Question

please help me only answer 11,12

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Answer 1

hey there!!

Answer:

Ans 10:

(i) 48, 980

(ii) 59, 535

Answer 11:

5631616

Explanation:

Ans 10:

Part 1:

Given,

population of place in 2003 = 54000

it has increased at the rate of 5% P.A.

here 5% is compounded rate.

So we use the formula

• A = P (1 + $\frac{r}{100}$) ^ n

Here,

A = population in year 2003 = 54000

P = population in year 2001.

R = 5%

N = number of years

= 2003 - 2001 = 2

• Putting values in formula,

54000 = P ( 1 + $\frac{5}{100}$)²

54000 = P ( 1 + $\frac{1}{20}$)²

54000 = P ( $\frac{20+1}{20}$)²

54000 = P ( $\frac{21}{20}$)²

54000 = P × $\frac{441}{400}$

$\frac{54000*400}{441}$ = P

P = $\frac{54*4*100000}{441}$

P = $\frac{21600000}{441}$

P = 48979.59

• Since population cannot be decimal. Thus population in year 2001 is around 48, 980

Part 2:

Given,

population in year 2003 (P) = 54000

rate (R) = 5% p.a.

Since

5% is compounded rate.

So we use the formula

• A = P (1 + $\frac{r}{100}$) ^ n

population in year 2005 = 54000 (1 + $\frac{5}{100}$)²

= 54000 × ( 1 + $\frac{1}{20}$)²

= 54000 × ($\frac{20 + 1}{20}$)²

= 54000 × ($\frac{21}{20}$)²

= 54000 × ($\frac{21*21}{20*20}$)

= 54000 × $\frac{441}{400}$

= $\frac{540}{4}$ × 441

= $\frac{270}{2}$ × 441

= 135 × 441

= 59535

• POPULATION IN YEAR 2005 = 59535

Ans 11:

Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time (n) = 2 hours

After 2 hours, number of bacteria,

Amount (A) = P ( 1 + $\frac{R}{100}$) ^ n

= 506000 ( 1 + $\frac{2.5}{100}$) ²

= 506000 ( 1 + $\frac{25}{1000}$ )²

= 506000 ( 1 + $\frac{1}{40}$ )²

= 506000 ($\frac{41}{40}$)²

= 506000 × $\frac{41}{40}$ × $\frac{41}{40}$

= 5,31,616.25

• Hence, number of bacteria after two hours are 531616 (approx.).

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