Total number of students = 25 Number of good students in mathematics = 72% of 25 = 72/100x25=18 Number of students not good in mathematics = 25 – 18 = 7 Hence percentage of students not good in mathematics = 7/25x100=28% Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video for more answers vist toRead more
Total number of students = 25
Number of good students in mathematics = 72% of 25 = 72/100×25=18
Number of students not good in mathematics = 25 – 18 = 7
Hence percentage of students not good in mathematics = 7/25×100=28%
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
Here, Principal (P) = ₹ 18,000, Time (n) = 2(1/2) years, Rate of interest (R) = 10% p.a. Amount (A) = P (1+R/100)ⁿ = 18000 (1+10/100)² = (1+1/10)² = 18000 (11/10)² = 18000 x11/10×11/10 = ₹ 21,780 Interest for 1/2 years on ₹ 21,780 at rate of 10% = 1/2 x21780x10x1/100=₹ 1,089 Total amount for 2(1/2)Read more
Here,
Principal (P) = ₹ 18,000, Time (n) = 2(1/2) years, Rate of interest (R) = 10% p.a.
Amount (A) = P (1+R/100)ⁿ = 18000 (1+10/100)² = (1+1/10)²
= 18000 (11/10)² = 18000 x11/10×11/10 = ₹ 21,780
Interest for 1/2 years on ₹ 21,780 at rate of 10% = 1/2 x21780x10x1/100=₹ 1,089
Total amount for 2(1/2) years = ₹ 21,780 + ₹ 1089 = ₹ 22,869
Compound Interest (C.I.) = A – P = ₹ 22869 – ₹ 18000 = ₹ 4,869
Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video
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
72% of 25 students are good in mathematics. How many are not good in mathematics?
Total number of students = 25 Number of good students in mathematics = 72% of 25 = 72/100x25=18 Number of students not good in mathematics = 25 – 18 = 7 Hence percentage of students not good in mathematics = 7/25x100=28% Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video for more answers vist toRead more
Total number of students = 25
Number of good students in mathematics = 72% of 25 = 72/100×25=18
Number of students not good in mathematics = 25 – 18 = 7
Hence percentage of students not good in mathematics = 7/25×100=28%
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/
Calculate the amount and compound interest on: ₹ 10,000 for 1 years at 8% per annum compounded half yearly.
Here, Principal (P) = ₹ 10,000, Time (n) = 1 years = 2 half-years (compounded half yearly) Rate of interest (R) = 8% = 4% (compounded half yearly) Amount (A) = P (1+R/100)ⁿ = 10000(1+4/100)² = 10000(1+1/25)² = 10000(26/25)=26/25 x 26/25 = ₹ 10,816 Compound Interest (C.I.) = A – P = ₹ 10,816 – ₹ 10,0Read more
Here, Principal (P) = ₹ 10,000,
Time (n) = 1 years = 2 half-years (compounded half yearly)
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) = P (1+R/100)ⁿ = 10000(1+4/100)² = 10000(1+1/25)²
= 10000(26/25)=26/25 x 26/25 = ₹ 10,816
Compound Interest (C.I.) = A – P = ₹ 10,816 – ₹ 10,000 = ₹ 816
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/
Calculate the amount and compound interest on: ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).
Here, Principal (P) = ₹ 8000, Time (n) = 1 years = 2 half-years(compounded half yearly) Rate of interest (R) = 9% = (9/2)% (compounded half yearly) Amount (A) = P (1+R/100)ⁿ = 8000(1+9/2×100)² = 8000(1+9/200)² = 8000(209/200)² = 8000 x 209/200 x 209/200 = ₹ 8,736.20 Compound Interest (C.I.) = A – PRead more
Here,
Principal (P) = ₹ 8000,
Time (n) = 1 years = 2 half-years(compounded half yearly)
Rate of interest (R) = 9% = (9/2)% (compounded half yearly)
Amount (A) = P (1+R/100)ⁿ = 8000(1+9/2×100)² = 8000(1+9/200)²
= 8000(209/200)² = 8000 x 209/200 x 209/200 = ₹ 8,736.20
Compound Interest (C.I.) = A – P = ₹ 8736.20 – ₹ 8000 = ₹ 736.20
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-9/
Calculate the amount and compound interest on: ₹ 62,500 for 1(1/2) % years at 8% per annum compounded annually.
Here, Principal (P) = ₹ 62500, Time (n) = 1(1/2) = 3/2 years = 3 half-years (compounded half yearly) Rate of interest (R) = 8% = 4% (compounded half yearly) Amount (A) = P (1+R/100)ⁿ = 62500(1+4/100)² = 62500(1+1/25)³ = 62500 (26/25)³ = 62500 x 26/25 x 26/25 x 26/25 = ₹ 70,304 Compound Interest (C.IRead more
Here, Principal (P) = ₹ 62500,
Time (n) = 1(1/2) = 3/2 years = 3 half-years (compounded half yearly)
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) = P (1+R/100)ⁿ = 62500(1+4/100)² = 62500(1+1/25)³
= 62500 (26/25)³ = 62500 x 26/25 x 26/25 x 26/25 = ₹ 70,304
Compound Interest (C.I.) = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804
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/
Calculate the amount and compound interest on: ₹ 18,000 for 2 (1/2) % years at 10% per annum compounded annually.
Here, Principal (P) = ₹ 18,000, Time (n) = 2(1/2) years, Rate of interest (R) = 10% p.a. Amount (A) = P (1+R/100)ⁿ = 18000 (1+10/100)² = (1+1/10)² = 18000 (11/10)² = 18000 x11/10×11/10 = ₹ 21,780 Interest for 1/2 years on ₹ 21,780 at rate of 10% = 1/2 x21780x10x1/100=₹ 1,089 Total amount for 2(1/2)Read more
Here,
Principal (P) = ₹ 18,000, Time (n) = 2(1/2) years, Rate of interest (R) = 10% p.a.
Amount (A) = P (1+R/100)ⁿ = 18000 (1+10/100)² = (1+1/10)²
= 18000 (11/10)² = 18000 x11/10×11/10 = ₹ 21,780
Interest for 1/2 years on ₹ 21,780 at rate of 10% = 1/2 x21780x10x1/100=₹ 1,089
Total amount for 2(1/2) years = ₹ 21,780 + ₹ 1089 = ₹ 22,869
Compound Interest (C.I.) = A – P = ₹ 22869 – ₹ 18000 = ₹ 4,869
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/
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/