Math, asked by shauryaasingh12, 8 months ago

An ore contains 8% zinc. How many kgs of the ore will be required to get 2.8 kg of zinc?

Answers

Answered by almir19
21

Answer:35kg

Step-by-step explanation:

Zinc in ore = 8%

Zinc in 1kg ore= 8/100x1kg=0.08kg

2.8/0.08 = 35

Therefore 35 kg ore is required

Answered by khuranaprisha2008
11

Answer:

35 kg

Step-by-step explanation:

Percentage of zinc present in the ore = 8%

Amount of zinc = 2.8 kg

Let the amount of ore be = x

Amount of ore required =

8% of x = 2.8 kg

⇒ 8/100 × x = 2.8

⇒ x = 28 × 100/8 × 10 = 35

∴ 35 kg of ore will be required to get 2.8 kg of zinc.

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