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Mathematics

Case Study III

In drilling world’s deepest hole, the Kola Superdeep Borehole, the deepest man made hole on the earth, it was found that the temperature T in degree Celsius, x km below the earth’s surface was given by, T = 30 + 25(x - 3) and 3 ≤ x ≤ 15. If the temperature lies between 180° C to 330° C, then based on this information, answer the following questions.

1. The linear inequation for the depth of the hole is:

  1. 180 < 30 + 25(x − 3) < 330

  2. 180 ≤ 30 + 25(x − 3) ≤ 330

  3. 330 < 30 + 25(x − 3) ≤ 180

  4. 330 < 30 + 25(x − 3) < 180

2. The solution set for the depth is :

  1. {x ∈ R : 6 ≤ x ≤ 12}

  2. {x ∈ R : 9 ≤ x ≤ 12}

  3. {x ∈ R : 3 ≤ x ≤ 15}

  4. {x ∈ R : 9 ≤ x ≤ 15}

3. The minimum possible depth of the hole for the given temperature range is:

  1. 3 km

  2. 6 km

  3. 9 km

  4. cannot be determined

4. The maximum possible depth of the hole for the given temperature range is:

  1. 9 km

  2. 12 km

  3. 15 km

  4. None of these

5. Which of the following is the graphical representation of the solution set for the depth of the hole for the given temperature range?

a.

Which of the following is the graphical representation of the solution set for the depth of the hole for the given temperature range? Linear Inequations, RSA Mathematics Solutions ICSE Class 10.

b.

Which of the following is the graphical representation of the solution set for the depth of the hole for the given temperature range? Linear Inequations, RSA Mathematics Solutions ICSE Class 10.

c.

Which of the following is the graphical representation of the solution set for the depth of the hole for the given temperature range? Linear Inequations, RSA Mathematics Solutions ICSE Class 10.

d.

Which of the following is the graphical representation of the solution set for the depth of the hole for the given temperature range? Linear Inequations, RSA Mathematics Solutions ICSE Class 10.

Linear Inequations

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Answer

1. Given,

T = 30 + 25(x - 3)

The temperature lies between 180° C to 330° C.

⇒ 180 ≤ 30 + 25(x − 3) ≤ 330

Hence, Option 2 is the correct option.

2. Solving,

⇒ 180 ≤ 30 + 25(x − 3) ≤ 330

Solving L.H.S of inequation,

⇒ 180 ≤ 30 + 25(x − 3)

⇒ 30 + 25(x − 3) ≥ 180

⇒ 30 + 25x - 75 ≥ 180

⇒ 25x - 45 ≥ 180

⇒ 25x ≥ 180 + 45

⇒ 25x ≥ 225

⇒ x ≥ 22525\dfrac{225}{25}

⇒ x ≥ 9 ……..(1)

Solving R.H.S of inequation,

⇒ 30 + 25(x − 3) ≤ 330

⇒ 30 + 25x - 75 ≤ 330

⇒ 25x - 45 ≤ 330

⇒ 25x ≤ 330 + 45

⇒ 25x ≤ 375

⇒ x ≤ 37525\dfrac{375}{25}

⇒ x ≤ 15 ……..(2)

From (1) and (2) we get,

⇒ 9 ≤ x ≤ 15

Since x ∈ R

Solution set = {x ∈ R : 9 ≤ x ≤ 15}

Hence, Option 4 is the correct option.

3. The minimum possible depth of the hole for the given temperature range is the minimum value in the interval 9 ≤ x ≤ 15 that is 9 km.

Hence, Option 3 is the correct option.

4. The maximum possible depth of the hole for the given temperature range is the maximum value in the interval 9 ≤ x ≤ 15 that is 15 km.

Hence, Option 3 is the correct option.

5. Solution set for depth of hole = {x ∈ R : 9 ≤ x ≤ 15}

Which of the following is the graphical representation of the solution set for the depth of the hole for the given temperature range? Linear Inequations, RSA Mathematics Solutions ICSE Class 10.

Hence, Option a is the correct option.

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