Class 10 Mathematics

Linear Equations

Exercise 3.5 (NCERT) Part 2

Question 5: Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(a) 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.

Solution: Let us assume that fixed charges is x and per day charges is y

Amount paid by student A = x + 20y = 1000

Amount paid by student B = x + 26y = 1180


`x+26y=1180`

`(a_1)/(a_2)=1` and `(b_1)/(b_2)=(20)/(26)=(10)/(13)`

It is clear that;

`(a_1)/(a_2)≠(b_1)/(b_2)`

Hence, there will be unique solution for the given pair of linear equations.

Subtracting first equation from second equation, we get;

`x + 26y – x – 20y = 1180 – 1000`

Or, `6y = 180`

Or, `y = 30`

Substituting the value of y in first equation, we get;

`x + 20y = 1000`

Or, `x + 20 x 30 = 1000`

Or, `x + 600 = 1000`

Or, `x = 1000 – 600 = 400`

Hence, fixed charges = Rs. 400 and per day charges = Rs. 30

(b) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ when 8 is added to its denominator. Find the fraction.

Solution: Let us assume numerator is x and denominator is y.

First condition:

`(x-1)/(y)=1/3`

Or, `3x-3=y`
Or, `3x-y-3=0` --------(1)

Second condition:

`(x)/(y+8)=1/4`

Or, `4x = y + 8`
Or, `4x – y – 8 = 0` ……… (2)

`(a_1)/(a_2)=3/4`

`(b_1)/(b_2)=1`

It is clear that;

`(a_1)/(a_2)≠(b_1)/(b_2)`

Hence, there will be unique solution for the given pair of linear equations.

Subtracting first equation from second equation, we get;

`4x – y – 8 – 3x + y + 3 = 0`
Or, `x – 5 = 0`
Or, `x = 5`

Substituting the value of x in second equation, we get;

`4x – y – 8 = 0`
Or, `4 xx 5 – y – 8 = 0`
Or, `20 – y – 8 = 0`
Or, `12 – y = 0`

Or, `y = 12`

Hence, `x = 5` and `y = 12`


(c) 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?

Solution: Let us assume that the number of correct answers = x and number of incorrect answers = y

First condition: `3x - y = 40`

Second condition: `4x - 2y = 50`

`(a_1)/(a_2)=3/4` and `(b_1)/(b_2)=1/2`

It is clear that;

`(a_1)/(a_2)≠(b_1)/(b_2)`

Hence, there will be unique solution for the given pair of linear equations.

Multiply the first equation by 2 and subtract second equation from the resultant:

`6x - 2y – 4x + 2y = 80 – 50`
Or, `2x = 30`
Or, `x = 15`

Substituting the value of x in second equation, we get;

`4x - 2y = 50`
Or, `4 xx 15 - 2y = 50`
Or, `60 – 2y = 50`
Or, `2y = 60 – 50 = 10`
Or, `y = 5`

Hence, number of correct answers = 15 and number of incorrect answers = 5


(d) 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?

Solution: Let us assume that the speed of car A is x and that of car B is y.

First condition: When both the cars are going in the same direction, their relative speed is; x – y

Hence,
`(100)/(x-y)=5`

Or, `5x-5y=100`
Or, `x-y=20`

Second condition: When the cars are moving towards each other, their relative speed is; x + y

Hence;

`(100)/(x+y)=1`

Or, `x+y=100`

Adding the two equations, we get;

`x – y + x + y = 20 + 100`
Or, `2x = 120`
Or, `x = 60`

Substituting the value of x in first equation, we get;

`x – y = 20`
Or, `60 – y = 20`
Or, `y = 60 – 20 = 40`

Hence, speed of car A = 60 km/h and speed of car B = 40 km/h

(e) 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.

Solution: Let us assume length is x and breadth is y.

Area of rectangle = xy

First condition: Length = x - 5 and breadth = y + 3

Area of rectangle:

`(x – 5)(y + 3) = xy – 9`
Or, `xy – 5y + 3x – 15 = xy – 9`
Or, `3x – 5y – 15 = - 9`
Or, `3x – 5y – 15 + 9 = 0`
Or, `3x – 5y – 6 = 0`

Second condition: Length = x +3 and breadth = y + 2

Area of rectangle:

`(x + 3)(y + 2) = xy + 67`
Or, `xy + 3y + 2x + 6 = xy + 67`
Or, `2x + 3y + 6 = 67`
Or, `2x + 3y – 61 = 0`

Multiplying the first equation by 2, we get;

`6x – 10y – 12 = 0`

Multiplying the second equation by 3, we get;

`6x + 9y – 183 = 0`

Subtracting these two equations, we get;

`6x + 9y – 183 – 6x + 10y + 12 = 0`
Or, `19y – 171 = 0`
Or, `19y = 171`
Or, `y = 9`

Substituting the value of y in any of the above equations, we get;

`6x + 9y – 183 = 0`
Or, `6x + 9 x 9 – 183 = 0`
Or, `6x + 81 – 183 = 0`
Or, `6x – 102 = 0`
Or, `6x = 102`
Or, `x = 17`

Hence, length = 17 unit and breadth = 9 unit


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Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3(Part 1)

Exercise:3.3 (part 2)

Exercise:3.4 (part 1)

Exercise:3.4 (part 2)

Exercise:3.5 (part 1)

Exercise:3.6 (part 1)

Exercise:3.6 (part 2)

Exercise:3.6 (part 3)