Question

Mr durani bought a plot of land for 18 lakh and a car for 320000 at the same time the value of the plot of a land grows uniformly at the rate of 30% while the value of the car depreciated by 20% in the first year and 15% p. A. thereafter if he sells the plot of land as well as the car after3 years what will be his profit or loss

Answer 1

Given:-

• Principal price of plot = Rs.18 lakh
• Principal price of car = Rs.3 lakh 20000

To Find :-

• The profit or loss after selling 3 years = ?

Solution:-

To solve this question at first we have to collect clue which given in the question then calculate total price of car and plot then calculate it's selling price. After that by applying formula to calculate it's profit or loss after selling.

Calculation for plot:-

• Rate grows uniformly = 30% p.a
• Principal price = Rs. 18 lakh
• Time = 3 years

Total price after 3 years :-

➞ Amount = P [ 1 + r/100 ] ^ t

➞ 1800000 [ 1 + 30/100 ] ³

➞ 1800000 [ 100 + 30/100 ] ³

➞ 1800000 [ 130/100 ] ³

➞ 1800000 × 13/10 × 13/10 × 13/10

➞ 1800 × 13 × 13 × 13

➞ Rs. 3954600

Calculation for car :-

• Principal price = Rs. 320000
• Rate depreciation = 20% 15% 15%

Total price after 3 years :-

➞ Amount = P [ 1 - r/100]^t

• For first year:-

➞ 320000 [ 1 - 20/100 ]

➞ 320000 × 80/100

➞ 3200 × 80

➞ Rs. 256000

• For second year :-

Principal price = 256000

Rate depreciation = 15%

➞ 256000 [ 1 - 15/100]

➞ 256000 × 85/100

➞ 2560 × 85

➞ Rs. 217600

• For third years :-

Principal price = 217600

Rate depreciation = 15%

➞ 217600 [ 1 - 15/100 ]

➞ 217600 × 85/100

➞ 2176 × 85

➞ Rs. 184960

Therefore, price after 3 years = Rs. 184960

• Now calculate total CP and SP :-

➞ Total principal price = Rs. Plot + Rs. car

➞ 1800000 + 320000 = Rs. 2120000

➞ Total selling price = Rs. Plot + Rs. car

➞ 3954600 + 184960 = Rs. 4139560

• Here SP is greator then CP :-

Total Profit = SP - CP :-

➞ Total Profit = 4139560 - 2120000

➞ Total Profit = Rs. 2019560

Hence,

• Mr Durani has got Profit = Rs. 2019560

Answer 2

Question :

Mr durani bought a plot of land for 18 lakh and a car for 320000 at the same time the value of the plot of a land grows uniformly at the rate of 30% while the value of the car depreciated by 20% in the first year and 15% p. A. thereafter if he sells the plot of land as well as the car after3 years what will be his profit or loss

To find :

The profit or loss

Solution :

Price of the plot → ₹ 1800000

Price of the plot grows by 30 %

$\sf \small A = P {(1 + \frac{r}{100} )}^{3}$

$\sf \small A = 1800000(1 + \frac{30}{100} )^{3}$

$\sf \small A = 180000 0{( \frac{13}{10}) }^{3}$

$\sf \small A =180000 0\times \frac{2197}{1000}$

$\sf \small A = 1800 \times 2197$

$\sf \small A = ₹ \: 3954600$

So the amount after 3 years is ₹ 3954600

Now,

Purchase price of the car → ₹ 320000

Depreciation of the car for three years = 20 % , 15 % and 15 %

$\sf \small A = P(1- \frac{ r_{1} }{100} )(1 - \frac{ r_{2} }{100})(1 - \frac{ r_{3} }{100} )$

$\sf \small A = P(1- \frac{ 20 }{100} )(1 - \frac{ 15 }{100})(1 - \frac{ 15 }{100} )$

$\sf \small A = 320000(\frac{4}{5} )( \frac{17}{20} )( \frac{17}{20} )$

$\sf \small A = 320000( \frac{4}{5} )( \frac{17}{20})^{2}$

$\sf \small A = 320000 \times \frac{4}{5} \times \frac{289}{400}$

$\sf \small A = ₹ \: 640 \times 289$

$\sf \small A = ₹ \: 184960$

Total money spent on car and plot → ₹ 320000 + ₹ 1800000

→ ₹ 2120000

Selling price of plot → ₹ 3954600

Selling price of plot car → ₹ 184960

By adding we get the the total S.P → ₹ 3954600 + ₹ 184960

→ ₹ 4139560

Total S.P. → ₹ 4139560

So profit becomes → Total S.P. – Total money spent

→ ₹ 4139560 – ₹ 2120000

→ ₹ 2019560

Therefore Mr. Durani earned a profit of ₹ 2019560