Question

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.

Answer 1

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

Given,

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

$x+1=y-1$

$x-y=-2$ .....(1)

Also,

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

$2x=y+1$

$2x-y=1$ .....(2)

Substracting (1) from (2)

$x=3$

Substituting value of x in (2),

$6-y=1$

$y=5$

Hence, the fraction is $\frac{3}{5}$

(ii) Let the age of Nuri be x and that of Sonu be y.

Given,

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

$3y-x=10$ ..... (1)

And,

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

$2y-x=-10$ ....(2)

Subtracting (2) from (1), we get

$y=20$

Substituting value of y in (1)

$60-x=10$

$x=50$

Hence, the ages of Nuri and Sonu are 50 and 20 respectively.

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

Given, $x+y=9$ .....(1)

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

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

$88x-11y=0$ .....(2)

Multiplying (1) by 11, we get

$11x+11y=99$ ....(3)

Adding (1) and (3)

$99x=99$

$x=1$

Substituting this value of x in (1),

$1+y=9$

$y=8$

Hence, the number is 18.

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

Given,

$x+y=25$ ...(1)

And,

$100x+50y=2000$ ...(2)

Multiplying (1) by 50, we get

$50x+50y=1250$ ....(3)

Subtracting (3) from (2),

$50x=750$

$x=15$

Substituting value of x in (1),

$15+y=25$

$y=10$

(v) $\text{Let the fixed charge for first 3days be}$ $Rsx$ $\text{and after 3days the charge for each day be}$ $Rsy$.

Given,

$x+4y=27$ ....(1) [first 3days + 4days]

And,

$x+2y=21$ ...(2) [first 3days + 2days]

Subtracting (2) from (1), we get

$2y=6$

$y=3$

Substituting value of y in (1)

$x+4\times 3=27$

$x+12=27$

$x=15$

Hence, the fixed charge is Rs 15 and the per day charge is Rs 3.