SSC Maharashtra Board Maths 1 Question Paper Dec 2020 with Answers

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

SSC Maths 1 Question Paper Dec 2020 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) Four alternative answers are given for every sub-question. Choose the correct alternative and write its alphabet with sub-question number: [4]
i. To draw graph of 4x + 5y = 19, what will be the value of y when x = 1: (B)
(A) 4
(B) 3
(C) 2
(D) -3
Answer:
(B) 3

ii. What is the sum of the first 10 natural numbers? (A)
(A) 55
(B) 20
(C) 65
(D) 11
Answer:
(A) 55

iii. From the following equations, which one is the quadratic equation? (B)
(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

iv. In the format of GSTIN there are _________ alpha-numerals. (D)
(A) 9
(B) 10
(C) 16
(D) 15
Answer:
(D) 15
Hints:
Substituting x = 1 in 4x + 5y = 19, we get
4(1) + 5y = 19
∴ 5y = 19 – 4 = 15
∴ y = 3

ii. First 10 natural numbers are 1, 2, 3,…, 9, 10.
The above sequence is an A.P.
∴ t₁ = 1, t₁₀ = 10
∴ S_n = \(\frac{n}{2}\)(t₁ + t₁₀)
= \(\frac{10}{2}\)(1 + 10)
= 5(11)
= 55

iii. x(x + 5) = 2
∴ x² + 5x – 2 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 5, c = -2 are real number and a ≠ 0

(B) Solve the following subquestions: [4]
i. For simultaneous equations in variable x and y, if Dx = 25, Dy = 40, D = 5, then what is the value of x?
ii. Find the first term and common difference for the following A.P.:
127, 135, 143, 151, …
iii. A die is rolled then write sample space ‘S’ and number of sample point n(S).
iv. If ∑f₁d₁ = 108, and ∑f₁ = 100, then find \(\bar{d}\) = ?
Answer:
i. x = \(\frac{D_x}{D}=\frac{25}{5}\) = 5

ii. The given A.P. is 127, 135, 143, 151, …
Here, t₁ = 127, t₂ = 135
∴ a = t₁ = 127 and
d = t₂ – t₁ = 135 – 127 = 8
∴ first term (a) = 127, common difference (d) = 8

iii. Sample space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6

iv. \(\overline{\mathrm{d}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{~d}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{108}{100}\) = 1.08

Question 2.
(A) Complete the following activities and rewrite it (any two):
i. Activity:

ii. One of the roots of quadratic equation 5m² + 2m + k = 0 is –\(\frac{7}{5}\).
Complete the following activity to find the value of k.
Activity:
–\(\frac{7}{5}\) is root of quadratic equation
5m² + 2m + k = 0

iii. Complete the activity to prepare a table showing the co-ordinates which are necessary to draw a frequency polygon:

(B) Solve the following sub-questions (any four):
i. Sum of two numbers is 7 and their difference is 5. Find the numbers.
ii. Solve the quadratic equation by factorisation method:
X² + x – 20 = 0
iii. Find the 19th term of the following A.P.:
7, 13, 19, 25, …….. .
iv. For the following experiments, write sample space ‘S’ and number of sample points n(S): Two digit numbers are formed using digits 2, 3 and 5 without repeating a digit.
v. The following table shows causes of noise pollution. Find the measure of central angles for each, to draw a pie diagram:

Answer:
(A) i.

(B)
i. Let the greater number be x and the smaller number be y.
According to the first condition, sum of the two numbers is 7.
∴ x + y = 7 …(i)
According to the second condition, difference of the two numbers is 5.
∴ x – y = 5 …(ii)
Adding equations (i) and (ii), we get

∴ x = 6
Substituting x = 6 in equation (i), we get
6 + y = 7
∴ y = 1
∴ The required numbers are 6 and 1.

ii. x² + x – 20 = 0
∴ x² + 5x – 4x – 20 = 0
∴ x (x + 5) – 4 (x + 5) = 0
∴ (x + 5) (x – 4) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 5 = 0 or x – 4 = 0
∴ x = -5 or x = 4
∴ The roots of the given quadratic equation are -5 and 4.

iii. The given A.P. is 7, 13, 19, 25, …
Here, a = 7, d = 13 – 7 = 6
Since t_n = a + (n – 1)d
∴ t₁₉ = 7 + (19 – 1)6
= 7 + 18 x 6 = 7 + 108
∴ t₁₉ = 115
∴ 19^th term of the given A.P. is 115.

iv. Sample space,
S = {23, 25, 32, 35, 52, 53}
∴ n(S) = 6

v.

Question 3.
(A) Complete the following activity and rewrite it (any one): [3]
i. In an A.P. the first term is -5 and last term is 45. If sum of ‘n’ terms in the A.P. is 120, then complete the activity to find n.
Activity:

ii. A card is drawn from a well shuffled pack of 52 playing cards.
Complete the activity to find the probability of the event that the card drawn is a red card. Activity:
‘S’ is the sample space.
n(S) = 52
Event A: Card drawn is a red card.


(B) Solve the following subquestions (any two): [6]
i. Solve the following simultaneous equations graphically:
x + y = 5; x – y = 1.
ii. Solve quadratic equation using formula method:
5m² + 13m + 8 = 0.
iii. A retailer sold 2 tins of lustre paint and taxable value of each tin is ‘ 2,800. If the rate of GST is 28%, then find the amount of CGST and SGST charged in the tax invoice.
iv. Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show this information:

Time (minutes)	No. of Students
60-80	14
80-100	20
100-120	24
120-140	22

Answer:
(A) i.

ii. ‘S’ is the sample space.
n(S) = 52
Event A: Card drawn is a red card.

(B) i. The given simultaneous equations are

The two lines intersect at point (3, 2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations x + y = 5 and x – y = 1.

ii. 5m² + 13m + 8 = 0
Comparing the above equation with am2 + bm + c = 0, we get
a = 5, b = 13, c = 8
b² – 4ac = (13)² – 4 × 5 × 8
= 169 – 160
= 9


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

iii. Taxable value of 1 tin = ₹2,800
∴ Taxable value of 2 tins = 2 × 2,800
= ₹5,600
Rate of GST = 28%
∴ Rate of CGST = Rate of SGST = 14%
CGST = 14% of taxable value
= \(\frac{14}{100}\) × 5,600
∴ CGST = ₹784
∴ SGST = CGST = ₹784
∴ The amount of CGST and SGST charged in the tax invoice is ₹784 each.

iv.

Question 4.
(A) Solve the following subquestions (any two): [8]
i. If one root of the quadratic equation ax² + bx + c = 0 is half of the other root, show that, b² = \(\frac{9ac}{2}\).
ii. Bhujangrao invested ₹2,50,590 in shares of F.V. ₹10 when M.V. is ₹250. Rate of brokerage is 0.2% and GST is 18%, then find:
a. the number of shares purchased,
b. the amount of brokerage paid, and
c. GST paid for the trading.
iii. The following table shows frequency distribution of number of trees planted by students in the school:

No. of Trees Planted	No. of Students
0-10	30
10-20	70
20-30	100
30-40	70
40-50	40

Find the mode of trees planted.
Answer:
i. Let α and β be the roots of the given quadratic equation.
According to the given condition, α = \(\frac{\beta}{2}\), i.e., β = 2α

ii. Amount invested = ₹2,50,590,
FV = ₹10, MV = ₹250, Rate of brokerage = 0.2%, Rate of GST = 18%

∴ 1000 shares were purchased for ₹2,50,590.

b. The amount of brokerage paid = Brokerage on 1 share × number of shares
= 0.5 × 1000 = ₹500
The amount of brokerage paid is ₹500.
c. GST paid = 18% of brokerage
= \(\frac{18}{100}\) × 500 = ₹90
∴ GST paid for trading is ₹90.

iii.

No. of trees planted	No. of students
0 – 10	30
10 – 20	70 → f0
20 – 30	100 → f1
30 – 40	70 → f2
40 – 50	40

Here, the maximum frequency is 100.
∴ The modal class is 20 – 30.
L = 20, h = 10, f₁ = 100, f₀ = 70, f₂ = 70

∴ The mode of trees planted is 25.

Question 5.
Solve the following subquestions (any one): [3]
i. Six faces of a die are as shown below:

If the die is rolled once, find the probability of event ‘M’ that ‘English vowel appears on upper face’.
ii. Construct any one linear equation in two variables. Obtain another equation by interchanging only coefficients of variables. Find the value of the variables.
Answer:
i. Sample space S = {A, B, C, D, E, O}
∴ n(S) = 6
Event M : English vowel appears on the upper face.
∴ M = {A, E, O}
∴ n(M) = 3
∴ P(M) = \(\frac{n(M)}{n(S)}=\frac{3}{6}\)
∴ P(M) = \(\frac{1}{2}\)

ii. Consider a linear equation in two variables:
3x + 2y = 5 …(i)
According to the given condition, another equation is
2x + 3y = 5 …(ii)
Adding equations (i) and (ii), we get

∴ x = 1
Substituting x = 1 in equation (iii), we get
x + y = 2
∴ 1 + y = 2
∴ y = 2 – 1 = 1
∴ x = 1 and y = 1 are the values of the variables.

SSC Maharashtra Board Maths 1 Question Paper with Solutions