Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Pair of Linear Equations in Two Variables

Short Answer Type

101.

Find the values of k for which the pair of equations :
kx + 3y = k - 2 and 12x + ky = k has no solution.

284 Views

102. Places A and B are 100 km a part on a highway. One car starts from A and another from B at the same time. If the cars travels 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 ?

94 Views

103. Write whether the following pair of linear equation is consistent or not. x + y = 14; x - y = 4

238 Views

104. Without drawing the graph, find whether the lines representing the following pair of linear equations intersect at a point are parallel or coincident.

18 straight x minus 7 straight y equals 24 semicolon space space 9 over 5 straight x minus 7 over 10 straight y equals 9 over 10

227 Views

105. Without drawing the graph, find whether the lines representing the following pair of linear equations intersect at a point are parallel or coincident.

5 straight x plus 3 straight y minus 6 equals 0 semicolon space space 9 over 5 straight x plus 3 straight y equals 6

132 Views

Long Answer Type

106. The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and graphically.

Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y
Case I. Cost of 3 bats = 3x
Cost of 6 balls = 6y
According to question,
3x + 6y = 3900
Case II. Cost of I bat = x
Cost of 3 more balls = 3y
According to question,
x + 3y = 1300
So, algebraically representation be
3x + 6y = 3900
x + 3y = 1300
Graphical representation :
We have,    3x + 6y = 3900
⇒    3(x + 2y) = 3900
⇒    x + 2y = 1300
⇒    a = 1300 - 2y
Thus, we have following table :

Let the cost of 1 bat be Rs. x and cost of I ball be Rs.yCase I. Cost

We have, x + 3y = 1300
⇒ x = 1300 - 3y
Thus, we have following table :

When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations.

Let the cost of 1 bat be Rs. x and cost of I ball be Rs.yCase I. Cost