Peter invested a certain sum of money in a simple interest bond whose

Question

Peter invested a certain sum of money in a simple interest bond whose value grew to $300 at the end of 3 years and to $ 400 at the end of another 5 years. What was the rate of interest in which he invested his sum?

A. 6.25%
B. 6 2/3%
C. 8 1/3%
D. 12%
E. 12.5%

A. 6.25%
B. 6 2/3%
C. 8 1/3%
D. 12%
E. 12.5%

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MorallyDeficient · Answer

Simple Interest = Principle * Rate * Years
Let x be the interest rate and S - the invested sum.
3xS + S = 300
8xS + S = 400
=> 5xS = 100
 xS = 20
=> S = 240
 => x = 1/12 = 8\frac{1}{3} %

Option C

HoudaSR · Answer

How did you get S= 240?

HoudaSR · Answer

Amount at the end of 3 years= S+(S x R x 3)= 300 --> S(1+3R)= 300
Amount at the end of 8 years= S + (S x R x 8)= 400--> S(1+8R)= 400

S= 300/(1+3R)= 400/(1+8R)
Cross myltiply: 2400R-1200R= 400-300
1200R=100
R=100/1200= 1/12=0,0833= 8,33%= 8 1/3%.

Correct answer C

Bunuel  is there a quicker method to solve for this?

SaquibHGMATWhiz · Answer

Solution:

We know that for simple interests  Interest=\frac{P	imes R	imes T}{100}  and  Amount=P+\frac{P	imes R	imes T}{100}=P(1+\frac{R	imes T}{100})

According to the question: 
  At the end of 3 years,  300=P(1+\frac{3R}{100})......(i)  
 At the end of NEXT 5 years,  400=P(1+\frac{8R}{100})......(ii)    
 Dividing the 2 equations, we get:
  \frac{3}{4}=\frac{1+\frac{3R}{100}}{1+\frac{8R}{100}} 
 ⇒\frac{3}{4}=\frac{100+3R}{100+8R} 
 ⇒300+24R=400+12R 
 ⇒12R=100 
 ⇒R=8.33\%

Hence the right answer is    Option C

HoudaSR · Answer

Here is a faster method to solve:

The amount of interest earned in 5 years= 400-300=100
Hence, the amount of interest earned in one year=100/5=20

The initial amount invested= 300 - (20x3)= 300-60= 240

Reminder: SI= PxRxT
In this case: 20=240xRx1
hence, R=20/240=1/12=8,33%=8 1/3

Correct answer is C

Kinshook · Answer

Given: Peter invested a certain sum of money in a simple interest bond whose value grew to $300 at the end of 3 years and to $ 400 at the end of another 5 years.

Asked: What was the rate of interest in which he invested his sum?

Let the Principal be p and rate of interest be r% per annum.

Peter invested a certain sum of money in a simple interest bond whose value grew to $300 at the end of 3 years and to $ 400 at the end of another 5 years. 
P + P*r*3/100 = $300
P + P*r*8/100 = $400

P*r*5/100 = $100
P = $2000/r

P(1+3r/100) = 300
2000/r (1 + 3r/100) = 300
2000/r + 60 = 300
2000/r = 240
r = 2000/240 = 100/12 %= 8 1/3 %

IMO C

rahul5657 · Answer

Annual simple interest = S
initial principle amount = P
At the end of 3 yrs => (1+3*S)*P = 300 ......(1)
at the end of another 5 yrs that means at the end of 8 yrs from the start => (1+8*S)*P = 400 .....(2)

Now (1)/(2) :
4+12*S = 3 + 24*S
S = 1/12
connect into % = S*100 %= 100/12 % = 8.33 %
Correct Option: C

PSKhore · Answer

@Bunuel
Why can't we use the principle as 300 and the interest as 100 and use the equation
(300*5*r)/100 = 100
This gives the rate of interest as 6.66%

Bunuel · Answer

Although the amount becomes 300 after 3 years, the next 5 years of interest are still calculated on the initial investment (240), not on 300.

Peter invested a certain sum of money in a simple interest bond whose

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