Time Principal (P) Simple Interest Amount after t years Solution Answer (Maturity Value) 1 (5,000) (0.25) (1) 1,250 5,000+1,250=6,250 2 5,000+2,500=7,250 P5,000 3 (5,000) (0.25) (3) 4 5,000 5,000+5,000=10,000 Time Principal Compound Interest Amount aftert years (P) Solution Answer (Maturity Value) 1 (5,000) (0.25) (1) 1,250 5,000 + 1,250 =6,250 2 (6,250) (0.25) (2) P5,000 3 (9,375) (0.25) (3) 9,375+7,031.25=16,406.25 16,406.25 4.
Time Principal (P) Simple Interest Amount after t years Solution Answer (Maturity Value) 1 (5,000) (0.25) (1) 1,250 5,000+1,250=6,250 2 5,000+2,500=7,250 P5,000 3 (5,000) (0.25) (3) 4 5,000 5,000+5,000=10,000 Time Principal Compound Interest Amount aftert years (P) Solution Answer (Maturity Value) 1 (5,000) (0.25) (1) 1,250 5,000 + 1,250 =6,250 2 (6,250) (0.25) (2) P5,000 3 (9,375) (0.25) (3) 9,375+7,031.25=16,406.25 16,406.25 4.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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ꜱɪᴍᴘʟᴇ ᴀɴᴅ ᴄᴏᴍᴘᴏᴜɴᴅ ɪɴᴛᴇʀᴇꜱᴛ
ɪ. ᴅɪʀᴇᴄᴛɪᴏɴ: ᴄᴏᴍᴘʟᴇᴛᴇ ᴛʜᴇ ᴛᴀʙʟᴇ ᴛʜᴀᴛ ɪʟʟᴜꜱᴛʀᴀᴛᴇꜱ ᴛʜᴇ ʙᴀʟᴀɴᴄᴇ ᴀꜰᴛᴇʀ 4 ʏᴇᴀʀꜱ ᴡɪᴛʜ
ꜱɪᴍᴘʟᴇ ᴀɴᴅ ᴄᴏᴍᴘʟᴇᴛᴇ ɪɴᴛᴇʀᴇꜱᴛ.
Transcribed Image Text:Time Principal
(P)
Simple Interest
Amount after t years
Solution
Answer
(Maturity Value)
1
(5,000) (0.25) (1)
1,250
5,000+1,250=6,250
2
5,000+2,500=7,250
P5,000
3
(5,000) (0.25) (3)
4
5,000
5,000+5,000=10,000
Time Principal
Compound Interest
Amount aftert years
(P)
Solution
Answer
(Maturity Value)
1
(5,000) (0.25) (1)
1,250
5,000 + 1,250 =6,250
2
(6,250) (0.25) (2)
P5,000
3
(9,375) (0.25) (3)
9,375+7,031.25=16,406.25
16,406.25
4.
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