Advanced Mathematics Chapter 3.1 Solutions

SEBA (New Book) Class 10 Maths Chapter 3.3 Solutions

(i) $x + y = 5$ and $2x – 3y = 4$

Solution:

Elimination Method:

$x + y = 5 \quad \text{— (i)}$

$2x – 3y = 4 \quad \text{— (ii)}$

$(i) \times 3 \implies 3x + 3y = 15 \quad \text{— (iii)}$

$(ii) \times 1 \implies 2x – 3y = 4 \quad \text{— (iv)}$

Adding (iii) & (iv) we get

$(3x + 3y) + (2x – 3y) = 15 + 4$

$\implies 5x = 19$

$\implies x = \frac{19}{5}$

Substitute $x = \frac{19}{5}$ into equation (i):

$\frac{19}{5} + y = 5$

$\implies y = 5 – \frac{19}{5}$

$\implies y = \frac{6}{5}$

Answer: $x = \frac{19}{5}, y = \frac{6}{5}$

Substitution Method:

The given Equations:

$x + y = 5 \quad \text{— (i)}$

$2x – 3y = 4 \quad \text{— (ii)}$

From Equation (i), we have

$x = 5 – y \quad \text{— (iii)}$

Substituting the value of $x = 5 – y$ in Equation (ii).

$2(5 – y) – 3y = 4$

$\implies 10 – 2y – 3y = 4$

$\implies 10 – 5y = 4$

$\implies -5y = 4 – 10$

$\implies -5y = -6$

$\implies y = \frac{-6}{-5}$

$∴ y = \frac{6}{5}$

Putting the value of $y = \frac{6}{5}$ in Equation (iii), we have

$x = 5 – \frac{6}{5}$

$\implies x = \frac{25 – 6}{5}$

$∴ x = \frac{19}{5}$

Final Answer:

The solution to the given pair of linear equations is:

$x = \frac{19}{5}, \quad y = \frac{6}{5}$

(ii) $3x + 4y = 10$ and $2x – 2y = 2$

Solution:

Elimination Method:

$3x + 4y = 10 \quad \text{— (i)}$

$2x – 2y = 2 \quad \text{— (ii)}$

$(i) \times 1 \implies 3x + 4y = 10 \quad \text{— (iii)}$

$(ii) \times 2 \implies 4x – 4y = 4 \quad \text{— (iv)}$

Adding (iii) & (iv) we get

$(3x + 4y) + (4x – 4y) = 10 + 4$

$\implies 7x = 14$

$\implies x = \frac{14}{7}$

$∴ x = 2$

Substitute $x = 2$ into equation (i):

$3(2) + 4y = 10$

$\implies 6 + 4y = 10$

$\implies 4y = 10 – 6$

$\implies 4y = 4$

$\implies y = 1$

Answer: $x = 2, y = 1$

Substitution Method:

The given Equations:

$3x + 4y = 10 \quad \text{— (i)}$

$2x – 2y = 2 \quad \text{— (ii)}$

From Equation (ii), we have

$2x = 2 + 2y$

$\implies x = 1 + y \quad \text{— (iii)}$

Substituting the value of $x = 1 + y$ in Equation (i).

$3(1 + y) + 4y = 10$

$\implies 3 + 3y + 4y = 10$

$\implies 3 + 7y = 10$

$\implies 7y = 10 – 3$

$\implies 7y = 7$

$\implies y = \frac{7}{7}$

$∴ y = 1$

Putting the value of $y = 1$ in Equation (iii), we have

$x = 1 + 1$

$∴ x = 2$

Final Answer:

The solution to the given pair of linear equations is:

$x = 2, \quad y = 1$

(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$

Solution:

Elimination Method:

Rearranging the equations into general form:

$3x – 5y = 4 \quad \text{— (i)}$

$9x – 2y = 7 \quad \text{— (ii)}$

$(i) \times 3 \implies 9x – 15y = 12 \quad \text{— (iii)}$

$(ii) \times 1 \implies 9x – 2y = 7 \quad \text{— (iv)}$

Subtracting (iv) from (iii) we get

$(9x – 15y) – (9x – 2y) = 12 – 7$

$\implies 9x – 15y – 9x + 2y = 5$

$\implies -13y = 5$

$\implies y = -\frac{5}{13}$

Substitute $y = -\frac{5}{13}$ into equation (i):

$3x – 5\left(-\frac{5}{13}\right) = 4$

$\implies 3x + \frac{25}{13} = 4$

$\implies 3x = 4 – \frac{25}{13}$

$\implies 3x = \frac{52 – 25}{13}$

$\implies 3x = \frac{27}{13}$

$\implies x = \frac{27}{13 \times 3}$

$∴ x = \frac{9}{13}$

Answer: $x = \frac{9}{13}, y = -\frac{5}{13}$

Substitution Method:

The given Equations:

$3x – 5y = 4 \quad \text{— (i)}$

$9x – 2y = 7 \quad \text{— (ii)}$

From Equation (i), we have

$3x = 4 + 5y$

$\implies x = \frac{4 + 5y}{3} \quad \text{— (iii)}$

Substituting the value of $x = \frac{4 + 5y}{3}$ in Equation (ii).

$9\left(\frac{4 + 5y}{3}\right) – 2y = 7$

$\implies 3(4 + 5y) – 2y = 7$

$\implies 12 + 15y – 2y = 7$

$\implies 12 + 13y = 7$

$\implies 13y = 7 – 12$

$\implies 13y = -5$

$∴ y = -\frac{5}{13}$

Putting the value of $y = -\frac{5}{13}$ in Equation (iii), we have

$x = \frac{4 + 5\left(-\frac{5}{13}\right)}{3}$

$\implies x = \frac{4 – \frac{25}{13}}{3}$

$\implies x = \frac{\frac{52 – 25}{13}}{3}$

$\implies x = \frac{27}{13 \times 3}$

$∴ x = \frac{9}{13}$

Final Answer:

The solution to the given pair of linear equations is:

$x = \frac{9}{13}, \quad y = -\frac{5}{13}$

(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$

Solution:

To simplify fractions, multiply the first equation by 6 (LCM of 2 and 3) and the second equation by 3 (LCM of 1 and 3).

$3x + 4y = -6 \quad \text{— (i)}$

$3x – y = 9 \quad \text{— (ii)}$

Elimination Method:

$3x + 4y = -6 \quad \text{— (i)}$

$3x – y = 9 \quad \text{— (ii)}$

$(i) \times 1 \implies 3x + 4y = -6 \quad \text{— (iii)}$

$(ii) \times 4 \implies 12x – 4y = 36 \quad \text{— (iv)}$

Adding (iii) & (iv) we get

$(3x + 4y) + (12x – 4y) = -6 + 36$

$\implies 15x = 30$

$\implies x = \frac{30}{15}$

$∴ x = 2$

Substitute $x = 2$ into equation (ii):

$3(2) – y = 9$

$\implies 6 – y = 9$

$\implies -y = 9 – 6$

$\implies -y = 3$

$\implies y = -3$

Answer: $x = 2, y = -3$

Substitution Method:

The given Equations:

$3x + 4y = -6 \quad \text{— (i)}$

$3x – y = 9 \quad \text{— (ii)}$

From Equation (ii), we have

$-y = 9 – 3x$

$\implies y = 3x – 9 \quad \text{— (iii)}$

Substituting the value of $y = 3x – 9$ in Equation (i).

$3x + 4(3x – 9) = -6$

$\implies 3x + 12x – 36 = -6$

$\implies 15x – 36 = -6$

$\implies 15x = -6 + 36$

$\implies 15x = 30$

$\implies x = \frac{30}{15}$

$∴ x = 2$

Putting the value of $x = 2$ in Equation (iii), we have

$y = 3(2) – 9$

$\implies y = 6 – 9$

$∴ y = -3$

Final Answer:

The solution to the given pair of linear equations is:

$x = 2, \quad y = -3$

(v) $\frac{3y}{2} – \frac{5x}{3} = -2$ and $\frac{y}{3} + \frac{x}{3} = \frac{13}{16}$

Solution:

Given Equations are:

$\frac{3y}{2} – \frac{5x}{3} = -2 \quad \text{— (i)}$

$\frac{y}{3} + \frac{x}{3} = \frac{13}{16} \quad \text{— (ii)}$

To clear the fractions, multiply equation (i) by 6 (LCM of 2 and 3) and equation (ii) by 48 (LCM of 3 and 16):

$\implies 3(3y) – 2(5x) = -12 \implies -10x + 9y = -12 \quad \text{— (1)}$

$\implies 16(y) + 16(x) = 3(13) \implies 16x + 16y = 39 \quad \text{— (2)}$

Elimination Method:

$-10x + 9y = -12 \quad \text{— (1)}$

$16x + 16y = 39 \quad \text{— (2)}$

$(1) \times 16 \implies -160x + 144y = -192 \quad \text{— (3)}$

$(2) \times 9 \implies 144x + 144y = 351 \quad \text{— (4)}$

Subtracting (3) from (4) we get

$(144x + 144y) – (-160x + 144y) = 351 – (-192)$

$\implies 144x + 144y + 160x – 144y = 351 + 192$

$\implies 304x = 543$

$\implies x = \frac{543}{304}$

Substitute $x = \frac{543}{304}$ into equation (2):

$16\left(\frac{543}{304}\right) + 16y = 39$

$\implies \frac{543}{19} + 16y = 39$

$\implies 16y = 39 – \frac{543}{19}$

$\implies 16y = \frac{741 – 543}{19}$

$\implies 16y = \frac{198}{19}$

$\implies y = \frac{198}{19 \times 16}$

$\implies y = \frac{99}{152}$

Answer: $x = \frac{543}{304}, y = \frac{99}{152}$

Substitution Method:

The given Equations:

$-10x + 9y = -12 \quad \text{— (1)}$

$16x + 16y = 39 \quad \text{— (2)}$

From Equation (1), we have

$9y = -12 + 10x$

$\implies y = \frac{10x – 12}{9} \quad \text{— (3)}$

Substituting the value of $y = \frac{10x – 12}{9}$ in Equation (2).

$16x + 16\left(\frac{10x – 12}{9}\right) = 39$

$\implies 16x + \frac{160x – 192}{9} = 39$

Multiplying the entire equation by 9 to clear the fraction:

$\implies 144x + 160x – 192 = 351$

$\implies 304x – 192 = 351$

$\implies 304x = 351 + 192$

$\implies 304x = 543$

$∴ x = \frac{543}{304}$

Putting the value of $x = \frac{543}{304}$ in Equation (3), we have

$y = \frac{10\left(\frac{543}{304}\right) – 12}{9}$

$\implies y = \frac{\frac{5430}{304} – 12}{9}$

$\implies y = \frac{\frac{5430 – 3648}{304}}{9}$

$\implies y = \frac{1782}{304 \times 9}$

$\implies y = \frac{198}{304}$

$∴ y = \frac{99}{152}$

Final Answer:

The solution to the given pair of linear equations is:

$x = \frac{543}{304}, \quad y = \frac{99}{152}$

(vi) $x – y = 3$ and $\frac{x}{3} + \frac{y}{2} = 6$

Solution:

Multiply the second equation by 6 to clear the fractions:

$2x + 3y = 36$

Elimination Method:

$x – y = 3 \quad \text{— (i)}$

$2x + 3y = 36 \quad \text{— (ii)}$

$(i) \times 3 \implies 3x – 3y = 9 \quad \text{— (iii)}$

$(ii) \times 1 \implies 2x + 3y = 36 \quad \text{— (iv)}$

Adding (iii) & (iv) we get

$(3x – 3y) + (2x + 3y) = 9 + 36$

$\implies 5x = 45$

$\implies x = \frac{45}{5}$

$∴ x = 9$

Substitute $x = 9$ into equation (i):

$9 – y = 3$

$\implies -y = 3 – 9$

$\implies -y = -6$

$\implies y = 6$

Answer: $x = 9, y = 6$

Substitution Method:

The given Equations:

$x – y = 3 \quad \text{— (i)}$

$2x + 3y = 36 \quad \text{— (ii)}$

From Equation (i), we have

$x = 3 + y \quad \text{— (iii)}$

Substituting the value of $x = 3 + y$ in Equation (ii).

$2(3 + y) + 3y = 36$

$\implies 6 + 2y + 3y = 36$

$\implies 6 + 5y = 36$

$\implies 5y = 36 – 6$

$\implies 5y = 30$

$\implies y = \frac{30}{5}$

$∴ y = 6$

Putting the value of $y = 6$ in Equation (iii), we have

$x = 3 + 6$

$∴ x = 9$

Final Answer:

The solution to the given pair of linear equations is:

$x = 9, \quad y = 6$

2. Form the pair of linear equations in the following problems, and find their solutions by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

Solution:-

Let the numerator be $x$ and the denominator be $y$.
So, the fraction is $\frac{x}{y}$.

According to the conditions 1:

1. $\frac{x + 1}{y – 1} = 1 $
   $ \implies x + 1 = y – 1 $
   $ ∴ x – y = -2 \quad \text{— (i)}$
   
2. and $\frac{x}{y + 1} = \frac{1}{2} $
   $ \implies 2x = y + 1 $
   $ ∴ 2x – y = 1 \quad \text{— (ii)}$

$(i) \times 1 \implies x – y = -2 \quad \text{— (iii)}$

$(ii) \times 1 \implies 2x – y = 1 \quad \text{— (iv)}$

Subtracting (iii) from (iv) we get

$(2x – y) – (x – y) = 1 – (-2)$

$\implies 2x – y – x + y = 1 + 2$

$\implies x = 3$

Substitute $x = 3$ into equation (i):

$3 – y = -2$

$\implies -y = -2 – 3$

$\implies -y = -5$

$\implies y = 5$

Answer: The fraction is $\frac{3}{5}$.

(ii) Five years ago, Nur was thrice as old as Sonu. Ten years later, Nur will be twice as old as Sonu. How old are Nur and Sonu?

Solution:- 

Let the present age of Nur be $x$ years and the present age of Sonu be $y$ years.

Now, five years ago,

Age of Nuri $ = (x – 5)$ years
Age of Sonu $ = (y – 5)$ years

Ten years later,

Age of Nuri $ = (x + 10)$ years
Age of Sonu $ = (y + 10)$ years

According to the Question:

1. Five years ago: $(x – 5) = 3(y – 5) $
   $ \implies x – 5 = 3y – 15 $
   $\implies x – 3y = -10 \quad \text{— (i)}$

2. Ten years later: $(x + 10) = 2(y + 10) $
   $ \implies x + 10 = 2y + 20 $
   $ \implies x – 2y = 10 \quad \text{— (ii)}$

$(i) \times 1 \implies x – 3y = -10 \quad \text{— (iii)}$

$(ii) \times 1 \implies x – 2y = 10 \quad \text{— (iv)}$

Subtracting (iii) from (iv) we get

$(x – 2y) – (x – 3y) = 10 – (-10)$

$\implies x – 2y – x + 3y = 10 + 10$

$\implies y = 20$

Substitute $y = 20$ into equation (ii):

$x – 2(20) = 10$

$\implies x – 40 = 10$

$\implies x = 10 + 40$

$\implies x = 50$

Answer: Nur’s age is 50 years and Sonu’s age is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the tens digit be $x$ and the units digit be $y$.
Therefore, the original number is $10x + y$.

According to the given conditions:

1. $x + y = 9 \quad \text{— (i)}$

2. $9(10x + y) = 2(10y + x) $
   $\implies 90x + 9y = 20y + 2x $
   $\implies 88x – 11y = 0 $
   $\implies 8x – y = 0 \quad \text{— (ii)}$

$(i) \times 1 \implies x + y = 9 \quad \text{— (iii)}$

$(ii) \times 1 \implies 8x – y = 0 \quad \text{— (iv)}$

Adding (iii) & (iv) we get

$(x + y) + (8x – y) = 9 + 0$

$\implies 9x = 9$

$\implies x = 1$

Substitute $x = 1$ into equation (i):

$1 + y = 9$

$\implies y = 9 – 1$

$\implies y = 8$

Answer: The number is $10(1) + 8 = 18$.

(iv) The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of Rs 50 notes be $x$ and the number of Rs 100 notes be $y$.

According to the question:

1. $x + y = 25 \quad \text{— (i)}$

2. $50x + 100y = 2000 $
   $ \implies x + 2y = 40 \quad \text{— (ii)}$

$(i) \times 1 \implies x + y = 25 \quad \text{— (iii)}$

$(ii) \times 1 \implies x + 2y = 40 \quad \text{— (iv)}$

Subtracting (iii) from (iv) we get

$(x + 2y) – (x + y) = 40 – 25$

$\implies x + 2y – x – y = 15$

$\implies y = 15$

Substitute $y = 15$ into equation (i):

$x + 15 = 25$

$\implies x = 25 – 15$

$\implies x = 10$

Answer: Number of Rs 50 notes = 10, Number of Rs 100 notes = 15.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Let the fixed charge for the first 3 days be Rs $x$ and the additional charge per day be Rs $y$.

According to the question:

1. Saritha (7 days $\implies$ 3 days fixed + 4 days extra):
   $x + 4y = 27 \quad \text{— (i)}$

2. Susy (5 days $\implies$ 3 days fixed + 2 days extra):
   $x + 2y = 21 \quad \text{— (ii)}$

$(i) \times 1 \implies x + 4y = 27 \quad \text{— (iii)}$

$(ii) \times 1 \implies x + 2y = 21 \quad \text{— (iv)}$

Subtracting (iv) from (iii) we get

$(x + 4y) – (x + 2y) = 27 – 21$

$\implies x + 4y – x – 2y = 6$

$\implies 2y = 6$

$\implies y = 3$

Substitute $y = 3$ into equation (ii):

$x + 2(3) = 21$

$\implies x + 6 = 21$

$\implies x = 21 – 6$

$\implies x = 15$

Answer: Fixed charge = Rs 15, Additional charge per day = Rs 3.

3. The monthly incomes of A and B are in the ratio of 9:7 and their monthly expenditures are in the ratio of 4: 3. If each saves Rs. 1600 per month, find the monthly income of each.

Solution:-

Let the monthly incomes of A and B be $9x$ and $7x$, and their expenditures be $4y$ and $3y$.

According to the question:

1. $9x – 4y = 1600 \quad \text{— (i)}$

2. $7x – 3y = 1600 \quad \text{— (ii)}$

$(i) \times 3 \implies 27x – 12y = 4800 \quad \text{— (iii)}$

$(ii) \times 4 \implies 28x – 12y = 6400 \quad \text{— (iv)}$

Subtracting (iii) from (iv) we get

$(28x – 12y) – (27x – 12y) = 6400 – 4800$

$\implies 28x – 12y – 27x + 12y = 1600$

$\implies x = 1600$

Now calculate their monthly incomes:

Income of A = $9 \times 1600 = \text{Rs } 14,400$

Income of B = $7 \times 1600 = \text{Rs } 11,200$

Answer: Monthly income of A is Rs 14,400 and B is Rs 11,200.

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of rows be $x$ and the number of students in each row be $y$. Total students = $xy$.

According to the given conditions:

1. $(x – 1)(y + 3) = xy $
   $\implies xy + 3x – y – 3 = xy $
   $ \implies 3x – y = 3 \quad \text{— (i)}$

2. $(x + 2)(y – 3) = xy $
   $ \implies xy – 3x + 2y – 6 = xy $
   $ \implies -3x + 2y = 6 \quad \text{— (ii)}$

$(i) \times 1 \implies 3x – y = 3 \quad \text{— (iii)}$

$(ii) \times 1 \implies -3x + 2y = 6 \quad \text{— (iv)}$

Adding (iii) & (iv) we get

$(3x – y) + (-3x + 2y) = 3 + 6$

$\implies y = 9$

Substitute $y = 9$ into equation (i):

$3x – 9 = 3$

$\implies 3x = 3 + 9$

$\implies 3x = 12$

$\implies x = 4$

Total students = $x \times y = 4 \times 9 = 36$.

Answer: The number of students in the class is 36.

5. A and B each have certain numbers of mangoes. A says to B, ‘if you give me 30 of your mangoes, I will have twice as many as left with you.’ ‘B replies, if you give me 10, I will have thrice as many as left with you.’ How many mangoes does each have?

Solution:

Let the number of mangoes with A be $x$ and with B be $y$.

According to the question:

1. $x + 30 = 2(y – 30) $
   $ \implies x + 30 = 2y – 60 $
   $ \implies x – 2y = -90$ \quad \text{— (i)}$

2. $y + 10 = 3(x – 10) $
   $ \implies y + 10 = 3x – 30 $
   $ \implies 3x – y = 40 \quad \text{— (ii)}$

$(i) \times 1 \implies x – 2y = -90 \quad \text{— (iii)}$

$(ii) \times 2 \implies 6x – 2y = 80 \quad \text{— (iv)}$

Subtracting (iii) from (iv) we get

$(6x – 2y) – (x – 2y) = 80 – (-90)$

$\implies 6x – 2y – x + 2y = 80 + 90$

$\implies 5x = 170$

$\implies x = 34$

Substitute $x = 34$ into equation (ii):

$3(34) – y = 40$

$\implies 102 – y = 40$

$\implies -y = 40 – 102$

$\implies -y = -62$

$\implies y = 62$

Answer: A has 34 mangoes and B has 62 mangoes.

6. If the pair of equations x+ay = b and ax + y = 1 has infinitely many solutions, then choose the correct option from the given alternatives.

(i) a=1, b =1
(ii) a=-1, b =-1
(iii) a=1, b=-1
(iv) a=-1, b=1
(a) Both (i) and (iii) are true
(b) Both (ii) and (iv) are true
(c) Both (i) and (ii) are true
(d) Both (iii) and (iv) are true

Solution:

$x + ay = b \quad \text{— (i)}$

$ax + y = 1 \quad \text{— (ii)}$

For a pair of linear equations to have infinitely many solutions:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $
$\implies \frac{1}{a} = \frac{a}{1} = \frac{b}{1}$

Taking first two parts:

$\frac{1}{a} = \frac{a}{1} $
$\implies a^2 = 1 $
$\implies a = 1 \text{ or } a = -1$

If $a = 1$:

$\frac{1}{1} = \frac{b}{1} $
$\implies b = 1 \quad \text{— Option (i)}$

If $a = -1$:

$\frac{1}{-1} = \frac{b}{1} $
$\implies b = -1 \quad \text{— Option (ii)}$

Answer: Both options (i) and (ii) are true.

Correct Option: (c)

7. For what value of K, the following pair of linear equations has infinite number of solutions?
$$kx + 3y – (k – 3) = 0$$
$$12x + ky – k = 0$$

(a) k=-6
(b) k= 4
(c) k=-4
(d) k =6

Solutions:

Given Equations:

$kx + 3y – (k – 3) = 0$
$12x + ky – k = 0$

For infinitely many solutions:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $
$ \implies \frac{k}{12} = \frac{3}{k} = \frac{k – 3}{k}$

From the first two parts:

$\frac{k}{12} = \frac{3}{k} $
$\implies k^2 = 36 $
$\implies k = 6 \text{ or } k = -6$

Testing $k = 6$ in the third ratio:

$\frac{3}{6} = \frac{6 – 3}{6} $
$\implies \frac{1}{2} = \frac{3}{6} = \frac{1}{2}$ (This holds true).

Answer: $k = 6$.

Correct Option: (d)

8. Following are the steps of solving a word problem of the linear equation in two variables.
Choose the correct sequence from the given options.
(i) Represent the unknowns using variables.
(ii) Solve the equations.
(iii) Form the required equations.
(iv) Interpret the result.

(a) (i)→(ii)→(iii)→(iv)
(b) (i)→(iii)→(ii)→(iv)
(c) (iv)→(iii)→(ii)→(i)
(d) (ii)→(i)→(iii)→(iv)

Solution:

The correct sequence from the given options are as follows:-

1. Represent the unknowns using variables. (i)

2. Form the required equations. (iii)

3. Solve the equations. (ii)

4. Interpret the result. (iv)

This gives the sequence: $(i) \rightarrow (iii) \rightarrow (ii) \rightarrow (iv)$ Correct Option: (b)

Question 9: Solve the following pair of linear equations.

$\frac{2}{x} + \frac{3}{y} = 2$ 
$\frac{4}{x} – \frac{9}{y} = -1$ 

Solution:

(i) Given Equations are:

$\frac{2}{x} + \frac{3}{y} = 2 \quad \text{— (1)}$
$\frac{4}{x} – \frac{9}{y} = -1 \quad \text{— (2)}$

Substituting these into equations (1) and (2) reduces them to:

$2a + 3b = 2 \quad \text{— (3)}$
$4a – 9b = -1 \quad \text{— (4)}$

$(3) \times 3 \implies 6a + 9b = 6 \quad \text{— (5)}$
$(4) \times 1 \implies 4a – 9b = -1 \quad \text{— (6)}$

Adding (5) and (6) we get
$(4a – 9b) + (6a + 9b) = -1 + 6$
$\implies 10a = 5$
$\implies a = \frac{5}{10} $
$\implies a = \frac{1}{2}$

Now putting $a = \frac{1}{2}$ in equation (3), we get

Since $\frac{1}{x} = a$:

$ ∴ \frac{1}{x} = \frac{1}{2} $
$\implies x = 2$

Since $\frac{1}{y} = b$:

Ans: $x = 2, y = 3$

9 (ii)
$\frac{1}{2x} + \frac{1}{3y} = 2 $
$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Solution:

Given Equations are:

$\frac{1}{2x} + \frac{1}{3y} = 2 \quad \text{— (1)}$
$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6} \quad \text{— (2)}$

Let $\frac{1}{x} = a$ and $\frac{1}{y} = b$.

Substituting these into equations (1) and (2) reduces them to:

• $\frac{a}{2} + \frac{b}{3} = 2 $
  \implies \frac{3a + 2b}{6} = 2 $
  $\implies 3a + 2b = 12 \quad \text{— (3)}$
• $\frac{a}{3} + \frac{b}{2} = \frac{13}{6} $
  $\implies \frac{2a + 3b}{6} = \frac{13}{6} $
  $\implies 2a + 3b = 13 \quad \text{— (4)}$

$(3) \times 3 \implies 9a + 6b = 36 \quad \text{— (5)}$

$(4) \times 2 \implies 4a + 6b = 26 \quad \text{— (6)}$

Subtracting (6) from (5) we get

$(9a + 6b) – (4a + 6b) = 36 – 26$

$\implies 5a = 10$

$\implies a = \frac{10}{5}$

$\implies a = 2$

Now putting $a = 2$ in equation (3), we get

$3(2) + 2b = 12$

$\implies 6 + 2b = 12$

$\implies 2b = 12 – 6$

$\implies 2b = 6$

$\implies b = \frac{6}{2}$

$\implies b = 3$

Now,

Since $\frac{1}{x} = a$:

$∴ \frac{1}{x} = 2$

$\implies x = \frac{1}{2}$

Since $\frac{1}{y} = b$:

$∴ \frac{1}{y} = 3$

$\implies y = \frac{1}{3}$

Ans: $x = \frac{1}{2}, y = \frac{1}{3}$

Question 9 (iii)

$\frac{5}{x-1} + \frac{1}{y-2} = 2 $
$\frac{6}{x-1} – \frac{3}{y-2} = 1 $

Solution:

Given Equations are:

$\frac{5}{x-1} + \frac{1}{y-2} = 2 \quad \text{— (1)}$
$\frac{6}{x-1} – \frac{3}{y-2} = 1 \quad \text{— (2)}$

Let $\frac{1}{x-1} = a$ and $\frac{1}{y-2} = b$.

Substituting these into equations (1) and (2) reduces them to:

$5a + b = 2 \quad \text{— (3)}$

$6a – 3b = 1 \quad \text{— (4)}$

$(3) \times 3 \implies 15a + 3b = 6 \quad \text{— (5)}$

$(4) \times 1 \implies 6a – 3b = 1 \quad \text{— (6)}$

Adding (5) and (6) we get

$(15a + 3b) + (6a – 3b) = 6 + 1$

$\implies 21a = 7$

$\implies a = \frac{7}{21}$

$\implies a = \frac{1}{3}$

Now putting $a = \frac{1}{3}$ in equation (3), we get

$5\left(\frac{1}{3}\right) + b = 2$

$\implies \frac{5}{3} + b = 2$

$\implies b = 2 – \frac{5}{3}$

$\implies b = \frac{6 – 5}{3}$

$\implies b = \frac{1}{3}$

Now,

Since $\frac{1}{x-1} = a$:

$∴ \frac{1}{x-1} = \frac{1}{3}$

$\implies x – 1 = 3$

$\implies x = 3 + 1$

$\implies x = 4$

Since $\frac{1}{y-2} = b$:

$∴ \frac{1}{y-2} = \frac{1}{3}$

$\implies y – 2 = 3$

$\implies y = 3 + 2$

$\implies y = 5$

Ans: $x = 4, y = 5$

Question 10: In a two digit number, the sum of the digit is 9. If the digits are reversed, the number is increased by 9. Find the number.

Solution:

Let the tens digit of the number be $x$ and the units digit be $y$.

Therefore, the original number is $10x + y$, and the reversed number is $10y + x$.

According to the question:

1. The sum of the digits is 9:

   $x + y = 9 \quad \text{— (1)}$

2. If the digits are reversed, the number is increased by 9:

   $(10y + x) – (10x + y) = 9$
   $\implies 9y – 9x = 9$
   $\implies y – x = 1 $
   $\implies -x + y = 1 \quad \text{— (2)}$

$(1) \times 1 \implies x + y = 9 \quad \text{— (3)}$

$(2) \times 1 \implies -x + y = 1 \quad \text{— (4)}$

Adding (3) and (4) we get

$(x + y) + (-x + y) = 9 + 1$

$\implies 2y = 10$

$\implies y = \frac{10}{2}$

$\implies y = 5$

Now putting $y = 5$ in equation (1), we get

$x + 5 = 9$

$\implies x = 9 – 5$

$\implies x = 4$

Therefore, the original number is:

$10x + y = 10(4) + 5 = 40 + 5 = 45$

Ans: The number is 45

Question 11: The sum of the digits of a two digit number is 15. The number decreased by 27 if the digits are reversed. Find the number.

Solution:

Let the tens digit of the number be $x$ and the units digit be $y$.

Therefore, the original number is $10x + y$, and the reversed number is $10y + x$.

According to the question:

1. The sum of the digits is 15:

   $x + y = 15 \quad \text{— (1)}$

2. The number is decreased by 27 if the digits are reversed:

   $(10x + y) – (10y + x) = 27$
   $\implies 9x – 9y = 27$
   $\implies x – y = 3 \quad \text{— (2)}$

$(1) \times 1 \implies x + y = 15 \quad \text{— (3)}$

$(2) \times 1 \implies x – y = 3 \quad \text{— (4)}$

Adding (3) and (4) we get

$(x + y) + (x – y) = 15 + 3$

$\implies 2x = 18$

$\implies x = \frac{18}{2}$

$\implies x = 9$

Now putting $x = 9$ in equation (1), we get

$9 + y = 15$

$\implies y = 15 – 9$

$\implies y = 6$

Therefore, the original number is:

$10x + y = 10(9) + 6 = 90 + 6 = 96$

Ans: The number is 96

Question 12: The difference between two numbers is 4. Twice the smaller number added to three times the larger number gives 82. Find the two numbers.

Solution:

Let the larger number be $x$ and the smaller number be $y$.

According to the question:

• $x – y = 4 \quad \text{— (1)}$
• $3x + 2y = 82 \quad \text{— (2)}$

$(1) \times 2 \implies 2x – 2y = 8 \quad \text{— (3)}$

$(2) \times 1 \implies 3x + 2y = 82 \quad \text{— (4)}$

Adding (3) and (4) we get

$(2x – 2y) + (3x + 2y) = 8 + 82$

$\implies 5x = 90$

$\implies x = \frac{90}{5}$

$\implies x = 18$

Now putting $x = 18$ in equation (1), we get

$18 – y = 4$

$\implies -y = 4 – 18$

$\implies -y = -14$

$\implies y = 14$

Ans: The two numbers are 18 and 14