Maharashtra Board Class 10 Maths 1 Sample Paper Set 7 with Answers

Maharashtra Board SSC Class 10 Maths 1 Sample Paper Set 7 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 1 Model Paper Set 7 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. 210 is the ________ term of the A.P. 21, 42, 63, 84, … (A)
(A) 10^th
(B) 11^th
(C) 12^th
(D) 13^th
Answer:
(A) 10^th

ii. To find the cost of one share at the time of buying, the amount of Brokerage and GST is to be MV of share. (A)
(A) added to
(B) subtracted from
(C) multiplied with
(D) divided by
Answer:
(A) added to

iii. The formula to find mean from a grouped frequency table is \(\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum f_i u_i}{\sum f_i} \times g\). (C)
In the formula u_i = ________.

Answer:
C

iv. To draw graph (A) 4
(B) 3
(C) 2
(D) -3
Answer:
(B) 3
Hints:
Let the n^th term be 210.
t_n = a + (n – 1) d
∴ 210 = 21 + (n – 1) 21
∴ 210 = 21 + 21n – 21
∴ 210 = 21n
∴ n = \(\frac{210}{21}\)
∴ n = 10

iv. Substituting x = 1 In 4x + 5y = 19,
4(1) + 5y = 19
∴ 5y = 19 – 4 = 15
∴ y = \(\frac{15}{5}\) = 3

Question 1.
(B) Solve the following sub-questions. [4]
i. Find the value of discriminant of the quadratic equation 5m2
ii. On certain article if rate of CGST is 9%, then what is the rate of SGST? and what is the rate of GST?
iii. Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure of ₹45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
iv. There are 15 tickets bearing the numbers from 1 to 15 in a bag and one ticket is drawn from this bag at random. Write the sample space (S) and n(S).
Answer:
i. Here, a = 5, b = -1, c = 0
∴b² – 4ac = (-1)² – 4 × 5 × 0
= 1 – 0
∴b² – 4ac = 1

ii. rate of SGST = rate of CGST
∴ Rate of SGST = 9%
Rate of GST = Rate of SGST + Rate of CGST
= 9% + 9%
∴ Rate of GST = 18%

iii.

iv. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
n(S) = 15

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
One of the roots of quadratic equation 2x² + kx – 2 = 0 is – 2. Complete the following activity to find the value of k.
-2 is one of the roots of the equation 2x² + kx -2 = 0.
∴ Putting x = [__] in the given equation, we get
2[__] +k[__] – 2 = 0
∴ k = [__]
ii. In a class of 42 students in Model High School, 3 students use spectacles. Fill in the following boxes to find the probability of a student selected at random is wearing spectacles.
The total number of students in the class is 42.
∴ n(S)= [__]
Let A be the event that a student uses spectacles.
∴ n(A) = [__]
∴ P(A) = [__]
∴ P(A) = [__]
iii. Complete the following activity to solve the simultaneous equations 4m – 2n = -4 and 4m + 3n = 16 by cramer’s method.

(B) Solve the following sub-questions (Any four): [8]
i. Compare the quadratic equation √x² + √2x – 2√3 = 0 with ax² + bx + c = 0 and find the value of discriminant and hence the nature of the roots.
ii. First term and common difference of an A.P. are 12 and 4 respectively. If t_n = 96, find n.
iii. Two numbers differ by 3. The sum of the greater number and twice the smaller number is 15. Find the smaller number.
iv. A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
v. Sachin invested in a national saving certificate scheme. In the first year he invested ₹500, in the second year ₹700, in the third year ₹900 and so on. Find the total amount that he invested in 12 years.
Answer:
(A) i. -2 is one of the roots of the equation 2x² + kx – 2 = 0.
∴ Putting x = [-2] in the given equation, we get
2[(-2)²] + k[-2] – 2 = 0
∴ 8 – 2k – 2 = 0
∴ 2k = 6
∴ k = [3]

ii. The total number of students in the class is 42.
∴ n(S) = [42]
Let A be the event that a student uses spectacles.

iii.

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.

(B) i. √3x² + √2x – 2√3 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = √3, b = √2 , c = -2√3
∴ ∆ = b² – 4ac
= (√2)² – 4 × √3 × (-2√3)
= 2 + 24
∴ ∆ = 26
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

ii. Given, first term (a) = 12,common difference (d) = 4, t_n = 96
Since, t_n = a + (n – 1)d
∴ 96 = 12 + (n – 1)(4)
∴ 96 – 12 = (n – 1)(4)
∴ 84 = (n – 1)4
∴ n – 1 = \(\frac{84}{4}\)
∴ n – 1 = 21
∴ n = 21 + 1 = 22
SMART TIP
To check our answer, verify if 22^nd term is 96 or not.
t_n = a + (n – 1)d
∴ t₂₂ = 12 + (22 – 1) 4
= 12 + 21 × 4 = 12 + 84 = 96
Hence, our answer is correct.

iii. Let the greater number be x and the smaller number be y.
According to the first condition,
x – y = 3 …(i)
According to the second condition,
x + 2y = 15 …(ii)
Subtracting equation (i) from (ii), we get

∴ The smaller number is 4.

iv. Total number of pens in the box = 5 + 8 + 3 = 16
∴ n(S) = 16
Let A be the event that Rutuja picks a blue pen.
Total number of blue pens = 8
∴ n(A) = 8
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{8}{16}\)
∴ P(A) = \(\frac{1}{2}\)

v. Amount invested by Sachin in each year are as follows:
500, 700, 900, …
The above sequence is an A.P.
∴ a = 500, d = 700 – 500 = 200, n = 12

∴ The total amount invested by Sachin in 12 years is ₹19,200.

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
i. To solve given equations, fill the below boxes suitably.

ii. A frequency distribution table for the production of oranges of some farm owners is given below. To find the mean production of oranges by ‘assumed mean’ method, fill in the blanks.

(B) Solve the following sub-questions (Any two): [6]
i. Joseph purchased following shares, Find his total investment.
Company A : 200 shares, FV = ₹2, Premium = ₹18.
Company B : 45 shares, MV = ₹500
Company C : 1 share, MV = ₹10,540
ii. If the 9^th term of an A.P. is zero, then show that the 29^th term is twice the 19^th term.
iii. Sum of the roots of a quadratic equation is double their product. Find k if equation is x² – 4kx + k + 3 = 0.
iv. In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at

a. an odd number.
b. a number greater than 2.
c. a number less than 9.
Answer:
(A) i.

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.

ii.

(B) i. For company A:
FV = ₹2, premium = ₹18, Number of shares = 200
∴ MV = FV + Premium
= 2 + 18
= ₹20
Sum invested = Number of shares × MV
= 200 × 20
= ₹4000

For company B:
MV = ₹500, Number of shares = 45
Sum invested = Number of shares × MV
= 45 × 500 = ₹22,500

For company C:
MV = ₹10,540, Number of shares = 1
∴ Sum invested = Number of shares × MV
= 1 × 10540 = ₹10,540
∴ Total investment of Joseph
= Investment for company A + Investment for company B + Investment for company C
= 4000 + 22,500 + 10,540
= ₹37040
∴ Total investment done by Joseph is ₹37,040.

ii. To prove: t₂₉ = 2 t₂₉
Proof: t₉ = 0 …[Given]
Since, t_n = a + (n – 1)d
∴ t₉ = a + (9 – 1)d
∴ 0 = a + 8d
∴ a = -8d …(i)
Also, t₁₉ = a + (19 – 1)d
= a + 18d
= -8d + 18d …[From (i)]
∴ t₁₉ = 10d
and t₂₉ = a + (29 – 1)d
= a + 28d
= -8d + 28d …[From (i)]
∴ t₂₉ = 20d = 2(10d)
∴ t₂₉ = 2(t₁₉) …[From (i)]
∴ The 29^th term is twice the 19^th term.

iii. x² – 4kx + k + 3 _ 0
Comparing the above equation with ax² + bx + c = 0, we get
a = 1, b = -4k, c = k + 3
Let α and β be the roots of the given quadratic equation.

iv. Sample space (S) = {1, 2, 3, 4, 5, 6, 7, 8}
∴ n(S) = 8
a. Let B be the event that the spinning arrow comes to rest at an odd number.
∴ B = {1, 3, 5, 7}
∴ n(B) = 4
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{8}\)
∴ P(B) = \(\frac{1}{2}\)

b. Let C be the event that the spinning arrow comes to rest at a number greater than 2.
∴ C = {3, 4, 5, 6, 7, 8}
∴ n(C) = 6
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{6}{8}\)
∴ P(C) = \(\frac{3}{4}\)

c. Let D be the event that the spinning arrow comes to rest at a number less than 9.
∴ D = {1, 2, 3, 4, 5, 6, 7, 8}
∴ n(D) = 8
∴ P(D) = \(\frac{n(D)}{n(S)}=\frac{8}{8}\)
∴ P(D) = 1

Question 4.
Solve the following sub-questions (Any two): [8]
i. From the same place at 7 am ‘A’ started walking in the north at the speed of 5 km/hr. After 1 hour, B started cycling in the east at a speed of 16 km/hr. At what time they will be at a distance of 52 km apart from each other?
ii. Draw a frequency polygon from the information given in the following table.

iii. When the son will be as old as his father today, the sum of their ages then will be 126. When the father was as old as his son is today, the sum of their ages then was 38. Find their present ages.
Answer:
i. Let the time taken be x hours.
∴ the distance travelled by A in x hours at the speed of 5 km/hr – 5x km.
and the distance travelled by B in A
(x – 1) hours at the speed of 16 km/hr – 16(x – 1) km.
Distance between them after x hours – 52 km
∴ According to the given condition and from the diagram,
∆ABC is a right angled triangle

∴ (5x)² + [16(x – 1)]² = (52)² …[By Pythagoras theorem]
∴ 25x² + 256(x² – 2x + 1) = 2704
∴ 25x² + 256x² – 512x + 256 = 2704
∴ 281x² – 512x – 2448 = 0
∴ 281x² – 1124x + 612x – 2448 = 0
∴ 281x(x – 4) + 612(x – 4) = 0
∴ (x – 4)(281x + 612) = 0
∴ x – 4 – 0 or 281x + 612 = 0
∴ x = 4 or x = –\(\frac{612}{281}\)
But time cannot be negative.
∴ x ≠ \(\frac{612}{281}\)
∴ x = 4
∴ the time taken – 4 hours.
∴ A and B are 52 kms apart from each other at 7 + 4 = 11 am.

ii.

iii. Let the present age of the father be x years and that of his son be y years.
After (x – y) years, son’s age will be x years. i.e. he will be as old as his father.
After (x – y) years, father’s age will be x + (x – y) years.
According to the first condition,
x + x + (x – y) = 126
i.e. 3x – y = 126 … (i)
(x – y) years ago, father’s age was y years. i.e. the father was as old as his son today.
(x – y) years ago, son’s age was
y – (x – y) = (2y – x)years
According to the second condition,
y + 2y – x = 38
i.e. -x + 3y = 38 …(ii)
Multiplying equation (ii) by 3,
-3x + 9y = 114 … (iii)
Adding equations (iii) and (i), we get

∴ y = 30
Substituting y = 30 in equation (i),
3x – 30 = 126
∴ 3x = 156
∴ x = 52
∴ The present age of the father and son are 52 years and 30 years respectively.

Question 5.
Solve the following sub-questions (Any one): [3]
i. The following frequency distribution table shows the distances travelled by some rickshaws in a day. Observe the table and answer the following questions:

a. Which is the modal class? Why?
b. Which is the median class and why?
c. Write the cumulative frequency (C.F.) of the class preceding the median class.
d. What is the class interval (h) to calculate median?
ii. There are three dealers A, B and C in Maharashtra. Suppose, the trade of each of them in September 2018 was as shown in the following table. The rate of GST on each transaction was 5%. Read the table and answer the questions below it.

a. How much amount did the dealer A get by sale?
b. For how much amount did the dealer B buy the articles?
c. How much is the balance of CGST and SGST left with the government that was paid by A?
Answer:
a. The modal class is 74.5 – 79.5 because the maximum frequency is 82.
b. Total frequency = N = 200
∴ \(\frac{N}{2}=\frac{200}{2}\) = 100
Cumulative frequency which is just greater than (or equal) to 100 is 102.
∴ The median class is 69.5-74.5.
c. 44 is the cumulative frequency of the class preceding the median class.
d. Class interval (h) = upper limit of median class – lower limit of median class
= 74.5 – 69.5 = 5

ii. a. The sale of dealer A = \(\frac{100}{5}\) × 5,000
= ₹1,00,000
b. The purchase of dealer B = \(\frac{100}{5}\) × 4000
= ₹80,000
c. Balance of CGST paid by A = \(\frac{1000}{2}\) = ₹500 and SGST = ₹500

SSC Maharashtra Board Maths 1 Question Paper with Solutions