RS Aggarwal Class 8 Solutions Chapter 11 Compound Interest Ex 11A
These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A.
Other Exercises
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11B
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11C
- RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11D
Question 1.
Solution:
Principal (p) = Rs. 2500
Rate (r) = 10% p.a.
Period (t) = 2 years
Interest for the first year = \(\\ \frac { prt }{ 100 } \)
= \(\\ \frac { 2500X10X1 }{ 100 } \)
= Rs. 250
Amount at the end of first year = Rs. 2500 + Rs. 250
= Rs 2750
Principal for the second year = Rs 2750
Interest for the second year = Rs \(\\ \frac { 2750X10X1 }{ 100 } \)
= Rs. 275
Amount at the end of second year = Rs 2750 + Rs. 275
= Rs. 305
and compound interest for the 2 years = Rs. 3025 – Rs. 2500
= Rs 525 Ans.
Question 2.
Solution:
Principal (P) = Rs. 15625 Rate
(R) = 12% p.a.
Period (n) = 3 years
Question 3.
Solution:
Principal (P) = Rs. 5000
Rate (R) = 9% p.a.
Time (n) = 2 years
Question 4.
Solution:
Amount of loan (p) = Rs. 25000
Rate of interest (r) = 8% p.a.
Period (t) = 2 years
Question 5.
Solution:
In case of Harpreet :
Amount borrowed by Harpreet (P) = Rs. 20000
Rate (r) = 12%
Period (t) = 2 Years
Question 6.
Solution:
Principal (p) = Rs. 64000
Rate (r) = \(7\frac { 1 }{ 2 } \) = \(\\ \frac { 15 }{ 2 } \)%
Period (t) = 3 years
Interest for the first year = \(\\ \frac { prt }{ 100 } \)
= Rs \(\\ \frac { 64000X15X1 }{ 100X2 } \)
= Rs 4800
= Rs 79507
Question 7.
Solution:
Principal (p) = Rs 6250
Rate (r) 8% p.a. or 4% half yearly
Period (t) = 1 year = 2 half years
Question 8.
Solution:
Principal (p) = Rs. = 16000
Rate (r) = 10% p.a. or 5% half yearly
Hope given RS Aggarwal Solutions Class 8 Chapter 11 Compound Interest Ex 11A are helpful to complete your math homework.
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