Exercise 11.1 (Revised) - Chapter 13 - Direct & Inverse Proportions - Ncert Solutions class 8 - Maths

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Chapter 11 - Direct & Inverse Proportions | NCERT Solutions for Class 8 Maths

Ex 11.1 Question 1.

Following are the car parking charges near a railway station upto:
4 hours Rs. 60
8 hours Rs. 100
12 hours Rs. 140
24 hours Rs. 180

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

Charges per hour:
$
\begin{aligned}
& C_1=\frac{60}{4}=\text { Rs. } 15 \\
& C_2=\frac{100}{8}=\text { Rs. } 12.50
\end{aligned}
$
$
\begin{aligned}
& C_3=\frac{140}{12}=\text { Rs. } 11.67 \\
& C_4=\frac{180}{24}=\text { Rs. } 7.50
\end{aligned}
$

Here, the charges per hour are not same, i.e., $C_1 \neq C_2 \neq C_3 \neq C_4$

Therefore, the parking charges are not in direct proportion to the parking time.
Ex 11.1 Question 2.

A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Answer.

Let the ratio of parts of red pigment and parts of base be $\frac{a}{b}$.
Here $a_1=1, b_1=8$
$
\Rightarrow \frac{a_1}{b_1}=\frac{1}{8}=k \text { (say) }
$

When $a_2=4, b_2=$ ?
$
k=\frac{a_2}{b_2} \Rightarrow b_2=\frac{a_2}{k}=\frac{4}{\frac{1}{8}}=4 \times 8=32
$

When $a_3=7, b_3=$ ?
$
k=\frac{a_3}{b_3} \Rightarrow b_3=\frac{a_3}{k}=\frac{7}{\frac{1}{8}}=7 \times 8=56
$

When $a_4=12, b_4=$ ?
$
k=\frac{a_4}{b_4} \Rightarrow b_4=\frac{a_4}{k}=\frac{12}{\frac{1}{8}}=12 \times 8=96
$

When $a_5=20, b_5=$ ?

$k=\frac{a_5}{b_5} \Rightarrow b_5=\frac{a_5}{k}=\frac{20}{\frac{1}{8}}=20 \times 8=160$

Ex 11.1 Question 3.

In Question 2 above, if 1 part of a red pigment requires $75 \mathrm{~mL}$ of base, how much red pigment should we mix with $1800 \mathrm{~mL}$ of base?

Ans. Let the parts of red pigment mix with $1800 \mathrm{~mL}$ base be $x$.

Since it is in direct proportion.
$
\begin{aligned}
& \therefore \frac{1}{75}=\frac{x}{1800} \\
& \Rightarrow 75 \times x=1 \times 1800 \\
& \Rightarrow x=\frac{1 \times 1800}{75}=24 \text { parts }
\end{aligned}
$

Hence with base $1800 \mathrm{~mL}, 24$ parts red pigment should be mixed.
Ex 11.1 Question 4.

A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Answer.

Let the number of bottles filled in five hours be $x$.

 Here ratio of hours and bottles are in direct proportion. 

$
\begin{aligned}
& \therefore \frac{6}{840}=\frac{5}{x} \\
& \Rightarrow 6 \times x=5 \times 840 \\
& \Rightarrow x=\frac{5 \times 840}{6}=700 \text { bottles }
\end{aligned}
$

Hence machine will fill 700 bottles in five hours.
Ex 11.1 Question 5.

A photograph of a bacteria enlarged 50,000 times attains a length of $5 \mathrm{~cm}$ as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Answer.

Let Actual length of bacteria be 'a'

It is enlarged 50,000 times so $50000 \times \mathrm{a}=5 \mathrm{~cm}$
Actual length of bacteria
$
=\frac{5}{50000}=\frac{1}{10000} \mathrm{~cm}=10^{-4} \mathrm{~cm}
$

Let enlarged length of bacteria be $x$

Here length and enlarged length of bacteria are in direct proportion.
$
\therefore \frac{5}{50000}=\frac{x}{20000}
$

$
\begin{aligned}
& \Rightarrow x \times 50000=5 \times 20000 \\
& \Rightarrow x=\frac{5 \times 20000}{50000}=2 \mathrm{~cm}
\end{aligned}
$

Hence the enlarged length of bacteria is $2 \mathrm{~cm}$.
Ex 11.1 Question 6.

In a model of a ship, the mast is $9 \mathrm{~cm}$ high, while the mast of the actual ship is $12 \mathrm{~m}$ high. If the length of the ship is $28 \mathrm{~m}$, how long is the model ship?

Answer.

$\text {Let the length of model ship be } x \text {. }$

Here length of mast and actual length of ship are in direct proportion.
$
\begin{aligned}
& \therefore \frac{12}{9}=\frac{28}{x} \\
& \Rightarrow x \times 12=28 \times 9 \\
& \Rightarrow x=\frac{28 \times 9}{12}=21 \mathrm{~cm}
\end{aligned}
$

Hence length of the model ship is $21 \mathrm{~cm}$.
Ex 11.1 Question 7.

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

Answer

$\text {(i) Let sugar crystals be } x \text {. }$

Here weight of sugar and number of crystals are in direct proportion.
$
\begin{aligned}
& \therefore \frac{2}{9 \times 10^6}=\frac{5}{x} \\
& \Rightarrow x \times 2=5 \times 9 \times 10^6 \\
& \Rightarrow x=\frac{5 \times 9 \times 10^6}{2} \\
& =22.5 \times 10^6=2.25 \times 10^7
\end{aligned}
$

Hence the number of sugar crystals is $2.25 \times 10^7$.

(ii) Let sugar crystals be $x$.

Here weight of sugar and number of crystals are in direct proportion.
$
\begin{aligned}
& \therefore \frac{2}{9 \times 10^6}=\frac{1.2}{x} \\
& \Rightarrow x \times 2=1.2 \times 9 \times 10^6 \\
& \Rightarrow x=\frac{1.2 \times 9 \times 10^6}{2} \\
& =0.6 \times 9 \times 10^6=5.4 \times 10^6
\end{aligned}
$

Hence the number of sugar crystals is $5.4 \times 10^{\circ}$.

Ex 11.1 Question 8.

Rashmi has a road map with a scale of $1 \mathrm{~cm}$ representing $18 \mathrm{~km}$. She drives on a road for $72 \mathrm{~km}$. What would be her distance covered in the map?

Answer.

Let distance covered in the map be $x$.

Here actual distance and distance covered in the map are in direct proportion.
$
\begin{aligned}
& \therefore \frac{18}{1}=\frac{72}{x} \\
& \Rightarrow x \times 18=72 \times 1 \\
& \Rightarrow x=\frac{72 \times 1}{18}=4 \mathrm{~cm}
\end{aligned}
$

Hence distance covered in the map is $4 \mathrm{~cm}$.
Ex 11.1 Question 9.

A $5 \mathrm{~m} 60 \mathrm{~cm}$ high vertical pole casts a shadow $3 \mathrm{~m} 20 \mathrm{~cm}$ long. Find at the same time (i) the length of the shadow cast by another pole $10 \mathrm{~m} 50 \mathrm{~cm}$ high (ii) the height of a pole which casts a shadow $5 \mathrm{~m}$ long.

Answer.

Here height of the pole and length of the shadow are in direct proportion.
And $1 \mathrm{~m}=100 \mathrm{~cm}$
$
\begin{aligned}
& 5 \mathrm{~m} 60 \mathrm{~cm}=5 \times 100+60=560 \mathrm{~cm} \\
& 3 \mathrm{~m} 20 \mathrm{~cm}=3 \times 100+20=320 \mathrm{~cm} \\
& 10 \mathrm{~m} 50 \mathrm{~cm}=10 \times 100+50=1050 \mathrm{~cm} \\
& 5 \mathrm{~m}=5 \times 100=500 \mathrm{~cm}
\end{aligned}
$

$\text { (i) Let the length of the shadow of another pole be } x \text {. }$

$
\begin{aligned}
& \therefore \frac{560}{320}=\frac{1050}{x} \\
& \Rightarrow x \times 560=1050 \times 320 \\
& \Rightarrow x=\frac{1050 \times 320}{560}=600 \mathrm{~cm}=6 \mathrm{~m}
\end{aligned}
$

Hence length of the shadow of another pole is $6 \mathrm{~m}$.
(ii) Let the height of the pole be $x$.

$
\begin{aligned}
& \therefore \frac{560}{320}=\frac{x}{500} \\
& \Rightarrow x \times 320=560 \times 500 \\
& \Rightarrow x=\frac{560 \times 500}{320} \\
& =875 \mathrm{~cm}=8 \mathrm{~m} 75 \mathrm{~cm}
\end{aligned}
$

Hence height of the pole is $8 \mathrm{~m} 75 \mathrm{~cm}$.
Ex 11.1 Question 10.

A loaded truck travels $14 \mathrm{~km}$ in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Answer.

Let distance covered in 5 hours be $x \mathrm{~km}$.

$\because 1$ hour $=60$ minutes
$\therefore 5$ hours $=5 \times 60=300$ minutes

Here distance covered and time are in direct proportion.
$
\begin{aligned}
& \therefore \frac{14}{25}=\frac{x}{300} \\
& \Rightarrow x \times 25=14 \times 300 \\
& \Rightarrow x=\frac{14 \times 300}{25}=168 \mathrm{~km}
\end{aligned}
$