 = 9(100), and the last payment P = a_n = 3^(n  1)*(100). In this geometric series, we see that the first term a_1 = 100, and the common ratio r = 3. Recall that the sum of the first n terms of a geometric series with first term a_1 and common ratio r is: S_n = a_1(1  r^n)/(1  r) So we can say: 328,000 = 100(1  3^n)/(1  3) 3280 = (1  3^n)/(2) 3280 = (3^n  1)/2 6560 = 3^n  1 3^n = 6561 Since 3^8 = 6561, n = 8. Answer: B
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