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Chapter 15

Inequalities - Exercise 15(B)

Class - 7 RS Aggarwal Mathematics Solutions



Multiple Choice Questions

Question 1

If a > b and m < 0, then which of the following is correct :

  1. am < bm
  2. am = bm
  3. am > bm
  4. am and bm cannot be compared

Answer

Given:

a > b

m < 0 means m is negative.

Let's evaluate each:

  1. am < bm

Multiplying an inequality by a negative number (m < 0) requires reversing the inequality sign. It is correct.

  1. am = bm

Multiplying by a non-zero number (m < 0) maintains a difference between unequal values; they cannot become equal. It is incorrect.

  1. am > bm

This is incorrect because, m is negative number and multiplying an inequality by a negative number requires reversing the inequality sign. But here the sign remains same.

  1. am and bm cannot be compared

Since the relationship between a, b and the sign of m is known, their products are strictly comparable. It is incorrect.

Among all four options, option 1 is correct.

Hence, option 1 is the correct option.

Question 2

Which one of the following is a solution to the inequality 3x - 5 < 6?

  1. 3
  2. 4
  3. 5
  4. 6

Answer

Given:

3x - 5 < 6

⇒ 3x < 6 + 5 \quad [Adding 5 on both sides]

⇒ 3x < 11

⇒ x < 113[Dividing both sides by 3]\dfrac{11}{3} \quad \text{[Dividing both sides by 3]}

⇒ x < 3.66

Among all four options only 3 is less than 3.66

Hence, option 1 is the correct option.

Question 3

The solution set of the inequality 17 - 4x < 7, x ∈ Z is

  1. {1, 2, 3, ...............}
  2. {2, 3, 4, ...............}
  3. {3, 4, 5, ...............}
  4. {4, 5, 6, ...............}

Answer

Given:

17 - 4x < 7

⇒ 17 - 7 < 4x \quad [Subtracting 7 and adding 4x on both sides]

⇒ 10 < 4x

104\dfrac{10}{4} < x \quad [Dividing both sides by 4]

⇒ 2.5 < x

⇒ x > 2.5

Integers greater than 2.5 are {3, 4, 5, ...}

Hence, option 3 is the correct option.

Question 4

Which one of the following is not a solution to the inequality 2x > 18 - 5x?

  1. 7
  2. 5
  3. 3
  4. 1

Answer

Given:

2x > 18 - 5x

⇒ 2x + 5x > 18 \quad [Adding 5x on both sides]

⇒ 7x > 18

⇒ x > 187[Dividing both sides by 7]\dfrac{18}{7} \quad \text{[Dividing both sides by 7]}

⇒ x > 2.57

The solution must be greater than 2.57. Since 1 is not greater than 2.57, therefore 1 is not a solution.

Hence, option 4 is the correct option.

Question 5

Which one of the following statements is incorrect?

  1. If a < b, then a - m < b - m

  2. If a > b and m > 0, then am > bm

  3. If a < b and m > 0, then am\dfrac{a}{m} > bm\dfrac{b}{m}.

  4. If a ≠ 0 and b ≠ 0, then a > b ⇒ 1a\dfrac{1}{a} < 1b\dfrac{1}{b}.

Answer

Let's evaluate each:

  1. If a < b, then a - m < b - m
    Adding/Subtracting doesn't change the sign. So, it is correct.

  2. If a > b and m > 0, then am > bm
    Multiplying by a positive number keeps the sign. So, it is correct.

  3. If a < b and m > 0, then am\dfrac{a}{m} > bm\dfrac{b}{m}.
    Incorrect. If we divide by a positive number, the sign should remain the same. It should be am\dfrac{a}{m} < bm\dfrac{b}{m}.

  4. If a ≠ 0 and b ≠ 0, then a > b ⇒ 1a\dfrac{1}{a} < 1b\dfrac{1}{b}.
    Generally correct for positive numbers (Reciprocal rule).

Option 3 is the incorrect statement.

Hence, option 3 is the correct option.

Mental Maths

Question 1

Fill in the blanks :

(i) A statement of inequality between two expressions is called an ............... .

(ii) The set from which the values of the variable satisfying a given inequality are chosen, is called the ............... .

(iii) Subset of the replacement set, consisting of all those values of the variable which satisfy the given inequation is called the ............... .

(iv) Multiplying each side of an inequality by a negative number, ............... the inequality.

(v) If a ≠ 0, b ≠ 0 and a < b, then 1a\dfrac{1}{a} ............... 1b\dfrac{1}{b}.

Answer

(i) A statement of inequality between two expressions is called an inequation.

(ii) The set from which the values of the variable satisfying a given inequality are chosen, is called the replacement set.

(iii) Subset of the replacement set, consisting of all those values of the variable which satisfy the given inequation is called the solution set.

(iv) Multiplying each side of an inequality by a negative number, reverses the inequality.

(v) If a ≠ 0, b ≠ 0 and a < b, then 1a\dfrac{1}{a} > 1b\dfrac{1}{b}.

Explanation

(i) While an "equation" uses an equals sign (=), a statement using symbols like <, >, ≤, or ≥ is termed an inequation.

(ii) Replacement set is the "universe" of numbers (like Natural numbers N or Integers Z) from which you are allowed to pick potential answers.

(iii) Solution set is the specific group of numbers that actually make the inequation true. It is always a subset of the replacement set.

(iv) Multiplying by a negative value changes the direction of the sign (> becomes <). This is the most important rule in inequalities.

(v) For non-zero numbers, taking the reciprocal reverses the inequality. This is the Reciprocal Rule.

For example, if 2 < 5, then 12\dfrac{1}{2} > 15\dfrac{1}{5} (0.5 > 0.2).

Question 2

Write true (T) or false (F) :

(i) 23x45\dfrac{2}{3}x - \dfrac{4}{5} ≥ 8 is an inequation.

(ii) If a < b and m < 0, then am\dfrac{a}{m} > bm\dfrac{b}{m}.

(iii) If a < b, m < 0, then a - m > b - m.

(iv) If a > b and m < 0, then am < bm.

(v) If a > b and m > 0, then am\dfrac{a}{m} < bm\dfrac{b}{m}.

Answer

(i) True
Reason — Any mathematical statement that uses inequality symbols like <, >, ≤, or ≥ to compare two expressions is defined as an inequation.

(ii) True
Reason — This follows the Negative Division Rule. When we divide both sides of an inequality by a negative number (m < 0), the direction of the inequality sign must be reversed (< becomes >).

(iii) False
Reason — Adding or subtracting any number (whether positive or negative) from both sides of an inequality never changes the direction of the sign.

If a < b, then a - m < b - m remains true regardless of the value of m.

(iv) True
Reason — Similar to the division rule, multiplying by a negative number (m < 0) requires flipping the sign.

(v) False
Reason — When we divide by a positive number (m > 0), the inequality sign stays the same. Therefore, if a > b, the result should be am>bm\dfrac{a}{m} \gt \dfrac{b}{m}.

Case Study Based Questions

Question 1

Madan Singh runs a rental car company. He charges ₹ 250 per day plus ₹ 15 for every kilometre the car is driven. Professor Dayal rents a car for 1 day, while his own car is being repaired. He assures Madan Singh that he will pay him more than ₹ 500 as rent for the day.

(1) The inequality for the rent paid by Dayal for 1 day is :

  1. 3x < 100
  2. x > 25
  3. 3x > 50
  4. x < 75

(2) The solution set for the inequality obtained above is given by :

  1. {16, 17, 18, ...............}
  2. {17, 18, 19, ...............}
  3. {19, 20, 21, ...............}
  4. {20, 21, 22, ...............}

(3) Dayal estimated that the rent for 1 day would be less than ₹ 600 as he calculated the distance he has to drive the car. The inequality for the rent in this case would be :

  1. y > 30
  2. 2y < 35
  3. 3y < 70
  4. 4y > 45

(4) The solution set for the above inequality is given by :

  1. {..............., 20, 21, 22, 23}
  2. {..............., 17, 18, 19, 20}
  3. {..............., 18, 19, 20, 21}
  4. {..............., 15, 16, 17, 18}

Answer

(1)

Total Rent = (Fixed Daily Charge) + (Charge per Kilometre x distance)

Let the distance driven be x km.

Total Rent = ₹ 250 + 15x

The total rent is more than 500:

250 + 15x > 500

15x > 500 - 250 \quad [Subtracting 250 from both sides]

15x > 250

Divide both sides by 5:

3x > 50

Hence, option 3 is the correct option.

(2)

Let's solve 3x > 50:

3x > 50

x > 503\dfrac{50}{3} \quad [Dividing 3 from both sides]

x > 16.66..

The solution must be greater than 16.66..

Solution set = {17, 18, 19, ....}

Hence, option 2 is the correct option.

(3)

Let the distance driven be y km. The total rent is less than 600:

250 + 15y < 600

15y < 600 - 250 \quad [Subtracting 250 from both sides]

15y < 350

Divide both sides by 5:

3y < 70

Hence, option 3 is the correct option.

(4)

Let's solve 3y < 70:

3y < 70

y < 703\dfrac{70}{3} \quad [Dividing 3 from both sides]

y < 23.33..

The distance must be 23 km or less.

Solution set = {..............., 20, 21, 22, 23}

Hence, option 1 is the correct option.

Assertions and Reasons

Question 1

Assertion: If 2x + 3 > 8, then 2x + 3 - 3 > 8 - 3.

Reason: Subtracting a number from each side of an inequality reverses the inequality.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true but Reason (R) is false.
  4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is true but Reason (R) is false.

Explanation

Assertion:

Given inequality:

2x + 3 > 8

Subtract 3 from both sides:

2x + 3 - 3 > 8 - 3

2x > 5

∴ Assertion is true.

Subtracting or adding a number never reverses an inequality. Only multiplying or dividing by a negative number does that. Therefore, the Reason is false.

Hence, option 3 is the correct option.

Question 2

Assertion: The solution set of the inequality 2x - 1 > 7, x ∈ N is {1, 2, 3}.

Reason: Taking the reciprocal of each side of an inequality, reverses the inequality.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true but Reason (R) is false.
  4. Assertion (A) is false but Reason (R) is true.

Answer

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion:

Let's solve the given inequality:

2x - 1 > 7

2x > 7 + 1 \quad [Adding 1 on both sides]

2x > 8

x > 82[Dividing both sides by 2]\dfrac{8}{2} \quad \text{[Dividing both sides by 2]}

x > 4

Since x must be a Natural number (N) greater than 4, the solution set should be {5, 6, 7, ...}. So, Assertion is false.

The statement in reason is a standard rule of inequalities. For example, if 2 < 4, then taking the reciprocal gives 12\dfrac{1}{2} > 14\dfrac{1}{4} (0.5 > 0.25). The sign reverses. Therefore, the Reason is True.

Hence, option 4 is the correct option.

Competency Focused Questions

Question 1

A school requires that students purchase at least 3 but no more than 10 books for the academic year. Each book costs ₹ 250. The minimum and maximum amounts a student may spend are:

  1. ₹ 750 and ₹ 2500
  2. ₹ 700 and ₹ 2000
  3. ₹ 650 and ₹ 2500
  4. ₹ 750 and ₹ 2000

Answer

Let the number of books purchased by a student be x.

Given:

At least 3 books and no more than 10 books, so :

3 ≤ x ≤ 10

Cost per book = ₹ 250.

Total amount spent = ₹ 250x.

When x = 3:
Minimum amount = ₹ 250 × 3 = ₹ 750 \quad

When x = 10:
Maximum amount = ₹ 250 × 10 = ₹ 2500 \quad

Hence, option 1 is the correct option.

Question 2

Given the inequation 3x − 5 > 7 and the replacement set {2, 3, 4, 5}, which values satisfy the inequation?

  1. {2, 3}
  2. {3, 4, 5}
  3. {4, 5}
  4. None of these

Answer

Given:

3x - 5 > 7

⇒ 3x > 7 + 5 \quad [Adding 5 to both sides]

⇒ 3x > 12

⇒ x > 123[Dividing both sides by 3]\dfrac{12}{3} \quad \text{[Dividing both sides by 3]}

⇒ x > 4

From the replacement set {2, 3, 4, 5}, the value greater than 4 is {5}.

Since the solution set is {5}, it does not match any of the given options (1, 2 or 3).

Hence, option 4 is the correct option.

Question 3

A school bus can carry 40 students. If each student occupies x square feet and the total space available is less than 120 square feet, what is the maximum space per student?

  1. 2.5 sq. ft.
  2. 3 sq. ft.
  3. 3.5 sq. ft.
  4. 4 sq. ft.

Answer

Given:

Number of students = 40

Space occupied by each student = x sq. ft.

Total space occupied = 40x sq. ft.

Total space available is 120 sq. ft, so:

40x < 120

⇒ x < 12040[Dividing both sides by 40]\dfrac{120}{40} \quad \text{[Dividing both sides by 40]}

⇒ x < 3

The maximum space per student is less than 3 sq. ft.

Hence, option 2 is the correct option.

Question 4

Two friends are collecting marbles. Friend A collects 2x − 3 marbles, and friend B collects 3x + 5 marbles, together they collect fewer than 50 marbles. What is the maximum number of marbles friend A can collect?

  1. 15
  2. 17
  3. 18
  4. 20

Answer

Given:

Marbles collected by Friend A = 2x - 3

Marbles collected by Friend B = 3x + 5

Total marbles collected by both friends together is fewer than 50 :

(2x - 3) + (3x + 5) < 50

⇒ 5x + 2 < 50

⇒ 5x < 50 - 2 \quad [Subtracting 2 from both sides]

⇒ 5x < 48

⇒ x < 485[Dividing both sides by 5]\dfrac{48}{5} \quad \text{[Dividing both sides by 5]}

⇒ x < 9.6

Since the numbers of marbles collected by both friends must be whole numbers, we take the greatest whole-number value of x satisfying x < 9.6, which is 9.

∴ Maximum marbles collected by Friend A = 2(9) - 3 = 18 - 3 = 15.

Hence, option 1 is the correct option.

Question 5

A gym offers a membership plan where members must attend at least 5 classes but fewer than 15 classes per month. Each class costs ₹200. If a member spends y rupees on classes, which of the following options correctly represents the range of values y can take?

  1. ₹ 1,000 < y < ₹ 3,500
  2. ₹ 1,000 < y < ₹ 3,000
  3. ₹ 1,500 < y < ₹ 3,000
  4. ₹ 1,500 < y < ₹ 3,500

Answer

Let the number of classes attended in a month be n.

Given:

At least 5 classes but fewer than 15 classes, so :

5 ≤ n < 15

Cost per class = ₹ 200.

Total amount spent y = 200n.

Multiplying each side of the inequality by 200 :

5 × 200 ≤ 200n < 15 × 200

⇒ 1000 ≤ y < 3000

So, the amount y lies between ₹ 1,000 and ₹ 3,000.

Hence, option 2 is the correct option.

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