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Class 10 Maths Important Questions for Linear equation



Given below are the Class 10 Maths Important Questions for Linear equation

a) Concepts questions
b) Calculation problems
c) Multiple choice questions
d) Long answer questions
e) Fill in the blank's

Question 1. Which of these equation have i) Unique solution ii) Infinite solutions iii) no solutions

  1. 152x-378y=-74  , -378x +152y=-604
  2. x+2y =10 , 3x+6y=30
  3. 3x+4y=6  , 12x+16x=30
  4. 7x-11y=53 ,  19x -17y=456
  5. x=7  ,y=-2
  6. ax+by=a-b , bx-ay=a+b
  7. 2x+3y=0, 124x+13y=0
  8.  y=11 ,y =-11

 

 

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

a1x+b1y+c1=0

a2x +b2y+c2=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

a1x+b1y+c1=0

a2x +b2y+c2=0

2) Find the LCM of a1 and a2 .Let it k.

3) Multiple the first equation by the value k/a1

4) Multiple the first equation by the value k/a2

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=a2-2ab-b2 , (a+b)(x+y)=a2+ b2

b)  x+y=5  ,2x-3y=4

 

Solution

3

Cross Multiplication method

1) Suppose the equation are

a1x+b1y+c1=0

a2x +b2y+c2=0

 

2) This can be written as

Cross multipication

 

3) This can be written as

Cross multipication

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-(a2-2ab-b2) =0

(a+b)x+(a+b)y-(a2+ b2)=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

  1. True
  2. True
  3. True
  4. True
  5. True
  6. False
  7. True

 

 

 

 

 

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



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