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How much more interest will maria receive if she invests 1000$ for one
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09 Apr 2022, 02:51
09 Apr 2022, 02:51
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How much more interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semianually, than if she invest 1000$ for one year at x percent annual interest, compounded annually?
A. 5x
B. 10x
C. \(\frac{x^2}{20}\)
D. \(\frac{x^2}{40}\)
E. \((10x+\frac{x^2}{40})\)
A. 5x
B. 10x
C. \(\frac{x^2}{20}\)
D. \(\frac{x^2}{40}\)
E. \((10x+\frac{x^2}{40})\)
Re: How much more interest will maria receive if she invests 1000$ for one
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10 Apr 2022, 18:08
10 Apr 2022, 18:08
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two cases
Interest (1): 1.000 (1+x)
Interest (2): 1.000 (1+x/2)²
1.000 [1²+2.1.(x/2)+(x/2)²
1.000 [1+2.1.(x/2)+(x/2)²]
1.000 [1+x+x²/4]
Comparing (1) and (2), you can notice that extra interest is x²/4
Interest (1): 1.000 (1+x)
Interest (2): 1.000 (1+x/2)²
1.000 [1²+2.1.(x/2)+(x/2)²
1.000 [1+2.1.(x/2)+(x/2)²]
1.000 [1+x+x²/4]
Comparing (1) and (2), you can notice that extra interest is x²/4
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Re: How much more interest will maria receive if she invests 1000$ for one
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06 May 2022, 01:14
06 May 2022, 01:14
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GeminiHeat wrote:
How much more interest will maria receive if she invests 1000$ for one year at x % annual interest, compounded semianually, than if she invest 1000$ for one year at x percent annual interest, compounded annually?
A. 5x
B. 10x
C. \(\frac{x^2}{20}\)
D. \(\frac{x^2}{40}\)
E. \((10x+\frac{x^2}{40})\)
A. 5x
B. 10x
C. \(\frac{x^2}{20}\)
D. \(\frac{x^2}{40}\)
E. \((10x+\frac{x^2}{40})\)
Its quicker if we assume some value of x
Let x = 10
C.I (Compounded semi-annually) = \(1000(1 + \frac{10}{200})^2 = 1000(1.05)^2 = $1102.5\)
C.I (Compounded annually) = \(1000(1 + \frac{10}{100})^1 = 1000(1.1)^1 = $1100\)
How much more?
$2.5 more
Now, put the value of x as 10 and see which option is giving us 2.5?
A. 5x
5(10) = 50
B. 10x
10(10) = 100
C. \(\frac{x^2}{20}\)
\(\frac{10^2}{20} = 5\)
D. \(\frac{x^2}{40}\)
\(\frac{10^2}{40} = 2.5\)
E. \((10x+\frac{x^2}{40})\)
\((10(10)+\frac{10^2}{40}) = 102.5\)
Hence, option D
