Exercise 4.3 - Chapter 4 Direct and Inverse Proportion Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.3$
Miscellaneous Practice Problems
Question $1 .$
If the cost of $7 \mathrm{~kg}$ of onions is ₹ 84 find the following :
(i) Weight of the onions bought for ₹ 180
(ii) The cost of $3 \mathrm{~kg}$ of onions
Solution:
(i) For ₹ 84 weight of onion bought
for $₹ 1$ weight of onion bought
$\therefore$ For $₹ 180$ weight of onion bought w
$\therefore$ For ₹ 180 weight of onion bought
(ii) Cost of $7 \mathrm{~kg}$ of onions $=15 \mathrm{~kg}$
Cost of $1 \mathrm{~kg}$ of onions $=\frac{84}{7}$
Cost of $3 \mathrm{~kg}$ of onions $=\frac{84}{7} \times 3=36$
Cost of $3 \mathrm{~kg}$ onions $=36$

Question $2 .$
If $\mathrm{C}=\mathrm{kd}$
(i) what is the relation between $\mathrm{C}$ and $\mathrm{d}$ ?
(ii) Find $\mathrm{k}$ when $\mathrm{C}=30$ and $\mathrm{d}=6$
(iii) Find $\mathrm{C}$, when $\mathrm{d}=10$
Solution:

Given $\mathrm{C}=k d \Rightarrow \frac{\mathrm{C}}{d}=k$
This is in the form $\frac{x}{y}=k$
As $\mathrm{C}$ increases $\mathrm{d}$ also increases
$\therefore$ It is direct proportion
(ii) When $\mathrm{C}=30$ and $d=6 ; k=\frac{\mathrm{C}}{d}$
$\begin{aligned}
&=\frac{30}{6} \\
&k=5
\end{aligned}$
(ii) When $d=6, \mathrm{C}=30$ [ from ii]
$\text { When } \begin{aligned}
d &=10, \mathrm{C}=? \\
\frac{\mathrm{C}_{1}}{d_{1}} &=\frac{\mathrm{C}_{2}}{d_{2}} \\
\frac{30}{6} &=\frac{\mathrm{C}_{2}}{10} \\
\mathrm{C}_{2} &=\frac{30 \times 10}{6}=50, \text { when } d=10 \\
\mathrm{C} &=50
\end{aligned}$

Question $3 .$
Every 3 months Tamilselvan deposits $₹ 5000$ as savings in his bank account. In how many years he can save ₹ $1,50,000$.
Solution:
Let the number of years required be $x$.

No. of years and deposit are direct proportion as they both increases simultaneously.

$\begin{aligned}
\frac{x_{1}}{y_{1}} &=\frac{x_{2}}{y_{2}} \\
\frac{\frac{3}{12}}{5000} &=\frac{x}{1,50,000} \\
\frac{\not 3}{12} \times 1,5,50,000 &=x \times 5000 \\
x &=\frac{37500}{5000} \\
x &=7 \frac{1}{2} \text { years }
\end{aligned}$
He can save $₹ 1,50,000$ in $7 \frac{1}{2}$ years.
 

Question $4 .$
A printer, prints a book of 300 pages at the rate of 30 pages per minute. Then, how long will it take to print the same book if the speed of the printer is 25 pages per minute?
Solution:
Let the required time taken to print be $x$
As the speed increases time taken to print decreases
$\therefore$ They are in inverse proportion
Time taken to print 30 pages $=1 \mathrm{~min}$
$\therefore$ Time taken to print 300 pages $=\frac{300}{30}=10 \mathrm{~min}$

$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
10 \times 30 &=x \times 25 \\
x &=\frac{30 \times 10}{25}=12 \mathrm{~min}
\end{aligned}$
Time taken to print $=12 \mathrm{~min}$
 

Question $5 .$
If the cost of 6 cans of juice in ₹ 210 , then what will be the cost of 4 cans of juice?
Solution:
Let the cost required be $x$

As number of cans increases cost also increases.
$\therefore$ They are in direct proportion
$\begin{aligned}
\therefore \quad \frac{x_{1}}{y_{1}} &=\frac{x_{2}}{y_{2}} \\
\frac{6}{210} &=\frac{4}{x} \\
6 \times x &=4 \times 210 \\
x &=\frac{2}{4 \times 210} \\
\therefore \quad 6
\end{aligned}$
$\mathrm{x}=140$
Cost of 4 cans of juice $=140$

Question $6 .$
$x$ varies inversely as twice of $y$. Given that when $y=6$, the value of $x$ is 4 . Find the value of $x$ when $y$ $=8$.
Solution:
Given x yaries inversely as twice of y.

$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
4 \times 6 &=x \times 8 \\
x &=\frac{4 \times 6}{8} \\
x &=3 \\
\text { When } y &=8 x=3
\end{aligned}$

Question $7 .$
A truck requires 108 litres of diesel for covering a distance of $594 \mathrm{~km}$. How much diesel will be required to cover a distance of $1650 \mathrm{~km}$ ?
Solution:
Let the required distance be $\mathrm{x}$

As the distance increases fuel quantity also increases.
$\therefore$ They are direct proportion.
$\begin{aligned}
\frac{x_{1}}{y_{1}} &=\frac{x_{2}}{y_{2}} \\
\frac{108}{594} &=\frac{x}{1650} \\
x &=\frac{108 \times 1650}{594} \\
x &=300
\end{aligned}$
$\therefore$ The diesel required $=300$ liters
 

Challenge Problems
Question 8 .
If the cost of a dozen soaps is ₹ 396 , what will be the cost of 35 such soaps?
Solution:
1 dozen $=12$
Cost of 12 soaps $=₹ 396$
$\therefore$ Cost of 1 soaps $=₹ \frac{396}{12}=₹ 33$
$\therefore$ Cost of 35 soaps $=₹ 33 \times 35=₹ 1155$
Cost of 35 soaps $=₹ 1155$

Question $9 .$
In a school there is 7 periods a day each of 45 minutes duration. How long each period is if the school has 9 periods a day assuming the number of hours to be the same?
Solution:
Number of periods increases as duration decreases, since the number of hours is same.
Let the duration of each period be $x$.

$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
7 \times 45 &=9 \times x \\
x &=\frac{7 \times 45}{9} \\
x &=35
\end{aligned}$
Duration of every period $=35 \mathrm{~min}$

Question 10 .
Cost of 105 note books is ₹ 2415 . How many notebooks can be bought for ₹ 1863 ?
Solution:
For 2415 number of notebooks bought $=105$
$\therefore$ For ₹ 1 number of note books bought $=\frac{105}{2415}=\frac{1}{23}$
$\therefore$ For ₹ 1863 number of note books bought $=\frac{1}{23} \times 1863=81$
81 Note books can be bought
 

Question $11 .$
10 farmers can plough a field in 21 days. Find the number of days reduced if 14 farmers ploughed the same field?
Solution:
Let the required number of days if 14 farmers ploughed $=x$

As number of farmers increases, number of days decreases.
$\therefore$ They are in inverse proportion
$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
21 \times 10 &=x \times 14 \\
x &=\frac{21 \times 10}{14} \\
x &=15
\end{aligned}$
Initially the farmers worked for 21 days. Now they worked for 15 days.
$\therefore$ The number of days reduced $=21-15=6$ days

Question 12 .
A flood relief camp has food stock by which 80 people can be benefited for 60 days. After 10 days 20 more people have joined the camp. Calculate the number of days of food shortage due to the addition of 20 more people?
Solution:

As number of people increases food last for less number of days.
$\begin{aligned}
80 \times 50 &=100 \times x \\
x &=\frac{80}{80 \times 50} \\
x =40 
\end{aligned}$
Remaining food is to be used for 50 days.
But it only last for 40 days.
No. of days shortage $=50-40=10$ days.
$\therefore 10$ days of food shortage due to the addition of 20 more people.

Question $13 .$
Six men can complete a work in 12 days. Two days later, 6 more men joined them. How many days will they take to complete the remaining work?
Solution:

As the number of men increases number of days increases.
$\therefore$ They are inversely proportional
$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
6 \times 10 &=12 \times x \\
x &=\frac{6 \times 10}{12} \\
x &=5 \text { days }
\end{aligned}$
$\mathrm{x}=5$ days
$\therefore$ Remaining work will be complete in 5 days