SSC Maharashtra Board Maths 1 Question Paper July 2023 with Answers

Maharashtra Board SSC Class 10 Maths 1 Question Paper July 2023 with Answers Solutions Pdf Download.

SSC Maths 1 Question Paper July 2023 with Answers Pdf Download Maharashtra Board

Time: 2 Hours
Total Marks: 40

Note:

1. All questions are compulsory.
2. Use of a calculator is not allowed.
3. The numbers to the right of the questions indicate full marks.
4. In case of MCQs [Q. No. 1(A)] only the first attempt will be evaluated and will be given credit.
5. For every MCQ, four alternatives (A), (B), (C), (D) of answers are given. Alternative of correct answer is to be written in front of the subquestion number.

Question 1.
(A) Choose the correct answer and write the alphabet of it in front of the subquestion number: [4]
i. Sum of first five multiples of 3 is __________. (A)
(A) 45
(B) 55
(C) 15
(D) 75
Answer:
(A) 45

ii. Find the value of determinant \(\left|\begin{array}{ll} 3 & 2 \\ 4 & 5 \end{array}\right|\): (B)
(A) 2
(B) 7
(C) -7
(D) 23
Answer:
(B) 7

iii. Which of the following quadratic equations has roots 3 and 5? (B)
(A) x² – 15x + 8 = 0
(B) x² – 8x + 15 = 0
(C) x² + 3x + 5 = 0
(D) x² + 8x – 15 = 0
Answer:
(B) x² – 8x + 15 = 0

iv. Two coins are tossed simultaneously. Write the number of sample points n(S): (C)
(A) 2
(B) 8
(C) 4
(D) 6
Answer:
(C) 4

Hints:
i. First five multiples of 3 are 3, 6, 9, 12, 15.
The above sequence is an A.P.
∴ t₁ = 3, t₅ = 15

iii. x² – (α + β)x + αβ = 0
∴ x² – (3 + 5)x + (3)(5) = 0
∴ x² – 8x + 15 = 0

iv. S = {HH, HT, TH, TT}, n(S) = 4

(B) Solve the following subquestions: [4]
i. If 15x + 17y = 21 and 17x + 15y = 11, then find the value of x + y.
ii. Given sequence is an A.P. Find the next two terms of this A.P.:
5, 12, 19, 26, …….
iii. On certain article if rate of CGST is 9%, then what is the rate of SGST and what is the rate of GST?
iv. If n(S) = 2 and n(A) = 1, then find P(A).
Answer:
i.

ii. The given A.P. is,
5, 12, 19, 26,…
Here, first term is 5 and common difference is 7.
∴ Next two terms in this A.P. are 26 + 7 = 33 and 33 + 7 = 40.
∴ Next two terms are 33 and 40.

iii. Rate of CGST = 9%
But, rate of SGST = rate of CGST
∴ Rate of SGST = 9%
Rate of GST = Rate of SGST + Rate of CGST
= 9% + 9%
∴ Rate of GST = 18%

iv. n(S) = 2 and n(A) = 1
∴ P(A) = \(\frac{n(A)}{n(S)}\)
∴ P(A) = \(\frac{1}{2}\)

Question 2.
(A) Complete the following activity and rewrite (any Wo): [4]
i. Complete the following table to draw the graph of the equation x + y = 3:

ii. Complete the following activity to find the value of discriminant of the equation x² + 10x – 7 = 0.
Solution:
Comparing x² + 10x – 7 = 0 with ax² + bx + c = 0

iii. Complete the following table using given information:

(B) Solve the following subquestions (any four): [8]
i. Solve the following simultaneous equations:
x + y = 6; x – y = 4
ii. Solve the following quadratic equation by factorisation method:
x² + 15x + 54 = 0
iii. The first term a = 8 and common difference d = 5 are given. Write an A.P.
iv. Mr. Rohit is a retailer. He paid GST of ₹ 6,500 at the time of purchase. He collected GST of ₹8,000 at the time of sale.
(a) Find his input tax and output tax.
(b) What is his input tax credit?
(c) Find his payable GST.
(d) Hence find the payable CGST and payable SGST.
v. Find the mean from the given values:
∑x₁f₂ = 1265; N = 50
Answer:
(A)
i. x + y = 3

iii. 1. Here, Premium = ₹7, FV = ₹10
∴ FV + Premium = MV
∴ MV = 10 + 7 = ₹17

2. Here, FV = ₹25, MV = ₹16
FV > MV
∴ Share is at discount
FV – Discount = MV
∴ 25 – Discount = 16
∴ 25 – 16 = Discount
∴ Discount = ₹9

3. Here, FV = ₹300, MV = ₹315
∴ FV < MV
∴ Share is at Premium
FV + Premium = MV
∴ Premium = 315 – 300
∴ Premium = ₹15

4. Here, share is at par.
∴ MV = FV
∴ FV = ₹5

(B)
i. x + y = 6 ….(i)
x – y = 4 ….(ii)
Adding equations (i) and (ii), we get

Substituting x = 5 in equation (i), we get
x + y = 6
∴ 5 + y = 6
∴ y = 6 – 5 = 1
∴ (x, y) = (5, 1) is the solution of the given simultaneous equations.
SMART TIP
To check our answer:
Substitute our answer (x, y) in the given equation.
If L.H.S. = R.H.S. (for both equation), then our answer is correct.
For x + y = 6, L.H.S. = 5 + 1 = 6 = R.H.S.
For x – y = 4, L.H.S. = 5 – 1 = 4 = R.H.S.
Hence, our answer is correct.

ii.

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
x + 9 = 0 or x + 6 = 0
∴ x = – 9 or x = – 6
∴ The roots of the given quadratic equation are – 9 and – 6.
SMART TIP
To check our answer, substitute the values of x in the given equation. If L.H.S. = R.H.S. then our answer is correct.
For x = -9, For x = -6,
L.H.S. = x2 + 15x + 54 L.H.S. = x2 + 15x + 54
= 81 – 135 + 54 = 36 – 90 + 54
= 0 = 0
= R.H.S. = R.H.S.
Hence, our answer is correct.

iii. a= 8, d = 5 .[Given]
∴ t₁ = a = 8
t₂ = t₁ + d = 8 + 5 = 13
t₃ = t₂ + d = 13 + 5 = 18
t₄ = t₃ + d = 18 + 5 = 23
∴ The required A.P. is 8, 13, 18, 23, …

iv. a. Input tax (Tax paid at the time of purchase) = ₹ 6500 and
output tax (Tax collected at the time of sale) = ₹ 8000.
b. Input tax credit = ₹ 6500
c. Payable GST = output tax – input tax credit
= 8000 – 6500
= ₹ 1500

∴ Payable CGST = ₹ 750 and payable SGST = ₹ 750.

v.

Question 3.
(A) Complete the following activity and rewrite (any one): [3]
Smita has invested ₹ 12,000 and purchased shares of FV ₹ 10 at a premium of ₹ 2. Find the number of shares she purchased. Complete the given activity to get the answer.
Solution:
FV = ₹ 10, Premium = ₹ 2

ii. If one die is rolled once, then find the probability of each of the following events:
(a) Number on the upper face is prime.
(b) Number on the upper face is even.
Solution:
‘S’ is the sample space
S = {1, 2, 3, 4, 5, 6} n(S) = ____
(a) Event A : Prime number on the upper face

(b) Event B : Even number on the upper face

(B) Solve the following subquestions (any two): [6]
i. Two numbers differ by 3. The sum of the twice the smaller number and thrice the greater number is 19. Find the numbers.
ii. Solve the given quadratic equation by using formula method: 5A² + 13x + 8 = 0
iii. A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets:
(a) a red balloon
(b) a blue balloon
(c) a green balloon.
iv. The following table shows the number of students of class X and the time they utilized daily for their studies. Find the mean time spent by 50 students for their studies by direct method:

Time (hrs.)	No. of students
0 – 2	7
2 – 4	18
4 – 6	12
6 – 8	10
8 – 10	3

Answer:
(A)
i. FV = ₹10, Premium= ₹2

ii. ‘S’is the sample space
S = {1, 2, 3, 4, 5, 6} n(S) = [6]
(a) Event A : Prime number on the upper face

(b) Event B : Even number on the upper face

(B) i. Let the greater number be x and the smaller number be y.
According to the first condition, two numbers differ by 3.
∴ x – y = 3 …(i)
According to the second condition, the sum of twice the smaller number and thrice the greater number is 19.
∴ 3x + 2y = 19 …..(ii)
Multiplying equation (i) by 2, we get
2x – 2y = 6 …(iii)
Adding equations (ii) and (iii), we get

∴ x = 5
Substituting x = 5 in equation (i), we get
x – y = 3
5 – y = 3
∴ 5 – 3 = y
∴ y = 2
∴ The required numbers are 5 and 2.
SMART TIP
To check our answer:
We apply given conditions to our answer. If the given conditions are satisfied, then our answer is correct. Note that, 5 and 2 differ by 3. Now, sum of twice the smaller number and thrice the greater number
= 2(2)+ 3(5)
= 4 + 15
= 19
Hence, our answer is correct.

ii. 5x² + 13x + 8 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = 5, b = 13, c = 8
∴ b² – 4ac = (13)² – 4 x 5 x 8
= 169 – 160 = 9

∴ The roots of the given quadratic equation are -1 and \(\frac{-8}{5}\).

iii. Let the 2 red balloon be R₁, R₂,
3 blue balloons be B₁, B₂, B₃, and
4 green balloons be G₁, G₂, G₃, G₄.
∴ Sample space
5 = {R₁, R₂, B₁, B₂, B₃, G₁, G₂, G₃, G₄}
∴ n(S) = 9
a. Let A be the event that Pranali gets a red balloon.
∴ A = {R₁, R₂}
∴ n(A) = 2
∴ P(A) = \(\frac{n(A)}{n(S)}\)
∴ P(A) = \(\frac{2}{9}\)
b. Let B be the event that Pranali gets a blue balloon.
∴ B = {B₁, B₂, B₃}
∴ n(B) = 3
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{1}{3}\)
∴ P(B) = \(\frac{1}{3}\)
c. Let C be the event that Pranali gets balloon.

iv.

∴ The mean of the time spent by the students for their studies is 4.36 hours.

Question 4.
Solve the following subquestions (any two): [8]
i. The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
ii. If p times the p^th term of an A.P. is equal to q times q^th term, then show that (p + q)^th term of that A.P. is zero (p ≠ q).
iii. Draw a pie diagram to represent the world population given in the following table :

Answer:
i. Let α and β be the roots of the quadratic equation.
According to the given conditions,
α + β = 5 and α³ + β³ = 35
Now, (α + β)³ = α³ + 3α²β + 3αβ² + β³
∴ (α + β)³ = α³ + β³ + 3αβ (α + β)
∴ (5)³ = 35 + 3αβ(5)
∴ 125 = 35 + 15αβ
∴ 125 – 35 = 15αβ
∴ 15αβ = 90
∴ αβ = \(\frac{90}{15}\)
∴ αβ = 6
∴ The required quadratic equation is
x² – (α + β)x + αβ = 0
∴ x² – 5x + 6 = 0

ii. p times the p^th term of an A.P. = pt_p.
q times the q^th term of an A.P. = qt_q.
According to the given condition,
pt_p = qt_q
∴ p[a + (p – 1)d] _ q[a + (q – 1)d]
∴ pa + pd(p – 1) = qa + qd(q – 1)
∴ pa + p²d – pd = qa + q²d – qd
∴ pa + p²d – pd – qa – q²d + qd = 0
∴ (pa – qa) + (p²d – q²d) – (pd – qd) = 0
∴ a(p – q) + d(p² – q²) – d(p – q) = 0
∴ a(p – q) + d(p + q) (p – q) – d(p – q) = 0
∴ (p – q) [a + d(p + q) – d] = 0
∴ (p – q)[a + (p + q – 1) d] = 0
∴ [a+ (p + q – 1)d] = 0 …[Dividing both sides by (p – q)]
∴ t_{(p + q)} = 0
∴ The (p + q)^th term of the A.P. is zero.

iii. Total percentage of world population = 100

Question 5.
Solve the following subquestions (any one): [3]
i. Represent the following data using histogram:

Daily Income (₹)	No. of Workers
130 – 135	4
135 – 140	7
140 – 145	14
145 – 150	16

ii. Observe the following flow chart and solve it:

Answer:
i.

ii. The given equations are
3x – y = 7 …(i)
x + 4y = 11 .(ii)
Equations (i) and (ii) are in ax + by = c form
Comparing the given equations with a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 3, b₁ = -1, c₁ = 7 and a₂ = 1, b₂ = 4, c₂ = 11

∴ x = 3 and y = 2
∴ (x, y) = (3, 2) is the solution of the given equations.

SSC Maharashtra Board Maths 1 Question Paper with Solutions