Maths Question Bank for Class 10 SSC 2022 Maharashtra Board

Maths Question Bank for Class 10 SSC 2022 Maharashtra Board

Standard 10 SSC Mathematics

Question Bank (Answers) 

1.Linear Equation In Two Variables 

Q.1(A) MCQ

 1. To draw the graph of 4x + 5y = 19, if x = 1 is taken then what will be the value of y?

 A) 4             B) 3              C) 2                 D) −3

ANS]-         B) 3

2) For the equations with variables x and y, if Dx = 26, Dy = −39 and D = 13 then x =?

 A)2             B) − 3            C) − 2              D) 3

ANS          -A)2

3) Which of the following is linear equation in two variables?

 A) 𝑥 3 + 5 𝑦 = 6             B) 2×2-3y=8-3y          

 C) x+2y=5-3y                 D) 3 x²+y

ANS           C) x+2y=5-3y 

4) which of the following is not the solution of 3x+6y=12?

 A) (-4,4)                      B) (0,2) 

 C) (8, -2)                    D) (3,1) 

ANS)  D) (3,1)

5) |3 5| = 2 ∴ 𝑥 = ——-

    |2 𝑥|

A)3           B)4 

C)3           D)4 

ANS) D)4

6) For equations 5x + 3y + 11 = 0 and 2x + 4y = – 10 find D.

A)14            B)-14 

C)26            D)26 

ANS) A)14

7) If 49 x – 57 y = 172 and 57 x – 49 y = 252 then x + y =? 

A) 80        B) 0 

C) 10        D) 8

ANS) C)10 

8) The solution of the equation 2x – y = 2 is ——.

 A) (2,2)           B) (5,2) 

C) (2,5)            D) (5,5)

ANS)A) (2,2)

9) The solution of the equation x-y=10 and x+y=70 is ——–. 

A) (40,30)                     B) (30,40) 

C) (10,60)                     D) (50,20)

ANS)A) (40,30) 

10) Find the value of Dx for the equation 4𝑥 + 3 𝑦 = 19 and 4 𝑥 − 3 𝑦 = −11 

A) 24           B) 0 

C) −24         D) 108 

ANS) C) −24 

Q. 1 B) Each of 1 mark 

1) State with reason whether the equation 3𝑥 2− 7y = 13 is a linear equation with two variables?

ANS) (No)

Reason-The degree of variable of X is 2

2) Show the condition using variable x and y: Two numbers differ by 3 

ANS)Answer: Let the bigger number be y and smaller number be x

ATQ

difference of the number is 3

y – x = 3 ( since the difference is positive 3 it should be bigger number – smaller number)

Hence, the required condition is y – x = 3

3) For the equation 4x + 5y = 20 find y when x = 0 

ANS) Substituting x=0

          4x+5y=20

          4 x 0 + 5y=20

          0 + 5y=20

          5y=20-0

          5y=20

          y =20

4) Write any two solutions of the equation x + y = 7. 

ANS) A linear equation is of the form ax + b = c.

Let us find two solutions of x + y = 7

  • if x = 1, then y = (7 – x) = (7 – 1) = 6, therefore, (1, 6) is one solution.
  • if x = 0, then y = (7 – x) = (7 – 0) = 7, therefore, (0, 7) is another solution.

Thus, the two solutions of the equation x + y = 7 are (1, 6) and (0, 7).

5) Decide whether (0, 2) is the solution of the equation 5x + 3y = 6 

ANS) 5x + 3y = 0

         =5 x 0 + 3 x 2  

         =0 + 6 

         =6

   Therefore (0.2) is the solution of the given equation

 6) Write any two solution of the equation a – b = – 3 

ANS  a – b = 3

          a = – 3 + b

        b = – 3 – a

7) If x+2y=5 and 2x+y=7 then find the value of x+y 

ANS) Suppose this is 

         x+2y=5_____(1)  equation

         2x+y=7 ______(2)  equation

          Adding equation (1) and (2)

             x+2y=5

          + 2x+y=7

            3x+3y=12

Take common 3 therefore 

3(x+y)=12

X+y =12 4 /  3  1

x + y =4

8) If Dx = 24 and x = – 3 then find the value of D. 

ANS)   x=Dx / D   

Therefore  x=Dx /

              24  8 / 3  1 

= -8

9) The cost of the book is 5 rupees more than twice the cost of a pen. Show this using 

linear equation by taking Cost of book(x) and cost of a pen(y).

ANS)

Cost of book =x

Cost of pen= y

Twice the cost of pen Means

x=2y+5

x-2y=5 

10) If 𝑎 4 + 𝑏 3 = 4, write the equation in standard form.

ANS)  a     b    

           4  + 3    = 4

Cross multiply 

3a + 4b =4

     12 

Taking 12 on right hand side 

3a+4b =48

3a+4b-48 =0

Q.2 A) Complete the activity (2 marks)

1) Complete the table to draw the graph of 2x – 3y = 3,

 __________________________

|  X                  −6                 3    | 
|  y                   -5                  1    |
| (x, y)            (-6,-5)      (3,1)   |  

—————————————— 

Solution- Substituting x= -6 

2x – 3y =3 

We get 

2 x (-6) -3y =3 

-12 – 3y = 3

-3y =3 +12 

-3y =15 

-y =15/3 

 y= -5 

Solution> y=1

2x – 3y =3

2 x x – 3 (1) = 3

2x – 3 = 3

2x = 3 +3 

2x = 6

x= 6/2 

x=3 

2. Solve the following to find the value of the following determinant.

|3 −2|

|4   5|  = 3 × (5) – (-2) × 4 =(15) + 8 =(23)


3) Complete the activity to find the value of x

3 x + 2y = 11  ____ (I)

and 

2 x + 3 y = 4 ____(II)

Solution: Multiply equation (I) by [3] and equation (II) by [2].

(3)× (3 x + 2y = 11)  ∴ 9 x + 6 y = 33   |     9 x + 6 y = 33  
(2) × (2 x + 3 y = 4)  ∴ 4 x + 6 y = 8     |      4 x + 6 y = 8  
subtract (II) from (I),                            |       –       –      – 
(5) x = 25                                              |  ———————–  
∴ x = (5)                                                |      5x + 0   = 25  
                                                                     x = 25/5 
                                                                     ∴x=5

4) If (2, 0) is the solution of 2x + 3y = k then finds the value of k by

completing the activity

Solution: (2,0) is solution of the equation 2x +3y = k

Putting x =(2) and y =(0)

∴ 2( 2 ) + 3 × 0 = k

∴ 4 + 0 = k

∴ k =(4 )

5) To find the values of x and y for the equations x- 2 y =5 and 2 x+ 3 y =10

complete the activity.

      |1     -2|

D =|2      3 |= 3 + 4 = 7

      |5     -2|

D =|10   3 |= ( )

      |1       5|

D =|2     10|= ( )

By Cramer’s rule

       Dx                 Dy
X = -— =  () , Y= -— = ( ) 
        D                  D 

Q. 2 B) Each of 2 marks

1) The difference between an angle and its complement is 10° find measure

of a larger angle.

Ans- Let the angle x and y

Sum of complementary angles is 90o

Means x + y =  90o

This is equation ____|

x + y =  90o_____|

According to the given condition The difference between an angle and its complement 10o 

x – y = 10o 

This is Equation 2 

X – y = 10o _____||

Adding equation | and ||

        x + y = 90o        

     + x – y = 10o 

________________

      2x + 0 = 100o

       2x= 100o 

        x= 100/2

 X = 50

2) Find the value of |5     2|

                                  |0  − 1|

Ans- Cross Multiply 

    5 x 1 – (-2) x (-3)

    5 – 6

      -1

∴Value of the determinant is -1

 2)Cross multiply 

5 x (-1) – 2 x 0

-5   –  0

 -5

 The value is -5

3) For the equation y + 2x = 19 and 2x − 3y = − 3 Find the value of D

Ans-The given simultaneous equations are 

y – 2x = 19

2x – 3y = – 3

∴y + 2x = 19 _____|

 In equation |

a|= 2

b|=1 

c|=19

Now 2x – 3y = -3 _____||

In equation ||

a2= 2

b2= -3

c2= -3

D=|a1     b1 |

     |a2     b2 |

= |2     1|

   |2    -3|

=2x – (-3) – 2 x 1

=       -6 -2 

=         -8 

       D = -8

3)2x – y = 2 

Solution 

Given x = 3

2x – y = 2

Substituting x = 3  

2 x 3 -y = 2

6 – y = 2 

-y = 2 – 6

– y = – 4

 Y = 4

4) In the equation 2x − y = 2 if x = 3 then find y=?

Ans-Solution

Give (2,-5) is the solution of the equation 2x – ky = 14

Substituting x = 2 , y = -5 in equation 

2x – ky =14 

2 x (2) – k x (-5) = 14

4 – (-5k) = 14

4 + 5k = 14 

∴5k = 14 – 4

∴5k=10

∴k =10/5

∴k=2

5) If (2, −5) is the solution of the equation 2x − k y = 14 then find k=?

Ans-Solution

Substituting b=4 in equation 

A + 2b = 7

a + 2b = 7

a + 2 x 4 = 7

a + 8 =7

a = 7 – 8

a = -1