Maharashtra State Board Class 10 Maths Solutions Digest Part 1 & 2

Maharashtra State Board Class 10 Maths Solutions

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Maharashtra State Board Class 10 Maths Solutions Algebra Pdf Part 1

Problem Set 1 Geometry 10th Maharashtra Board Chapter 1 Linear Equations in Two Variables

Maharashtra State Board Class 10 Maths Solutions Chapter 2 Quadratic Equations

Problem Set 3 Algebra Class 10 Chapter 3 Arithmetic Progression

Financial Planning Class 10 Pdf Chapter 4

Practice Set 5.1 Geometry Class 10 Chapter 5 Probability

Problem Set 6 Algebra Class 10 Chapter 6 Statistics

Maharashtra State Board Class 10 Maths Solutions Geometry Pdf Part 2

Maharashtra Board Class 10 Maths Chapter 1 Similarity

Maharashtra Board Class 10 Maths Chapter 2 Pythagoras Theorem

Maharashtra Board Class 10 Maths Chapter 3 Circle

Problem Set 6 Geometry Class 10 Chapter 4 Geometric Constructions

Maharashtra State Board Class 10 Maths Solutions Geometry Chapter 5 Co-ordinate Geometry

Trigonometry 10th Class Problem Set 6 Maharashtra Board Class 10 Maths Chapter 6

Problem Set 7 Geometry 10th Maharastra Board Chapter 7 Mensuration

Maharashtra State Board Class 10 Textbook Solutions

ebalbharti

Maharashtra Board 10th Class Maths Part 1 Problem Set 4B Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 4B Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Problem Set 4B Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Financial Planning Class 10 Problem Set 4b
Question 1.
Write the correct alternative for the following questions.

i. If the Face Value of a share is ₹ 100 and Market value is ₹ 75, then which of the following statement is correct?
(A) The share is at premium of ₹ 175
(B) The share is at discount of ₹ 25
(C) The share is at premium of ₹ 25
(D) The share is at discount of ₹ 75
Answer:
(B)

ii. What is the amount of dividend received per share of face value ₹ 10 if dividend declared is 50%.
(A) ₹ 50
(B) ₹ 5
(C) ₹ 500
(D) ₹ 100
Answer:
Dividend = 10 × \(\frac { 50 }{ 100 } \) = ₹ 5
(B)

iii. The NAV of a unit in mutual fund scheme is ₹ 10.65, then find the amount required to buy 500 such units.
(A) 5325
(B) 5235
(C) 532500
(D) 53250
Answer:
(A)

iv. Rate of GST on brokerage is _______
(A) 5%
(B) 12%
(C) 18%
(D) 28%
Answer:
(C)

v. To find the cost of one share at the time of buying the amount of Brokerage and GST is to be ______ MV of share.
(A) added to
(B) subtracted from
(C) Multiplied with
(D) divided by
Answer:
(A)

Problem Set 4b Algebra Class 10 Question 2. Find the purchase price of a share of FV ₹ 100 if it is at premium of ₹ 30. The brokerage rate is 0.3%.
Solution:
Here, Face Value of share = ₹ 100,
premium = ₹ 30, brokerage = 0.3%
MV = FV + Premium
= 100 + 30
= ₹ 130
Brokerage = 0.3% of MV
= \(\frac { 0.3 }{ 100 } \) × 130 = ₹ 0.39
Purchase price of a share = MV + Brokerage
= 130 + 0.39
= ₹ 130.39
Purchase price of a share is ₹ 130.39.

Question 3.
Prashant bought 50 shares of FV ₹ 100, having MV ₹ 180. Company gave 40% dividend on the shares. Find the rate of return on investment.
Solution:
Here, Number of shares = 50, FV = ₹ 100,
MV = ₹ 180, rate of dividend = 40%
∴ Sum invested = Number of shares × MV
= 50 × 180
= ₹ 9000
Dividend per share = 40% of FV
= \(\frac { 40 }{ 100 } \) × 100
Dividend = ₹ 40
∴ Total dividend on 50 shares = 50 × 40
= ₹ 2000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 1
∴ Rate of return on investment is 22.2%.

Question 4.
Find the amount received when 300 shares of FV ₹ 100, were sold at a discount of ₹ 30.
Solution:
Here, FV = ₹ 100, number of shares = 300,
discount = ₹ 30
MV of 1 share = FV – Discount
= 100 – 30 = ₹ 70
∴ MV of 300 shares = 300 × 70
= ₹ 21,000
∴ Amount received is ₹ 21,000.

Question 5.
Find the number of shares received when ₹ 60,000 was invested in the shares of FV ₹ 100 and MV ₹ 120.
Solution:
Here, FV = ₹ 100, MV = ₹ 120,
Sum invested = ₹ 60,000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 2
∴ Number of shares received were 500.

Question 6.
Smt. Mita Agrawal invested ₹ 10,200 when MV of the share is ₹ 100. She sold 60 shares when the MV was ₹ 125 and sold remaining shares when the MV was ₹ 90. She paid 0.1% brokerage for each trading. Find whether she made profit or loss? and how much?
Solution:
For purchasing shares:
Here, sum invested = ₹ 10,200, MV = ₹ 100
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 3
For selling shares:
60 shares sold at MV of ₹ 125.
∴ MV of 60 shares = 125 × 60
= ₹ 7500
Brokerage = \(\frac { 0.1 }{ 100 } \) × 7500 = ₹ 7.5
∴ Sale value of 60 shares = 7500 – 7.5 = ₹ 7492.5
Now, remaining shares = 102 – 60 = 42
But 42 shares sold at MV of ₹ 90.
∴ MV of 42 shares = 42 × 90 = ₹ 3780
∴ Brokerage = \(\frac { 0.1 }{ 100 } \) × 3780 = ₹ 3.78
∴ Sale value of 42 shares = 3780 – 3.78 = ₹ 3776.22
Total sale value = 7492.5 + 3776.22 = ₹ 11268.72
Since, Purchase value < Sale value
∴ Profit is gained.
∴ Profit = Sale value – Purchase value
= 11268.72 – 10210.2
= ₹ 1058.52
∴ Smt. Mita Agrawal gained a profit of ₹ 1058.52.

Question 7. Market value of shares and dividend declared by the two companies is given below.
Face value is same and it is 7 100 for both the shares. Investment in which company is more profitable?
i. Company A – ₹ 132,12%
ii Company B – ₹ 144,16%
Solution:
For company A:
FV = ₹ 100, MV = ₹ 132,
Rate of dividend = 12%
Dividend = 12% of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 4
∴ Rate of return of company B is more.
∴ Investment in company B is more profitable.

Question 8. 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?
Solution:
Here, FV = ₹ 100, MV = ₹ 50
Purchase value of shares = ₹ 50118,
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = 0.2% of MV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 5
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 6
∴ 1000 shares were purchased for ₹ 50,118.

Question 9. Shri. Batliwala sold shares of ₹ 30,350 and purchased shares of ₹ 69,650 in a day. He paid brokerage at the rate of 0.1% on sale and purchase. 18% GST was charged on brokerage. Find his total expenditure on brokerage and tax.
Solution:
Total amount = sale value + Purchase value
= 30350 + 69650
= ₹ 1,00,000
Rate of Brokerage = 0.1 %
Brokerage = 0.1 % of 1,00,000
= \(\frac { 0.1 }{ 100 } \) × 1,00,000
= ₹ 100
Rate of GST = 18%
∴ GST = 18 % of brokerage
= \(\frac { 18 }{ 100 } \) × 100
∴ GST = ₹ 18
Total expenditure on brokerage and tax
= 100 + 18 = ₹ 118
∴ Total expenditure on brokerage and tax is ₹ 118.

Alternate Method:
Brokerage = 0.1 %, GST = 18%
At the time of selling shares:
Total sale amount of shares = ₹ 30,350
Brokerage = 0.1% of 30,350
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 7
For purchasing shares:
Total purchase amount of shares = ₹ 69,650
Brokerage = 0.1% of 69,650
= \(\frac { 0.1 }{ 100 } \) × 69650
= ₹ 69.65
GST = 18% of 69.65
= \(\frac { 18 }{ 100 } \) × 69.65
= ₹ 12.537
∴ Total expenditure on brokerage and tax = Brokerage and tax on selling + Brokerage and tax on purchasing
= (30.35 + 5.463) + (69.65 + 12.537)
= ₹ 118
∴ Total expenditure on brokerage and tax is ₹ 118.

Question 10. Sint. Aruna Thakkar purchased 100 shares of FV 100 when the MV is ₹ 1200. She paid brokerage at the rate of 0.3% and 18% GST on brokerage. Find the following –
i. Net amount paid for 100 shares.
ii. Brokerage paid on sum invested.
iii. GST paid on brokerage.
iv. Total amount paid for 100 shares.
Solution:
Here, FV = ₹ 100,
Number of shares = 100, MV = ₹ 1200
Brokerage = 0.3%, GST = 18%
i. Sum invested = Number of shares × MV
= 100 × 1200 = ₹ 1,20,000
∴ Net amount paid for 100 shares is ₹ 1,20,000.

ii. Brokerage = 0.3% of sum invested
= \(\frac { 0.3 }{ 100 } \) × 1,20,000 = ₹ 360
∴ Brokerage paid on sum invested is ₹ 360.

iii. GST = 18% of brokerage
= \(\frac { 18 }{ 100 } \) × 360 = ₹ 64.80
∴ GST paid on brokerage is ₹ 64.80.

iv. Total amount paid for 100 shares
= Sum invested + Brokerage + GST
= 1,20,000 + 360 + 64.80
= ₹ 1,20,424.80
∴ Total amount paid for 100 shares is ₹ 1,20,424.80.

Question 11. Smt. Anagha Doshi purchased 22 shares of FV ₹ 100 for Market Value of ₹ 660. Find the sum invested. After taking 20% dividend, she sold all the shares when market value was ₹ 650. She paid 0.1% brokerage for each trading done. Find the percent of profit or loss in the share trading. (Write your answer to the nearest integer)
Solution:
For purchasing shares:
Here, FV = ₹ 100, MV = ₹ 660, Number of shares = 22, rate of brokerage = 0.1%
Sum invested = MV × Number of shares
= 660 × 22
= ₹ 14,520
Brokerage = 0.1 % of sum invested
= \(\frac { 0.1 }{ 100 } \) × 14520 = ₹ 14.52
∴ Amount invested for 22 shares
= Sum invested + Brokerage
= 14520 + 14.52
= ₹ 14534.52
For dividend:
Rate of dividend = 20%
∴ Dividend per share = 20 % of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4B 8
∴ Percentage of profit in the share trading is 1 % (nearest integer).

Alternate Method:
For purchasing share:
Here, FV = ₹ 100, MV = ₹ 660, Number of shares = 22, rate of brokerage = 0.1%
Sum invested = MV × Number of shares
= 660 × 22
= ₹ 14,520
Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 660 = ₹ 0.66
Amount invested for 1 share = 660 + 0.66
= ₹ 660.66
For dividend:
Rate of dividend = 20%
Dividend = 20% of FV = \(\frac { 20 }{ 100 } \) × 100 = ₹ 20
For selling share:
MV = ₹ 650, rate of brokerage = 0.1%
Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 650 = ₹ 0.65 100
Amount received after selling 1 share
= 650 – 0.65 = 649.35
∴ Amount received including divided
= selling price of 1 share + dividend per share
= 649.35 + 20
= ₹ 669.35
Since, income > Amount invested
∴ Profit is gained.
∴ profit = 669.35 – 660.66 = ₹ 8.69
Profit Percentage = \(\frac { 8.69 }{ 660.66 } \) × 100= 1.31%
∴ Percentage of profit in the share trading is 1 % (nearest integer).

Maharashtra Board 10th Class Maths Part 1 Problem Set 3 Solutions Chapter 3 Arithmetic Progression

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Problem Set 3 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1.
Choose the correct alternative answer for each of the following sub questions.

i. The sequence – 10,- 6,- 2, 2, …
(A) is an A.P. Reason d = – 16
(B) is an A.P. Reason d = 4
(C) is an A.P. Reason d = – 4
(D) is not an A.P.
Answer:
(B)

ii. First four terms of an A.P. are …, whose first term is -2 and common difference is -2.
(A) -2, 0, 2, 4
(B) -2, 4,- 8, 16
(C) -2, -4, -6, -8
(D) -2,-4, -8, -16
Answer:
(C)

iii. What is the sum of the first 30 natural numbers?
(A) 464
(B) 465
(C) 462
(D) 461
Answer:
(B)

iv. For an given A.P. t7 = 4, d = – 4, then a = ………
(A) 6
(B) 7
(C) 20
(D) 28
Answer:
(D)

v. For an given A.P. a = 3.5, d = 0, n = 101, then tn = ….
(A) 0
(B) 3.5
(c) 103.5
(D) 104.5
Answer:
(B)

vi. In an A.P. first two terms are – 3, 4, then 21st term is ….
(A) -143
(B) 143
(C) 137
(D) 17
Answer:
(C)

vii. If for any A.P. d = 5, then t18 – t13 = ….
(A) 5
(B) 20
(C) 25
(D) 30
Answer:
(C)

viii. Sum of first five multiples of 3 is …
(A) 45
(B) 55
(C) 15
(D) 75
Answer:
(A)

ix. 15, 10, 5, … In this A.P. sum of first 10 terms is…
(A) -75
(B) -125
(C) 75
(D) 125
Answer:
(A)

x. In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is 399, then n = ….
(A) 42
(B) 38
(C) 21
(D) 19
Answer:
(B)

Hints:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 1

Question 2.
Find the fourth term from the end in an
A.P.: -11, -8, -5, …, 49.
Solution:
The given A.P. is
-11,-8,-5, ……. 49
Reversing the A.P., we get 49, …, -5, -8, -11
Here, a = 49, d = -11 -(-8) = -11 + 8 = -3
Since, tn = a + (n – 1)d
∴ t4 = 49 + (4 – 1)(-3)
= 49 + (3) (-3)
= 49 – 9
= 40
∴ Fourth term from the end in the given A.P. is 40.
[Note: If an AY. is reversed, then the resulting sequence is also an A.P.]

Question 3.
In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.
Solution:
For an A.P., let a be the first term and d be the common difference.
t10 = 46, t5 + t7 = 52 …[Given]
Since, tn = a + (n – 1)d
∴ t10 = a + (10 – 1)d
∴ 46 = a + 9d
i. e. a + 9d = 46 …(i)
Also, t5 + t7 = 52
∴ a + (5 – 1)d + a + (7 – 1)d = 52
∴ a + 4d + a + 6d = 52
∴ 2a + 10d = 52
∴ 2 (a + 5d) = 52
∴ a + 5d = \(\frac { 52 }{ 2 } \)
∴ a + 5d = 26 …(ii)
Subtracting equation (ii) from (i), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 2
Substituting d = 5 in equation (ii), we get
a + 5(5) = 26
∴ a + 25 = 26
∴ a = 26 – 25 = 1
t1 = a = 1
t2 = t1 + d = 1 + 5 = 6
t3 = t2 + d = 6 + 5 = 11
t4 = t3 + d = 11 + 5 = 16
The required A.P. is 1,6,11,16,….

Question 4.
The A.P. in which 4th term is -15 and 9th term is -30. Find the sum of the first 10 numbers.
Solution:
t4 = -15, t9 = – 30 …[Given]
Since, tn = a + (n – 1)d
∴ t4 = a + (4 – 1)d
∴ – 15 = a + 3d
i. e. a + 3d = -15 …(i)
Also, t9 = a + (9 – 1)d
∴ -30 = a + 8d
i.e. a + 8d = -30 …(ii)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 3
∴ The sum of the first 10 numbers is -195.

Question 5.
Two given A.P.’s are 9, 7, 5, … and 24, 21, 18, … If nth term of both the .progressions are equal then find the value of n and n,h term.
Solution:
The first A.P. is 9, 7, 5,…
Here, a = 9, d = 7- 9 = -2
∴nth term = a + (n – 1)d
= 9 + (n – 1) (-2)
= 9 – 2n + 2
= 11 – 2n
The second A.P. is 24, 21, 18, …
Here, a = 24, d = 21 – 24 = – 3
∴ nth term = a + (n – 1)d
= 24 + (n – 1) (-3)
= 24 – 3n + 3
= 27 – 3n
Since, the nth terms of the two A.P.’s are equal.
∴ 11 – 2n = 27 – 3n
∴ 3n – 2n = 27 – 11
∴ n = 16
∴ t16 = 9 + (16 – 1) (-2)
= 9 + 15 × (-2)
= 9 – 30
∴ t16 = -21
∴ The values of n and nth term are 16 and -21 respectively.

Question 6.
If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is -3, then find the 10th term.
Solution:
for an A.P., let a be the first term and d be the common difference.
According to the first condition,
t3 + tg = 7
∴ a + (3 – 1) d + a + (8 – 1)d = 7 …[∵ tn = a + (n – 1)d]
∴ a + 2d + a + 7d = 7
∴ 2a + 9d = 7 …(i)
According to the second condition,
t7 + t14 = -3
∴ a + (7 – 1)d + a + (14 – 1 )d = -3
∴ a + 6d + a + 13d = -3
∴ 2a + 19 d = – 3 …(ii)
Subtracting equation (i) from (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 4

Question 7.
In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Solution:
Let the number of terms in the A.P. be n and the common difference be d.
Then, a = -5, tn = 45, Sn = 120
Since, tn = a + (n – 1)d
∴ 45 = -5 + (n – 1)d
∴ 45 + 5 = (n – 1)d
∴ (n – 1)d = 50 …(i)
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 5
Substituting n = 6 in equation (i), we get
(6 – 1)d = 50
∴ 5d = 50
∴ d = \(\frac { 50 }{ 5 } \) = 10
∴ There are 6 terms in the A.P. and the common difference is 10.

Alternate Method:
Let the number of terms in the A.P. be n.
Then, t1 = a = -5, tn = 45, Sn = 120
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 6
∴ There are 6 terms in the A.P. and the common difference is 10.

Question 8.
Sum of 1 to n natural numbers is 36, then find the value of n.
Solution:
The natural numbers from 1 to n are
1,2, 3, ……, n.
The above sequence is an A.P.
∴ a = 1, d = 2 – 1 = 1
Sn = 36 …[Given]
Now, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ 36 = \(\frac { n }{ 2 } \) [2(1) + (n – 1)(1)]
∴ 36 = \(\frac { n }{ 2 } \) (2 + n – 1)
∴ 36 × 2 = n (n + 1)
∴ 72 = n (n + 1)
∴ 72 = n2 + n
∴ n2 + n – 72 = 0
∴ n2 + 9n – 8n – 72 = 0
∴ n(n + 9) – 8 (n + 9) = 0
∴ (n + 9) (n – 8) = 0
∴ n + 9 = 0 or n – 8 = 0
∴ n = -9 or n = 8
But, n cannot be negative.
∴ n = 8
∴ The value of n is 8.

Question 9.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Solution:
Let the three parts of 207 that are in A.P. be
a – d, a, a + d
According to the first condition,
(a – d) + a + (a + d) = 207
∴ 3a = 207
∴ a = \(\frac { 207 }{ 3 } \)
∴ a = 69 …(i)
According to the second condition,
(a – d) × a = 4623
∴ (69 – d) × 69 = 4623 …[From (i)]
∴ 69 – d = \(\frac { 4623 }{ 69 } \)
∴ d = 69 – 67
∴ d = 2
∴ a – d = 69 – 2 = 67
a = 69
a + d = 69 + 2 = 71
∴ The three parts of 207 that are in A.P. are 67, 69 and 71.

Question 10.
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
Solution:
Since, there are 37 terms in the A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 7
Substituting d = 4 in equation (i), we get
3a + 54(4) = 225
∴ 3a + 216 = 225
∴ 3a = 225 – 216
∴ 3a = 9
∴ a = \(\frac { 9 }{ 3 } \) = 3
∴The required A. P. is
a, a + d, a + 2d, a + 3d, …., a + (n – 1)d
i.e. 3, 3 + 4,3 + 2 × 4, 3 + 3 × 4,…, 3 + (37 – 1)4
i.e. 3, 7,11,15, …,147

Question 11.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 8
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 9

Question 12.
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)
Solution:
For an A.P., let a be the first term and d be the common difference.
The sum of first n terms of an A.P. is given by
Sn = [2a + (n – 1)d]
According to the given condition,
Sp = Sq
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 10
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 11
∴ The sum of the first (p + q) terms is zero

Question 13.
If m times the mth term of an A.P. is equal to n times nth term, then show that the (m + n)th term of the A.P. is zero.
Solution:
According to the given condition,
mtm = ntn
∴ m[a + (m – 1)d] = n[a + (n – 1)d]
∴ ma + md(m – 1) = na + nd(n- 1)
∴ ma + m2d – md = na + n2d – nd
∴ ma + m2d – md – na – n2d + nd = 0
∴ (ma – na) + (m2d – n2d) – (md – nd) = 0
∴ a(m – n) + d(m2 – n2) – d(m – n) = 0
∴ a(m – n) + d(m + n) (m – n) – d(m – n) = 0
∴ (m – n)[a + (m + n – 1) d] = 0
∴ [a+ (m + n – 1)d] = 0 …[Dividing both sides by (m – n)]
∴ t(m+n) = 0
∴ The (m + n)th term of the A.P. is zero.

Question 14.
₹ 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 12
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Problem Set 3 13

Maharashtra Board 9th Class Maths Part 1 Practice Set 2.5 Solutions Chapter 2 Real Numbers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.5 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Find the value.
i. | 15 – 2|
ii. | 4 – 9|
iii. | 7| x | -4|
Solution:
i. |15 – 2| = |13| = 13
ii. |4 – 9| = |-5| = 5
iii. |7| x |- 4| = 7 x 4 = 28

Question 2.
Solve.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 1
Solution:
i. |3x – 5| = 1
∴ 3x – 5 = 1 or 3x – 5 = -1
∴ 3x = 1 + 5 or 3x = -1 + 5
∴ 3x = 6 or 3x = 4
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 2

ii. |7 – 2x| = 5
∴ 7 – 2x = 5 or 7 – 2x = -5
∴ 7 – 5 = 2x or 7 + 5 = 2x
∴ 2x = 2 or 2x = 12
∴ x = \(\frac { 2 }{ 2 }\) or x = \(\frac { 12 }{ 2 }\)
∴ x = 1 or x = 6

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 3
∴ 8 – x = 10 or 8 – x = -10 .. [Multiplying both the sides by 2]
∴ 8 – 10 = x or 8 + 10 = x
∴ x = -2 or x = 18

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.5 4

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.3 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.3 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
Write sample space ‘S’ and number of sample points n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).

i. One die is rolled,
Event A: Even number on the upper face.
Event B: Odd number on the upper face.
Event C: Prime number on the upper face.

ii. Two dice are rolled simultaneously,
Event A: The sum of the digits on upper faces is a multiple of 6.
Event B: The sum of the digits on the upper faces is minimum 10.
Event C: The same digit on both the upper faces.

iii. Three coins are tossed simultaneously.
Condition for event A: To get at least two heads.
Condition for event B: To get no head.
Condition for event C: To get head on the second coin.

iv. Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A: The number formed is even.
Condition for event B: The number is divisible by 3.
Condition for event C: The number formed is greater than 50.

v. From three men and two women, environment committee of two persons is to be formed.
Condition for event A: There must be at least one woman member.
Condition for event B: One man, one woman committee to be formed.
Condition for event C: There should not be a woman member.

vi. One coin and one die are thrown simultaneously.
Condition for event A: To get head and an odd number.
Condition for event B: To get a head or tail and an even number.
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
Solution:
i. Sample space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Condition for event A: Even number on the upper face.
∴ A = {2,4,6}
∴ n(A) = 3
Condition for event B: Odd number on the upper face.
∴ B = {1, 3, 5}
∴ n(B) = 3
Condition for event C: Prime number on the upper face.
∴ C = {2, 3, 5}
∴ n(C) = 3

ii. Sample space,
S = {(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
Condition for event A: The sum of the digits on the upper faces is a multiple of 6.
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
∴ n(A) = 6

Condition for event B: The sum of the digits on the upper faces is minimum 10.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 6

Condition for event C: The same digit on both the upper faces.
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(C) = 6

iii. Sample space,
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ n(S) = 8

Condition for event A: To get at least two heads.
∴ A = {HHT, HTH, THH, HHH}
∴ n(A) = 4

Condition for event B: To get no head.
∴ B = {TTT}
∴ n(B) = 1

Condition for event C: To get head on the second coin.
∴ C = {HHH, HHT, THH, THT}
∴ n(C) = 4

iv. Sample space (S) = {10, 12, 13, 14, 15,
20, 21, 23, 24, 25,
30, 31, 32, 34, 35,
40, 41, 42, 43,
45, 50, 51, 52, 53, 54}
∴ n(S) = 25
Condition for event A: The number formed is even
∴ A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54)
∴ n(A) = 13
Condition for event B: The number formed is divisible by 3.
∴ B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n(B) = 9
Condition for event C: The number formed is greater than 50.
∴ C = {51,52, 53,54}
∴ n(C) = 4

v. Let the three men be M1, M2, M3 and the two women be W1, W2.
Out of these men and women, a environment committee of two persons is to be formed.
∴ Sample space,
S = {M1M2, M1M3, M1W1, M1W2, M2M3, M2W1, M2W2, M3W1, M3W2, W1W2}
∴ n(S) = 10
Condition for event A: There must be at least one woman member.
∴ A = {M1W1, M1W2, M2W1, M2W2, M3W1, M3W2, W1W2}
∴ n(A) = 7
Condition for event B: One man, one woman committee to be formed.
∴ B = {M1W1, M1W2, M2W1, M2W2, M3W2, M3W2}
∴ n(B) = 6
Condition for event C: There should not be a woman member.
∴ C = {M1M2, M1M3, M2M3}
∴ n(C) = 3

vi. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) = 12
Condition for event A: To get head and an odd number.
∴ A = {(H, 1), (H, 3), (H, 5)}
∴ n(A) = 3
Condition for event B: To get a head or tail and an even number.
∴ B = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}
∴ n(B) = 6
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
The greatest number on the upper face of a die is 6.
∴ Event C is an impossible event.
∴ C = { }
∴ n(C) = 0

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.2 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.2 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
For each of the following experiments write sample space ‘S’ and number of sample Point n(S)
i. One coin and one die are thrown simultaneously.
ii. Two digit numbers are formed using digits 2,3 and 5 without repeating a digit.
Solution:
i. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) =12
ii. Sample space,
S = {23,25,32, 35, 52, 53}
∴ n(S) = 6

Question 2.
The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 1
Solution:
There are total six colours on the disc.
Sample space,
S = {Red, Orange, Yellow, Blue, Green, Purple}
∴ n(S) = 6
∴ Arrow may stop on any one of the six colours.

Question 3.
In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar).
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 2
Solution:
Dates which are multiple of 5:
5,10, 15,20,25,30
∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Question 4.
Form a ‘Road safety committee’ of two, from 2 boys (B1 B2) and 2 girls (G1, G2). Complete the following activity to write the sample space.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 3

Question 1.
Sample Space

  • The set of all possible outcomes of a random experiment is called sample space.
  • It is denoted by ‘S’ or ‘Ω’ (omega).
  • Each element of a sample space is called a sample point.
  • The number of elements in the set S is denoted by n(S).
  • If n(S) is finite, then the sample space is called a finite sample space.

Some examples of finite sample space. (Textbook pg. no, 117)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 4 Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 5

Maharashtra Board 10th Class Maths Part 1 Practice Set 5.1 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

Practice Set 5.1 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
How many possibilities are there in each of the following?
i. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
ii. Any day of a week is to be selected randomly.
iii. Select one card from the pack of 52 cards.
iv. One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
i. Here, 8 sites of Maharashtra are given.
∴ There are 8 possibilities in a random experiment of visiting a site out of 8 sites in Maharashtra.

ii. There are 7 days in a week.
∴ There are 7 possibilities in a random experiment of selecting a day of the week.

iii. There are 52 cards in a pack of cards.
∴ There are 52 possibilities in a random experiment of selecting one card from the pack of 52 cards.

iv. There are 11 cards numbered from 10 to 20.
∴ There are 11 possibilities in a random experiment of selecting one card from the given set of cards.

Question 1.
In which of the following experiments possibility of expected outcome is more? (Textbook pg, no. 116)
i. Getting 1 on the upper face when a die is thrown.
ii. Getting head by tossing a coin.
Solution:
i. On a die there are 6 numbers.
∴ There are 6 possibilities of getting any one number from 1 to 6 on the upper face i.e. \(\frac { 1 }{ 6 } \) is the possibility.

ii. There are two possibilities (H or T) on tossing a coin i.e. \(\frac { 1 }{ 2 } \) possibility.
∴ In the second experiment, the possibility of expected outcome is more.

Question 2.
Throw a die, once. What are the different possibilities of getting dots on the upper face? (Textbook pg. no. 114)
Answer:
There are six different possibilities of getting dots on the upper face. They are
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Maharashtra Board 10th Class Maths Part 1 Practice Set 4.4 Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Practice Set 4.4 Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Question 1.
Market value of a share is ₹ 200. If the brokerage rate is 0.3% then find the purchase value of the share.
Solution:
Here, MV = ₹ 200, Brokerage = 0.3%
Brokerage = 0.3% of MV
= \(\frac { 0.3 }{ 100 } \) × 200
= ₹ 0.6
∴ Purchase value of the share = MV + Brokerage
= 200 + 0.6
= ₹ 200.60
∴ Purchase value of the share is ₹ 200.60.

Question 2.
A share is sold for the market value of ₹ 1000. Brokerage is paid at the rate of 0.1%. What is the amount received after the sale?
Solution:
Here, MV = ₹ 1000, Brokerage = 0.1%
∴ Brokerage = 0.1 % of MV
= \(\frac { 0.1 }{ 100 } \) × 1000
∴ Brokerage = ₹ 1
∴ Selling value of the share = MV – Brokerage
= 1000 – 1
= ₹ 999
∴ Amount received after the sale is ₹ 999.

Question 3.
Fill in the blanks given in the contract note of sale-purchase of shares.
(B – buy S – sell)
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 1
Solution:
For buying shares:
Here, Number of shares = 100,
MV of one share = ₹ 45
∴ Total value = 100 × 45
= ₹ 4500
Brokerage= 0.2% of total value 0.2
= \(\frac { 0.2 }{ 100 } \) × 4500
CGST = 9% of brokerage
= \(\frac { 9 }{ 100 } \) × 9 = ₹ 0.81
But, SGST = CGST
∴ SGST = ₹ 0.81
∴ Purchase value of shares
= Total value + Brokerage
= 4500 + 9 + 0.81 + 0.81
= ₹ 4510.62

ii. For selling shares:
Here, Number of shares = 75,
MV of one share = ₹ 200
∴ Total value = 75 × 200
= ₹ 15000
Brokerage = 0.2% of total value
= \(\frac { 0.2 }{ 100 } \) × 15000
= ₹ 30
CGST = 9% of brokerage
= \(\frac { 9 }{ 100 } \) × 30 = ₹ 2.70
But, SGST = CGST
∴ SGST = ₹ 2.70
∴ Selling value of shares = Total value – (Brokerage + CGST + SGST)
= 15000 – (30 + 2.70 + 2.70)
= 15000 – 35.40
= ₹ 14964.60
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 2

Question 4.
Smt. Desai sold shares of face value ₹ 100 when the market value was ₹ 50 and received ₹ 4988.20. She paid brokerage 0.2% and GST on brokerage 18%, then how many shares did she sell?
Solution:
Here, face value of share = ₹ 100,
MV = ₹ 50,
Selling price of shares = ₹ 4988.20,
Rate of brokerage = 0.2%, Rate of GST = 18%
Brokerage = 0.2% of MV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 3

Question 5.
Mr. D’souza purchased 200 shares of FV ₹ 50 at a premium of ₹ 100. He received 50% dividend on the shares. After receiving the dividend he sold 100 shares at a discount of ₹ 10 and remaining shares were sold at a premium of ₹ 75. For each trade he paid the brokerage of ₹ 20. Find whether Mr. D’souza gained or incurred a loss? By how much?
Solution:
For purchasing shares:
Here, FV = ₹ 50, Number of shares = 200,
premium = ₹ 100
MV of 1 share = FV + premium
= 50 + 100
= ₹ 150
∴ MV of 200 shares = 200 × 150 = ₹ 30,000
∴ Mr. D’souza invested amount
= MV of 200 shares + brokerage
= 30,000 + 20
= ₹ 30,020
For selling shares:
Rate of dividend = 50 %, FV = ₹ 50,
brokerage = ₹ 20
Number of shares = 200
Dividend per share = 50% of FV
= \(\frac { 50 }{ 100 } \) × 50
= ₹ 25
∴ Dividend of 200 shares = 200 × 25 = ₹ 5,000
Now, 100 shares are sold at a discount of ₹ 10.
∴ Selling price of 1 share = FV – discount
= 50 – 10
= ₹ 40
∴ Selling price of 100 shares = 100 × 40
= ₹ 4000
∴ Amount obtained by selling 100 shares
= selling price – brokerage
= 4000 – 20
= ₹ 3980
Also, remaining 100 shares are sold at premium of ₹ 75.
∴ selling price of 1 share = FV + premium
= 50 + 75
= ₹ 125
∴ selling price of 100 shares = 100 × 125
= ₹ 12,500
∴ Amount obtained by selling 100 shares
= selling price – brokerage
= 12,500 – 20
= ₹ 12,480
∴ Mr D’souza income = 5000 + 3980 + 12480
= ₹ 21460
Now, Mr D’souza invested amount > income
∴ Mr D’souza incurred a loss.
∴ Loss = amount invested – income
= 30020 – 21460
= ₹ 8560
∴ Mr. D’souza incurred a loss of ₹ 8560.

Question 1.
Nalinitai invested ₹ 6024 in the shares of FV ₹ 10 when the Market Value was ₹ 60. She sold all the shares at MV of ₹ 50 after taking 60% dividend. She paid 0.4% brokerage at each stage of transactions. What was the total gain or loss in this transaction? (Textbook pg. no. 106)
Solution:
Rate of GST is not given in the example, so it is not considered.
For Purchased Shares:
FV = ₹ 10, MV = ₹ 60
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.4 4

Question 2.
In the above example if GST was paid at 18% on brokerage, then the loss is ₹ 451.92. Verify whether you get the same answer. (Textbook pg, no. 107)
Solution:
For Purchased Shares:
FV = ₹ 10, MV = ₹ 60, sum invested = ₹ 6024, brokerage = 0.4 %, GST = 18%
Brokerage per share = \(\frac { 0.4 }{ 100 } \) × 60 = ₹ 0.24 100
GST per share = \(\frac { 18 }{ 100 } \) × 0.24 = ₹ 0.0432
∴ Cost of one share = 60 + 0.24 + 0.0432
= ₹ 60.2832
∴ Cost of 100 shares = 100 × 60.2832 = ₹ 6028.32
For sold shares:
FV = ₹ 10, MV = ₹ 50, brokerage = 0.4 %,
GST = 18%, Number of shares = 100
Brokerage per share = \(\frac { 0.4 }{ 100 } \) × 50 = ₹ 0.20
GST per share = \(\frac { 18 }{ 100 } \) × 0.20 = ₹ 0.036
Selling price per share = 50 – 0.2 – 0.036
= ₹ 49.764
Selling price of 100 shares = 100 × 49.764
= ₹ 4976.4
Dividend received 60 %
∴ Dividend per share = \(\frac { 60 }{ 100 } \) × 10 = ₹ 6
Dividend on 100 shares = 6 × 100 = ₹ 600
∴ Nalinitai’s income = 4976.4 + 600 = ₹ 5576.4
∴ Cost of 100 shares = ₹ 6028.32
∴ Loss = 6028.32 – 5576.4 = ₹ 451.92
∴ Nalinitai’s loss is ₹ 451.92.

Maharashtra Board 10th Class Maths Part 1 Problem Set 4A Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 4A Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Problem Set 4A Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Financial Planning Class 10 Problem Set 4a Question 1.
Write the correct alternative for each of the following.

i. Rate of GST on essential commodities is ______
(A) 5%
(B) 12%
(C) 0%
(D) 18%
Answer:
(C)

ii. The tax levied by the central government for trading within state is ______
(A) IGST
(B) CGST
(C) SGST
(D) UTGST
Answer:
(B)

iii. GST system was introduced in our country from ______
(A) 31st March 2017
(B) 1st April 2017
(C) 1st January 2017
(D) 1st July 2017
Answer:
(D)

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

v. In the format of GSTIN there are ______ alpha-numerals.
(A) 15
(B) 10
(C) 16
(D) 9
Answer:
(A)

vi. When a registered dealer sells goods to another registered dealer under GST, then this trading is termed as ______
(A) BB
(B) B2B
(C) BC
(D) B2C
Answer:
(B)

10th Class Algebra Problem Set 4a Question 2.
A dealer has given 10% discount on a showpiece of ₹ 25,000. GST of 28% was charged on the discounted price. Find the total amount shown in the tax invoice. What is the amount of CGST and SGST.
Solution:
Printed price of showpiece = ₹ 25,000,
Rate of discount = 10%
∴ Amount of discount = 10% of printed price
= \(\frac { 10 }{ 100 } \) × 25000
= ₹ 2500
∴ Taxable value
= Printed price – discount
= 25,000 – 2500 = ₹ 22,500
Rate of GST = 28%
∴ Rate of CGST = 14% and
Rate of SGST = 14%
CGST = 14% of taxable value
= \(\frac { 14 }{ 100 } \) × 22500
= ₹ 3150
∴ CGST = SGST = ₹ 3150
∴ Total amount of tax invoice
= Taxable value + CGST + SGST
= 22500 + 3150 + 3150
= ₹ 28,800
∴ The total amount shown in the tax invoice is ₹ 28,800, and the amount of CGST and SGST is ₹ 3150 each.

Financial Planning Problem Set 4a Question 3.
A ready-made garment shopkeeper gives 5% discount on the dress of ₹ 1000 and charges 5% GST on the remaining amount, then what is the purchase price of the dress for the customer?
Solution:
Printed price of dress = ₹ 1000
Rate of discount = 5%
∴ Amount of discount = 5% of printed price
= \(\frac { 5 }{ 100 } \) × 1000
= ₹ 50
∴ Taxable value = Printed price – discount
= 1000 – 50
= ₹ 950
Rate of GST = 5%
∴ GST = 5% of taxable value
= \(\frac { 5 }{ 100 } \) × 950
∴ GST = ₹ 47.5
Purchase price of the dress
= Taxable value + GST
= 950 + 47.5 = ₹ 997.50
∴ Purchase price of the dress for the customer is ₹ 997.50.

Question 4.
A trader from Surat, Gujarat sold cotton clothes to a trader in Rajkot, Gujarat. The taxable value of cotton clothes is ₹ 2.5 lacs. What is the amount of GST at 5% paid by the trader in Rajkot?
Solution:
Taxable amount of cotton clothes = ₹ 2.5 lacs,
Rate of GST = 5%
GST = 5% of taxable amount
= \(\frac { 5 }{ 100 } \) × 2,50,000
= ₹ 12500
∴ Trader of Rajkot has to pay GST of ₹ 12,500.

Question 5.
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?
Solution:
Output tax = 5% of 90000
= \(\frac { 5 }{ 100 } \) × 90000
= ₹ 4500
Input tax = 5% of 85000
= \(\frac { 5 }{ 100 } \) × 85000
= ₹ 4250
ITC = ₹ 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.

Question 6.
A company provided Z-security services for the taxable value of ₹ 64,500. Rate of GST is 18%. Company had paid GST of ₹ 1550 for laundry services and uniforms etc. What is the amount of ITC (input Tax Credit)? Find the amount of CGST and SGST payable by the company.
Solution:
Output tax = 18% of 64500
= \(\frac { 18 }{ 100 } \) × 64500
= ₹ 11610
Input tax = ₹ 1550
GST payable = Output tax – ITC
= 11610 – 1550
∴ GST payable = ₹ 10060
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 1
∴ Amount of ITC is ₹ 1550. Amount of CGST and SGST payable by the company is ₹ 5030 each.

Question 7.
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.
Solution:
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
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 2
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 3
∴ Amount of state and central GST charged by the dealer is ₹ 4,500 each. Taxable value of the set is ₹ 75,000.

Question 8.
A wholesaler purchased electric goods for the taxable amount of ₹ 1,50,000. He sold it to the retailer for the taxable amount of ₹ 1,80,000. Retailer sold it to the customer for the taxable amount of ₹ 2,20,000. Rate of GST is 18%. Show the computation of GST in tax invoices of sales. Also find the payable CGST and payable SGST for wholesaler and retailer.
Solution:
For Wholesaler:
Output tax = 18% of ₹ 1,80,000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 4
Statement of GST payable at each stage of trading:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 5

Question 9.
Anna Patil (Thane, Maharashtra) supplied vacuum cleaner to a shopkeeper in Vasai (Mumbai) for the taxable value of ₹ 14,000, and GST rate of 28% . Shopkeeper sold it to the customer at the same GST rate for ₹ 16,800 (taxable value). Find the following:
i. Amount of CGST and SGST shown in the tax invoice issued by Anna Patil.
ii. Amount of CGST and SGST charged by the shopkeeper in Vasai.
iii. What is the CGST and SGST payable by shopkeeper in Vasai at the time of filing the return.
Solution:
i. For Anna Patil:
Output tax = 28% of 14,000
= \(\frac { 18 }{ 100 } \) × 14000
= ₹ 3920
∴ CGST = SGST = \(\frac { GST }{ 2 } \)
= \(\frac { 3920 }{ 2 } \)
= ₹ 1960
∴ Amount of CGST and SGST shown in the tax invoice issued by Anna Patil is ₹ 1960 each.

ii. For Shopkeeper in Vasai:
Output tax = 28% of 16,800
= \(\frac { 28 }{ 100 } \) × 16,800
= ₹ 4704
∴ CGST = SGST = \(\frac { GST }{ 2 } \)
= \(\frac { 4704 }{ 2 } \)
= ₹ 2352
∴ Amount of CGST and SGST charged by the shopkeeper in Vasai is ₹ 2352 each.

iii. ITC = ₹ 3920
GST payable by shopkeeper in Vasai
= Output tax – ITC
= 4704 – 3920
= ₹ 784
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 6
∴ CGST and SGST payable by shopkeeper in Vasai at the time of filing the return is ₹ 392 each.

Question 10.
For the given trading chain prepare the tax invoice I, II, III. GST at the rate of 12% was charged for the article supplied.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 7
i. Prepare the statement of GST payable under each head by the wholesaler, distributor and retailer at the time of filing the return to the government.
ii. At the end what amount is paid by the consumer?
iii. Write which of the invoices issued are B2B and B2C.
Solution:
i. For wholesaler:
Output tax = 12% of 5000
= \(\frac { 12 }{ 100 } \) × 5000 = ₹ 600
For Distributor:
Output Tax = 12% of 6000
= \(\frac { 12 }{ 100 } \) × 6000 = ₹ 720
ITC = ₹ 600
∴ GST payable = Output tax – ITC
= 720 – 600
= ₹ 120
For Retailer:
Output tax = 12 % of 6500
= \(\frac { 12 }{ 100 } \) × 6500 = ₹ 780
ITC = ₹ 720
∴ GST payable = Output tax – ITC
= 780 – 720 = ₹ 60
Statement of GST payable at each stage of trading:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Problem Set 4A 8

ii. ITC for consumer = ₹ 780
∴ Amount paid by consumer
= taxable value + ITC
= 6500 + 780
= ₹ 7280
∴ Amount paid by the consumer is ₹ 7280.

iii. B2B = Wholesaler to Distributor
B2B = Distributor to Retailer
B2C = Retailer to Consumer

Maharashtra Board 10th Class Maths Part 1 Practice Set 4.2 Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Practice Set 4.2 Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Question 1. ‘Chetana Store’ paid total GST of ₹ 1,00,500 at the time of purchase and collected GST ₹ 1,22,500 at the time of sale during 1st of July 2017 to 31st July 2017. Find the GST payable by Chetana Stores.
Answer:
Output tax (Tax collected at the time of sale)
= ₹ 1,22,500
Input tax (Tax paid at the time of purchase)
= ₹ 1,00,500
ITC (Input Tax credit) = ₹ 1,00,500.
GST payable = Output tax – ITC
= 1,22,500 – 1,00,500
= ₹ 22,000
GST payable by Chetana stores is ₹ 22,000.

Question 2. Nazama is a proprietor of a firm, registered under GST. She has paid GST of ₹ 12,500 on purchase and collected ₹ 14,750 on sale. What is the amount of ITC to be claimed? What is the amount of GST payable?
Solution:
Output tax = ₹ 14,750
Input tax = ₹ 12,500
∴ ITC for Nazama = ₹ 12,500.
∴ GST payable = Output tax – ITC
= 14750 – 12500
= ₹ 2250
∴ Amount of ITC to be claimed is ₹ 12,500 and amount of GST payable is ₹ 2250.

Question 3. Amir Enterprise purchased chocolate sauce bottles and paid GST of ₹ 3800. He sold those bottles to Akbari Bros, and collected GST of ₹ 4100. Mayank Food Corner purchased these bottles from Akbari Bros, and paid GST of ₹ 4500. Find the amount of GST payable at every stage of trading and hence find payable CGST and SGST.
Solution:
For Amir Enterprise:
Output tax = ₹ 4100
Input tax = ₹ 3800
ITC for Amir enterprise = ₹ 3800.
∴ GST payable = Output tax – ITC
= 4100 – 3800
= ₹ 300
For Akbari Bros.:
Output tax = ₹ 4500
Input tax = ₹ 4100
ITC for Akbari Bros = ₹ 4100.
GST payable = Output tax – ITC
= 4500 – 4100 = ₹ 400
∴ Statement of GST payable at every stage of trading:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 1

Question 4. Malik Gas Agency (Chandigarh Union Territory) purchased some gas cylinders for industrial use for ₹ 24,500, and sold them to the local customers for ₹ 26,500. Find the GST to be paid at the rate of 5% and hence the CGST and UTGST to be paid for this transaction, (for Union Territories there is UTGST instead of SGST.)
Solution:
For Malik Gas Agency:
Output tax = 5% of 26500
= \(\frac { 5 }{ 100 } \) × 26500
= ₹ 1325
Input tax = 5% of 24500
= \(\frac { 5 }{ 100 } \) × 24500
= ₹ 1225
ITC for Malik Gas Agency = ₹ 1225.
∴ GST payable = Output tax – ITC
= 1325 – 1225
= ₹ 100
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 2
∴ CGST = UTGST = ₹ 50
∴ The GST to be paid at the rate of 5% is ₹ 100 and hence, CGST and UTGST paid for the transaction is ₹ 50 each.

Question 5.
M/s Beauty Products paid 18% GST on cosmetics worth ₹ 6000 and sold to a customer for ₹ 10,000. What are the amounts of CGST and SGST shown in the tax invoice issued?
Solution:
Output tax = 18% of 10,000
= \(\frac { 18 }{ 100 } \) × 10,000
= ₹ 1800
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 3
∴ Amount of CGST and SGST shown in the tax invoice issued is ₹ 900 each.

Question 6.
Prepare Business to Consumer (B2C) tax invoice using given information. Write the name of the supplier, address, state, Date, Invoice number, GSTIN etc. as per your choice.
Supplier: M/s ______ Address _______ State _______ Date _______ Invoice No. _______ GSTIN _______
Particulars
Rate of Mobile Battery ₹ 200 Rate of GST 12% HSN 8507 1 PC
Rate of Headphone ₹750 Rate of GST 18% HSN 8518 1 Pc
Solution:
Rate of Mobile Battery = ₹200
CGST = 6% of 200
= \(\frac { 6 }{ 100 } \) × 200
= ₹ 12
∴ CGST = SGST = ₹ 12

Rate of Headphone = ₹ 750
COST = 9% of 750
= \(\frac { 9 }{ 100 } \) × 750
= ₹ 67.5
∴ CGST = SGST = ₹ 67.5
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 4

Question 7.
Prepare Business to Business (B2B) Tax Invoice as per the details given below, name of the supplier, address, Date etc. as per your choice.
Supplier – Name, Address, State, GSTIN, Invoice No., Date
Recipient – Name, Address, State, GSTIN,
Items:
i. Pencil boxes 100, HSN – 3924, Rate – ₹ 20, GST 12%
ii. Jigsaw Puzzles 50, HSN 9503, Rate – ₹ 100 GST 12%.
Solution:
Cost of 100 Pencil boxes
= 20 × 1oo
= ₹ 2000
CGST = 6% of 2000
= \(\frac { 6 }{ 100 } \) × 2000
= ₹ 120
∴ CGST = SGST = ₹ 120

Cost of 50 Jigsaw Puzzles = 100 × 50
= ₹ 5000
CGST = 6% of 5000
= \(\frac { 6 }{ 100 } \) × 5000
= ₹ 300
CGST – SGST = ₹ 300
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 5

Question 1.
Suppose a manufacturer sold a cycle for a taxable value of ₹ 4000 to the wholesaler. Wholesaler sold it to the retailer for ₹ 4800 (taxable value). Retailer sold it to a customer for ₹ 5200 (taxable value). Rate of GST is 12%. Complete the following activity to find the payable CGST and SGST at each stage of trading. (Textbook pg. no. 92)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 6
GST payable by manufacturer = ₹ 480
Output tax of wholesaler
= 12% of 4800 = \(\frac { 12 }{ 100 } \) × 4800 = ₹ 576
∴ GST payable by wholesaler
= Output tax – Input tax
= 576 – 480
= ₹ 96
Output tax of retailer = 12% of 5200
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.2 7

Question 2. Suppose in the month of July the output tax of a trader is equal to the input tax, then what is his payable GST?(Textbook pg. no. 93)
Answer:
Here, output tax is same as input tax.
∴ Trader payable GST will be zero.

Question 3.
Suppose in the month of July output tax of a trader is less than the input tax then how to compute his GST? (Textbook pg. no. 93)
Answer:
If output tax of a trader in a particular month is less than his input tax, then he won’t be able to get entire credit for his input tax. In such a case his balance credit will be carried forward and adjusted against the subsequent transactions.