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

Published June 8, 2017 · Updated June 13, 2017

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 $26/2 = \boxed{ \mathrm{Php} \, 13.00}$
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 $8 \times 43 = \boxed{344}$ .
What is $(4 + 4 - 4 \times 4 \div 4 )^4$ ?[Sol]
If $12.8 \times 3.4 = 43.52$ , what is $0.128 \times 0.34$ ?[Sol]
How many positive divisors does the product $(2^3)(3^4)(5^6)$ have?
[Sol] $(3 + 1) \times (4 + 1) \times (6 + 1) = \boxed{140}$
What is the sum of all prime numbers from 1 to 20?
[Sol] $2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = \boxed{77}$
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 $\boxed{\mathrm{Saturday}}$ .
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 $37 - 10 = \boxed{27}$ .
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
$60 \left( \dfrac{1}{3} \right) + 45 \left( \dfrac{5}{9} \right) + 64 \left( \dfrac{3}{8} \right) = 20 + 25 + 24 = \boxed{69}$
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,
$\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} \dots \cdot \dfrac{9}{10} = \boxed{\dfrac{1}{10}}$