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

## SSC Maths 1 Question Paper 2023 with Solutions Pdf Download Maharashtra Board

Time : 2 Hours

Max. Marks : 40

General Instructions :

- All questions are compulsory.
- Use of calculator is not allowed.
- The numbers to the right of the questions indicate full marks.
- In case of MCQs [Q. No. 1(A)] only the first attempt will be evaluated and will be given credit.
- 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) To draw the graph of 4x + 5y = 19, find y when x = 1:

(A) 4

(B) 3

(C) 2

(D) – 3

Answer:

(B) 3

(ii) Out of the following equations which one is not a quadratic equation?

(A) x^{2} + 4x = 11 + x^{2}

(B) x^{2} = 4x

(C) 5x^{2} = 90

(D) 2x – x^{2} = x^{2} + 5

Answer:

(A) x^{2} + 4x = 11 + x^{2}

(iii) For the given A.P. a = 3.5, d = 0 then t_{n} = ____

(A) 0

(B) 3.5

(C) 103.5

(D) 104.5

Answer:

(B) 3.5

(iv) If n(A) = 2, P(A) = \(\frac{1}{5}\) ,then n(S) = ?

(A) 10

(B) \(\frac{5}{2}\)

(C) \(\frac{2}{5}\)

(D) \(\frac{1}{3}\)

Answer:

(A) 10

(B) Solve the following subquestions : [4]

(i) Find the value of the following determinant:

\(\left|\begin{array}{ll}

4 & 3 \\

2 & 7

\end{array}\right|\)

Solution:

\(\left|\begin{array}{ll}

4 & 3 \\

2 & 7

\end{array}\right|\) = 4(7) – 3(2)

= 28 – 6

= 22

∴ \(\left|\begin{array}{ll}

4 & 3 \\

2 & 7

\end{array}\right|\) = 22.

(ii) Find the common difference of the following A.P:

2, 4, 6, 8, …….

Solution:

Let a = 2

t_{1} = 2

t_{2} = 4

t_{3} = 6

t_{4} = 8

∴ d = t_{2} – t_{1} = 4 – 2 = 2

d = t_{3} – t_{2} = 6 – 4 = 2

d = t_{4} – t_{3} = 8 – 6 = 2

∴ Common difference = 2

d = 2

(iii) On certain article if rate of CGST is 9%, then what is the rate of SGST?

Solution:

Rate of CGST = 9%

as Rate of CGST = Rate of SGST

∴ Rate of SGST = 9%

(iv) If one coin is tossed, write the sample space ‘S’.

Solution:

Let, S be the sample space of one coin tossed.

∴ S = {H, T}.

Sample space, S = {HH, HT, TH, TT}

Question 2.

(A) Complete any two given activities and rewrite it:

(i) Complete the following activity; find the value of x:

5x + 3y = 9 …… (I)

2x – 3y = 12 …… (II)

Add equations (I) and (II)

Solution:

5x + 3y = 9 ……. (I)

2x – 3y = 12 …… (II)

Add equations (I) and (II)

(ii) Complete the following activity to determine the nature of the roots of the quadratic equation x^{2} + 2x – 9 = 0:

Compare x^{2} + 2x – 9 = 0 with ax^{2} + bx + c = 0

∴ The roots of the equation are real and unequal.

Solution:

x^{2} + 2x – 9 = 0

∴ The roots of the equation are real and unequal.

(iii) Complete the following table using given information:

Solution:

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

(i) Solved the following simultaneous equations:

x + y = 4;

2x – y = 2

Solution:

Let, x + y = 4 …… (1)

2x + y = 2 ……(2)

Add equations (1) and (2)

3x = 6

x = \(\frac{6}{3}\)

∴ x = 2

Put x = 2 in equation (1) we get

∴ Solution: (x, y) = (2,2)

(ii) Write the following equation in the form ax^{2} + bx + c =0, then write the values of a, b, c

2y = 10 – y^{2}

Solution:

2y = 10 – y^{2}

y^{2} + 2y – 10 = 0

Compare the given equation with

ax^{2} + bx + c = 0

∴ a = 1, b = 2, c = -10.

(iii) Write an A.P. whose first term is a = 10 and common difference d 5.

Solution:

a =10

d = 5

∴ t_{1} = a = 10

t_{2} = 10 + 5 = 15

t_{3} = t_{2} + d = 15+ 5 = 20

t_{4} = t_{3} + d = 20 + 5 = 25

∴ A.P. = 10, 15, 20, 25,………..

(iv) Courier service agent charged total ₹ 590 to courier a parcel from Nashik to Nagpur. In the tax invoice taxable value is ₹ 500 on which CGST is ₹ 45 and SGST is ₹ 45. Find the rate of GST charged for this service.

Solution:

Given:

CSGT = 45

SGST = 45

∴ Total GST = 45 + 45

= ₹ 90

∴ Rate of GST = \(\frac{90}{500}\) × 100

= \(\frac{90}{5}\)

= 80%

∴ Rate of GST is 18%

(v) Observe the following table and find Mean :

Assumed mean A = 300

Solution:

\(\Sigma f_i\) = 50

\(\Sigma f_i d_i\) = 320

Assumed mean = A = 30

\(\bar{d}\) = \(\frac{\sum f_i d_i}{\sum f_i}\)

= \(\frac{320}{50}\)

= 6.4

∴ Mean \(\overline{\mathrm{X}}\) = A + \(\bar{d}\)

= 300 + 6.4

= 306.4

∴ Mean = 306.4

Question 3.

(A) Complete any one activity and rewrite it: [3]

(i) Form a ‘Road Safety Committee’ of two, from 2 boys (B_{1}, B_{2}) and 2 girls (G_{1}, G_{2}). Complete the following activity to write the sample space:

Solution:

(ii) Fill in the boxes with the help of given information:

Solution:

(B) Solve the following sub-questions (any two): [6]

(i) Solve the following simultaneous equations using Cramer’s rule:

4m + 6n = 54; 3m + 2n = 28

Solution:

The given simultaneous equations are

4m + 6 n = 54 ………(i)

3m + 2n = 28 ………(ii)

Comparing the given equation with

a_{1}m + b_{1}n = c_{1} and a_{2}m + b_{2}n = c_{2}, we get

a_{1} = 4, b_{1} = 6, c_{1} = 54

a_{2} = 3, b_{2} = 2, c_{2} = 28

∴ D_{m} = \(\left|\begin{array}{ll}

c_1 & b_1 \\

c_2 & b_2

\end{array}\right|\) = \(\left|\begin{array}{ll}

54 & 6 \\

28 & 2

\end{array}\right|\) = 108 – 168 = -60

∴ D_{n} = \(\left|\begin{array}{ll}

a_1 & c_1 \\

a_2 & c_2

\end{array}\right|\) = \(\left|\begin{array}{ll}

4 & 54 \\

3 & 28

\end{array}\right|\) = 112 – 162 = -50

∴ By Cramer’s rule, we get

m = \(\frac{\mathrm{D}_m}{\mathrm{D}}\) = \(\frac{-60}{-10}\) = 6

n = \(\frac{\mathrm{D}_n}{\mathrm{D}}\) = \(\frac{-50}{-10}\) = 6

∴ Solution = (m, n) = (65)

(ii) Solve the following quadratic equation by formula method:

x^{2} + 10x + 2 = 0

Solution:

x^{2} + 10x + 2 = 0

Compare the given equation with

ax^{2} + bx + c = 0

∴ a = 1, b = 10, c = 2

∆ = b^{2} – 4ac

= 10^{2} – 4(1)(2)

= 100 – 8

= 92

= 2\(\sqrt{23}\)

(iii) A two digit number is formed with digits 2, 3, 5, 7,9 without repetition. What is the probability of the following events?

Event A : The number formed is an odd number.

Event B : The number formed is a multiple of 5.

Solution:

Let S be the sample space of two digit number formed by 2, 3, 5, 7, 9.

∴ S = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}.

∴ n(S) = 20

Event A : The number formed is an odd number.

A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75, 79, 93, 95, 97}

n(A) = 16

∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{16}{20}\) = \(\frac{4}{5}\)

∴ P(A) = \(\frac{4}{5}\)

Event B: The number formed is multiple of 5.

= {25, 35, 75, 95}.

n(B) = 4.

∴ P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\)

= \(\frac{4}{20}\)

P(B) = \(\frac{1}{5}\)

(iv) The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data :

No. of Mangoes | No. of Trees |

50-100 | 33 |

100-150 | 30 |

150-200 | 90 |

200-250 | 80 |

250-300 | 17 |

Solution:

L = Lower class limit of the median

= 150

N = sum of frequencies = 250

h = 50

f = 90

cf = 63

∴ Hence, the median of data is 184.

Question 4.

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

(i) If the first term of an A.P. is p, second term is q and last term is r, then show that sum of all terms is (q + r – 2p) × \(\frac{(p+r)}{2(q-p)}\).

Solution:

Given:

p – First term of AP.

q – Second term of A.P.

r – Last term of A.P.

Common difference = q – p

t_{n} = a + (n – 1)d.

r = p + (n – 1)(q – p)

r – p = (n – 1)(q – p)

\(\frac{r-p}{q-p}\) = n – 1

\(\frac{r-p}{q-p}\) + 1 = n

\(\frac{r-p+q-p}{q-p}\) = n

\(\frac{r+q-2 p}{q-p}\) = n

Now S_{n} = \(\frac{n}{2}\)[First trem + Last term]

S = \(\frac{r+q-2 p}{2(q-p)}\)(p + r)

S = \(\frac{(p+r)(r+q-2 p)}{2(q-p)}\)

Hence prove

(ii) Show the following data by a frequency polygon:

Electricity bill (₹) | Families |

200-400 | 240 |

400-600 | 300 |

600-800 | 450 |

800-1000 | 350 |

1000-1200 | 160 |

Solution:

(iii) The sum of the squares of five consecutive natural numbers is 1455. Find the numbers.

Solution:

Let, n, n + 1, n + 2, n + 3, n + 4 are five consecutive natural numbers

Then,

n^{2} + (n + 2)^{2} + (n + 2)^{2} + (n + 3)^{2} + (n + 4)^{2} = 1455

n^{2} + n^{2} +1 + 2n + n^{2} + 4n + 4 + n^{2} + 6n + 9 + n^{2} + 8n + 16 = 1455

∵ [(a + b)^{2} = a^{2} + 2ab + b^{2}]

5n^{2} + 20 n + 30 = 1455

5n^{2} + 20n + 30 – 1455 = 0

5n^{2} + 20n – 1425 = 0

n^{2} + 4n – 285 = 0

∴ Compare with ax^{2} + 2bx + c = 0

a = 1, b = 4, c = – 285

∴ b^{2} – 4ac = 4^{2} – 4(1)(-285)

= 16 + 1140 = 1156

As, n = – 19 is not natural

Hence, the five consecutive numbers are 15, 16, 17, 18, 19

Question 5.

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

(i) Draw the graph of the equation x + 2y = 4. Find the area of the triangle formed by the line intersecting to X-axis and Y-axis.

Solution:

x + 2y = 4

x | 0 | 2 | 4 |

y | 2 | 1 | 0 |

(x, y) | (0,2) | (2,1) | (4,0) |

(ii) A survey was conducted for 180 people in a city. 70 ate Pizza, 60 ate burgers and 50 ate chips. Draw a pie diagram for the given information.

Solution:

Pizza = \(\frac{70}{180}\) × 360°

= 70 × 2

= 140°

Burger = \(\frac{60}{180}\) × 360°

= 60 × 2

= 120°

Chips = \(\frac{50}{180}\) × 360°

= 50 × 2

= 100°