High School

10. Write 27,707,807 in words.

11. Use either <, >, or = to relate the fractions below.

a) \(\frac{2}{4}\) and \(\frac{3}{6}\)

b) \(\frac{60}{25}\) and \(\frac{60}{132}\)

12. State the place values of the following digits in the number 201.789.

a) 1

b) 8

c) 7

13. Use a factor tree to decompose 256 into prime factors.

14. Evaluate \(\left[\left(\frac{1}{4} - \frac{2}{3}\right) - \frac{5}{12} + \frac{1}{4}\right]\).

15. A car consumes \(8\frac{1}{8}\) liters of petrol to cover \(5\frac{1}{3}\) km. What average distance does it travel per liter?

16. Use a number line to perform the following operations.

a) \((-10) - (-3)\)

b) \((-3) - (-4)\)

c) \(+1 - (-8)\)

17.

a) A two-digit number is such that the ones digit is \(1\frac{1}{4}\) times greater than the tens digit. If the sum of the digits is 9, find the number.

b) Find the product of the positive difference and the maximum quotient between the digits.

Answer :

  1. The number 27,707,807 in words is: twenty-seven million, seven hundred seven thousand, eight hundred seven.

  2. To compare the fractions, we need to express them with a common denominator or simplify them.

a) [tex]\frac{2}{4}[/tex] and [tex]\frac{3}{6}[/tex]:
- Simplify both fractions:
[tex]\frac{2}{4} = \frac{1}{2}[/tex] and [tex]\frac{3}{6} = \frac{1}{2}[/tex].
- So, [tex]\frac{2}{4} = \frac{3}{6}[/tex].

b) [tex]\frac{60}{25}[/tex] and [tex]\frac{60}{132}[/tex]:
- Simplify [tex]\frac{60}{25} = \frac{12}{5}[/tex] and [tex]\frac{60}{132} = \frac{5}{11}[/tex].
- To compare, [tex]\frac{12}{5}[/tex] (which is greater than 2) is greater than [tex]\frac{5}{11}[/tex] (which is less than 1).
- So, [tex]\frac{60}{25} > \frac{60}{132}[/tex].

  1. Determining the place value for each digit in the number 201.789:

a) 1: The place value is the thousandths place.

b) 8: The place value is the hundredths place.

c) 7: The place value is the tenths place.

  1. To use a factor tree to decompose 256 into prime factors:

  • Start with 256 and keep dividing by 2 (the smallest prime number):
    • [tex]256 \div 2 = 128[/tex]
    • [tex]128 \div 2 = 64[/tex]
    • [tex]64 \div 2 = 32[/tex]
    • [tex]32 \div 2 = 16[/tex]
    • [tex]16 \div 2 = 8[/tex]
    • [tex]8 \div 2 = 4[/tex]
    • [tex]4 \div 2 = 2[/tex]
    • [tex]2 \div 2 = 1[/tex]
  • So, 256 = [tex]2^8[/tex].

  1. Evaluate [tex][\left(\frac{1}{4} - \frac{2}{3}\right) - \frac{5}{12} + \frac{1}{4}][/tex]:

  • First, compute [tex]\frac{1}{4} - \frac{2}{3}[/tex]:
    [tex]\frac{1}{4} = \frac{3}{12}, \quad \frac{2}{3} = \frac{8}{12}[/tex]
    [tex]\frac{1}{4} - \frac{2}{3} = \frac{3}{12} - \frac{8}{12} = -\frac{5}{12}[/tex]
  • Then, together with [tex]-\frac{5}{12}[/tex]:
    [tex]-\frac{5}{12} - \frac{5}{12} = -\frac{10}{12} = -\frac{5}{6}[/tex]
  • Finally add [tex]\frac{1}{4}[/tex]:
    [tex]-\frac{5}{6} + \frac{1}{4} = -\frac{10}{12} + \frac{3}{12} = -\frac{7}{12}[/tex]

  1. To find the average distance a car travels for every distance:

  • Given: [tex]8\frac{1}{8} = \frac{65}{8}[/tex] litres covers [tex]5\frac{1}{3} = \frac{16}{3}[/tex] km.
  • Average distance per litre = [tex]\frac{\frac{16}{3}}{\frac{65}{8}} = \frac{16}{3} \times \frac{8}{65} = \frac{128}{195}[/tex] km per litre.

  1. Performing operations using a number line:

a) [tex](-10) - (-3) = -10 + 3 = -7[/tex].

b) [tex](-3) - (-4) = -3 + 4 = 1[/tex].

c) [tex](+1) - (-8) = 1 + 8 = 9[/tex].

  1. a) Let's find the two-digit number:

    • Let the tens digit be [tex]x[/tex] and the ones digit [tex]1.25x[/tex].
    • Equation: [tex]x + 1.25x = 9[/tex].
    • [tex]2.25x = 9[/tex] so [tex]x = 4[/tex]. Therefore, the ones digit is [tex]5[/tex] (as [tex]1.25 \times 4 = 5[/tex]).
    • The number is 45.

b) Positive difference between the digits: [tex]5 - 4 = 1[/tex].

  • Maximum quotient by division: [tex]\frac{5}{4} = 1.25[/tex].
  • Product: [tex]1 \times 1.25 = 1.25[/tex].

Other Questions