SSC Maharashtra Board Maths 1 Question Paper 2019 with Solutions

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:

1. All questions are compulsory.
2. Use of calculator is not allowed.
3. 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 :
2x³ + x² – 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 2x³ + 2x, when x = -1.
Solution:
Given, P(x) = 2x³ + 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 x² + 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:
x² + 8x + 15 = 0.
Solution:
Given, x² + 8x + 15 = 0
On splitting the middle term
⇒ x² + 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
t₁ = 7, t₂ = 13,
d = t₂ – t₁ = 13 – 7 = 6
According to formula,
t_n = a + (n – 1)d
∴ 19^th term is:
t₁₉ = 7 + (19 – 1) 6
= 7 + (18) × 6
= 7 + 108
∴ t₁₉ = 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.
x² – (α + β) x + αβ = 0
∴ x² – (-10)x + 21 = 0
x² + 10x + 21 = 0
Hence, x² + 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 = t₁ = 200, t₂ = 250
d = t₂ – t₁ = 250 – 200 = 50 and n = 17
To find his total savings in 17 months, we have to find S₁₇.

∴ 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(x² + 3x) = 40(2x + 3)
⇒ 13x² + 39x = 80x + 120
⇒ 13x² + 39x – 80x – 120 = 0
⇒ 13x² – 41x – 120 = 0
⇒ 13x² – 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

SSC Maharashtra Board Maths 1 Question Paper with Solutions