Investment of P250,000.00 per year | CE Board Problem in Mathematics, Surveying and Transportation Engineering

Date of Exam

November 2004

Subject

Situation
An investment of P250,000 is made at the end of each year with interest of 2.5% compounded annually.

Determine the equal-payment-series compound-amount factor after 10 years.

A. 11.203 C. 9.632

B. 10.578 D. 8.736
Determine the total amount of the investment after 10 years.

A. P2,800,000.00 C. P2,400,000.00

B. P2,600,000.00 D. P2,200,000.00
How long (in years) will it take for the investment to amount to P10,000,000.00?

A. 25 C. 15

B. 18 D. 28

Answer Key

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Part 1: [ A ]
Part 2: [ A ]
Part 3: [ D ]

Solution

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Part (1)

Equal-payment-series compound-amount factor

$(F/A, \, i, \, n) = \dfrac{(1 + i)^n - 1}{i}$

$(F/A, \, 2.5\%, \, 10) = \dfrac{(1 + 0.025)^{10} - 1}{0.025}$

$(F/A, \, 2.5\%, \, 10) = 11.2034$

Part (2)

Amount of investment after 10 years

$F = A(F/A, \, i, \, n)$

$F = 250,000(F/A, \, 2.5\%, \, 10)$

$F = \text{P}2,800,845.44$

Part (3)

Number of years for the investment to grow to P10,000,000.00

$F = \dfrac{A [ \, (1 + i)^n - 1 \, ]}{i}$

$10,000,000 = \dfrac{250,000 [ \, (1 + 0.025)^t - 1 \, ]}{0.025}$

$1.025^t = 2$

$t \log 1.025 = \log 2$

$t = 28.07 \text{ years}$

A. P2,800,000.00	C. P2,400,000.00
B. P2,600,000.00	D. P2,200,000.00

A. 11.203	C. 9.632
B. 10.578	D. 8.736

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