Chapter 3: Pair of Linear Equations in Two Variables

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Pair of linear equations in two variables CBSE NCERT Solutions

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 frac{1}{3}   when 1 is subtracted from the numerator and it becomes  frac{1}{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:

Solutions:

(i) Let fixed monthly charge is Rs x and charge of food for one day is Rs y

According to given conditions,

+ 20y = 1000 … (1),

and x + 26y = 1180 … (2)

Subtracting equation (1) from equation (2), we get

6y = 180

 = 30

Putting the value of y in (1), we get

+ 20 (30) = 1000

 = 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,

\frac{x-1}{y}=\frac{1}{3} ....(1)               \frac{x}{y+8}=\frac{1}{4} ....(2)

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

 = 5

Putting the value of x in (1), we get

3 (5) – y = 3

15 – = 3

 = 12

Therefore, numerator = 5 and, denominator = 12

It means fraction   \frac{x}{y} = \frac{5}{12}

(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

 = 15

Putting the value of x in (1), we get

3 (15) – y = 40

45 – = 40

 = 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,

\frac{100}{x-y} = 5  (Assuming x > y)

5x − 5y = 100

 – y = 20 … (1)

And,

\frac{100}{x+y} =1

 + y = 100 … (2)

Adding (1) and (2), we get

2x = 120

 = 60 km/hr

Putting the value of x in (1), we get

60 – y = 20

 = 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

\Rightarrow x = \frac{6 + 5y}{3}

Putting this in (2), we get

\Rightarrow 2\left ( \frac{6+5y}{3} \right ) + 3y = 61

12 + 10y + 9y = 183

19y = 171

 = 9 units

Putting the value of y in (2), we get

      2x + 3 (9) = 61

2x = 61 – 27 = 34

 = 17 units

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

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