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 1/2 if we only add 1 to the denominator. What is the fraction? (ii) The age of Nuri is 5 years more than three times the age of 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) The ratio of the number of Rs 100 notes to the number of Rs 50 notes is 1:2. The total value of these notes is Rs 2000. Find the number of each type of notes. (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.

(i) Let the fraction be x/y
Given, (x+1)/(y-1) = 1
x+1 = y-1
x - y = -2 (1)
Also, x/(y+1) = 1/2
2x = y+1
2x - y = 1 (2)
Subtracting (1) from (2)
x = 3
Substituting value of x in (2), 6 - y = 1
y = 5
Hence, the fraction is 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) Let the fixed charge for first 3 days be Rs x and after 3 days the charge for each day be Rs y.
Given, x + 4y = 27 (1) [first 3 days + 4 days]
And, x + 2y = 21.. (2) [first 3 days + 2 days]
Subtracting (2) from (1), we get
2y = 6
y = 3
Substituting value of y in (1)
x + 4 × 3 = 27
x + 12 = 27
x = 15
Hence, the fixed charge is Rs 15 and the per day charge is Rs 3.