Rate of investment

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Vincelauret: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Fri Mar 13, 2015 11:07 pm; Followed by:1 members

Rate of investment

by Vincelauret » Fri Mar 20, 2015 6:57 am

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An investment of $1,500 was made in a certain bank account and it earned interest that was compounded annually; the annual interest rate was fixed for the entire duration of the investment. If after 12 years the $1,500 increased to $24,000 by earning interest, in how many years after the initial investment was made would the $1,500 have increased to $96,000 by earning interest at the same rate?
A. 15
B. 18
C. 20
D. 21
E. It cannot be determined from the information given

I found the answer by testing the different options, but does it exist an algebric formula?

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Source: Beat The GMAT — Problem Solving | Fri Mar 20, 2015 6:57 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Fri Mar 20, 2015 7:49 am

Vincelauret wrote:An investment of $1,500 was made in a certain bank account and it earned interest that was compounded annually; the annual interest rate was fixed for the entire duration of the investment. If after 12 years the $1,500 increased to $24,000 by earning interest, in how many years after the initial investment was made would the $1,500 have increased to $96,000 by earning interest at the same rate?
A. 15
B. 18
C. 20
D. 21
E. It cannot be determined from the information given

First, notice that $24,000 is SIXTEEN TIMES $1500
In other words, the investment DOUBLED 4 times during those 12 years.
Here's what I mean:
Start with $1500 ..... double to get $3000
Now have $3000 ..... double to get $6000
Now have $6000 ..... double to get $12,000
Now have $12,000 ..... double to get $24,000

If the investment doubled 4 times during those 12 years, we can conclude that the investment doubles once every 3 years.
So, to summarize, we have:
Year 0: $1500
Year 3: $3000
Year 6: $600
Year 9: $12,000
Year 12: $24,000
To answer our question, let's continue the pattern...
Year 15: $48,000
[spoiler]Year 18: $96,000[/spoiler]

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Fri Mar 20, 2015 8:56 am

Vincelauret wrote:An investment of $1,500 was made in a certain bank account and it earned interest that was compounded annually; the annual interest rate was fixed for the entire duration of the investment. If after 12 years the $1,500 increased to $24,000 by earning interest, in how many years after the initial investment was made would the $1,500 have increased to $96,000 by earning interest at the same rate?
A. 15
B. 18
C. 20
D. 21
E. It cannot be determined from the information given

I found the answer by testing the different options, but does it exist an algebric formula?

We could use the formula for EXPONENTIAL GROWTH:

Final amount = (original amount) * (multiplier)^(number of changes).

After 12 years the $1,500 increased to $24,000.
In this case:
Final amount = 24000
Original amount = 1500
Multiplier = m
Number of changes = 12.

Plugging these values in the formula, we get:
24000 = 1500 * mÂ¹Â²
16 = mÂ¹Â²
(2â�´)^(1/12) = m
2^(1/3) = m.

How many years after the initial investment was made would the $1,500 have increased to $96,000?
In this case:
Final amount = 96000
Original amount = 1500
Multiplier = 2^(1/3)
Number of changes = x.

Plugging these values in the formula, we get:
96000 = 1500 * (2^1/3)^x
64 = 2^(x/3)
2â�¶ = 2^(x/3).
Since the bases are equal, the exponents must also be equal.
Thus:
x/3 = 6
x = 18.

The correct answer is B.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Fri Mar 20, 2015 10:49 am

Vincelauret wrote:An investment of $1,500 was made in a certain bank account and it earned interest that was compounded annually; the annual interest rate was fixed for the entire duration of the investment. If after 12 years the $1,500 increased to $24,000 by earning interest, in how many years after the initial investment was made would the $1,500 have increased to $96,000 by earning interest at the same rate?
A. 15
B. 18
C. 20
D. 21
E. It cannot be determined from the information given

Slight variation of Brent's approach:

After 12 years the $1,500 increased to $24,000.
24000/1500 = 16.
Since 16 = 2â�´, the implication is that 12 YEARS are required for the investment to DOUBLE 4 TIMES.

How many years after the initial investment was made would the $1,500 have increased to $96,000?
96000/1500 = 64.
Since 64 = 2â�¶, we must determine HOW MANY YEARS are required for the investment to DOUBLE 6 TIMES.

Thus, we could set up the following proportion:
12/4 = x/6
3 = x/6
x = 18.

The correct answer is B.

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piyushdabomb: Master | Next Rank: 500 Posts; Posts: 115; Joined: Sun Sep 28, 2008 5:31 pm; Thanked: 4 times

Re: Rate of investment

by piyushdabomb » Wed Mar 09, 2022 9:02 pm

After painfully trying to solve this equation, no one had a good solution out there using the formula!

I finally figured it out using a structured easy to remember method. I have a more structured way to solve this problem using exponents and simultaneous equations.

Here it is.

Compound interest formula = P(1+r/n)^nt

P = initial amount
r = interest rate
n = number of times the interest is applied in a year
t = number of periods in question

Here's what know:
1. P = 1,500 since that's what's originally invested
2. The first relationship tells us that the resulting investment yields 24,000, so the first relationship will use the equation with 24,000.
3. The second relationship tells us that the resulting investment yields 96,000, so the second relationship will use the equation with 96,000.
n = 1 since the story tells us that the interest was compounded "annual" (instead of bi-annually, etc...)

(1) 24,000 = 1,500*(1+r/1)^(1*12)
(2) 96,000 = 1500*(1+r/1)^(1*t)

Let's solve for t.

Steps:
1. Simplify (1) and (2):
(1) 16 = (1+r)^12
(2) 64 = (1+r)^t

2. Let's put (1+r) to one side in (1), so: 12th_root(16) = 1+r, which is also the same as 1+r = 16^(1/12)
3. Now, let's plug in 1+r into (2) and solve for t.
When you plug 1+r into (2), you will realize that exponent (1/12) will multiply into t to give you t/12
64 = 16*[(1/12)*t] = 16^(t/12)
4. Now, remember that to solve for t, the bases must be the same so you can drop the bases completely like this:

64 = 4^3
16 = 4^2
Therefore:
4^3 = 4^[2*(t/12)] = 4^(2t/12) = 4^(t/6)
4^3 = 4^(t/6)

5. Finally, since the bases are 4s, drop the 4s and solve the equation:
3 = t/6
t = 18.

Answer is 18!

-------------------
Sincerely,

Piyush A.