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case /data -based (4 Marks)

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Question 14 Marks
Ramlal is the incharge of a laboratory. In a bacteria culture under observation in his laboratory,the population of bacteria doubles every hour. When Ramlal started the observation, there were100 bacteria.
Q.1. Which of the following expressions gives the bacterial population after n hours?
(a)$\frac{100}{2^n}$$\quad$(b) $\frac{100}{2^{n-1}}$$\quad$(c)$2^n \times 100$$\quad$(d) $2^{n-1} \times 100$
Q.2.  The population size of the bacteria after 3 hours will bе
(a) 800$\quad$ (b) 300$\quad$(c) 600$\quad$(d) 120О
Q.3.  How many bacteria will be there in the culture after 1 day?
(a) $\frac{100}{2^{12}}$$\quad$(b) $2^{24} \times 100$$\quad$(c) $2^{12} \times 100$$\quad$(d) $\frac{100}{2^{24}}$
Q.4. Ramlal observes the culture after every one hour. Find the number of hours after which the population size of the bacteria will be larger than 3000.
(a) 5 hours$\quad$(b) 6 hours$\quad$(c) 7 hours$\quad$(d) 8 hours
Answer
1.(c): The population of bacteria doubles every hour.
Bacterial population after 1 hour $=2 \times 100$.
Bacterial population after 2 hours $=2 \times(2 \times 100)=2^2 \times 100$.
Bacterial population after 3 hours $=2 \times\left(2^2 \times 100\right)=2^3 \times 100$.
$\therefore$ the bacterial population after $n$ hours $=2^n \times 100$.
2. (a): Bacterial population after 3 hours $=2^3 \times 100=8 \times 100=800$.
3. (b): After 1 day, i.e., after 24 hours, the bacterial population will be $=2^{24} \times 100$.
4. (a): Bacterial population after 1 hour $=2 \times 100=200$;
after 2 hours $=2^2 \times 100=400$;
after 3 hours $=2^3 \times 100=800$;
after 5 hours $=2^5 \times 100=3200$.
So, the bacterial population will be larger than 3000 in 5 hours.
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Question 24 Marks
A motorcycle purchased for 218700 loses two thirds of its value every year. Rakesh purchased this motorcycle and he evaluates its value at the end of every year.
Q.1. Which of the following expressions gives the value of the motorcycle (in) after n years?
(a) $\frac{218700}{\left(\frac{2}{3}\right)^n}$$\quad$(b) $\frac{2^n \times 218700}{3^n}$$\quad$(c) $\frac{218700}{3^n}$$\quad$(d) $\frac{218700}{2^n \times 3^n}$
Q.2. Find the value of the motorcycle after 3 years.
(a) ₹8100$\quad$(b) ₹24300$\quad$(c) ₹16200$\quad$(d) ₹32400
Q.3. In how many years will the value of the motorcycle be less than500?
(a) 9 years$\quad$(b) 8 years$\quad$(c) 7 years$\quad$(d) 6 years
Q.4. By how much will the value of the motorcycle decrease in 4 years?
(a) ₹2700$\quad$(b) ₹ 8100$\quad$(c) ₹210600$\quad$(d) ₹216000
Answer
1. (c): The motorcycle loses two thirds of its value every year. So, its value becomes $\frac{1}{3}$ of its value in the previous year.
$\therefore$ value after 1 year $=₹\left(\frac{1}{3} \times 218700\right)$,
Value after 2 years $=₹\left(\frac{1}{3} \times \frac{1}{3} \times 218700\right)=₹\left(\frac{1}{3^2} \times 218700\right)$,
Value after 3 years $=₹\left(\frac{1}{3} \times \frac{1}{3^2} \times 218700\right)=₹\left(\frac{1}{3^3} \times 218700\right)$.
Its value after $n$ years $=₹\left(\frac{1}{3^n} \times 218700\right)=₹\left(\frac{218700}{3^n}\right)$.
2. (a): The value of the motorcycle after 3 years $=₹\left(\frac{218700}{3^3}\right)=₹\left(\frac{218700}{27}\right)=₹ 8100$.
3. (d): Value after 1 year $=₹\left(\frac{218700}{3}\right)=₹ 72900$. Value after 2 years $=₹\left(\frac{72900}{3}\right)=₹ 24300$.
Value after 3 years $=₹\left(\frac{24300}{3}\right)=₹ 8100$. Value after 4 years $=₹\left(\frac{8100}{3}\right)=₹ 2700$.
Value after 5 years $=₹\left(\frac{2700}{3}\right)=₹ 900$. Value after 6 years $=₹\left(\frac{900}{3}\right)=₹ 300$.
$\therefore$ the value will be less than ₹ 500 in 6 years.
4. (d): The value of the motorcycle after 4 years $=₹\left(\frac{218700}{3^4}\right)=₹\left(\frac{218700}{81}\right)=₹ 2700$.
$\therefore$ decrease in value after 4 years $=₹(218700-2700)=₹ 216000$.
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case /data -based (4 Marks) - MATHS STD 7 Questions - Vidyadip