Maharashtra Board Class 10 Maths 1 Sample Paper Set 4 with Answers

Maharashtra Board SSC Class 10 Maths 1 Sample Paper Set 4 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 1 Model Paper Set 4 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. First four terms of an A.P. are whose first term is -2 and common difference is -2. (C)
(A) -2, 0,2,4
(B) -2, 4,-8, 16
(C) -2,-4,-6,-8
(D) -2,-4,-8,-16
Answer:
(C) -2,-4,-6,-8

ii. The rate of GST on stainless steel utensils is 18%, then the rate of state GST is _________. (B)
(A) 18%
(B) 9%
(C) 36%
(D) 0.9%
Answer:
(B) 9%

iii. What is the probability of the event that a number chosen from 1 to 10 is a prime number? (C)

Answer:
C

iv. Which of the following will give the solution of simultaneous equations by Cramer’s rule? (D)

Answer:
D

Hints:
i. First term (a) = -2, common difference (d) = -2
The first four terms are
-2,
-2 – 2 = -4,
-4 – 2 = -6,
-6 – 2 = -8

iii. n(S)= 10
Let A be the event that the number chosen is a prime number.
∴ A = {2, 3, 5, 7}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{10}=\frac{2}{5}\)

Question 1.
(B) Solve the following sub-questions.
i. The sale of salesman in a week is given below in pie diagram. If the total sale is ₹72000, then find the sale of salesman B.

ii. Write the following quadratic equation in standard form ax² + bx + c = 0 :
x² + 5x = – (3 – x).
iii. Three coins are tossed simultaneously. Write the sample space S.
iv. Find the value of \(\left|\begin{array}{cc} -1 & 7 \\ 2 & 4 \end{array}\right|\)
Answer:
i.

ii. x² + 5x = -(3 – x)
∴ x² + 5x = -3 + x
∴ x² + 5x – x + 3 = 0
∴ x² + 4x + 3 = 0

iii. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
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 3 coins are tossed. So the number of elements in the sample space is 2³ = 8

iv.

Question 2.
(A) Complete the following activities and rewrite it (Any Two): [4]
i. The monthly expenditure of a family on different items is given in the following table. Calculate the related central angles.

ii. Complete the following table according to the given information.

FV	Share is at	MV
1. ₹ 10	Premium is  ₹ 7
2. ₹25		₹16
3.	AT PAR	₹5
4. ₹20		₹30

iii. Complete the following activity to find the number of natural numbers between 1 and 171, which are divisible by 5:

(B) Solve the following sub-questions (Any four):
i. Solve the following quadratic equation by factorisation. 5m² = 22m +15
ii. 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?
iii. Which term of the following A.P. is 560?
2,11, 20, 29, …
iv. Form a quadratic equation if its roots are 3 and -10.
v. Find the mode from the following information:
L = 10, h = 2, f₀ = 58, f₁ = 70, f₂ = 42.
Answer:
(A) i.

ii.

FV	Share is at	MV
1. ₹10	Premium is  ₹7	₹17
2. ₹25	Discount is ₹9	₹16
3. ₹5	AT PAR	₹5
4. ₹20	Premium is₹10	₹30

iii.

(B) i. 5m² = 22m + 15
∴ 5m² – 22m – 15 = 0
∴ 5m² – 25m + 3m – 15 = 0
∴ 5m(m – 5) + 3(m – 5) = 0
∴ (m – 5) (5m + 3) = 0
∴ m – 5 = 0 or 5m + 3 = 0
∴ m = 5 or 5m = -3
∴ m = 5 or m = \(\frac{-3}{5}\)
∴ The roots of the given quadratic equation are 5 and –\(\frac{3}{5}\).

ii. Output tax = 5% of 90000 = \(\frac{5}{100}\) × 90000 = ₹4500
Input tax = 5% of 85000 = \(\frac{5}{100}\) × 85000 = ₹4250
GST payable = Output tax – ITC
= 4500 – 4250
∴ GST payable = ₹250
∴ ITC of Smt. Malhotra is ₹4250 and amount of GST payable by her is ₹250.

iii. The given A.P. is 2, 11, 20, 29, …
Here, a = 2, d = 11 – 2 = 9
Let the n^th term of the given A.P. be 560.
Then, t_n = 560
Since, t_n = a + (n – 1) d
∴ 560 = 2 + (n – 1) 9
∴ 560 = 2 + 9n – 9
∴ 560 = 9n – 7
∴ 9n = 567
∴ n = \(\frac{567}{9}\) = 63
SMART TIP
To check our answer, verify if 63^rd term is 560 or not.
t_n = a + (n – 1)d
= 2 + (63 – 1) × 9
= 2 + 62 × 9
= 2 + 558
= 560
Hence, our answer is correct.
∴ 63^rd term of the given A.P. is 560.

iv. Let α = 3 and β = -10
∴ α + β = 3 – 10 = -7
and α × β = 3 × -10 = -30
∴ The required quadratic equation is
∴ x² – (α + β) x + αβ = 0
∴ x² – (-7) x + (-30) = 0
∴ x² + 7x – 30 = 0

v. Given L = 10, h = 2, f₀ = 58, f₁ = 70, f₂ = 42

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
i. In the given figure, □ ABCD is a trapezium, AB || CD and its area is 33 cm². From the information given in the figure, find the length of side AB. Fill in the empty boxes to get the solution. □ABCD is a trapezium. AB || CD

Area of trapezium = \(\frac{1}{2}\) × (Sum of parallel sides) x Height

But, length is never negative.

ii. Complete the following table.

(B) Solve the following sub-questions (Any two): [6]
i. Solve the following simultaneous equations graphically.
x + y = 0; 2x – y = 9
ii. If two dice fire rolled simultaneously, find the probability of the following events.
a. The sum of the digits on the upper faces is 33.
b. The digit on the second die is greater than the digit on first die.
iii. A dealer supplied Walky-Talky set of ₹84,000 (with GST) to police control room. Rate of GST is 12%. Find the amount of state and central GST charged by the dealer. Also find the taxable value of the set.
iv. In an A.P. 19^th term is 52 and 38^th term is 128, find sum of first 56 terms.
Answer:
i.

ii.

(B) i. The given simultaneous equations are

The two lines intersect at point (3, -3).
∴ x = 3 and y = -3 is the solution of the simultaneous equations x + y = 0 and 2x – y = 9.
SMART TIP
To check the answer, Substitute our answer (x, y) in the given equations. If L.H.S. = R.H.S. (for both the equations), then our answer is correct.
For x + y = 0, L.H.S. = 3 + (-3) = 0 = R.H.S.
For 2x – y = 9, L.H.S. = 2(3) – (-3) = 6 + 3 = 9 = R.H.S. Hence, our answer is correct.

ii. Sample space,
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36

a. Let B be the event that the sum of the digits on the upper faces is 33.
The sum of the digits on the upper faces can be maximum 12.
∴ Event B is an impossible event.
∴ B = { }
∴ n(B) = 0
∴ P(B)= \(\frac{n(B)}{n(S)}=\frac{0}{36}\)
∴ P(B) = 0

b. Let C be the event that the digit on the second die is greater than the digit on the first die.
∴ C = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3),
(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6),
(4, 5), (4, 6), (5, 6)}
∴ n(C)= 15
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{15}{36}\)
∴ P(C) = \(\frac{5}{12}\)

iii. Let the amount of GST be ₹x.
Price of walky talky with GST = ₹84,000
Taxable value of walky talky = ₹(84,000 – x)
Now, GST = 12% of taxable value
∴ x = \(\frac{5}{12}\)
∴ 100x = (84,000 – x) × 12
∴ 100x = 84000 × 12 – 12x
∴ 100x + 12x = 84000 × 12
∴ 112x = 84000 × 12 84000 x 12
∴ x = \(\frac{84000\times12}{112}\) = 750 × 12 = ₹9000
∴ GST = ₹9000
∴ Taxable value of walky talky = ₹(84,000 – x)
= ₹(84,000 – 9000) = ₹75,000
Now, CGST = SGHT = \(\frac{GST}{2}=\frac{9000}{2}\) = ₹4,500
∴ Amount of state and central GST charged by the dealer is ₹4,500 each.
Taxable value of the set is ₹75,000.

iv. t₁₉ = 52, t₃₈ = 128 ….[Given]
Since, t_n = a + (n – 1)d
∴ t₁₉ = a + (19 – 1)d
∴ 52 = a + 18d
i.e. a + 18d = 52
Also, t₃₈ = a + (38 – 1)d
∴ 128 = a + 37d
i.e. a + 37d = 128 ….(ii)
Adding equations (i) and (ii), we get

∴ The sum of first 56 terms is 5040.

Question 4.
Solve the following sub-questions (Any two): [8]
i. The following table gives the result of certain examination for 180 students, a. Find the value of x. b. Draw histogram.

ii. The radius of a circle is greater than the radius of other circle by 3 m. The sum of their areas is 89π m². Find the radius of each circle.
iii. Out of the 900 km distance, Vasantrao travelled some distance by bus at the speed of 60 km/hr and the remaining distance by railway at the speed of 90 km/hr. Totally he travelled for 13 hrs. Find the distance travelled by bus. (Use two variables)
Answer:
a. Total number of students = 180
10 + x + 25 + 2x + 55 + 30 = 180
∴ 3x + 120 = 180
∴ 3x = 60
∴ x = 20

b.

ii. Let the radius of the first circle be x m.
∴ the radius of the second circle = (x + 3) m
According to the given condition,
πx² + π(x + 3)² = 89π
∴ π[x² + π(x + 3)²] = 89π
∴ x² + (x + 3)² = 89
∴ x² + x² + 6x + 9 = 89
∴ 2x² + 6x + 9 – 89 = 0
∴ 2x² + 6x – 80 = 0
∴ 2(x² + 3x – 40) = 0
∴ x² + 3x – 40 = 0
∴ x² + 8x – 5x – 40 = 0
∴ x(x + 8) – 5(x + 8) = 0
∴ (x + 8) (x – 5) = 0
∴ x + 8 = 0 or x – 5 = 0
∴ x = -8 or x = 5
But, x = -8 is not possible because radius cannot be negative.
∴ x = 5
Radius of the other circle = (x + 3) = 5 + 3 = 8 m
∴ Radius of first circle = 5 m,
Radius of second circle = 8 m

iii. Let Vasantrao travel for x hours by bus at the speed of 60 km/hr and y hours by railway at the speed of 90 km/hr.
Distance = speed × time
Distance travelled by bus = 60x
Distance travelled by railway = 90y
According to the first condition,
60x + 90y = 900
∴ 2x + 3y = 30     …(i)[Dividing both sides by 30]
According to the second condition
x + y = 13 …(ii)
Multiplying equation (ii) by 3, we get
3x + 3y = 39 ….(iii)
Subtracting equation (i) from (iii), we get

∴ Distance travelled by bus = 60x = 60 * 9
= 540 km

Question 5.
Solve the following sub-questions (Any one): [3]
i. Solve the following simultaneous equations:
ax + by = a – b,
bx = ay + a + b
where ‘a’ and ‘b’ are non – zero constants.
ii. An analysis of particular information is given in the following table.

Age group	Frequency
0-10	3
10-20	5
20-30	8
30-40	5
40-50	3

For this data, mode = median = 25. Calculate the mean. Observing the given frequency distribution and values of the central tendency interpret your observation.
Answer:
i. The given simultaneous equations are
ax + by = a – b …(i)
and bx = ay + a + b
∴ bx – ay = a + b …(ii)
Equations (i) and (ii) are in ax + by = c form.

∴ x = 1, y = -1 is the solution of the given simultaneous equations.

ii.

Since, the frequency distribution is symmetric about the central class (20 – 30 years), the mean, median and mode are equal to 25.

SSC Maharashtra Board Maths 1 Question Paper with Solutions