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# 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 $${{1}\over {3}}$$ 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, 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.

Ans.(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y

According to given conditions,

$$x + 20y = 1000$$… (1),
$$x + 26y = 1180$$… (2)

Subtracting equation (1) from equation (2),
$$6y = 180$$
=>$$y = 30$$

Putting value of y in (1),
$$x + 20 (30) = 1000$$
=>$$x = 1000 – 600 = 400$$

Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

(ii) Let numerator = x and let denominator = y

According to given conditions,

$${{x - 1}\over {y}} = {{1}\over {3}}$$… (1) and $${{x}\over {y + 8}} = {{1}\over {4}}$$… (2)
=>$$3x – 3 = y$$… (1) $$4x = y + 8$$ … (2)
=>$$3x – y = 3$$… (1) $$4x – y = 8$$… (2)

Subtracting equation (1) from (2),
$$4x – y - (3x - y) = 8 – 3$$
$$x = 5$$

Putting value of x in (1),
$$3 (5) – y = 3$$
$$=> 15 – y = 3$$
$$=> y = 12$$

Therefore, numerator = 5 and, denominator = 12

It means fraction = $${{5}\over {12}}$$

(iii) Let number of correct answers = x and let number of wrong answers = y

According to given conditions,
$$3x – y = 40$$ … (1)
And, $$4x - 2y = 50$$ … (2)

From equation (1), $$y = 3x - 40$$
Putting this in (2),
$$4x – 2 (3x - 40) = 50$$
=>$$4x - 6x + 80 = 50$$
=>$$-2x = -30$$
=>$$x = 15$$

Putting value of x in (1),
$$3 (15) – y = 40$$
=>$$45 – y = 40$$
=>$$y = 45 – 40 = 5$$

Therefore, number of correct answers = x = 15 and number of wrong answers = y = 5

Total questions = x + y = 15 + 5 = 20

(iv)Let speed of car which starts from part A = x km/hr

Let speed of car which starts from part B = y km/hr

According to given conditions,(Assuming x > y)

=>$${{100} \over {x-y}} = 5$$
=>$$5x - 5y = 100$$
=>$$x – y = 20$$… (1)

And, $${{100} \over {x+y}} = 1$$
=>$$x + y = 100$$ … (2)

$$2x = 120$$
=>$$x = 60$$

Putting value of x in (1),
$$60 – y = 20$$
=>$$y = 60 – 20 = 40$$ km/hr

Therefore, speed of car starting from point A = 60 km/hr

And, Speed of car starting from point B = 40 km/hr

(v) Let length of rectangle = x units and Let breadth of rectangle = y units

Area =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),
=>$$3x = 6 + 5y$$
=>$$x = {{6 + 5y} \over {3}}$$

Putting this in (2),
=>$$2({{6 + 5y} \over {3}}) + 3y = 61$$
=>$$12 + 10y + 9y = 183$$
=>$$19y = 171$$
=>$$y = 9$$ units

Putting value of y in (2),
$$2x + 3 (9) = 61$$
=>$$2x = 61 – 27 = 34$$
=>$$x = 17$$ units

Therefore, length = 17 units and, breadth = 9 units