A full tub of water weighs 3 1/8 pounds. If the tub is filled up only 1/7 full how much would it weigh?

Answers

Answer 1

Answer:

[tex]\frac{25}{56}[/tex] pounds

Step-by-step explanation:

By using unitary method,

∵ 1 full tub of water weighs = [tex]3\frac{1}{8}[/tex] pounds

∴ [tex]\frac{1}{7}[/tex] of a water tub will weigh = [tex]3\frac{1}{8}\times \frac{1}{7}[/tex]

                                               = [tex]\frac{(24+1)}{8}\times \frac{1}{7}[/tex]

                                               = [tex]\frac{25}{56}[/tex] pounds

Therefore, water filled in [tex]\frac{1}{7}[/tex]th part of the tub will weigh [tex]\frac{25}{56}[/tex] pounds.

Answer 2

The weight of the tub if filled up to 1/7 full is 25/56 pounds

How to find product of fraction

Given:

Full weight of tub = 31/8 pounds

If filled halfway:

= 1/7 × 3 1/8

= 1/7 × 25/8

= (1 × 25) / (7 × 8)

= 25/56 pounds

Therefore, the weight of the tub if filled to 1/7 of the capacity is 25/56 pounds

Learn more about fraction:

https://brainly.com/question/11562149


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