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A sum of $12 million was divided between two investments. The smaller portion was invested at 6%, and the remainder at 4%. The earnings from the 4% investment exceeded the earnings from the 6% investment by $20,000. Find the amount invested at each rate.


Suppose,

smaller portion (@6%) = 100x
larger portion (@4%) = 100y

Putting all values in '000s, we get

100x + 100y = 12000
or, x + y = 120.....................(1)
-6x + 4y = 20................................(2)

Solving (1) & (2) equations, we get
100x = 4600 ('000s) = 4.6 mn
100y = 7400 ('000s) = 7.4 mn

A. Amount invested at 6%

$4.6 million


B. Amount invested at 4%

$7.4 million
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­Lets assume x as a small investment therefore remaining amount is 12 - x
The smaller portion was invested at 6%, And larger portion is at 4%
Return on 4% - Return on 6% = 20k = 0.02 millions
(12-x)*4/100 - 6x/100 = 0.02
(48-4x - 6x) / 100 = 0.02
48-10x = 2
46 = 10x
x=4.6 mil
y=12 - 4.6 = 7.4 mil.
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Sajjad1994
Data Insights (DI) Butler 2023-24 [Question #90, Date: Nov-03-2023] [Click here for Details]

A sum of $12 million was divided between two investments. The smaller portion was invested at 6%, and the remainder at 4%. The earnings from the 4% investment exceeded the earnings from the 6% investment by $20,000. Find the amount invested at each rate.

we calculate: 12,000,000 * 0.04 = 480,000 and 12,000,000 * 0.06 = 720,000

now take the equation: 480,000 * x = 720,000 * (1-x) + 20,000
solve for x: x = 7,400,000 which is the sum invested ate 4%.

The sum invested at 6% follows as 12,000,000 - 7,400,000 = 4,600,000
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