SOLUTIONS TO ACTIVITY #1
II. Solve the following problems. Show clean and complete details of your work for
each number.
1.
The price in pesos for a certain product is p( x ) = 900 – 20x – x2 when x units is
demanded. Also p( x ) = x2 + 10x is the price when the supply is x units. Compute the
equilibrium price.
Solution:
y
p( x ) = 900 – 20x – x2 -------------- ( 1 )
p( x ) = x2 + 10x --------------------- ( 2 )
100
O
x
5 10 15 20
Substitute ( 1 ) to ( 2 ):
x2 + 10x = 900 – 20x – x2
2x + 30x – 900 = 0
2
x2 + 15x – 450 = 0
( x – 15 )( x + 30 ) = 0
x = 15
and
x = – 30
Therefore x = 15.
Substitute to ( 2 ):
p( x ) = x2 + 10x
p( 15 ) = 152 + 10( 15 )
p( 15 ) = P 375
,
2.
You borrow from your friend an amount of P 80,000 with simple interest of 8%. How
much will you pay to your friend after 8 months?
Solution:
F = P( 1 + rt )

 8 
F = 80,000  1  0.08 
 
 12  

F = P 84,266.67
,
1
3.
Compute the interest of P 15,000 invested at 12% simple interest from February 16 to
October 24, 2008.
Solution:
The period of the loan is tabulated below.
February
March
April
May
June
July
August
September
October
16 - 28
1 – 31
1 – 30
1 – 31
1 – 30
1 – 31
1 – 31
1 - 30
1 - 24
13 days ( excluding Feb. 16 )
31 days
30 days
31 days
30 days
31 days
31 days
30 days
24 days ( including Oct. 24 )
Total number of days ------------- 251 days
I = Prt
 251 
I = 15,000( 0.12 ) 

 366 
I =P 1,234.43
,
4.
Mr. Ybanez borrowed money from a loan firm at a simple interest of 10%. At the end
of 3 years he will pay P 82,354 which consist of the principal plus the total interest.
How much is the interest?
Solution:
P=
F
1  rt
P=
82,354
1  0.10  3 
P = P 63,349.23
I=F–P
I = 82,354 - 63,249.23
I = P 19,004.77
,
2
5.
It is a common practice in lending companies that the interest is deducted at the time
money is borrowed. Mr. Anito applied a loan amount of P 50,000 with an interest rate
of 16% per year. During the time he received the money, the 16% interest is being
deducted and he received only P 42,000. At the end of one year, Mr. Anito has to
pay P 50,000. What is the actual rate of interest?
Solution:
F = P( 1 + rt )
50,000 = 42,000( 1 + r( 1 ))
r = 0.1905
r = 19.05%
6.
In buying a cellphone, the buyer was offered the options of paying P 17,500 cash or
P 22,200 at the end of 152 days. At what rate is the buyer paying simple interest if he
agrees to pay at the end of 152 days?
Solution:
P 22,200
,
P 17,500
,
152 days
0
152 days
The amount of P17,500 will increase to an amount of P 22,200 for a period of
152 days. Thus we have
F = P( 1 + rt )

 152  
22,200 = 17,500  1  r 
 
 360  

r = 0.6361
r = 63.61 %
7.
A bill for a cellphone specifies the cost as P 25,400 due at the end of 100 days but
offers a 4% discount for cash in 24 days. What is the highest simple interest rate at
which the buyer can afford to borrow money in order to take advantage of the discount?
Solution:
P25,400
,
P24,384
,
100 days
24 days
76 days
3
Amount paid in 30 days = 25,400 – 25,400( 0.04 ) = P 24,384.
,
F = P( 1 + rt )

 76  
25,400 = 24,384  1  r 
 
 360  

r = 0.1974
r = 19.74 %
8.
By the condition of a will, the sum of P 200,000 is left to a child to be held in a trust
fund by her father until it amounts to P 1,000,000. When will the child received the
money if the fund is invested at 12% compounded quarterly?
Solution:
r 

F  P  1  
n 

nt
0.12

1,000,000  200,000  1 
4

5 = ( 1.03 )4t



4t
ln5 = ln( 1.03 )4t
ln5 = 4tln( 1.03 )
t = 13.61 years
9.
A certain amount is invested 5 years and 6 months ago at an interest rate of 8%
compounded semi – annually. The accrued amount now is P 347,054. How much
was the amount invested?
Solution:
r 

P  F  1  
n 

n t
0.08

P  347,054  1 
2

P = P 225,440
,
4



2  5.5 
10. If the principal is P 124,860 and the future value is P 182,568, what is the value of the
single payment present worth factor?
Solution:
P = F( 1 + i )– N
124,860 = 182,568( 1 + i )– N
( 1 + i )– N = 0.6839 ( single payment present worth factor )
11. You have used your credit card to purchase automobile tires for P 28,000. Unable to
make payments for six months, you then write a letter of apology and enclose a check
to pay your bill in full. The credit card’s nominal interest rate is 16% compounded
monthly. For what amount should you write the check?
Solution:
r 

F  P  1  
n 

nt
0.16

F  28,000  1 
12




6
F = P 30,316
,
12. Compute the interest rate if the single payment compound amount factor is 1.601 in 5
years.
Solution:
( 1 + i ) 5 = 1.601
1 + i = 1.0987
i = 0.0987
i = 9.87%
13. At what rate compounded continuously will money triple its amount in five years?
Solution:
F = Pern
3P = Per( 5 )
3 = e5r
ln3 = lne5r
r = 0.2197
r = 21.97%
5
14. Find the nominal rate compounded continuously is equivalent to a nominal rate of 6%
compounded semiannually?
Solution:
n
r 

er – 1 =  1  1   1
n 

2
0.06 

er – 1 =  1 
 1
2 

er = 1.0609
lner = ln1.0609
r = 0.0591
r = 5.91%
15. Compute the effective rate of interest of 15% compounded monthly.
Solution:
n
r 

EIR   1    1
n 

0.15

EIR   1 
12




12
1
EIR = 0.1675
EIR = 16.75%
16. A man invested P100,000 at an interest rate of 10% compounded annually. What will
,
be the final amount
of his investment, in terms of today’s peso, after 10 years
assuming inflation at a rate of 8% remains the same within the 10 – year period?
Solution:
 1 i 
FC = PC 

 1 f 
N
10
 1  0.10 
FC = 100,000 

 1  0.08 
FC = P 120,140
,
17. The uninflated present worth of P3,000 in two years is P 2,308. What is the rate of
, 10%?
,
inflation if the real rate of interest is
Solution:
F = P( 1 + i ) N
6
3,000 = 2,308( 1 + i ) 2
1.3 = ( 1 + i ) 2
1.3 = ( 1 + i ) 2
i = 0.1402
i = 14.02%
ir =
0.10 =
if
1 f
0.1402  f
1 f
0.10 + 0.10f = 0.1402 – f
1.10f = 0.0402
f = 0.0365
f = 3.65%
18. If the inflation rate is 8% and cost of money is 12%, what interest rate will take care of
the inflation and the cost of money?
Solution:
ir =
if
1 f
i  0.08
1  0.08
0.1296 = i – 0.08
0.12 =
i = 0.2096
i = 20.96%
19. An amount of P 300,000 is invested for 3 years at 5 % compounded continuously.
What is the future value of the investment after 3 years?
Solution:
F = Pern
F = 300,000e0.05( 3 )
F = 348,550.27
20. Mr. Cruz invested P 20,000 for 1 year at 12% compounded monthly. At the end of 1
year, he reinvested all ( the principal and the interest ) at 6 % compounded
continuously for 1 year. How much will the investment amount?
Solution:
F1 = P( 1 + i )n
7
F1 = 20,000( 1 + 0.12/12 )12
F1 = 22,536.50
F2 = F1ern
F2 = 22,536.50e0.06( 1 )
F2 = P 23,930.08
8