Question

Ms. Jamison plans to invest $3,800 in two accounts for one year. The first account earns a yearly interest of 3.5%. The second account earns a yearly interest of 6%. How much money does Ms. Jamison need to invest in the second account in order to earn $188 in interest from both accounts? Show all of your work.

Answer 1

Answer:

The Amount investment in second account is $2200

Step-by-step explanation:

Given as :

Total money invested in two accounts =  $3800

The time period of investment = t = 1 year

Let The amount investment in first account = $p

Let The amount investment in second account = $3800 - $p

The rate of interest for first account = $r_1$ = 3.5%

The rate of interest for second account = $r_2$ = 6%

The total interest earn from both account = s.i = $188

Let The interest earn from first account = $s.i_1$

Let The interest earn from second account = $s.i_2$

Now, From simple interest method

Simple interest = $\dfrac{\textrm principal\times \textrm rate\times \texrm time}{100}$

For First account

$s.i_1$ = $\frac{p\times r_1\times t}{100}$

i.e $s.i_1$ = $\frac{p\times 3.5\times 1}{100}$

Again

For second account

$s.i_2$ = $\frac{(3800-p)\times r_2\times t}{100}$

$s.i_2$ = $\frac{(3800-p)\times 6\times 1}{100}$

Since  total interest earn from both account = s.i = $188

i.e s.i = $s.i_2$ + $s.i_2$

Or, s.i = $\frac{p\times 3.5\times 1}{100}$ + $\frac{(3800-p)\times 6\times 1}{100}$

Or, $188 = $\frac{p\times 3.5\times 1}{100}$ + $\frac{(3800-p)\times 6\times 1}{100}$

Or, 188 × 100 = 3.5 p + 22800 - 6 p

Or, 6 p - 3.5 p = 22800 - 18800

Or, 2.5 p = 4000

∴ p = $\frac{4000}{2.5}$

i.e p = $1600

So, The amount investment in first account = p = $1600

So, The amount investment in second account = $3800 - $p

i.e The amount investment in second account = $3800 - $1600

Or, The amount investment in second account = $2200

Hence, The Amount investment in second account is $2200  Answer