Selina solutions

Selina solutions

Grade 7

Fractions (Including Problems) | Exercise 3(C)

Question 10

Q10)A field is 16\ \frac{1}{2} m long and 12\frac{2}{5} m wide. Find the perimeter of the field.

Solution :

Length of the field = 16\frac{1}{2}m=\frac{33}{2}m

Breadth of the field = 12\frac{2}{5}m=\frac{62}{5}m

Now according to the question,

Perimeter of the field = 2(length + breadth)

=2\left(\frac{33}{2}+\frac{62}{2}\right)

=2\left(\frac{33\times5}{2\times5}+\frac{62\times2}{5\times2}\right) (∵ L.C.M. of 2 and 5 is 10)

=2\left(\frac{165}{10}+\frac{124}{10}\right)\

=2\left(\frac{165+124}{10}\right)\

=2\times\frac{289}{10}\

=\frac{289}{5}\ \

=57\frac{4}{5}m

Therefore, the perimeter of the field is 57\frac{4}{5}m.

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subject-cta

Question 10

Q10)A field is 16\ \frac{1}{2} m long and 12\frac{2}{5} m wide. Find the perimeter of the field.

Solution :

Length of the field = 16\frac{1}{2}m=\frac{33}{2}m

Breadth of the field = 12\frac{2}{5}m=\frac{62}{5}m

Now according to the question,

Perimeter of the field = 2(length + breadth)

=2\left(\frac{33}{2}+\frac{62}{2}\right)

=2\left(\frac{33\times5}{2\times5}+\frac{62\times2}{5\times2}\right) (∵ L.C.M. of 2 and 5 is 10)

=2\left(\frac{165}{10}+\frac{124}{10}\right)\

=2\left(\frac{165+124}{10}\right)\

=2\times\frac{289}{10}\

=\frac{289}{5}\ \

=57\frac{4}{5}m

Therefore, the perimeter of the field is 57\frac{4}{5}m.

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