Respuesta :
Answer:
[tex]\frac{13}{64}\ in[/tex] is the smallest but still larger than [tex]\frac{3}{16}\ in[/tex]
Step-by-step explanation:
Given:
Plumber needs to drill a hole slightly larger than [tex]\frac{3}{16}\ in[/tex] in diameter.
The fractions of lengths to compare are : [tex]\frac{5}{32}\ in,\frac{5}{16}\ in, \frac{13}{64}\ in, \frac{9}{32}\ in[/tex]
Solution:
Comparing all fractions by taking LCD.
[tex]\frac{3}{16}\ in,\frac{5}{32}\ in,\frac{5}{16}\ in, \frac{13}{64}\ in, \frac{9}{32}\ in[/tex]
Taking LCD = 64 and multiplying numerator and denominator of each fraction to make the denominators common.
[tex]\frac{3\times 4}{16\times 4}\ in,\frac{5\times 2}{32\times 2}\ in,\frac{5\times 4}{16\times 4}\ in, \frac{13}{64}\ in, \frac{9\times 2}{32\times 2}\ in[/tex]
[tex]\frac{12}{64}\ in,\frac{10}{64}\ in,\frac{20}{64}\ in, \frac{13}{64}\ in, \frac{18}{64}\ in[/tex]
Arranging the fractions in ascending order by comparing the numerators.
[tex]\frac{10}{64}\ in< \frac{12}{64}\ in<\frac{13}{64}\ in<\frac{18}{64}\ in<\frac{20}{64}\ in[/tex]
This can be arranged in the original form as:
[tex]\frac{5}{32}\ in<\frac{3}{16}\ in<\frac{13}{64}\ in< \frac{9}{32}\ in< \frac{5}{16}\ in[/tex]
Thus, we find out that [tex]\frac{13}{64}\ in[/tex] is the smallest but still larger than [tex]\frac{3}{16}\ in[/tex]
