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Q1. If one of the angle of the triangle is  equal to the sum of the other two angles .Express this information into equivalent linear equation of two variables  and  also find the angles of triangle id  measure of one of the angle is 60 degree.

Solution

  Let  the first angle = x             Let the second angle=y             According to the given information, Third angle=x+y                         We know that sum of angles in  a triangle is equal to 180 degree.             So,             X+y+(x+y)=180             2x+2y=180             X+y=90                                    (i)              Now measure of one angle=60 degree.             Substituting in (i) we get             60+y=90             y=90-60=30             Third angle=x+y=60+30=90             Thus angles of triangle are 60, 30 and 90 degrees

Q2. Sita bought 18 apples for Rs 45 and 12 oranges for Rs 25. What will be the linear equation  of above statement ?

Solution

Let the apple be denoted by  A  Oranges be denoted by  O.   Sita bought 18 apples which is equal to 18A and  12 oranges  which is equal to 12O Total cost of 18 apples and  12 oranges=Rs45+Rs25=Rs70 Thus, the given information can be represented as a linear equation as: 18A+12O=70

Q3. Dhoni made century  only by hitting four’s and  sixes . Write  a linear equation in two variables to represent the above statement.

Solution

Let the number of fours hited  by Dhoni be x             And number of sixes hited by him be y. We know that he made century only by hitting fours and sixes So , 4x+6y=100 Thus the given information can be represented as  a linear equation  as   4x+6y=100

Q4. If x = 2 and y = 1 is the solution of the linear equation 2x + 3y + k = 0, find the value of k.

Solution

2x + 3y + k = 0 2(2) + 3(1) + k = 0 k = -7

Q5. John age is 10 years less than three times his son’s age. If John son is 20 years old, then how old is John?

Solution

Let us assume that John age is x years and his son’s age is y years. A linear equation in two variables can be formed to represent the given statement as  x = 3y – 10   We know that the age of John’s son is 20 years. This implies that the value of y is 20.             Substituting this value of y in the equation x = 3y – 10, we get:             x = 3 × 20 – 10             Þ x = 60 – 10             Þ x = 50             Thus, John’s age is 50 years.

Q6. After 5 years, the age of father will be two times the age of the son. Write a linear equation in two variables to represent this statement.

Solution

Let father's present age be = x years Son's present age = y years After 5 years father's age will be = (x + 5) years After 5 years son's age will be = (y + 5) years According to the question: x + 5 = 2(y + 5) x + 5 = 2y + 10 x - 2y = 10 - 5 x - 2y = 5

Q7. A part of monthly expenses of a family on milk is fixed, which is Rs.700 and remaining varies with quantity of extra milk taken at the rate of Rs. 25 per litre. Assuming quantity of extra milk required as x litres and total expenditure on milk Rs. y, write a linear equation representing above information.

Solution

Fixed monthly expenses on milk = Rs 700 Extra milk required = x Rate per litre for extra milk = Rs 25 Amount for extra milk = Rs 25x Total expenditure on milk = Rs y Therefore, 700 + 25x = y 25x - y + 700 = 0

Q8. Find four solutions of the equation 2x - y = 4.

Solution

2x - y = 4 Or, y = 2x - 4 …………equation (1) Putting x = 0 in equation (1), y = (2 x 0) - 4 = 0 - 4 = -4 Putting x = 1 in equation (1), y = (2 x 1) - 4 = 2 - 4 = -2 Putting x = 2 in equation (1), y = (2 x 2) - 4 = 4 - 4 = 0 Putting x = 3 in equation (1), y = (2 x 3) - 4 = 6 - 4 = 2 Hence, four solutions of equation 2x - y = 4 are (0, -4), (1, -2), (2, 0), (3, 2).

Q9. The cost of six cups is equal to  48 rupees  and  the cost of 5 plates  is 30 rupees. Express the above information in the form of linear equation.

Solution

Let the  the Cup be denoted by C and plate be denoted by P   Six Cup is equal to   6C and  5 plate is equal to  5P   Total cost of six cups and five plates=Rs48+Rs30=Rs78   So, 6C+5P=78

Q10. Find four solutions of the following linear equation in two variables. 2 (x + 3) -3 (y - 1) = 0

Solution

2 (x + 3) -3 (y - 1) = 0 2x + 6 - 3y + 3 = 0 2x - 3y = -9 Solutions can be calculated as below: (i) When x = 0, we have: -3y = -9 y = 3 (x, y) = (0, 3) (ii) When y = 0, we have: 2x = -9 x = -4.5 (x, y) = (-4.5, 0) (iii) When y = 1, we have: 2x - 3 = -9 2x = -6 x = -3 (x, y) = (-3, 1) (iv) When y = -1, we have: 2x - (-3) = -9 2x + 3 = -9 2x = -12 x = -6 (x, y) = (-6, -1) The four solutions of the given equation are (0, 3), (-4.5, 0), (-3, 1), (-6, -1).