RS Aggarwal Class 10 Solutions Chapter 3 - Linear Equations In Two Variables - Ex 3I(3.9)

RS Aggarwal Class 10 Chapter 3 - Linear Equations In Two Variables - Ex 3I(3.9) Solutions Free PDF

The RS Aggarwal Class 10 solutions of Chapter 3 comprise of all the exercise question solutions presented in the RS Aggarwal maths textbook. These solutions are prepared by subject experts in sync to the latest syllabus of CBSE. These solutions help student while solving the exercise questions and can refer to it whenever they have any doubt. These solutions assist students to solve the difficult questions with effortless ease so that they can perform well in the exam.

Students with the help of these solutions can prepare effectively and score good percentile in the Class 10 exam. It explains different methods to solve difficult questions in a simple way so that students don’t get confused while solving the final question paper. It is important for students to understand the RS Aggarwal Solutions Class 10 Chapter 3 – Linear Equations In Two Variables Ex 3.1 for scoring good marks in the exam.

Download PDF of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3I(3.9)

Question 1: A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

We know,

Dividend = (divisor) (quotient) + remainder

According to the given question:

10x + y = 6(x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 ……..(1)

Number obtained by reversing the digits is 10y + x

10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) = 9

(x – y) = 1 ……..(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 ……..(3)

5x – 5y = 5 ……..(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1),we get

(4)(5) – 5y = 0

-5y = -20

y =4

Hence, the required number = 54.

Question 2: A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

Then,

xy = 35

Required number = 10x + y

Also,

10x + y + 18 = 10y + x

9x – 9y = -18

y – x = 2 ……..(1)

Now,

(y + x)2-(y – x)2= 4xy

\( Rightarrow y + x = \sqrt{(y – x)^{2} + 4xy} \) \( = sqrt{4 + 4 \times 35} \) \( = sqrt{144} \)

 

= 12

y + x = 12 …….. (2)

Adding (1) and (2), we get

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5(10) + 7 = 57

Question 3: A two-digit number is such that the product of its digits is 14. When 45 is added from the number, the digits interchange their places. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

Then,

xy = 14

Required number = 10x + y

Number obtained on reversing = 10y + x

Also,

(10x + y) + 45 = 10y + x

9(y – x) = 45

y – x = 5 ……..(1)

Now,

(y + x)2 – (y – x)2 = 4xy

\( \Rightarrow y + x = \sqrt{(y – x)^{2} + 4xy} \) \( = \sqrt{25 + 4 \times 14} \) \( = \sqrt{81} \) = 9

y + x = 9 …….. (2) [digits cannot be negative, hence -9 is not possible]

On adding (1) and (2), we get

2y = 14

y = 7

Putting y = 7 in (2), we get

7 + x = 9

x = (9 -7) = 2

x = 2 and y = 7

Hence, the required number is (2) (10) + 7 = 27.

Question 4: A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

Then,

xy = 18

Required number = 10x + y

Number obtained on reversing = 10y + x

Also,

(10x + y) – 63 = 10y + x

9(y – x) = 63

y – x = 7 ……..(1)

Now,

(y + x)2 – (y – x)2 = 4xy

\( \Rightarrow y + x = \sqrt{(y – x)^{2} + 4xy} \) \( = \sqrt{7^{2} + 4 \times 14} \) \( = \sqrt{121} \) = 11

y + x = 11 …….. (2) [Digits cannot be negative, hence -11 is not possible]

On adding (1) and (2), we get

2x = 18

x = 9

Putting x = 9 in (1), we get

9 – y = 7

y = (9 -7) = 2

x = 9 and y = 2

Hence, the required number is (9) (10) + 2 = 92.

Question 5: The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes 3/4. Find the number.

Solution:

Let the numerator and denominator of the fraction be x and y respectively.

According to the question:

x + y = 8 …….. (1)

And

\( \frac{x + 3}{y + 3} = \frac{3}{4} \) \( \Rightarrow 4x + 12 = 3y + 9 \) \( \Rightarrow 4x – 3y = -3 \) …….. (2)

Multiplying (1) by 3 and (2) by 1

3x + 3y = 24 …….. (3)

4x – 3y = -3 …….. (4)

Adding (3) and (4), we get

7x = 21

x = 3

Putting x = 3 in (!), we get

3 + y = 8

y = 5

x = 3, y = 5

Hence, the fraction is >\( \frac{x}{y} = \frac{3}{5} \).

Question 6: Find a fraction which becomes (1/2) when 1 is subtracted from the numerator and 2 is added to the denominator.

Solution:

Let the numerator and denominator of fraction be x and y respectively.

Then, the fraction is (x/y)

Therefore, >\( \frac{x – 1}{y + 2} = \frac{1}{2} \Rightarrow 2x – 2 = y + 2 \Rightarrow 2x – y = 4 \) ……..(2)

Subtracting (1) from (2), we get

(2)(15) – y = 4

30 – y = 4

y = 26

x = 15 and y = 26

Hence, the given fraction is >\( \frac{x}{y} = \frac{15}{26}\)

Question 7: The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes 3/4. Find the fraction.

Solution:

Let the numerator and denominator of the fraction be x and y respectively.

Then, the fraction is (x/y).

According to the given equation:

y = x + 11

y – x = 11 ……..(1)

\( \frac{x + 8}{y + 8} = \frac{3}{4} \Rightarrow 4x + 32 = 3y + 24 \Rightarrow 4x – 3y = -8 \)

-3y + 4x = -8 …….. (2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44 …….. (3)

-3y + 4x = -8 …….. (4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11

x = 25

x = 25, y = 36

Hence, the fraction is >\( \frac{25}{36} \)

Question 8: If 2 is added to the numerator of a fraction, it reduces to (1/2) and if 1 is subtracted from the denominator, it reduces to (1/3). Find the fraction.

Solution:

Let the numerator and denominator of fraction be x and y respectively.

Then, the fraction is (x/y).

According to the given equation:

\( \frac{x + 2} {y} = \frac{1}{2} \Rightarrow 2x + 4 = y \Rightarrow 2x – y = -4 \) …….(1)

And

\( \frac{x}{y – 1} = \frac{1}{3} \Rightarrow 3x = y – 1 \Rightarrow 3x – y = -1 \) …….. (2)

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

y = 10

x = 3, y = 10

Hence, the fraction is >\( \frac{3}{10} \)

Question 9: The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2:3. Determine the fraction.

Solution:

Let the fraction be >\( \frac{x}{y} \)

When 2 is added to both the numerator and denominator, the fraction becomes:

\( \frac{x + 2}{y + 2} = \frac{1}{3} \) or 3x + 6 = y + 2

3x – y = -4 …….. (1)

When 3 is added to both the numerator and the denominator, the fraction becomes:

\( \frac{x + 3}{y + 3} = \frac{2}{5} \) or 5x + 15 = 2y + 6

5x – 2y = -9 …….. (2)

Multiplying (1) by 2 and (2) by 1, we get

6x – 2y = -8 …….. (3)

Subtracting (4) from (3), we get

x = 1

Putting x = 1 in (1),

(3)(1) – y = 4

y =7

Required fraction = >\( \frac{x}{y} = \frac{1}{7} \)

Question 10: The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.

Solution:

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16 ……..(1)

And

\( \frac{1}{x} + \frac{1}{y} = \frac{1}{3} \) ……..(2)

From (2),

\( \frac{x + y} {xy} = \frac{1}{3} \, or \, \frac{16}{xy} = \frac{1}{3} \) [x + y = 16]

xy = 48

We know,

(x – y)2 = (x + y)2 – 4xy

= 162 – 4(48) = 256 – 192 = 64

x – y = 8 ……..(3)

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1).

y = 16 – x

y = 4

Hence, the required numbers are 12 and 4.

Question 11: Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages.

Solutions:

Let the present ages of the man and his son be x years and y rears respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)

x – 2 = 5y – 10

x – 5y = -8 ……..(1)

Two years later:

(x + 2) = 3(y + 2) + 8

x = 2 = 3y + 6 + 8

x – 3y = 12 ……..(2)

Subtracting (2) from (1), we get

-2y = -20

y =10

Putting y = 10 in (1), we get

x – 5(10) = -8

x = 42

Hence, the present ages of the man and his are 42 years and 10 years respectively.

Question 12: Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.

Solution:

Let the present ages of A and B be x and y respectively.

Five years ago:

(x – 5) = 3(y – 5)

x – 5 = 3y – 15

x – 3y = -10 ……..(1)

Ten years later:

(x + 10) = 2(y + 10)

x + 10 = 2y + 20

x – 2y = 10 ……..(2)

Subtracting (2) from (1), we get

y = 20

Putting y = 20 in (1), we get

x – 3y = -10

x – 3(20) = -10

x = -10 + 60 = 50

x =50, y = 20

Hence, the present ages of A and B are 50 years and 20 years respectively.

Question 13: The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.

Solution:

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3 ……..(1)

Three years later:

(x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

x – 2y = 13 ……..(2)

Subtracting (2) from (1), we get

y =10

Putting y = 10 in (1), we get

x – 3(10) = 3

x =33

x =33, y = 10

Hence, the present ages of the mother and her daughter are 33 and 10 years.

Key Features of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3I (3.9)

  • All the solutions are prepared by subject experts in a proper stepwise manner.
  • Students are advised to practice more and more questions from the RS Aggarwal maths solution.
  • It is one of the best resource study material for revision purpose as it covers the entire syllabus.
  • It enhances the problem-solving skills of students.

Practise This Question

 ______ is mating between two individuals with different alleles at one genetic locus of interest.

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