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Question

Walter earns a year-end bonus of $5000 and puts it in 3 one-year investments that pay $243 in simple interest.

Part is invested at 3%, part at 4% and part at 6%.

There is $1500 more invested at 6% than at 3%.

Find the amount invested at each rate.


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Solution

Explanation:

Step-1: Framing two equations:

Walter earns a year-end bonus of $5000 .

Let x,y,z be the amount invested at rate 3%, 4%, 6% respectively.

x+y+z=5000.....(1)

The total interest he gets is $243.

Formula for calculating simple interest is,

SI=PRT100

Therefore,

3x100+4y100+6z100=2433x+4y+6z=24300.....(2)

The amount invested at 6% is $1500 more than the amount invested at rate 3%.

z=x+1500

Substituting this in equation 1and 2.

x+y+x+1500=50002x+y=3500.......(3)

and

3x+4y+6(x+1500)=243003x+4y+6x+9000=243009x+4y=15300....(4)

Step-2: Finding the value of x

Multiply equation (3) with 4.

42x+y=35008x+4y=4(3500)8x+4y=14000.....(5)

Subtracting equation 5 from equation 4

x=15300-14000x=1300

Step-3: Calculating y,z

Substitute the value of x in equation (3)

2(1300)+y=35002600+y=3500y=3500-2600y=900

Substituting the value of x in equation z=x+1500.

z=(1300)+1500z=2800

Therefore, Walter invests $1300, $900, $2800 in one-year investment with rates 3%, 4%, 6% respectively.


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