Math, asked by raghustudio2006, 2 months ago


24.The population of a city was 120000 in the year 2009. During next year it increased by 6%
but due to an epidemic it decreased by 5% in the following year. What is its population in the
year 20112
25. The count of bacteria in a certain experiment was increasing at the rate of 2% per hour.
Find the bacteria at the end of 2 hours if the count was initially 500000.
26. The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second
hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if
the count was initially 20000.
27. A machine is purchased for ? 625000. Its value depreciates at the rate of 8% per annum.
What will be its value after 2 years?
28. Ascooter is bought at 56000. Its value depreciates at the rate of 10% per annum. What will
be its value after 3 years?
19. A car is purchased for * 348000. Its value depreciates at 10% per annum during the first
year and at 20% per annum during the second year. What will be its value after 2 years?
30. The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years
If its present value is * 291600, for how much was it purchased?

--------Please Solve the question
25 and 26?????

Answers

Answered by RvChaudharY50
23

Question 25) :- The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.

Answer :-

→ Initial count of bacteria = 500000

→ Rate of increasing = 2% per hour .

so,

→ Final count of bacteria = Initial count[1 + (rate/100)]^time

→ Final count of bacteria = 500000[1 + (2/100)]²

→ Final count of bacteria = 500000[1 + (1/50)]²

→ Final count of bacteria = 500000 * (51/50)²

→ Final count of bacteria = (500000 * 51 * 51)/50 * 50

→ Final count of bacteria = 520200 (Ans.)

Question 26) :- The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.

Answer :-

→ Final count of bacteria = Initial count[1 + (rate of first hour /100)] * Initial count[1 - (rate of second hour /100)] * Initial count[1 + (rate of third hour /100)]

{ since rate increase in 1st and 3rd hour we took + , and rate decrease in 2nd hour we took negative. }

then,

→ Final count of bacteria = 20000[1 + (10/100)] * [1 - (10/100)] * [1 + (10/100)]

→ Final count of bacteria = 20000 * [1 + (1/10)] * [1 - (1/10)] * [1 + (1/10)]

→ Final count of bacteria = 20000 * (11/10) * (9/10) * (11/10)

→ Final count of bacteria = (20000 * 121 * 9)/1000

→ Final count of bacteria = 20 * 1089

→ Final count of bacteria = 21780 (Ans.)

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Answered by Anonymous
7

25]

Given :-

The count of bacteria in a certain experiment was increasing at the rate of 2% per hour.

To Find :-

bacteria at end of 2 hours if the count was initially 500000.

Solution :-

Here we have to use the compound interest formula

CI = P × [1 + R/100]ⁿ

Here

CI = Bacteria

P = No of bacteria

n = Time

CI = 5,00,000 × [1 + 2/100]²

CI = 5,00,000 × [100 + 2/100]²

CI = 5,00,000 × [102/100]²

CI = 5,00,000 × 102/100 × 102/100

CI = 50 × 102 × 102

CI = 520200

Now

26]

Given :-

The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second  hour and again increases by 10% in the third hour.

To Find :-

Bacteria at the end of 3 hours if  the count was initially 20000.

Solution :-

At first increase in 1st hour

10% of 20000

10/100 × 20000

1/10 × 20000

1 × 2000

2000

Increase bacteria = 20000 + 2000 = 22000

In the second hour

10% of 22000

10/100 × 22000

1/10 × 22000

1 × 2200

2200

In the third hour the bacteria remains

22,000 - 2,200 = 19,800

Increase in bacteria in third hour

10% of 19800

10/100 × 19800

1/10 × 19800

1 × 1980

1980

At the end of 4 th hour

19,800 + 1,980 = 21,780

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