Selina solutions

Selina solutions

Grade 6

Perimeter and Area of Plane Figures | Exercise 32(A)

Question 13

Q13) Mohit makes 8 full rounds of a rectangular field with length = 120 m and breadth = 75 m. John makes 10 full rounds of a square field with each side 100 in. Find who covers larger distance and by how much?

Solution 13:

Mohit

Length of the rectangular field = 120

Breadth of the rectangular field = 75 m

∴ Distance covered in one round (perimeter) = 2(1 + b)

= 2(120 + 75) = 390 m Hence, distance covered in 8 rounds = 390 x 8 m = 3120 m

John

Side of the field = 100 m

∴ Distance covered in one round = 4 x a = 4 x 100 = 400 m

Hence, Distance covered in 10 rounds = 400 x 10 m = 400 m

John a covers greater distance then Mohit by = (4000-3120) m = 880

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subject-cta

Question 13

Q13) Mohit makes 8 full rounds of a rectangular field with length = 120 m and breadth = 75 m. John makes 10 full rounds of a square field with each side 100 in. Find who covers larger distance and by how much?

Solution 13:

Mohit

Length of the rectangular field = 120

Breadth of the rectangular field = 75 m

∴ Distance covered in one round (perimeter) = 2(1 + b)

= 2(120 + 75) = 390 m Hence, distance covered in 8 rounds = 390 x 8 m = 3120 m

John

Side of the field = 100 m

∴ Distance covered in one round = 4 x a = 4 x 100 = 400 m

Hence, Distance covered in 10 rounds = 400 x 10 m = 400 m

John a covers greater distance then Mohit by = (4000-3120) m = 880

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subject-cta
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