Respuesta :
Answer:
[tex]5[/tex] such strips of [tex]\frac{7}{10}\ m[/tex] can be cut and [tex]\frac{6}{10}\ m[/tex] would be left over.
Step-by-step explanation:
Given is[tex]4.1= \frac{41}{10}\ m[/tex] length of a wire.
We have to cut strips of [tex]\frac{7}{10}\ m[/tex]
If we factorize [tex]41 \ by \ 7[/tex]
We get [tex]5[/tex] full and [tex][/tex] and [tex]\frac{6}{7}[/tex]
Similarly , if we factorize [tex]\frac{41}{10} \ by \ \frac{7}{10}[/tex]
We get full [tex]5 \ strips[/tex] and another [tex]\frac{6}{10} \ m[/tex] would be left.
From 4.1 meters wire, we can make 5 stripes with 0.6 meters wire being left.
Solution:
Given that, an electrician has 4.1 meters of wire.
We have to find
1) Number of stripes 7/10m long can he cut:
Now, we know that, number of stripes he can make [tex]=\frac{\text {available length of wire}}{\text {Length of each stripe}}=\frac{4.1}{\frac{7}{10}}[/tex]
[tex]\Rightarrow \frac{4.1}{\frac{7}{10}}=4.1 \times \frac{10}{7}=\frac{41}{7}=5.857[/tex]
So, he can make 5 full stripes. We have to neglect fractional value as that is not considered as stripe.
2) Measure of left over wire:
No, we know that, remaining length of wire = total wire length-used length wire
[tex]\begin{array}{l}{\text { Length of left over wire }=4.1 \text { meters- number stripes used }\times \text {length of each stripe }} \\\\ {\text { Length of left over wire }=4.1-5 \times \frac{7}{10}=4.1-\frac{7}{2}=4.1-3.5=0.6 \text { meters }}\end{array}[/tex]
So, 0.6 meters of wire is left.
