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

## Maharashtra Board Class 10 Maths 1 Model Paper Set 1 with Solutions

Question 1.

(A) For every subquestion 4 alternative answers are given. Choose the correct answer and write the alphabet of it :

(i) For an A.P., if a = 7 and d = 2.5 then t_{12} = ?

(A) 37.5

(B) 34.5

(C) 28.2

(D) 44.5

Answer:

Given : a = 7 and d = 2.5

t_{12} = a + (12 – 1)d

= 7 + (12 – 1) (2.5)

= 7 + (11) (2.5)

= 7 + 27.5

= 34.5

Thus, the value of t_{12} is 34.5.

Hence, the correct option is (B).

(ii) 28% GST was charged on the scooter having cost ₹ 50,000. Find the amount of CGST charged.

(A) ₹ 8,000

(B) ₹ 7,500

(C) ₹ 14,000

(D) ₹ 7,000

Answer:

Given: Cost = ₹ 50,000

and GST = 28%

So, Rate of CGST = \(=\frac{\text { Rate of GST }}{2}\) = 14%

∴ CGST = \(\frac{14}{100}\) × 50000 = 7000

Thus, the amount of CGST is ₹ 7,000.

Hence, the correct option is (D).

(iii) If α and β are the roots of the equation is 3x^{2} + x – 10 = 0, then the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is

(A) 10

(B) –\(\frac{1}{10}\)

(C) \(\frac{1}{10}\)

(D) \(\frac{1}{3}\)

Answer:

Given equation is,

3x^{2} + x – 10 = 0

Since, α and β are the roots of given equation

∴ Sum of roots (α + β) = \(\frac{-b}{a}\) = \(\frac{-1}{3}\)

Product of roots (αβ) = \(\frac{c}{a}\) = \(\frac{-10}{3}\)

Now,

Thus, the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is \(\frac{1}{10}\)

Hence, the correct option is (C).

(iv) Two coins are tossed simultaneously. The probability of getting at the most one head is ……..

(A) \(\frac{1}{2}\)

(B) \(\frac{1}{4}\)

(C) \(\frac{3}{4}\)

(D) 1

Answer:

When two coins are tossed simultaneously

Sample Space = (HH, HT, TH, TH)

Thus,

Probability of getting at most one head = \(\frac{\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}\}}{\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}}\)

= \(\frac{3}{4}\)

Thus, the probability of getting at most one head is \(\frac{3}{4}\)

Hence, the correct option is (C).

Question 1.

(B) Solve the following sub questions.

(i) If x + y = 14 and 2x – y = 16 then x = ?

Answer:

Given: x + y = 14

and 2x – y = 16

Adding equations (i) and (ii)

Hence, the value of x is 10.

(ii) If (5\(\sqrt{2}\) + 3\(\sqrt{3}\)) – (6\(\sqrt{2}\) – 7\(\sqrt{3}\)) = a\(\sqrt{2}\) + b\(\sqrt{3}\), then find a and b.

Answer:

Comparing both sides, a = -1, b = 1o

Hence, a = -1, b = 1o.

(iii) From one footwear shop, 12 pairs of chappals were sold. The sizes of these chappals are given below.

7, 8, 6, 7, 7, 5, 9, 7, 6, 7, 8, 7

Find their mode.

Answer:

Given data is 7, 8, 6, 7, 7, 5, 9, 7, 6, 7, 8, 7.

The observatîon repeated maximum number of times = 7.

∴ Mode = 7

Hence, the mode is 7.

(iv) Mr. Dubey’s payable amount of income tax is ₹ 7000. He needs to pay education cess at 3% on income tax. Hence how much total income tax will he have to pay ?

Answer:

Given: Payable amount = ₹ 7,000

Education cess = 3% of income tax

⇒ Education cess = 3% of 7,000

= \(\frac{3}{100}\) × 7,000 = \(\frac{21,000}{100}\) = ₹ 210

Total income tax = ₹ 7,000 + 210

= ₹ 7,210

Hence, the total income tax which he will pay is ₹ 7,210.

Question 2.

(A) Complete and write any two activities from the following :

(i) In the adjoining figure, the arrow rests on any number, after the rotation of the disc. The probability that it will rest on any of the numbers on the disc is equal. Let A be any random event. To find the probability of A, fill in the boxes.

Answer:

(i)

(1) S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(2) n(S) = 12

(3) A = {1, 8} ∴ n(A) = __2__

(4) ∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = __\(\frac{2}{12}\)__ = __\(\frac{1}{6}\)__

(ii) Complete the following table.

Answer:

(iii) In the pie-diagram, the data of 720 students who opted for their favourite literature type is shown. The data is expressed fiction in the percentages. Using this diagram complete the following table.

Answer:

Question 2.

(B) Solve any four subquestions from the following.

(i) If \(\left|\begin{array}{ll}

1 & y \\

4 & x

\end{array}\right|\) = 12 and \(\left|\begin{array}{cc}

-1 & 2 \\

y & x

\end{array}\right|\) = 6, from the given determinants, from the two simultaneous equations in x and y and solve them.

Answer:

Given: \(\left|\begin{array}{ll}

1 & y \\

4 & x

\end{array}\right|\) = 12 and \(\left|\begin{array}{rr}

-1 & 2 \\

y & x

\end{array}\right|\) = 6

x – 4y = 12 ….. (i)

and -x – 2y = 6 …… (ii)

Adding equations (i) and (ii), we get

Putting the value of y in equation (i), we get

x – 4 (-3) = 12

⇒ x+ 12 = 12

⇒ x = 12 – 12

⇒ x = 0

Hence, x = 0 and y = -3.

(ii) Find the sum of all odd numbers between 351 and 373.

Answer:

Odd numbers between 351 and 373 are 353, 355, 357, 359, ……., 371.

The above series is an A.P.

So, a = 353, d = 2 and a_{n} = 371

We know, a_{n} = a + (n – 1)d

⇒ 371 = 353 + (n – 1)2

⇒ 371 – 353 = (n – 1)2

⇒ 18 = (n – 1)2

⇒ n – 1 = 9

n = 10

Now, S_{n} = \(\frac{n}{2}\)(a + a_{n})

= \(\frac{10}{2}\)(353 + 371)

= 5(724)

= 3620

Hence, the required sum is 3620.

(iii) Solve : 7x^{2} – 30x – 25 = 0

Answer:

Given equation is,

7x^{2} – 30x – 25 = 0

The roots of a quadratic equation are given as,

Hence, the roots of given quadratic equation are 5 and \(\frac{-5}{7}\).

(iv) Solve : 4m + 6n = 54, 3m + 2n = 28.

Answer:

The given system of equations is

4m + 6n = 54 …….. (i)

and 3m + 2n = 28 ……… (ii)

Multiplying equation (i) by 3 and equation (ii) by 4, we get

12m + 18n = 162 ……. (iii)

and 12m + 8n = 112 ….. (iv)

Now, subtracting equation (iv) from equation (iii), we get

⇒ n = 5

Putting the value of n in equation (i)

4m + 6(5) = 54

⇒ 4m = 54 – 30

⇒ 4m = 24

⇒ m = 6

Hence, the value of m = 6 and n = 5.

(v) Compare the quadratic equation x^{2} + 9\(\sqrt{3}\)x + 24 = 0 to ax2 + bx + c = 0 and find the value of discriminant and hence write the nature of the roots.

Answer:

Given equation is,

x^{2} + 9\(\sqrt{3}\)x + 24 = 0

Comparing the given equation with ax^{2} + bx + c = 0, we get

a = 1, b = 9\(\sqrt{3}\), c = 24

Discriminant, D = b^{2} – 4ac

= (9\(\sqrt{3}\))^{2} – 4(1)(24)

= 243 – 96 = 147

Here, D > 0.

Hence, the roots of quadratic equation are real and unequal.

Question 3.

(A) Complete and write any one activity from the following :

(i) Determine whether (x – 3) is a factor of polynomial x^{3} – 19x + 30.

Let P(x) = x^{3} – 19x + 30

Answer:

Let P(x) = x^{3} – 19x + 30

By remainder theorem, x – 3 will be a factor of P(x), if P(3) = 0

Now,

Hence, __x – 3__ is a factor of polynomial x^{3} – 19x + 30.

(ii) First term and common difference of an A.P. are 1 and 2 respectively; find S_{10}

Given : a = 1, d = 2

Answer:

Given: a = 1, d = 2

find S_{10} = ?

Question 3.

(B) Attempt any two subquestions from the following :

(i) Using the digits 0, 2, 3, 5 the two-digit numbers are constructed without repetition of digits. Find the probability of the following events :

(i) Condition for event A : The number formed is even.

(ii) Condition for event B : The number formed is prime number.

Answer:

Total number of two-digit numbers = 4 × 3 = 12

(i) Let A be the event that the number formed is even.

Even two-digit numbes = {32, 52, 20, 30, 50}

⇒ n(A) = 5

Thus, P (A) \(=\frac{\text { Number of even digits numbers }}{\text { Total number of two-digit numbers }}\)

= \(\frac{5}{12}\)

(ii) Let B be the event that the number formed is prime.

Prime two-digit numbers = {23, 53}

⇒ n(B) = 2

Thus, P(B) \(=\frac{\text { Number of prime digits numbers }}{\text { Total number of two-digit numbers }}\)

= \(\frac{2}{12}\) = \(\frac{1}{6}\)

(ii) Solve the given simultaneous equations using graphical method.

3x + 2y = 6, 5x + y = 10

Answer:

The given equations are:

3x + 2y = 6 …… (i)

and 5x + y = 10 …… (ii)

For equation (i) : 3x + 2y = 6

For equation (ii) : 5x + y = 10

The graphical representation of the two lines is shown below:

The intersection point of two equations is (2, 0).

Hence, x = 2 and y = 0

(iii) A businessmen supplied CCTV camera sets for police control room, worth of ₹ 1,77,000. The rate of GST is 18%. Then find the amount of SGST and CGST. Also find the taxable price of CCTV sets.

Answer:

Given : Price of CCTV camera set = ₹ 1,77,000

Rate of GST = 18%

The taxable value is rate excluding GST.

Let the rate excluding GST = x

Then, ₹ 1,77,000 = Rate excluding GST + 18% of rate excluding GST

⇒ ₹ 1,77,000 = x + 18% × x

₹ 1,77,000 = x + \(\frac{18}{100}\) × x = \(\frac{100 x+18 x}{100}\) = \(\frac{118 x}{100}\)

⇒ ₹ 1,77,000 = 1.18x

⇒ x = 1,50,000

Thus, the taxable value is ₹ 1,50,000.

So, GST = ₹ 1,77,000 – ₹ 1,50,000

= ₹ 27,000

So, CGST = SGST = \(=\frac{\text { GST }}{2}\)

= \(\frac{27,000}{2}\) = ₹ 13,500

Hence, the taxable value is ₹ 1,50,000 and CGST and SGST is ₹ 13,500.

(iv) For arithmetic progression, first term is – 8 and last term is 55. If sum of all these terms is 235, find the number of terms and common difference.

Answer:

Let a and d the first term and the common difference respectively.

So, a = -8, a_{n} = 55 , and S_{n} = 235

We know, S_{n} = \(\frac{n}{2}\)(a + a_{n})

⇒ 235 = \(\frac{n}{2}\)(-8 + 55)

⇒ 470 = n(47)

⇒ n = 10

Now, a_{n} = a + (n – 1)d

⇒ 55 = -8 + (10 – 1) d

⇒ 55 + 8 = 9d

⇒ 63 = 9d

⇒ d = 7

Hence, the number of terms is 10 and common difference is 7.

Question 4.

Attempt any two subquestions from the following :

(i) The difference between two numbers is 2 and their product is 1443. Find the numbers.

(ii) An electric company producing electric bulb, has packed 100 bulbs in each box. Some bulbs from 16 such boxes are tested defective. The information of number of defective bulbs in 16 boxes is given below.

No. of defective bulbs | No. of boxes |

0-2 | 3 |

2-4 | 4 |

4-6 | 5 |

6-8 | 3 |

8-10 | 1 |

(i) How many boxes contains maximum number of defective bulbs ?

Answer:

Let one of the number be x.

Thus, second number will be x + 2.

According to the question,

(x) (x + 2) = 1443

⇒ x^{2} + 2x = 1443

⇒ x^{2} + 2x – 1443 = 0

⇒ x^{2} + (39 – 37)x – 1443 = 0

⇒ x^{2} + 39x – 37x – 1443 = 0

⇒ x(x + 39) – 37(x + 39) = 0

⇒ (x – 37) (x + 39) = 0

⇒ (x – 37) = 0 or (x + 39) = 0

⇒ x = 37 or x = – 39

If x = 37,

Then, x + 2 = 37+ 2 = 39

So, two numbers are 37 and 39.

If x = – 39

Then, x + 2= -39 + 2 = -37

So, two numbers are – 39, – 37 .

Hence, the numbers are 37, 39 or – 39, – 37.

(ii) Find the mean of defective bulbs.

Answer:

(1) 1 box contain maximum number of defective bulbs.

(2)

So, the mean of the defective bulbs is 4.375.

(iii) The box is selected at random. What is the probability that it will contain average 2 to 4 defective bulbs ?

Answer:

Total boxes = 16

Boxes with average of 2 to 4 defective bulb = 4

Thus, P(E) = \(\frac{4}{16}\) = \(\frac{1}{4}\)

(iv) Find the probability of getting highest number of defective bulbs,

Answer:

Total boxes = 16

Boxes with highest number of defective bulbs = 1.

Thus, P(E) = \(\frac{1}{16}\)

(iii) Solve the following simultaneous equations.

\(\frac{2}{x+y}\) – \(\frac{3}{x-y}\) = 15; \(\frac{8}{x+y}\) + \(\frac{5}{x-y}\) = 77

Answer:

\(\frac{2}{x+y}\) – \(\frac{3}{x-y}\) = 15 ……..(i)

\(\) + \(\frac{5}{x-y}\) = 77 ……. (ii)

Replacing \(\left(\frac{1}{x+y}\right)\) by a and \(\frac{1}{x-y}\) by b in eq.(i) and eq.(ii), we get,

2a – 3b = 15 …….. (iii)

8 a + 5b = 77 ……. (iv)

Multiplying eq. (iii) by 5 and eq. (iv) by 3

10a – 15b = 75 …….. (v)

24a + 15b = 231 ……. (vi)

Now adding eq. (v) and (vi), we get

⇒ 10a + 24a = 75 + 231

⇒ 34a = 306

∴ a = \(\frac{306}{34}\)

∴ a = 9

Substituting, a = 9 in eq. (iii), we get

⇒ 2(9) – 3b = 15

⇒ 3b = 18 – 15

∴ 3b = 3

∴ b = 1

Resubstituting the values of a and b, we get,

\(\frac{1}{x+y}\) = 9 and \(\frac{1}{x-y}\) = 1

∴ x + y = 9 ……. (vii)

and x – y = 1 ……. (viii)

Adding eq. (vii) and (viii), we get

∴ 2x = 10

x = 5

Substituting, x = 5 in eq. (vii), we get

5 + y = 9

y = 4

Hence, the solution of the given eq. is x = 5, y = 4.

Question 5.

Attempt any one subquestion from the following:

(i) The following determinants are obtained from the simultaneous equations in variable x and y.

If D_{x} = \(\left|\begin{array}{cc}

24 & a \\

16 & -1

\end{array}\right|\)‚ D_{y} = \(\left|\begin{array}{ll}

5 & 24 \\

b & 16

\end{array}\right|\) D = \(\left|\begin{array}{cc}

5 & 1 \\

3 & -1

\end{array}\right|\) the solution for this equations are x = 5 and

y = -1. Find the value of ‘a’ and ‘b’. Also form the òriginal simultaneous equations having this solution.

Answer:

By Cramer’s rule,

x = \(\frac{D_x}{D}\)

⇒ 5 = \(\frac{-24-16 a}{-8}\)

⇒ 5 = 3 + 2a

⇒ 5 – 3 = 2a

⇒ a = 1

Also, y = \(\frac{\mathrm{D}_y}{\mathrm{D}}\)

⇒ -1 = \(\frac{80-24 b}{-8}\)

⇒ -1 = -10 + 3b

⇒ -1 + 10 = 3b

⇒ b = 3

So, a = 1 and b = 3.

So, D_{x} = \(\left|\begin{array}{cc}

24 & 1 \\

16 & -1

\end{array}\right|\) and D_{y} = \(\left|\begin{array}{ll}

5 & 24 \\

3 & 16

\end{array}\right|\)

Hence, the original simultaneous equations having this solution are 5x + y = 24 and 3x – y = 16

(ii) An analysis of particular information is given in the following table.

Age Group | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequency | 2 | 5 | 6 | 5 | 2 |

For this data, mode = median = 25. Calculate the mean. Observing the given frequency distribution and values of the central tendency interpret your observation.

Answer:

Given : Mode = Median = 25

We know, 3 Median = Mode + 2 Mean

⇒ 3(25) = 25 + 2 Mean

⇒ 75 – 25 = 2 Mean

⇒ Mean = 25

Now,

Thus, the mode of given data is 25.

Now, maximum frequency = 6.

Class corresponds to maximum frequency = 20 – 30

∴ Modal class = 20 – 30.

So, L = Lower limit of the modal class = 20

h = Class interval of the modal class = 10

f_{1} = Frequency of modal class = 6

f_{0} = Frequency of class preceeding modal class = 5

f_{2} = Frequency of class succeeding modal class = 5

Now, Mode = L + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h

= 20 + \(\left(\frac{6-5}{2 \times 6-5-5}\right)\) × 10

= 20 + \(\frac{1}{2}\) × 10

= 20 + 5 = 25

Thus, the mode of given data is 25.

From the table, N = 20

∴ \(\frac{\mathrm{N}}{2}\) = \(\frac{20}{2}\) = 10

Cumulative frequency just greater than 10 is 13, which belongs to class interval 20 – 30.

∴ Median class = 20 – 30

So, L = Lower limit of median class = 20

h = Class interval of the median class = 10

f = Frequency of median class = 6

c.f. = Cumulative frequency of class preceeding median class = 7

Thus, the median of given data is 25.

Hence, the median, mode and mean of the data are 25, 25 and 25 respectively.