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Please help

by dhruvbadruka » Wed Mar 26, 2014 1:02 am
A man invested 60% of his saving to a municipal bond paid annually compound interest for 2 years and received $ 330 as interest. He deposited the rest in a bank at 6% simple interest and received $ 220 as interest. What is the bond's interest rate?

3%
4%
5%
sqrt(1.1) - 1 %
sqrt(1.2)- 1 %
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by [email protected] » Wed Mar 26, 2014 10:47 am
Hi dhruvbadruka,

Did you transcribe this question correctly? I only ask because the answer choices seem to point to an answer if the simple interest was 5% instead of 6%.

Here's how to solve:

We don't know how much money the man has, so let's call it X.

Based on the first sentence (and the formula for compound interest) we have:

.6X(1+r)^2 = 330

Based on the second sentence (and the formula for simple interest) we have:

.4X[1 +(.05)2] = 220

Here we can solve for X:

.4X(1.1) = 220
.44X = 220
X = 500

Plugging X = 500 back into the first equation gets us...

.6(500)(1.+r)^2 = 330

300(1+r)^2 = 330

(1+r)^2 = 1.1

1 + r = sqrt(1.1)

r = sqrt(1.1) - 1

Final Answer: D

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by dhruvbadruka » Wed Mar 26, 2014 11:19 pm
Hi rich,

I did copy the question correctly.

Here's what i did, please tell me if there's a faster alternative:

Simple interest= PNR/100

P=.4x
N=2
R=6

220=(.4x*2*6)/100

therefore x=(220*100)/(12*.4)
since we need .6x for the next calculation I multiplied the above with .6
hence .6x=2750


Compound interest = P(1+r/n)^nt - P

since P= .6x=2750
r= unknown
n=1
t=2

I arrived at

330= 2750(1+r)^2 - 2750

(330+2750)/2750 =(1+r)^2

(1+r)^2= 1.1( approximated from 1.12)

thus r= sqrt(1.1)-1

But the answer says option e, so is it possible that the source is wrong

Dhruv
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