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18 Apr 2022, 05:45
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
81% (02:32) correct
19%
(02:42)
wrong
based on 95
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Kyle invested an unknown amount of money in an account that pays 12% annually. Later, he invested an additional $50,000 for the same amount of time at 6% interest. If Kyle's two investments earn a total of 8% annual interest, how much was his original investment?
A. $25,000
B. $30,000
C. $35,000
D. $40,000
E. $45,000
A. $25,000
B. $30,000
C. $35,000
D. $40,000
E. $45,000
03 May 2022, 23:31
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From the question, we have
(i) Kyle has invested some amount at 12% interest per annum.
(ii) He has also invested $50,000 at 6% interest per annum for the same period of time.
(iii) His investments earn a total of 8% annual interest.
To find: Amount invested at 12% interest per annum.
Let us assume 'x' be the amount invested by Kyle at 12% interest.
Method-1: Using the interest formula
We can calculate the interests and equate them.
The formula for interest = \(\frac{PNR}{100}\), where P = Amount invested, R = Rate of interest and N = Number of years.
Interest from x at 12% p.a. + Interest from $50,000 at 6% p.a. = Interest from (x + $50,000) at 8% p.a.
\(\frac{(x*N*12)}{100}\) + \(\frac{(50,000*N*6)}{100}\) = \(\frac{((x+50,000)*N*8)}{100}\)
Multiplying the above equation by \(\frac{100}{N}\) on both sides, we get
x*12 + 50,000*6 = (x + 50,000)*8
12x + 300,000 = 8x + 400,000
4x = 100,000
x = \(\frac{100,000}{4}\)
x = $25,000
Method-2: Using weighted average.
The second method is a more intuitive method and helps us solve the problem within 30 seconds.
The resulting rate of interest will be equal to the weighted average of the rates with amounts invested as weights.
Let x and y be the amounts invested at 12% and 6% rates of interest respectively. Then,
\(\frac{(12*x + 6*y)}{(x+y)}\) = 8
12x + 6y = 8x + 8y
4x = 2y
\(\frac{x}{y}\) = \(\frac{1}{2}\)
Thus amounts invested at 12% and 6% rates of interest will be in the ratio of 1 : 2
We know that the amount invested at 6% rate of interest is $50,000. So, the amount invested at 12% rate of interest will be half of it.
x = \(\frac{50,000$}{2}\) = $25,000
Choice (A) $25,000 is the answer
(i) Kyle has invested some amount at 12% interest per annum.
(ii) He has also invested $50,000 at 6% interest per annum for the same period of time.
(iii) His investments earn a total of 8% annual interest.
To find: Amount invested at 12% interest per annum.
Let us assume 'x' be the amount invested by Kyle at 12% interest.
Method-1: Using the interest formula
We can calculate the interests and equate them.
The formula for interest = \(\frac{PNR}{100}\), where P = Amount invested, R = Rate of interest and N = Number of years.
Interest from x at 12% p.a. + Interest from $50,000 at 6% p.a. = Interest from (x + $50,000) at 8% p.a.
\(\frac{(x*N*12)}{100}\) + \(\frac{(50,000*N*6)}{100}\) = \(\frac{((x+50,000)*N*8)}{100}\)
Multiplying the above equation by \(\frac{100}{N}\) on both sides, we get
x*12 + 50,000*6 = (x + 50,000)*8
12x + 300,000 = 8x + 400,000
4x = 100,000
x = \(\frac{100,000}{4}\)
x = $25,000
Method-2: Using weighted average.
The second method is a more intuitive method and helps us solve the problem within 30 seconds.
The resulting rate of interest will be equal to the weighted average of the rates with amounts invested as weights.
Let x and y be the amounts invested at 12% and 6% rates of interest respectively. Then,
\(\frac{(12*x + 6*y)}{(x+y)}\) = 8
12x + 6y = 8x + 8y
4x = 2y
\(\frac{x}{y}\) = \(\frac{1}{2}\)
Thus amounts invested at 12% and 6% rates of interest will be in the ratio of 1 : 2
We know that the amount invested at 6% rate of interest is $50,000. So, the amount invested at 12% rate of interest will be half of it.
x = \(\frac{50,000$}{2}\) = $25,000
Choice (A) $25,000 is the answer
06 Jun 2024, 11:01
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