Compound Interest


  1. A sum becomes ₹ 1,352 in 2 years at 4% per annum compound interest. The sum is









  1. View Hint View Answer Discuss in Forum

    Let the sum be ₹ p.
    Here , A = ₹ 1352 , r = 4% , n = 2 years
    Using the given formula ,

    A = P1 +
    r
    n
    100

    ∴  1352 = p 1 +
    4
    2
    100

    ⇒  1352 = p 1 +
    1
    2
    25

    Correct Option: D

    Let the sum be ₹ p.
    Here , A = ₹ 1352 , r = 4% , n = 2 years
    Using the given formula ,

    A = P1 +
    r
    n
    100

    ∴  1352 = p 1 +
    4
    2
    100

    ⇒  1352 = p 1 +
    1
    2
    25

    ⇒  1352 = p
    26
    2
    25

    ⇒  p =
    1352 × 25 × 25
    = ₹ 1250
    26 × 26


  1. At what rate percent per annum will ₹ 2304 amount to ₹ 2500 in 2 years at compound interest ?









  1. View Hint View Answer Discuss in Forum

    Let the rate percent per annum be r.
    Given that , P = ₹ 2304 , A = ₹ 2500 , n = 2 years
    Using the given formula ,

    A = P1 +
    r
    n
    100

    2500 = 23041 +
    r
    2
    100

    ⇒ 1 +
    r
    2 =
    2500
    =
    50
    2
    100230448

    ⇒  1 +
    r
    =
    50
    =
    25
    1004824

    ⇒ 
    r
    =
    25
    − 1 =
    1
    1002424

    Correct Option: C

    Let the rate percent per annum be r.
    Given that , P = ₹ 2304 , A = ₹ 2500 , n = 2 years
    Using the given formula ,

    A = P1 +
    r
    n
    100

    2500 = 23041 +
    r
    2
    100

    ⇒ 1 +
    r
    2 =
    2500
    =
    50
    2
    100230448

    ⇒  1 +
    r
    =
    50
    =
    25
    1004824

    ⇒ 
    r
    =
    25
    − 1 =
    1
    1002424

    ⇒  r =
    100
    =
    25
    = 4
    1
    %
    2466



  1. A sum of money on compound interest amounts to ₹ 10648 in 3 years and ₹ 9680 in 2 years. The rate of interest per annum is :









  1. View Hint View Answer Discuss in Forum

    Let the sum be ₹ P and rate of interest be R% per annum.
    Here , A1 = ₹ 10648 , A2 = ₹ 9680
    and t1 = 3 years , t2 = 2 years
    Using the given formula ,

    A = P1 +
    R
    n
    100

    Then,
    P1 +
    R
    2 = 9680    ...(i)
    100

    P1 +
    R
    3 = 10648    ...(ii)
    100

    On dividing equation (ii) by (i)
    1 +
    R
    =
    10648
    1009680

    ⇒ 
    R
    =
    10648
    − 1
    1009680

    =
    10648 − 9680
    9680

    Correct Option: B

    Let the sum be ₹ P and rate of interest be R% per annum.
    Here , A1 = ₹ 10648 , A2 = ₹ 9680
    and t1 = 3 years , t2 = 2 years
    Using the given formula ,

    A = P1 +
    R
    n
    100

    Then,
    P1 +
    R
    2 = 9680    ...(i)
    100

    P1 +
    R
    3 = 10648    ...(ii)
    100

    On dividing equation (ii) by (i)
    1 +
    R
    =
    10648
    1009680

    ⇒ 
    R
    =
    10648
    − 1
    1009680

    =
    10648 − 9680
    9680

    ⇒ 
    R
    =
    968
    =
    1
    100968010

    ⇒  R =
    1
    × 100 = 10%
    10


  1. The principal, which will amount to ₹ 270.40 in 2 years at the rate of 4% per annum compound interest, is









  1. View Hint View Answer Discuss in Forum

    Let the principal be ₹ P.
    Here , A = ₹ 270.40 , r = 4%, n = 2 years
    Using the given formula ,

    A = P1 +
    r
    n
    100

    ∴  270.40 = P1 +
    4
    2
    100

    ⇒  270.40 = P (1 + 0.04)2

    Correct Option: C

    Let the principal be ₹ P.
    Here , A = ₹ 270.40 , r = 4%, n = 2 years
    Using the given formula ,

    A = P1 +
    r
    n
    100

    ∴  270.40 = P1 +
    4
    2
    100

    ⇒  270.40 = P (1 + 0.04)2
    ⇒  P =
    270.40
    = ₹ 250
    1.04 × 1.04



  1. In what time will ₹ 1000 becomes ₹ 1331 at 10% per annum compounded annually ?









  1. View Hint View Answer Discuss in Forum

    Here , P = ₹ 1000, P1 = ₹ 1331 , r = 10%
    Let the required time be n years. Then,
    Using the given formula ,

    ∴  P1 = P 1 +
    r
    n
    100

    1331 = 1000 1 +
    10
    n
    100

    ⇒ 
    1331
    =
    10 + 1
    n
    100010

    Correct Option: A

    Here , P = ₹ 1000, P1 = ₹ 1331 , r = 10%
    Let the required time be n years. Then,
    Using the given formula ,

    ∴  P1 = P 1 +
    r
    n
    100

    1331 = 1000 1 +
    10
    n
    100

    ⇒ 
    1331
    =
    10 + 1
    n
    100010

    ⇒ 
    11
    n =
    11
    3
    1010

    Equating on both sides , we get
    ⇒  n = 3