Ex 11.1, 7 - Chapter 11 Class 8 Direct and Inverse Proportions

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 11.1, 7 Suppose 2 kg of sugar contains 9 × 10^6 crystals. How many sugar crystals are there in (i) 5 kg of sugar?Given,
2 kg of sugar contains 9 × 10^6 crystals
Let 5 kg of sugar contains y crystals.
Thus, our table looks like
As quantity of sugar increases,
the number of crystals also increases.
∴ Quantity of Sugar and Number of Crystals are in direct proportion.
𝟐/(𝟗 × 〖𝟏𝟎〗^𝟔 ) = 𝟓/𝒚
𝑦 × 2 = 5 × 9 × 106
𝑦 × 2 = 45 × 106
𝒚 = 𝟒𝟓/𝟐 × 106
𝑦 = 22.5 × 106
𝑦 = 2.25 × 101 × 106
𝑦 = 2.25 × 101+6
𝒚 = 2.25 × 107
∴ 5kg of sugar contains 2.25 × 107 crystals
Ex 11.1, 7 Suppose 2 kg of sugar contains 9 × 10^6 crystals. How many sugar crystals are there in (ii) 1.2 kg of sugar?Given,
2 kg of sugar contains 9 × 10^6 crystals
Let 1.2 kg of sugar contains y crystals.
Thus, our table looks like
Since
Quantity of Sugar and Number of Crystals are in direct proportion.
𝟐/(𝟗 × 〖𝟏𝟎〗^𝟔 ) = (𝟏.𝟐)/𝒚
𝑦 × 2 = 1.2 × 9 × 106
𝑦 × 2 = 12/10 × 9 × 106
𝒚 = 𝟏/𝟐 × 𝟏𝟐/𝟏𝟎 × 9 × 106
𝑦 = 6/10 × 9 × 106
𝑦 = 54/10 × 106
𝒚 = 5.4 × 106
∴ 1.2 kg of sugar contains 5.4 × 106 crystals

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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