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.

Maharashtra Board 10th Class Maths Part 1 Practice Set 4.3 Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Practice Set 4.3 Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Practice Set 4.3 Financial Planning Question 1. Complete the following table by writing suitable numbers and words.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 1
Solution:
i. Here, share is at par.
∴ MV = FV
∴ MV = ₹ 100

ii. Here, Premium = ₹ 500, MV = ₹ 575
∴ FV + Premium = MV
∴ FV + 500 = 575
∴ FV = 575 – 500
∴ FV = ₹ 75

iii. Here, FV = ₹ 10, MV = ₹ 5
∴ FV > MV
Share is at discount.
FV – Discount = MV
∴ 10 – Discount = 5
∴ 10 – 5 = Discount
₹ Discount = ₹ 5
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 2

Practice Set 4.3 Question 2. Mr. Amol purchased 50 shares of Face value ₹ 100 when the Market value of the share was ₹ 80. Company had given 20% dividend. Find the rate of return on investment.
Solution:
Here, MV = ₹ 80, FV = ₹ 100,
Number of shares = 50, Rate of dividend = 20%
∴ Sum invested = Number of shares × MV
= 50 × 80
= ₹ 4000

Dividend per share = 20% of FV
= \(\frac { 20 }{ 100 } \) × 100 = ₹ 20
∴ Total dividend of 50 shares = 50 × 20
= ₹ 1000
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 3
∴ Rate of return on investment is 25%.

Practice Set 4.3 Question 3.
Joseph purchased following shares, Find his total investment.
Company A : 200 shares, FV = ₹ 2, Premium = ₹ 18.
Company B : 45 shares, MV = ₹ 500
Company C : 1 share, MV = ₹ 10,540
Solution:
For company A:
FV = ₹ 2, premium = ₹ 18,
Number of shares = 200
∴ MV = FV+ Premium
= 2 + 18
= ₹ 20
Sum invested = Number of shares × MV
= 200 × 20
= ₹ 14000

For company B:
MV = ₹ 500, Number of shares = 45
Sum invested = Number of shares × MV
= 45 × 500 = ₹ 22,500

For company C:
MV = ₹ 10,540, Number of shares = 1
∴ Sum invested = Number of shares × MV
= 1 × 10540
= ₹ 10,540
∴ Total investment of Joseph
= Investment for company A + Investment for company B + Investment for company C
= 4000 + 22,500 + 10,540
= ₹ 37040
∴ Total investment done by Joseph is ₹ 37,040.

Practice Set 4.3 Class 7th Question 4.
Smt. Deshpande purchased shares of FV ₹ 5 at a premium of ₹ 20. How many shares will she get for ₹ 20,000?
Solution:
Here, FV = ₹ 5, Premium = ₹ 20,
Sum invested = ₹ 20,000
∴ MV = FV + Premium
= 5 + 20
∴ MV = ₹ 25
Now, sum invested = Number of shares × MV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 4
∴ Smt. Deshpande got 800 shares for ₹ 20,000.

Question 5.
Shri Shantilal has purchased 150 shares of FV ₹ 100, for MV of ₹ 120. Company has paid dividend at 7%. Find the rate of return on his investment.
Solution:
Here, FV = ₹ 100, MV = ₹ 120
Dividend = 7%, Number of shares = 150
∴ Sum invested = Number of shares × MV
= 150 × 120 = ₹ 18000
Dividend per share = 7% of FV
= \(\frac { 7 }{ 100 } \) × 100 = ₹ 7
∴ Total dividend of 150 shares
= 150 × 7 = ₹ 1050
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 5
∴ Rate of return on investment is 5.83%.

4.3 Class 10 Question 6. If the face value of both the shares is same, then which investment out of the following is more profitable?
Company A : dividend 16%, MV = ₹ 80,
Company B : dividend 20%, MV = ₹ 120.
Solution:
Let the face value of share be ₹ x.
For company A:
MV = ₹ 80, Dividend = 16%
Dividend = 16% of FV
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 6
∴ Rate of return of company A is more.
∴ Investment in company A is profitable.

Question 1.
Smita has invested ₹ 12,000 and purchased shares of FV ₹ 10 at a premium of ₹ 2. Find the number of shares she purchased. Complete the given activity to get the answer. (Textbook pg. no. 101.)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.3 7

Maharashtra Board 10th Class Maths Part 1 Practice Set 4.1 Solutions Chapter 4 Financial Planning

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 4.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 4 Financial Planning.

Practice Set 4.1 Algebra 10th Std Maths Part 1 Answers Chapter 4 Financial Planning

Financial Planning Class 10 Practice Set 4.1 Question 1.
‘Pawan Medical’ supplies medicines. On some medicines the rate of GST is 12%, then what is the rate of CGST and SGST?
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 1

Question 2.
On certain article if rate of CGST is 9% then what is the rate of SGST? and what is the rate of GST?
Solution:
Rate of CGST = 9%
But, rate of SGST = rate of CGST
∴ Rate of SGST = 9%
Rate of GST = Rate of SGST + Rate of CGST = 9% + 9%
∴ Rate of GST = 18%

Financial Planning Class 10 Question 3.
‘M/s. Real Paint’ sold 2 tins of lustre paint and taxable value of each tin is ₹ 2800. If the rate of GST is 28%, then find the amount of CGST and SGST charged in the tax invoice.
Solution:
Taxable value of 1 tin = ₹ 2800
∴ Taxable value of 2 tins = 2 × 2800
= ₹ 5600
Rate of GST = 28 %
∴ Rate of CGST = Rate of SGST = 14 %
CGST = 14% of taxable value 14
= \(\frac { 14 }{ 100 } \) × 5600
∴ CGST = ₹ 784
∴ SGST = CGST = ₹ 784
∴ The amount of CGST and SGST charged in the tax invoice is ₹ 784 each.

Question 4.
The taxable value of a wrist watch belt is 7 586. Rate of GST is 18%. Then what is price of the belt for the customer?
Solution:
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.

Question 5.
The total value (with GST) of a remote-controlled toy car is ₹ 1770. Rate of GST is 18% on toys. Find the taxable value, CGST and SGST for this toy-car.
Solution:
Let the amount of GST be ₹ x.
Total value of remote controlled toy car = ₹ 1770
∴ Taxable value of remote controlled toy car = ₹ (1770 – x)
Now, GST = 18% of taxable value
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 2
∴ Taxable value of toy car is ₹ 1500 and CGST and SGST for it is ₹ 135 each.

Question 6.
‘Tiptop Electronics’ supplied an AC of 1.5 ton to a company. Cost of the AC supplied is ₹ 51,200 (with GST). Rate of CGST on AC is 14%. Then find the following amounts as shown in the tax invoice of Tiptop Electronics.
i. Rate of SGST
ii. Rate of GST on AC
iii. Taxable value of AC
iv. Total amount of GST
v. Amount of CGST
vi. Amount of SGST
Solution:
i. Rate of CGST = 14%
But, Rate of SGST = Rate of CGST
∴ Rate of SGST = 14%

ii. Rate of GST on AC
= Rate of SGST + Rate of CGST
= 14% + 14% = 28%
∴ Rate of GST on AC is 28%.

iii. Let the cost (Taxable value) of AC be ₹ 100.
Given, GST = 28%
∴ The cost of AC with GST is ₹ 128.
For the total value of ₹ 128, the taxable value is ₹ 100.
For the total value of ₹ 51200, let the taxable value be ₹ x
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 3
∴ Taxable value of AC is ₹ 40,000.

iv. Total amount of GST = 28% of taxable value
= \(\frac { 28 }{ 100 } \) × 40000
= ₹ 11,200
∴ Total amount of GST is ₹ 11,200.

∴ Amount of CGST is ₹ 5600.

Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 4

vi. Amount of SGST = Amount of CGST
= ₹ 5600
Amount of SGST is ₹ 5600.

Question 7.
Prasad purchased a washing-machine from ‘Maharashtra Electronic Goods’. The discount of 5% was given on the printed price of ₹ 40,000. Rate of GST charged was 28%. Find the purchase price of washing machine. Also find the amount of CGST and SGST shown in the tax invoice.
Solution:
Printed price of washing machine = ₹ 40,000
Rate of discount = 5%
Amount of discount = 5% of printed price
= \(\frac { 5 }{ 100 } \) × 40000 = ₹ 2000
∴ Taxable value = Printed price – Discount
= 40000 – 2000 = ₹ 38000
Rate of GST = 28%
∴ Rate of CGST = 14% and
Rate of SGST = 14%
CGST = 14% of taxable value
= \(\frac { 14 }{ 100 } \) × 38000
∴ CGST = ₹ 5320
∴ CGST = SGST = ₹ 5320
Purchase price of washing machine
= Taxable value + CGST + SGST
= 38000 + 5320 + 5320
= ₹ 48,640
∴ Purchase price of washing machine is ₹ 48640. Amount of CGST and SGST in tax invoice is ₹ 5320 each.

Question 1.
Observe the given bill and fill in the boxes with the appropriate number. (Textbook pg. no. 82 and 83)
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 5
Solution:
i. Price of 1 kg of Pedhe is ₹ 400, therefore cost of 500 gm. of Pedhe is ₹ 200.
CGST for pedhe at the rate of 2.5% is ₹ [5] and SGST at the rate of [2.5| % is ₹ 5.00. It means that the rate of GST on Pedhe is 2.5% + 2.5% = 5% and hence the total GST is ₹ 10.
ii. The rate of GST on chocolate is [28] % and hence the total GST is ₹ [22.40]
iii. Rate of GST on Ice-cream is [18] %, hence the total cost of ice-cream is ₹ 236
iv. On butter CGST rate is [6] % and SGST rate is also [6] %. So GST rate on butter is [12]%.

Question 2.
Fill in the blanks with the help of given information for the table given below. (Textbook pg. no. 83)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 6

Question 3.
Make a list of ten things you need in your daily life. Find the GST rates with the help of GST rate chart given in the textbook, news papers or books, internet, or the bills of purchases. Verify these rates with the list prepared by your friends. (Textbook pg. no. 85)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 7

Question 4.
Make a list of ten services and their GST rates as per activity 1. (e.g. Railway and ST bus booking services etc.) You can also collect service bills and complete the given information (Textbook pg. no. 85)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 8

Question 5.
Complete the given table by writing remaining SAC and HSN codes with rates and add some more items in the list. (Textbook pg, no. 85)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Financial Planning Practice Set 4.1 9
[Note : The above Activities has many answers students may write answers other than the ones given]

Maharashtra Board 10th Class Maths Part 1 Practice Set 3.3 Solutions Chapter 3 Arithmetic Progression

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

Practice Set 3.3 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Arithmetic Progression Practice Set 3.3 Question 1.
First term and common difference of an A.P. are 6 and 3 respectively; find S27.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 1

Arithmetic Progression Class 10 Practice Set 3.3 Question 2.
Find the sum of first 123 even natural numbers.
Solution:
The even natural numbers are 2, 4, 6, 8,…
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, n = 123
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 2
∴ The sum of first 123 even natural numbers is 15252.

Practice Set 3.3 Question 3.
Find the sum of all even numbers between 1 and 350.
Solution:
The even numbers between 1 and 350 are 2, 4, 6,…, 348.
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, tn = 348
Since, tn = a + (n – 1)d
∴ 348 = 2 + (n – 1)2
∴ 348 – 2 = (n – 1)2
∴ 346 = (n – 1)2
∴ n – 1 = \(\frac { 346 }{ 2 } \)
∴ n – 1 = 173
∴ n = 173 + 1 = 174
Now, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ S174 = \(\frac { 174 }{ 2 } \) [2 (2) + (174 – 1)2]
= 87(4 + 173 × 2)
= 87(4 + 346)
= 87 × 350
∴ S174 = 30450
∴ The sum of all even numbers between 1 and 350 is 30450.

Arithmetic Progression 3.3 Question 4.
In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
Solution:
For an A.P., let a be the first term and d be the common difference.
t19 = 52, t38 = 128 …[Given]
Since, tn = a + (n – 1)d
∴ t19 = a + (19 – 1)d
∴ 52 = a + 18d
i.e. a + 18d = 52 …(i)
Also, t38 = a + (38 – 1)d
∴ 128 = a + 37d
i.e. a + 37d = 128 …(ii)
Adding equations (i) and (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 3

3 Arithmetic Progression Question 5.
Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 4

10th Algebra Practice Set 3.3 Question 6.
Sum of first 55 terms in an A.P. is 3300, find its 28th term.
Solution:
For an A.P., let a be the first term and d be the common difference.
S55 =3300 …[Given]
Since, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
∴ S55 = \(\frac { 55 }{ 2 } \) [2a + (55 – 1)d]
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 5

Arithmetic Practice Set Question 7.
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a – d, a, a + d.)
Solution:
Let the three consecutive terms in an A.P. be
a – d, a and a + d.
According to the first condition,
a – d + a + a + d = 27
∴ 3a = 27
∴ a = \(\frac { 27 }{ 3 } \)
∴ a = 9 ….(i)
According to the second condition,
(a – d) a (a + d) = 504
∴ a(a2 – d2) = 504
∴ 9(a2 – d2) = 504 …[From (i)]
∴ 9(81 – d2) = 504
∴ 81 – d2 = \(\frac { 504 }{ 9 } \)
∴ 81 – d2 = 56
∴ d2 = 81 – 56
∴ d2 = 25
Taking square root of both sides, we get
d = ± 5
When d = 5 and a =9,
a – d 9 – 5 = 4
a = 9
a + d 9 + 5 = 14
When d = -5 and a = 9,
a – d = 9 – (-5) = 9 + 5 = 14
a = 9
a + d = 9 – 5 = 4
∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.

10th Maths 1 Practice Set 3.3 Question 8.
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. (Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)
Solution:
Let the four consecutive terms in an A.P. be
a – d, a, a + d and a + 2d.
According to the first condition,
a – d + a + a + d + a + 2d = 12
∴ 4a + 2d =12
∴ 2(2a + d) = 12
∴ 2a + d = \(\frac { 12 }{ 2 } \)
∴ 2a + d = 6 …(i)
According to the second condition,
a + d + a + 2d = 14
∴ 2a + 3d = 14 …(ii)
Subtracting equation (i) from (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 6
∴ The four consecutive terms are -3,1,5 and 9.

Math 1 Practice Set 3.3 Question 9.
If the 9th term of an A.P. is zero, then show that the 29th term is twice the 19th term.
To prove: t29 = 2t19
Proof:
For an A.P., let a be the first term and d be the common difference.
t9 = 0 …[Given]
Since, tn = a + (n – 1)d
∴ t9 = a + (9 – 1)d
∴ 0 = a + 8d
∴ a = -8d …(i)
Also, t19 = a + (19 – 1)d
= a + 18d
= -8d + 18d … [From (i)]
∴ t19 = 10d …(ii)
and t29 = a + (29 – 1)d
= a + 28d
= -8d + 28d …[From (i)]
∴ t29 = 20d = 2(10d)
∴ t29 = 2(t19) … [From (ii)]
∴ The 29th term is twice the 19th term.

10 Class Math Part 1 Practice Set 3.3 Question 1.
Find the sum of all odd numbers from 1 to 150. (Textbook pg, no. 71)
Solution:
Odd numbers from 1 to 150 are 1,3, 5, 7,…, 149
Here, difference between any two consecutive terms is 2.
∴ It is an A.P.
∴ a = 1, d = 2
Let us find how many odd numbers are there from 1 to 150, i.e. find the value of n if
tn = 149
tn = a + (n – 1)d
∴ 149 = 1 + (n – 1)2
∴ 149 – 1 = (n – 1)2
∴ \(\frac { 148 }{ 2 } \) = n – 1
∴ 74 = n – 1
∴ n = 74 + 1 = 75

ii. Now, let’s find the sum of 75 numbers
i. e. 1 + 3 + 5 + 7 + … + 149
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.3 7

Maharashtra Board 10th Class Maths Part 1 Problem Set 2 Solutions Chapter 2 Quadratic Equations

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Problem Set 2 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.
Choose the correct answers for the following questions.

i. Which one is the quadratic equation?
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 1
Answer:
(B)

ii. Out of the following equations which one is not a quadratic equation?
(A) x2 + 4x = 11 + x2
(B) x = 4x
(C) 5x2 = 90
(D) 2x – x2 = x2 + 5
Answer:
(A)

iii. The roots of x2 + kx + k = 0 are real and equal, find k.
(A) 0
(B) 4
(C) 0 or 4
(D) 2
Answer:
(C)

iv. For √2 x2 – 5x + √2 = 0, find the value of the discriminant.
(A) -5
(B) 17
(C) √2
(D) 2 √2 – 5
Answer:
(B)

v. Which of the following quadratic equations has roots 3,5?
(A) x2 – 15x + 8 = 0
(B) x2 – 8x + 15 = 0
(C) x2 + 3x + 5 = 0
(D) x2 + 8x – 15 = 0
Answer:
(B)

vi. Out of the following equations, find the equation having the sum of its roots -5.
(A) 3x2 – 15x + 3 = 0
(B) x2 – 5x + 3 = 0
(C) x2 + 3x – 5 = 0
(D) 3x2 + 15x + 3 = 0
Answer:
(D)

vii. √5m2 – √5 m + √5 =0 which of the following statement is true for this given equation?
(A) Real and unequal roots
(B) Real and equal roots
(C) Roots are not real
(D) Three roots
Answer:
(C)

viii. One of the roots of equation x2 + mx – 5 = 0 is 2; find m.
(A) -2
(B) – \(\frac { 1 }{ 2 } \)
(C) \(\frac { 1 }{ 2 } \)
(D) 2
Answer:
(C)

Question 2.
Which of the following equations is quadratic
i. x2 + 2x + 11 = 0
ii. x2 – 2x + 5 = x2
iii. (x + 2)2 = 2x2
Solution:
i. The given equation is
x2 + 2x + 11 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 11 are real numbers and
a ≠ 0.
The given equation is a quadratic equation.

ii. The given equation is
x2 – 2x + 5 = x2
∴ x2 – x2 + 2x – 5 = 0
∴ 2x – 5 = 0
Here, x is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.

iii. The given equation is
(x + 2)2 = 2x2
∴ x2 + 4x + 4 = 2x2
∴ 2x2 – x2 – 4x – 4 = 0
∴ x2 – 4x – 4 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = -4, c = —4 are real numbers and
a ≠ 0.
∴ The given equation is a quadratic equation.

Question 3.
Find the value of discriminant for each of the following equations.
i. 2y2 – y + 2 = 0
ii. 5m2 – m = 0
iii. √5 x2 – x – √5 = 0
Solution:
i. 2y2 – y + 2 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 2, b = -1, c = 2
∴ b2 – 4ac = (-1)2 – 4 × 2 × 2
= 1 – 16
∴ b2 – 4ac = -15

ii. 5m2 – m = 0
∴ 5m2 – m + 0 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 5, b = -1, c = 0
∴ b2 – 4ac = (-1)2 – 4 × 5 × 0
= 1 – 0
∴ b2 – 4ac = 1

iii. √5x2 – x – √5 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = √5, b = -1, c = -√5
∴ b2 – 4ac = (-1)2 – 4 × √5 × √5
= 1 + 20
∴ b2 – 4ac = 21

Question 4.
One of the roots of quadratic equation 2x2 + kx – 2 = 0 is – 2, find k.
Solution:
-2 is one of the roots of the equation
2x2 + kx – 2 = 0.
∴ Putting x = – 2 in the given equation, we get
2(-2)2 + k(-2) -2 = 0
∴ 8 – 2k – 2 = 0
∴ 6 – 2k = 0
∴ 2k = 6
∴ k = \(\frac { 6 }{ 2 } \)
∴ k = 3

Question 5.
Two roots of quadratic equations are given; frame the equation.
i. 10 and -10
ii. 1 – 3√5 and 1 + 3√5
iii. 0 and 7
Solution:
i. Let α = 10 and β = -10
∴ α + β = 10 – 10 = 0
and α × p = 10 × -10 = -100
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 0x + (-100) = 0
∴ x2 – 100 = 0

ii. Let α = 1 – 3 √5 and β = 1 + 3 √5
α + β = 1 – 3 √5 + 1 + 3 √5 = 2
and α × β = (1 – 3√5) (1 + 3 √5)
= (1)2 – (3√5)2
= 1 – 45
= -44
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 2x – 44 = 0

iii. Let α = 0 and β = 7
∴ α + β = 0 + 7 = 7
and α × β = 0 × 7 = 0
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 7x + 0 = 0
∴ x2 – 7x = 0

Question 6.
Determine the nature of roots for each of the quadratic equation.
i. 3x2 – 5x + 7 = 0
ii. √3 x2 + √2 x – 2 √3 = 0
iii. m2 – 2m + 1 = 0
Solution:
i. 3x2 – 5x + 7 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 3, b = -5, c = 7
∴ ∆ = b2 – 4ac
= (-5)2 -4 × 3 × 7
= 25 – 84
∴ ∆ = -59
∴ ∆ < 0
∴ Roots of the given quadratic equation are not real.

ii. √3 x2 + √2 x – 2 √3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = √3 , b = √2, c = -2√3
∴ ∆ = b2 – 4ac
= (√2)2 – 4 × √3 × (-2√3)
= 2 + 24
∴ ∆ = 26
∴ ∆ > 0
∴ Roots of the given quadratic equation are real and unequal.

iii. m2 – 2m + 1 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 1, b = -2, c = 1
∴ ∆ = b2 – 4ac
= (-2)2 – 4 × 1 × 1
= 4 – 4
∴ ∆ = 0
∴ Roots of the given quadratic equation are real and equal

Question 7.
Solve the following quadratic equations.
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 2
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 3
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 4

ii. x2 – \(\frac { 3x }{ 10 } \) – \(\frac { 1 }{ 10 } \) = 0
∴ 10x2 – 3x – 1 = 0
…[Multiplying both sides by 10]
∴ 10x2 – 5x + 2x – 1 = 0
∴ 5x(2x – 1) + 1(2x – 1) = 0
∴ (2x – 1)(5x + 1) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴2x – 1 = 0 or 5x + 1 = 0
∴2x = 1 or 5x = -1
∴ x = –\(\frac { 1 }{ 2 } \) or x = \(\frac { -1 }{ 5 } \)
∴ The roots of the given quadratic equation are \(\frac { 1 }{ 2 } \) and \(\frac { -1 }{ 5 } \)

iii. (2x + 3)2 = 25
∴ (2x + 3)2 – 25 = 0
∴ (2x + 3)2 – (5)2 = 0
∴ (2x + 3 – 5) (2x + 3 + 5) = 0 ….. [∵ a2 – b2 = (a – b) (a + b)]
∴ (2x – 2) (2x + 8) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 2x – 2 = 0 or 2x + 8 = 0
∴ 2x = 2 or 2x = -8
∴ x = \(\frac { 2 }{ 2 } \) or x = \(\frac { -8 }{ 2 } \)
∴ x = 1 or x = -4
∴ The roots of the given quadratic equation are 1 and -4.

iv. m2 + 5m + 5 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 1, b = 5, c = 5
∴ b2 – 4ac = (5)2 – 4 × 1 × 5
= 25 – 20 = 5
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 5

v. 5m2 + 2m+1 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 5, b = 2, c = 1
∴ b2 – 4ac = (2)2 -4 × 5 × 1
= 4 – 20
= -16
∴ b2 – 4ac < 0
∴ Roots of the given quadratic equation are not real.

vi. x2 – 4x – 3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = -4, c = -3
∴ b2 – 4ac = (-4)2 – 4 × 1 × -3
= 16 + 12
= 28
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 6

Question 8.
Find m, if (m – 12) x2 + 2(m – 12) x + 2 = 0 has real and equal roots.
Solution:
(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) (m – 14) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ 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

Question 9.
The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
Solution:
Let α and β be the roots of the quadratic equation.
According to the given conditions,
α + β = 5 and α3 + β3 = 35
Now, (α + β)3 = α3 + 3α2β + 3αβ2 + β3
∴ (α + β)3 = α3 + β3 + 3αβ (α + β)
∴ (5)3 = 35 + 3αβ(5)
∴ 125 = 35 + 15αβ
∴ 125 – 35 = 15αβ
∴ 15αβ = 90
∴ αβ = \(\frac { 90 }{ 15 } \)
∴ αβ = 6
∴ The required quadratic equation is
x2 – (α + β)x + αβ = 0
∴ x2 – 5x + 6 = 0

Question 10.
Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation
2x2 + 2(p + q)x + p2 + q2 = 0.
Solution:
The given quadratic equation is
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 7
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 8
According to the given condition,
Roots of the required quadratic equation are
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 9

Question 11.
Mukund possesses ₹ 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.
Solution:
Let the amount Sagar possesses be ₹ x.
∴ the amount Mukund possesses = ₹ (x + 50)
According to the given condition,
x(x +50)= 15000
∴ x2 + 50x – 15000 = 0
∴ x2 + 150x- 100x- 15000 = 0
∴ x(x + 150) – 100(x + 150) = 0
∴ (x + 150)(x – 100) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 150 = 0 or x – 100 = 0
∴ x = -150 or x = 100
But, amount cannot be negative.
∴ x= 100 and x + 50 = 100 + 50 = 150
∴ The amount possessed by Sagar and Mukund are ₹ 100 and ₹150 respectively.

Question 12.
The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
Solution:
Let the numbers be x and y (x > y).
According to the given condition,
x2 – y2 = 120 …(i)
y2 = 2x …(ii)
Substituting y2 = 2x in equation (i), we get
x2 – 2x = 120
∴ x2 – 2x – 120 = 0
∴ x2 – 12x + 10x – 120 = 0
∴ x(x – 12) + 10(x – 12) = 0
∴ (x – 12)(x + 10) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 12 = 0 or x + 10 = 0
∴ x = 12 or x = -10
But x ≠ -10
as, y2 = 2x = 2(-10) = -20 …[Since, the square of number cannot be negative]
∴ x = 12
Smaller number = y2 =2x
∴ y2 = 2 × 12
∴ y2 = 24
∴ y = ± √24 …[Taking square root of both sides]
∴ The smaller number is √24 and greater number is 12 or the smaller number is – √24 and greater number is 12.

Question 13.
Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.
Solution:
Let the number of students be x.
Total number of oranges = 540
∴ the number of oranges each student gets = \(\frac { 540 }{ x } \)
If there were 30 more students, the total number of students = (x + 30) and the total number of oranges each student gets
= (\(\frac { 540 }{ x+30 } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 10Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 11
∴ 30 × 540 = 3x2 + 90 x
∴ 3x2 + 90x= 16200
∴ x2 + 30x – 5400 = 0
…[Dividing both sides by 3]
∴ x2 + 90x – 60x – 5400 = 0
∴ x(x + 90) – 60(x + 90) = 0
∴ (x + 90) (x – 60) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x + 90 = 0 or x – 60 = 0
∴ x = – 90 or x = 60
But, number of students cannot be negative,
x = 60
∴ The total number of students is 60.

Question 14.
Mr. Dinesh owns an rectangular agricultural farm at village Talvel. The length of the farm is 10 metre more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is \(\frac { 1 }{ 3 } \) of the breadth of the farm. The
area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and side of the pond.
Solution:
Let the breadth of the rectangular farm be x m.
∴ Length of rectangular farm = (2x + 10) m
Area of rectangular farm = Length × Breadth
= (2x + 10) × x
= (2x2+ 10x) sq. m
Now ,side of square shaped pond = \(\frac { x }{ 3 } \) m
∴ Area of square shaped pond = (side)2
= (\(\frac { x }{ 3 } \))2 m
= \(\frac { { x }^{ 2 } }{ 9 } \) m
According to the given condition,
Area of rectangular farm = 20 × Area of pond
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 11
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x = 0 or x – 45 = 0
x = 0 or x = 45
But, breadth of the rectangular farm cannot be zero,
∴ x = 45
Length of rectangular farm
= 2x + 10 = 2(45) + 10 = 100 m
Side of the pond = \(\frac { x }{ 3 } \) = \(\frac { 45 }{ 3 } \) = 15 m
∴ Length and breadth of the farm and the side of the pond are 100 m, 45 m and 15 m respectively.

Question 15.
A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?
Solution:
Let the larger tap take x hours to fill the tank completely.
∴ Part of tank filled by the larger tap in 1 hour = \(\frac { 1 }{ x } \)
Also, the smaller tap takes (x + 3) hours to fill the tank completely.
∴ Part of tank filled by the smaller tap in 1 hour = \(\frac { 1 }{ x+3 } \)
∴Part of tank filled by both the taps in 1 hour
= (\(\frac { 1 }{ x } \) + \(\frac { 1 }{ x+3 } \))
But, the tank gets filled in 2 hours by both the taps.
∴ Part of tank filled by both the taps in 1 hour = \(\frac { 1 }{ 2 } \)
According to the given condition,
Maharashtra Board Class 10 Maths Solutions Chapter 2 Quadratic Equations Problem Set 2 12
∴ 2(2x + 3) = x(x + 3)
∴ 4x + 6 = x2 + 3x
∴ x2 + 3x – 4x – 6 = 0
∴ x2 – x – 6 = 0
∴ x2 – 3x + 2x – 6 = 0
∴ x(x – 3) + 2(x – 3) = 0
∴ (x – 3)(x + 2) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
∴ x – 3 = 0 or x + 2 = 0
∴ x = 3 or x = -2
But, time cannot be negative.
∴ x = 3 and x + 3 = 3 + 3 = 6
∴ The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.

Maharashtra Board 10th Class Maths Part 1 Practice Set 1.1 Solutions Chapter 1 Linear Equations in Two Variables

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.1 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.
Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 …(i)
2x-3y=12 …(ii)
Solution:
5x + 3y = 9 …(i)
2x-3y=12 …(ii)
Add equations (i) and (ii).
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 1

Question 2.
Solve the following simultaneous equations.
i. 3a + 5b = 26; a + 5b = 22
ii. x + 7y = 10; 3x – 2y = 7
iii. 2x – 3y = 9; 2x + y = 13
iv. 5m – 3n = 19; m – 6n = -7
v. 5x + 2y = -3;x + 5y = 4
vi. \(\frac { 1 }{ 3 } \) x+ y = \(\frac { 10 }{ 3 } \) ; 2x + \(\frac { 1 }{ 4 } \) y = \(\frac { 11 }{ 4 } \)
vii. 99x + 101y = 499 ; 101x + 99y = 501
viii. 49x – 57y = 172; 57x – 49y = 252
Solution:
i. 3a + 5b = 26 …(i)
a + 5b = 22 …(ii)
Subtracting equation (ii) from (i), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 2
Substituting a = 2 in equation (ii), we get
2 + 5b = 22
∴ 5b = 22 – 2
∴ 5b = 20
∴ b = \(\frac { 20 }{ 5 } \) =4
∴ (a, b) = (2, 4) is the solution of the given simultaneous equations.

ii. x + 7y = 10
∴ x = 10 – 7y …(i)
3x – 2y = 7 …1(ii)
Substituting x = 10 – ly in equation (ii), we get
3 (10 – 7y) – 2y = 7
∴ 30 – 21y – 2y = 7
∴ -23y = 7 – 30
∴ -23y = -23
∴ y = \(\frac { -23 }{ -23 } \)
Substituting y = 1 in equation (i), we get
x = 10 – 7 (1)
= 10 – 7 = 3
∴ (x, y) = (3, 1) is the solution of the given simultaneous equations.

iii. 2x – 3y = 9 …(i)
2x + y = 13 …(ii)
Subtracting equation (ii) from (i), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 3
∴ (x, y) = (6, 1) is the solution of the given simultaneous equations.

iv. 5m – 3n = 19 …(i)
m – 6n = -7
∴ m = 6n – 7 …(ii)
Substituting m = 6n – 7 in equation (i), we get
5(6n – 7) – 3n = 19
∴ 30n – 35 – 3n = 19
∴ 27n = 19 + 35
∴ 27n = 54
∴ n = \(\frac { 54 }{ 27 } \) = 2
Substituting n = 2 in equation (ii), we get
m = 6(2) – 7
= 12 – 7 = 5
∴ (m, n) = (5, 2) is the solution of the given simultaneous equations.

v. 5x + 2y = -3 …(i)
x + 5y = 4
∴ x = 4 – 5y …(ii)
Substituting x = 4 – 5y in equation (i), we get
5(4 – 5y) + 2y = -3
∴ 20 – 25y + 2y = -3
∴ -23y = -3 – 20
∴ -23y = -23
∴ y = \(\frac { -23 }{ -23 } \) = 1
Substituting y = 1 in equation (ii), we get
x = 4 – 5(1)
= 4 – 5 = -1
∴ (x, y) = (-1, 1) is the solution of the given simultaneous equations.

Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 4
Substituting y = 3 in equation (i), we get
x = 10 – 3(3)
= 10 – 9 = 1
∴ (x, y) = (1, 3) is the solution of the given simultaneous equations.

vii. 99x + 101 y = 499 …(i)
101 x + 99y = 501 …(ii)
Adding equations (i) and (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 5
Substituting x = 3 in equation (iii), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

viii. 49x – 57y = 172 …(i)
57x – 49y = 252 …(ii)
Adding equations (i) and (ii), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 6
Substituting x = 7 in equation (iv), we get
7 + y = 10
∴ y = 10 – 7 = 3
∴ (x, y) = (7, 3) is the solution of the given simultaneous equations.

Complete the following table. (Textbook pg. no. 1)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 7

Question 1.
Solve: 3x+ 2y = 29; 5x – y = 18 (Textbook pg. no. 3)
Solution:
3x + 2y = 29 …(i)
and 5x- y = 18 …(ii)
Let’s solve the equations by eliminating ‘y’.
Fill suitably the boxes below.
Multiplying equation (ii) by 2, we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Ex 1.1 8

Maharashtra Board 10th Class Maths Part 2 Problem Set 4 Solutions Chapter 4 Geometric Constructions

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 4 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 4 Geometric Constructions.

Problem Set 4 Geometry 10th Std Maths Part 2 Answers Chapter 4 Geometric Constructions

Question 1.
Select the correct alternative for each of the following questions.

i. The number of tangents that can be drawn to a circle at a point on the circle is ______
(A) 3
(B) 2
(C) 1
(D) 0
Answer:
(C)

ii. The maximum number of tangents that can be drawn to a circle from a point outside it is ______
(A) 2
(B) 1
(C) one and only one
(D) 0
Answer:
(A)

iii. If ∆ABC ~ ∆PQR and \(\frac { AB }{ PQ } \) = \(\frac { 7 }{ 5 } \), then ______
(A) AABC is bigger.
(B) APQR is bigger.
(C) both triangles will be equal
(D) can not be decided
Answer:
(A)

Question 2.
Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.
Solution:
Analysis:
As shown in the figure, let P be a point in the exterior of circle at a distance of 5.7 cm.
Let PQ and PR be the tangents to the circle at points Q and R respectively.
∴ seg OQ ⊥ tangent PQ …[Tangent is perpendicular to radius]
∴ ∠OQP = 90°
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 1
∴ point Q is on the circle having OP as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly, point R also lies on the circle having OP as diameter.
∴ Points Q and R lie on the circle with OP as diameter.
On drawing a circle with OP as diameter, the points where it intersects the circle with centre O, will be the positions of points Q and R respectively.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 2

Question 3.
Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.
Solution:
Analysis:
As shown in the figure, line l is a tangent to the circle at point A.
seg BA is a chord of the circle and ∠BCA is an inscribed angle.
By tangent secant angle theorem,
∠BCA = ∠BAR
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 3
By converse of tangent secant angle theorem,
If we draw ∠BAR such that ∠BAR = ∠BCA, then ray AR (i.e. line l) is a tangent at point A.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 4

Question 4.
Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.
Solution:
Diameter of circle = 6.4 cm 6.4
Radius of circle = \(\frac { 6.4 }{ 2 } \) = 3.2 cm
Analysis:
As shown in the figure, let R be a point in the exterior of circle at a distance of 6.4 cm.
Let RQ and RS be the tangents to the circle at points Q and S respectively.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 5
∴ seg PQ ⊥tangent RQ …[Tangent is perpendicular to radius]
∴ ∠PQR = 90°
∴ point Q is on the circle having PR as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly,
Point S also lies on the circle having PR as diameter.
∴ Points Q and S lie on the circle with PR as diameter.
On drawing a circle with PR as diameter, the points where it intersects the circle with centre P, will be the positions of points Q and S respectively.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 6

Question 5.
Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.
Solution:
m(arc AB) = ∠APB = 100°
Analysis:
seg PA ⊥ line l
seg PB ⊥line m … [Tangent is perpendicular to radius]
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 7
The perpendicular to seg PA and seg PB at points A and B respectively will give the required tangents at A and B.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 8
Steps of construction:
i. With centre P, draw a circle of any radius and take any point A on it.
ii. Draw ray PA.
iii. Draw ray PB such that ∠APB = 100°.
iv. Draw line l ⊥ray PA at point A.
v. Draw line m ⊥ ray PB at point B.
Lines l and m are tangents at points A and B to the circle.

Question 6.
Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E – F – A and FA = 4.1 cm. Draw tangents to the circle from point A.
Solution:
Analysis:
Draw a circle of radius 3.4 cm
As shown in the figure, let A be a point in the exterior of circle at a distance of (3.4 + 4.1) = 7.5 cm.
Let AP and AQ be the tangents to the circle at points P and Q respectively.
∴ seg EP ⊥ tangent PA … [Tangent is perpendicular to radius]
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 9
∴ ∠EPA = 90°
∴ point P is on the circle having EA as diameter. …[Angle inscribed in a semicircle is a right angle]
Similarly, point Q also lies on the circle having EA as diameter.
∴ Points P and Q lie on the circle with EA as diameter.
On drawing a circle with EA as diameter, the points where it intersects the circle with centre E, will be the positions of points P and Q respectively.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 10

Question 7.
∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \). Construct ∆ABC and ∆LBN.
Solution:
Analysis:
As shown in the figure,
Let B – C – N and B – A – L.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 11
∆ABC ~ ∆LBN …[Given]
∴ ∠ABC ≅ ∠LBN …[Corresponding angles of similar triangles]
\(\frac { AB }{ LB } \) = \(\frac { BC }{ BN } \) = \(\frac { AC }{ LN } \) …(i)[Corresponding sides of similar triangles]
But. \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \) …(ii)[Given]
∴ \(\frac { AB }{ LB } \) = \(\frac { BC }{ BN } \) = \(\frac { AC }{ LN } \) = \(\frac { 4 }{ 7 } \) …[From(i)and(ii)]
∴ sides of ∆LBN are longer than corresponding sides of ∆ABC.
∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC.
So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B.
Now, point L is the point of intersection of ray BA and a line through N, parallel to AC.
∆LBN is the required triangle similar to ∆ABC.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 12
Steps of construction:
i. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.
ii. Taking convenient distance on compass, mark 7 points B1, B2, B3, B4, B5, B6 and B7 such that
BB1 = B1B2 = B2B3 B3= B44 = B4B5 = B5B6 = B6B7.
iii. Join B4C. Draw line parallel to B4C through B7 to intersects ray BC at N.
iv. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L.
∆LBN is the required triangle similar to ∆ABC.

Question 8.
Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If = \(\frac { YZ }{ YQ } \) = \(\frac { 6 }{ 5 } \) then construct ∆XYZ similar to ∆PYQ.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 13
Analysis:
As shown in the figure,
Let Y – Q – Z and Y – P – X.
∆XYZ ~ ∆PYQ …[Given]
∴ ∠XYZ ≅ ∠PYQ …[Corresponding angles of similar triangles]
\(\frac { XY }{ PY } \) = \(\frac { YZ }{ YQ } \) = \(\frac { XZ }{ PQ } \) …(i)[Corresponding sides of similar triangles]
But, \(\frac { YZ }{ YQ } \) = \(\frac { 6 }{ 5 } \) ,..(ii)[Given]
∴ \(\frac { XY }{ PY } \) = \(\frac { YZ }{ YQ } \) = \(\frac { XZ }{ PQ } \) = \(\frac { 6 }{ 5 } \) …[From (i) and (ii)]
∴ sides of ∆XYZ are longer than corresponding sides of ∆PYQ.
∴ If seg YQ is divided into 5 equal parts, then seg YZ will be 6 times each part of seg YQ.
So, if we construct ∆PYQ, point Z will be on side YQ, at a distance equal to 6 parts from Y.
Now, point X is the point of intersection of ray YP and a line through Z, parallel to PQ.
∆XYZ is the required triangle similar to ∆PYQ.
Maharashtra Board Class 10 Maths Solutions Chapter 4 Geometric Constructions Problem Set 4 14
Steps of construction:
i. Draw ∆ PYQ of given measure. Draw ray YT making an acute angle with side YQ.
ii. Taking convenient distance on compass, mark 6 points Y1, Y2, Y3, Y4, Y5 and Y6 such that
YY1 = Y1Y2 = Y2Y3 = Y3Y4 = Y4Y5 = Y5Y6.
iii. Join Y5Q. Draw line parallel to Y5Q through Y6 to intersects ray YQ at Z.
iv. Draw a line parallel to side PQ through Z. Name the point of intersection of this line and ray YP as X.
∆XYZ is the required triangle similar to ∆PYQ.