mahts

1)Peter claims that when a number x is increased by 10% and then decreased by 10%,

the value of x is unchanged. His calculation : x(1+10%-10%)=x Peter is wrong. Point out the mistake.

2)Find a two-digit number that satisfies the following condition:

(a) The tens-digit is greater than the unit-digit by 2.

(b)The square of the unit-digit is larger than twice the tens-digit.

A possible solution is 64. Would you suggest other solutions?

Must have step. If no step no mark will be given.

2 個解答

評分
  • 1 十年前
    最愛解答

    1.

    When a number x is increased by 10%, the new value will be (1+10%) x = 1.1x

    When a number y is decreased by 10%, the new value will be (1-10%) y = 0.9y

    So when a number x is increased by 10% then decreased by 10%, the new value should be (1+10%) (1-10%) x = 1.1 * 0.9 x = 0.99x

    The correct answer should be decreased by 1% instead of no change.

    2.

    Let the ten-digit be x and the unit-digit be y.

    x = y + 2 ………….equation 1

    y ^2 > 2x => y^2 – 2x > 0 …………….equation 2

    Let’s substitute equation 1 into equation 2

    y^2 – 2(y+2) > 0

    y^2 – 2y – 4 > 0 …………..equation 3

    From the question we know that both x and y are within the range 1 to 9. Since x = y + 2, so x must be in the range 3 to 9 while y must be in the range 1 to 7.

    By substituting various value of y in the range 1 to 7 into equation 3, we note that

    When y = 4, y^2 – 2y – 4 = 16 – 8 – 4 = 4 >0 …. Satisfies equation 3

    When y = 5, y^2 – 2y – 4 = 25 – 10 – 4 = 11 >0 …. Satisfies equation 3

    When y = 6, y^2 – 2y – 4 = 36 – 12 – 4 = 20 >0 …. Satisfies equation 3

    When y = 7, y^2 – 2y – 4 = 49 – 14 – 4 = 31 >0 …. Satisfies equation 3

    When y = 4, x = 6

    When y = 5, x = 7

    When y = 6, x = 8

    When y = 7, x = 9

    So the possible numbers are 64, 75, 86 and 97.

  • 1 十年前

    1) It is because when x increased by10%= x(1+10%)

    and then decreased by 10%=1.1x(1-10%)

    so the ans. should be 0.99x...not x

    2)a.42

    b.33

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