Number System


  1. The sum of all the 3-digit numbers, each of which on division by 5 leaves remainder 3, is









  1. View Hint View Answer Discuss in Forum

    According to the question,
    First number = a = 103
    Last number = l = 998
    ∴  If the number of such numbers be n, then,
    998 = 103 + (n – 1) × 5
    ⇒  (n – 1) × 5= 998 – 103 = 895

    ⇒  n − 1 =
    895
    = 179
    5

    ⇒  n = 180
    ∴  S =
    n
    (a + 1)
    2

    =
    180
    (103 + 998)
    2

    = 90 × 1101 = 99090

    Correct Option: D

    According to the question,
    First number = a = 103
    Last number = l = 998
    ∴  If the number of such numbers be n, then,
    998 = 103 + (n – 1) × 5
    ⇒  (n – 1) × 5= 998 – 103 = 895

    ⇒  n − 1 =
    895
    = 179
    5

    ⇒  n = 180
    ∴  S =
    n
    (a + 1)
    2

    =
    180
    (103 + 998)
    2

    = 90 × 1101 = 99090


  1. The sum of first 50 odd natural numbers is









  1. View Hint View Answer Discuss in Forum

    S = 1 + 3 + 5 + ..... to 50 terms Here, a = 1
    d = 3 – 1 = 2
    n = 50

    ∴  S =
    n
    [2a + (n − 1)d]
    2

    =
    50
    [2 × 1 + (50 − 1) × 2]
    2

    = 25 (2 + 98) = 25 × 100
    = 2500

    Correct Option: D

    S = 1 + 3 + 5 + ..... to 50 terms Here, a = 1
    d = 3 – 1 = 2
    n = 50

    ∴  S =
    n
    [2a + (n − 1)d]
    2

    =
    50
    [2 × 1 + (50 − 1) × 2]
    2

    = 25 (2 + 98) = 25 × 100
    = 2500



  1. The sum of all the natural numbers from 51 to 100 is









  1. View Hint View Answer Discuss in Forum

    As we know that the sum of N natural Numbers 1 + 2 + 3 ........ + n

    S =
    n ( n + 1)
    2

    Given in the question,
    ∴  51 + 52 + ..... + 100
    We can write as below
    ⇒  51 + 52 + ..... + 100 = (1 + 2 + 3 + 4 + 5 + ............ + 100) – ( 1 + 2 + 3 +..........+ 50)

    Correct Option: D

    As we know that the sum of N natural Numbers 1 + 2 + 3 ........ + n

    S =
    n ( n + 1)
    2

    Given in the question,
    ∴  51 + 52 + ..... + 100
    We can write as below
    ⇒  51 + 52 + ..... + 100 = (1 + 2 + 3 + 4 + 5 + ............ + 100) – ( 1 + 2 + 3 +..........+ 50)
    =
    100 × 101
    50 × 51
    22

    = 5050 – 1275 = 3775


  1. The sum of three consecutive numbers is 87. The middle number is









  1. View Hint View Answer Discuss in Forum

    Let us assume the three consecutive natural numbers be N, N + 1, N + 2.
    According to the question,
    N + N + 1 + N + 2 = 87

    Correct Option: B

    Let us assume the three consecutive natural numbers be N, N + 1, N + 2.
    According to the question,
    N + N + 1 + N + 2 = 87
    ⇒  3N + 3 = 87

    ⇒ 3N = 84 ⇒  N =
    84
    = 28
    3

    ∴  Middle number = 28 + 1 = 29



  1. Sum of three consecutive even integers is 54. Find the least among them.









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    Let us assume the 3 consecutive even integers be 2A, 2A + 2 and 2A + 4 respectively.
    According to given question,
    ∴  2A + 2x + 2 + 2x + 4 = 54

    Correct Option: D

    Let us assume the 3 consecutive even integers be 2A, 2A + 2 and 2A + 4 respectively.
    According to given question,
    ∴  2A + 2x + 2 + 2x + 4 = 54
    ⇒  6A + 6 = 54
    ⇒  6A = 54 – 6 = 48
    ⇒  A = 8
    ∴  The least even number = 2A = 2 × 8 = 16