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"Investment" Word Problems: Examples (page 2 of 2) If you set up your investment problems so everything is labeled and well-organized, they should all work out fairly easily. Just take your time and do things in an orderly fashion. I've done the set-up (but not the complete solutions) for a few more examples:
First fill in the P, r, and t columns with the given values. Then multiply across the rows to the left in order to fill in the I column. Then add down the I column to get the equation 150 + 0.09 x = (3,000 + x)(0.075). Solve for the value of x.
In this problem, you don't actually need the "total" row at all. First fill in the P, r, and t columns, and multiply to the left to fill in the I column. From the interest column, you get the equation 0.09x = ($6,000 – x)(0.06), because the yields are required to be equal. Solve for the value of x. Back-solve to find the value invested in the 6% account. (This problem used that "how much is left" construction, mentioned earlier.)
Let "x" be the amount invested in the mutual fund.
In this problem, you don't actually need the "total" for the "rate" or "time" columns. First fill in the P, r, and t columns, multiplying to the left to fill in the I column. Then add down the I column to get the equation 0.09x + (x + 2,500)(0.05) = 475. Solve for the value of x. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
For this problem, let "x" stand for the total amount invested.
Once 30% and 25% was accounted for in the 6% and 8% accounts, then there is 100% – 30% – 25% = 45% left for the third account. Use this information to fill in the "principal" column. Then fill out the "rate" and "time" columns, and multiply to the left to fill in the "interest" column. From the interest column, you get the equation 0.018x + 0.02x + 0.03375x = 25,875. Solve for the value of x. << Previous Top | 1 | 2 | Return to Index
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Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
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