Maharashtra Board Class 10 Maths 1 Sample Paper Set 6 with Answers

Maharashtra Board SSC Class 10 Maths 1 Sample Paper Set 6 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 1 Model Paper Set 6 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 of the following quadratic equations has roots 2, -1? (B)
(A) x² + x + 2 = 0
(B) x² – x – 2 = 0
(C) x² + 2x – 2 = 0
(D) x² – 2x + 2 = 0
Answer:
(B) x² – x – 2 = 0

ii. In an A.P., the first term is -5 and last term is 45. If the sum of all the terms is 120, then n is equal to (C)
(A) 4
(B) 5
(C) 6
(D) 7
Answer:
(C) 6

iii. Cumulative frequencies in a grouped frequency table are useful to find _________. (B)
(A) Mean
(B) Median
(C) Mode
(D) All of these
Answer:
(B) Median

iv. When a registered dealer sells goods to another registered dealer under GST, then this trading is termed as _________. (B)
(A) BB
(B) B2B
(C) BC
(D) B2C
Answer:
(B) B2B
Hints:
i. Sum of roots = 2 – 1 = 1
Product of roots = 2 × (-1) = -2
∴ The required quadratic equation is
x² – (Sum of roots)x + Product of roots = 0
x² – x – 2 = 0

ii.

Question 1.(B)
Solve the following sub-questions. [4]
i. Write any two quadratic equations.
ii. ‘Pawan Medical’ supplies medicines. On some medicines the rate of GST is 12%, then what is the rate of CGST and SGST?
iii. Two digit numbers are formed using digits 2, 3 and 5 without repeating a digit. Write the sample space ‘S’.
iv. If D_x = 36, D_y = 45 and D = 3 are the values of the determinants for certain simultaneous equations x and y, find x.
Answer:
i. a. y² – 7y + 12 = 0
b. x² – 8 = 0

ii. Rate of GST = 12 %
∴ Rate of CGST = Rate of SGST = \(\frac{Rate of GST}{2}=\frac{12}{2}\) = 6%
∴ Rate of CGST = Rate of SGST = 6%

iii. S = {23, 25, 32, 35, 52, 53}

iv. Using Cramer’s rule, we get
x = \(\frac{D_x}{D}=\frac{36}{3}\) = 12

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
i. Given below is the distribution of turnover (In lakh X) of 50 departmental stores in a year.

By completing the following table, find mean turnover of the stores.

ii. The six faces of a die are marked.

The event M is getting a vowel on the upper face of the die when it is tossed. Complete the following activity.

iii. Complete the following table to draw the graph of 2x – 6y = 3.

(B) Solve the following sub-questions (Any four): [8]
i. In the year 2010 in the village there were 4000 people who were literate. Every year the number of literate people increases by 400. How many people will be literate in the year 2020?
ii. The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.

iii. A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is
a. an ace.
b. a spade.
iv. Find the first four terms of an A.P. when a = 8 and d = -5.
v. Solve the following simultaneous equations.
x + 7y = 10; 3x – 2y = 7
Answer:
(A) i.

(B)
i. The number of literate people increases every year by 400.
∴ The resulting sequence will be an A.P.
with a = 4000, d = 400, n = 11.
Since, t_n = a + (n – 1)d
∴ t₁₁ = 4000 + (11 – 1) 400
= 4000 + 10 × 400
= 4000 + 4000 = 8000
∴ 8000 people will be literate in the year 2020.

ii.

∴ The mean of the time spent by the students for their studies is 4.36 hours.

iii. There are 52 playing cards.
∴ n(S) = 52
a. Let A be the event that the card drawn is an ace.
∴ n(A) = 4
∴ P(A) =\(\frac{n(A)}{n(S)}=\frac{4}{52}\)
∴ P(A)= \(\frac{1}{13}\)

b. Let B be the event that the card drawn is a spade. n(B)= 13
∴ n(B) = 13
∴ P(B) =\(\frac{n(B)}{n(S)}=\frac{13}{52}\)
∴ P(B)= \(\frac{1}{4}\)

iv. a = 8, d = -5 …[Given]
t₁ = a = 8
t₂ – t₁ + d = 8 – 5 = 3
t₃ – t₂ + d = 3 – 5 = -2
t₄ – t₃ + d = -2 – 5 = -7
∴ The first four terms of the A.P. are 8, 3, -2, -7.

v. x + 7y – 10
∴ x = 10 – 7y …(i)
3x – 2y – 7 …(ii)
Substituting x – 10 – 7y in equation (ii), we get
3(10 – 7y) – 2y = 7
∴ 30 – 21y – 2y = 7
∴ -23y = 7 – 30
∴ -23y = -23
∴ y = \(\frac{-23}{-23}\) = 1
Substituting y – 1 in equation (i), we get
x = 10 – 7(1) = 10 – 7 = 3
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) = (3, 1) is the solution of the given simultaneous equations.

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
The total value (with GST) of a remote-controlled toy car is ₹ 1770. Rate of GST is 18% on toys. Find the taxable value, CGST and SGST for this toy-car by completing the following activity.
Let the amount of GST be ₹x.
Total value of remote controlled toy car = ₹ 1770
Taxable value of remote controlled toy car = ₹(1770 -x)

ii. To solve the quadratic equation √2x² + 7x + 5√2 = 0 by factorisation, complete the following activity.

(B) Solve the following sub-questions (Any two): [6]
i. Sum of first 55 terms in an A.P. is 3300, find its 28^th term.
ii. Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
iii. Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.

iv. 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 prime number.
Answer:
(A) i. Let the amount of GST be ₹x.
Total value of remote controlled toy car = ₹1770
Taxable value of remote controlled toy car = ₹(1770 – x)

ii.

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.

(B) i.

∴ 28^th term of the A.P. is 60.

ii. Let the present ages of Manish and Savita be x years and y years respectively.
According to the first condition,
x + y = 31 …(i)
3 years ago,
Manish’s age = (x – 3) years, Savita’s age = (y – 3) years
According to the second condition,
(x – 3) = 4 (y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = -12 + 3
∴ x – 4y = -9 …(ii)
Subtracting equation (ii) from (i), we get

Substituting y = 8 in equation (i), we get
x + 8 = 31
∴ x = 31 – 8
∴ x = 23
∴ The present ages of Manish and Savita are 23 years and 8 years respectively.

iii.

iv. Sample space (S) =
{10, 11, 12, 13, 14,
20, 21, 22, 23, 24,
30, 31, 32, 33, 34,
40, 41, 42, 43, 44}
∴ n(S) = 20
Let A be the event that the number so formed is a prime number.
∴ A = {11, 13, 23, 31, 41, 43}
∴ n(A) = 6
∴ P(A) \(\frac{n(A)}{n(S)}=\frac{6}{20}\)
∴ P(A) = \(\frac{3}{10}\)

Question 4.
Solve the following sub-questions (Any two): [8]
i. If the mean of the following data is 14.7, find the values of p and q.

ii. A travels a distance of 24 km with a speed of 4km/hr. B start 1 hour late than A and he covers the same distance in 4 hours. Draw the graphs of their journey and find when and where B will meet A.
iii. If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has real and equal roots, find k.
Answer:

∴ 588 = 408 + 9p + 27q
∴ 180 = 9p + 27q
∴ p + 3q = 20 …(ii)
Subtracting equation (i) from (ii), we get

∴ q = 3
Substituting q = 3 in equation (i), we get
p + 3 = 14
∴ p = 11
∴ p = 11 and q = 3

ii. Speed of A = 4km/hr
Speed of B = \(\frac{24}{4}\) = 6 km/hr.
Let A and B meet after x hours at a distance of y km.
Then y = 4x + 4 …(i)
y = 6x …(ii)


The point of intersection is (2, 12).
∴ A and B will meet 2 hrs after B starts his journey.
B will meet A at a distance of 12 km from the starting point.

iii. 2 is a root of the equation 3X² + px – 8 = 0.
∴ 3(2)² + p(2) – 8 = 0
∴ 12 + 2p – 8 = 0
∴ 2p + 4 = 0
∴ 2p = – 4
∴ p = -2
Substituting the value of p in 4x² – 2px + k = 0, we get
4×2 – 2(-2) x + k = 0
∴ 4×2 + 4x + k = 0
Comparing with ax2 + bx + c = 0, we get
a = 4, b = 4, c = k
Since, the roots of the equation are real and equal.
∴ b² – 4ac = 0
∴ 4² – 4(4)(k) = 0
∴ 16 – 16k = 0
∴ 16k = 16
∴ k = 1

Question 5.
Solve the following sub-questions (Any one): [3]
i. Construct a word problem on quadratic equation such that one of its answers is 20 (years, mpees, centimetres, etc.). Also solve it.
ii. The market value of a mutual fund is 400 crore rupees, which is divided into 8 crore units.
a. Suppose you invest ₹10,000 in the units, how many units will you get?
b. While selling the units if their market value is increased by 10%, how much amount will you get by selling them?
Answer:
i. Word problem:
The product of two consecutive natural numbers is 420. Find the numbers.
Solution:
Let the first natural number be x.
∴ The second consecutive natural number = x + 1
According to the given condition,
x (x + 1) = 420
∴ x2 + x = 420
∴ x2 + x – 420 = 0
∴ x2 + 21x – 20x – 420 = 0
∴ x(x + 21) – 20(x + 21) = 0
(x + 21)(x – 20) = 0
∴ x + 21 = 0 or x – 20 = 0
∴ x = -21 or x = 20
But, natural number cannot be negative.
∴ x = 20 and x + 1 = 20 + 1 = 21
∴ The two consecutive natural numbers are 20 and 21.

ii. The price of one unit = \(\frac{400crores}{8crores}\) = ₹50
a. No. of units by investing ₹10,000 = \(\frac{10,000}{50}\) = 200
b. If the market value is increased by 10% by selling one unit, the profit will be
50 × \(\frac{10}{100}\) = ₹5
∴ By selling 200 units, the profit will be 200 × 5 = ₹1,000

SSC Maharashtra Board Maths 1 Question Paper with Solutions