2 Marks Questions

Question 12 Marks

Which of the following are in inverse proportion?
The number of workers on a job and the time to complete the job.

Answer

If the number of workers increases, then time to complete the job would decrease.
So, it is the case of inverse proportion.

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Question 22 Marks

A loaded truck travels 14 km in 25 min . If the speed remains the same, how far can it travel in 5h?

Answer

Let distance covered and time taken by truck be $x km$ and $y h$, respectively.
We know that
$
\begin{aligned}
\text { Speed } & =\frac{\text { Distance }}{\text { Time }} \\
\Rightarrow \quad \text { Distance } & =\text { Speed } \times \text { Time }
\end{aligned}
$
Hence, the speed remains the same, so the distance covered would be directly proportional to time.
Here, $x_1=14 km$ and $y_1=25 min=25 \times \frac{1}{60} h$
$\left[\because 1 \min =\frac{1}{60} h\right]$
$
\Rightarrow \quad y_1=\frac{25}{60}=\frac{5}{12} h \text { and } y_2=5 h
$
Now, using the relation $\frac{x_1}{y_1}=\frac{x_2}{y_2}$, we get
$
\begin{aligned}
& \frac{14}{5 / 12}=\frac{x_2}{5} \Rightarrow x_2 \times \frac{5}{12}=5 \times 14 \\
\Rightarrow \quad x_2 & =\frac{5 \times 14 \times 12}{5}=14 \times 12=168 km
\end{aligned}
$
Hence, the loaded truck can travel upto 168 km in 5 h .

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Question 32 Marks

Suppose 2 kg of sugar contains $9 \times 10^6$ crystals. How many sugar crystals are there in
(i) 5 kg of sugar? $\qquad$ (ii) 1.2 kg of sugar?

Answer

Let the amount of sugar and number of crystals be $x$ and $y$. As the amount of sugar increases, the number of crystals also increases in the same ratio. So, it is a case of direct proportion.
(i) Here, $x_1=2, y_1=9 \times 10^6$ and $x_2=5, y_2=$ ?
Now, using the relation $\frac{x_1}{y_1}=\frac{x_2}{y_2}$, we get
$
\begin{aligned}
& \frac{2}{9 \times 10^6}=\frac{5}{y_2} \Rightarrow 2 \times y_2=5 \times 9 \times 10^6 \\
\Rightarrow \quad y_2 & =\frac{45 \times 10^6}{2}=225 \times 10^6 \Rightarrow y_2=2.25 \times 10^7
\end{aligned}
$
Hence, there are $2.25 \times 10^7$ crystals of sugar in 5 kg of sugar.
(ii) Ans. $5.4 \times 10^{\circ}$

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Question 42 Marks

A photograph of a bacteria enlarged 50000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20000 times only, what would be its enlarged length?

Answer

Let the number of times a photograph of a bacteria is enlarged be $x$ and attained length be $y$.
Length of bacteria when enlarged 50000 times = 5 cm
$\begin{aligned} \text { Length of bacteria when enlarged } 1 \text { time } & =\frac{5}{50000} cm \\ & =\frac{1}{10000}=10^{-4} cm\end{aligned}$
So, the actual length of the bacteria is $10^{-4} cm$.
Now, the number of times a photograph of a bacteria enlarged is decreasing, so enlarged length will also decrease. So, the number of times a photograph of a bacteria is enlarged and the length attained are directly proportional to each other.
Here, $x_1=50000, y_1=5$ and $x_2=20000, y_2=$ ?
Now, by using the relation $\frac{x_1}{y_1}=\frac{x_2}{y_2}$, we get
$
\begin{aligned}
\frac{50000}{5} & =\frac{20000}{y_2} \Rightarrow 50000 \times y_2=5 \times 20000 \\
\Rightarrow \quad y_2 & =\frac{5 \times 20000}{50000}=2 \Rightarrow y_2=2
\end{aligned}
$
Hence, its enlarged length would be 2 cm .

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Question 52 Marks

In question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Answer

Let the number of parts of red pigment be $x$.
and the amount of base $=y mL$
Here, $x_1=1, y_1=75$ and $y_2=1800, x_2=$ ?
$
\Rightarrow \frac{1}{75}=\frac{x_2}{1800} \Rightarrow x_2 \times 75=1 \times 1800
$
$\Rightarrow \quad x_2=\frac{1800}{75}=24 \Rightarrow x_2=24$
Hence, 24 parts of red pigment should be mixed with 1800 mL of base.

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Question 62 Marks

Following are the car parking charges near a railway station upto

4 h	₹ 60
8 h	₹ 100
12 h	₹ 140
24 h	₹ 180

Check, if the parking charges are in direct proportion to the parking time.

Answer

Let the parking charges be $x$ and time be $y$.
If $x=$ ₹ $ 60$ and $y=4 h$, then $\frac{x}{y}=\frac{60}{4}=\frac{60÷4}{4÷4}=\frac{15}{1}$
If $x=$ ₹ $ 100$ and $y=8 h$, then $\frac{x}{y}=\frac{100}{8}=\frac{100÷4}{8÷4}=\frac{25}{2}$
If $x=$ ₹ $ 140$ and $y=12 h$, then $\frac{x}{y}=\frac{140}{12}=\frac{140÷4}{12÷4}=\frac{35}{3}$
If $$ ₹ $ x=180$ and $y=24 h$, then $\frac{x}{y}=\frac{180}{24}=\frac{180÷12}{24÷12}=\frac{15}{2}$
Since, all the values of ratio $\frac{x}{y}$ are not same, so the parking charges are not in direct proportion to the parking time.

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Question 72 Marks

Amrita takes 18 h to travel 720 km . Find the time taken by her to travel 360 km if speed of Amrita is constant.

Answer

If speed is constant, then
Distance $=$ Speed $\times$ Time $\Rightarrow$ Distance $\propto$ Time
If we denote distance $=d$, Time $=t$
$
\Rightarrow \quad \frac{d_1}{t_1}=\frac{d_2}{t_2}
$
Here, $t_1=18 h, d_1=720 km$
and $d_2=360 km, t_2=$ ?
$
\therefore \frac{720}{18}=\frac{360}{t_2} \Rightarrow t_2=\frac{360 \times 18}{720}=9
$
Hence, time taken is 9 h .

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Question 82 Marks

If $x$ varies inversely as $y$ and $x=20$ when $y=600$, then find $y$ when $x=400$.

Answer

It is given that $x$ varies inversely as $y$.
Hence, $x_1 y_1=x_2 y_2$
Here, $x_1=20, y_1=600, x_2=400, y_2=$ ?
$
\begin{aligned}
20 \times 600 & =400 \times y_2 \\
\Rightarrow \quad y_2 & =\frac{20 \times 600}{400}=30
\end{aligned}
$

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Question 92 Marks

A car is travelling 60 km in 2 h . Find the distance travelled by the car in 6 h if the speed remains constant.

Answer

If the speed of car remains constant.
Then, distance is directly proportional to time
i.e. $\qquad d \propto t$
Here, $d_1=60 km, t_1=2$ and $d_2=?, t_2=6$
$
\begin{array}{ll}
\Rightarrow & \frac{d_1}{t_1}=\frac{d_2}{t_2}\qquad [\because d= distance,t= time ]\\
\therefore & \frac{60}{2}=\frac{d_2}{6} \\
\Rightarrow & d_2=\frac{60 \times 6}{2}=180 km
\end{array}
$
So, 180 km travelled by the car in 6 h .

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Question 102 Marks

Justify the statement.
For fixed time period and rate of interest, the simple interest is directly proportional to the principal.

Answer

We know that
$
\text { Simple Interest }(SI)=\frac{P \times R \times T}{100}
$
where, $\frac{R \times T}{100}$ is constant. $\qquad$ [given]
$
\begin{array}{rlrl}
\therefore & SI & =P \times \text { constant } . \\
SI & \propto P
\end{array}
$
Hence, it is clear that for fixed time period and rate of interest, the simple interest is directly proportional to the principal.

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Question 112 Marks

If the cost of 10 pencils is ₹ 90. Find the cost of 19 pencils?

Answer

It is the case of direct proportion.
Let number of pencils be $x$ and cost of pencil be ₹ $y$.
Here,
$x \propto y$
$x_1=10, y_1=$ ₹ $90, x_2=19, y_2=?$
Using formula,
$\frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{10}{90}=\frac{19}{y_2}$
$
\begin{array}{l}
\Rightarrow \quad y_2 \times 10=90 \times 19 \\
\therefore \quad y_2=\frac{90 \times 19}{10}=171
\end{array}
$
So, cost of 19 pencils is ₹ 171.

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