Maharashtra Board Class 10 Maths 1 Sample Paper Set 5 with Answers

Maharashtra Board SSC Class 10 Maths 1 Sample Paper Set 5 with Answers Solutions Pdf Download.

Maharashtra Board Class 10 Maths 1 Model Paper Set 5 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. The persons of O – blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O – blood group? (D)
(A) 114°
(B) 140°
(C) 104°
(D) 144°
Answer:
(D) 144°

ii. A die is rolled. What is the probability that the number appearing on upper face is less than 3? (B)
(A) \(\frac{1}{6}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{2}\)
(D) 0
Answer:
(B) \(\frac{1}{3}\)

iii. For √3x² – 2√2x- 2√3 = 0, the value of the discriminant is (C)
(A) 24
(B) -24
(C) 32
(D) – 32
Answer:
(C) 32

iv. ax + by = c and mx + nv = d and an ≠ bm, then these simultaneous equations have (A)
(A) only one common solution.
(B) no solution.
(C) infinite number of solutions.
(D) only two solutions.
Answer:
(A) only one common solution.
Hints:
i. Measure of the central angle = \(\frac{40}{100}\) × 360°
= 144°

iii. √3x² – 2√2x – 2√3 = 0
Comparing the above equation with ax² + bx + c = 0, we get
a = √3, b = -2√2, c = -2√3
∴ ∆ = b² – 4ac
= (-2√2)² – 4(√3)(-2√3)
= 8 + 24
∴ ∆ = 32

Question 1.
(B) Solve the following sub-questions. [4]
If n (S) = 36, P (A) = \(\frac{5}{12}\), find n (A).
ii. Find the common difference of the A. P. 3, 5, 7, ………..
iii. If D_x = -21, = 14 and D_y = -7 are the values of the determinants for certain simultaneous equations in x and y, find y.
iv. Find the number of shares received when ₹60,000 was invested in the shares of FV ₹100 and MV ₹120.
Answer:
i.

ii. The gIven A.P. Is 3, 5, 7, …
∴ t₁ = 3, t₂ = 5
d = t₂ – t₁ = 5 – 3
∴ d = 2

iii. By Cramer’s rule, we get
y = \(\frac{D_y}{D}=\frac{14}{-7}\) = -2

iv. Here, FV = ₹100, MV = ₹120,
Sum Invested = ₹60,000

∴ Number of shares received were 500.

Question 2.
(A) Complete the following activities and rewrite it (Any two): [4]
i. Fill up the boxes and find out the number of terms in the A.P. 2,4,6, … ,148.

ii. Complete the following activity to form a quadratic equation.
Activity:

iii.

Observe the above frequency polygon and fill in the following boxes,
a. The class [__] has the maximum number of students.
b. The classes 20 – 30 and [__] have frequency zero.
c. The class mark of the class having 50 students is [__]
d. There are [__] students in the class 80 – 90.
(B) Solve the following sub-questions (Any four). [8]
i. The taxable value of a wrist watch belt is ₹586. Rate of GST is 18%. Then what is price of the belt for the customer?
ii. The annual investments of a family are shown in the given pie diagram. Answer the following questions based on it.
a. If the investment in shares is ₹2000, find the total investment.
b. How much amount is deposited in bank?

iii. Find the A.P. whose first term is -19 and common difference is -4.
iv. Smt. Deshpande purchased shares of FV ₹5 at a premium of? 20. How many shares will she get for ₹20,000?
v. Obtain a quadratic equation whose roots are -3 and -7.
Answer:
(A) i. a = 2, d = 4 – 2 = 1, t_n = 148
t_n = a + (n – 1)d
∴ 148 = 2 + (n-1)2
∴ 148 = 2 + 2n – 2
∴ 146 = 2n – 2
∴ 2n = 148
∴ n = \(\frac{148}{2}\) = 74

ii.

iii. a. The class [60 – 70] has the maximum number of students.
b. The classes 20 – 30 and [90 – 100] have frequency zero.
c. The class mark of the class having 50 students is [55].
d. There are [15] students in the class 80 – 90.

(B) i. Taxable value of wrist watch belt = ₹586
Rate of GST = 18%
∴ GST = 18% of taxable value = \(\frac{18}{100}\) × 586
∴ GST = ₹105.48
∴ Amount paid by customer = Taxable value of wrist watch belt + GST
= 586 + 105.48 = ₹691.48
∴ The price of the belt for the customer is ₹691.48.

ii.

∴ The total investment is ₹12000.

b. central angle for deposit in bank (θ) = 90°

∴ The amount deposited in bank is ₹3000.

iii. a = -19, d = -4 …[Given]
∴ t₁ = a = -19
t₂ = t₁ + d = -19 – 4 = -23
t₃ = t₂ + d = -23 – 4 = -27
t₄ = 13 + d = -27 – 4 = -31
∴ The required A.P. is -19, -23, -27, -31, …

iv. Here, FV= ₹5, Premium = ₹20,
Sum invested = ₹20,000
∴ MV = FV + Premium
= 5 + 20
∴ MV = ₹25
Now, sum invested = Number of shares × MV

∴ Smt. Deshpande will get 800 shares for ₹20,000.

v. Let α = -3 and β = -7
∴ α + β = – 3 – 7= -10
and αβ = (-3) (-7) = 21
∴ The required quadratic equation is
x² – (α + β) x + αβ = 0
∴ x² – (-10)x + 21 = 0
∴ x² + 10x + 21 = 0
SMART TIP
In order to find out if our answer is correct or not, substitute the values of α and β in the equation that we have obtained.
If L.H.S. = R.H.S., then the answer is correct.

Question 3.
(A) Complete the following activity and rewrite it (Any one): [3]
i. To solve the simultaneous equations by determinant method, fill in the blanks.
y + 2x-19 = 0;2x-3y + 3 = 0
Write the given equations in the form
ax + by = c.
2x + y = 19
2x – 3y = -3

ii. Shri. Aditya Sanghavi invested ₹50,118 in shares of FV ₹100, when the market value is ₹50. Rate of brokerage is 0.2% and Rate of GST on brokerage is 18%, then how many shares were purchased for ₹50,118? Complete the following activity.
Here, FV = ₹100, MV = ₹50
Purchase value of shares = [__]
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = [__] of MV
∴ Brokerage = [__]
GST = 18% of brokerage = [__]
Purchase value of a share = MV + Brokerage + GST

(B) Solve the following sub-questions (Any two): [6]
i. A man borrows ₹8000 and agrees to repay with a total interest of ₹1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹40. Find the amount of the first instalment.
ii. There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
a. a natural number.
b. a number less than 1.
c. a whole number.
iii. The following table shows the classification of percentages of marks of students and the number of students. Draw a frequency polygon from the table.

iv. Find m, if (m – 12) x² + 2(m – 12) x + 2 = 0 has real and equal roots.
Answer:
(A)
i. Write the given equations in the form ax + by = c.
2x + y = 19
2x – 3y = -3

∴ By Cramer’s rule, we get

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. Here, FV = ₹100, MV = ₹50
Purchase value of shares = [₹50118]
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = [0.2%] of MV
= \(\frac{0.2}{100}\) × 50
∴ Brokerage = ₹0.1

(B) i. The instalments are in A.P.
Amount repaid in 12 instalments (S₁₂) = Amount borrowed + total interest
= 8000 + 1360
∴ S₁₂ = 9360
Number of instalments (n) = 12
Each instalment is less than the preceding one by ₹40.
∴ d = -40

∴ Amount of the first instalment is ₹1000.

ii. Sample space (S) = {0, 1, 2, 3, 4, 5}
∴ n(S) = 6
a. Let A be the event that the card drawn shows a natural number.

b. Let B be the event that the card drawn shows a number less than 1.

c. Let C be the event that the card drawn shows a whole number. C = {0, 1, 2, 3, 4, 5} n(C) = 6

iii.

iv. (m – 12)x² + 2(m – 12)x + 2 = 0
Comparing the above equation with
ax² + bx + c = 0, we get
a = m – 12, b = 2(m – 12), c = 2
∆ = b² – 4ac
= [2(m -12)]² – 4 × (m – 12) × 2
= 4(m – 12)² – 8(m – 12)
= 4(m – 12) (m – 12 – 2)
∴ ∆ = 4(m – 12) (m – 14)
Since, the roots are real and equal,
∴ ∆ = 0
∴ 4(m – 12) (m – 14) = 0
∴ m – 12 = 0 or m – 14 = 0
∴ m = 12 or m = 14
But, if m = 12, then quadratic coefficient becomes zero.
∴ m ≠ 12
∴ m = 14
SMART TIP
To verify our answer, substitute m = 14 in the given equation. If the given condition holds, then our answer is correct.
For m = 14, given equation is
2x² + 4x + 2 = 0
i.e., x² + 2x + 1 = 0
i.e., (x = 1)² = 0
∴ x = -1, -1
Hence, our answer is correct.

Question 4.
Solve the following sub-questions (Any two): [8]
i. Sum of areas of two squares is 244 cm2 and the difference between their perimeter is 8 cm. Find the ratio of their diagonals.
ii. Draw the graphs representing the equations 4x + 3y = 24 and 3y = 4x + 24 on the same graph paper. Find the area of the triangle formed by these lines and the X-axis.
iii. The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.

Marks	Frequency
20-30	p
30-40	15
40-50	25
50-60	20
60-70	q
70-80	8
80-90	10

Answer:
i. Difference between the perimeters of two squares is 8 cm.
∴ Difference between their sides is 2 cm.
Let the side of the smaller square be x cm.
∴ The side of the bigger square is (x + 2) cm.
Area of square = (side)2
∴ Area of smaller square = x² cm²
Area of bigger square = (x + 2)² cm2
According to the given condition,
x² + (x + 2)² = 244
∴ x² + x² + 4x + 4 = 244
∴ 2x² + 4x – 240 = 0
∴ x² + 2x – 120 = 0
∴ x² + 12x – 10x – 120 = 0
∴ x(x + 12) – 10(x + 12) = 0
∴ (x + 12) (x -10) = 0
∴ x + 12 = 0 or x – 10 = 0
∴ x = -12 or x = 10
But, x ≠ -12 as the side of a square cannot be negative.
∴ x = 10
∴ Side of smaller square is 10 cm and side of bigger square is 12 cm.
As diagonal of a square = √2 × side
∴ Length of the diagonal of smaller square is 10√2cm and length of the diagonal of bigger square is 12√2 cm.
Ratio of the diagonals = 10√2 : 12√2 = 5 : 6
∴ Ratio of their diagonals is 5 : 6.

ii.


∴ The point of intersection of given lines is (0, 8).
From the graph, we get ∆ABC, where BO is the height of the triangle and AC is the base.
Now, /(AC) = 12 cm and /(BO) = 8 cm
∴ Area of ∆ABC = \(\frac{1}{2}\) × /(AC) × /(BO) = \(\frac{1}{2}\) × 12 × 8
∴ Area of ∆ABC = 48 cm²

iii.

Class Marks	Frequency fi	Cumulative frequency (less than)
20 – 30	p	p
30 – 40	15	15 + p
40 – 50	25	40 + p → cf
50 – 60	20 → f	60 + p
60 – 70	q	60 + p + q
70 – 80	8	68 + p + q
80 – 90	10	78 + p + q
Total	78 + p + q

N = 90 …[Given]
Since, 50 (median) lies in the class 50 – 60.
∴ The median class is 50 – 60.
Now, L = 50, cf = 40 + p, f = 20, h = 10

∴ O = \(\frac{5-p}{2}\)
∴ 0 = 5 – p
∴ p = 5
N = 90
∴ 78 + p + q = 90 ….[Given]
∴ p + q = l2 …(i)
Substituting p = 5 in equation (i), we get
5 + q = 12
∴ q = 7
∴ p = 5 q = 7

Question 5.
Solve the following sub-questions (Any one): [3]
i. A student made a cube shaped die from a card sheet. Instead of writing numbers 1, 2, 3, 4, 5, 6 on its faces, he wrote letters a, b, c, d, e, f; one on each face, randomly. If he rolls the die twice, find the probability that he gets a vowel on the upper face both times.
ii. The following determinants are obtained from the simultaneous equations in variables x and y.

The solutions of the equations are x = -1 andy = -4. Find the values of a and b. Also find the original simultaneous equations having this solution.
Answer:
i. A student makes a die and writes a, b, c, d, e and f on its faces. He rolls the die twice.
Sample space,
S = {(a, a), (a, b), (a, c), (a, d), (a, e), (a, f),
(b, a), (b, b), (b, c), (b, d), (b, e), (b, f),
(c, a), (c, b), (c, c), (c, d), (c, e), (c, f),
(d, a), (d, b), (d, c), (d, d), (d, e), (d, f),
(e, a), (e, b), (e, c), (e, d), (e, e), (e, f),
(f, a), (f, b), (f, c), (f, d), (f, e), (f, f)}
∴ n(S) = 36
Let A be the event that the student gets a vowel on the upper faces both times.
∴ A = {(a, a), (a, e), (e, a), (e, e)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

ii.

a₁ = 3, b₁ = 2, c₁ = -11 and
a₂ = 7, b₂ = -4 , c₂ = 9
∴ The required equations are
3x + 2y = -11 and 7x – 4y = 9
SMART TIP
To verify our answer, substitute x = -1 and y = -4 in the equation that we have obtained. If L.H.S. = R.H.S., then the answer is correct.
For 3x + 2y = -11
L.H.S. = 3(-1) + 2(-4) = -3 – 8 = -11 = R.H.S.
For 7x – 4y = 9,
L.H.S. = 7(-1) – 4(4) = -7 + 16 = 9 = R.H.S.
Hence, our answer is correct.

SSC Maharashtra Board Maths 1 Question Paper with Solutions