Last updated at Dec. 28, 2018 by Teachoo

Transcript

Ex3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : 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. Let the fixed charge = Rs x & Charge per day taken food = Rs y Given that charge paid by student A for 20 days is Rs 1000 Fixed charge + 20 × (Charge per day) = 1000 x + 20y = 1000 Also, given that charge paid by student B for 26 days is Rs 1180 Fixed charge + 26 × (Charge per day) = Rs 1180 x + 26y = 1180 Hence, our equations are x + 20y = 1000 …(1) x + 26y = 1180 …(2) From (1) x + 20y = 10000 x = 1000 – 20y Putting value of x in (2) x + 26y = 1180 (1000 – 20y) + 26y = 1180 – 20y + 26y = 1180 – 1000 6y = 180 y = 180/6 y = 30 Putting y = 30 in equation (3) x = 1000 – 20y x = 1000 – 20 (30) x = 1000 – 600 x = 400 Hence, x = 400, y = 30 is the solution of the equations So, the fixed charges = x = Rs 400 Change per km = y = Rs 30 Ex 3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction Let numerator be x and denominator be y So, fraction is 𝑥/𝑦 Given that, if 1 is subtracted from numerator fraction becomes 1/3. (𝑁𝑢𝑛𝑒𝑟𝑎𝑡𝑜𝑟 −1)/(𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 )=1/3 (𝑥 − 1)/(𝑦 )=1/3 3(x – 1) = 1(y) 3x – 3 = y 3x – y = 3 Also, if 8 is added to the denominator, fraction becomes 1/4. (𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 )/(𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 + 8)=1/4 (𝑥 )/(𝑦 + 8)=1/4 4(x) = 1(y + 8) 4x = y + 8 4x – y = 8 From (1) 3x – y = 3 3x = y + 3 x = ((𝑦 + 3)/3) Putting value of x in (2) 4x – y – 8 = 0 4 ((𝑦 + 3)/3 ) – y – 8 = 0 Multiplying the equation by 3 3 ×"4 (" (𝑦 + 3)/3 ")"−3×𝑦−3×8=0×8 4(y + 3 ) – 3y – 24 = 0 4y + 12 – 3y – 24 = 0 4y – 3y – 24 + 12 = 0 y – 12 = 0 y = 12 Putting y = 12 in equation (1) 3x – y = 3 3x – 12 = 3 3x = 12 + 3 3x = 15 x = 15/3 x = 5 Therefore x = 5, y = 12 is the solution So, Numerator = x = 5 Denominator = y = 12 Hence, original fraction = 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟/𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 "= " 𝑥/𝑦=5/12 Ex 3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (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? Let the number of right answers be x And the number of wrong answers be y Given that , Yash scored 40 marks if we get 3 marks for right answer and lose 1 mark for wrong answer 3x – y = 40 Also, Yash scored 50 marks if he gets 4 marks for correct answer and loses 2 mark for wrong answer 4x – 2y = 50 2(2x – y)= 50 (2x – y) = 50/2 2x – y = 25 Hence, our equations are 3x – y = 40 …(1) 2x – y = 25 From (1) 3x – y = 40 3x – 40 = y y = 3x – 40 Putting value of y in (2) 2x – y = 25 2x – (3x – 40) = 25 2x – 3x + 40 = 25 2x – 3x = 25 - 40 -x = -15 x = 15 Putting x in (1) 3x – y = 40 3(15) – y = 40 3(15) – y = 40 45 – y = 40 45 – 40 = y 5 = y y = 5 Therefore x = 15, y = 5 is the solution So, Number of right answers = x = 15 Number of wrong answers = y = 5 Total questions in the test = x + y = 15 + 5 = 20 Ex3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (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? Let the speed of first car be x km/hr & let the speed of second car be y km/hr If travelling in same direction Distance travelled by 1st car = AC = AB + BC Distance travelled by 2nd car = BC Difference of distance travelled = AB + BC – BC = AB = 100 km Distance travelled by 1st car – Distance travelled by 2nd car = 100 km (Speed of first car × 5 hours) – (Speed of 2nd car × 5 hours) = 100 km 5x – 5y = 100 5x – 5y = 100 5(x – y) = 100 (x – y) = 100/5 x – y = 20 If travelling in opposite direction Distance travelled by 1st car = AD Distance travelled by 2nd car = BD Sum of distance travelled = AD + BD = AB = 100 km Distance travelled by 1st car + Distance travelled by 2nd car = 100 km (Speed of first car × 1 hours) – (Speed of 2nd car × 1 hours) = 100 km x + y = 100 So, our two equations are x – y = 20 …(1) x + y = 100 …(2) From (1) x – y = 20 x = y + 20 Putting value of x in (2) x + y = 100 (y + 20) + y = 100 2y + 20 = 100 2y = 100 – 20 2y = 80 y = 80/2 y = 40 Putting y = 40 in equation (1) x – y = 20 x – 40 = 20 x = 40 + 20 x = 60 Therefore x = 60, y = 40 is the solution Thus, Speed of first car = x km/hr = 60 km/hr Speed of second car = y km/hr = 20 km/hr Ex 3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (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. Let length of rectangle be x units And breadth of rectangle be y units Hence, Area = Length × Breadth Area = xy Given that, Area gets reduced by 9 square units, if length is reduced by 5 units and breadth increased by 3 units and Area increase by 67 square units, if length is increased by 3 units and breadth increased by 2 units (x – 5) (y + 3) = xy – 9 x (y + 3) – 5 (y + 3) = xy – 9 x (y + 3) – 5 (y + 3) = xy – 9 xy + 3x – 5y - 15 = xy – 9 xy + 3x – 5y - 15 - xy + 9 = 0 3x – 5y – 6 = 0 (x + 3) (y + 2) = xy + 67 x(y + 2) + 3(y + 2) = xy + 67 xy + 2x + 3y + 6 = xy + 67 xy + 2x + 3y + 6 - xy – 67 = 0 2x + 3y – 61 = 0 Hence, our equations are 3x – 5y – 6 = 0 …(1) 2x + 3y – 61 = 0 …(2) From (1) 3x – 5y – 6 = 0 3x = 6 + 5y x = (6 + 5𝑦)/3 Putting value of x in (2) 2x + 3y – 61 = 0 2((6 + 5𝑦)/3) + 3y – 61 = 0 Multiplying both sides by 3 3 × 2(((6+5𝑦))/3) + 3 × 3y – 3 × 61 = 0 2(6 + 5y) + 9y – 183 = 0 12 + 10y + 9y - 183 = 0 19y – 171 = 0 19y = 171 y = 171/19 y = 9 Putting y = 9 in equation (1) 3x – 5y – 6 = 0 3x – 5(9) – 6 = 0 3x – 45 – 6 = 0 3x – 51 = 0 3x = 51 x = 51/3 x = 17 Therefore x = 17, y = 9 is the solution Hence, Length of rectangle = x = 17 units Breadth of rectangle = y = 9 units

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.