Respuesta :
Answer:
Yes, Tegan buys enough ribbon.
Step-by-step explanation:
Given:
Length of ribbon bought by Tegan = [tex]\frac{11}{6}[/tex] = [tex]1.83[/tex] meters
Ribbon needed to wrap her brother's present = [tex]\frac{2}{5}[/tex] meters
Ribbon needed to wrap her sister's present = [tex]\frac{3}{4}[/tex] meters
Let the length of ribbon needed for wrapping be 'x' meters
So,
⇒ [tex]x=\frac{2}{5} +\frac{3}{4}[/tex]
⇒ Taking LCD of [tex]5[/tex] and [tex]4[/tex] that is [tex]20[/tex].
⇒ [tex]x=\frac{(2\times 4)+(3\times 5)}{20}[/tex]
⇒ [tex]x=\frac{8+15}{20}[/tex]
⇒ [tex]x=\frac{23}{20}[/tex] meters
Comparing [tex]\frac{23}{20}[/tex] and [tex]\frac{11}{6}[/tex].
LCD of [tex]20[/tex] and [tex]6[/tex] is [tex]60[/tex].
Now equating the denominators.
⇒ [tex]\frac{23\times 3}{20\times 3} = \frac{69}{60}[/tex] and ⇒ [tex]\frac{11\times 10}{6\times 10} =\frac{110}{60}[/tex]
Now we can compare the above like fractions.
⇒ [tex]\frac{69}{60}[/tex] < [tex]\frac{110}{60}[/tex]
Shortcut:
We can directly compare the factions by cross multiplying.
⇒ [tex]\frac{23}{20} \ \ \ \ \ \ \ \ \ \frac{11}{6}[/tex]
⇒ [tex]23\times 6[/tex] , [tex]20\times 11[/tex]
⇒ [tex]138[/tex] and [tex]220[/tex]
⇒ [tex]138[/tex] < [tex]220[/tex]
Then
⇒ [tex]\frac{23}{20}[/tex] < [tex]\frac{11}{6}[/tex]
She has enough ribbon to wrap the presents as ribbon bought by Tegan is greater in length than the ribbon used.
