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Playing with Numbers Test - 11

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Playing with Numbers Test - 11
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Weekly Quiz Competition
  • Question 1
    1 / -0
    How many numbers are divisible by $$9$$?
    $$863, 267, 129, 774, 981, 997, 936$$
    Solution
    Here, 
    $$774=7+7+4=18, 981= 9+8+1=18, 936= 9+3+6=18$$ are divisible by $$9$$ because the sum of the digits of those numbers are a multiple of $$9$$.
    Therefore, there are $$3$$ numbers divisible by $$9$$.

    So, option B is correct.
  • Question 2
    1 / -0
    Determine the number into general form: $$90$$
    Solution
    The general form of any two digit number is, $$ab = a \times 10 + b$$
    So, $$9 \times 10 + 0$$ is in general form
    So, option A is correct.
  • Question 3
    1 / -0
    Subtract and find the value of $$B$$.
     $$1 0$$
    -$$ B$$
    ___
     $$ B$$
    Solution
       $$1 0$$
    $$- B$$
    ___
      $$ B$$
    The subtraction of $$10$$ and $$B$$ is giving $$5$$ i.e., $$10 - B = B$$
    The above condition is possible only when digit $$B$$ is $$5$$, because $$10 - 5 = 5.$$
    Therefore, the subtraction is as follows:
      $$ 1 0$$
      -$$ 5$$
     ____
      $$  5 $$
    The value of $$B = 5.$$
  • Question 4
    1 / -0
    Choose the correct answer if $$6 \times 10 + 9$$ is in general form. 
    Find its usual form.
    Solution
    The usual form of $$6 \times 10 +9 $$ $$=60+9 = 69$$
    So, option B is correct.
  • Question 5
    1 / -0
    The general form of $$302$$ is
    Solution
    The general form of any three digits number will be, $$abc = a \times 100 + b \times 10 + c$$
    Therefore, $$302= 3\times 100 +0\times10+ 2 \times 1$$
                              $$=3\times100+2\times1$$
  • Question 6
    1 / -0
    If $$4 \times 100 + 5 \times 10 + 0$$ is in generalised form. Find its usual form.
    Solution
    The general form of any three digit number will be, $$abc = a \times 100 + b \times 10 + c$$
    Here, $$4 \times 100 + 5 \times 10 + 0 = 450 $$
    So, option C is correct.
  • Question 7
    1 / -0
    Find the number for the generalised form: $$3 \times 100 + 0 \times 10 + 0$$
    Solution
    The general form of any three digits number is, $$abc = a \times 100 + b \times 10 + c$$
    Here, $$300 = 3 \times 100 + 0 \times 10 + 0$$
    So, option A is correct.
  • Question 8
    1 / -0
    Find the reverse number of the three digit number: $$276$$
    Solution
    The reversed digit will be $$672$$.
  • Question 9
    1 / -0
    In a two digit number, if the digit in units place is $$8$$ and the digit in tens place is $$y$$ then that number is __________.
    Solution

    Lets take an example of $$23$$
    The digit at units place$$=3$$
    The digit at tens place$$=2$$
    The number$$=2\times10+3=23$$
    In the question
    The digit at units place is $$8$$
    Thus, the number $$=y\times10+8=10y+8$$
  • Question 10
    1 / -0
    Which of the following is divisible by $$9$$?
    Solution
    Number are divisible by  $$9 $$ if sum of the all the digits present in it , is divisible by  $$9 $$.

    $$(a)75636$$
    $$\Rightarrow 7+5+6+3+6=27$$  and $$27 $$ is divisible by 9.
    Hence,  $$75636 $$ is divisible by  $$9 $$

    $$(b)89321$$
    $$\Rightarrow 8+9+3+2+1=23$$ and  $$23 $$ is not divisible  by  $$9 $$.
    So, $$ 89321 $$ is not divisible by  $$9 $$.

    $$(c)75637$$
    $$\Rightarrow 7+5+6+3+7=28$$ and  $$28 $$ is not divisible  by  $$9 $$.
    So,  $$75637 $$ is not divisible by  $$9 $$.

    $$(c)75632$$
    $$\Rightarrow 7+5+6+3+2=26$$ and  $$26 $$ is not divisible  by  $$9 $$.
    So,  $$75632 $$ is not divisible by  $$9 $$
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