Question 2

Let's form pairs of linear equations for each problem and solve them using the elimination method.

Step 1 — Fraction Problem Setup

Let the numerator of the fraction be $x$. Let the denominator of the fraction be $y$. The fraction is $\frac{x}{y}$.

The first condition states that adding 1 to the numerator and subtracting 1 from the denominator makes the fraction 1.

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

Let's simplify this equation.

$x+1 = y-1$

$x - y = -1 - 1$

$\boxed{x - y = -2 \quad (1)}$

The second condition states that adding 1 to the denominator makes the fraction $\frac{1}{2}$.

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

Let's simplify this equation.

$2x = y+1$

$2x - y = 1$

$\boxed{2x - y = 1 \quad (2)}$

Step 2 — Solve Fraction Problem

We have two equations: (1) $x - y = -2$ (2) $2x - y = 1$

Let's subtract equation (1) from equation (2) to eliminate $y$.

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

$2x - y - x + y = 1 + 2$

$x = 3$

$\boxed{x = 3}$

Now, let's substitute the value of $x$ into equation (1).

$3 - y = -2$

$-y = -2 - 3$

$-y = -5$

$y = 5$

$\boxed{y = 5}$

The numerator is 3 and the denominator is 5. The fraction is $\frac{3}{5}$.

Step 3 — Age Problem Setup

Let Nuri's current age be $x$ years. Let Sonu's current age be $y$ years.

Five years ago, Nuri's age was $x-5$ years. Five years ago, Sonu's age was $y-5$ years. Nuri was thrice as old as Sonu then.

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

Let's simplify this equation.

$x-5 = 3y - 15$

$x - 3y = -15 + 5$

$\boxed{x - 3y = -10 \quad (1)}$

Ten years later, Nuri's age will be $x+10$ years. Ten years later, Sonu's age will be $y+10$ years. Nuri will be twice as old as Sonu then.

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

Let's simplify this equation.

$x+10 = 2y + 20$

$x - 2y = 20 - 10$

$\boxed{x - 2y = 10 \quad (2)}$

Step 4 — Solve Age Problem

We have two equations: (1) $x - 3y = -10$ (2) $x - 2y = 10$

Let's subtract equation (2) from equation (1) to eliminate $x$.

$(x - 3y) - (x - 2y) = -10 - 10$

$x - 3y - x + 2y = -20$

$-y = -20$

$y = 20$

$\boxed{y = 20}$

Now, let's substitute the value of $y$ into equation (1).

$x - 3(20) = -10$

$x - 60 = -10$

$x = -10 + 60$

$x = 50$

$\boxed{x = 50}$

Nuri's age is 50 years. Sonu's age is 20 years.

Step 5 — Two-Digit Number Problem Setup

Let the unit's digit of the number be $x$. Let the ten's digit of the number be $y$. The number can be written as $10y + x$.

The sum of the digits is 9.

$x + y = 9$

$\boxed{x + y = 9 \quad (1)}$

The number obtained by reversing the digits is $10x + y$. Nine times the original number is twice the reversed number.

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

Let's simplify this equation.

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

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

$88y = 11x$

Let's divide both sides by 11.

$8y = x$

$x - 8y = 0$

$\boxed{x - 8y = 0 \quad (2)}$

Step 6 — Solve Two-Digit Number Problem

We have two equations: (1) $x + y = 9$ (2) $x - 8y = 0$

Let's subtract equation (2) from equation (1) to eliminate $x$.

$(x + y) - (x - 8y) = 9 - 0$

$x + y - x + 8y = 9$

$9y = 9$

$y = 1$

$\boxed{y = 1}$

Now, let's substitute the value of $y$ into equation (1).

$x + 1 = 9$

$x = 9 - 1$

$x = 8$

$\boxed{x = 8}$

The unit's digit is 8 and the ten's digit is 1. The number is $10(1) + 8 = 18$.

Step 7 — Bank Notes Problem Setup

Let the number of ₹50 notes be $x$. Let the number of ₹100 notes be $y$.

Meena got 25 notes in total.

$x + y = 25$

$\boxed{x + y = 25 \quad (1)}$

The total amount withdrawn was ₹2000. The value from ₹50 notes is $50x$. The value from ₹100 notes is $100y$.

$50x + 100y = 2000$

Let's divide the entire equation by 50 to simplify.

$\frac{50x}{50} + \frac{100y}{50} = \frac{2000}{50}$

$x + 2y = 40$

$\boxed{x + 2y = 40 \quad (2)}$

Step 8 — Solve Bank Notes Problem

We have two equations: (1) $x + y = 25$ (2) $x + 2y = 40$

Let's subtract equation (1) from equation (2) to eliminate $x$.

$(x + 2y) - (x + y) = 40 - 25$

$x + 2y - x - y = 15$

$y = 15$

$\boxed{y = 15}$

Now, let's substitute the value of $y$ into equation (1).

$x + 15 = 25$

$x = 25 - 15$

$x = 10$

$\boxed{x = 10}$

Meena received 10 notes of ₹50. Meena received 15 notes of ₹100.

Step 9 — Library Charges Problem Setup

Let the fixed charge for the first three days be ₹$x$. Let the additional charge for each day thereafter be ₹$y$.

Saritha kept the book for 7 days and paid ₹27. The fixed charge covers 3 days. The extra days are $7 - 3 = 4$ days. The charge for extra days is $4y$.

$x + 4y = 27$

$\boxed{x + 4y = 27 \quad (1)}$

Susy kept the book for 5 days and paid ₹21. The fixed charge covers 3 days. The extra days are $5 - 3 = 2$ days. The charge for extra days is $2y$.

$x + 2y = 21$

$\boxed{x + 2y = 21 \quad (2)}$

Step 10 — Solve Library Charges Problem

We have two equations: (1) $x + 4y = 27$ (2) $x + 2y = 21$

Let's subtract equation (2) from equation (1) to eliminate $x$.

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

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

$2y = 6$

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

$y = 3$

$\boxed{y = 3}$

Now, let's substitute the value of $y$ into equation (2).

$x + 2(3) = 21$

$x + 6 = 21$

$x = 21 - 6$

$x = 15$

$\boxed{x = 15}$

The fixed charge is ₹15. The additional charge for each extra day is ₹3.

Answer

(i) The fraction is $\frac{3}{5}$. (ii) Nuri's age is 50 years and Sonu's age is 20 years. (iii) The two-digit number is 18. (iv) Meena received 10 notes of ₹50 and 15 notes of ₹100. (v) The fixed charge is ₹15 and the additional charge for each extra day is ₹3.