Maharashtra (MSBSHSE) SSC Board Question Paper Class 10 Maths Paper-1 - 2018

Q: 1. If the sum of zeroes of the quadratic polynomial $3x^2 – kx + 6$ is 3, then find the value of k.

Here a = 3, b = -k, c = 6Sum of the zeroes,$\alpha +\beta=\dfrac{-b}{a}= 3$ (given)$\Rightarrow \dfrac{−(−k)}{3} = 3$$\Rightarrow k = 9$

Q: 2. Two cubes each of volume $27 cm^3$ are joined end to end to form a solid. Find the surface area of the resulting cuboid.

Volume of a cube $= 27 cm^3$$\Rightarrow (side)^3=3^3$$\therefore (side)=3$Length of resulting cuboid, $l = 2 \times 3 = 6 cm$Breadth of resulting cuboid, $b = 3 cm$Height of resulting cuboid, $h = 3 cm$Surface area of resulting cuboid $= 2(lb + bh + hl)$$= 2(6 \times 3 + 3 \times 3 + 3 \times 6)$$= 2(18 + 9 + 18) = 2(45) $$= 90 cm^2$

Q: 3. Three distinct coins are tossed together. Find the probability of getting (i) at least 2 heads(ii) at most 2 heads.

Total number of possible outcomes $ = 2^3 = 8$(HHH, TTT, HHT, THH, THT, HTH, TTH, HTT)(i) Possible outcomes of at least two heads = 4(HHT, THH, HHH, HTH)$\therefore$ P(at least two heads)$ =\dfrac{ 4}{8}=\dfrac{1}{2}$(ii) Possible outcomes of at most two heads = 7(HHT, TTT, THH, THT, HTH, TTH, HTT)$\therefore$ P(at most two heads) = $ \dfrac{7}{8}$

Q: 5. If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.

HCF of 408 and 1032 is 24.$1032 \times 2 + 408 \times (p) = 24$$408p = 24 – 2064$$ p = -5$

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Maharashtra (MSBSHSE) SSC Board Question Paper Class 10 Maths Paper-1 - 2018

Question Paper for Maharashtra SSC Board Class 10 Maths Paper-1 2018 - Free PDF Download

Free PDF download of Maharashtra SSC Board Class 10 Maths Paper-1 question paper 2018 with solutions solved by expert teachers on Vedantu.com. By practising Class 10 Maths Paper-1 Maharashtra board question paper 2018 to score more marks in your examination.

Maharashtra SSC Board Class 10 Maths Paper

Question Paper for Maharashtra SSC Board Class 10 Maths Paper-1 2018 consists of Algebra part. The paper is of two hours duration and carries 40 marks. In the first question, there are six parts and students have to attempt any five. Each question is for one mark. In the second question, there are five subparts and students have to attempt only four. Each question carries 2 marks. In the third question, there are five subparts and students have to attend only 3 questions and each question carries 3 marks. In the fourth question, there are 3 questions and students have to attend 2 questions and each question carries four marks. In the fifth question, there are 3 questions and students have to attend two questions and each question carries 5 marks.

Paper 1 mostly covers the topics related to Algebra, therefore students must practice algebra questions carefully to score high marks in the exam. Students can get the questions as given below:

a) Find the next two terms of an A.P. 4, 9, 14—

Given: 4,9,14

First term= a= 4

Difference = d= 9-4 = 5

Fourth term = a + 3d= 4 + 3 (5) = 19

Fifth term = a + 4 (d) = 4 + 4 (5) = 24

Thus, the next two terms are 19 and 24

Thus, students can prepare similar types of questions for the final exam. Students can download the complete Question Paper for Maharashtra SSC Board Class 10 Maths Paper-1 2018 from Vedantu. It is available along with the solutions so that students can solve the question paper easily. Solving previous year question papers can help students to score high marks in their exams. They can understand the syllabus and pattern for the exam. The question paper can also help students to understand the syllabus for Maharashtra SSC Board Class 10 Maths.

FAQs on Maharashtra (MSBSHSE) SSC Board Question Paper Class 10 Maths Paper-1 - 2018

1. If the sum of zeroes of the quadratic polynomial $3x^2 – kx + 6$ is 3, then find the value of k.

Here a = 3, b = -k, c = 6
Sum of the zeroes,$\alpha +\beta=\dfrac{-b}{a}= 3$ (given)
$\Rightarrow \dfrac{−(−k)}{3} = 3$
$\Rightarrow k = 9$

2. Two cubes each of volume $27 cm^3$ are joined end to end to form a solid. Find the surface area of the resulting cuboid.

Volume of a cube $= 27 cm^3$

$\Rightarrow (side)^3=3^3$

$\therefore (side)=3$
Length of resulting cuboid, $l = 2 \times 3 = 6 cm$
Breadth of resulting cuboid, $b = 3 cm$
Height of resulting cuboid, $h = 3 cm$
Surface area of resulting cuboid $= 2(lb + bh + hl)$
$= 2(6 \times 3 + 3 \times 3 + 3 \times 6)$
$= 2(18 + 9 + 18) = 2(45) $

$= 90 cm^2$

3. Three distinct coins are tossed together. Find the probability of getting
(i) at least 2 heads
(ii) at most 2 heads.

Total number of possible outcomes $ = 2^3 = 8$
(HHH, TTT, HHT, THH, THT, HTH, TTH, HTT)
(i) Possible outcomes of at least two heads = 4
(HHT, THH, HHH, HTH)
$\therefore$ P(at least two heads)$ =\dfrac{ 4}{8}=\dfrac{1}{2}$
(ii) Possible outcomes of at most two heads = 7
(HHT, TTT, THH, THT, HTH, TTH, HTT)
$\therefore$ P(at most two heads) = $ \dfrac{7}{8}$

4. If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, calculate the diameter of the larger circle (in cm).

Let R be radius of larger circle and $r_1$, $r_2$ be radius of smaller circle.

So, as per question.
$\pi R^2 = \pi r_1^2 +\pi r_2^2 $

$\pi R^2 = \pi( r_1^2 + r_2^2 ) [ r_1 = \dfrac{10}{2} = 5cm, r_2 = \dfrac{24}{2} = 12 cm]$

$ R^2 =5^2 +12^2 $

$ R^2 =169$

$ R =13 cm$

So, Diameter $=2 \times R$

Diameter $=2 \times 13$

Diameter $=26 cm$

5. If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.

HCF of 408 and 1032 is 24.
$1032 \times 2 + 408 \times (p) = 24$
$408p = 24 – 2064$
$ p = -5$

6. From where we can download Maharashtra (MSBSHSE) SSC Board Question Paper Class 10 Maths Paper?

We can download Maharashtra (MSBSHSE) SSC Board Question Paper Class 10 Maths Paper from Vedantu site. Also, these are available offline in a bookstore.