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# Ngoc invests a total of \$22,000 in two accounts\

Ngoc invests a total of \$22,000 in two accounts paying 11% and 14% annual interest, respectively. How much was invested in each account if, after one year, the total interest was \$2,810.00.

how much in each account? hmm...let's say a in account 1 and b in account 2. we have 2 things we don't know, so it would be nice to have 2 equations to help us out.. 1) a + b = 22000 (we know the total invested is 22,000) 2) 0.11a + 0.14b = 2810 there! 2 equations and 2 unknowns. you can use any method u want. let's try substitution.. from (1) we get a = 22000-b we put this in (2) and get 0.11(22000-b) + .14b = 2810 2420-.11b+.14b = 2810, then we get .03b = 390 so b = 13000, and a = 22000-13000 = 9000

In this case you are given two accounts, lets call them x and y, which pay out 11% and 14% annual interest. Let x be the amount of money in the account paying 11% interest and y be the amount of money in the account paying 14% interest.

Now we know that x + y =22000 because 22000 is the total amount of money you can split between the two accounts. Now the interest on the amount invested is (11%) * x + (14%) * y which is equal to 2810.

We now have a system of equations with two variables and two unknowns. Pick a variable in one equation and solve the equation for that variable. Then plug that equation into the other equation. In this case we will pick the first equation and solve for x.

x + y = 22000

x = 22000 - y  # solved for x

(11%) * x + (14%) * y = 2810 # this is the second equation

(11%) * (22000 - y) + (14%) * y = 2810 # plug in the first equation for x

(.11)*(22000 - y) + (.14)*y = 2810 # changed percents to decimals to make the adding easier

Now solve for y:

2420 - .11y +.14y = 2810

.03y = 390

y = 390 / 0.03 = 13000

Now plug 13000 into y for the first equation to find x:

x + y = 22000

x + 13000 = 22000

x = 9000

So \$9000 was put into the account with 11% interest and \$13000 was put into the account with 14% interest.

Let's say x is the amount invested at 11%, and y is the amount invested at 14%. We know that the two together total 22,000, so:
x+y = 22,000
y = 22,000 - x

Now write an equation representing the information given in the problem, remembering that 11% = 0.11, and 14% = 0.14:
0.11x + 0.14y = 2810
0.11x + 0.14(22,000 - x) = 2810

Expand the equation:
0.11x + 3080 - 0.14x = 2810                 combine like terms
-0.03x + 3080 = 2810                           subtract 3080 from both sides
-0.03x = -270                                       divide both sides by -0.03
x = 9000

Solve for y:
y = 22,000 - x = 22,000 - 9000 = 13,000

So the amount invested at 11% is \$9000, and the amount invested at 14% is \$13,000

(Amount 1),   A1 = C1expk1t                                (Amount 2)A2 = C2expk2t

Interest 1,  I1  = C1(expk1t  -  1)                          Interest 2,  I2  =  C2(expk2t -1)

And C1 = 22000-C2                                             I1  =  2810 - I2

t = 1. and k1 = .11, k2 = .14

Two equations with two unknowns.  Substitute and after some huffing and puffing,

C2 = 7409.24

C1 = 14590.76