From the pair of linear equation in the following problems, and find their solutions (if they exist) by the elimination method

Question : From the pair of linear equation 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 \(1\over 2\) 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 number. Find the number.

(iv)  Meena went to a bank to withdraw Rs 2000. She asked the cashier to give Rs 50 and Rs 1000 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and 100 she recieved ?

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

Solution :

(i)  Let the numerator be x and the denominator be y of the fraction. Then, fraction is \(x\over y\).

According to Question,

\(x + 1\over y – 1\) = 1          \(\implies\)       x + 1 = y – 1

\(\implies\)   x – y = -2          ……..(1)

and  \(x\over y + 1\) = \(1\over 2\)   \(\implies\)    2x = y + 1

\(\implies\)  2x – y = 1           ………(2)

Now, subtract equation (2) from equation (1), we get

(x – y) – ( 2x – y) = -2 – 1

\(\implies\)    x – y – 2x + y = -3

\(\implies\)   x = 3

Put the value of x in equation (1), we get

3 – y = – 2      \(\implies\)   y = 5

Hence, the required fraction is \(3\over 5\).

(ii)  Let Nuri’s present age be x years and Sonu’s present age be y years.

Five years ago,

Nuri’s age = (x – 5)  years

and  Sonu’s age = (y – 5) years

According to Question,

(x – 5) = 3(y – 5)   \(\implies\)   x – 5 = 3y – 15

\(\implies\)   x – 3y = – 15 + 5   \(\implies\)   x – 3y = – 10         ……..(1)

Ten years later,

Nuri’s age = (x + 10) years

and  Sonu’s age = (y + 10) years

According to Question,

x + 10 = 2(y + 10)   \(\implies\)   x + 10 = 2y + 20

\(\implies\)  x – 2y = 10         ……….(2)

Now, Subtract equation (2) form equation (1), we get

y = 20

Put the value of y in equation (2), we get

x – 2(20) = 10       \(\implies\)    x = 40

\(\therefore\)    Present age of Nuri is 50 years and present age of sonu is 20 years.

(iii)  Let x and y be the ten’s and unit’s digit in the number respectively. So, the number may be written as 10x + y.

By reversing the order of the number, unit’s digit becomes x and ten’s digit becomes y.

Then, the number can be written as 10y + x.

According to Question,

x + y = 9             …….(1)

We are given in the question that nine times the number i.e. 9(10x + y) is twice the number obtained by reversing the order of number i.e. 2(10y + x).

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

\(\implies\)  90x + 9y = 20y + 2x

\(\implies\)  8x – y = 0      ……(2)

Adding the equation (1) and (2), we get

9x = 9      \(\implies\)    x = 1

Now, Put the value of x in equation (1), we get

y = 9 – 1 = 8

Thus, the number is \(10 \times 1) + 8 = 10 + 8 = 18

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

According to Question(ATQ),

x + y = 25            …….(1)

50x + 100y = 2000            …….(2)

Divide equation (2) by 50, we get

x + 2y = 40        ………(3)

Subtract equation (1) from equation (3), we get

y = 15

Now, put the value of y in equation (1), we get

x + 15 = 25     \(\implies\)   x = 10

Hence,  x = 10  and   y = 15.

(v)  Let x  be the fixed charges for 3 days and y be the charges per day.

According to the Question,

x + 4y = 27          ……(1)

and  x + 2y = 21       ……..(2)

Subtracting the equation (2)  from equation (1),  we get

x + 4(3)  = 27   \(\implies\)    x = 27 – 12 = 15

Put the value of x in equation (2), we get

y = 3

Hence, x = 15 and y = 3.

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