SSC Maharashtra Board Maths 1 Question Paper 2022 with Solutions

Maharashtra Board SSC Class 10 Maths 1 Question Paper 2022 with Solutions Answers Pdf Download.

SSC Maths 1 Question Paper 2022 with Solutions Pdf Download Maharashtra Board

Time : 2 Hours Max.
Marks : 40

General Instructions :

1. All questions are compulsory.
2. Use of calculator is not allowed.
3. The numbers to the right of the questions indicate full marks.
4. In case ofMCQs [Q. No. 1(A)] only the first attempt will be evaluated and will be given credit.
5. For every MCQ, the correct alternative (A), (B), (C) or (D) with subquestion number is to be written as an answer.

Question 1.
(A) Four alternative answers are given for every subquestion. Choose the correct alternative and write its alphabet with subquestion number. [4]

(i) Which one is the quadratic equation ?
(A) \(\frac{5}{x}\) – 3 = x²
(B) x(x + 5) = 2
(C) n – 1 = 2n
(D) \(\frac{1}{x^2}\)(x + 2) = x
Answer:
(B) x(x + 5) = 2.

(ii) First four terms of A.P. are ___, whose first term is -2 and common
difference is -2.
(A) -2, 0, 2, 4
(B) -2, 4, -8, 16
(C) -2, -4, -6, -8
(D) -2, -4, -8, -16
Answer:
(C) – 2, – 4, – 6, – 8

(iii) For simultaneous equations in variables x and y, D_x = 49, D_y = – 63, D = 7, then what is the value of y?
(A) 9
(B) 7
(C) -7
(D) -9
Answer:
(D) – 9

(iv) Which number cannot represent a probability ?
(A) 1.5
(B) \(\frac{2}{3}\)
(C) 15%
(D) 0.7
Answer:
(A) 1.5

(B) Solve the following subquestions : [4]

(i) To draw a graph of 4x + 5y -19, find y when x = 1.
Solution:
y = 3

(ii) Determine whether 2 is a root of quadratic equation 2m² – 5m = 0.
Solution:
The given equation is
2m² – 5m = 0 ……(i)
Putting m = 2 in L.H.S. of equation (i), we get
L.H.S. = 2 × (2)² – 5 × (2)
= 2 × 4 – 10
= 8 – 10
= -2
∴ L.H.S. ≠ R.H.S.
∴ 2 is not the root of given equation.

(iii) Write second and third term of an A.P. whose first term is 6 and common difference is – 3.
Solution:
a = 6, d = – 3 [Given]
∴ t₁ = a = 6
t₂ = t₁ + d
∴ t₂ = 6 + (-3)
∴ t₂ = 6 – 3
∴ t₂ = 3
t₃ = t₂ + d
∴ t₃ = 3 + (- 3)
∴ t₃ = 3 – 3
∴ t₃ = 0
∴ t₂ = 3 and t∴ t₃ = 0

(iv) Two coins are tossed simultaneously. Write the sample space ‘S’.
Solution:
If two coins are tossed simultaneously,
Sample space, S = {HH, HT, TH, TT}

Question 2.
(A) Complete the following activities and rewrite it (any two)
(i) Complete the activity to find the value of the determinant.
Activity:

Solution:

(ii) Complete the following activity to find the 19th term of an A.P. 7, 13, 19, 25, …… :
Activity:
Given A.P.: 7, 13, 19, 25………
Here first term a = 7; t₁₉ = ?
Solution:
Given, A.P.: 7, 13, 19, 25, ………
Here, first term a = 7; t₁₉ = ?

Solution:
Given, A.P. : 7, 13, 19, 25,………..
Here, first term a = 7; t₁₉ = ?

(iii) If one die is rolled, then to find the probability of an event to get prime number on upper face, complete the following activity.
Activity:
One die is rolled.
‘S’ is sample space.

Solution:
One die is rolled.
‘S’ is sample spaces.

(B) Solve the following subquestions (any four): [8]

(i) To solve the following simultaneous equations by Cramer’s rule, find the values of D_x and D_y.
3x + 5y = 26
x + 5y =22
Solution:
3x + 5y = 26
x + 5y = 22
By Cramer’s rule
D_x = \(\left|\begin{array}{ll}
26 & 5 \\
22 & 5
\end{array}\right|\)
= 26 × 5 – 22 × 5
= 130 – 110
= 20
D_y = \(\left|\begin{array}{ll}
3 & 26 \\
1 & 22
\end{array}\right|\)
= 3 × 22 – 1 × 26
= 66 – 26
= 40
∴ D_x = 20 and D_y = 40

(ii) A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue ?
Solution:
Let 5 red pens be R₁, R₂, R₃, R₄, R₅; 8 blue pens be B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈ and 3 green pens be G₁, G₂, G₃.
Sample space
S = {R₁, R₂, R₃, R₄, R₅, B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈, G₁, G₂, G₃}
∴ n(S) = 16
Let, A be event Rutuja picks a blue pen.
∴ A = {B₁, B₂, B₃, B₄, B₅, B₆, B₇, B₈}
∴ n(A) = 8
P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{8}{16}\)
∴ P(A) = \(\frac{1}{2}\)
∴ The probability that Rutuja picks a blue pen is \(\frac{1}{2}\)

(iii) Find the sum of first ‘n’ even natural numbers.
Solution:
First n even natural numbers are 2, 4, 6, 8, …….. 2n
t₁ = first term = 2
t_n = last term = 2n
S_n = \(\frac{n}{2}\)[t₁ + t_n] [formula]
= \(\frac{n}{2}\)[2 + 2n]
= \(\frac{n}{2}\) × 2(1 + n)
= n(1 + n)
∴ The sum of first n even natural numbers is n(1 + n).

(iv) Solve the following quadratic equation by factorisation method :
x² + x – 20 = 0
Solution:
x² + x – 20 = 0
∴ x + 5x – 4x – 20 = 0
∴ x(x + 5) – 4(x + 5) = 0
∴ (x + 5)(x – 4) = 0
∴ x + 5 = 0 or x – 4 = 0
∴ x = – 5 or x = 4
∴ -5 and 4 are roots of given quadratic equation.

(v) Find the values of (x + y) and (x – y) of the following simultaneous equations :
49x – 57y = 172
57x – 49y = 252
Solution:
49x – 57y = 172 ……. (i)
57x – 49y = 252 …….. (ii)
Adding equations (i) and (ii),

Subtracting equation (ii) from equation (i),

Question 3.
(A) Complete the following activity and rewrite it (any one). [4]

(i) One of the roots of equation kx² – 10x + 3 = 0 is 3. Complete the following activity to find the value of k.
Activity:
One of the roots of equation
kx² – 10x + 3 = 0 is 3.

Solution:
One of the roots of equation
kx² – 10x + 3 = 0 is 3.

(ii) A card is drawn at random from a pack of well shuffled 52 playing cards. Complete the following activity to find the probability that the card drawn is :
Event A : The card drawn is an ace.
Event B : The card drawn is a spade. Activity:
‘S’ is the sample space.
∴ n(S) = 52

Solution:
‘S’ is the sample space.
∴ n(S) = 52

(B) Solve the following subquestions (any two):

(i) Solve the simultaneous equations by using graphical method
x + 3y = 7
2x + y = -1
Solution:
x + 3y = 7


Two lines of equation,
x + 3y = 7
and 2x + y = -1
intersect at point (-2, 3).
∴ x = -2 and y = 3 satisfies the equation x + 3y = 7 and 2x + y = -1

(ii) There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find how many total seats are there in the auditorium?
Solution:
The number of seats arranged row-wise are as follows:
20, 22, 24 ………..
This sequence is in A.P.
∴ a = 20, d = 22 – 20 = 2, n = 27
Now,
S_n = \(\frac{n}{2}\)[2a + (n – 1)d] [formula]
S₂₇ = \(\frac{27}{2}\)[2 × 20 + (27 – 1) × 2]
= \(\frac{27}{2}\)[40 + 26 × 2]
= \(\frac{27}{2}\)[40 + 52]
= \(\frac{27}{2}\) × 92
= 27 × 46
∴ S₂₇ = 1242
∴ Total seats in auditorium are 1242.

(iii) Sum of the present ages of Manish and Savita is 31 years. Manish’s age 3 years ago was 4 times the age of Savita at that time. Find their present ages.
Solution:
Suppose the present age of Manish be x years and Savita be y years. According to first condition, sum of their present ages is 31
∴ x + y = 31 …….. (i)
Three years ago
Age of Manish = (x – 3) years
Age of Savita = (y – 3) years
According to second condition, 3 years ago Manish’s age was 4 times the age of Savita.
∴ (x – 3) = 4(y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = -12 + 3
∴ x – 4y = – 9 ……..(ii)
Subtracting equation (ii) from equation (i), we get

Substituting y = 8 in equation (i),
x + y = 31
x + 8 = 31
∴ x = 31 – 8
∴ x = 23
∴ The present age of Manish is 23 years and Savita is 8 years.

(iv) Solve the following quadratic equation using formula :
x² + 10x + 2 = 0
Solution:
x² + 10x + 2 = 0
Compare the given equation with
ax² + bx + c = 0
∴ a = 1, b = 10, c = 2
∆ = b² – 4ac
= 10² – 4(1)(2)
= 100 – 8
= 92
= 2\(\sqrt{23}\)

Question 4.
Solve the following subquestions (any two): [8]

(i) If 460 is divided by a natural number, then quotient is 2 more than nine times the divisor and remainder is 5. Find the quotient and divisor.
Solution:
Suppose divisor = x
Quotient is 2 more than 9 times the divisor.
∴ Quotient = 9x + 2
Dividend = 460 and remainder = 5
Dividend = Divisor × Quotient + Remainder
∴ 460 = x × (9x + 2) + 5
∴ 460 = 9x² + 2x + 5
∴ 9x² + 2x + 5 – 460 = 0
∴ 9x² + 2x – 455 = 0
∴ 9x² – 63x + 65x – 455 = 0
∴ 9x(x – 7) + 65(x – 7) = 0
∴ (9x + 65)(x – 7) = 0
∴ 9x + 65 = 0 or x – 7 = 0
∴ x = \(\frac{-65}{9}\) or x = 7
But, 460 is divided by natural number and natural number cannot be negative.
∴ x = 7
∴ Divisor = 7
∴ Quotient = 9x⁷ + 2 = 63 + 2 = 65
∴ Quotient is 65 and divisor is 7.

(ii) If the 9^th term of an A.P. is zero, then prove that the 29^th term is double fbe 19^th term.
Solution:
To prove : t₂₉ = 2t₁₉
Proof: Let ‘a’ be the first term and ‘d’ be the common difference of an A.P.
t₉ = 0 [Given]
t_n = a + (n – 1)d [Formula]
∴ t₉ = a + (9 – 1)d
∴ 0 = a + 8d
∴ a = -8d ……..(1)
Now, t₁₉ = a + (19 – 1)d
t₁₉ = -8d + 18d [From (1)]
∴ t₁₉ = 10d …….(2)
t₂₉ = a + (29 – 1)d
∴ t₂₉ = -8d + 28d
∴ t₂₉ = 20d …….(3)
∴ t₂₉ = 2(10d)
∴ t₂₉ = 2(t₁₉) [From (2)]
∴ Proved that 29^th term is double the 19^th term.

(iii) The perimeter of an isosceles triangle is 24 cm. The length of its congruent sides is 13 cm less than twice the length of its base. Find the lengths of all sides of the triangle.
Solution:
Length of base of isosceles triangle = x cm
∴ Length of congruent side = 2x – 13 cm
Perimeter of isosceles triangle = 24 cm
∴ Perimeter of isosceles triangle = Base + Congruent sides
∴ x + 2x – 13 + 2x – 13 = 24
∴ 5x – 26 = 24
∴ 5x = 26 + 24
∴ 5x = 50
∴ x = \(\frac{50}{5}\)
∴ x = 10
∴ Base = 10 cm
Length of congruent side = 2x – 13
= 2 × 10 – 13
= 20 – 13
∴ Base is 10 cm and each congruent side is 7 cm in length.

Question 5.
Solve the following subquestions (any one): [3]

(i) A bag contains 8 red and some blue balls. One ball is drawn at random from the bag. If ratio of probability of getting red ball and blue ball is 2: 5, then find the number of blue balls.
Solution:
Suppose number of blue balls = x
∴ n(Blue ball) = x
number of red balls = 8
∴ n(Red ball) = 8
Total number of balls = 8 + x
n(T) = 8 + x
P(Blue ball drawn) \(=\frac{n(\text { Blue ball })}{n(\mathrm{~T})}\)
= \(\frac{x}{8+x}\)
P(Red ball drawn) \(=\frac{n(\text { Red ball })}{n(\mathrm{~T})}\)
= \(\frac{8}{8+x}\)
According to given condition

(ii) Measures of angles of a triangle are in A.P. The measure of smallest angle is five times of common difference. Find the measures of all angles of a triangle.
(Assume the measures of angles as a, a + d, a + 2d)
Solution:
Let the angles of triangles are a, a + d, a + 2d
We know that, sum of angles of triangle = 180°
∴ a + a + d + a + 2d = 180°
∴ 3a + 3d = 180°
∴ 3(a + d) = 180°
∴ (a + d) = \(\frac{180}{3}\)
a + d = 60°
According to given condition,
Smallest angle, a = 5 × d
∴ a = 5d
Putting the value of a in equation (i)
∴ 5d + d = 60°
∴ 6d = 60°
∴ d = \(\frac{60}{6}\)
d = 10°
Putting the value of d in (i)
a + d = 60°
∴ a + 10 = 60°
∴ a = 60° – 10°
∴ a = 50°
∴ a + 2d = 50° + 2(10)
= 50° + 20°
= 70°
∴ Angles of triangle are 50°, 60°, 70° respectively.

SSC Maharashtra Board Maths 1 Question Paper with Solutions