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Question: Betty invested some money at 15% interest. Betty also invested \$237 more than 2 times that amount at 13%. How much is invested at each rate if Betty receives \$1688.03 in interest after one year? (Round to two decimal places if necessary)

Detailed Solution:

From the given problem, it is evident that there are two different investments. One investment earns 15% interest and the other investment earns 13%.

Let us say that the amount X is invested at 15% interest,  amount Y is invested at 13% interest.

Since there are two variables X & Y, We need to frame two equations to solve for the two given variables.

Amount of interest earned by the  amount X at 15% interest = x\cdot \dfrac {15}{100}=0\cdot 15x

Amount of interest earned by the amount Y at 13% interest =y\cdot \dfrac {13}{100}=0.13y

Given \$1668.03 is the total interest earned from the two investments.

Therefore 0.15x+0.13y=1668.03

Multiplying with 100 throughout the equation to avoid the decimals.

15x+13y=168803—->Equation 1

Given that the “Betty invested $237 more than 2 times that amount at 13%”

Therefore we can say

y= 2x+237 —->Equation 2

By using the substitution method, we can replace y as  2x +237 in Equation 1.




x=\dfrac {165722}{41}


Therefore y= 2x+237 =2(4042)+237


Therefore,In conclusion,the  amount invested at  15% is \$ 4042 and the amount invested at 13% interest is  \$ 8321 .


We can cross check the  answers we got for the X =4042 and  Y =8321 values.

When the values of X and Y are plugged back into the original equations, they must hold good.

Let us take up  the  equation1




168803=168803.Therefore, the solutions we  arrived work for the equation1

Let us cross-check with the equation2.



8321=8321 Therefore, the solutions we arrived work for the equation2 as well.

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