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(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.

Answer :

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)

Adding (1) and (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

- Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method. (i) \(x – 3y – 3 = 0,3x – 9y – 2 = 0\)(ii) \(2x + y = 5,3x + 2y = 8\)(ii) \(3x - 5y = 20,6x - 10y = 40\)(ii) \(x - 3y – 7 = 0,3x - 3y – 15 = 0\)
- (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7(a - b) x + (a + b) y = 3a + b – 2 (ii) For which value of k will the following pair of linear equations have no solution? 3x + y = 1(2k - 1) x + (k - 1) y = 2k + 1
- Solve the following pair of linear equations by the substitution and cross-multiplication methods: \(8x + 5y = 9\) \(3x + 2y = 4\)

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