More solved problems on compound interest using formula are shown below.
Solution:
Given, SI = $ 6750, R = \(\frac{20}{3}\)% p.a. and T = 3 years.
sum = 100 × SI / R × T
= $ (100 × 6750 × ³/₂₀ × 1/3 ) = $ 33750.
Now, P = $ 33750, R = \(\frac{20}{3}\)% p.a. and T = 3 years.
Therefore, amount after 3 years
= $ {33750 × (1 + (20/3 × 100)}³ [using A = P (1 + R/100)ᵀ]
= $ (33750 × 16/15 × 16/15 × 16/15) = $ 40960.
Thus, amount = $ 40960.
Hence, compound interest = $ (40960 - 33750) = $ 7210.
Solution:
Let the sum be $ 100. Then,
SI = $ (100 × 6 × 2/100) = $ 12
and compound interest = $ {100 × (1 + 6/100)² - 100}
= $ {(100 × 53/50 × 53/50) - 100} = $ (2809/25 - 100) = $ 309/25
Therefore, (CI) - (SI) = $ (309/25 – 100) = $ 9/25
If the difference between the CI and SI is $ 9/25, then the sum = $ 100.
If the difference between the CI and SI is $ 18, then the sum = $ (100 × 25/9 × 18 )
= $ 5000.
Hence, the required sum is $ 5000.
Alternative method
Let the sum be $ P.
Then, SI = $ (P × 6/100 × 2) = $ 3P/25
And, CI = $ {P × (1 + 6/100)² - P}
= $ {(P × 53/50 × 53/50) - P} = $ (\(\frac{2809}{2500}\)P - P) = $ (309P/2500)
(CI) - (SI) = $ (309P/2500 – 3P/25) = $ (9P/2500)
Therefore, 9P/2500 = 18
⇔ P = 2500 × 18/9
⇔ P = 5000.
Hence, the required sum is $ 5000.
Solution:
Let the sum be $ 100. Then,
amount = $ {100 × (1 + 8/100)²}
= $ (100 × 27/25 × 27/25) = $ (2916/25)
If the amount is $ 2916/25 then the sum = $ 100.
If the amount is $ 72900 then the sum = $ (100 × 25/2916 × 72900) = $ 62500.
Hence, the required sum is $ 62500.
Alternative method
Let the sum be $ P. Then,
amount = $ {P × (1 + 8/100)²}
= $ {P × 27/25 × 27/25} = $ (729P/625)
Therefore, 729P/625 = 72900
⇔ P = (72900 × 625)/729
⇔ P = 62500.
Hence, the required sum is $ 62500.
Solution:
Let the required rate be R% per annum.
Here, A = $ 2205, P = $ 2000 and n = 2 years.
Using the formula A = P(1 + R/100)ⁿ,
2205 = 2000 × ( 1 + R/100)²
⇒ (1 + R/100)² = 2205/2000 = 441/400 = (21/20)²
⇒ ( 1 + R/100) = 21/20
⇒ R/100 = (21/20 – 1) = 1/20
⇒ R = (100 × 1/20) = 5
Hence, the required rate of interest is 5% per annum.
Solution:
Let the required time be n years. Then,
amount = $ {1000 × (1 + 10/100)ⁿ}
= $ {1000 × (11/10)ⁿ}
Therefore, 1000 × (11/10)ⁿ = 1331 [since, amount = $ 1331 (given)]
⇒ (11/10)ⁿ = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)³
⇒ (11/10)ⁿ = (11/10)³
⇒ n = 3.
Thus, n = 3.
Hence, the required time is 3 years.
● Compound Interest
Compound Interest with Growing Principal
Compound Interest with Periodic Deductions
Compound Interest by Using Formula
Compound Interest when Interest is Compounded Yearly
Compound Interest when Interest is Compounded Half-Yearly
Compound Interest when Interest is Compounded Quarterly
Variable Rate of Compound Interest
Difference of Compound Interest and Simple Interest
Practice Test on Compound Interest
Uniform Rate of Growth and Depreciation
● Compound Interest - Worksheet
Worksheet on Compound Interest
Worksheet on Compound Interest when Interest is Compounded Half-Yearly
Worksheet on Compound Interest with Growing Principal
Worksheet on Compound Interest with Periodic Deductions
Worksheet on Variable Rate of Compound Interest
Worksheet on Difference of Compound Interest and Simple Interest
Worksheet on Uniform Rate of Growth
Worksheet on Uniform Rate of Depreciation
Worksheet on Uniform Rate of Growth and Depreciation
8th Grade Math Practice
From Problems on Compound Interest to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Dec 12, 24 09:20 AM
Dec 09, 24 10:39 PM
Dec 09, 24 01:08 AM
Dec 08, 24 11:19 PM
Dec 07, 24 03:38 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.