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

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Bunuel
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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink] New post   29 Jan 2019, 01:58
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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
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B
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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink] New post   29 Jan 2019, 03:53
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Bunuel wrote:
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


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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The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink] New post   Updated on: 30 Jan 2019, 04:06
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Bunuel wrote:
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


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

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 $225 and th [#permalink] New post   29 Jan 2019, 02:54
As simple interest is the same for each year. SI per year = 225/3 = 75$.
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] New post   29 Jan 2019, 03:26
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Solution


Given:
    • The simple interest on a certain sum of money for 3 years = $225
    • The compound interest on the same sum at the same rate for 2 years = $153

To find:
    • The principal amount invested

Approach and Working:
Let the principal amount invested = $P
    • Simple interest earned for 3 years =\(P * 3 * \frac{r}{100} = $225\)
      o Implies, \(\frac{r}{100}= \frac{75}{P}\)

    • Compound interest earned for 2 years = P(1 + r/100)^2 – P = $153
      o Implies, \(P(1 + \frac{75}{P})^2 – P = $153\)
      o \((P + 75)^2 – P^2 = 153P\)
      o \(150P + 75^2 = 153P\)
      o \(3P = 75^2\)
    • Therefore, P = 25 * 75 = 1875

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] New post   29 Jan 2019, 03:32
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EgmatQuantExpert wrote:

Solution


Given:
    • The simple interest on a certain sum of money for 3 years = $225
    • The compound interest on the same sum at the same rate for 2 years = $153

To find:
    • The principal amount invested

Approach and Working:
Let the principal amount invested = $P
    • Simple interest earned for 3 years =\(P * 3 * \frac{r}{100} = $225\)
      o Implies, \(\frac{r}{100}= \frac{75}{P}\)

    • Compound interest earned for 2 years = P(1 + r/100)^2 – P = $153
      o Implies, \(P(1 + \frac{75}{P})^2 – P = $153\)
      o \((P + 75)^2 – P^2 = 153P\)
      o \(150P + 75^2 = 153P\)
      o \(3P = 75^2\)
    • Therefore, P = 25 * 75 = 1875

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] New post   29 Jan 2019, 03:37
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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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Re: The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink] New post   07 Aug 2019, 13:11
Bunuel wrote:
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


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).
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Re: The simple interest on a certain sum of money for 3 yrs is $225 and th [#permalink] New post   16 Oct 2020, 05:53
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