Maharashtra Board SSC Class 10 Maths 1 Sample Paper Set 2 with Solutions Answers Pdf Download.
Maharashtra Board Class 10 Maths 1 Model Paper Set 2 with Solutions
Question 1.
For every subquestion 4 alternative answers are given. Choose the correct answer and write the alphabet of it :
(i) If the sum of the roots of the quadratic equation x2 + kx + 6 = 0 is 6, then the value of k is
(A) -12
(B) 6
(C) 12
(D) – 6
Answer:
(D) – 6
Given quadratic equation is,
x2 + kx + 6 = 0
We know,
Sum of roots (α + β) = –\(\frac{b}{a}\)
6 = –\(\frac{k}{1}\)
-k = 6
k = -6
Hence, the correct option is (D).
(ii) Which of the following sequence form an A.P. ?
(A) 3, 6, 9, 15, ……
(B) 1, 8, 15, 22, 30, ….
(C) 2, 5, 9, 14, 20, ….
(D) 11, 22, 33, 44, 55, ….
Answer:
(D) 11, 22, 33, 44, 55, ….
In the sequence : 11, 22, 33, 44, 55,
a2 – a1 = 22 – 11 = 11
a3 – a2 = 33 – 22 = 11
a4 – a3 = 44 – 33 = 11
a5 – a4 = 55 – 44 = 11
i.e., the common difference is same.
Thus, the sequence is an A.P.
Hence, the correct option is (D).
(iii) Let A be an event. For event A, which probability cannot be possible?
(A) \(\frac{2}{3}\)
(B) 1
(C) \(\frac{3}{2}\)
(D) 0
Answer:
(C) \(\frac{3}{2}\)
We know, the probability of any event cannot be more than 1.
Thus, \(\frac{3}{2}\) cannot be the probability of an event A.
Hence, the correct option is (C).
(iv) Ayush buys a share of FV ₹ 200 for MV of ₹ 250. A company confirmed 20% dividend on the share. What will be the rate of return ?
(A) 16%
(B) 15%
(C) 20%
(D) 5%
Answer:
(A) 16%
Given:
FV = ₹ 200
MV = ₹ 250
Dividend = 20%
Let the rate of return be x%.
We know, \(\frac{\mathrm{FV} \times \text { Dividend }}{\mathrm{MV}}\) = Rate of return
⇒ x = \(\frac{20 \times 200}{250}\)
⇒ x = 16%
Hence, the correct option is (A).
Question 1.
(B) Solve the following subquestions.
(i) Find the value of x, if 3x-7 × 4x – 4 = 768.
Answer:
Given: 3x – 7 × 4x – 4 = 768
⇒ 3x-7 × 4x-4 = 3 × 44
Equating the exponents, we get
x – 7 = 1 and x – 4 = 4
x = 1 + 7 and x = 4 + 4
x = 8 and x = 8
Hence, the value of x is 8.
(ii) Evaluate : (1004)3
Answer:
(1004)3 = (1000 + 4)3
= (1000)3 + (4)3 + 3 × 1000 × 4(1000 + 4)
[∵ (a + b)3 = a3 + b3 + 3ab (a + b)]
= 1000000000 + 64 + 12000 (1000 + 4)
= 1000000000 + 64 + 12000000 + 48000
= 1012048064
Hence, the value of (1004)3 is 1012048064.
(iii) Find the range of given data : 46, 35, 78, 90, 20, 56, 45, 76.
Answer:
Given data is 46, 35, 78, 90, 20, 56, 45, 76.
Here, Highest value = 90
Lowest value = 20
We know,
Range = Highest Value – Lowest Value
= 90 – 20
= 70
Hence, the range is 70.
(iv) Find the remainder when p (x) = 3x2 + 2x – 7 is divided by 2x + 1.
Answer:
Given polynomial is,
p(x) = 3x2 + 2x – 7
⇒ x = –\(\frac{1}{2}\) (Put in (ii))
By the remainder theorem, we know that when p(x) is divided by x = – \(\frac{1}{2}\), the remainder is P(-\(\frac{1}{2}\))
∴ Putting x = –\(\frac{1}{2}\) in the given polynomial, we get
Hence, the remainde is –\(\frac{29}{4}\).
Question 2.
(A) Complete and write any two activities from the following :
(i) The following table shows the percentage of demand of different plants.
| Plants | Rose | Lily | Tulip | Daisy | Dahlia |
| No. of Plants | 31 | 25 | 12 | 17 | 15 |
Complete the following table :
Answer:
(ii) Fill in the boxes :
Sunita choose a card from a well-shuffled deck of 52 cards. Total number of cards, S =
(1) Let E be the event that the choosen card is a queen. Number of queens Number of queens =
Thus, P(E) \(=\frac{\text { Number of queens }}{\text { Total number of cards }}\) =
(2) Let F be the event that the choosen card is an eight of the spade
Number of the eight of the spade =
Thus,
Answer:
Sunita chooses a card from a well shuffled deck of 52 cards.
Total number of cards, S = 52
(1) Let E be the event that the choosen card is a queen.
Number of queens = 4
= \(\frac{4}{52}\) = \(\frac{1}{13}\)
(2) Let F be the event that the choosen card is an eight of the spade. Number of the eight of the spade = 1
(iii) Complete the following information :
Answer:
Question 2.
(B) Solve any four subquestions from the following :
(i) Check whether 422 is in the sequence 10, 13, 16, 19, 22, ….
Answer:
Given sequence is 10, 13, 16, 19, 22,………..
Here, t1 = 10, t2 = 13, t3 = 16
Now, t2 – t1 = 13 – 10 = 3
and t3 – t2 = 16 – 13 = 3
∵ t2 – t1 = t3 – t2
∵ This sequence is an A.P.
So, First term, a = lo
Common Difference, d = 3
Let 422 be its nth term.
Then, tn = a + (n – 1)d
⇒ 422 = 10 + (n – 1)3
⇒ 422 = 10 + 3n – 3
⇒ 422 = 7 + 3n
⇒ 422 – 7 = 3n
⇒ 415 = 3n
⇒ n = 138.333
But it is not an integer.
Hence, 422 is not in the given sequence.
(ii) Solve the quadratic equation : 16x2 + 24x + 9 = 0.
Answer:
Given equation is,
16x2 + 24x + 9 = O
Compare the equation to ax2 + bx + c, we get
a = 16, b = 24, c = 9
Now, we know
Hence, the value of x is –\(\frac{3}{4}\).
(iii) Write the two points of the equation 4x – 5y = 10.
Answer:
Given equation is,
4x – 5y = 10
For x = 0,
4(0) – 5y = 1o
0 – 5y = 1o
-5y = 10
y = -2
Thus, the point is (0, -2).
For x = 2.5,
4(2.5) – 5y = 10
1o – 5y = 10
-5y = 10
-5y = 10 – 1o
y = 0
Hence, the two points of the equation are (0, -2) and (2.5, 0).
(iv) A wholeseller buys pens having taxable value ₹ 12,000 and sold it to the retailer for the taxable value ₹ 14,000. Rate of GST is 20%. Find the CGST and SGST payable by wholeseller.
Answer:
Wholeseller taxable value = ₹ 12,000
∴ Input tax = \(\frac{20}{100}\) × 12,000 = ₹ 2,400
Retailer taxable value = ₹ 14,000
∴ Output tax = \(\frac{20}{100}\) × 14,000 = ₹ 2,800
GST payable = Output tax – Input tax
= ₹ (2,800 – 2,400) = ₹ 400
Hence, CGST payable by wholeseller = ₹ 200 and SGST payable by wholeseller = ₹ 200.
(v) If the roots of the quadratic equation x2 + 12x + a = 0 are real and equal, then find the value of a.
Answer:
Given equation is,
x2 + 12x + a = 0
Since the roots of quadratic equation are real and equal,
∴ Discriminant, D = O
⇒ B2 – 4AC = 0
⇒ (12)2 – 4 × 1 × a = 0
⇒ 4a= 144
⇒ a = 36
Hence, the value of a is 36.
Question 3.
(A) Complete and write any one activity from the following :
(i) Roshan spends 80% of the money that he receives every month and saves ₹ 2500. How much money does he get monthly ?
Let the total money be x.
Answer:
Let the total money be x.
Hence, Roshan gets ₹ 12,500 as monthly salary.
(ii) Nitin is younger than Nishant by 5 years. Sum of their ages is 45. What is Nishants age ? Let the ages of Nitin and Nishant be x years and y years respectively.
Then, according to the question,
Answer:
Let the ages of Nitin and Nishant be x year and y year respectively
Then, according to the question,
x = y – 5
and x + y = 45 ……(ii)
Substituting the value of x from equation (i) to equation (ii), we get
Hence, the age of Nishant is 25 years.
Question 3.
(B) Attempt any two subquestions from the following :
(i) Two straight paths are represented by the linear equations 2x – 4y – 4 and – 6x + 12y = 6. Check whether the paths cross each other or not, by using the graphical representation.
Answer:
The given equations are,
2x – 4y = 4 ……. (i)
and -6x + 12y = 6 …….(ii)
From equation (i), we have
The graphical representation is as follows:
From the above graph, it can be observed that the two straight paths do not cross each other.
Hence, the given equations have no common solution.
(ii) Calculate the sum of 35 terms in an A.P., whose fourth term is 16 and ninth term is 31.
Answer:
Let a and d be the first term and the common difference, respectively.
Now, a4 = 16 …[Given
⇒ a + 3d = 16 ….. (i)
Also, a9 = 31 ……[Given]
⇒ a + 8d = 31 ……(ii)
Subtracting the equation (i) from equation (ii), we get
⇒ d = 3
Substituting the value of d in equation (i), we get
a +3 (3) = 16
⇒ a + 9 = 16
⇒ a = 16 – 9 = 7
Now, Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S35 = \(\frac{35}{2}\)[2(7) + (35 – 1)3]
= \(\frac{35}{2}\)[14 + (34)3]
= \(\frac{35}{2}\)(14 + 102)
= \(\frac{35}{2}\)(116)
= 35 × 58
= 2030
Hence, the sum of 35 terms of the A.P. is 2030.
(iii) A handbag contained fifty ten rupees note, thirty-five fifty rupees note and fifteen hundred rupees note. One note is drawn from a handbag. What is the probability of getting :
(i) Ten rupees note
(ii) Fifty rupees note
(iii) Hundred rupees note
(iv) Himanshu invested ₹ 1,55,91,490 in the shares of FV ₹ 100 when MV is ₹ 1,550. Also, the rate of brokerage is 0.5% and GST is 18%. Then find how many shares were buy. Also find the amount of brokerage paid and GST paid for the trading.
Answer:
Total number of notes = 50 + 35 + 15 = 100
(i) Let E1 be the event that the choosen note is ten rupees note.
Number of ten rupees note = 50
Thus, P(E1) \(=\frac{\text { Number of ten rupees note }}{\text { Total number of notes }}\)
= \(\frac{50}{100}\) = \(\frac{1}{2}\)
(ii) Let E2 be the event that the choosen note is fifty rupees note.
Number of fifty rupees note = 35
Thus P(E2) \(=\frac{\text { Number of fifty rupees note }}{\text { Total number of notes }}\)
= \(\frac{35}{100}\) = \(\frac{7}{20}\)
(iii) Let E3 be the event that the choosen note is hundred rupees note.
Number of hundred rupees note = 15
Thus P(E3) \(=\frac{\text { Number of hundred rupees note }}{\text { Total number of notes }}\)
= \(\frac{15}{100}\) = \(\frac{3}{20}\)
(iv) Money invested = ₹ 1,55,91,490
Brokerage = 0.5%
GST = 18%
∴ Brokerage per share = 1550 × \(\frac{0.5}{100}\)
= 7.75
So, GST per share = 18% of 7.75
= 1.395
Cost of 1 share = MV + Brokerage + GST
= 1550 + 7.75 + 1.395
= 1559.145
Number of shares = \(\frac{15591490}{1559.145}\)
= 10000
Total brokerage = Brokerage per share × Number of shares
= 7.75 × 10000
= ₹ 77500
Total GST = 10000 × 1.395
= ₹ 13950
Hence, Himanshu buy 10000 shares, paid ₹ 77500 as brokerage and ₹ 13950 as GST.
Question 4.
Attempt any two subquestions from the following :
(i) The ratio of two numbers is 3 : 2 and the difference of their square is 500. Find the numbers.
Answer:
Given Ratio of two numbers = 3 : 2
So, let the two numbers be 3x and 2x.
Now, according to the question,
(3x)2 – (2x)2 = 500
⇒ 9x2 – 4x2 = 500
⇒ 5x2 = 500
⇒ x2 = \(\frac{500}{5}\) = 100
⇒ x = \(\sqrt{100}\)
⇒ x = ±10
For x = -10,
First number = 3 × (-10) = -30
and second number = 2 × (-10) = -20
For x = 10,
First number = 3 × (10) = 30
and second number = 2 × (10) = 20
Hence, the numbers are -30, -20 or 30, 20.
(ii) Find the values of x and y if the mean and total frequency of the distribution are 25 and 50 respectively.
tablee 3
Answer:
Given: Mean = 25
Mean = 25
Also, Mean, \(\bar{x}\) = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)
25 = 25 + \(\frac{10 y-10 x}{18+x+y}\)
25 – 25 = \(\frac{10 y-10 x}{18+x+y}\)
0 = \(\frac{10 y-10 x}{18+x+y}\)
0(18 + x + y) = 10y – 10x
10y – 10x = 0
y – x = 0 …… (i)
Also, 50 = 18 + x + y
x + y = 50 – 18
x + y = 32 …….(ii)
Adding equations (i) and (ii), we get
2y = 32
y = 16
Putting the value of y in equation (ii), we get
x + 16 = 32
x = 32 – 16 = 16
Hence, the values of x and y are 16 and 16 respectively.
(iii) If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero, (p ≠ q)
Answer:
Let ‘a’ be the first term and ‘d’ be the common difference.
Given, Sp = Sq
We know that, Sum (Sn) = \(\frac{n}{2}\)[2a + (n – 1)d]
Question 5.
Attempt any one subquestion from the following :
(i) Find the mean, median and mode of the given data :
tablee 7
Answer:
We know, Mean, \(\overline{\mathrm{X}}\) = A + \(\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right)\) × h
= 135 + \(\left(\frac{-28}{75}\right)\) × 20
= 135 + \(\left(\frac{-28}{15}\right)\) × 4
= 135 – 7.47
= 127.53
Thus, the mean of the data is 127.53.
Now, the maximum frequency is 22 belonging to class interval 105 — 125.
∴ Modal class = 105 – 125
So, L = 105, f1 = 22, f0 = 7, f2 = 17, h = 20.
∴ \(\left(\frac{\mathrm{N}}{2}\right)^{\mathrm{th}}\) term lies in the class interval 125 – 145
So, L = 125, f = 17, \(\frac{N}{2}\) = 37.5, c.f. = 37, h = 20
Median = L + \(\left(\frac{\frac{\mathrm{N}}{2}-c . f .}{f}\right)\) × h
= 125 + \(\left(\frac{37.5-37}{17}\right)\) × 20 = 125 + \(\left(\frac{0.5}{17}\right)\) × 20
= 125 + \(\left(\frac{5}{17}\right)\) × 2
= 125 + 0.588
= 125.588
Thus, the median is 125.588
Hence, the mean, median and mode of the data are 127.53, 125.588 and 120 respectively.
(ii) X takes 3 hours more than Y to walk 30 km. But, if X doubles his pace, he is ahead of Y by 1\(\frac{1}{2}\) hours. Find their speed of walking.
Answer:
Let the speed of X and Y be x km/h and y km/h respectively.
Then,
Time taken by X to travel 30 km = \(\frac{30}{x}\)
and time taken by Y to travel 30 km = \(\frac{30}{y}\)
According to the question,
\(\frac{30}{x}\) = \(\frac{30}{y}\) + 3
⇒ \(\frac{30}{x}\) – \(\frac{30}{y}\) = 3
⇒ 3(\(\frac{30}{x}\) – \(\frac{30}{y}\)) = 3
\(\frac{10}{x}\) – \(\frac{10}{y}\) = 1 ……… (i)
If X double his speed, then X’s speed = 2x
Then, according to the question,
\(\frac{30}{y}-\frac{30}{2 x}\) = 1\(\frac{1}{2}\)
\(\frac{30}{y}-\frac{15}{x}\) = \(\frac{3}{2}\)
⇒ \(3\left(\frac{10}{y}-\frac{5}{x}\right)\) = \(\frac{3}{2}\)
⇒ \(\frac{20}{y}-\frac{10}{x}\) = 1 ……. (ii)
Let \(\frac{1}{x}\) = p and \(\frac{1}{y}\) = q.
Then, the equations become
10p – 10q = 1 …….(iii)
and 20q – 10p = 1 ……(iv)
Adding equation (iii) and (iv), we get
10q = 2
⇒ q = \(\frac{2}{10}\) = \(\frac{1}{5}\)
Putting q = \(\frac{1}{5}\) in equation (iii), we get
10p – 10 × \(\frac{1}{5}\) = 1
⇒ 10p – 2 = 1
⇒ 10p = 1 + 2 = 3
⇒ p = \(\frac{3}{10}\)
Now, p = \(\frac{1}{x}\)
⇒ \(\frac{3}{10}\) = \(\frac{1}{x}\) ⇒ x = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)
and q = \(\frac{1}{y}\)
⇒ \(\frac{1}{5}\) = \(\frac{1}{y}\) ⇒ y = 5
Hence, X’s speed = 3\(\frac{1}{3}\) km/h and Y’s speed = 5 km/h.