Q. A total of $1000 was invested at compounded annual interest rate. At the end of 12 years, the total value will be $4000. How many years are needed to reach a total of $8,000?
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Compound Interest
Source: Beat The GMAT — Problem Solving |
The key here is to find the rate of interest using the values 1000, 4000 in CI formula. From finding the rate of interest, as a second step, find the number of years requires using the values 1000, 8000 and r% in CI formula. This requires a lot of calculations.
I dont think you need to do lot of calculations
Consider this
A = P*( 1 + r/100)^n formula we all know for compound interest .
Substitue now
4000 = 1000*(1 + r/100)^12
Thus ( 1 + r/100) = 12th root of 4. -- eqn (1)
Now you got to find no of yrs for getting to 8000
So , 8000 = 1000*( 1 + r/100) ^n ---> 8 = (1+r/100)^n -- eqn (2)
Using eqn q and eqn 2 , you simpy get 8 = ( 12th root of 4) ^ n ==> 2 ^ 3 ==> n/6th root of 2 ==> 3 = n/6 and hence n = 18
Consider this
A = P*( 1 + r/100)^n formula we all know for compound interest .
Substitue now
4000 = 1000*(1 + r/100)^12
Thus ( 1 + r/100) = 12th root of 4. -- eqn (1)
Now you got to find no of yrs for getting to 8000
So , 8000 = 1000*( 1 + r/100) ^n ---> 8 = (1+r/100)^n -- eqn (2)
Using eqn q and eqn 2 , you simpy get 8 = ( 12th root of 4) ^ n ==> 2 ^ 3 ==> n/6th root of 2 ==> 3 = n/6 and hence n = 18
First of all, it is wery interesting question.
Here is my explanation:
1- equotion for a compounded procent is S=D(1+%)^n, where n is teh number of time periods, D is the starting deposit, and S is the total sum.
2- we have 4000=1000(1+%)^12, from this we can find %
% = 2^1/6 - 1
3- now use the sasme reasoning to find solution, if
8000 = 4000(1+2^1/6 - 1)^x
2 = (2^1/6)^x
x = 6
My answer is 6+12(first timeperiod) = 18
Here is my explanation:
1- equotion for a compounded procent is S=D(1+%)^n, where n is teh number of time periods, D is the starting deposit, and S is the total sum.
2- we have 4000=1000(1+%)^12, from this we can find %
% = 2^1/6 - 1
3- now use the sasme reasoning to find solution, if
8000 = 4000(1+2^1/6 - 1)^x
2 = (2^1/6)^x
x = 6
My answer is 6+12(first timeperiod) = 18
I don't understand how you get the interest in the first part
4000= 1000(1+%)^12, I can simplify that to 4= (1+%)^12, but how do you solve from there?
4000= 1000(1+%)^12, I can simplify that to 4= (1+%)^12, but how do you solve from there?
here
4000 = 1000*(1 + r/100)^12
Thus ( 1 + r/100) = 12th root of 4. this can be written as
(1+r/100) = 6ht root of 2 [2^2 = 4]--- I
8000 = 1000*(1 + r/100)^n then
( 1 + r/100) = nth root of 8 (8 = 2^3)
so 1+r/100 = 2^3/n ----- II
from I & II
2^1/6 = 2^3/n then 1/6 = 3/n and n = 18
4000 = 1000*(1 + r/100)^12
Thus ( 1 + r/100) = 12th root of 4. this can be written as
(1+r/100) = 6ht root of 2 [2^2 = 4]--- I
8000 = 1000*(1 + r/100)^n then
( 1 + r/100) = nth root of 8 (8 = 2^3)
so 1+r/100 = 2^3/n ----- II
from I & II
2^1/6 = 2^3/n then 1/6 = 3/n and n = 18
Gmat710,, Hyd
Guys,
Do we have to do all those calculations? Please let me know if the reasoning below is incorrect.
Investment quadrupled in 12 years
Investment will double in 6 years
18 years is needed
Do we have to do all those calculations? Please let me know if the reasoning below is incorrect.
Investment quadrupled in 12 years
Investment will double in 6 years
18 years is needed
In this problem, you can get away with such reasoning, because sqrt(4) = 4/2okigbo wrote:Guys,
Do we have to do all those calculations? Please let me know if the reasoning below is incorrect.
Investment quadrupled in 12 years
Investment will double in 6 years
18 years is needed
What you are seeing a logarithmic proportion
Assume that investment triped in 6 years.
Does the investment double in 2 years or 3 years?
