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Question: 4 - One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number? Solution: Let one of the digit , which is at ones place, of a two digit number = a Therefore, other digit of the two digit number = 3a Therefore, number = (10 x 3a) + a = 30a + a = 31a After interchange, the digit = 10a + 3a = 13a Now, since sum of the original and resulting number = 88 Therefore, 31a + 13a = 88 ⇒ 44 a = 88 Now, after dividing both sides by 44, we get By substituting the value of ‘a’ in original number we get Since, original number = 31a = 31 x 2 = 62 Thus, the number = 62 Answer- One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number? ...

Linear Equation Exercise 3.1 (NCERT) Question 1: Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically. Solution: Let us assume Aftab’s current age = x and his daughter’s current age = y Seven years ago: Aftab’s age = x - 7 and daughter’s age = y - 7 As per question; x – 7 = 7(y – 7) Or, x – 7 = 7y – 49 Or, x = 7y – 49 + 7 Or, x = 7y – 42 Or, 7y – x – 42 = 0 ……(1) 9 12 15 18 21 24 27 30 33 36 39 42 45 y 1 2 3 4 5 6 7 8 9 10 11 12 13 The following graph is plotted for the given pair of linear equation

Question - 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent the above statement. Solution:-  Let the cost of a notebook be x and that of a pen be y. According to question, The cost of a notebook is twice the cost of a pen Or, x = 2y Or, x - 2y = 0.

Q2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: (iii) -2x + 3y = 6 (iv) x = 3y (v) 2x = -5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x (iii) -2x + 3y = 6 Or, -2x + 3y – 6 = 0 Comparing the equation with ax + by + c = 0, we have a = -2, b = 3, c = -6 (iv) x = 3y Or, x - 3y = 0 Comparing the equation with ax + by + c = 0, we have a = 1, b = -3, c = 0 (v) 2x = -5y Or, 2 x + 5y = 0 Comparing the equation with ax + by + c = 0, we have a = 2, b = 5, c = 0 (vi) 3x + 2 = 0 3 x + 0y + 2 = 0 Comparing the equation with ax + by + c = 0, we have a = 3, b = 0, c = 2 (vii) y – 2 = 0 Or, 0 x + 1y - 2 = 0 Comparing the equation with ax + by + c = 0, we have a = 0, b = 1, c = -2 (viii) 5 = 2x Or, -2x - 0y + 5 = 0 Comparing the equation with ax + by + c = 0, we have a = -2, b = 0, c = 5

Question: 3 - Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number? Solution: Let the number at ones place of two digit number = a According to question, the sum of digits of given two digit number = 9 i.e. Digit at tens place + digit at ones place = 9 Or, Digit at tens place + a = 9 By transposing ‘a’ to RHS, we get Digit at tens place = 9 – a Thus, the number = 10(9 – a) + a After interchange of digit, the number = 10a + (9 – a) Since, number obtained after interchange of digit is greater than the original number by 27 Therefore, New number – 27 = Original number Here, we have original number = 10(9 – a) + a And, new number = 10a + (9 – a) ⇒ 10a + (9 – a) – 27 = 10 (9 – a) + a ⇒ 10a + 9 – a – 27 = 90 – 10a + a ⇒ 10a – a + 9 – 27 = 90 – 9a ⇒ 9a – 18 = 90 – 9a By transposing 18 to RHS, we get 9a = 90 – 9a + 18 By transposing – 9...

Exercise 2.4 Question: 2 - A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? Solution: Let the given positive number = a Therefore, another number which is 5 times of it = 5a Now, after adding 21 to both of the number, First number = a + 21 Second number = 5a + 21 According to question, one new number becomes twice of the other new number Therefore, Second number = 2 x first number i.e. 5a + 21 = 2 (a + 21) ⇒ 5a + 21 = 2a + 42 By transposing ‘2a’ to LHS, we get ⇒ 5a + 21 – 2a = 42 Now, after transposing 21 to RHS, we get ⇒ 5a – 2a = 42 – 21 ⇒ 3a = 21 After dividing both sides by 3, we get 3a\3 = 21\3 we get a=7 Therefore, another number 5a = 5 x 7 = 35 Thus, required numbers are 7 and 35 Answer