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# Form the pair of linear equations in the following problems, and find their solutions (if they exist) 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?

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

**(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.

**(iv)** Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.

**(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.

To do:

We have to form the pair of linear equations and find the solutions by the elimination method.

Solution:

(i) Let the numerator and denominator of the original fraction be $x$ and $y$ respectively.

The original fraction$=\frac{x}{y}$

The fraction becomes $1$ if 1 is added to the numerator and 1 is subtracted from the denominator.

This implies,

New fraction$=\frac{x+1}{y-1}$

According to the question,

$\frac{x+1}{y-1}=1$

$x+1=1(y-1)$ (On cross multiplication)

$x+1=y-1$

$y=x+1+1$

$y=x+2$.....(i)

When 1 is added to only the denominator, it becomes $\frac{1}{2}$.

This implies,

$\frac{x}{y+1}=\frac{1}{2}$

$2(x)=1(y+1)$ (On cross multiplication)

$2x=y+1$

$2x-y-1=0$

$2x-(x+2)-1=0$ (From (i))

$2x-x-2-1=0$

$x=3$

$\Rightarrow y=x+2$

$y=3+2$

$y=5$

Therefore, the original fraction is $\frac{3}{5}$.

(ii) Let the ages of Nuri and Sonu be $x$ and $y$ respectively.

This implies,

Age of Nuri 5 years ago $= x-5$ years.

Age of Sonu 5 years ago $= y-5$ years.

Age of Nuri after 10 years $= x+10$ years.

Age of Sonu after 10 years $= y+10$ years.

According to the question,

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

$x+10=2y+20$

$x=2y+20-10$

$x=2y+10$.....(i)

$x-5=3(y-5)$

$x-5=3y-15$

$3y=(2y+10)+15-5$ (From (i))

$3y=2y+10+10$

$3y-2y=20$

$y=20$

$\Rightarrow x=2(20)+10=40+10=50$

The present ages of Nuri and Sonu are 50 years and 20 years respectively.

(iii) Let the two-digit number be $10x+y$.

$x + y = 9$

$x=9-y$.....(i)

The number formed on reversing the digits is $10y+x$.

Therefore,

$9(10x+y) = 2(10y+x)$

$90x+9y=20y+2x$

$90x-2x+9y-20y=0$

$88x-11y=0$

$11(8x-y) = 0$

$8x-y = 0$

$y=8x$

Substituting $y = 8x$ in equation (i), we get,

$x =9-8x $

$x+8x = 9$

$9x = 9$

$x=1$

This implies,

$y = 8x = 8(1)=8$

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

The original number is 18.

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

The total number of notes $= 25$

This implies,

$x + y = 25$

$x = 25-y$....(i)

Total amount only in Rs. 50 notes $= 50x$

Total amount only in Rs. 100 notes $=100y$

Total amount $=Rs.\ 2000$

$50x + 100y = 2000$

$50(25-y) + 100y = 2000$ (From (i))

$50\times25 - 50y + 100y = 2000$

$1250 +50 y = 2000$

$50y = 2000-1250 = 750$

$y = \frac{750}{50} = 15$

$ x = 25-15 = 10$ (From (i))

Therefore, she received $10$ Rs. 50 notes and $15$ Rs. 100 notes.

(v) Let the fixed charge for the first three days and the additional charge for each day thereafter be $x$ and $y$ respectively.

According to the question,

$x + 4y = 27$.....(i)

$x + 2y = 21$.....(ii)

Subtracting equation (ii) from equation (i), we get,

$(x+4y)-(x+2y)=27-21$

$x-x+4y-2y=6$

$2y=6$

$y=\frac{6}{2}$

$y=3$

Substituting $y=3$ in equation (ii), we get,

$x+2(3)=21$

$x+6=21$

$x=21-6$

$x=15$

The fixed charge is Rs. 15 and the charge for each extra day is Rs. 3.

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