Here, Principal (P) = ₹ 42,000, Rate of Interest (R) = 8%, Time (n) = 1 years Amount (A) = P(1-R/100)ⁿ = 42000(1-8/100)¹ = 42000(1+2/25)¹ = 42000(27/25)¹ = 42000 x 27/25 = ₹ 38,640 Hence, the value of scooter after one year is ₹ 38,640 Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video for moreRead more
Here, Principal (P) = ₹ 42,000, Rate of Interest (R) = 8%, Time (n) = 1 years
Amount (A) = P(1-R/100)ⁿ
= 42000(1-8/100)¹
= 42000(1+2/25)¹
= 42000(27/25)¹
= 42000 x 27/25 = ₹ 38,640
Hence, the value of scooter after one year is ₹ 38,640
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours After 2 hours, number of bacteria, Amount (A) =P(1+R/100)ⁿ = 506000(1+ 2.5/100)² = 506000(1+ 25/100)² = 506000(1+ 25/40)² = 506000(41/40)² = 506000 x 41/40 x 41/40 = 5,31,616.25 Hence, number of bacteria after two hoursRead more
Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours
After 2 hours, number of bacteria,
Amount (A) =P(1+R/100)ⁿ
= 506000(1+ 2.5/100)²
= 506000(1+ 25/100)²
= 506000(1+ 25/40)²
= 506000(41/40)²
= 506000 x 41/40 x 41/40 = 5,31,616.25
Hence, number of bacteria after two hours are 531616 (approx.).
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
(i) Here, A₂₀₀₃ = 54,000, R = 5%, n = 2 years Population would be less in 2001 than 2003 in two years. Here population is increasing. ∴ A₂₀₀₃ = P₂₀₀₁ (1+R/100)ⁿ ⇒ 54000 = P₂₀₀₁(1+5/100)² ⇒ 54000 = P₂₀₀₁ (1+ 1/20)² ⇒ 54000 = P₂₀₀₁(21/20)² ⇒ 54000 = P₂₀₀₁ x 21/20 x 21/20 ⇒ P₂₀₀₁= 54000x20X20/21x21 ⇒ PRead more
(i) Here, A₂₀₀₃ = 54,000, R = 5%, n = 2 years
Population would be less in 2001 than 2003 in two years.
Here population is increasing.
(ii) According to question, population is increasing.
Therefore population in 2005,
A₂₀₀₅ = P(1+R/100)ⁿ = 54000 (1+5/100)² = 54000(1+ 1/20)²
= 54000(21/20)² = 54000 x 21/20 x 21/20 = 59,535
= Hence population in 2005 would be 59,535.
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
Calculate the amount and compound interest on: ₹ 10,800 for 3 years at 12 (1/2) % per annum compounded annually.
Here, Principal (P) = ₹ 10800, Time (n) = 3 years, Rate of interest (R) = 12 (1/2) % = 25/2 % Amount (A) = P(1+ R/10)ⁿ = 10800(1+25/2×100)³ = 10800 (1+ 1/2×4)³ = 10800 (1+1/8)³ = 10800(9/8)³ = 10800 x 9/8 x 9/8 x 9/8 = ₹ 15,377.34 Compound Interest (C.I.) = A – P = ₹ 10800 – ₹15377.34 = ₹4,577.34 ClRead more
Here, Principal (P) = ₹ 10800, Time (n) = 3 years,
Rate of interest (R) = 12 (1/2) % = 25/2 %
Amount (A) = P(1+ R/10)ⁿ = 10800(1+25/2×100)³ = 10800 (1+ 1/2×4)³
= 10800 (1+1/8)³ = 10800(9/8)³ = 10800 x 9/8 x 9/8 x 9/8 = ₹ 15,377.34
Compound Interest (C.I.) = A – P = ₹ 10800 – ₹15377.34 = ₹4,577.34
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-9/
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Here, Principal (P) = ₹ 42,000, Rate of Interest (R) = 8%, Time (n) = 1 years Amount (A) = P(1-R/100)ⁿ = 42000(1-8/100)¹ = 42000(1+2/25)¹ = 42000(27/25)¹ = 42000 x 27/25 = ₹ 38,640 Hence, the value of scooter after one year is ₹ 38,640 Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video for moreRead more
Here, Principal (P) = ₹ 42,000, Rate of Interest (R) = 8%, Time (n) = 1 years
Amount (A) = P(1-R/100)ⁿ
= 42000(1-8/100)¹
= 42000(1+2/25)¹
= 42000(27/25)¹
= 42000 x 27/25 = ₹ 38,640
Hence, the value of scooter after one year is ₹ 38,640
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/
In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours After 2 hours, number of bacteria, Amount (A) =P(1+R/100)ⁿ = 506000(1+ 2.5/100)² = 506000(1+ 25/100)² = 506000(1+ 25/40)² = 506000(41/40)² = 506000 x 41/40 x 41/40 = 5,31,616.25 Hence, number of bacteria after two hoursRead more
Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours
After 2 hours, number of bacteria,
Amount (A) =P(1+R/100)ⁿ
= 506000(1+ 2.5/100)²
= 506000(1+ 25/100)²
= 506000(1+ 25/40)²
= 506000(41/40)²
= 506000 x 41/40 x 41/40 = 5,31,616.25
Hence, number of bacteria after two hours are 531616 (approx.).
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/
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. (i) Find the population in 2001. (ii) What would be its population in 2005?
(i) Here, A₂₀₀₃ = 54,000, R = 5%, n = 2 years Population would be less in 2001 than 2003 in two years. Here population is increasing. ∴ A₂₀₀₃ = P₂₀₀₁ (1+R/100)ⁿ ⇒ 54000 = P₂₀₀₁(1+5/100)² ⇒ 54000 = P₂₀₀₁ (1+ 1/20)² ⇒ 54000 = P₂₀₀₁(21/20)² ⇒ 54000 = P₂₀₀₁ x 21/20 x 21/20 ⇒ P₂₀₀₁= 54000x20X20/21x21 ⇒ PRead more
(i) Here, A₂₀₀₃ = 54,000, R = 5%, n = 2 years
Population would be less in 2001 than 2003 in two years.
Here population is increasing.
∴ A₂₀₀₃ = P₂₀₀₁ (1+R/100)ⁿ ⇒ 54000 = P₂₀₀₁(1+5/100)²
⇒ 54000 = P₂₀₀₁ (1+ 1/20)² ⇒ 54000 = P₂₀₀₁(21/20)²
⇒ 54000 = P₂₀₀₁ x 21/20 x 21/20 ⇒ P₂₀₀₁= 54000x20X20/21×21
⇒ P₂₀₀₁ = 48,980 (approx.)
(ii) According to question, population is increasing.
Therefore population in 2005,
A₂₀₀₅ = P(1+R/100)ⁿ = 54000 (1+5/100)² = 54000(1+ 1/20)²
= 54000(21/20)² = 54000 x 21/20 x 21/20 = 59,535
= Hence population in 2005 would be 59,535.
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/
Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at (12) ½ % per annum, interest being compounded half yearly.
Here, Principal (P) = ₹ 4096, Rate of Interest (R) = (12)½ = 25/2% = 25/2% (compounded half yearly) Time (n) = 18 months = 3/2 years = 3 half-years (compounded half yearly) Amount (A) = P(1+R)ⁿ = 4096 (1+25/4x100)³ = 4096(1+1/4x4)³ = 4096(17/16)³ = 4096x 17/16 x 17/16 x 17/16 =₹ 4,913 Class 8 MathsRead more
Here, Principal (P) = ₹ 4096,
Rate of Interest (R) = (12)½ = 25/2% = 25/2% (compounded half yearly)
Time (n) = 18 months = 3/2 years = 3 half-years (compounded half yearly)
Amount (A) = P(1+R)ⁿ = 4096 (1+25/4×100)³ = 4096(1+1/4×4)³
= 4096(17/16)³ = 4096x 17/16 x 17/16 x 17/16 =₹ 4,913
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/