Maharashtra Board SSC Class 10 Maths 1 Question Paper 2020 with Solutions Answers Pdf Download.

## SSC Maths 1 Question Paper 2020 with Solutions Pdf Download Maharashtra Board

Time allowed : 2 Hours

Maximum 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 MCQ’s Q. No. 1(A) only the first attempt will be evaluated and will be given credit.
- For every MCQ, the correct alternative (A), (B), (C) or (D)of answers with subquestion number is to be written as an answer.

Section – A

Question 1.

(A) For every subquestion 4 alternative answers are given. Choose the correct answer and write the alphabet of it: [4]

(i) In the format of GSTIN there are ___ alpha-numerals.

(A) 15

(B) 10

(C) 16

(D) 9

Answer:

(A) 15

(ii) From the following equations, which one is the quadratic equation ?

(A) \(\frac{5}{x}\) – 3 = x^{2}

(B) x(x + 5) = 4

(C) n – 1 = 2n

(D) \(\frac{1}{x^2}\)(x + 2) = x

Answer:

(B) x (x + 5) = 4

(iii) For simultaneous equations in variables x and y, if D_{x} = 49, D_{y} = -63, D = 7, then what is the value of x ?

(A) 7

(B) -7

(C) \(\frac{1}{7}\)

(D) –\(\frac{1}{7}\)

Answer:

(A) 7

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

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

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

(C) 10

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

Answer:

(C) 10

Question 1.

(B) Solve the following subquestions : [4]

(i) Find second and third term of an A.P. whose first term is – 2 and common difference is -2.

Answer:

Given, First term, a = – 2

Common difference, d = – 2

We know that,

Second term = a + d

= -2 + (-2)

= -4

and, Third term = a + 2d

= – 2 + 2 (-2)

= -2 – 4

= -6

∴ Second term is – 4 and third term is – 6.

(ii) ‘Pawan Medicals’ supplies medicines. On some medicines the rate of GST is 12%, then what is the rate of CGST and SGST ?

Answer:

Rate of CGST = 12%

Rate of CGST = Rate of SGST \(=\frac{\text { Rate of GST }}{2}\)

= \(\frac{12}{2} \%\)

[∵ Rate of the GST is equal to the GST Rate]

= 6%

∴ Rate of CGST = Rate of SGST = 6%

(iii) Find the values of a and b from the quadratic equation 2x^{2} – 5x + 7 = 0.

Answer:

The given quadratic equation is

2x^{2} – 5x + 7 = 0

Comparing the given quadratic equation with

ax^{2} + bx + c = 0

∴ The values of a = 2 and b = – 5

(iv) If 15x + 17y = 21 and 17x + 15y = 11, then find the value of x + y.

Answer:

The given equations are

15x + 17y = 21 ……… (1)

17x + 15y = 11 ……… (2)

Adding equations (1) and (2)

Dividing both sides by 32, we get

x + y = 1

Question 2.

(A) Complete and write any two activities from the following : [4]

(i) Complete the following table to draw the graph of 2x – 6y = 3 :

Solution:

(ii) First term and common difference of an A.P. are 6 and 3 respectively. Find S_{27}.

Solution:

First term a = 6,

common difference d = 3,

S_{27} = ?

Solution :

First term a = 6,

common difference d = 3,

(iii) A card is drawn from a well shuffled pack of 52 playing cards. Find the probability of the event, the card drawn is a red card.

Solution:

Suppose ‘S’ is sample space.

∴ n(S) = 52

Event A : Card drawn is a red card.

Solution:

Suppose ‘S’ is sample space.

∴ n(S) = 52

∴ Event A : Card drawn is a red card.

∴ Total red cards = __13__ hearts + 13 diamonds

Question 2.

(B) Solve any four subquestions from the following

(i) Find the value of the determinant:

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

\frac{7}{5} & \frac{5}{3} \\

\frac{3}{2} & \frac{1}{2}

\end{array}\right|\)

Solution:

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

\frac{7}{5} & \frac{5}{3} \\

\frac{3}{2} & \frac{1}{2}

\end{array}\right|\) = \(\frac{7}{10}\) – \(\frac{5}{2}\)

= \(\frac{7-5(5)}{10}\)

= \(\frac{-18}{10}\)

= \(\frac{-9}{5}\)

(ii) Solve the quadratic equation by factorisation method: x^{2} – 15x + 54 = 0.

Solution:

The given quadratic equation is

x^{2} – 15x + 54 = 0

⇒ x^{2} – 9x – 6x + 54 = 0

⇒ x (x – 9) – 6(x – 9) = 0

⇒ (x – 9) (x – 6) = 0

⇒ (x – 9) = 0 or (x – 6) = 0

∴ x = 9 or x = 6

∴ 9 and 6 are the roots of the given quadratic equation.

(iii) Decide whether the following sequence is an A.P. If so, find the 20^{th} term of the progression:

-12, -5, 2, 9, 16, 23, 30, ……….

Solution:

Here a = t_{1} = first term = – 12, t_{2} = -5,

Common difference = d = t_{2} – t_{1}

d = – 5 – (- 12)

= -5 + 12

∴ d = 7

We know that, t_{n} = a + (n – 1)d

Here, n = 20, a = -12, d = 7

∴ t_{20} = – 12 + (20 – 1)7

= – 12 + 133

t_{20} = 121

∴ 20^{th} term of the progression is 121.

(iv) A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is an odd number?

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) = n(A)n(S) = 1620 = 45

∴ P(A) = 45

Event B: The number formed is multiple of 5.

= {25, 35, 75, 95}.

n(B) = 4.

∴ P(B) =n(B)n(S)

=420

P(B) = 15

(v) If L = 10, f_{1} = 70, f_{o} = 58, f_{2} = 42, h = 2, then find the mode by using formula.

Solution:

Question 3.

(A) Complete and write any one activity from the following:

(i)

Solution:

(ii) Shri Shantilal has purchased 150 shares of FV ₹ 100, for MV of ₹ 120, Company has paid dividend at 7%, then to find the rate of return on his investment, complete the following activity :

Solution:

FV = ₹ 100; Number of shares = 150

Market value = ₹ 120

Solution:

FV = ₹100; Number of shares = 150

Market value = ₹120

Question 3.

(B) Attempt any two subquestions from the following : [6]

(i) 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 :

1. a red balloon.

2. a blue balloon.

Solution:

Available ballons are 2 red, 3 blue and 4 green. Sample space S : One ballon to be choose on random basis,

∴ n(S) = 2 + 3 + 4 = 9

1. Event A: Probability that a red ballon is chosen

∴ n(A) = 2

∴ p(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\)

∴ p(A) = \(\frac{2}{9}\)

2. Event B : Probability that a blue ballon is chosen

∴ n(B) = 3

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

= \(\frac{3}{9}\) = \(\frac{1}{3}\)

∴ p(B) = \(\frac{1}{3}\)

Probability that a red ballon is choosen is \(\frac{2}{9}\) and probability that a blue ballon is chosen is \(\frac{1}{3}\).

(ii) The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6, find the fraction.

Solution:

Suppose numerator is x, then denominator will be 2x + 4

∴ Fraction is \(\frac{x}{2 x+4}\)

According to the given information we can write,

\(\frac{x-6}{(2 x+4)-6}\) = \(\frac{1}{12}\)

\(\frac{x-6}{2 x-2}\) = \(\frac{1}{12}\)

∴ 12(x – 6) = 2x – 2

∴ 12x – 72 = 2x – 2

∴ 12x – 2x – 72 + 2 = 0

10x – 70 = 0

∴ x = \(\frac{70}{10}\) = 7

∴ x = 7

But fractions

∴ The fraction is \(\frac{7}{18}\).

(iii) A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method :

Milk sold (litre) | No. of customers |

1-2 | 17 |

2-3 | 13 |

3-4 | 10 |

4-5 | 7 |

5-6 | 3 |

Solution:

(iv) In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)

Answer:

Assume that the three consecutive terms are a- d, a, and a + d According to first condition,

∴ (a – d) + a + (a + d) = 27

∴ 3a = 27

∴ a = 9

According to second condition,

(a – d) (a) (a + d) = 504

Putting the value of a = 9 in above equation, we get

∴ (9 – d) (9) (9 + d) = 504

∴ (9^{2} – d^{2}) × 9 = 504

∴ (81 – d^{2}) = 56

∴ 81 – d^{2} = 56

∴ d^{2} = 81 – 56

∴ d^{2} = 25

∴ d = 5

∴ First term = a – d = (9 – 5) = 4

Second term = a = 9

Third term = a + d = 9 + 5 = 14

∴ The three terms are 4, 9, 14.

Question 4.

Attempt any two sub-questions from the following :

(i) Represent the following data by histogram :

Price of Sugar (per kg in ₹) | Number of weeks |

18-20 | 4 |

20-22 | 8 |

22-24 | 22 |

24-26 | 12 |

26-28 | 6 |

28-30 | 8 |

Answer:

(ii) One person borrows ₹ 4,000 and agrees to repay with a total interest of ₹ 500 in 10 instalments. Each instalment being less than the preceding instalment by ₹ 10. What should be the first and the last instalments ?

Answer:

Number of instalments, n = 10

Let the first instalment be ₹ a

As per the given data each further instalment is less than the preceding one by ₹ 10.

∴ These instalments are in A.P.

∴ First term = a,

and common difference, d = – 10

Here the negative sign indicates that the next term of A.P. is less than that the preceding term.

∴ Repayment of loan is a below :

∴ S_{n} = Loan + Total interest

S_{n} = 4000 + 500

S_{n} = 4500

Here n = 10

We know that, S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d]

∴ 4500 = \(\frac{10}{2}\)[2a + (10 – 1) (-10)]

∴ 4500 = 5 [2a -90]

∴ 4500 = 10a – 450

∴ 10a = 4500 + 450

∴ a = \(\frac{4950}{10}\) = 495

∴ first instalment = a = ₹ 495

The last instalment is the 10^{th} instalment

∴ n = 10

We know that, a_{n} = a + (n – 1)d

Here n = 10, a = 495, d = – 10

∴ a_{10} = 495 + (10 – 1)(-10)

= 495 – 90

a_{10} = 405

Last instalment = a_{10} = ₹ 405

∴ First instalment is ₹ 495 and the last instalment is ₹ 405.

(iii) The sum of the areas of two squares is 400 sq.m. If the difference between their perimeters is 16 m, find the sides of two squares.

Answer:

Let the side of first square be x metre and the side of second square be y metre As per the first given condition,

∴ x^{2} + y^{2} = 400 ………(i)

As per the second given condition,

4x – 4y = 16

∴ x – y = 4

∴ x = y + 4 ……(ii)

Put the value of x = y + 4 in equation (i), we get

∴ (y + 4)^{2} + y^{2} = 400

∴ y^{2} + 8y + 16 + y^{2} = 400

∴ 2y^{2} + 8y + 16 – 400 = 0

∴ 2y^{2} + 8y – 384 = 0

Dividing both sides by 2 we get

∴ y^{2} + 4y – 192 = 0

∴ y^{2} + 16y – 12y – 192 = 0

∴ y(y +16) – 12 (y + 16) = 0

∴ (y + 16) (y – 12) = 0

∴ (y + 16) = 0 or (y -12) = 0

∴ y = -16 or y = 12

But side of square is never negative

∴ y ≠ -16

∴ y = 12

Putting the value of y = 12 in equation (ii), we get

x = 12 + 4 = 16

∴ Side of first square, x = 16 m and side of second square, y = 12 m.

Question 5.

Attempt any one subquestion from the following:

(i) Convert the following equations into simultaneous equations and solve:

\(\sqrt{\frac{x}{y}}\) = 4, \(\frac{1}{x}+\frac{1}{y}\) = \(\frac{1}{x y}\)

Answer:

The given eqautions are \(\sqrt{\frac{x}{y}}\) = 4, \(\frac{1}{x}+\frac{1}{y}\) = \(\frac{1}{x y}\)

\(\sqrt{\frac{x}{y}}\) = 4

Taking square on both the sides

\(\left(\sqrt{\frac{x}{y}}\right)^2\) = (4)^{2}

∴ \(\frac{x}{y}\)

∴ x = 16y

∴ x – 16y = 0 ……. (i)

Now, \(\frac{1}{x}+\frac{1}{y}\) = \(\frac{1}{x y}\)

∴ \(\frac{y+x}{x y}\) = \(\frac{1}{x y}\)

∴ y + x = 1 (multiplying both sides by xy)

∴ x + y = 1 …….. (ii)

The two simultaneous equations formed are

x – 16y = 0

and x + y = 1

Substracting equation (ii) from equation (i), we get

Putting the value of y in equation (ii), we get

x + \(\frac{1}{17}\) = 1

∴ 17x + 1 = 17

∴ 17x = 17 – 1 = 16

∴ x = \(\frac{16}{17}\)

∴ (x, y) = (\(\frac{16}{17}\), \(\frac{1}{17}\)) is the solution of the simultaneous equation.

(ii) A dealer sells a toy for ₹ 24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.

Answer:

Selling price of the toy = ₹ 24

Let the cost price of the toy be ₹ x

Gain% = x% (Given)

Gain% \(=\left(\frac{\text { Selling price }- \text { Cost price }}{\text { Cost price }}\right)\) × 100

∴ x = \(\left(\frac{24-x}{x}\right)\) × 100

∴ x^{2} = 2400 – 100x

∴ x^{2} + 100x – 2400 = 0

∴ x^{2} + 120x – 20x – 2400 = 0

∴ x(x + 120) – 20 (x + 120) = 0

∴ (x + 120) (x – 20) = 0

∴ (x + 120) = 0 or (x – 20) = 0

∴ x = -120 or x = 20

x ≠ -120, because cost cannot be negative

∴ x = 20

∴ Cost price = 20

∴ The cost price of the toy is ₹ 20.