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2017 MMC National Finals Grade 6 Individual Competition 15-second Questions

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  1. Four mangoes and 3 oranges cos Php 62.00. Four mangoes and 5 oranges cos Php 88.00. How much does one orange cost?
    [Sol]By subtracting 5 oranges and 4 mangoes to 3 oranges and 4 mangoes, the result will be 5 – 3 = 2 oranges. Thus, 2 \, \mathrm{oranges} = 88 -62 = \, \mathrm{Php}\, 26. Therefore, 1 orange is 26/2 = \boxed{ \mathrm{Php} \, 13.00}
  2. Forty three is 12.5% of what number? 

    [Sol]Notice that 12.5\% = \dfrac{1}{8}.
    Thus, 43 = \dfrac{1}{8} \times n .
    The number, therefore, is 8 \times 43 = \boxed{344}.

  3.  What is (4 + 4 - 4 \times 4 \div 4 )^4 ?[Sol]

  4.  If 12.8 \times 3.4 = 43.52 , what is 0.128 \times 0.34?[Sol]


  5.  How many positive divisors does the product (2^3)(3^4)(5^6) have? 

    [Sol](3 + 1) \times (4 + 1) \times (6 + 1) = \boxed{140}

  6.  What is the sum of all prime numbers from 1 to 20? 

    [Sol]2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = \boxed{77}

  7.  If January 1 is a Wednesday, what day is February 1? 

    [Sol]There are 31 days between January 1 and February 1. Thus, 31 divided by 7 gives remainder 3. Wednesday + 3 days is a \boxed{\mathrm{Saturday}}.

  8.  The average of 11 positive consecutive odd integers is 37. What is the smallest number? 

    [Sol]Since the difference of 2 odd numbers is 2, then the middle number should be subtracted by 5 \times 2 = 10. Since 11 is odd, then the middle number is its average. Therefore the smallest number is 37 - 10 = \boxed{27}.

  9.  What is 33 \frac{1}{3} \% of 60 + 55 \frac{5}{9} \%  of 45 + 37 \frac{1}{2} \% of 64 ? 

    [Sol]Converting the percentages into decimals, we get
    60 \left( \dfrac{1}{3} \right) + 45 \left( \dfrac{5}{9} \right) + 64 \left( \dfrac{3}{8} \right) = 20 + 25 + 24 = \boxed{69}

  10.  What is the product of \left(1 - \dfrac{1}{2} \right)\left(1 - \dfrac{1}{3} \right)\left(1 - \dfrac{1}{4} \right) \cdots \left(1 - \dfrac{1}{10} \right) ?[Sol]

    Simplifying the values inside the parentheses and by cancellation,
    \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} \dots \cdot \dfrac{9}{10} = \boxed{\dfrac{1}{10}}

 

Daniel James Agsaullo Molina

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