Mrs. B. invested $30,000; part at 5%, and part at 8%. The total interest on the investment was $2,100. How much did she invest at each rate?
A. x = $1,000, y = $29,000.
B. x = $5,000, y = $25,000.
C. x = $10,000, y = $20,000.
D. x = $15,000, y = $15,000.
E. x = $20,000, y = $10,000.
The total interest on investment: $2,100 represents 7% of the total investment ($30,000).
[2,100/30,000 = 7/100 or 7%]
This means that part of the investment at 5% plus part at 8% add up to 7% average.
Since 7% is closer to 8% than 5%, this means that the 5% part weighs less than the 8% part.
Looking at the answer choices, I can eliminate D and E. I can also elmininate A (it seems too extreme).
A. x = $1,000, y = $29,000.B. x = $5,000, y = $25,000.
C. x = $10,000, y = $20,000.
D. x = $15,000, y = $15,000.
E. x = $20,000, y = $10,000.The next step is to figure out the exact weight of the two investments.
Putting it together; 5% part x+ 8% part y = 7% (part X+ Part Y)
I need to find the ratio. I can make more calculations by looking at the difference between the two parts and the percentage of the final investment.
5% is 2% lower than the final investment, and 8% is 1% higher than the final investment.
We can say that Part x has a -2 differential (5-7=-2) and Part y has a +1 (8-7=1) differential.
We need to make these differentials cancel out therefore we should multiply both differntials by the different numbers so that the positive will cancel out the negative.
X(-2) + Y(+1) = 0
if X = 1 and Y = 2, then 1(-2) + 2 (1) = 0
Therefore the ratio is 1:2
This means there is ONE part at 5% for TWO parts at 8% to produce an average of 7%.
The only answer that matches this ratio is C
C. x = $10,000, y = $20,000
I hope this helps.
For more information: look for lessons on Data points and weighted averages.
I'd also recommend
MGMAT's Word problem book p55 to p58. Took me a litte while to get my head around it.
But once you do, you gain a couple of seconds.