Respuesta :

The given ratios expresses the number of time a value is larger or smaller

than another value.

The correct responses are;

  • 3. (B) 12 and 24
  • 4. (D) 6, 9, 21
  • 5. [tex]\underline{(E) \ \displaystyle \frac{11}{2}}[/tex]
  • 6. [tex]\underline{(E) \ \displaystyle \frac{5}{14}}[/tex]
  • 7. (B) [tex]\overline{EF}[/tex] = 6, [tex]\overline{AC}[/tex] = 9·√5
  • 8. (C) The values are equal
  • 9. (A) The value in column A is greater

Reasons:

3. Given that the perimeter of the rectangle = 72

The ratio of the lengths of the sides = 1:2

Let a and b represent the sides, we have;

2·a + 2·b = 72

[tex]\displaystyle \frac{a}{b} = \mathbf{\frac{1}{2}}[/tex]

Which gives;

2·a = b

2·a + 2·(2·a) = 72

6·a = 72

a = 72 ÷ 6 = 12

b = 2·a = 2 × 12 = 24

The lengths of the sides are; (B) 12 and 24

4. Extended ratio = 2:3:7

The perimeter = 36

The lengths of the sides are;

[tex]\displaystyle \frac{2}{2 + 3 + 7} \times 36 = \mathbf{6}[/tex]

[tex]\displaystyle \frac{3}{2 + 3 + 7} \times 36 = \mathbf{9}[/tex]

[tex]\displaystyle \frac{7}{2 + 3 + 7} \times 36 = \mathbf{21}[/tex]

The lengths are; (D) 6, 9, 21

5. The given equation is presented as follows;

[tex]\displaystyle \frac{5}{x + 7} = \mathbf{\frac{3}{x + 2}}[/tex]

5 × (x + 2) = 3 × (x + 7)

5·x + 10 = 3·x + 21

5·x - 3·x  = 21 - 10

2·x = 11

[tex]\displaystyle x = \mathbf{ \frac{11}{2}}[/tex]

The correct option is; [tex]\displaystyle \underline{(E) \ \frac{11}{2}}[/tex]

6. The width to length ratio is [tex]\displaystyle \mathbf{\frac{2.5}{7.0}}[/tex]

The simplified ratio is therefore;

[tex]\displaystyle \frac{2.5}{7.0} = \frac{2 \times 2.5}{2 \times 7.0} = \mathbf{\frac{5}{14}}[/tex]

The correct option is (E) [tex]\displaystyle \underline{(E) \ \frac{5}{14}}[/tex]

7. The given ratio of the lengths is 3:1

Therefore;

[tex]\overline{BC}[/tex]:[tex]\overline{EF}[/tex] = 3:1

Which gives;

[tex]\displaystyle \mathbf{\frac{\overline{BC}}{\overline{EF}}} = \frac{3}{1}[/tex]

[tex]\overline{BC}[/tex] = 18

Therefore;

[tex]\displaystyle \frac{18}{\overline{EF}} = \frac{3}{1}[/tex]

18 × 1 = 3 × [tex]\overline{EF}[/tex]

[tex]\displaystyle \overline{EF} = \frac{18 \times 1}{3} = 6[/tex]

[tex]\overline{EF}[/tex] = 6

By Pythagorean theorem, we have;

[tex]\overline{DF}[/tex]² = [tex]\mathbf{\overline{DE}}[/tex]² + [tex]\mathbf{\overline{EF}}[/tex]²

Which gives;

[tex]\overline{DF}[/tex]² = 3² + 6² = 45

[tex]\overline{DF}[/tex] = √(45) = 3·√5

Using the given ratio, we have;

[tex]\overline{AC}[/tex] = 3 × [tex]\mathbf{\overline{DF}}[/tex]

Which gives;

[tex]\overline{AC}[/tex] = 3 × 3·√5 = 9·√5

[tex]\overline{AC}[/tex] = 9·√5

The correct option is; (B) [tex]\overline{EF}[/tex] = 6, [tex]\overline{AC}[/tex] = 9·√5

8. EF = 1, AB = 2

CD = 2, CE = 4

Therefore;

[tex]\displaystyle \mathbf{\frac{EF}{AB}} =\frac{1}{2}[/tex]

[tex]\displaystyle \mathbf{\frac{CD}{CE}} =\frac{2}{4} = \frac{1}{2}[/tex]

Which gives;

[tex]\displaystyle \frac{EF}{AB} =\displaystyle \mathbf{\frac{CD}{CE}}[/tex]

(C) The values are equal

9. AC = 6, BE = 8

DF = 3, BD = 6

Column A

[tex]\displaystyle \frac{AC}{BE} =\displaystyle \mathbf{\frac{6}{8}} = \frac{3}{4}[/tex]

Column B

[tex]\displaystyle \frac{DF}{BD} =\displaystyle \mathbf{\frac{3}{6}} = \frac{1}{2}[/tex]

[tex]\displaystyle \frac{3}{4} > \mathbf{\frac{1}{2}}[/tex]

Therefore;

Column A is greater than column B

(A) The value in column A is greater

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