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Joan's mortgage

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Joan's mortgage

by rohit_gmat » Mon Oct 17, 2011 4:58 am
Joan took out a mortgage from her 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 $328,000, how many months will it take Joan to pay back her mortgage?

A) 6
B) 8
C) 10
D) 11
E) 13

OA B

IMO A
Source : Veritas Prep

plz advise on why my answer/approach is incorrect...
heres what i did : [spoiler] listed everything down (1) 100 ; (2) 300 ; (3) 900 ; (4) 2700 ; (5) 81,000 ; (6) 243,000 ..... added them all up & got 328,000 .. 6 months so picked A[/spoiler]
Last edited by rohit_gmat on Mon Oct 17, 2011 7:15 pm, edited 1 time in total.
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by shankar.ashwin » Mon Oct 17, 2011 5:08 am
After 2700 you would not get 81,000. You get 8,100.

But an easier way would be to consider this as a GP and do.

1,3,9,27 .... = 3280.

Sum of GP (3280) = (3^n - 1)/3-1

3^n = 6561

n=8 B
Last edited by shankar.ashwin on Mon Oct 17, 2011 5:47 am, edited 1 time in total.
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by Abhishek009 » Mon Oct 17, 2011 5:44 am
rohit_gmat wrote:Joan took out a mortgage from her 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 $328,0000, how many months will it take Joan to pay back her mortgage?

A) 6
B) 8
C) 10
D) 11
E) 13

OA B

IMO A
Source : Veritas Prep

plz advise on why my answer/approach is incorrect...
heres what i did : [spoiler] listed everything down (1) 100 ; (2) 300 ; (3) 900 ; (4) 2700 ; (5) 81,000 ; (6) 243,000 ..... added them all up & got 328,000 .. 6 months so picked A[/spoiler]
Month 1 = 100

Month 2 = 300

Month 3 = 900

Month 4 = 2700


So it's a GP series , with the common ratio of 3 ( Coz every time 3 is multiplied )

Now sum of a GP series is -

S_n=a(1-r^n)/ (1-r)

-328,0000 = 100( 1 - 3^n ) / (1 - 3)

-6560 = 1 - 3^n

3^n = 6561

Now calculate -

3^8 = 6561

Hence n = 8...........
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by enjoylife1788 » Mon Oct 17, 2011 5:45 am
I went mad trying to solve this question. And then realise the amount mentioned in the question is wrong. I guess according to the OA, the amount of loan should be 3,28,000 and not 32,28,000 as mentioned in the question stem.
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by rohit_gmat » Mon Oct 17, 2011 7:15 pm
enjoylife1788 wrote:I went mad trying to solve this question. And then realise the amount mentioned in the question is wrong. I guess according to the OA, the amount of loan should be 3,28,000 and not 32,28,000 as mentioned in the question stem.
ouch! my bad...sorry.. the official source says it is : $ 328,000

(i corrected my original post)
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by rohit_gmat » Mon Oct 17, 2011 7:18 pm
Abhishek009 wrote: S_n=a(1-r^n)/ (1-r)
hi guys, do we really need to know the formula for all the series on the GMAT? (GP, AP ?? etc.)
Im not really used to these.. do u recommend i learn them anyway?
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by GmatKiss » Tue Oct 18, 2011 2:55 am
Joan took out a mortgage from her 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 $328,0000, how many months will it take Joan to pay back her mortgage?

100 - 1st payment, first month!
300
900
2700
8100
24300
72900
218700 - 8th month (How is this the answer)
between 8th and 9th month
656100 - 9th month

Please help!
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by Anurag@Gurome » Tue Oct 18, 2011 3:43 am
Joan took out a mortgage from her 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 $328,000, how many months will it take Joan to pay back her mortgage?
The monthly payments will be $100, $300, $900,...
Hence the payments are in geometric progression with first term 100 and common ratio 3.
Sum of first n terms of a geometric progression with first term a and common ratio r is given by,
  • S = a*(r^n - 1)/(r - 1)
Let us assume it will take n months to pay back the mortgage.
So, 100*(3^n - 1)/(3 - 1) = 328,000
--> (3^n - 1) = 6560
--> 3^n = 6561
--> n = 8

The correct answer is B.
GmatKiss wrote:Please help!
You need to add those figures.
100 + 300 + 900 + ... + 218700 = 328000
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by rohit_gmat » Tue Oct 18, 2011 4:49 am
GmatKiss wrote:100 - 1st payment, first month!
300
900
2700
8100
24300
72900
218700 - 8th month (How is this the answer)
between 8th and 9th month
656100 - 9th month

Please help!
hey GmatKiss .. they r asking when will the TOTAL pay back amount reach 328,000 (so we need to keep adding the pyayments n stop when it reaches 328,000)
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