Maharashtra Board SSC Class 10 Maths 1 Question Paper 2019 with Solutions Answers Pdf Download.

## SSC Maths 1 Question Paper 2019 with Solutions Pdf Download Maharashtra Board

Time : 2 Hours

Max. Marks : 40

General Instructions:

- All questions are compulsory.
- Use of calculator is not allowed.
- Figures to the right of questions indicate full marks.

Question 1.

(A) Solve the following questions (Any four): [4]

(i) Find the median of :

66, 98, 54, 92, 87, 63, 72.

Solution:

66, 98, 54, 92, 87, 63, 72.

On arranging the terms in ascending order, we get;

54, 63, 66, 72, 87, 92, 98

∵ The number of terms are 7.

∴ Median = Middle term = 72.

(ii) Multiply and write the answer in the simplest form :

5\(\sqrt{7}\) × 2\(\sqrt{7}\)

Solution:

5\(\sqrt{7}\) × 2\(\sqrt{7}\) = (5 × 2) × (\(\sqrt{7}\) × \(\sqrt{7}\)) = 10 × 7 = 70

(iii) If 3x + 5y = 9 and 5x + 3y = 7, the find the value of x + y.

Solution:

Given,

3x + 5y = 9 ……… (i)

5x + 3y = 7 ………. (ii)

On adding equation (i) and (ii), we get

∴ 8 (x + y) = 16

∴ x + y = \(\frac{16}{8}\) = 2

∴ x + y = 2

(iv) Write the ratio of second quantity to first quantity in the reduced form 5 dozen pens, 120 pens.

Solution:

Given,

First quantity = 5 Dozen Pens = 5 × 12 = 60 Pens

Second quantity = 120 Pens

∴ Ratio \(=\frac{\text { Second quantity }}{\text { First quantity }}\) = \(\frac{120}{5 \times 12}\) = \(\frac{120}{60}\) = \(\frac{2}{1}\)

(v) Write the following polynomial in coefficient form :

2x^{3} + x^{2} – 3x + 4.

Solution:

Polynomial in coefficient form is: (2, 1, -3, 4)

(vi) For computation of income tax which is the assessment year of financial year 01-04-2016 to 31-03-2017 ?

Solution:

Assessment year is : 2017 – 18

Question 1.

(B) Solve the following questions (Any two): [4]

(i) Find the value of the polynomial 2x^{3} + 2x, when x = -1.

Solution:

Given, P(x) = 2x^{3} + 2x

Put x = -1

P(-1) = 2(-1)^{3} + 2(-1)

= -2 – 2

= -4

(ii) If A = {11, 21, 31, 41,}, B = {12, 22, 31, 32}, then find :

(1) A ∪ B

(2) A ∩ B.

Solution:

Given, A = {11, 21, 31, 41}, B = {12, 22, 31, 32}

(1) A ∪ B = {11, 12, 21, 22, 31, 32, 41}

(2) A ∩ B = {31}

(iii) Sangeeta’s monthly income is ₹ 25,000. She spent 90% of her income and donated 3% for socially useful causes. How much money did she save ?

Answer:

Given, Sangeeta’s monthly income = ₹ 25,000

Her expenditure = 90% of ₹ 25,000

= \(\frac{90}{100}\) × 25,000 = ₹ 22,500

Donation = 3% of ₹ 25,000 = \(\frac{3}{100}\) × 25,000

= ₹ 750

∴ Sangeeta’s Per month saving = (Total monthly income) – (Total expenses)

= 25,000 – (22,500 + 750)

= ₹ 1,750

∴ Sangeeta’s total saving per month is ₹ 1,750.

Question 2.

(A) Choose the correct alternative: [4]

(i) In the A. P. 2, -2, -6, -10,………. common difference (d) is:

(A) -4

(B) 2

(C) -2

(D) 4

Answer:

(A) – 4

(ii) For the quadratic equation x^{2} + 10x – 7 = 0,

(A) a = -1, b = 10, c = 7

(B) a = 1, b = -10, c = -7

(C) a = 1, b = 10, c = -7

(D) a = 1, b = 10, c= -7

Answer:

(C) a = 1, b = 10, c = -7

(iii) The tax levied by Central Government for

(A) IGST

(B) CGST

(C) SGST

(D) UTGST

Answer:

(B) CGST

(iv) If a die is rolled, what is the probability that number appearing on upper face is less than 2 ?

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

(B) \(\frac{1}{2}\)

(C) 1

(D) \(\frac{1}{6}\)

Solution:

(D) \(\frac{1}{6}\)

Question 2.

(B) Solve the following questions (Any two): [4]

(i) First term and common difference of an A.P. are 12 and 4 respectively.

If t_{n} = 96, find n.

Solution:

Given, first term = 12,

common difference = 4 and t_{n} = 96.

According to the formula,

t_{n} = a + (n – 1)d

∵ t_{n} = 96

∴ 96 = 12 + (n – 1)4

⇒ 96 = 12 + 4n – 4

⇒ 96 = 8 + 4n

⇒ 4n = 96 – 8

⇒ 4n = 88

⇒ n = \(\frac{88}{4}\)

∴ n = 22

(ii) If \(\left|\begin{array}{ll}

4 & 5 \\

m & 3

\end{array}\right|\) = 22, then find the value of m.

Solution:

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

4 & 5 \\

m & 3

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

⇒ (4 × 3) – (m × 5) = 22

⇒ 12 – 5m = 22

⇒ -5m = 10

⇒ m = \(-\frac{10}{5}\)

∴ m = -2

(iii) Solve the following quadratic equation:

x^{2} + 8x + 15 = 0.

Solution:

Given, x^{2} + 8x + 15 = 0

On splitting the middle term

⇒ x^{2} + 5x + 3x + 15 = 0

⇒ x(x + 5) + 3(x + 5) = 0

⇒ (x + 3)(x + 5) = 0

⇒ (x + 3) = 0 or (x + 5) = 0

⇒ x = -3 or x = -5

∴ -3 and -5 are the roots of the given quadratic equation.

Question 3.

(A) Complete the following activities (Any two): 14]

(i) Smita has invested ₹ 12,000 to purchase shares of FV ₹ 10 at a premium of 2. Find the number of shares she purchased. Complete the given activity to get the answer.

Activity : FV = ₹ 10, Premium = ₹ 2

Solution:

(ii) The following table shows the daily supply of electricity to different places in a town. To show the information by a pie diagram, measures of central angles of sectors are to be decided.

Complete the following activity to find the measures :

Solution:

(iii) Two coins are tossed simultaneously. Complete the following activity of writing the sample space (S) and expected outcomes of the events :

(i) Event A : to get at least one head.

(ii) Event B : to get no head.

Activity : If two coins are tossed simultaneously

(i) Event A : at least getting one head.

(ii) Event B : to get no head.

Solution:

Question 3.

(B) Solve the following questions (Any two): [4]

(i) Find the 19th term of the A.P. 7, 13, 19, 25, ………

Solution:

Given, A. P. is 7, 13, 19, 25,……..

Here, a = 7

t_{1} = 7, t_{2} = 13,

d = t_{2} – t_{1} = 13 – 7 = 6

According to formula,

t_{n} = a + (n – 1)d

∴ 19^{th} term is:

t_{19} = 7 + (19 – 1) 6

= 7 + (18) × 6

= 7 + 108

∴ t_{19} = 115

∴ The 19^{th} term is 115.

(ii) Obtain a quadratic equation whose roots are -3 and -7.

Solution:

Given, roots of quadratic equation are -3 and -7.

Let, α = -3 and

β = -7

∴ α + β = -3 + (-7) = -3 – 7 = -10

and αβ = (-3)(-7) = 21

∵ The quadratic equation is given by.

x^{2} – (α + β) x + αβ = 0

∴ x^{2} – (-10)x + 21 = 0

x^{2} + 10x + 21 = 0

Hence, x^{2} + 0x + 21 = 0 is the required quadratic equation.

(iii) Two numbers differ by 3. The sum of the greater number and twice the smaller number is 15. Find the smaller number.

Solution:

Let, the greater number be x and the smaller number be y.

From the given condition’.

x – y = 3 …….. (i)

and x – = 15 ……(ii)

Now, multiplying equation (i) by 2 and adding both the equations, we get. ∵

x = \(\frac{21}{3}\) = 7

Now, substituting the value of x in equation (i), we get.

⇒ 7 – y = 3

⇒ -y = 3 – 7

⇒ -y = – 4

⇒ y = 4

The smaller number is 4.

Question 4.

Solve the following questions (Any three): [9]

(i) Amit saves certain amount every month in a specific way, In the first month he saves ₹ 200, in the second month ₹ 250, in the third month ₹ 300 and so on. How much will be his total savings in 17 months ?

Solution:

Given, Amit’s saving in first month is ₹ 200, in second month is ₹ 250, in third month is ₹ 300 and so on.

200, 250, 300,…….., forms an A. P.

Here, a = t_{1} = 200, t_{2} = 250

d = t_{2} – t_{1} = 250 – 200 = 50 and n = 17

To find his total savings in 17 months, we have to find S_{17}.

∴ Amit’s total savings in 17 months will be ₹ 10,200.

(ii) A two digit number is to be formed using the digits 0,1, 2, 3. Repetition of the digits is allowed. Find the probability that a number so formed is a prime number.

Solution:

Sample space (S) = Two digit numbers formed out of 0,1, 2, 3, with repetition.

∴ S = {10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33}

∴ n(S) = 12

Event A : Number formed is a prime number.

∴ A = {11, 13, 23, 31}

∴ n(A) = 4

∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)

∴ P(A) = \(\frac{1}{3}\)

(iii) Smt. Malhotra purchased solar panels for the taxable value of ₹ 85,000. She sold them for ₹ 90,000. The rate of GST is 5%. Find the ITC of Smt. Malhotra. What is the amount of GST payable by her ?

Solution:

Given, rate of GST = 5% and taxable value at the time of purchase is ₹ 85,000.

∴ Input tax (Tax paid at the time of purchase)

= 5% of ₹ 85,000

∴ ITC = \(\frac{5}{100}\) × 85,000 = ₹ 4,250

∴ ITC is ₹ 4,250 of Smt. Malhotra.

Taxable value at the time of sale = ₹ 90,000

∴ Output tax = 5% of ₹ 90,000

= \(\frac{5}{100}\) × 90000 = 5 × 900

Output tax = ₹ 4,500

∴ GST payable by Smt. Malhotra = Output tax – ITC

= 4,500 – 4230 = ₹ 250

∴ ITC is ₹ 4,250 and amount of GST payable by Smt. Malhotra is ₹ 250.

(iv) Solve the following simultaneous equations graphically :

x + y = 6, 2x – y = 9.

Solution:

Given, x + y = 0 and 2x – y = 9.

x + y = 0

x | 1 | 2 | 3 | 5 |

y | – 1 | – 2 | -3 | – 5 |

(x, y) | (1, – 1) | (2, – 2) | (3, – 3) | (5, – 5) |

2x – y = 9

x | 0 | 2 | 3 | 4 |

y | – 9 | – 5 | – 3 | – 1 |

(x, y) | (0, – 9) | (2, – 5) | (3,-3) | (4, – 1) |

Solution of given equations : p(3, -3)

Question 5. (A)

Solve the following questions (Any one): [4]

(i) The following frequency distribution table shows marks obtained by 180 students in Mathematics examination :

Marks | Number of Students |

0 – 10 | 25 |

10 – 20 | x |

20 – 30 | 30 |

30 – 40 | 2x |

40 – 50 | 65 |

Find the value of x.

Also draw a histogram representing the above information.

Solution:

Given, total number of students = 180

⇒ 25 + x + 30 + 2x + 65 = 180

⇒ 3x + 120 = 180

⇒ 3x = 180 – 120

⇒ 3x = 60

⇒ x = \(\frac{60}{3}\) = 20

∴ x = 20 and 2x = 2 × 20 = 40

Marks | Number of Students |

0 – 10 | 25 |

10 – 20 | 20 |

20 – 30 | 30 |

30 – 40 | 40 |

40 – 50 | 65 |

(ii) Two taps together can fill a tank completely in 3\(\frac{1}{13}\) minutes. The smaller tap takes 3 minutes more than the bigger tap to fill the tank. How much time does each tap take to fill the tank completely ?

Solution:

Let, the bigger tap alone fills the tanks in x minutes.

Then, the smaller tap alone fills the tank in (x + 3) minutes.

Therefore, In one minute the bigger tap will fill \(\frac{1}{x}\) part of the tank and in one minute smaller tap will fill \(\frac{1}{x+3}\) part of the tank.

∴ Both the taps together fill \(\left(\frac{1}{x}+\frac{1}{x+3}\right)\) part of the tank in one minute.

But, both the taps together fill the tank in 3\(\frac{1}{13}\) minutes = \(\frac{40}{13}\) minutes

⇒ \(\frac{1}{x}+\frac{1}{x+3}\) = \(\frac{1}{40 / 13}\)

⇒ \(\frac{x+3+x}{x(x+3)}\) = \(\frac{13}{40}\)

⇒ \(\frac{2 x+3}{x^2+3 x}\) = \(\frac{13}{40}\)

⇒ 13(x^{2} + 3x) = 40(2x + 3)

⇒ 13x^{2} + 39x = 80x + 120

⇒ 13x^{2} + 39x – 80x – 120 = 0

⇒ 13x^{2} – 41x – 120 = 0

⇒ 13x^{2} – 65x + 24x – 120 = 0

⇒ 13x(x – 5) + 24(x – 5) = 0

⇒ (13x + 24) (x – 5) = 0

⇒ 13x + 24 = 0 or x – 5 = 0

⇒ x = \(\frac{-24}{13}\) or x = 5

But, the time cannot be negative

∴ x = \(\frac{-24}{13}\) is unacceptable

∴ x = 5 and x + 3 = 5 + 3 = 8

Bigger tap fills the tank in 5 minutes and smaller tap takes 8 minutes to fill the tank.

Question 6.

Solve the following questions (Any one): [3]

(A) The co-ordinates of the point of intersection of lines ax + by = 9 and bx + ay = 5 is (3, -1). Find the values of a and b.

Solution:

Given,

ax + by = 9 …….. (i)

bx + ay = 5 …… (ii)

The point of intersection of equation (i) and (ii) is (3, -1)

∴ x = 3 and y = – 1

Substituting the values of x and y in equations (i) and (ii), we get

3a – b = 9 …….(iii)

3b – a = 5

or – a + 3b = 5 …….(iv)

Multiplying equation (iv) by 3 and adding to equation (iii), we get

Substituting the value of b = 3 in equation (iii), We get

∴ 3a – 3 = 9

∴ 3a = 9 + 3 = 12

a = \(\frac{12}{3}\) = 4

The values of a is 4 and b is 3.

(B) The following frequency distribution table shows the distances travelled by some rickshaws in a day. Observe the table and answer the following questions :

Class (Daily distance travelled in km) | Continuous Classes | Frequency (Number of Rickshaws) | Cumulative Frequency less than type |

60 – 64 | 59.5 – 64.5 | 10 | 10 |

65 – 69 | 64.5 – 69.5 | 34 | 10 + 34 = 44 |

70 – 74 | 69.5 – 74.5 | 58 | 44 + 58 = 102 |

75 – 79 | 74.5 – 79.5 | 82 | 102 + 82 = 184 |

80 – 84 | 79.5 – 84.5 | 10 | 184 + 10 = 194 |

85 – 89 | 84.5 – 89.5 | 6 | 194 + 6 = 200 |

(i) Which is the modal class ? Why ?

(ii) Which is the median class and why ?

(iii) Write the cumulative frequency (C. F.) of the class preceding the median class.

(iv) What is the class interval (h) to calculate median ?

Solution:

(i) Modal class is 74.5 – 79.5

Because from the table maximum number of rickshaws (82) is for the class interval 74.5 – 79.5. Ans.

(ii) Median class is 69.5 – 74.5

Because here total number of rickshaws are 200

∴ \(\frac{N}{2}\) = \(\frac{200}{2}\) = 100

Cumulative frequency which is just greater than 100 is 102. And it is in the class interval 69.5 – 74.5.

(iii) The cumulative frequency of the class preceding the median class 44.

(iv) Class interval h = 64.5 – 59.5.

∴ h = 5