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Updated on: 06 Mar 2026, 19:32
Originally posted by MartyMurray on 17 Feb 2026, 20:42.
Last edited by MartyMurray on 06 Mar 2026, 19:32, edited 4 times in total.
Last edited by MartyMurray on 06 Mar 2026, 19:32, edited 4 times in total.
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D
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Difficulty:
85%
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Question Stats:
57% (02:35) correct
43%
(02:29)
wrong
based on 84
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An investment account growing at a constant rate doubles in value every 4 years, while a second account growing at a constant rate quadruples in value every 6 years. The first account starts with $8,000, and, at the same time, the second starts with $1,000. After how many years will the two accounts have the same value?
(A) 8
(B) 12
(C) 24
(D) 36
(E) 48
Source: Marty Murray Coaching
(A) 8
(B) 12
(C) 24
(D) 36
(E) 48
Source: Marty Murray Coaching
Updated on: 24 Feb 2026, 19:34
Originally posted by MartyMurray on 18 Feb 2026, 08:06.
Last edited by MartyMurray on 24 Feb 2026, 19:34, edited 3 times in total.
Last edited by MartyMurray on 24 Feb 2026, 19:34, edited 3 times in total.
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An investment account growing at a constant rate doubles in value every 4 years, while a second account growing at a constant rate quadruples in value every 6 years. The first account starts with $8,000, and, at the same time, the second starts with $1,000. After how many years will the two accounts have the same value?
This is an exponential growth question.
To answer it, let \(y\) be the number of years after which the two accounts will have the same value.
The value of the first account is multiplied by \(2\) every \(4\) years. So, after \(y\) years, the first account will be multiplied by \(2\) \(\frac{y}{4}\) times.
The value of the second account is multiplied by \(4\) every \(6\) years. So, after \(y\) years, the second account will be multiplied by \(4\) \(\frac{y}{6}\) times.
Thus, we have the following:
\(8,000 × 2^{\frac{y}{4}} = 1,000 × 4^{\frac{y}{6}}\)
\(8 × 2^{\frac{y}{4}} = 4^{\frac{y}{6}}\)
\(2^3 × 2^{\frac{y}{4}} = 2^{\frac{2y}{6}}\)
\(3 + \frac{y}{4} = \frac{y}{3}\)
\(3 = \frac{y}{12}\)
\(36 = y\)
(A) 8
(B) 12
(C) 24
(D) 36
(E) 48
Correct answer: D
This is an exponential growth question.
To answer it, let \(y\) be the number of years after which the two accounts will have the same value.
The value of the first account is multiplied by \(2\) every \(4\) years. So, after \(y\) years, the first account will be multiplied by \(2\) \(\frac{y}{4}\) times.
The value of the second account is multiplied by \(4\) every \(6\) years. So, after \(y\) years, the second account will be multiplied by \(4\) \(\frac{y}{6}\) times.
Thus, we have the following:
\(8,000 × 2^{\frac{y}{4}} = 1,000 × 4^{\frac{y}{6}}\)
\(8 × 2^{\frac{y}{4}} = 4^{\frac{y}{6}}\)
\(2^3 × 2^{\frac{y}{4}} = 2^{\frac{2y}{6}}\)
\(3 + \frac{y}{4} = \frac{y}{3}\)
\(3 = \frac{y}{12}\)
\(36 = y\)
(A) 8
(B) 12
(C) 24
(D) 36
(E) 48
Correct answer: D
23 Feb 2026, 20:15
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Couldn't figure out the most efficient approach, so I ended up testing answer choices using: [principal]*[growth factor]^(years/periods)
The "years" exponent being how many periods of doubling or quadrupling have taken place
A) 8 years
8,000(2)(2) ≠ 1,000(4)(4)
B) 12 years
8,000(2)(2)(2) ≠ 1,000(4)(4)
C) 24 years
\(8,000(2)^6\) ≠ \(1,000(4)^4\)
D) 36 years
\(8,000(2)^9\) = \(2^{12} *1000\)
\(1,000(4)^6\) = \(2^{12} *1000\)
Answer
The "years" exponent being how many periods of doubling or quadrupling have taken place
- Example: $8,000 will double every four years, so in 8 years, it will double twice (i.e., 8 yrs/4 years per period = exponent of 2): \($8,000(2)^2\)
- Example: $1,000 will quadruple every 6 years, so in 12 years, it will quadruple twice: \($1,000(4)^2\)
A) 8 years
8,000(2)(2) ≠ 1,000(4)(4)
B) 12 years
8,000(2)(2)(2) ≠ 1,000(4)(4)
C) 24 years
\(8,000(2)^6\) ≠ \(1,000(4)^4\)
D) 36 years
\(8,000(2)^9\) = \(2^{12} *1000\)
\(1,000(4)^6\) = \(2^{12} *1000\)
Answer
