1. 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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/

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    • 7
  2. 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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/

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    • 6
  3. 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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/

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    • 6
  4. 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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/

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    • 6
  5. 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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/

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    • 5