The simple interest on a certain sum of money for 3 yrs is $225 and th : Problem Solving (PS)

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The simple interest on a certain sum of money for 3 yrs is $225 and th

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Math Expert

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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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29 Jan 2019, 01:58

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54% (03:04) correct

46% (02:57) wrong

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The simple interest on a certain sum of money for 3 yrs is $225 and the compound interest on the same sum at the same rate for 2yrs is $153. What is the principal invested?

A. $1500

B. $1875

C. $2250

D. $2500

E. $2750

Spoiler: OA

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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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Updated on: 30 Jan 2019, 04:06

Bunuel wrote:

SI=p*r*t/100
225=p*r*3/100
r/100= 75/p

CI= P * ( 1+r/100) ^t

Compounded amount in 2 years = P(1+r100)2P(1+r100)2

interest earned = Compounded amount in 2 years - Invested amount = P(1+r100)^2 - P

153=P*(1+r/100)^2-P
153=P*(1+75/P)^2-P
solve for P
3P=75*75
p= 75*25
p= 1875
IMOB
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Originally posted by Archit3110 on 29 Jan 2019, 02:18.
Last edited by Archit3110 on 30 Jan 2019, 04:06, edited 2 times in total.

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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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29 Jan 2019, 02:54

As simple interest is the same for each year. SI per year = 225/3 = 75$.
SI = PTR/100
225 = PR*3/100
PR = 7500

Also, $CI_2-SI_2 = \frac{PR^2}{100^2}$ = 153-150 = 7500*R/10000

$CI_2$ and $SI_2$ are compound and simple interest for 2 years.
R = 4%
75 = P*4*1/100
P = 1875

B is the answer.
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Re: The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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29 Jan 2019, 03:26

Solution

Given:

To find:

• The principal amount invested
Approach and Working:
Let the principal amount invested = $P

o Implies, $\frac{r}{100}= \frac{75}{P}$

Hence, the correct answer is Option B

Answer: B

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Re: The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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29 Jan 2019, 03:32

EgmatQuantExpert wrote:

Solution

Given:

To find:

• The principal amount invested
Approach and Working:
Let the principal amount invested = $P

o Implies, $\frac{r}{100}= \frac{75}{P}$

P(1 + r/100)^2 – P = $153

Hence, the correct answer is Option B

Answer: B

EgmatQuantExpert
could you please explain the highlighted part , how did you get that? why is CI subtracted with P?
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Re: The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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29 Jan 2019, 03:37

Archit3110 wrote:

EgmatQuantExpert
could you please explain the highlighted part , how did you get that? why is CI subtracted with P?

Hi Archit3110,

Compounded amount in 2 years = $P(1 + \frac{r}{100})^2$

Now, the interest earned = Compounded amount in 2 years - Invested amount = $P(1 + \frac{r}{100})^2$ - P

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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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29 Jan 2019, 03:53

Bunuel wrote:

Simple interest for 3 yrs is $225

i.e. Simple Interest for 1 year = 225/3 = $75 (Simple interest on a fixed sum remains same every year as the principle sum remains static)

i.e. Compound interest for first year = 75 (For first compounding Simple interest and compound interests are same... they start differing from second compounding)
Compound interest for Second year = 153 - 75 = 78

i.e. Coumpound interest for second year is 78-75 = $3 more which is interest charged on interest earned in previous compounding

i.e. Interest earned on $75 = $3

i.e. interest rate = (3/75)*100 = 4%

i.e. interest at Principle P for 1 year @4% = $75

i.e. (4/100)*P = 75

i.e. P = $1875

Answer: Option B
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Re: The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink]

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07 Aug 2019, 13:11

Bunuel wrote:

simple interest: principal • rate • years - principal = interest
compound interest: principal • (1+rate)^years - principal = interest

given: p(1+r/100)(3)-p=225 and p(1+r/100)^2-p=153;
isolate (r/100): p(1+r/100)(3)-p=225 is the same as p(r/100)(3)=225,…(r/100)=(75/p)
substitue (r/100): p(1+r/100)^2-p=153,… p(1+(75/p))^2-p=153,…p=1875

Answer (B).