Mathematics in investment

1. Key Answers with solutions (Quiz 2 Math110)
1. Pilar split her Php1, 000,000 and invested Php500, 000 to each of two different companies.
Company X offers interest at 8% compounded monthly while company Y offers interest at 12%
compounded quarterly. How many more years will it take the accumulated amount in company
X the same as the accumulated amount in company Y for 6 years?
Solution:
Data for company X: j = 0.08, m = 12, p =Php 500,000 t = ?
Data for company Y: j=0.12, m = 4, p =Php 500,000 ty = 6
Steps
a. Solve for F using data for company Y. Thus,
� = 500,000(1.0324
) = 1,016,397.053
b. Solve for the time, ��, that it will take the investment in company X to accumulate the amount
yielded in company Y.
�� = ���(1,016,397.053
500,000
)/12���(1 + . 02/3) = 8.9 years
c. Compute for the difference in time between investment in company X and in company Y to
accumulate the same amount.
t = tx − ty = 8.9 − 6 = 2.9 years
Therefore, investment of Php500,000 in company X needs 2.9 years more to
accumulate the amount of 1,016,397.053 that is yielded in company Y in 6 years.
2. A capital of Php15, 000 was invested for 5 years to earn money at 10% simple interest. How
much more money could have been earned if the computation of income is based on compound
interest using the same rate and converting annually or once a year?
Steps
a. Solve for the accumulated amount �� using simple interest.
�� = �(1 + ��) = 15,000(1 + 0.1(5)) = 22,500
b. Solve for the accumulated amount �� using compound interest.
�� = �(1 + �)� = 15,000(1 + 0.1)5
= 24,157.65
c. Compute for the difference, D, between �� and ��.
� = �� − �� = 2424,157.65 − 22,500 = 1,657.65
Therefore, the amount of 1,657.65 more could have been earned using compound
interest.
3. Consider these transactions: a) accumulating an amount of Php50, 000 at 7% simple interest
for 15 years and then discounting the result at 7% simple discount for 5 years, and b)
accumulating the given amount at 7% simple interest for 10 years. Which transaction yields
more money? How much more compared to the other?
Solution:
�) � = 50,000; � = 0.07 �1 = 15 �����; �2 = 5 �����; � = 0.07; P*=?.
Steps
a. Compute for F using simple interest as follows
� = �(1 + ��1) = 50,000(1 + 0.07(15)) = 102,500
b. Compute for P* (or discounted value using simple discount, �).
�∗ = �(1 − ��2) = 102,500(1 − . 07(5)) = 66,625
�) � = 50,000; � = 0.07; �3 = 10 ����� ; F’ (accumulated amount in 10 years using simple
interest)=?.
a. Compute for F’
�' = 50,000(1 + 0.07(10) = 85,000
b. Find the difference , D, between P* and F’.
� = �' − �∗ = 85,000 − 66,625 = 18,375
4. Consider the following ways of discounting the worth (Php10,500) of an appliance so that its
final price is obtained: a) Discounting the price at 4% compounded annually for 3 years and b)
discounting the same amount at 4% simple discount for 3 years. In which mode of discounting is
more beneficial to the seller? (a or b). How much more money is discounted in one mode
compared to the other mode?

2. Solution: � = 10,500, � = 0.04, � = 1, � = 3, � = 0.04, �� = ? , �� = ?
Steps
a. Compute for the discounted value, Pj by way of compound discount.
a) �� = �(1 + �)−�
= 10,500(1 + 0.04)−3
= 9,334.46
b. Compute for the discounted value, Pd by way of simple discount.
b) �� = �(1 − ��) = 10,500(1 − 0.04(3)) = 9,240
c. Compute for the difference, D, between Pj and Pd.
� = �� − �� = 9,334.46 − 9,240 = 94.46
Therefore, the discount in a is higher by 94.46 than in b.
5. A capital of Php50, 500 has accumulated to Php65, 800 at a certain interest rate
compounded quarterly for 5 years. The same accumulated amount from the same capital was
yielded using a new interest rate compounded semiannually for 4 years. What is the new
interest rate applied? How many percent is the new rate bigger or smaller than the old rate?
Solution: � = 50,500, � = 65,800, �1 = 4, �1 = 5, �1 = ? , �1 = 20,
�2 = 2, �2 = 4, �2 = 8, �2 = ? , �� = ?
Steps
a. Compute for �1 and �2.
For �1: � = �(1 + �1/�1)�1
65,800 = 50,500(1 + �1/4)20
(65,800/50,500)1/20
= (1 + �1/4)
4[(65,800/50,500)1/20 − 1] = �1
0.0533 = �1
For �2: � = �(1 + �2/�2)�2
(65,800 − 50,500) = (1 + �2/2)8
(65,800/50,500)1/8
= (1 + �2/2)
2[(65,800/50,500)1/8 − 1] = �2
0.0673 = �2
b. Compute for the difference, D, between �1 and �2.
� = �2 − �1 = 0.0673 − 0.0533 = 0.014
Therefore, the new rate,�2 , is bigger by 0.014 than the old rate,�1.
6. At a certain effective rate, a net gain of Php30, 900 at the end of 6 years can be generated
from an investment involving a capital of Php40, 000. a) What new rate compounded
semiannually can yield the same net gain at the end of the given investment period? b) How
many percent is the new rate bigger or smaller than the effective rate?
Solution: � = 30,900, � = 6, � = 40,000, �� = ? (��� ����), �� = ? (��� ����),
�� = 1(��� ���������� ������), �� = 2(��� ���������� ������)
Steps
a. Solve for the old rate ��
��: � = �(1 + ��/��)�
��= ��[(�/�)
1
� − 1]
= (1)[(70,900/40,000)
1
6 − 1]
= 0.1001
b. Solve for the new rate, jn
��= ��[(�/�)
1
� − 1]
= (2)[(70,900/40,000)
1
12 − 1]
= 0.0977
c. Solve for the difference, D, between jo and jn.
� = �� − �� = 0.1001 − 0.0977 = 0.0024
Therefore, a) the new rate that yield the same income is 0.0977 and b) it is smaller than the old
rate by 0.0024.