Question

1. A machinery worth ₹10,500 depreciated by 5%. Find it’s value after one year.
2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.

Answer 1

Answer:

Step-by-step explanation:

Solution 1

Given, principal = 10500, R = 5%, Time n = 1 year.

We know that A = P(1 + r/100)^n

⇒ 10500(1 + 5/100)^1

⇒ 10500(1 + 1/20)

⇒ 10500(21/20)

⇒ 11025.

Solution 2

Present population = 12 lakh = 1200000

Time = 2 years

Rate of increase = 4%

Therefore, population after 2 years is

= present population(1 + r/100)²

= 1200000 (1 + 4/100)²

= 1200000 * (104/100)²

= 1200000 * 104/100 * 104/100

= 120 * 104 * 104

= 1297920

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Answer 2

Answer:

Q1 ₹1995

Q2. 1297920

Step-by-step explanation:

1) Principal (P)= ₹10500

Rate of depreciation= 5%

Time Taken= 1yr

Amount = P[1+R/100]

= 10500×[1-5/100]. [it will be subracted because (its written) Rate of depreciation]

= 10500×[1-1/20]

= 10500×[20-1/100]

= 10500×[19/100]

= 10500×19/100

= 105×19

= ₹1995

2) Present population= 12 lakh

Rate of increase =4%

Time Taken = 2yrs

Amount = P[1+R/100]

=. 1200000[1+4/100]²

= 1200000[1+1/25]²

= 1200000[25+1/25]²

= 1200000[26/25]²

= 1200000×26/25×26/25

= 1920×26×26

= 1297920

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