Grade 10 math problem - linear equations?

Question

I have an exam next week and have no idea how to answer questions like this, though I've been told that there is one on the exam ... it says:

Richard has $10,000. He invested part of it in a term deposit paying 4% per annum, and the remainder in bonds paying 5% per annum. If the total interest after one year is $440, how much did he invest at each rate?

I know the answer is $6,000 at 4% and $4,000 at 5%, but could someone explain to me how to figure out how to do these questions? I would really appreciate it, I have no idea how to label them so I can solve it ... thanks! :)

Answer 1

let us say that the amount he invested in the term deposit(4%) = x

therefore the amount deposited in 5% = 10000 -x

simple interest =
Principal * num of years * rate / 100

therefore interest from term deposit = x * 1 * 4/100
and interest from bonds = (10000 -x) * 1* 5/100

we know that the total = 440

therefore 4x/100 + (50000 - 5x)/100 = 440

therefore 4x + 50000 - 5x = 44000
therefore x = 6000

so, now he invested x i.e. 6000 in the 4% term deposit
and invested (10000 - x) i.e. 4000 in the 5% term deposit

Answer 2

It is clear that:

If all 10000 is invested with 4% interest rate, then the the total interest is 10000 * 0.04 = 400.

If all 10000 is invested with 5% interest rate, then the the total interest is 10000 * 0.05 = 500.

Now let “a” is a number between “0” and “1” representing that part of investment, which was invested with 4% interest rate.

If “a” part ot the 10000 was invested with 4% interest rate, then the total interest from this part of investment is a*10000 * 0.04.

The remainig part of the capital is (1-a)*10000.

This part of capital was invested with 5% interest rate, and the total interest from this second part of investment is (1-a)*10000 * 0.05.

Then the total interest from both parts of investment is

a*10000 * 0.04 + (1-a)*10000 * 0.05.

But we know, that the total interest is 440.

So now we got the equation to be solved:

a*10000 * 0.04 + (1-a)*10000 * 0.05 = 440.

Let us solve it:

400 a + 500 – 500 a = 440

- 100 a = - 60

The solution is:

a = 0.6

and

1-a = 0.4.

So the answer to the question is

a * 10000 = 0.6 * 10000 = 6000 was invested with 4% interest rate

and

(1-a) * 10000 = 0.4 * 10000 = 4000 was invested with 5% interest rate.

Answer 3

Let Richard invest $x in term deposit @4% and the remaining $10.000-x in bonds paying 5%
annual interest on $x @4%
=x*0.04*1
=0.04x
Annual interest of $10000-x @5%
=(10000-x)*0.05*1
=500-0.05x
Therefore by the problem,
0.04x+500-0.05x=440
or, -0.01x=440-500= -60
or,x=60/0.01=6000
so he invested $6000 in term deposit and the remaining $4000 in bonds
Now,as you can calculate the interest,here is an easier way to get the answer
if he invests the whole amount @4% he gets an annual interest of $400
But actually he gets $40 more.This is because he has invested a part of the amount @5%
So for every $1 more ,he has to invest $100 in the bonds.
Therefore for $40 more,he has to invest $40*100 or $4000 in the bonds.
got it?

Answer 4

Let x be the amount he invested at 5%. Then he invested 10000 - x at 4%. Since the total interest is 440, you have:

x*0.05 + (10000 - x)*0.04 = 440.
(this is because after investing $x at 5% per year, you make $x * 0.05 in the first year)

Solve this for x:

x*0.05 - x*0.04 = 440 - 10000*0.04
x*(0.05 - 0.04) = 40
x = 4000.

Answer 5

set x = the amount invested at 4%, and y = the amount invested at 5%

x + y = 10000
y = 10000 - x

0.04x + 0.05y = 440
0.04x + 0.05(10000 - x) = 440
0.04x + 500 - 0.05x = 440
-0.01x = -60
x = -60/-0.01 = 60/0.01 = 6000

y = 10000 - 6000 = 4000

Answer 6

Let x dollars were invested in bonds with 3 % interest Let y dollars were invested in term deposits with 5 % interest 0.03x+0.05y = 760 --- (1) x+y=20,000 --- (2) Solve for x and y from (2) y=20,000-x plug this into (1) 0.03x + 0.05(20,000-x) = 760 0.03x-0.05x+1000 = 760 -0.02x = - 240 x=12,000 (at 3 %) y=8,000 (at 5 %)