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vivado [14]
4 years ago
9

Thomas invested $8,500 for one year. Part of the money was invested at6% and the rest at 9%. The total interest earned was $667.

50. How much did Thomas invest at the 6% rate?
Mathematics
1 answer:
oee [108]4 years ago
4 0
Okay so say that x=amount invested with 6% and y=amount invested with 9%
x+y=8,500 so > x=8500-y
6%=0.06        <span>9%=0.09</span>
0.06x +0.09y=667.5 (Substitute in x=8500-y so only numbers and y)
0.06(8500-y)+0.09y=667.5 (expand brackets)
510-0.06y+0.09y=667.5 (-510)
0.03y=117.5 (/0.03)
$3916.67=Y  ->9%
X=8500-Y 
x=$4583.33 ->6%
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Answer:

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Step-by-step explanation:

Let Mrs. grub bought total number of shirts = x

Let Mrs. grub bought total number of hats = y

As given,

Price of 1 shirt = $4

Price of 1 hat = $12

⇒ Price of x shirts = $ 4x

   Price of y hats = $ 12y

Given that,

Mrs. Grubb bought total piece of clothes = 9

⇒ x + y = 9     .......(1)

Also given,

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⇒ 4x + 12 y = 100       ........(2)

Divide equation (2) by 4 we get

x + 3y = 25     .......(3)

Now,

Subtract equation (1) from equation (3) , we get

x + 3y  - ( x + y ) = 25 - 9

⇒ x + 3y - x - y = 16

⇒2y = 16

⇒y = \frac{16}{2} = 8

⇒ y = 8

Put value of y in equation (3), we get

x + 3(8) = 25

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Mrs. grub bought total number of shirts = x = 1

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