## Question :

From the pair of linear equations in the following problem and find their solutions (if they exist) by any algebraic method :

**(i)** A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student, A takes food for 20 days, A has to pay Rs 1000 as hostel charges, whereas a student B, who takes food for 26 days pays Rs 1180 as hostel charges. Find the fixed charge and cost of food per day.

**(ii)** A fraction becomes \(1\over 3\) when 1 is subtracted from the numerator and it becomes \(1\over 4\) when 8 is added to its denominator. Find the fraction.

**(iii)** Yash scored 40 marks in a test, receiving 3 marks for each correct answer and loosing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test ?

**(iv)** Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the car travel in the same direction at a different speed, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speed of the two cars ?

**(v)** The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

## Solution :

**(i)** Let Rs x be the fixed hostel charges

And Rs y be the cost of food per day.

ATQ,

A’s hostel charges are

Fixed hostel charges + cost of food for 20 days = Rs 1000

x + 20y = 1000 ……..(1)

B’s hostel charges are

Fixed hostel charges + cost of food for 26 days = Rs 1180

x + 26y = 1180 ……….(2)

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

6y = 180 \(\implies\) y = 30

Put y = 30 in equation (1),

x + 20(30) = 1000 \(\implies\) x + 600 = 1000

\(\implies\) x = 400

Hence, **x = 400 and y = 30.**

**(ii)** Let \(x\over y\) be the fraction,

According to Question,

\(x – 1\over y\) = \(1\over 3\) \(\implies\) 3x – 3 = y \(\implies\) 3x – y – 3 = 0 ………(1)

and \(x\over y + 8\) \(\implies\) \(1\over 4\) \(\implies\) 4x = y + 8 \(\implies\) 4x – y – 8 = 0 ……….(2)

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

x – 5 = 0 \(\implies\) x = 5

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

y = 12

Hence, the **fraction is \(5\over 12\).**

**(iii)** Let x be the number of correct answers and y be the number of wrong answers given by yash.

Then, ATQ,

**Case 1** : He got 40 marks if 3 marks are given for correct answer and 1 mark is deducted for every incorrect answers.

3x – y = 40 ……..(1)

**Case 2** : He gets 50 marks if 4 marks are given for correct answer and 2 marks are deducted for every incorrect answers.

4x – 2y = 50 …….(2)

Multiplying equation (1) by (2) and subtracting from equation (2), we get

-2x = -30 \(\implies\) x = 15

Putting x = 15 in equation (1), we get

45 – y = 40 \(\implies\) y = 5

Hence, the total number of questions = 15 + 5 = **20**.

**(iv)** Let x km/hr be the speed of the first car, starting from A.

and y km/hr is the speed of second car, starting from B.

Distance travelled by first car in 5 hours = 5x km

Distance travelled by second car in 5 hours = 5y km

ATQ,

When they are moving in the same direction,

5x = 100 + 5y

\(\implies\) x = 20 + y

\(\implies\) x – y = 20 ……..(1)

When they are moving in the opposite direction,

Distance travelled by first car in 1 hour = x km

Distance travelled by second car in 1 hour = y km

\(\implies\) x + y = 100 ………..(2)

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

2x = 120 \(\implies\) x = 60

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

60 – y = 20 \(\implies\) y = 40

Hence, the speed of first car is **60 km/hr** and the speed of second car is **40 km/hr.**

**(v)** Let x be the length of rectangle and y be the breadth of the rectangle.

Then, area of rectangle = xy

According to Question,

**Case 1** : Reduced Length = (x – 5) units

Increased breadth = (y + 3) units

Reduced area = (x – 5)(y + 3)

Reduction in area = 9

Original area – Reduced area = 9

xy – [(x – 5)(y + 3)] = 9

\(\implies\) 3x – 5y = 6 ……….(1)

**Case 2** : Increased length = (x + 3) units

Increased breadth = (y + 2) units

Increased area = (x + 3)(y +2)

Increase in area = 67

Increased area – Original area = 67

(x + y) (y + 2) – xy = 67

\(\implies\) 2x + 3y = 61 …….(2)

On solving equation (1) and equation (2), we get

x = 17 and y = 9

Hence, Length = **17 units** and breadth = **9 units**