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Test: Linear Inequalities- Case Based Type Questions - Commerce MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Linear Inequalities- Case Based Type Questions

Test: Linear Inequalities- Case Based Type Questions for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Linear Inequalities- Case Based Type Questions questions and answers have been prepared according to the Commerce exam syllabus.The Test: Linear Inequalities- Case Based Type Questions MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Linear Inequalities- Case Based Type Questions below.
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Test: Linear Inequalities- Case Based Type Questions - Question 1

A beaker contains 640 litres of 8% solution of Boric acid. This is to be diluted by adding a 2% Boric acid solution to it.

Based on the above data, answer the following question.

Q. If x litres of 2% Boric acid solution is added to the beaker, the amount of acid becomes ........... litres.

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 1

Total acid =

Test: Linear Inequalities- Case Based Type Questions - Question 2

A beaker contains 640 litres of 8% solution of Boric acid. This is to be diluted by adding a 2% Boric acid solution to it.

Based on the above data, answer the following question.

Q. If the resulting mixture is to be more than 4% but less than 6% boric acid, then the range of x is ..........

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 2

4(640 + x) ≤ 8 × 640 + 2x ≤ 6(640 + x)

2560 + 4x ≤ 5120 + 2x ≤ 3840 + 6x

2560 + 4x ≤ 5120 + 2x

⇒ 2x ≤ 2560

⇒ x ≤ 1280 ...(i)

5120 + 2x ≤ 3840 + 6x

⇒ – 4x ≤ –1280

⇒ 320 ≤ x ...(ii)

From (i) and (ii),

320 ≤ x ≤ 1280

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Test: Linear Inequalities- Case Based Type Questions - Question 3

A beaker contains 640 litres of 8% solution of Boric acid. This is to be diluted by adding a 2% Boric acid solution to it.

Based on the above data, answer the following question.

Q. The initial quantity of water in the beaker is ........... litres.

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 3
Amount of water = 640 – 51.2 = 588.8 litres
Test: Linear Inequalities- Case Based Type Questions - Question 4

A beaker contains 640 litres of 8% solution of Boric acid. This is to be diluted by adding a 2% Boric acid solution to it.

Based on the above data, answer the following question.

Q. The amount of water after the mixing is .......... litres.

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 4

Total water =

=

Test: Linear Inequalities- Case Based Type Questions - Question 5

A beaker contains 640 litres of 8% solution of Boric acid. This is to be diluted by adding a 2% Boric acid solution to it.

Based on the above data, answer the following question.

Q. The initial amount of acid in the beaker is ........... litres.

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 5

Amount of acid =

= 51.2 litres

Test: Linear Inequalities- Case Based Type Questions - Question 6

Kelvin(K), degree Celsius(°C) and degree Fahrenheit(°F) are three units of temperature. The conversion formula for them is as follows:

F = 9 / 5C + 32 and K = C + 273.15

Based on the above data, answer the following question.

Q. When F = 68°F, C = ..............

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 6

F = 68°F

C = 5 / 9 (F – 32)

= 5 / 9 (68 – 32) = 20°C

Test: Linear Inequalities- Case Based Type Questions - Question 7

Kelvin(K), degree Celsius(°C) and degree Fahrenheit(°F) are three units of temperature. The conversion formula for them is as follows:

F = 9 / 5C + 32 and K = C + 273.15

Based on the above data, answer the following question.

Q. For part (iv), the range of temperature in Kelvin(K) is ................

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 7
20° < C < 25°

20 + 273.15 < K < 25 + 273.15

⇒ 293.15 < K < 298.15

Test: Linear Inequalities- Case Based Type Questions - Question 8

Kelvin(K), degree Celsius(°C) and degree Fahrenheit(°F) are three units of temperature. The conversion formula for them is as follows:

F = 9 / 5C + 32 and K = C + 273.15

Based on the above data, answer the following question.

Q. When C = 120° F = .................

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 8
C = 120° F

= 9 / 5 × 120+ 32 = 248°F

Test: Linear Inequalities- Case Based Type Questions - Question 9

Kelvin(K), degree Celsius(°C) and degree Fahrenheit(°F) are three units of temperature. The conversion formula for them is as follows:

F = 9 / 5C + 32 and K = C + 273.15

Based on the above data, answer the following question.

Q. A solution is kept between 68°F and 77°F. What is the range in degree Celsius(C)?

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 9
Given 68° < F < 77 °

i.e; 68° < 9/5C + 32 < 77°

36° < 9/5C < 45°

20° < C < 25°

Test: Linear Inequalities- Case Based Type Questions - Question 10

Kelvin(K), degree Celsius(°C) and degree Fahrenheit(°F) are three units of temperature. The conversion formula for them is as follows:

F = 9 / 5C + 32 and K = C + 273.15

Based on the above data, answer the following question.

Q. The formula for C in terms of F is C = ...............

Detailed Solution for Test: Linear Inequalities- Case Based Type Questions - Question 10
Given F = 9 /5C + 32

9/5 C = F – 32

C = 5/9 (F – 32)

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