MATH 115 Quiz 3 Summer 2018

MATH 115 Quiz 3 Summer 2018
1. Solve the following: 21−x=42 x
Solution: 21−x=42 x
⟹ 21−x= ( 22 ) 2 x
⟹ 21−x=24 x
If base term is same then power will be equal.
Therefore, ⟹ 1−x=4 x
⟹ 1=5 x
⟹ x= 1
5
2. How long will it take an investment to double if it is invested at 7.8%, compounded
continuously?
Solution: Given r = 7.8 % = 7.8
100 = 0.078
if the interest rate is compounded continuously, then the general form is:
a=b ert
a → Amount after time ‘t’
b → Initial amount
r → Rate of interest compounded continuously
t → Time in years
Let, Initial amount = b
Amount after time ‘t’ = 2b (doubled)
Then equation looks like
⟹ 2 b=b e0.078 t
⟹ 2=e0.078 t
⟹ 0.078 t=ln 2

⟹ 0.078 t=0.693147
⟹ t= 0.693147
0.078
⟹ t=8.8865 Years
3. An initial investment of $1000 is appreciated for 2 years in an
account that earns 6% interest, compounded semiannually.
Find the amount of money in the account at the end of the
period.
Solution: Given
Initial investment P = $ 1000
Time ‘ t ‘ = 2 years
Rate of interest = r = 6 % = 0.06
Semiannually = N = 2
Let us suppose at the end of period, amount of money be A
Then A=P(1+ r
N )
t N
Where A=1000 (1+ 0.06
2 )2∗2
A=1000 ( 1+ 0.03 ) 4
A=1000 (1.03 )4
A=$ 1125.5
4. A company begins a radio advertising campaign in Chicago to
market a new soft drink. The percentage of the target market
that buys a soft drink is estimated by the function
Solution:
Given:
P ( t ) =100−100 e−0.01 t
Where P(t) is the percentage of the target market that buys soft drink.
“t” is the number of days of campaign.

Given,P ( t ) =75 %
⟹ 75=100−100 e−0.01 t
⟹−25=−100 e−0.01 t
⟹ 1=4 e−0.01 t
⟹ 1
4 =e−0.01t
Take natural log both sides, ⟹ ln 1
4 =−0.01t
⟹ ln 0.25=−0.01 t
⟹−1.38629=−0.01 t
t=138.63 days
5. A certain radioactive isotope decays at a rate of 0.15%
annually. Determine the half-life of this isotope, to the nearest
year.
Solution:
Radioactive isotope decay takes place exponentially
N=N 0 ekt (1)
Where N is the concentration of isotope left after time ‘t’
and No is the Initial concentration
k is the order of decay
Therefore,
Given , t = 1 year
Concentration of isotope left after 1year,
N=N 0−(0.15 % of N0 )
N=N 0− 0.15
100 N0
N=N 0 (1−0.0015)