On a certain date, Pat invested $10,000 at x percent annual : Problem Solving (PS)

arjtryarjtry

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On a certain date, Pat invested $10,000 at x percent annual [#permalink]

Updated on: 02 Jul 2014, 10:15

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On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?

A. 15
B. 16
C. 18
D. 20
E. 24

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Originally posted by arjtryarjtry on 22 Jul 2008, 18:59.
Last edited by Bunuel on 02 Jul 2014, 10:15, edited 1 time in total.

Renamed the topic, edited the question, added the OA and moved to PS forum.

lsmv479

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Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

03 Sep 2008, 12:17

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$40,000 is 4 times the original amount.

Therefore it doubled twice in 12 years.

Therefore its doubling time is 6.

$80,000 is double $40,000 so another 6 years will get us to $80,000.

Therefore 12+6=18 is the necessary amount of time.

DavidArchuleta

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Joined: 05 Jul 2008

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GMAT 1: 740 Q51 V38

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

22 Jul 2008, 21:13

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arjtryarjtry wrote:

On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

:twisted:

x- interest rate
80.000= 10.000 (1+x)^year
=> 8=(1+x)^year

40.000=10.000 . (1+x)^12
=> 4= (1+x)^12
=>2= (1+x)^6
=> 8 = (1+x)^ 18
So, after 18 years, the total value of the investment plus interest will increase to $80,000.

arjtryarjtry

Current Student

Joined: 11 May 2008

Posts: 377

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

03 Sep 2008, 01:22

2

Kudos

given 40000=10000(1+x/100)^12
or 4 ^1/12 = (1+x/100)
or x/100 = 4^1/12-1........A

asked 80000=10000(1+x/100)^n..............b
n=?

substitute value of x/100 from A in B.
8 = (1+4^1/12-1)^n
or 8= 4^n/12
2^3=2^2n/12
n=18

ans = C

x2suresh

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Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

03 Sep 2008, 06:31

2

Kudos

arjtryarjtry wrote:

On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???

40k = 10k(1+x/100)^12 --> 4=(1+x/100)^12
--> (1+x/100)^6=2
80k=40k(1+x/100)^n
--> (1+x/100)^n =2 =(1+x/100)^6
n=6

Total years = 12+6=18

GMAT TIGER

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Joined: 29 Aug 2007

Posts: 1021

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

02 Feb 2009, 08:51

2

Kudos

GMAT TIGER wrote:

arjtryarjtry wrote:

On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???

your second post blocked my post.

i cannot do it without calculator or computer and also beleieve this is not real gmat type question cuz its very difficult to get the value without those machines.

if i were to choose during the test, would go for 18 or 20. probably 18 cuz it wont take too long to get the value doubled....

I found a method: Rule of 72.

Given an x% return, it takes 10,000 to quadralope 12 years.

So according to the rule: 72/x is the no of years 10,000.00 took to double 20,000.00.
Again, 20,000.00 took to double 40,000.00 same (72/x) no of years.

72/x+ 72/x = 12
x = 12% (though rate here is not very much required).

Again, 40,000.00 takes the same (72/x) no of years to double 80,000.00.

72/x = 6 years.

So altogather:
10,000 - 20,000 = 6 years
20,000 - 40,000 = 6 years
40,000 - 80,000 = 6 years
total 18 years.

devilmirror

Manager

Joined: 04 Jan 2006

Posts: 102

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

03 Sep 2008, 12:47

1

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arjtryarjtry wrote:

On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???

Let's simplify this question.

First equation; 40,000 = 10,000(1 + x)^12
Given that (1 + x) = a and rearrange the equation above we get
4 = a^12 = 2^2
So now we know that a^12 = (a^6)^2 = 2^2
Therefore, (a^6) = 2

Second equation; 80,000 = 10,000(1 + x)^n
Or 8 = a^n
2^3 = a^n
(a^6)^3 = a^n = a^18 ---- (since a^6 = 2)

therefore, n = 18. C is the answer

LoyalWater

Intern

Joined: 20 May 2008

Posts: 30

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

02 Feb 2009, 02:35

1

Kudos

$40,000 is 4 times the original amount.

Therefore it doubled twice in 12 years.

Therefore its doubling time is 6.

$80,000 is double $40,000 so another 6 years will get us to $80,000.

Therefore 12+6=18 is the necessary amount of time.

________________________

is this the right method???

arjtryarjtry

Current Student

Joined: 11 May 2008

Posts: 377

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

02 Sep 2008, 19:50

On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???

GMAT TIGER

VP

Joined: 29 Aug 2007

Posts: 1021

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

02 Sep 2008, 23:24

arjtryarjtry wrote:

On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000?
A. 15
B. 16
C. 18
D. 20
E. 24

for compound interest problems, when the time is large, then how does one approach???

dont just give the steps...
also mention, for variations in the problems, how does one approach???

your second post blocked my post.

i cannot do it without calculator or computer and also beleieve this is not real gmat type question cuz its very difficult to get the value without those machines.

if i were to choose during the test, would go for 18/ or 20. probably 18 cuz not it wont take too long to get the value doubled....

ssandeepan

Manager

Joined: 09 Jul 2007

Posts: 117

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

03 Sep 2008, 13:42

Even though it looks like a tough question, it is actually not. 4 times in 12 yesrs = > 2 times in 6 years ( from 40K to 80K ).
So total year = 12 +6 =18

sagarag

Intern

Joined: 13 May 2014

Posts: 16

Schools: IIMA ISB '17 IIMC IIM Indore'17 IIMB

GMAT Date: 11-01-2014

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

26 Jun 2015, 23:29

The key is to identify how quickly the money gets doubled..

for the current problem money gets doubled in every 6 years..

t=0 Investment = 10000
t=6 Investment= 20000
t=12 Investment = 40000..

Therefore for money to double from 40,000 to 80,000 it will take another 6 years. i.e 18years

P

mvictor

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Posts: 2163

Location: United States (IL)

Concentration: Finance, Economics

Schools: Stanford '20 (WD)

GMAT 1: 650 Q49 V30 Verified score

GPA: 3.92

WE:General Management (Transportation)

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

12 Dec 2015, 11:33

i tried different approaches, but I believe this is the most straight forward one.
in 12y, the amount is 4x, where x is the initial amount.
it means that if we start with 40k in year 12, in another 12 years, we'll have 160k. we can eliminate right away 24 years.
now, if the amount 4x in 12y, it might be true that the amount will be 2x in 6y.
ok, so we have 12y initial, then another 6 years, so total 18 years to reach 80k.
C.

B

law258

Senior Manager

Joined: 05 Sep 2016

Status:DONE!

Posts: 274

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

01 Dec 2016, 20:46

Tricky little bugger...

Compound Interest Formula: P[1+(r/n)]^(nt)

10,000(1+x)^12 = 40,000
(1+x)^12 = 4
(1+x) = 4^(1/12)

10,000(1+x)^t = 80,000
(4^(1/12))^t = 8
(2^2)^(1/12) = 2^3
(2^(1/6)^t = 2^3

Bases are equal, so we can set the exponents equal to each other --> (1/6)t = 3 --> t = 18

C.

B

uttonio

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Joined: 03 Nov 2021

Posts: 4

Location: Viet Nam

Concentration: Finance, Healthcare

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

27 Dec 2021, 22:55

Hi guys. I have two questions related to this question. I would really appreciate it if anyone can help me understand the answers of these questions. The questions are from Brightlink by the way.

1. If $1,000 is invested at 15 percent annual interest, compounded semiannually, what is the approximate balance after 1 year?
A) $1,350
B) $1,320
C) $1,300
D) $1,200
E) $1,150

I chose E: 1000 *(1+0.075)^2 = 1155.6. But the answer is B.

2. On a certain date, Pat invested $X in Panda account at a simple annual interest rate of 2.5 percent. On the same date, he invested another $Y in Zebra account at a simple annual interest rate of 1.5 percent. The sum of the amounts that he invested in the two accounts was $6,400. If each investment earned him the same interest after two years, what was his average rate of interest from the two investments?
A) 3.5%
B) 3.75%
C) 4%
D) 4.25%
E) 4.5%

I didn't understand the average interest here. I got 1.875% or and logically, I think the answer seems wrong because X has 2.5% interest and Y has 1.5% interest. The average interest should be between 1.5 and 2.5. I just don't understand what they mean.

If you agree with me, I assume that Brightlink needs to be more meticulous in uploading questions. Thanks a lot!

B

ABHISEKGMAT123

Intern

Joined: 19 Dec 2018

Posts: 16

Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

19 Sep 2022, 11:41

1----4 (4times) in 12 years
Therefore,
Root 1---Root 4 time in 6 years i.e.
1----2 in 6 years
Therefore further 6 years after 12 years it will take
Ans: 18

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Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

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Re: On a certain date, Pat invested $10,000 at x percent annual [#permalink]

29 Sep 2023, 08:30

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