ES 123-18 - Moris, Jennifer Ingrid F. - Seatwork#1 PDF

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engineering economics seatwork for industrial engineering...

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Name: Moris, Jennifer Ingrid F. ENGINEERING ECONOMICS Seatwork No. 1 Solve the following problems. Show your solutions: Submit your answers (vle). 1.

What is the annual rate of interest if P 565 – is earned in 6 months on an investment of P 16,000- ? 𝑛= 𝑖= 𝑖=

6 1 = 2 12 𝐼 𝑃𝑛 𝑃565 1 (𝑃16,000)( ) 2

= 0.070625 𝑜𝑟 70 .6 %

Thus, the annual rate of interest is 70.6%

2.

Determine the exact simple interest on P 11,000 – for the period from Jan. 15 – Nov. 28, 2008, if the rate of interest is 24 %? Given: P= 11,000 i= 24% Solution: January 15 = 16 (excluding Jan.15) February = 29 March = 31 April = 30 May = 31 June = 30 July = 31 August = 31 September = 30 October = 31 November = 28 (including Nov. 28) 318 days Exact simple interest = Pin = P11,000 x 318/366 days x 0.22% Answer = P2,293.77

3.

By the conditions of a will, the sum of P 25,000- is left to a girl to be held in trust by her guardian until it amounts to P 85,000 -. When will the girl receive the money , if the fund is invested at 8% compounded quarterly? Given: P= P25, 000 F= P85, 000 i= 8%/4 =2% Solution: 𝑭 = 𝑷 (𝟏 + 𝒊)𝒏 𝑃85, 000 = 𝑃25 , 000 (1 + 0.02)4𝑛 𝑃85, 000/𝑃25 , 000 (1.02)4𝑛 𝑃3.4 (1.02)4𝑛 𝐼𝑛 (𝑃3.4) = 4𝑛𝑙𝑛 (1.02) 27,060 = 4𝑛 𝑛 = 6.77 𝑦𝑒𝑎𝑟𝑠

4.

At a certain interest rate compounded semiannually, P 6000- will amount to P35,000 - after 10 years. What is the amount at the end of 15 years? What is the effective rate of interest? Given: 𝑃 = 𝑃6000 𝑛1 = 10 𝐹1 = 𝑃35, 000 𝑛2 = 15 𝐹2 = ? Solution: 𝐴𝑡 𝑛1 = 10, 𝐹1 = 𝑃35, 000 𝑖 𝐹1 = 𝑃 (1 + 2 )2 (𝑛1 ) 𝑖

𝑃35,000 = 𝑃6,000 (1 + )2 (10) 2 𝑖 = 14.35%

𝐴𝑡 𝑛2 = 15

𝑖

𝐹2 = 𝑃 (1 + )2 (𝑛2 ) 2 𝐹2 = 𝑃6000 (1 + 𝐹2 = 𝑃6000 (1 + 𝐹2 = 𝑃47,968.49 Answer: P47,968.49

0.1435 2 (15) ) 2 0.1435 2 (15) 2

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