Question 1189020: 1. Train A has a speed 20 miles per hour greater than that of train B. If train A travels 300 miles in the same times train B travels 220 miles, what are the speeds of the two trains?
2. A woman has a total of $13,000 to invest. She invests part of the money in an account that pays 11% per year and the rest in an account that pays 12% per year. If the interest earned in the first year is $1490 how much did she invest in each account?
Found 3 solutions by ikleyn, greenestamps, josgarithmetic: Answer by ikleyn(52773) (Show Source):
You can put this solution on YOUR website! .
A woman has a total of $13,000 to invest. She invests part of the money in an account that pays 11% per year
and the rest in an account that pays 12% per year. If the interest earned in the first year is $1490
how much did she invest in each account?
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interest + interest = total interest
0.12*x + 0.11*(13000-x) = 1490 dollars
0.12x + 1430 - 0.11x = 1490
0.01x = 1490 - 1430 = 60 ====> x = = 6000.
Answer. $6000 was invested at 12%. The rest 13000-6000 = 7000 dollars were invested at 11%.
Check. 0.12*6000 + 0.11*7000 = 1490 dollars, total annual interest. ! Correct !
Solved.
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Answer by greenestamps(13198) (Show Source):
You can put this solution on YOUR website! 1. The difference in speeds of the two trains is 20mph; the difference in distances over the same amount of time is 80 miles. So the time traveled is 80/20 = 4 hours.
Train A travels 300 miles in 4 hours, a speed of 75mph; train B travels 220 miles in 4 hours, a speed of 55mph.
ANSWERS: 75 and 55mph
2. Here is an informal method for solving "mixture" problems like this that is nearly always faster and easier than the standard formal algebraic method shown by the other tutor.
(a) All $13,000 invested at 11% would yield $1430 interest; all invested at 12% would yield $1560 interest.
(b) View the three interest amounts $1430, $1490, and $1560 on a number line and observe/calculate that $1490 is $60/$130 = 6/13 of the way from $1430 to $1560.
(c) That means 6/13 of the total was invested at the higher rate.
6/13 of $13,000 is $6000, so
ANSWERS: $6000 invested at 12%, the other $7000 at 11%
CHECK: .12(6000)+.11(7000) =720+770=1490
Answer by josgarithmetic(39615) (Show Source):
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