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.
(i) Let Rs x be the fixed hostel charges
And Rs y be the cost of food per day.
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.
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
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