On a certain date, Pat invested $10,000 at x percent annual

Question

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:

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

lsmv479 · Accepted Answer

$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 · Answer

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 · Answer

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 · Answer

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....

arjtryarjtry · Answer

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 · Answer

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

devilmirror · Answer

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

ssandeepan · Answer

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

LoyalWater · Answer

$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???

GMAT TIGER · Answer

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.

sagarag · Answer

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

mvictor · Answer

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.

law258 · Answer

Tricky little bugger...

Compound Interest Formula: P ^(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.

uttonio · Answer

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!

ABHISEKGMAT123 · Answer

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

bumpbot · Answer

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On a certain date, Pat invested $10,000 at x percent annual

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