Given below are the Class 10 Maths Important Questions for Linear equation
a) Concepts questionsQuestion 1. Which of these equation have i) Unique solution ii) Infinite solutions iii) no solutions
Solution
Condition |
Algebraic interpretation |
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ |
One unique solution only. |
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ |
Infinite solution. |
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ |
No solution |
One unique solution: (a), (d), (e), (f), (d)
Infinite solution :( b)
No solution: (c), (h)
Question 2. Using substitution method solve the below equation
a) x-2y+300=0, 6x-y-70=0
b) 5x-y=5, 3x-y=3
Solution
1 |
Method of elimination by substitution |
1) Suppose the equation are a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0 2) Find the value of variable of either x or y in other variable term in first equation 3) Substitute the value of that variable in second equation 4) Now this is a linear equation in one variable. Find the value of the variable 5) Substitute this value in first equation and get the second variable |
a) From Ist equation
x=2y-300
Substituting this in second equation
6(2y-300) –y-70=0 => y=170
Putting this Ist equation
x=> 40
b) ( 1,0)
Question 3. Using elimination method ,solve the following
a) x+y-40=0 ,7x+3y=180
b) x+10y =68 , x+15y=98
Solution
2 |
Method of elimination by equating the coefficients |
1) Suppose the equation are a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0 2) Find the LCM of a_{1} and a_{2} .Let it k. 3) Multiple the first equation by the value k/a_{1} 4) Multiple the first equation by the value k/a_{2} 4) Subtract the equation obtained. This way one variable will be eliminated and we can solve to get the value of variable y 5) Substitute this value in first equation and get the second variable |
a) x+y-40=0 ---(1)
7x+3y=180 ---(2)
Multiplying equation (1) by 7
7x+7y-280=0 ---(3)
Subtracting equation 2 from equation 7
7x+7y-280=0
7x+3y=180
We get
4y=100 => y=25
Substituting this in (1) ,we get x=15
b) ( 8,6)
Question 4 Solve the below linear equation using cross-multiplication method
a) (a-b)x + (a=b) y=a^{2}-2ab-b^{2} , (a+b)(x+y)=a^{2}+ b^{2}
b) x+y=5 ,2x-3y=4
Solution
3 |
Cross Multiplication method |
1) Suppose the equation are a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0
2) This can be written as
3) This can be written as 4) Value of x and y can be find using the x => first and last expression y=> second and last expression |
a) The equation can be written as
(a-b)x + (a+b) y-(a^{2}-2ab-b^{2}) =0
(a+b)x+(a+b)y-(a^{2}+ b^{2})=0
By cross multiplication we have
x=a+b, y=-2ab(a+b)^{-1}
b) (19/5,6/5)
Question 6 -True or False statement
a) Line 4x+5y=0 and 11x+17y=0 both passes through origin
b) Pair of lines 117x+14y=30 , 65x+11y=19 are consistent and have a unique solution
c) There are infinite solution for equation 17x+12y=30
d) x=0 ,y=0 has one unique solution
e) Lines represented by x-y=0 and x+y=0 are perpendicular to each other
f) 2x+6y=12 and 8x+24y=65 are consistent pair of equation
g) x+6y=12 and 4x+24y=64 are inconsistent pair of equation
Solution
Multiple choice Questions
Question 7 find the value of p for which the linear pair has infinite solution
12x+14y=0
36x+py=0
a) 14
b) 28
c) 56
d) 42
Solution (d)
For infinite solution
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$
So 12/36=14/p => p=42
Question 8. There are 10 students in XII class. Some are maths and some bio student.. The no of bio students are 4 more then math’s students?. Find the no of math’s and bio students
a) 1,9
b) 4,6
c) 2,8
d) 3,7
Solution (d)
Let x be math’s students
y be bio students
Then
x+y=10
y=x+4
Solving these linear pair through any method we get
x=3 and y=7
Question 9 which of the below pair are consistent pair?
a) x-3y=3 . 3x-9y=2
b) 51x +68y=110 , 3x+4y=99
c) 2x+3y=10 , 9x+11y=12
d) None of these
Solution c
For consistent pair
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$
Or
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Analyzing all of them ,we get c as the answer
Question 10. There are two numbers. Two conditions are there for them
i) Sum of these two numbers are 100
ii) One number is four time another number.
What are these numbers?
a) 20,80
b) 30,70
c)40,60
d) 25,75
Solution (a)
Let x and y are the number
x+y=100
y=4x
Solving them we get x=20 and y=80
Question 11 The number of solution of the linear pair
x+37y=123
21x-41y=125
a) No solution
b) One solution
c) infinite many
d) None of these
Solution (d)
Short answer question
Question 12 The sum of a 2 digit number and number obtained by reversing the order of the digits is 99. If the digits of the number differ by 3. Find the number
Solution 63 or 36
Question 13. Rajdhani train covered the distance between Lucknow and Delhi at a uniform speed. It is observed that if rajdhani would have run slower by 10 km/hr,it would have taken 3 hours more to reach the destination and if rajdhani would have run faster by 10 km/hr,it would have taken 2 hours less. Find the distance Lucknow and Delhi?
Solution
Let x be the speed and t be the original timing ,then distance between Lucknow and Delhi
Distance =speed X time =xy
Now from first observation
xy=(x-10)(y+3) => 3x-10y-30=0
From second observation
xy=(x+10)(y-2) => 2x-10y+20=0
Solving both we get
x=50km/hr
y=12hours
So distance between Lucknow and Delhi=50X12=600 Km