(i) Here, Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 2 years Amount (A) = P (1+R/100)ⁿ = 8000(1+5/100)² = 8000(1+1/20)² = 8000(21/20)² = 8000 x 21/20 x21/20 =₹8,820 (ii) Here, Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 3 years Amount (A) = P(1+R/100)ⁿ = 8000(1+5Read more
(i) Here,
Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 2 years
Amount (A) = P (1+R/100)ⁿ = 8000(1+5/100)² = 8000(1+1/20)²
= 8000(21/20)² = 8000 x 21/20 x21/20 =₹8,820
(ii) Here, Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 3 years
Amount (A) = P(1+R/100)ⁿ = 8000(1+5/100)³ = 8000(1+ 1/20)³
= 8000 (21/20)³ = 8000x 21/20 x21/21 x 21/20 = ₹ 9,261
Interest for 3rd year = A – P = ₹ 9,261 – ₹ 8,820 = ₹ 441
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
Here, Principal (P) = ₹12,000, Time (T) = 2 years, Rate of interest (R) = 6% p.a. Simple Interest = PxRxT/100 = 12000x6x2/100 = ₹ 1,440 Had he borrowed this sum at 6% p.a., then Compound Interest = P (1+R/100)ⁿ - P=12000(1+6/100)²-12000 = 12000(1+3/50)² - 12000 = 12000(53/50)² - 12000 = 12000 x 53/5Read more
Here, Principal (P) = ₹12,000, Time (T) = 2 years, Rate of interest (R) = 6% p.a.
Simple Interest = PxRxT/100 = 12000x6x2/100 = ₹ 1,440
Had he borrowed this sum at 6% p.a., then
Compound Interest = P (1+R/100)ⁿ – P=12000(1+6/100)²-12000
= 12000(1+3/50)² – 12000 = 12000(53/50)² – 12000
= 12000 x 53/50 x53/50 – 12000 = ₹ 13,483.20 – ₹ 12,000
= ₹ 1,483.20
Difference in both interests = ₹ 1,483.20 – ₹ 1,440.00 = ₹ 43.20
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
Find the amount and the compound interest on ₹ 10,000 for 3/2 years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Here, Principal (P) = ₹ 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly) Time (n) = 3/2 years = 3 half-years (compounded half yearly) Amount (A) = P(1+R/100)ⁿ = 10000(1+ 5/100)³ = 10000(1+1/20)³ = 10000(21/20)³ = 10000 x 21/20 x 21/20 x 21/20 = ₹ 11,576.25 Compound Interest (C.I.) = ARead more
Here, Principal (P) = ₹ 10000,
Rate of Interest (R) = 10% = 5% (compounded half yearly)
Time (n) = 3/2 years = 3 half-years (compounded half yearly)
Amount (A) = P(1+R/100)ⁿ = 10000(1+ 5/100)³ = 10000(1+1/20)³
= 10000(21/20)³ = 10000 x 21/20 x 21/20 x 21/20 = ₹ 11,576.25
Compound Interest (C.I.) = A – P = ₹ 11,576.25 – ₹ 10,000 = ₹ 1,576.25
If it is compounded annually, then
Here, Principal (P) = ₹ 10000, Rate of Interest (R) = 10%, Time (n) = 3/5 years
Amount (A) for 1 year = P (1+R/100)ⁿ = 10000(1+ 10/100)¹ = 10000(1+ 1/10)¹
= 10000(11/10)¹ = 10000x 11/10 = ₹ 11,000
Interest for 1/2 year = 11000x1x10/2×100 =₹ 550
∴ Total amount = ₹ 11,000 + ₹ 550 = ₹ 11,550
Now, C.I. = A – P = ₹ 11,550 – ₹ 10,000 = ₹ 1,550
Yes, interest ₹ 1,576.25 is more than ₹ 1,550.
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find: (i) The amount credited against her name at the end of the second year. (ii) The interest for the third year.
(i) Here, Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 2 years Amount (A) = P (1+R/100)ⁿ = 8000(1+5/100)² = 8000(1+1/20)² = 8000(21/20)² = 8000 x 21/20 x21/20 =₹8,820 (ii) Here, Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 3 years Amount (A) = P(1+R/100)ⁿ = 8000(1+5Read more
(i) Here,
Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 2 years
Amount (A) = P (1+R/100)ⁿ = 8000(1+5/100)² = 8000(1+1/20)²
= 8000(21/20)² = 8000 x 21/20 x21/20 =₹8,820
(ii) Here, Principal (P) = ₹ 8000, Rate of Interest (R) = 5%, Time (n) = 3 years
Amount (A) = P(1+R/100)ⁿ = 8000(1+5/100)³ = 8000(1+ 1/20)³
= 8000 (21/20)³ = 8000x 21/20 x21/21 x 21/20 = ₹ 9,261
Interest for 3rd year = A – P = ₹ 9,261 – ₹ 8,820 = ₹ 441
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 3/2 years if the interest is: (i) compounded annually. (ii) compounded half yearly.
(i) Here, Principal (P) = ₹ 80,000, Time (n) = 1(1/2) years, Rate of interest (R) = 10% Amount for 1 year (A) = P (1+R/100)ⁿ = 80000 (1+10/100)¹ = 80000(1+1/10)¹ = 80000 (11/10)¹ = ₹ 88,000 Interest for 1/2 year = 88000x10x1/100x2 = ₹ 4,400 Total amount = ₹ 88,000 + ₹ 4,400 = ₹ 92,400 (ii) Here, PriRead more
(i) Here,
Principal (P) = ₹ 80,000, Time (n) = 1(1/2) years, Rate of interest (R) = 10%
Amount for 1 year (A) = P (1+R/100)ⁿ = 80000 (1+10/100)¹ = 80000(1+1/10)¹
= 80000 (11/10)¹ = ₹ 88,000
Interest for 1/2 year = 88000x10x1/100×2 = ₹ 4,400
Total amount = ₹ 88,000 + ₹ 4,400 = ₹ 92,400
(ii) Here, Principal (P) = ₹ 80,000,
Time (n) = 1(1/2) year = 3 half-years (compounded half yearly)
Rate of interest (R) = 10% = 5% (compounded half yearly)
Amount (A) = P(1+R/100)ⁿ = 80000(1+5/100)³ = 80000(1+1/20)³
= 80000(21/20)³ = 80000 x 21/20 x 21/20 x 21/20 = ₹ 92,610
Difference in amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get: (i) after 6 months? (ii) after 1 year?
Here, Principal (P) = ₹ 60,000, Time (n) = 6 months = 1 half-year (compounded half yearly) Rate of interest (R) = 12% = 6% (compounded half yearly) Amount (A) = P(1+R/100)ⁿ = 60000(1+6/100)¹ = 60000(1+3/50)¹ = 60000(53/50)¹ = 60000 x 53/50 x 53/50 = ₹ 67,416 After 1 year Vasudevan would get amount ₹Read more
Here, Principal (P) = ₹ 60,000,
Time (n) = 6 months = 1 half-year (compounded half yearly)
Rate of interest (R) = 12% = 6% (compounded half yearly)
Amount (A) = P(1+R/100)ⁿ = 60000(1+6/100)¹ = 60000(1+3/50)¹
= 60000(53/50)¹ = 60000 x 53/50 x 53/50 = ₹ 67,416
After 1 year Vasudevan would get amount ₹ 67,416.
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/
I borrows ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Here, Principal (P) = ₹12,000, Time (T) = 2 years, Rate of interest (R) = 6% p.a. Simple Interest = PxRxT/100 = 12000x6x2/100 = ₹ 1,440 Had he borrowed this sum at 6% p.a., then Compound Interest = P (1+R/100)ⁿ - P=12000(1+6/100)²-12000 = 12000(1+3/50)² - 12000 = 12000(53/50)² - 12000 = 12000 x 53/5Read more
Here, Principal (P) = ₹12,000, Time (T) = 2 years, Rate of interest (R) = 6% p.a.
Simple Interest = PxRxT/100 = 12000x6x2/100 = ₹ 1,440
Had he borrowed this sum at 6% p.a., then
Compound Interest = P (1+R/100)ⁿ – P=12000(1+6/100)²-12000
= 12000(1+3/50)² – 12000 = 12000(53/50)² – 12000
= 12000 x 53/50 x53/50 – 12000 = ₹ 13,483.20 – ₹ 12,000
= ₹ 1,483.20
Difference in both interests = ₹ 1,483.20 – ₹ 1,440.00 = ₹ 43.20
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/