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Question:
Form the pair of linear equations in the following problems 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 she 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 charges and the cost of food per day.
(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes
when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 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 cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds 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 the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Answer:
Solutions:
(i) Let fixed monthly charge is Rs x and charge of food for one day is Rs y
According to given conditions,
x + 20y = 1000 … (1),
and x + 26y = 1180 … (2)
Subtracting equation (1) from equation (2), we get
6y = 180
⇒ y = 30
Putting the value of y in (1), we get
x + 20 (30) = 1000
⇒ x = 1000 – 600 = 400
Therefore, fixed monthly charges are Rs 400 and, charges of food for one day is Rs 30.
(ii) Let numerator = x and let denominator = y
According to given conditions,
⇒ 3x – 3 = y … (1) 4x = y + 8 … (1)
⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)
Subtracting equation (1) from (2), we get
4x – y − (3x − y) = 8 – 3
⇒ x = 5
Putting the value of x in (1), we get
3 (5) – y = 3
⇒ 15 – y = 3
⇒ y = 12
Therefore, numerator = 5 and, denominator = 12
It means fraction
(iii) Let number of correct answers are x and number of wrong answers is y.
According to given conditions,
3x – y = 40 … (1)
And, 4x − 2y = 50 … (2)
From equation (1), y = 3x − 40
Putting this in (2), we get
4x – 2 (3x − 40) = 50
⇒ 4x − 6x + 80 = 50
⇒ −2x = −30
⇒ x = 15
Putting the value of x in (1), we get
3 (15) – y = 40
⇒ 45 – y = 40
⇒ y = 45 – 40 = 5
Therefore, the number of correct answers = x = 15and number of wrong answers = y = 5
Total questions = x + y = 15 + 5 = 20
(iv)Let the speed of the car which starts from part A is x km/hr
Let the speed of car which starts from part B is y km/hr
According to given conditions,
(Assuming x > y)
⇒ 5x − 5y = 100
⇒ x – y = 20 … (1)
And,
⇒ x + y = 100 … (2)
Adding (1) and (2), we get
2x = 120
⇒ x = 60 km/hr
Putting the value of x in (1), we get
60 – y = 20
⇒ y = 60 – 20 = 40 km/hr
Therefore, the speed of car starting from point A is 60 km/hr
And, Speed of car starting from point B = 40 km/hr
(v) Let the length of rectangle is x units and breadth of rectangle is y units.
Area is xy square units.
According to given conditions,
xy – 9 = (x − 5) (y + 3)
⇒ xy – 9 = xy + 3x − 5y – 15
⇒ 3x − 5y = 6 … (1)
And, xy + 67 = (x + 3) (y + 2)
⇒ xy + 67 = xy + 2x + 3y + 6
⇒ 2x + 3y = 61 … (2)
From equation (1),
3x = 6 + 5y
Putting this in (2), we get
⇒ 12 + 10y + 9y = 183
⇒ 19y = 171
⇒ y = 9 units
Putting the value of y in (2), we get
2x + 3 (9) = 61
⇒ 2x = 61 – 27 = 34
⇒ x = 17 units
Therefore, length = 17 units and, breadth = 9 units.
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