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Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 2 Insurance and Annuity Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Insurance and Annuity Ex 2.1

Question 1.
Find the premium on a property worth ₹ 25,00,000 at 3% if
(i) the property is fully insured
(ii) the property is insured for 80% of its value.
Solution:
Case-1
Property value = ₹ 25,00,000
Rate of Premium = 3%
Policy Value = ₹ 25,00,000
∴ Amount of Premium = 3% × 25,00,000 = ₹ 75,000
Case-2
Property Value = ₹ 25,00,000
Policy value = 80% × 25,00,000 = ₹ 20,00,000
Rate of Premium = 3%
∴ Amount of Premium = 3% × 20,00,000 = ₹ 60,000

Question 2.
A shop is valued at ₹ 3,60,000 for 75% of its value. If the rate of premium is 0.9%, find the premium paid by the owner of the shop. Also, find the agents commission if the agent gets commission at 15% of the premium.
Solution:
Property Value = ₹ 3,60,000
Policy Value = 75% × 3,60,000 = ₹ 2,70,000
Rate of Premium = 0.9%
∴ Amount of Premium = 0.9% × 2,70,000 = ₹ 2,430
Rate of Commission = 15%
∴ Amount of Commission = 15% × 2,430 = ₹ 364.5

Question 3.
A person insures his office valued at ₹ 5,00,000 for 80% of its value. Find the rate of premium if he pays ₹ 13,000 as premium. Also, find agent’s commission at 11%.
Solution:
Property Value = ₹ 5,00,000
Policy Value = 80% × 5,00,000 = ₹ 4,00,000
Amount of Premium = ₹ 13000
Let the rate of Premium be x%
Amount of premium = Rate × Policy Value
∴ 13000 = x% × 4,00,000
∴ \(\frac{13,000}{4,00,000}=\frac{x}{100}\)
∴ \(\frac{13,000 \times 100}{4,00,000}\) = x
∴ x = 3.25%
Rate of commission = 11%
∴ Amount of Commission = 11% × 13,000 = ₹ 1,430

Question 4.
A building is insured for 75% of its value. The annual premium at 0.70 percent amounts to ₹ 2625. If the building is damaged to the extent of 60% due to fire, how much can be claimed under the policy?
Solution:
Let the Property Value of building be ₹ x
Policy Value = 75% × x = 0.75x
Rate of Premium = 0.70%
Amount of Policy = Rate × Policy Value
2625 = 0.70% × 0.75x
\(\frac{2625}{0.75}\) = 0.70% × x
3520 = \(\frac{0.70}{100}\) × x
\(\frac{3500 \times 100}{0.70}\) = x
x = ₹ 5,00,000
∴ Damage = 60% × Property Value
= \(\frac{60}{100}\) × 5,00,000
= ₹ 3,00,000
∴ Policy Value = 0.75 × 3,00,000 = ₹ 2,25,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{2,25,000}{5,00,000}\) × 3,00,000
= ₹ 1,35,000

Question 5.
A stock worth ₹ 7,00,000 was insured for ₹ 4,50,000. Fire burnt stock worth ₹ 3,00,000 completely and damaged there remaining stock to the extent of 75% of its value. What amount can be claimed undertaken policy?
Solution:
Property Value = ₹ 7,00,000
Policy Value = ₹ 4,50,000
Complete Loss = 3,00,000
Partial loss = 75% × [7,00,000 – 3,00,000]
= \(\frac{75}{100}\) × 4,00,000
= ₹ 3,00,000
∴ Total loss = ₹ 3,00,000 + ₹ 3,00,000 = ₹ 6,00,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{4,50,000}{7,00,000}\) × 6,00,000
= ₹ 3,85,714.29

Question 6.
A cargo of rice was insured at 0.625 % to cover 80% of its value. The premium paid was ₹ 5,250. If the price of rice is ₹ 21 per kg. find the quantity of rice (in kg) in the cargo.
Solution:
Let Property Value be ₹ x
Policy Value = 80% × x = ₹ 0.8x
Rate of Policy = 0.625%
Amount of Premium = Rate × Policy value
∴ 5250 = 0.625% × 0.8x
∴ 5250 = 0.005x
∴ x = \(\frac{5250}{0.005}\)
∴ x = ₹ 10,50,000
Rate of Rice = ₹ 21/kg
∴ Quantity of Rice (in kg) = \(\frac{\text { Total value }}{\text { Rate of Rice }}\)
= \(\frac{10,50,000}{21}\)
= 50,000 kgs

Question 7.
60,000 articles costing ₹ 200 per dozen were insured against fire for ₹ 2,40,000. If 20% of the articles were burnt and 7,200 of the remaining articles were damaged to the extent of 80% of their value, find the amount that can be claimed under the policy.
Solution:
No of articles = 60,000
Cost of articles = ₹ 200/dozen
∴ Property of Value = \(\frac{60,000}{12}\) × 200 = ₹ 1o,oo,ooo
∴ Policy Value = ₹ 2,40,000
Complete Loss = 20% × 10,00,000 = ₹ 2,00,000
Partial loss = \(\frac{7200}{12}\) × 200 × 80% = ₹ 96,000
∴ Total loss = 2,00,000 + 96,000 = ₹ 2,96,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{2,40,000}{10,00,000}\) × 2,96,000
= ₹ 71,040

Question 8.
The rate of premium is 2% and other expenses are 0.075%. A cargo worth ₹ 3,50,100 is to be insured so that all its value and the cost of insurance will be recovered in the event of total loss.
Solution:
Let the Policy Value of Cargo be ₹ 100 which includes insurance and other expenses
∴ Property Value = 100 – [2 + 0.075] = ₹ 97.925
If Policy Value is ₹ 100, then Property Value is ₹ 97.925
If Property Value is ₹ 3,50,100
Then policy Value = \(\frac{100 \times 3,50,100}{97.925}\) = ₹ 3,57,518.51

Question 9.
A property worth ₹ 4,00,000 is insured with three companies. A, B, and C. The amounts insured with these companies are ₹ 1,60,000, ₹ 1,00,000 and ₹ 1,40,000 respectively. Find the amount recoverable from each company in the event of a loss to the extent of ₹ 9,000.
Solution:
Property Value = ₹ 4,00,000
Loss = ₹ 9,000
Total Value of Policies = 1,60,000 + 1,00,000 + 1,40,000 = ₹ 4,00,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
Claim of company A = \(\frac{1,60,000}{40,000}\) × 9,000 = ₹ 3,600
Claim of company B = \(\frac{1,00,000}{4,00,000}\) × 9,000 = ₹ 2,250
Claim of company C = \(\frac{1,40,000}{4,00,000}\) × 9,000 = ₹ 3,150

Question 10.
A car valued at ₹ 8,00,000 is insured for ₹ 5,00,000. The rate of premium is 5% less 20%. How much will the owner bear including the premium if value of the ear is reduced to 60% of its original value.
Solution:
Property Value = ₹ 8,00,000
Policy Value = ₹ 5,00,000
Rate of Premium = 5% less 20%
= 5% – 20% × 5%
= (5 – 1)%
= 4%
Amount of Premium = 4% × 5,00,000 = ₹ 20,000
Loss = [100 – 60]% × Property Value
= 40% × 8,00,000
= ₹ 3,20,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{5,00,000}{8,00,000}\) × 3,20,000
= ₹ 2,00,000
Loss bear by owner = Loss – claim + Premium
= 3,20,000 – 2,00,000 + 20,000
= ₹ 1,40,000

Question 11.
A shop and a godown worth ₹ 1,00,000 and ₹ 2,00,000 respectively were insured through an agent who was paid 12% of the total premium. If the shop was insured for 80% and the godown for 60% of their respective values, find the agent’s commission, given that the rate of premium was 0.80% less 20%.
Solution:
Rate of Premium = 0.80% Less 20%
= 0.80% – 20% × 0.80%
= (0.80 – 0.16)%
= 0.64%
For Shop
Property Value = ₹ 1,00,000
Policy Value = 80% × 1,00,000 = ₹ 80,000
Premium = 0.64% × 80,000 = ₹ 512
For Godown
Property Value = ₹ 2,00,000
Policy Value = 60% × 2,00,000 = ₹ 1,20,000
Premium = 0.64% × 1,20,000 = ₹ 768
∴ Total Premium = 512 + 768 = ₹ 1,280
Rate of Commission = 12%
∴ Agent Commission = 12% × 1,280 = ₹ 153.6

Question 12.
The rate of premium on a policy of ₹ 1,00,000 is ₹ 56 per thousand per annum. A rebate of ₹ 0.75 per thousand is permitted if the premium is paid annually. Find the net amount of premium payable if the policy holder pays the premium annually.
Solution:
Policy Value = ₹ 1,00,000
Rate of Premium = ₹ 56 per thousand p.a
Rate of Rebate = ₹ 0.75 per thousand p.a
Premium is paid annually
∴ Net rate of = 56 – 0.75 = ₹ 55.25 per thousand p.a.
∴ Net Amount ot Premium = \(\frac{1,00,000}{1000}\) × 55.25 = ₹ 5,525

Question 13.
A warehouse valued at ₹ 40,000 contains goods worth ₹ 2,40,000. The warehouse is insured against fire for ₹ 16,000 and the goods to the extent of 90% of their value. Goods worth ₹ 80,000 are completely destroyed, while the remaining goods are destroyed to 80% of their value due to a fire. The damage to the warehouse is to the extent of ₹ 8,000. Find the total amount that can be claimed.
Solution:
For Warehouse
Property Value = ₹ 40,000
Policy Value = ₹ 16,000
Loss = ₹ 8,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{16,000}{40,000}\) × 8,000
= ₹ 3,200
For Goods
Property Value = ₹ 2,40,000
Policy Value = 90% × 2,40,000 = ₹ 2,16,000
Complete Loss = 80,000
Partial Loss = 80% × (2,16,000 – 80,000)
= 80% × 1,36,000
= ₹ 1,08,800
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{2,16,000}{24,000}\) × 1,08,800
= ₹ 97,920
∴ Total Claim = 3,200 + 97,920 = ₹ 1,01,120

Question 14.
A person takes a life policy for ₹ 2,00,000 for a period of 20 years. He pays premium for 10 years during which bonus was declared at an average rate of ₹ 20 per year per thousand. Find the paid up value of the policy if he discontinuous paying premium after 10 years.
Solution:
Policy Value = ₹ 2,00,000
Rate of Bonus = ₹ 20 Per thousand p.a.
Total Bonus = \(\frac{2,00,000 \times 20}{1,000}\) = ₹ 4,000
∴ Bonus for 10 years = 4,000 × 10 = ₹ 40,000
Period of Policy = 20 years
∴ Amount of Premium = \(\frac{2,00,000}{20}\) = ₹ 10,000 p.a.
∴ Total Premium for 10 years = 10,000 × 10 = ₹ 1,00,000
∴ Paid up Value of Policy = Total premium + Total Bonus
= 1,00,000 + 40,000
= ₹ 1,40,000

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Commission, Brokerage and Discount Miscellaneous Exercise 1

(I) Choose the correct alternative.

Question 1.
An agent who gives a guarantee to his principal that the party will pay the sale price of goods is called
(a) Auctioneer
(b) Del Credere Agent
(c) Factor
(d) Broker
Answer:
(b) Del Credere Agent

Question 2.
An agent who is given the possession of goods to be sold is known as
(a) Factor
(b) Broker
(c) Auctioneer
(d) Del Credere Agent
Answer:
(a) Factor

Question 3.
The date on which the period of the bill expires is called
(a) Legal Due Date
(b) Grace Date
(c) Nominal Due Date
(d) Date of Drawing
Answer:
(c) Nominal Due Date

Question 4.
The payment date after adding 3 days of grace period is known as
(a) The legal due date
(b) The nominal due date
(c) Days of grace
(d) Date of drawing
Answer:
(a) The legal due date

Question 5.
The sum due is also called as
(a) Face value
(b) Present value
(c) Cash value
(d) True discount
Answer:
(a) Face value

Question 6.
P is the abbreviation of
(a) Face value
(b) Present worth
(c) Cash value
(d) True discount
Answer:
(b) Present worth

Question 7.
Banker’s gain is the simple interest on
(a) Banker’s discount
(b) Face Value
(c) Cash value
(d) True discount
Answer:
(d) True discount

Question 8.
The marked price is also called as
(a) Cost price
(b) Selling price
(c) List price
(d) Invoice price
Answer:
(c) List price

Question 9.
When only one discount is given then
(a) List price = Invoice price
(b) Invoice price = Net selling price
(c) Invoice price = Cost price
(d) Cost price = Net selling price
Answer:
(b) Invoice price = Net selling price

Question 10.
The difference between the face value and present worth is called
(a) Banker’s discount
(b) True discount
(c) Banker’s gain
(d) Cash value
Answer:
(b) True discount

(II) Fill in the blanks.

Question 1.
A person who draws the bill is called ____________
Answer:
Drawee

Question 2.
An ____________ is an agent who sells the goods by auction.
Answer:
Auctioneer

Question 3.
Trade discount is allowed on the ____________ price.
Answer:
Catalogue/List

Question 4.
The banker’s discount is also called ____________.
Answer:
Commercial Discount

Question 5.
The banker’s discount is always ____________ than the true discount.
Answer:
higher

Question 6.
The diffrence between the banker’s discount and the true discount is called ____________.
Answer:
bankers gain

Question 7.
The date by which the buyer is legally allowed to pay the amount is known as ____________.
Answer:
legal due date

Question 8.
A ____________ is an agent who brings together the buyer and the seller.
Answer:
broker

Question 9.
If buyer is allowed both trade and cash discounts, ____________ discount is fist calculated on ____________ price.
Answer:
Trade, Catalogue/List

Question 10.
____________ = List price (catalogue Price) – Trade Discount.
Answer:
Invoice Price

(III) State whether each of the following is True or False.

Question 1.
A broker is an agent who gives a guarantee to the seller that the buyer will pay the sale price of goods.
Answer:
False

Question 2.
A cash discount is allowed on the list price.
Answer:
False

Question 3.
Trade discount is allowed on catalogue price.
Answer:
True

Question 4.
The buyer is legally allowed 6 days grace period.
Answer:
False

Question 5.
The date on which the period of the bill expires is called the nominal due date.
Answer:
True

Question 6.
The difference between the banker’s discount and true discount is called sum due.
Answer:
False

Question 7.
The banker’s discount is always lower than the true discount.
Answer:
False

Question 8.
The banker’s discount is also called a commercial discount.
Answer:
True

Question 9.
In general cash, the discount is more than trade discount.
Answer:
False

Question 10.
A person can get both, trade discount and a cash discount.
Answer:
True

(IV) Solve the following problems.

Question 1.
A salesman gets a commission of 6.5% on the total sales made by him and a bonus of 1% on sales over ₹ 50,000. Find his total income on a turnover of ₹ 75,000.
Solution:
Rate of commission = 6.5% on the total sales
∴ Commission on a turnover of ₹ 75,000
= \(\frac{6.5}{100}\) × 75,000
= ₹ 4,875
Rate of bonus = 1% on sales over ₹ 50,000
∴ Amount of bonus = \(\frac{1}{100}\) × (75,000 – 50,000) = ₹ 250
∴ Total income of the sales man = ₹ 4,875 + ₹ 250 = ₹ 5,125

Question 2.
A shop is sold at 30% profit, the amount of brokerage at the rate of \(\frac{3}{4}\)% amounts to ₹ 73,125. Find the cost of the shop.
Solution:
Rate of brokerage = \(\frac{3}{4}\)%
Amount of brokerage = ₹ 73,125
Let the selling price of the shop be ₹ 100 then the brokerage = ₹ \(\frac{3}{4}\)
Thus, if the amount of brokerage is ₹ \(\frac{3}{4}\) then the selling price of the shop is ₹ 100
If the amount of brokerage is ₹ 73,125, then the selling price of the shop is = 73125 × \(\frac{4}{3}\) × 100 = ₹ 97,50,000
The shop is sold at 30% profit
∴ If the cost of the shop is ₹ 100, then it is sold at ₹ 130
Thus, if the shop is sold at ₹ 130, then its cost price is ₹ 100
If the shop is sold at ₹ 97,50,000 then its cost price is = \(\frac{97,50,000 \times 100}{130}\) = ₹ 75,00,000
Then, the cost of the shop is ₹ 75,00,000

Question 3.
A merchant gives 5% commission and 1.5% delcredere to his agents. If the agent sells goods worth ₹ 30,600 how much does he get? How much does the merchant receive?
Solution:
Rate of commission = 5%
Total sales = ₹ 30,600
Amount of commission = \(\frac{5}{100}\) × 30,600
Rate of delcredere = 1.5%
= \(\frac{1.5}{100}\) × 30,600
= ₹ 459
Thus, the agents gets 1,530 + 459 = ₹ 1,989
And the merchant receives = 30,600 – 1,989 = ₹ 28,611

Question 4.
After deducting commission at 7\(\frac{1}{2}\)% on first ₹ 50,000 and 5% on the balance of sales made by him, an agent remits ₹ 93,750 to his principal. Find the value of goods sold by him.
Solution:
Rate of commission = 7\(\frac{1}{2}\)% on first ₹ 50,000
= \(\frac{7.5}{100}\) × 50,000
= ₹ 3,750
Let the total sales be ₹ x
Rate of commission on the balance sales = 5%
Commission on the balance sales = \(\frac{5}{100}\) × (x – 50000) = \(\frac{x}{20}\) – 2,500
Total commission = 3750 + \(\frac{x}{20}\) – 2,500 = \(\frac{x}{20}\) + 1,250
Now, the amount to be remitted to the principal = Value of goods sold – Commission of the agent
= x – (\(\frac{x}{20}\) + 1250)
= \(\frac{19x}{20}\) – 1250
The agents remits ₹ 93,750 to his principal
∴ \(\frac{19x}{20}\) – 1,250 = 93,750
∴ \(\frac{19x}{20}\) = 95,000
∴ x = ₹ 1,00,000
Thus, the value of the goods sold by the agent is ₹ 1,00,000

Question 5.
The present worth of ₹ 11,660 due 9 months hence is ₹ 11,000. Find the rate of interest.
Solution:
Given, PW = ₹ 11,000, SD = ₹ 11,660
n = \(\frac{9}{12}\) year = \(\frac{3}{4}\) year
We have,

∴ The rate of interest is 8% p.a.

Question 6.
An article is marked at ₹ 800, a trader allows a discount of 2.5% and gains 20% on the cost. Find the cost price of the article?
Solution:
Marked price of the article = ₹ 800
Rate of discount = 2.5%
Amount of discount = \(\frac{2.5}{100}\) × 800 = ₹ 20
∴ Selling price of the article = 800 – 20 = ₹ 780
Now, given, gain = 20%
Let cost price of the article be ₹ 100, then
The selling price of the article is ₹ 120
Thus if cost price of the articles is ₹ x
Then the selling price is ₹ 780
∴ x = \(\frac{780 \times 100}{120}\)
∴ x = 650
∴ Cost price of the article is ₹ 650

Question 7.
A salesman is paid a fixed monthly salary plus commission on the sales. If on sale of ₹ 96,000 and ₹ 1,08,000 in two successive months he receives in all ₹ 17,600 and ₹ 18,800 respectively. Find his monthly salary and rate of commission paid to him.
Solution:
Let the monthly salary of the salesman be ₹ x
And the rate of commission be y%
Income = monthly salary + commission on the sales
17600 = x + \(\frac{y}{100}\) × 96,000
∴ 17600 = x + 960y ………(1)
and 18800 = x + \(\frac{y}{100}\) × 108000
∴ 18,800 = x + 1080y ………(2)
Subtracting equation (1) from equation (2), we get
1,200 = 120y
∴ y = 10
Substituting y = 10 in (1), we get
17,600 = x + 960(10)
∴ x = 17,600 – 9,600 = 8,000
∴ Salary of the salesman = ₹ 8,000
Rate of commission = 10%

Question 8.
A merchant buys some mixers at a 15% discount on catalogue price. The catalogue price is ₹ 5,500 per price of the mixer. The freight charges amount to 2\(\frac{1}{2}\)% on the catalogue price. The merchant sells each mixer at a 5% discount on the catalogue price. His net profit is ₹ 41,250, Find the number of mixers.
Solution:
Catalogue price of a mixer = ₹ 5,500
Trade discount = 15% on catalogue price
= \(\frac{15}{100}\) × 5,500
= ₹ 825
Freight charges = 2\(\frac{1}{2}\)% of the catalogue price
= \(\frac{5}{2} \times \frac{1}{100} \times 5,500\)
= ₹ 137.5
∴ Cost price of a mixer for the merchant = 5,500 – 825 + 137.5 = 4,812,5
Catalogue price = ₹ 5,500
Rate of discount = 5%
∴ Selling price of one mixer = 5500 – \(\frac{5}{100}\) × 5,500 = ₹ 5,225
∴ Profit on one mixer = 5,225 – 4,812.5 = ₹ 412.5
Now, total profit = ₹ 41,250
∴ Number of mixers = \(\frac{41,250}{412.5}\) = 100
Thus the number of mixers is 100.

Question 9.
A bill is drawn for ₹ 7,000 on 3rd May for 3 months and is discounted on 25th May at 5.5% Find the present worth.
Solution:
Face value of the bill = ₹ 7,000
Date of drawing = 3rd May
Period = 3 months
Normal due date = 3rd August
Legal due date = 6th August
Rate of interest = 5.5%
Date of discounting = 25th May
Unexpired period (number of days from date of discounting to legal due date)

∴ Bankers discount = 7,000 × \(\frac{73}{365} \times \frac{5.5}{100}\) = ₹ 77
Also PW = SD – BD
= 7,000 – 77
= ₹ 6,923
∴ Present worth is ₹ 6,923

Question 10.
A bill was drawn on 14th April 2005 for ₹ 3,500 and was discounted on 6th July 2005 at 5% per annum. The banker paid ₹ 3,465 for the bill. Find the period of the bill.
Solution:
Face value of the bill = ₹ 3,500
Date of drawing = 14/04/2005
Date of discount = 06/07/2005
Rate of interest = 5%
Cash value = ₹ 3,465
Bankers discount = Face value – Cash value
= 3,500 – 3,465
= ₹ 35
Let the unexpired days be n days
∴ BD = \(\frac{\mathrm{FV} \times n \times r}{365 \times 100}\)
∴ 35 = \(\frac{3,500 \times n \times 5}{365 \times 100}\)
∴ n = 73 days
Thus, legal due date is 73 days from the date of discounting

∴ Legal due date = 17/09/2005
∴ Nominal due date = 14/09/2005
∴ The period of the bill is 5 months

Question 11.
The difference between true discount and banker’s discount on 6 months hence at 4% p.a. is ₹ 80. Find the true discount, banker’s discount, and amount of the bill.
Solution:
BG = BD – TD
∴ BG = ₹ 80
Also BG = \(\frac{\mathrm{TD} \times n \times r}{100}\)
∴ 80 = \(\frac{\mathrm{TD} \times 6 \times 4}{12 \times 100}\)
∴ TD = \(\frac{80 \times 100}{2}\)
∴ TD = ₹ 4,000
Now BD = TD + BG
= 4,000 + 80
= ₹ 4,080
Also, BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 4,080 = \(\frac{\mathrm{FV} \times 6 \times 4}{12 \times 100}\)
∴ FV = \(\frac{4,080 \times 100}{2}\)
∴ FV = ₹ 2,04,000
Amount of the bill = ₹ 2,04,000

Question 12.
A manufacturer makes a clear profit of 30% on the cost after allowing a 35% discount. If the cost of production rises by 20%, by what percentage should he reduce the rate of discount so as to make the same rate of profit keeping his list prices unaltered.
Solution:
Rate of discount = 35%
Let the list price be ₹ 100.
Then discount at 35% = ₹ 35
∴ Net selling price = 100 – 35 = ₹ 65 ……..(1)
The manufacturer makes a clear profit of 30% on the cost after allowing a 35% Discount.
Let the cost be ₹ 100.
Then selling price at 30% profit is 100 + 30 = ₹ 130.
Thus, if the net selling price is ₹ 130, then the cost price is ₹ 100.
But, the net selling price is ₹ 130, then the cost price is ₹ 65 ……[from (1)]
∴ The cost price is \(\frac{65}{130} \times 100\) = ₹ 50
Hence, we have,

Now, the cost of production has increased by 20%.
Let the old cost price be ₹ 100.
∴ The new cost price is ₹ 120.
But, the old cost price is ₹ 50.
∴ The new cost price is = \(\frac{50}{100} \times 120\) = ₹ 60.
The old net price is ₹ 65.
Now 20% of ₹ 65 = \(\frac{20}{100} \times 65\) = ₹ 13
∴ New net price = 65 + 13 = ₹ 78
Hence, we have

Now, 100 – 78 = ₹ 22
Thus, the rate of discount should be reduced by 22%, The original rate of discount is 35%.
Hence, the reduction in discount should be (35 – 22)% = 13%
so as to make the same rate of profit, keeping the list price unaltered.

Question 13.
A trader offers a 25% discount on the catalogue price of the radio and yet makes a 20% profit. If he gains ₹ 160 per radio, what must be the catalogue price of the radio?
Solution:
Rate of discount = 25% on the catalogue price of a radio.
Let the catalogue price of the radio be ₹ 100.
Then, the discount on a radio = ₹ 25.
Net selling price = 100 – 25 = ₹ 75.
He makes a profit of 20%.
Let the cost price be ₹ 100.
Then, at 20% profit, net selling price = ₹ 120.
Thus, if net SP is ₹ 120, then cost price is ₹ 100.
But, the net SP is ₹ 75.
∴ The cost price is \(\frac{75}{120}\) × 100 = \(\frac{750}{12}\) = ₹ 62.50
∴ Profit on a radio set = 75 – 62.5 = ₹ 12.50
Thus, if the profit on a radio set is ₹ 12.50 then its catalogue price is ₹ 100.
But the profit on a radio set is ₹ 160.
∴ The catalogue price of radio = \(\frac{160}{12.50}\) × 100
= 12.80 × 100
= ₹ 1,280
∴ Thus, the catalogue price of the radio is ₹ 1280

Question 14.
A bill of ₹ 4,800 was drawn on 9th March 2006 at 6 months and was discounted on 19th April 2006 for 6\(\frac{1}{4}\)% p.a. How much does the banker charge and how much does the holder receive?
Solution:
Face value of the bill = ₹ 4.800
Date of drawing = 09/03/2006
Period of the bill = 6 months
Normal due date = 09/09/2006
Legal due date = 12/09/2006
Rate of discount = 6\(\frac{1}{4}\)% = 6.25%
Now, for the unexpired

Thus the banker charges ₹ 120
Amount received by the holder = 4,800 – 120 = ₹ 4,680

Question 15.
A bill of ₹ 65,700 drawn on July 10 for 6 months was discounted for ₹ 65,160 at 5% p.a. On what day was the bill discounted?
Solution:
BD = FV – Cash value
= 65,700 – 65,160
= ₹ 540
Let the unexpired days be x days
BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 540 = \(\frac{65,700 \times x \times 5}{365 \times 100}\)
∴ x = 60 days
The unexpired days = 60 days
Date-of drawing = 10th July
Period of the bill = 6 months
Nominal due date = 10th January (next year)
Legal due date = 13th January (next year)
Then the date of discount is 60 days before, the legal due date

∴ The date of discounting is 14th November

Question 16.
An agent sold a car and charged a 3% commission on the sale value. If the owner of the car received ₹ 48,500, find the sale value of the car. If the agent charged 2% from the buyer, find his total remuneration.
Solution:
Let the sale value of the car be ₹ x
Rate of commission of the agent = 3%
Since the owner received ₹ 48,500 after agent has charged his commission
x – \(\frac{3 x}{100}\) = 48500
∴ \(\frac{97 x}{100}\) = 48500
∴ x = \(\frac{48,500 \times 100}{97}\)
∴ x = ₹ 50,000
∴ Sale value of the car = ₹ 50,000
Against commission received from the owner = \(\frac{3}{100}\) × 50,000 = ₹ 1500
Against commission received from the buyer = \(\frac{2}{100}\) × 50,000 = ₹ 1000
∴ Agents total remuneration = 1,500 + 1,000 = ₹ 2,500

Question 17.
An agent is paid a commission of 4% on cash sales and 6% on credit sales made by him. If on the sale of ₹ 51,000 the agent claims a total commission of ₹ 2,700, find the sales made by him for cash and on credit.
Solution:
Total sales = ₹ 51,000
Let eash sales be ₹ x
∴ Credit sales = ₹ (51,000 – x)
Agent’s commission on cash sales = 4%
= \(\frac{4}{100}\) × x
= \(\frac{4x}{100}\)
Commission on credit sales = 6%
= \(\frac{6}{100}\)(51,000 – x)
Given total commission = ₹ 2,700

∴ Cash sales = ₹ 18,000
∴ Credit sales = 51,000 – 18,000 = ₹ 33,000

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Commission, Brokerage and Discount Ex 1.2

Question 1.
What is the present worth of a sum of ₹ 10,920 due six months hence at 8% p.a simple interest?
Solution:
Given, SD = ₹ 10,920
n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
r = 8%
We have,

Thus the present worth is ₹ 10,500

Question 2.
What is the sum due of ₹ 8,000 due 4 months at 12.5% simple interest?
Solution:
Given, PW = ₹ 8,000, n = \(\frac{4}{12}\) year = \(\frac{1}{3}\) year, r = 12.5%
We have,

Thus, the sum due is ₹ 8,333.33

Question 3.
The true discount on the sum due 8 months hence at 12% p.a. is ₹ 560. Find the sum due and present worth of the bill.
Solution:
Given, TD = ₹ 560, n = \(\frac{8}{12}\) year = \(\frac{2}{3}\) year, r = 12%
We have,
TD = \(\frac{\mathrm{PW} \times n \times r}{100}\)
∴ 560 = \(\frac{\mathrm{PW} \times 2 \times 12}{3 \times 100}\)
∴ PW = 560 × \(\frac{25}{2}\) = ₹ 7,000
Now, SD = PW + TD
= 7,000 + 560
= ₹ 7,560

Question 4.
The true discount on a sum is \(\frac{3}{8}\) of the sum due at 12% p.a. Find the period of the bill.
Solution:

8 × n × 12 = 3(100 + n × 12)
96n = 300 + 36n
60n = 300
∴ n = 5
∴ Period of the bill = 5 years.

Question 5.
20 copies of a book can be purchased for a certain sum payable at the end of 6 months and 21 copies for the same sum in ready cash. Find the rate of interest.
Solution:
Given, n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
Let the sum payable be ₹ x
Let the rate of interest be r%
According to given condition,
PW of one book = \(\frac{x}{21}\)
SD of one book = \(\frac{x}{20}\)

Thus, the rate of interest is 10%.

Question 6.
Find the true discount, Banker’s discount, and Banker’s gain on a bill of ₹ 4,240 due 6 months hence at 9% p.a.
Solution:

And, Banker’s Gain (BG) = BD – TD
= 190.80 – 182.58
= ₹ 8.22

Question 7.
The true discount on a bill is ₹ 2,200 and bankers discount is ₹ 2,310. If the bill is due 10 months, hence, find the rate of interest.
Solution:
Given, TD = ₹ 2,200, BD = ₹ 2,310
n = \(\frac{10}{12}=\frac{5}{6}\) year

∴ \(\frac{r}{120}=\frac{1}{20}\)
∴ r = 6%
Thus, rate of interest is 6%

Question 8.
A bill of ₹ 6,395 drawn on 19th January 2015 for 8 months was discounted on 28th February 2015 at 8% p.a. interest. What is the banker’s discount? What is the cash value of the bill?
Solution:
Face value = ₹ 6,395
Date of drawing = 19/01/2015
Period of the bill = 8 months
Nominal Due date = 19/09/2015
Legal due date = 22/09/2015
Date of discounting = 28/02/2015
Now, the unexpired period = Legal due date – Date of discounting
= 22/09/2015 – 28/02/2015
= days (as shown below)

Cash Value = FV – BD
= 6,395 – 313.12
= ₹ 6,621.38

Question 9.
A bill of ₹ 8,000 drawn on 5th January 1998 for 8 months was discounted for ₹ 7,680 on a certain date. Find the date on which it was discounted at 10% p.a.
Solution:
Bankers discount (BD) = FV – cash value
= 8,000 – 7,680
= ₹ 320
Let the unexpired period be x days
∴ BD = \(\frac{\mathrm{FV} \times x \times r}{365 \times 100}\)
∴ 320 = \(\frac{8,000 \times x \times 10}{365 \times 100}\)
∴ x = 146 days
∴ The unexpired days = 146 days
Date of drawing = 05/01/1998
Period of bill = 8 months
Nominal due date = 05/09/1998
Legal due date = 08/09/1998
Thus, the date of discounting is 146 days before the legal due date

∴ Date of discounting of the bill is 15th April 1998

Question 10.
A bill drawn on 5th June for 6 months was discounted at the rate of 5% p.a. on 19th October. If the cash value of the bill is ₹ 43,500, find the face value of the bill.
Solution:
Date of drawing = 5th June
Period of bill = 6 months
Nominal due date = 5th December
Legal due date = 8th December
Date of discounting = 19th October
Rate of interest = 5% p.a.
Let the face value of the bill be ₹ x
The unexpired period

Question 11.
A bill was drawn on 14th April for ₹ 7,000 and was discounted on 6th July at 5% p.a. The Banker paid ₹ 6,930 for the bill. Find the period of the bill.
Solution:
Face value = ₹ 7,000, cash value = ₹ 6,930
∴ Banker’s discount = 7,000 – 6,930 = ₹ 70
Date of drawing = 14/04
Date of discounting = 06/07
Rate of interest = 5%
Let the unexpired period = x days
∴ BD = \(\frac{7,000 \times x \times 5}{365 \times 100}\)
∴ 70 = \(\frac{70 \times x}{73}\)
∴ x = 73 days
∴ Legal due date of the bill is 73 days after the date of discounting.

∴ Legal due date = 17/09
∴ Nominal due date = 14/09
∴ Period of the bill = 5 months

Question 12.
If the difference between true discount and banker’s discount on a sum due 4 months hence is ₹ 20. Find true discount, banker’s discount and amount of bill, the rate of simple interest charged is 5% p.a.
Solution:
Banker’s gain (BG) = Banker’s discount (BD) – True Discount (TD)
∴ BG = ₹ 20
Also, BG = \(\frac{\mathrm{TD} \times n \times r}{100}\)
∴ 20 = \(\frac{\mathrm{TD} \times 4 \times 5}{12 \times 100}\)
∴ 20 = \(\frac{\mathrm{TD}}{60}\)
∴ TD = ₹ 1200
Now, BD = BG + TD
= 20 + 1,200
= ₹ 1,220
Also, BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 1,220 = \(\frac{\mathrm{FV} \times 4 \times 5}{12 \times 100}\)
∴ FV = 1,200 × 60 = ₹ 73,200
∴ Amounting the bill = ₹ 73,200

Question 13.
A bill of ₹ 51,000 was drawn on 18th February 2010 for 9 months. It was encashed on 28th June 2010 at 5% p.a. Calculate the banker’s gain and true discount.
Solution:
Face Value = ₹ 51,000
Date of drawing = 18/02/2010
Period of the bill = 9 months
Nominal due date = 18/11/2010
Legal due date = 21/11/2010
Date of discounting = 28/06/2010
Unexpired period

∴ TD = ₹ 1,000
∴ BG = BD – TD
= 1,020 – 1,000
= ₹ 20

Question 14.
A certain sum due 3 months hence is \(\frac{21}{20}\) of the present worth, what is the rate of interest.
Solution:

Question 15.
A bill of a certain sum drawn on 28th February 2007 for 8 months was encashed on 26th March 2007 for ₹ 10,992 at 14% p.a. Find the face value of the bill.
Solution:
Date drawing = 28/02/2007
Period of the bill = 8 months
Nominal due date = 28/10/2007
Legal due date = 31/10/2007
Date of discounting = 26/03/2007
Cash value = ₹ 10,992
Rate of interest = 14%
Let face value of the bill = ₹ x
Bankers discount = Face value – Cash value = x – 10,992
Also, Banker s discount = \(\frac{F V \times n \times r}{365 \times 100}\)
Where n is the unexpired days

Thus face value of the bill = ₹ 12,000

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Commission, Brokerage and Discount Ex 1.1

Question 1.
An agent charges a 12% commission on the sales. What does he earn if the total sale amounts to ₹ 48,000? What does the seller get?
Solution:
Rate of commission = 12%
Total sales = ₹ 48,000
Agent’s commission = \(\frac {12}{100}\) × 48,000
= ₹ 5,760
Amount received by the seller = Total sales – commission
= ₹ 8,000 – ₹ 5760
= ₹ 2,240

Question 2.
A salesman receives a 3% commission on sales up to ₹ 50,000 and a 4% commission on sales over ₹ 50,000. Find his total income on the sale of ₹ 2,00,000.
Solution:
Total sales = ₹ 2,00,000
Rate of commission upto ₹ 50,000 = 3%
= \(\frac{3}{100}\) × 50,000
= ₹ 1,500
Rate of commission on the sales over ₹ 50,000 = 4%
Sales over ₹ 50,000 is 2,00,000 – 50,000 = ₹ 1,50,000
Commission on sales over ₹ 50,000 = \(\frac{4}{100}\) × 1,50,000 = ₹ 6,000
His total income = ₹ 1,500 + ₹ 6,000 = ₹ 7,500

Question 3.
Ms. Saraswati was paid ₹ 88,000 as commission on the sale of computers at the rate of 12.5%. If the price of each computer was ₹ 32,000, how many computers did she sell?
Solution:
Total commission = ₹ 88,000
Rate of commission = 12.5%
Let the number of computers sold be x
since price of each computer = ₹ 32,000
Total sales = ₹ 32,000x
Total commission = 12.5% of total sales
88,000 = \(\frac{12.5}{100}\) × 32,000x
= \(\frac{125}{1000}\) × 32,000x
x = \(\frac{88,000}{125 \times 32}\)
x = 22

Question 4.
Anita is allowed 6.5% commission on the total sales made by her, plus, a bonus of \(\frac{1}{2}\)% on the sale over ₹ 20,000. If her total commission amounts to ₹ 3,400. Find the sales made by her.
Solution:
Let the total sales made by Anita be ₹ x
Rate of commission = 6.5% of total sales
= \(\frac{6.5}{100} \times x\)
= \(\frac{65 x}{1,000}\)
= \(\frac{13 x}{200}\)

Question 5.
Priya gets a salary of ₹ 15,000 per month and a commission of 8% on sales over ₹ 50,000. If she gets ₹ 17,400 in a certain month. Find the sales made by her in that month.
Solution:
Let the total sales made by Priya be ₹ x
Salary of Priya = ₹ 15,000
Commission = Total earning – salary
= ₹ 17,400 – ₹ 15,000
= ₹ 2,400
Commission = 8% on the sales over ₹ 50,000
2400 = \(\frac{8}{100}\) (x – 50000)
\(\frac{2,400 \times 100}{8}\) = x – 50,000
30,000 = x – 50,000
30,000 + 50,000 = x
∴ x = ₹ 80,000

Question 6.
The income of the broker remains unchanged though the rate of commission is increased from 4% to 5%. Find the percentage reduction in the value of the business.
Solution:
Let the original value of business be ₹ 100
Original rate of commission = 4%
∴ Original commission = \(\frac{4}{100}\) × 100 = ₹ 4
Let the new value of business be ₹ x
The new rate of commission = 5%
∴ New commission = \(\frac{5}{100}\) × x = \(\frac{x}{20}\)
Given, original income = New income
4 = \(\frac{x}{20}\)
∴ x = ₹ 80
Thus there is 20% reduction in the value of the business.

Question 7.
Mr. Pavan is paid a fixed weekly salary plus commission based on a percentage of sales made by him. If on the sale of ₹ 68,000 and ₹ 73,000 in two successive weeks, he received in all ₹ 9,880 and ₹ 10,180. Find his weekly salary and the rate of commission paid to him.
Solution:
Let the weekly salary of Mr. Pavan be ₹ x and the rate of commission paid to him be y%
Income = Weekly salary + Commission on the sales
∴ 9,880 = x + \(\frac{y}{100}\) × 68,000
i.e. 9,880 = x + 680y …….(1)
Also, 10,180 = x + \(\frac{y}{100}\) × 73,000
i.e 10,180 = x + 730y ………(2)
Subtracting (1) from (2), we get
50y = 300
∴ y = 6
Substituting y = 6 in equation (1)
9,880 = x + 680(6) ‘
∴ 9,880 – 4,080 = x
∴ x = 5,800
Weekly salary = ₹ 5,800
Rate of commission = 6%

Question 8.
Deepak’s salary was increased from ₹ 4,000 to ₹ 5,000. The sales being the same, due to a reduction in the rate of commission from 3% to 2%, his income remained unchanged. Find his sales.
Solution:
Let Deepak’s total sales be ₹ x
Original salary of Deepak = ₹ 4,000
Original rate of commission = 3%
His new salary = ₹ 5,000
New rate of commission = 2%
Original income = New income (given)
4000 + \(\frac{3 x}{100}\) = 5000 + \(\frac{2 x}{100}\)
\(\frac{3 x}{100}-\frac{2 x}{100}\) = 5,000 – 4,000
\(\frac{x}{100}\) = 1000
x = ₹ 1,00,000
∴ His total sales = ₹ 1,00,000

Question 9.
An agent is paid a commission of 7% on cash sales and 5% on credit sales made by him. If on the sale of ₹ 1,02,000 the agent claims a total commission of ₹ 6,420, find his cash sales and credit sales.
Solution:
Total Sales = ₹ 1,02,000
Let cash sales ₹ x
∴ Credit sales = ₹ (1,02,000 – x)
Agent’s commission on cash sales = 7%
= \(\frac{7}{100}\) × x
= \(\frac{7x}{100}\)
Commission on credit sales = 5%
= \(\frac{5}{100}\)(1,02,000 – x)
Given, Total commission = ₹ 6,420
∴ \(\frac{7x}{100}\) + \(\frac{5}{100}\)(1,02,000 – x) = 6420
∴ \(\frac{7x}{100}\) + 5100 – \(\frac{5x}{100}\) = 6,420
∴ \(\frac{2x}{100}\) = 6,420 – 5,100
∴ \(\frac{2x}{100}\) = 1320
∴ x = ₹ 66,000
∴ Cash sales = ₹ 66,000
∴ Credit sales = 1,02000 – 66,000 = ₹ 36,000

Question 10.
Three cars were sold through an agent for ₹ 2,40,000, ₹ 2,22,000 and ₹ 2,25,000 respectively. The rates of the commission were 17.5% on the first, 12.5% on the second. If the agent overall received 14% commission on the total sales, find the rate of commission paid on the third car.
Solution:
Total selling price of three cars = 2,40,000 + 2,22,000 + 2,25,000 = ₹ 6,87,000
Commission on total sales = 14%
= \(\frac{14}{100}\) × 6,87,000
= ₹ 96,180
Selling price of first car = ₹ 2,40,000
Rate of commission = 17.5% = \(\frac{17.5}{100}\) × 2,40,000
∴ Commission on first car = ₹ 42,000
Selling price of second car = ₹ 2,22,000
Rate of commission = 12.5% = \(\frac{12.5}{100}\) × 2,22,000
∴ Commission on second car = ₹ 27,750
Selling price of third car = ₹ 2,25,000
Let the rate of commission be x%
Commission on third car = \(\frac{x}{100}\) × 2,25,000
96,180 – (42,000 + 27,750) = \(\frac{x}{100}\) × 2,25,000
\(\frac{26,430 \times 100}{2,25,000}\) = x
∴ x = 11.75
∴ Rate of commission on the third car = 11.75%

Question 11.
Swatantra Distributors allows a 15% discount on the list price of the washing machines. Further 5% discount is giver for cash payment. Find the list price of the washing machine if it was sold for the net amount of ₹ 38,356.25.
Solution:
Let the list price of the washing machine be ₹ 100
Trade discount = 15% = \(\frac{15}{100}\) × 100 = ₹ 15
∴ Invoice price =100 – 15 = ₹ 85
Cash discount = 5% = \(\frac{5}{100}\) × 85 = ₹ 4.25
∴ Net price = 85 – 4.25 = ₹ 80.75
Thus if List price is 100 than Net price is 80.75
if List price is x than Net price is 38,356.25.
∴ x = \(\frac{38356.25 \times 100}{80.75}\)
∴ x = ₹ 47,500
The list price of the washing machine is ₹ 47,500

Question 12.
A bookseller received ₹ 1,530 as a 15% commission on the list price. Find the list price of the books.
Solution:
Let the list price of the books be ₹ x
Rate of commission = 15%
Book seller’s commission = ₹ 1,530
∴ \(\frac{15}{100}\) × x = 1,530
∴ x = \(\frac{1,530 \times 100}{15}\)
∴ x = ₹ 10,200

Question 13.
A retailer sold a suit for ₹ 8,832 after allowing an 8% discount on market price and a further 4% cash discount. If he made 38% profit, find the cost price and the market price of the suit.
Solution:
Let the marked price of the suit be ₹ 100
Trade discount = 8% = \(\frac{8}{100}\) × 100 = ₹ 8
Invoice price = 100 – 8 = ₹ 92
Cash discount = 4% = \(\frac{4}{100}\) × 92 = ₹ 3.68
∴ Net price = 92 – 3.68 = ₹ 88.32
Thus if list price is 100 then net price is 88.32, if list price is x then net price is 8,832
∴ x = \(\frac{8,832 \times 100}{88.32}\)
∴ x = ₹ 10,000
The retailer made 38% profit.
Let the CP of the suit be ₹ 100
∴ SP of the suit = 100 + 38 = ₹ 138
Thus if the SP of the suit is ₹ 138 then its CP is ₹ 100
If the SP of the suit is 88.32 then its
CP = \(\frac{88.32 \times 100}{138}\) = ₹ 6400

Question 14.
An agent charges 10% commission plus 2% delcredere. If he sells goods worth ₹ 37,200, find his total earnings.
Solution:
Total sales = ₹ 37,200
Rate of commission = 10%
Agents commission = \(\frac{4}{100}\) × 37200 = ₹ 3720
Rate of delcredere = 2%
Amount of delcredere = \(\frac{2}{100}\) × 37,200 = ₹ 744
Total earning of the agent = ₹ 3,720 + ₹ 744 = ₹ 4,464

Question 15.
A whole seller allows a 25% trade discount and 5% cash discount. What will be the net price of an article marked at ₹ 1600?
Solution:
Marked price of the article = ₹ 1,600
Trade discount = 25%
= \(\frac{25}{100}\) × 1,600
= ₹ 400
∴ Invoice price = 1,600 – 400 = ₹ 1,200
Cash discount = 5%
= \(\frac{5}{100}\) × 1,200
= ₹ 60
∴ Net price = 1,200 – 60 = ₹ 1,140

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of \(\left(\frac{d y}{d x}\right)^{3}-\frac{d^{3} y}{d x^{3}}+y e^{x}=0\) are respectively.
(a) 3, 1
(b) 1, 3
(c) 3, 3
(d) 1, 1
Answer:
(a) 3, 1

Question 2.
The order and degree of \(\left[1+\left(\frac{d y}{d x}\right)^{3}\right]^{\frac{2}{3}}=8 \frac{d^{3} y}{d x^{3}}\) are respectively
(a) 3, 1
(c) 3, 3
(b) 1, 3
(d) 1, 1
Answer:
(c) 3, 3

Question 3.
The differential equation of y = k₁ + \(\frac{k_{2}}{x}\) is
(a) \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
(b) \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}=0\)
(d) \(x \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}=0\)
Answer:
(b) \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

Question 4.
The differential equation of y = k₁ e^x + k₂ e^-x is
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
(b) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=0\)
(d) \(\frac{d^{2} y}{d x^{2}}+y=0\)
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)

Question 5.
The solution of \(\frac{d y}{d x}\) = 1 is
(a) x + y = c
(b) xy = c
(c) x² + y² = c
(d) y – x = c
Answer:
(d) y – x = c

Question 6.
The solution of \(\frac{d y}{d x}+\frac{x^{2}}{y^{2}}=0\) is
(a) x³ + y³ = 7
(b) x² + y² = c
(c) x³ + y³ = c
(d) x + y = c
Answer:
(c) x³ + y³ = c

Question 7.
The solution of x \(\frac{d y}{d x}\) = y log y is
(a) y = ae^x
(b) y = be^2x
(c) y = be^-2x
(d) y = e^ax
Answer:
(d) y = e^ax

Question 8.
Bacterial increases at a rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in
(a) 4 hours
(b) 6 hours
(c) 8 hours
(d) 10 hours
Answer:
(b) 6 hours

Question 9.
The integrating factor of \(\frac{d y}{d x}\) – y = e^x is
(a) x
(b) -x
(c) e^x
(d) e^-x
Answer:
(c) e^x

Question 10.
The integrating factor of \(\frac{d y}{d x}\) – y = e^x is e^-x, then its solution is
(a) ye^-x = x + c
(b) ye^x = x + c
(c) ye^x = 2x + c
(d) ye^-x = 2x + c
Answer:
(a) ye^-x = x + c

(II) Fill in the blanks:

Question 1.
The order of highest derivative occurring in the differential equation is called ________ of the differential equation.
Answer:
order

Question 2.
The power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called ________ of the differential equation.
Answer:
degree

Question 3.
A solution of differential equation that can be obtained from the general solution by giving particular values to the arbitrary constants is called _________ solution.
Answer:
particular

Question 4.
Order and degree of a differential equation are always _________ integers.
Answer:
positive

Question 5.
The integrating factor of the differential equation \(\frac{d y}{d x}\) – y = x is _________
Answer:
e^-x

Question 6.
The differential equation by eliminating arbitrary constants from bx + ay = ab is _________
Answer:
\(\frac{d^{2} y}{d x^{2}}=0\)

(III) State whether each of the following is True or False:

Question 1.
The integrating factor of the differential equation \(\frac{d y}{d x}\) – y = x is e^-x.
Answer:
True

Question 2.
The order and degree of a differential equation are always positive integers.
Answer:
True

Question 3.
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
Answer:
True

Question 4.
The order of highest derivative occurring in the differential equation is called the degree of the differential equation.
Answer:
False

Question 5.
The power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called the order of the differential equation.
Answer:
False

Question 6.
The degree of the differential equation \(e^{\frac{d y}{d x}}=\frac{d y}{d x}+c\) is not defined.
Answer:
True

(IV) Solve the following:

Question 1.
Find the order and degree of the following differential equations:
(i) \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given differential equation is \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}\)
∴ \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3
∴ order = 3 and degree = 3

(ii) \(x+\frac{d y}{d x}=1+\left(\frac{d y}{d x}\right)^{2}\)
Solution:
The given differential equation is \(x+\frac{d y}{d x}=1+\left(\frac{d y}{d x}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ order = 1, degree = 2.

Question 2.
Verify that y = log x + c is a solution of the differential equation \(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\).
Solution:
y = log x + c
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{x}+0=\frac{1}{x}\)
∴ x\(\frac{d y}{d x}\) = 1
Differentiating again w.r.t. x, we get
\(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \times 1=0\)
∴ \(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
This shows that y = log x + c is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)

Question 3.
Solve the following differential equations:
(i) \(\frac{d y}{d x}\) = 1 + x + y + xy
Solution:
\(\frac{d y}{d x}\) = 1 + x + y + xy
∴ \(\frac{d y}{d x}\) = (1 + x) + y(1 + x) = (1 + x)(1 + y)
∴ \(\frac{1}{1+y}\) dy = (1 + x) dx
Integrating, we get
∫\(\frac{1}{1+y}\) dy = ∫(1 + x) dx
∴ log|1 + y| = x + \(\frac{x^{2}}{2}\) + c
This is the general solution.

(ii) \(e^{d y / d x}=x\)
Solution:

∴ from (1), the general solution is
y = x log x – x + c, i.e. y = x(log x – 1) + c.

(iii) dr = ar dθ – θ dr
Solution:
dr = ar dθ – θ dr
∴ dr + θ dr = ar dθ
∴ (1 + θ) dr = ar dθ
∴ \(\frac{d r}{r}=\frac{a d \theta}{1+\theta}\)
On integrating, we get
\(\int \frac{d r}{r}=a \int \frac{d \theta}{1+\theta}\)
∴ log |r| = a log |1 + θ| + c
This is the general solution.

(iv) Find the differential equation of the family of curves y = e^x (ax + bx²), where a and b are arbitrary constants.
Solution:
y = e^x (ax + bx²)
ax + bx² = ye^-x …….(1)
Differentiating (1) w.r.t. x twice and writing \(\frac{d y}{d x}\) as y₁ and \(\frac{d^{2} y}{d x^{2}}\) as y₂, we get


This is the required differential equation.

Question 4.
Solve \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) when x = \(\frac{2}{3}\) and y = \(\frac{1}{3}\).
Solution:

Question 5.
Solve y dx – x dy = -log x dx.
Solution:
y dx – x dy = -log x dx
∴ y dx – x dy + log x dx = 0
∴ x dy = (y + log x) dx
∴ \(\frac{d y}{d x}=\frac{y+\log x}{x}=\frac{y}{x}+\frac{\log x}{x}\)
∴ \(\frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x}\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 6.
Solve y log y \(\frac{d x}{d y}\) + x – log y = 0.
Solution:

Question 7.
Solve (x + y) dy = a² dx
Solution:

Question 8.
Solve \(\frac{d y}{d x}+\frac{2}{x} y=x^{2}\)
Solution:
\(\frac{d y}{d x}+\frac{2}{x} y=x^{2}\) ……..(1)
This is a linear differential equation of the form

This is the general solution.

Question 9.
The rate of growth of the population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lakh, when will the city have a population of 400000?
Solution:
Let P be the population at time t years.
Then the rate of growth of the population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\)= k dt
On integrating, we get
\(\int \frac{d P}{P}=k \int d t\)
∴ log P = kt + c
The population doubled in 25 years and present population is 1,00,000.
∴ initial population was 50,000
i.e. when t = 0, P = 50000
∴ log 50000 = k × 0 + c
∴ c = log 50000
∴ log P = kt + log 50000
When t = 25, P = 100000
∴ log 100000 = k × 25 + log 50000
∴ 25k = log 100000 – log 50000 = log(\(\frac{100000}{50000}\))
∴ k = \(\frac{1}{25}\) log 2
∴ log P = \(\frac{t}{25}\) log 2 + log 50000
If P = 400000, then
log 400000 = \(\frac{t}{25}\) log 2 + log 50000
∴ log 400000 – log 50000 = \(\frac{t}{25}\) log 2
∴ log(\(\frac{400000}{50000}\)) = \(\log (2)^{t / 25}\)
∴ log 8 = \(\log (2)^{t / 25}\)
∴ 8 = \((2)^{t / 25}\)
∴ \((2)^{t / 25}\) = (2)³
∴ \(\frac{t}{25}\) = 3
∴ t = 75
∴ the population will be 400000 in (75 – 25) = 50 years.

Question 10.
The resale value of a machine decreases over a 10 years period at a rate that depends on the age of the machine. When the machine is x years old, the rate at which its value is changing is ₹ 2200(x – 10) per year. Express the value of the machine as a function of its age and initial value. If the machine was originally worth ₹ 1,20,000 how much will it be worth when it is 10 years old?
Solution:
Let V be the value of the machine after x years.
Then rate of change of the value is \(\frac{d V}{d x}\) which is 2200(x – 10)
∴ \(\frac{d V}{d x}\) = 2200(x – 10)
∴ dV = 2200(x – 10) dx
On integrating, we get
∫dV = 2200∫(x – 10) dx
∴ V = 2200[\(\frac{x^{2}}{2}\) – 10x] + c
Initially, i.e. at x = 0, V = 120000
∴ 120000 = 2200 × 0 + c = c
∴ c = 120000
∴ V = 2200[\(\frac{x^{2}}{2}\) – 10x] + 120000 …….(1)
This gives value of the machine in terms of initial value and age x.
We have to find V when x = 10.
When x = 10, from (1)
V = 2200[\(\frac{100}{2}\) – 100] + 120000
= 2200 [-50] + 120000
= -110000 + 120000
= 10000
Hence, the value of the machine after 10 years will be ₹ 10000.

Question 11.
Solve y² dx + (xy + x²) dy = 0
Solution:

Question 12.
Solve x²y dx – (x³ + y³) dy = 0
Solution:

Question 13.
Solve yx \(\frac{d y}{d x}\) = x² + 2y²
Solution:

Question 14.
Solve (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:
(x + 2y³) \(\frac{d y}{d x}\) = y
∴ x + 2y³ = y \(\frac{d x}{d y}\)

This is the general solution.

Question 15.
Solve y dx – x dy + log x dx = 0
Solution:
y dx – x dy + log x dx = 0
∴ (y + log x) dx = x dy


This is the general solution.

Question 16.
Solve \(\frac{d y}{d x}\) = log x dx
Solution:
\(\frac{d y}{d x}\) = log x dx
∴ dy = log x dx
On integrating, we get
∫dy = ∫log x . 1 dx
∴ y = (log x) ∫1 dx – \(\int\left[\left\{\frac{d}{d x}(\log x)\right\} \cdot \int 1 d x\right] d x\)
∴ y = (log x) . x – \(\int \frac{1}{x} \cdot x d x\)
∴ y = x log x – ∫1 dx
∴ y = x log x – x + c
This is the general solution.

Question 17.
y log y \(\frac{d x}{d y}\) = log y – x
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.6

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
∫\(\frac{d x}{x}\) = k∫dt
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x₀
∴ log x₀ = k × 0 + c
∴ c = log x₀
∴ log x = kt + log x₀
∴ log x – log x₀ = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ……(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀

∴ the number of bacteria will be 8 times the original number in 12 hours.

Question 2.
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? (Given: \(\sqrt{\frac{3}{2}}\) = 1.2247)
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population, is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, k is a constant
∴ \(\frac{d P}{P}\) = k dt
Integrating, we get
∫\(\frac{d P}{P}\) = k∫dt
∴ log P = kt + c
Initially, i.e. when t = 0, P = 40000
∴ log 40000 = 0 + c
∴ c = log 40000
∴ log P = kt + log 40000
∴ log P – log 40000 = kt
∴ log(\(\frac{P}{40000}\)) = kt ………(1)
When t = 40, P = 60000

∴ population after 60 years will be 73482.

Question 3.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after \(\frac{5}{2}\) hours. [Given: √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
∫\(\frac{d x}{x}\) = k∫dt
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt …….(1)
Now, when t = 1, x = 2 × 1000 = 2000

∴ number of bacteria after \(\frac{5}{2}\) hours = 5656.

Question 4.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population, is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
∫\(\frac{1}{P}\)dP = k∫dt
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k x 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt

∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Question 5.
The rate of depreciation \(\frac{d V}{d t}\) of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹ 1,00,000 in the first year. Find the value after 6 years.
Solution:
Let V be the value of the machine at the end of t years.
Then \(\frac{d V}{d t}\), the rate of depreciation, is inversly proportional to (t + 1)².

Initially, i.e. when t = 0, V = 800000
∴ 800000 = \(\frac{k}{1}\) + c = k + c ………(1)
Now, when t = 1, V = 800000 – 100000 = 700000
∴ 700000 = \(\frac{k}{1+1}\) + c = \(\frac{k}{2}\) + c ……(2)
Subtracting (2) from (1), we get
100000 = \(\frac{1k}{2}\)
∴ k = 200000
∴ from (1), 800000 = 200000 + c
∴ c = 600000 200000
∴ V = \(\frac{200000}{t+1}\) + 600000
When t = 6,
V = \(\frac{200000}{7}\) + 600000
= 28571.43 + 600000
= 628571.43 ~ 628571
Hence, the value of the machine after 6 years will be ₹ 6,28,571.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.5

Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}+y=e^{-x}\)
Solution:
\(\frac{d y}{d x}+y=e^{-x}\) …….(1)
This is the linear differential equation of the form

This is the general solution.

Question 2.
\(\frac{d y}{d x}\) + y = 3
Solution:
\(\frac{d y}{d x}\) + y = 3
This is the linear differential equation of the form

This is the general solution.

Question 3.
x\(\frac{d y}{d x}\) + 2y = x² . log x.
Solution:
x\(\frac{d y}{d x}\) + 2y = x² . log x
∴ \(\frac{d y}{d x}+\left(\frac{2}{x}\right) \cdot y=x \cdot \log x\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 4.
(x + y)\(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 5.
y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 6.
\(\frac{d y}{d x}\) + 2xy = x
Solution:
\(\frac{d y}{d x}\) + 2xy = x ………(1)
This is the linear differential equation of the form

This is the general solution.

Question 7.
(x + a) \(\frac{d y}{d x}\) = -y + a
Solution:
(x + a) \(\frac{d y}{d x}\) + y = a
∴ \(\frac{d y}{d x}+\left(\frac{1}{x+a}\right) y=\frac{a}{x+a}\) ……..(1)
This is the linear differential equation of the form

This is the general solution.

Question 8.
dy + (2y) dx = 8 dx
Solution:
dy + (2y) dx = 8 dx
∴ \(\frac{d y}{d x}\) + 2y = 8 …….(1)
This is the linear differential equation of the form

This is the general solution.

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.4

Solve the following differential equations:

Question 1.
x dx + 2y dy = 0
Solution:
x dx + 2y dy = 0
Integrating, we get
∫x dx + 2 ∫y dy = c₁
∴ \(\frac{x^{2}}{2}+2\left(\frac{y^{2}}{2}\right)=c_{1}\)
∴ x² + 2y² = c, where c = 2c₁
This is the general solution.

Question 2.
y² dx + (xy + x²) dy = 0
Solution:
y² dx + (xy + x²) dy = 0
∴ (xy + x²) dy = -y² dx
∴ \(\frac{d y}{d x}=\frac{-y^{2}}{x y+x^{2}}\) ………(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting these values in (1), we get



This is the general solution.

Question 3.
x²y dx – (x³ + y³) dy = 0
Solution:
x²y dx – (x³ + y³) dy = 0
∴ (x³ + y³) dy = x²y dx
∴ \(\frac{d y}{d x}=\frac{x^{2} y}{x^{3}+y^{3}}\) ……(1)
Put y = vx
∴ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)


This is the general solution.

Question 4.
\(\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0\)
Solution:



This is the general solution.

Question 5.
(x² – y²) dx + 2xy dy = 0
Solution:
(x² – y²) dx + 2xy dy = 0
∴ 2xy dy = -(x² – y²) dx = (y² – x²) dx
∴ \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\) ………(1)

Question 6.
xy\(\frac{d y}{d x}\) = x² + 2y²
Solution:

Question 7.
x²\(\frac{d y}{d x}\) = x² + xy – y²
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.3

Question 1.
Solve the following differential equations:
(i) \(\frac{d y}{d x}\) = x²y + y
Solution:
\(\frac{d y}{d x}\) = x²y + y
∴ \(\frac{d y}{d x}\) = y(x² + 1)
∴ \(\frac{1}{y}\) dy = (x² + 1) dx
Integrating, we get
∫\(\frac{1}{y}\) dy = ∫(x² + 1) dx
∴ log |y|= \(\frac{x^{3}}{3}\) + x + c
This is the general solution.

(ii) \(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)
Solution:
\(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)

This is the general solution.

(iii) (x² – yx²) dy + (y² + xy²) dx = 0
Solution:
(x² – yx²) dy + (y² + xy²) dx = 0
∴ x²(1 – y) dy + y²(1 + x) dx = 0
∴ \(\frac{1-y}{y^{2}} d y+\frac{1+x}{x^{2}} d x=0\)
Integrating, we get


This is the general solution.

(iv) \(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)
Solution:
\(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)

∴ 2y² log |x + 1| = 2cy² – 1 is the required solution.

Question 2.
For each of the following differential equations find the particular solution:
(i) (x – y²x) dx – (y + x²y) dy = 0, when x = 2, y = 0.
Solution:
(x – y²x) dx – (y + x²y) dy = 0
∴ x(1 – y²) dx – y(1 + x²) dy = 0

∴ the general solution is
log |1 + x²| + log |1 – y²| = log c, where c₁ = log c
∴ log |(1 + x²)(1 – y²) | = log c
∴ (1 + x²)(1 – y²) = c
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x²)(1 – y²) = 5.

(ii) (x + 1) \(\frac{d y}{d x}\) -1 = 2e^-y, when y = 0, x = 1.
Solution:

∴ log |2 + e^y| = log |c(x + 1)|
∴ 2 + e^y = c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰ = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is
2 + e^y = \(\frac{3}{2}\)(x + 1)
∴ 4 + 2e^y = 3x + 3
∴ 3x – 2e^y – 1 = 0

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, when x = e, y = e².
Solution:

∴ from (1), the general solution is
log |x log x| – log |y| = log c, where c₁ = log c
∴ log |\(\frac{x \log x}{y}\)| = log c
∴ \(\frac{x \log x}{y}\) = c
∴ x log x = cy
This is the general solution.
Now, y = e², when x = e
e log e = ce²
1 = ce ……[∵ log e = 1]
c = \(\frac{1}{e}\)
∴ the particular solution is x log x = (\(\frac{1}{e}\)) y
∴ y = e^x log x

(iv) \(\frac{d y}{d x}\) = 4x + y + 1, when y = 1, x = 0.
Solution:
\(\frac{d y}{d x}\) = 4x + y + 1
Put 4x + y + 1 = v

∴ log |v + 4| = x + c
∴ log |4x + y + 1 + 4| = x + c
i.e. log |4x + y + 5| = x + c
This is the general solution.
Now, y = 1 when x = 0
∴ log|0 + 1 + 5| = 0 + c,
i.e. c = log 6
∴ the particular solution is
log |4x + y + 5| = x + log 6
∴ \(\log \left|\frac{4 x+y+5}{6}\right|\) = x

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2

Question 1.
Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) y = Ae^3x + Be^-3x
Solution:
y = Ae^3x + Be^-3x ……(1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

(ii) y = \(c_{2}+\frac{c_{1}}{x}\)
Solution:
y = \(c_{2}+\frac{c_{1}}{x}\)
∴ xy = c₂x + c₁
Differentiating w.r.t. x, we get

(iii) y = (c₁ + c₂x) e^x
Solution:
y = (c₁ + c₂x) e^x


This is the required D.E.

(iv) y = c₁ e^3x+ c₂ e^2x
Solution:



This is the required D.E.

(v) y² = (x + c)³
Solution:
y² = (x + c)³
Differentiating w.r.t. x, we get

This is the required D.E.

Question 2.
Find the differential equation by eliminating arbitrary constant from the relation x² + y² = 2ax.
Solution:
x² + y² = 2ax
Differentiating both sides w.r.t. x, we get
2x + 2y\(\frac{d y}{d x}\) = 2a
Substituting value of 2a in equation (1), we get
x² + y² = [2x + 2y \(\frac{d y}{d x}\)]x = 2x² + 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) = y² – x² is the required D.E.

Question 3.
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution:
bx + ay = ab
∴ ay = -bx + ab
∴ y = \(-\frac{b}{a} x+b\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{b}{a} \times 1+0=-\frac{b}{a}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 0 is the required D.E.

Question 4.
Find the differential equation whose general solution is x³ + y³ = 35ax.
Solution:

Question 5.
Form the differential equation from the relation x² + 4y² = 4b².
Sol ution:
x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4(2y\(\frac{d y}{d x}\)) = 0
i.e. x + 4y\(\frac{d y}{d x}\) = 0 is the required D.E.