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ICSE-VIII-Mathematics

05: Playing with Number Class 8 Maths

with Solutions - page 2

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  • Qstn #6
    If a = 6, show that abc = bac.
    Ans : Given: a = 6
    To show: abc = bac
    Proof:abc = 100a + 106 + c ...(i)
    (By using property 3)
    bac = 1006 + 10a + c ...(ii)
    (By using property 3)
    Since, a = 6
    Substitute the value of a = 6 in equation (i) and (ii), we get
    abc = 1006 + 106 + c ...(iii)
    bac = 1006 + 106 + c ...(iv)
    Subtracting (iv) from (iii) abc - bac = 0
    abc = bac
    Hence, proved.
  • Qstn #7
    if a > c; show that abc - cba = 99(a - c).
    Ans : Given, a > c
    To show : abc - cba = 99(a - c)
    Proof:abc = 100a + 10b + c ...(i)
    (By using property 3)
    cba = 100c + 10b + a ...(i)
    (By using property 3)
    Subtracting, equation (ii) from (i), we get
    abc - cba = 100a + c - 100c - a
    ⇒ abc - cba = 99a - 99c
    ⇒ abc - cba = 99(a - c)
    Hence, proved.
  • Qstn #8
    If c > a; show that cba - abc = 99(c - a).
    Ans : Given, c > a
    To show : cba - abc = 99(c - a)
    Proof:
    cba = 100c + 106 + a ...(i)
    (by using property 3)
    abc = 100a + 106 + c ...(ii)
    (by using property 3)
    Subtracting (ii) from (i)
    cba - abc = 100c + 106 + a - 100a - 106c
    ⇒ cba - abc = 99c - 99a
    ⇒ cba - abc = 99(c - a)
    Hence proved.
  • Qstn #9
    If a = c, show that cba- abc = 0
    Ans : Given: a = c
    To show: cba - abc = 0
    Proof:
    cba = 100c + 106 + a ...(i)
    (By using property 3)
    abc = 100a + 106 + c ...(ii)
    (By using property 3)
    Since, a = c,
    Substitute the value of a = c in equation (i) and (ii), we get
    cba = 100c + 10b + c ...(iii)
    abc = 100c + 10b + c ...(iv)
    subtracting (iv) from (iii), we get
    cba - abc - 100c + 106 + c - 100c - 106 - c
    ⇒ cba - abc = 0
    ⇒ cba = abc
    Hence, proved.
  • Qstn #10
    Show that 954 - 459 is exactly divisible by 99.
    Ans : To show: 954 - 459 is exactly divisible by 399, where a = 9, b = 5, c = 4,abc = 100a + 10b + c
    ⇒ 954 = 100 × 9 + 10 × 5 + 4
    ⇒ 954 = 900 + 50 + 4 ...(i)
    And 459 = 100 × 4 + 10 × 5 + 9
    ⇒ 459 = 400 + 50 + 9 ...(ii)
    Subtracting (ii) from (i), we get
    954 - 59 = 900 + 50 + 4 - 400 - 50 - 9
    ⇒ 954 - 459 = 500 - 5
    ⇒ 954 - 459 = 495
    ⇒ 954 - 458 = 99 × 5
    Hence, 954 - 459 is exactly divisible by 99
    Hence proved.
  • #
    Section : B
  • Qstn #1
    Answer
    A = 7 as 7 + 5 = 12.We want 2 at units place and 1 is carry over. Now 3 + 2 + 1 = 6
    B = 6
    Hence, A = 7 and B = 6
  • Qstn #2
    dlt26083680321513838123">AnswerA = 5 as 8 + 5 = 13.We want 3 at units place and 1 is carry over. Now 9 + 4 + 1 = 14.
    B = 4 and C = 1
    Hence A = 5 and B = 4 and C = 1
  • Qstn #3
    Answer
    B = 9 as 9 + 1 = 10.We want 0 at units place and 1 is carry over. Now B - 1 - 1 = A.
    ∴ A = 9 - 2 = 7
    Hence A = 7 and B = 9
  • Qstn #4
    AnswerB = 7 as 7 + 1 = 8. We want 8 at unit place.
    Now,
    7 + A = 11
    ∴ A = 11 - 7 = 4
    Hence A = 4 and B = 7
  • Qstn #5
    Answer
    A + B = 9
    And 2 + A = 10
    ∴ A = 10 - 2= 8
    And 8 + B = 9
    ∴ B = 9 - 8 = 1
    Hence, A = 8 and B = 1
  • Qstn #6
    Answer
    As we need A at unit place and 9 at ten’s place,
    A = 6 as 6 × 6 = 36
  • Qstn #7
    Answer
    As we need B at unit place and B at ten’s place,
    ∴ B = 4 as 6 × 4 = 24
    Now, we want to find A, 6 × A + 2 = 4 (at unit’s place)
    ∴ A = 7
  • Qstn #8
    AnswerAs we need B at unit place and A at ten’s place,
    ∴ B = 0 as 3 × 0 = 0
    Now, we want to find A, 3 × A (at unit’s place)
    ∴ A = 5, as 3 × 5 = 15
    ∴ C = 1
  • Qstn #9
    Answer
    As we need B at unit place and A at ten's place,
    B = 0 as 5 × 0 = 0
    Now, we want to find A, 5 × A = A (at unit's place)
    A = 5, as 5 × 5 = 25
    C = 2
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