Simple interest is calculated on the interest every year. So, if the interest at the end of 2yrs is $600 then interest at the end of 1year is 300

The formula for s.i for 1 year is (dxkx1)/100 and this = 300 (1)

Now need new D such that (Dxkx1)/100 = 800 (again, si for 3yrs = 2400, therefore for 1yr = 800) (2)

From (1) we get k = 30,000/D

Substitute this in (2)

bbuisson wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2 yr period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 yr period?

a. 2d/3
b. 3d/4
c. 4d/3
d. 3d/2
e. 8d/3

Answer E

please please help...i am lost

I'd pick numbers on this question.

We know that we get $600 simple interest over 2 years, which means $300 interest per year.

So, let's pick:

principal (d) = 1000
interest rate (k) = 30%

Now, let's look at the rest of the question. We want to determine what amount of principal at 30% simple interest will give us $2400 interest over a 3 year period, or to convert to an annual basis, $800 per year.

So, 30%(principal) = 800

(3/10)p = 800

p = 800(10/3) = 8000/3

(note: no need to reduce further, since the answers are all expressed as fractions)

Now, let's plug our d value (1000) into the answers until we get a match.

We can quickly narrow it down to (a), (c) and (e) based on the denominator 3.

(a) 2d/3 = 2000/3
(c) 4d/3 = 4000/3
(e) 8d/3 = 8000/3 - BINGO! Choose (e).

we have a simple annual interest yields $600 interest over a 2 yr period ( not compounded), so we can break it down at $300 for the first year and $300 for the second year.

also, we can give name to the animals; if is $300 in each year, so d can be $3,000 and k= 10%, so 3,000 x 0.10= 300.

The question is " what dollar amount invested at the same rate will yield $ 2,400 interest over a 3 year".

Using 8d/3, we're going to have 8 x 3,000/3 = $ 8,000; now 8,000 at the same rate, $8,000 x 10%= $8,000 x 0.10 = $800 per year, in 3 years it will be 2,400.

Good question

Picking numbers is a good way like Stuart said.

If u dont feel comfortable with picking numbers use the formula for simple interest

Put this on your flashcard(comes in handy even in DataSufficiency problem wiht simple interest):

SI = P*N*R /100
where p-principal n- number of years r-rate
SI- SIMPLE INTEREST

GIVEN:

600 = d * 2 * k/100
kd = 600*50 (1)

Lets say x is the principal we need to find in terms of d

2400 = x * 3 *k/100
x = 2400 *100/3k
= 2400 * 100 * d / 3 * 600 *50 (from (1) k = 600*50/d)

8d/3

Could somebody explain this question again in simpler terms? I'm familiar with simple and compound interest, but I can not wrap my brain around any of these explanations as I find them hard to follow. I'm not a fan of these "In Terms of: questions as they get me all the time, even when i plug in numbers.

vitapup wrote:Could somebody explain this question again in simpler terms? I'm familiar with simple and compound interest, but I can not wrap my brain around any of these explanations as I find them hard to follow. I'm not a fan of these "In Terms of: questions as they get me all the time, even when i plug in numbers.

I would say check the explanation given by cramya again.. it doesnt' get simpler than that..

I would put it this way.

If interst for 2 years is 600 then interest for 3 years is 900 for d dollars.

now if I want interest of 2400 I need to increase d i.e. the principal, much it is d *2400/900. d8/3.

lets try another short cut

600 to become 2400

so you need 4 times the Initial Capital(As rate is unchanged)

Now Duration wise you get 1.5 times(from 2 to 3 yrs)

hence you need 4/1.5 times the Initial Capital

thats 40/15 = 8/3

bbuisson wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2 yr period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 yr period?

a. 2d/3
b. 3d/4
c. 4d/3
d. 3d/2
e. 8d/3

Answer E

please please help...i am lost

This is a very simple question

One thing to remember , Its Simple Interest = Simple problem

We know that the interest is proportional to the amount invested if the rate is constant (SI)

So that solves the problem for 2 years we get 600 , then per year we must get 300 that is @ k% interest

So if we want to get 2400 for 3 years @ K% , we need to get 800 per year

But we know that if we invest $d we get 300 , hence we have to invest

800*d/300 or 8d/3

This is so easy because we are dealing with SI and the interest rate doesn't change for the 2 situations

So Interest is proportional to the Money invested


Mathematically we can write I = kM for some positive integer k

Then I(1) = KM(1) and I(2) = KM(2)

Then I(1)/I(2) = M(1)/M(2) or I(1)M(2) = I(2)M(1)

But all the calculation is unnecessary if one understand the main point expressed above


I don't know why we have to substitute numbers , its simple test of a concept ...............if you understand the concept the answer is so easy to pick

vineetbatra wrote:I would put it this way.

If interst for 2 years is 600 then interest for 3 years is 900 for d dollars.

now if I want interest of 2400 I need to increase d i.e. the principal, much it is d *2400/900. d8/3.

Thanks for the simple and meaningful explanation.

Simple Interest= Principal * Rate * Time

2*d*r=600

r=300/d

3*300/d*p=2,400

p=2400d/900=8d/3 E

bbuisson wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2 yr period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3 yr period?

a. 2d/3
b. 3d/4
c. 4d/3
d. 3d/2
e. 8d/3

Answer E

please please help...i am lost

600=d1(k/100)2
2400=d2(k/100)3

problem says same rate so solve for k/100 at d1 investment then plug into simple interest equation to get d2.

300/d1=k/100
therefore
2400=d2(300/d1)3
d2=8d1/3
answer 3

I went for a simpler one

I = P*r* t

Therfore
1) 600 = d*k*2

and for assuming x times of d
2) 2400 = (xd)*k*3


Divide one with other
4 = x3/2

x=8/3

and hence 8/3d