Form the pair of linear equations for the following problems and

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.
Advertisement

 Multiple Choice QuestionsShort Answer Type

41.

Solve the following pairs of linear equations by the substitution method.


space space space space space space space space fraction numerator 3 straight x over denominator 2 end fraction minus fraction numerator 5 straight y over denominator 3 end fraction equals negative 2

space space space space straight x over 3 plus straight y over 2 equals 13 over 6


    
     

 

160 Views

42.

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

347 Views

43. Form the pair of linear equations for the following problems and find their solutions by substitution method.

The difference between two numbers is 26 and one number is three times the other. Find them.
111 Views

44. Form the pair of linear equations for the following problems and find their solutions by substitution method.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
108 Views

Advertisement
45.

Form the pair of linear equations for the following problems and find their solution by substitution method.

The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball.

202 Views

46.

Form the pair of linear equations for the following problems and find their solution by substitution method.

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

154 Views

Advertisement

47.

Form the pair of linear equations for the following problems and find their solution by substitution method.

A fraction becomes 9 over 11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes space 5 over 6 . Find the fraction.


Let the numerator of the fraction be x and denominator bey.


Let the numerator of the fraction be x and denominator bey.
Thus, we

Thus, we have following equations
11x - 9y = -4    ...(i)
6x - 5y = -3    ...(ii)
From (ii), we have
6x - 5y=-3

rightwards double arrow space space space 6 straight x space equals space 5 straight y minus 3
rightwards double arrow space space space straight x equals fraction numerator 5 straight y minus 3 over denominator 6 end fraction space space space space space space... left parenthesis iii right parenthesis
Substituting the value of (iii) in (i), we get

rightwards double arrow space space 11 straight x minus 9 straight y equals negative 4
rightwards double arrow space space space 11 open parentheses fraction numerator 5 straight y minus 3 over denominator 6 end fraction close parentheses minus 9 straight y equals negative 4
rightwards double arrow space space space fraction numerator 11 left parenthesis 5 straight y minus 3 right parenthesis minus 54 straight y over denominator 6 end fraction minus 4
rightwards double arrow space space space 55 straight y minus 33 minus 54 straight y equals negative 24
rightwards double arrow space space space space space straight y minus 33 equals negative 24
rightwards double arrow space space space space space straight y equals negative 24 plus 33
rightwards double arrow space straight y space equals space 9
Now, substituting the value ofy in (iii), we get

straight x equals fraction numerator 5 straight y minus 3 over denominator 6 end fraction
rightwards double arrow space space straight x equals fraction numerator 5 cross times 9 minus 3 over denominator 6 end fraction
equals fraction numerator 45 minus 3 over denominator 6 end fraction equals 42 over 6 equals 7
Hence, the required fraction.

171 Views

Advertisement
48.

Form the pair of linear equations for the following problems and find their solution by substitution method.

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

183 Views

Advertisement
49.

Solve the following pair of linear equations by the elimination method and the substitution method

x + y = 5 and 2x - 3y = 4

113 Views

50.

Verify that the following system of equations has a unique solution and find the solution
ax + by + m = 0; ax - cy - n = 0

147 Views

Advertisement