Respuesta :
Answer:
Amount invested at 10% = $47,400
Amount invested at 6% = $94,800
Step-by-step explanation:
Let
Amount invested at 10% = X
Amount invested at 6% = 2X
According to given condition
Interest received at 10% + Interest received at 6% = 10428
X10% + 2X6% = 10428
0.1X + 0.12X = 10428
0.22X = 10428
X = 10428/0.22
X = 47400
So
Amount invested at 10% = X = 47,400
Amount invested at 6% = 2X = 2*47400 = 94,800
Check
47400*10% + 94800*6% = 10428
4740 + 5688 = 10428
10428 = 10428
Answer:
The amount invested at 6% is $94,800
&
The amount invested at 10% is $47,400.
Step-by-step explanation:
Total yearly interest for the two accounts is: $10,428
Let x be the amount invested at 6%
& y be the amount invested at 10%
From the question we can get 2 equations as;
x = 2y --------------------------Equation 1
0.06x + 0.10y = 10428 ----------Equation 2
Substitute for x in Equation 2 we get;
0.06 (2y) + 0.10y = 10428
0.12y + 0.10y = 10428
0.22y = 10428
Divide the above equation by 0.22, we get;
y = [tex]\frac{10428}{0.22}[/tex]
y = $47,400
Let us substitute the value of y in Equation 1 we get;
x = 2(47400)
x = $94,800
Now to check our answer let us put in the simple interest formula. If we get the sum of the two interests equal to 10428 then our answers are correct:
0.06 x 94800 + 0.10 x 47400
= 5688 + 4740
= $10,428
Hence the amount invested at 6% is $94,800 and the amount invested at 10% is $47,400.