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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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Posts: 43889
Re: Need help with exponents problem (bothering me) [#permalink]

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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: 221 Schools: Johnson '15 Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] ### Show Tags 21 Apr 2012, 05:59 3 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: 17 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) 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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Joined: 02 Sep 2009
Posts: 43889
Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink]

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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: 43889 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) _________________ "When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas Director Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 502 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] ### Show Tags 06 Jan 2013, 08:40 bunuel, do we really need to know ap gp & hp for gmat? _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992 Intern Joined: 02 Nov 2012 Posts: 34 Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] ### Show Tags 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 Intern Joined: 25 Mar 2014 Posts: 35 Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] ### Show Tags 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7951 Location: Pune, India Re: Joan took out a mortgage from hel local bank. Each monthly [#permalink] ### Show Tags 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 [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! Check out this post on Geometric progressions: https://www.veritasprep.com/blog/2012/0 ... gressions/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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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

[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!

Let $$x = 100$$. As per the question,

$$328,000 = x + 3x + 9x + ....................$$.. Its an GP Series.
Sum of GP Series = $$[a(r^n - 1)/r-1$$. 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 [#permalink]

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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.

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Re: Joan took out a mortgage from hel local bank. Each monthly   [#permalink] 02 Feb 2018, 11:04
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