Maharashtra Board Class 10 Maths 1 Sample Paper Set 3 with Answers

Maharashtra Board SSC Class 10 Maths 1 Sample Paper Set 3 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 1 Model Paper Set 3 with Answers

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. Which number cannot represent a probability? (B)
(A) \(\frac{2}{3}\)
(B) 1.5
(C) 15%
(D) 0.7
Answer:
(B) 1.5

ii. What is the amount of dividend received per share of face value ₹50 if dividend declared is 40%? (B)
(A) ₹10
(B) ₹20
(C) ₹30
(D) ₹40
Answer:
(B) ₹20

iii. For a given A.P., t₇ = 13, d = -2, then a = (D)
(A) 10
(B) 15
(C) 20
(D) 25
Answer:
(D) 25

iv. Find the value of \(\left|\begin{array}{cc} 5 & 3 \\ -7 & -4 \end{array}\right|\) (D)
(A) -1
(B) -41
(C) 41
(D) 1
Answer:
(D) 1
[Note: For this question, student should write the correct option along with its contents in the answer sheet. (See above)
Here hints provided for understanding purpose of the students.]
Hints:
i. The probability of any event is from 0 to 1 or 0% to 100%.

ii. Dividend = \(\frac{1}{2}\) × 50 = ₹20

iii. t₇ = a + (7 – 1) (-2)
∴ 13 = a – 12
∴ a = 25

iv. \(\left|\begin{array}{cc} 5 & 3 \\ -7 & -4 \end{array}\right|\) = (5 x – 4) – (3 x – 7)
= -20 – (-21)
= -20 + 21
= 1

Question 1.
(B) Solve the following sub-questions. [4]
i. If ∑f_i = 15 and ∑f_ix_i = 240, find the value of mean (\(\bar{X}\))
ii. Verify whether -1 is the root of quadratic equation x² + 4x – 5 = 0.
iii. Two coins are tossed simultaneously. Write the sample space S.
iv. To solve x + 2y = -1, 2x – 3y = 12 by determinant method, find D.
Answer:
(B)
i.

ii. Putting x = – 1 in L.H.S. of x² + 4x – 5 = 0 (i), we get
L.H.S. = (-1)² + 4(-l) – 5
= 1 – 4 – 5 = -8
∴ L.H.S. ≠ R.H.S.
∴ x – -1 is not the root of the given quadratic equation.

iii. S = {HH. HT, TH, TT}
SMART TIP
If ‘n’ coins are tossed or a coin is tossed ‘n’ times, then the possible number of outcomes (number of elements in smaple space) will be 2ⁿ.
In this case, n = 2. So the number of elements in the sample space is 2² = 4

iv. Here a₁ = 1, b₁ = 2
a₂ = 2, b₂ = -3

= (l × -3) – (2 × 2)
= -3 – 4 = -7

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
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

ii. Fill in the blanks.

iii. Observe the given frequency polygon and complete the table.

(B) Solve the following sub-questions (Any four): [8]
i. The pie diagram in the given figure shows the proportions of different workers in a town. Answer the following questions with its help.
a. If the total workers is 10,000, how many of them are in the field of construction?
b. How many workers are working in the administration?

ii. Given Arithmetic Progression is 12, 16, 20, 24,… Find the 24^th term of this progression.
iii. Solve the following quadratic equation by using formula.
m² + 5m + 5 = 0
iv. Market value of a share is ₹200. If the brokerage rate is 0.3% then find the purchase value of the share.
v. On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
Answer:
(A) i. FV = ₹10, Premium = ₹2

(B)
i.

∴ There are 1000 workers working in the administration.

ii. The given A.P. is 12,16, 20, 24,…
Here, a = 12, d = 16 – 12 = 4
Since, t_n = a + (n – 1)d
t₂₄ = 12 + (24 – 1)4
= 12 + 23 × 4
= 12 + 92
∴ t₂₄ = 104
∴ 24^th term of the given A.P. is 104.

iii. m² + 5m + 5 = 0
Comparing the above equation with
am² + bm + c = 0, we get
a = 1, b = 5,c = 5

∴ The roots of the given quadratic equation are \(\frac{-5+\sqrt{5}}{2}\) and \(\frac{-5-\sqrt{5}}{2}\).
SMART TIP
In order to find out if our answer is correct or not, for these type of equations, substitute the value of x in the equation.
If L.H.S. = R.H.S., then the answer is correct.

iv. Here, MV = ₹200, Brokerage = 0.3%
Brokerage = 0.3% of MV
= \(\frac{0.3}{100}\) × 200
= ₹0.6
∴ Purchase value of the share = MV + Brokerage
= 200 + 0.6
= ₹200.60
∴ Purchase value of the share is ₹200.60.

v. The number of trees planted row-wise are as follows:
1,2,3,…
The above sequence is an A.P.
a = 1, d = 2 – 1 = 1, n = 25

∴ The total number of trees in 25 rows are 325.

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
i. Complete the following activity.

ii. Out of 200 students from a school, 135 like Kabaddi and the remaining students do not like the game. If (pie student is selected at random from all the students, fill in the following boxes to find the probability that the student selected doesn’t like Kabaddi.
Total number of students in the school = [__]
∴ n(S) = [__]
Number of students who like Kabaddi = 135
∴ Number of students who do not like Kabaddi = [__]
Let A be the event that the student selected does not like Kabaddi.

(B) Solve the following sub-questions (Any two): [6]
i. A dealer has given 10% discount on a showpiece of ₹25,000. GST of 28% was charged on the discounted price. Find the total amount shown in the tax invoice. What is the amount of CGST and SGST.
ii. Solve the following simultaneous equations.

iii. A two digit number is to be formed from the digits 0, 1,2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a multiple of 11.
iv. If m times the m^th term of an A.P. is equal to n times the n^th term, then show that (m + n)^th term of the A.P. is zero.
Answer:
(A) i.

ii. Total number of students in the school = [200]
∴ n(S) = [200]
Number(of students who like Kabaddi = 135
∴ Number of students who do not like Kabaddi = 200 – 135 = [65]
Let A be the event that the student selected does not like Kabaddi.

(B) i. Printed price of showpiece = ₹25,000, Rate of discount = 10%
Amount of discount = 10% of printed price
= \(\frac{10}{100}\) × 25000 = ₹2500
∴ Taxable value = Printed price – discount
= 25,000 – 2500 = ₹22,500
Rate of GST = 28%
∴ Rate of COST = 14% and
Rate of SGST = 14%
CG5T = 14% of taxable value
= \(\frac{14}{100}\) × 22500 100
= ₹3150
∴ CGST = SGST = ₹3150
Total amount of tax invoice= Taxable value + CGST + SSST
= 22500 + 3150 + 3150 = ₹28,800
∴ The total amount shown in the tax invoice is ₹28,800, and the amount of CGST and S6ST is ₹3150 each.

ii.

Equations (i) and (ii) become
2p – 3q = 15 ….(iii)
8p + 5q = 77 ….(iv)
Multiplying equation (iii) by 4, we get
8p – 12q = 60 …(v)
Subtracting equation (v) from (iv), we get

Substituting q = 1 in equation (iii), we get
2p – 3(1) = 15
2p – 3 = 15
2p = 15 + 3 = 18
P = \(\frac{18}{2}\) = 9
(p, q) = (9, 1)
Resubstituting the values of p and q, we get
3 = \(\frac{1}{x}\) and 1 = \(\frac{1}{y}\)
∴ x = \(\frac{1}{x}\) and y = 1
SMART TIP
In order to find out if our answer is correct or not, substitute the values of (x, y) in the given equations.
If L.H.S = R.H.S, then the answer is correct.
∴ (x, y) = [ \(\frac{1}{9}\), 11 is the solution of the given simultaneous equations.

iii.

Let B be the event that the number so formed is a multiple of 11.
∴ B = {11, 22,33,44}
∴ n(B) = 4
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{20}\)
∴ P(5) = \(\frac{1}{5}\)

iv. According to the given condition,
mt_m = nt_n
∴ m[a+ (m – 1)d] = n[a + (n – l)d]
∴ ma + md(m – 1) = na + nd(n – 1)
∴ ma + m2d – md = na + n2d – nd
∴ ma + m2d – md – na – n2d + nd = 0
∴ (ma – na) + (m2d – n2d) – (md – nd) = 0
∴ a(m – n) + d(m2 – n2) – d(m – n) = 0
∴ (m – n)[a + (m + n – 1) d] = 0
∴ [a + (m + n – l)d] = 0 …[Dividing both sides by (m – n)]
∴ t_{(m + n)} = 0
∴ The (m + n)^th term of the A.P. is zero.

Question 4.
Solve the following sub-questions (Any two): [8]
i. An electric company producing electric bulb, has packed 100 bulbs in each box. Some bulbs from 16 such boxes are tested defective. The information of number of defective bulbs in 16 boxes is given below.

No. of defective bulbs	No. of boxes
0 – 1	3
2 – 4	4
4 – 6	5
6 – 8	3
8 – 10	1

a. How many boxes contain maximum number of defective bulbs?
b. Find the mean of defective bulbs.
c. The box is selected at random, what is the probability that it will contain average 2 to 4 defective bulbs?
d. Find the probability of getting a box with most number of defective bulbs?
ii. Draw the graphs representing the equations 2x = y + 2 and 4x + 3y = 24 on the same graph paper. Find the area of the triangle formed by these lines and the X-axis.
iii. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then prove that 2a²c = c²b + b²a.
Answer:
a. Only 1 box contain maximum number of defective bulbs.
b.

∴ Mean of defective bulbs is 4.38.
c. n(S) = 16
Let A the event that box contain average 2 to 4 defective bulbs.
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)
d. Let B the event of getting highest number of defective bulbs. n(B) = 1
n(B) = 1
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{1}{16}\)

ii.


From the graph, we get AABC, where BE is the height of the triangle and AC is the base.
Now, /(AC) = 5 cm and /(BE) = 4 cm
∴ Area of triangle = \(\frac{1}{2}\) × base × height
∴ Area of ∆ABC = × /(AC) × /(BE)
= \(\frac{1}{2}\) × 5 × 4 = 2 × 5 = 10 cm².

iii. Let α and β be the roots of the equation
ax² + bx + c = 0
∴ α + β = \(\frac{-b}{a}\) and αβ = \(\frac{c}{a}\)
According to the given condition,

Question 5.
Solve the following sub-questions (Any one): [3]
i. Following is the age wise distribution of the members of a family.

Suppose there are total 60 members in the family, then find x. Find the median age of family member.
ii. Find the quadratic equation whose roots exceed the roots of the quadratic equation x² – 2x + 3 = 0 by 2.
Answer:
i. The total number of family members = 60
∴ 5 + 2x + 12 + 3x + 18 = 60
∴ 35 + 5x = 60
∴ 5x = 25
∴ x = 5
The following table shows cumulative frequency of less than type.

Class interval	Frequency fi	Cumulative frequency (less than)
0 – 10	5	5
10 – 20	10	15
20 – 30	12	27 → cf
30 – 40	15 → f	42
40 – 50	18	60
Total	∑fi = 60

Here, total frequency = ∑f_i = N = 60
∴ \(\frac{N}{2}=\frac{60}{2}\) = 30
Cumulative f requency just greater than (or equal) to 30 is 42.
∴ Median class is 30 – 40.
Now, L = 30, f = 15, cf = 27, h = 10

∴ Median = 32
∴ Median age of a family member is 32 years.

ii. Let α and β be the roots of the quadratic equation x² – 2x + 3 = 0.
α + β = 2 and αβ = 3
We have to find a quadratic equation whose roots are (α + 2) and (β + 2).
Now, sum of roots = (α + 2) + (β + 2)
= α + β + 4 = 2 + 4
= 6
product of roots = (α + 2) (β + 2)
i = aβ + 2(α + β) + 4
= 3 + 2 x 2 + 4
= 11
The required quadratic equation whose roots are (α + 2) and (β + 2) is
x² – (sum of roots)x + product of roots = 0
x² – 6x + 11 = 0

SSC Maharashtra Board Maths 1 Question Paper with Solutions