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Ex 3.5, 4 - Form and solve (i) A part of monthly hostel - Cross Multiplication Method

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise
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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 …(2) 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 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 Total questions in the test = x + y = 15 + 5 = 5 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 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 = 20 units

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