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 \[\dfrac{1}{2}\] 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 the first 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

Hint: Form linear equations in two variables for two varying parameters and then solve them by elimination method, i.e., adding or subtracting two equations in a way that eliminates one of the variables.

Complete step-by-step solution -
We will begin by forming linear equations for each case using the variables x and y.
(i) Let’s assume that our fraction is of the form \[\dfrac{x}{y}\] where x is for the numerator and y is for the denominator.
Adding 1 to the numerator and subtracting 1 from the denominator, we get the fraction as 1. Thus, we have \[\dfrac{x+1}{y-1}=1\].
Solving this equation, we have \[x+1=y-1\Rightarrow x-y+2=0\].
If we add 1 to the denominator, the fraction becomes \[\dfrac{1}{2}\]. Thus, we have \[\dfrac{x}{y+1}=\dfrac{1}{2}\].
Solving the equation by cross-multiplying, we have \[2x=y+1\Rightarrow 2x-y-1=0\].
Thus, we have the equations \[2x-y-1=0\] and \[x-y+2=0\].
We will solve them by elimination method. Subtracting equations from one another, we have \[2x-y-1-\left( x-y+2 \right)=0-0\].
Further simplifying the equations, we get \[x-3=0\].
Thus, we have \[x=3\].
Substituting the value \[x=3\] in the equation \[x-y+2=0\], we get \[3-y+2=0\].
Further simplifying the equation, we get \[5-y=0\]. Thus, we have \[y=5\].
Hence, the fraction is of the form \[\dfrac{x}{y}\] where \[x=3,y=5\], i.e., \[\dfrac{3}{5}\].
(ii) Let’s assume Nuri’s age to be x years and Sonu’s age to be y years.
Five years ago, Nuri’s age was \[x-5\] years and Sonu’s age was \[y-5\] years. Five years ago, Nuri was thrice as old as Sonu. Thus, we have \[x-5=3\left( y-5 \right)\].
Further solving the equation, we get \[x-5=3y-15\Rightarrow x-3y+10=0\].
Ten years later, Nuri’s age will be \[x+10\] and Sonu’s age will be \[y+10\] years. Ten years later, Nuri’s age will be twice of Sonu’s age. Thus, we have \[x+10=2\left( y+10 \right)\].
Further solving the equation, we get \[x+10=2y+20\Rightarrow x-2y-10=0\].
Thus, we have the equations \[x-3y+10=0\] and \[x-2y-10=0\].
We will solve them by elimination method.
Subtracting the two equations, we get \[x-3y+10-\left( x-2y-10 \right)=0-0\].
Further simplifying the equation, we get \[-y+20=0\].
Thus, we have \[y=20\].
Substituting the value \[y=20\] in equation \[x-2y-10=0\], we get \[x-2\left( 20 \right)-10=0\Rightarrow x=50\].
Hence, Nuri’s age is \[x=50\] years and Sonu’s age is \[y=20\] years.
(iii) Let’s assume that the two digits of a two-digit number are x and y.
Thus, the two-digit number is of the form \[10x+y\].
The sum of digits of the number is 9. Thus, we have \[x+y=9\].
The number obtained by reversing the digits is \[10y+x\].
Nine times the first number is twice the number obtained by reversing the digits. Thus, we have \[9\left( 10x+y \right)=2\left( 10y+x \right)\].
Simplifying the equation, we have \[90x+9y=20y+2x\Rightarrow 88x-11y=0\Rightarrow 8x-y=0\].
Thus, we have the equations \[x+y=9\] and \[8x-y=0\]. We will solve them by elimination method.
Adding the two equations, we get \[8x+x-y+y=9\].
Further simplifying the equations, we get \[9x=9\Rightarrow x=1\].
Substituting the value \[x=1\] in the equation \[x+y=9\], we get \[1+y=9\Rightarrow y=8\].
Hence, the number is of form \[10x+y\] where \[x=1,y=8\], i.e., 18.
(iv) Let’s assume there are x notes of value \[Rs.50\] and y notes of value \[Rs.100\].
As there are 25 notes in total, we have \[x+y=25\].
We know the total value of notes is \[Rs.2000\]. Thus, we have \[50x+100y=2000\].
Further simplifying the equation, we have \[x+2y=40\].
Thus, we have the equations \[x+y=25\] and \[x+2y=40\].
Subtracting the two equations, we get \[x+2y-\left( x+y \right)=40-25\].
Further simplifying the equations, we get \[y=15\].
Substituting the value \[y=15\] in the equation \[x+y=25\], we get \[x+15=25\Rightarrow x=10\].
Hence, the number of \[Rs.50\] notes are \[x=10\] and the number of \[Rs.100\] notes are \[y=15\].
(v) Let’s assume that the fixed charge is \[Rs.x\] and the additional charge for each day is \[Rs.y\].
As Saritha paid \[Rs.27\] for a book kept for seven days, she paid the additional charge for 4 days. Thus, we have \[x+4y=27\].
As Susy paid \[Rs.21\] for a book kept for five days, she paid the additional charge for 2 days. Thus, we have \[x+2y=21\].
Hence, we have the equations \[x+4y=27\] and \[x+2y=21\]. We will solve them by elimination method.
Subtracting the two equations, we get \[x+4y-\left( x+2y \right)=27-21\].
Further simplifying the equation, we have \[2y=6\Rightarrow y=3\].
Substituting the value \[y=3\] in the equation \[x+2y=21\], we get \[x+2\left( 3 \right)=21\].
Simplifying the equation, we have \[x=21-6\Rightarrow x=15\].
Hence, we have the fixed charge as \[Rs.15\] and the additional charge for each day as \[Rs.3\].

Note: We can also solve this question by forming linear equations in one variable and writing the other variable in terms of the first one and then solving the equation in one variable to get the answer.