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Pair of Linear Equations in Two Variables : Exercise - 3.1 (Mathematics NCERT Class 10th)


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Q.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". (Isn't this interesting)? Represent this situation algebraically and graphically.
Sol.       Let's denote the present age of daughter and her father Aftab as x years and y years respective. Then, algebraic representation is given by the following equations :
             7(x – 7) = y – 7
              \Rightarrow 7x – 49 = y – 7
              \Rightarrow 7x – y = 42
             and, 3(x + 3) = y + 3
              \Rightarrow 3x + 9 = y + 3
              \Rightarrow 3x – y = – 6
             To obtain the equivalent graphical representation, we find two points on the line representing each equation. That is,  we find two solutions of each equation. That is, we find two solutions of each equation.
            These solutions are given below in the tables:
            For 7x – y = 42
1            For 3x – y = – 6
2
            To represent these equations graphically, we plot the points A(6, 0) and B(5, –7) to get the graph of (i) and the points C(0, 6) and D(–2, 0) give the graph of (ii).
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Q.2      The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Sol.       Let us denote the cost of bat be Rs. x and one ball be Rs. y. Then, the algebraic representation is given by the following equations :
             3x + 6y = 3900  \Rightarrow x + 2y = 1300 ...(1)
             and, x + 3y = 1300 ...(2)
             To obtain the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation.
             These solutions are given below in the table.
             For x + 2y = 1300
4

             For x + 3y = 1300
5
             We plot the points A(0, 650), B(1300, 0) to obtain the geometric representation of x + 2y = 1300 and C\left( {0,{{1300} \over 3}} \right) and B (1300, 0) to obtain the geometric representation of x + 3y = 1300.

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              We observe that these lines intersect at B (1300, 0).


Q.3      The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a months, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Sol.        Let us denote the cost of 1 kg of apple by Rs. x and cost of 1 kg grapes by Rs.y.
              Then the algebraic representation is given by the following equations :
              2x + y = 160 ...(1)
              4x + 2y = 300 ...(2)
               \Rightarrow 2x + y = 150
              To find the equivalent geometric representation, we find two points on the line representing each equation. That is, we find two solutions of each equation.
              2x + y = 160 

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              2x + y = 150
100
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