## Maths Question Bank for Class 10 SSC 2022 Maharashtra Board

Maths Question Bank for Class 10 SSC 2022 Maharashtra Board

## Standard 10 SSC Mathematics

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  and equation (II) by .

(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