Exercise 4.2 - 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 $4.2$
Question $1 .$
Fill in the blanks
(i) 16 taps can fill a petrol tank in 18 minutes. The time taken for 9 taps to fill the same tank will be minutes.
(ii) If 40 workers can do a project work in 8 days, then workers can do it in 4 days.
Solutions:
(i) 32
(ii) 80
 

Question $2 .$
6 pumps are required to fill a water sump in $1 \mathrm{hr} 30$ minutes. What will be the time taken to fill the sump if one pump is switched off?
Solution:
Let $x$ be the required time taken

As the number of pumps increases the time taken to fill the sump will be less $\therefore$ They are in inverse proportion
$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
90 \times 6 &=x \times 5 \\
x &=\frac{90 \times 6}{\$}=108 \text { minutes }=\frac{108}{60}=1 \mathrm{hr} 48 \mathrm{~min}
\end{aligned}$

Question $3 .$
A farmer has enough food for 144 ducks for 28 days. If he sells 32 ducks how long will the food last?
Solution:
Let the required number of days be $\mathrm{X}$.

As the number of ducks decreases the food will last for more days.
$\therefore$ They are in inverse proportion. $\mathrm{x}_{1} \mathrm{y}_{1}=\mathrm{x}_{2} \mathrm{y}_{2}$
$\begin{aligned}
28 \times 144 &=x \times 112 \\
x &=\frac{28 \times 144}{112}
\end{aligned}$
$x=36$
The food lasts for 36 days
 

Question $4 .$
It takes 60 days for 10 machines to dig a hole. Assuming that all machines work at the same speed, how long will it take 30 machines to dig the same hole?
Solution:
Let the number of days required be $x$.

As the number of machines increases it takes less days to complete the work $\therefore$ They are in inverse proportion, $\mathrm{x}_{1} \mathrm{y}_{1}=\mathrm{x}_{2} \mathrm{y}_{2}$
$\begin{aligned}
10 \times 60 &=30 \times x \\
x &=\frac{10 \times 60}{30} ; x=20
\end{aligned}$
It takes 20 days to dig the hole

Question $5 .$
Forty students stay in a hostel. They had food stock for 30 days. If the students are doubled then for how many days the stock will last?
Solution:
Let the required number of days be $x$.

As the number of students increases the food last for less number of days
$\therefore$ They are in inverse proportion.
$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
30 \times 40 &=x \times 80 \\
x &=\frac{30 \times 40}{80} \\
x &=15
\end{aligned}$
The food stock lasts for 15 days
 

Question $6 .$
Meena had enough money to send 8 parcels each weighing 500 grams through a courier service. What would be the weight of each parcel, if she has to send 40 parcel for the same money?
Solution:
Let the required weight of the parcel be $x$ grams.

As the number of parcels increases weight of a parcel decreases.
$\therefore$ They are in inverse proportion.
$\begin{aligned}
x_{1} y_{1} &=x_{2} y_{2} \\
8 \times 500 &=40 \times x \\
x &=\frac{8 \times 500}{40} \\
x &=100
\end{aligned}$
Weight of each parcel $=100$ grams
 

Question $7 .$
It takes 120 minutes to weed a garden with 6 gardeners. If the same work is to be done in 30 minutes, how many more gardeners are needed?
Solution:
Let the, number of gardeners needed be $x$.

As the number of gardeners increases the time decreases. They are in inverse proportion,
$\begin{aligned}
\mathrm{x}_{1} \mathrm{y}_{1}=\mathrm{x}_{2} \mathrm{y}_{2} & \\
6 \times 120 &=x \times 30 \\
x &=\frac{6 \times 120}{30} \\
x &=24
\end{aligned}$
$\therefore$ To complete the work in $30 \mathrm{~min}$ gardeners needed $=24$
Already existing gardeners $=6$
$\therefore$ More gardeners needed $=24-6=18$
18 more gardeners are needed
 

Question $8 .$
Neelaveni goes by bicycle to her school every day. Her average speed is $12 \mathrm{~km} / \mathrm{hr}$ and she reaches
school in 20 minutes. What is the increase in speed, If she reaches the school in 15 minutes?
Solution:
Let the speed to reach school in $15 \mathrm{~min}$ be $\mathrm{x}$

$\therefore$ They are in inverse proportion $\mathrm{x}_{1} \mathrm{y}_{1}=\mathrm{x}_{2} \mathrm{y}_{2}$
$\begin{aligned}
12 \times 20 &=x \times 15 \\
x &=\frac{12 \times 20}{15} \\
x &=16
\end{aligned}$
If she reaches in $15 \mathrm{~min}$ the speed $=16 \mathrm{~km} / \mathrm{hr}$
Already running with $12 \mathrm{~km} / \mathrm{hr}$
$\therefore$ Increased speed $=16-12=4 \mathrm{~km} / \mathrm{hr}$
Increase in speed $=4 \mathrm{~km} / \mathrm{hr}$
 

Question 9 .
A toy company requires 36 machines to produce car toys in 54 days. How many machines would be required to produce the same number of car toys in 81 days?
Solution:
Let the required number of machines be $x$

As the number of machines increases number of days required decreases.
$\therefore$ They are in inverse proportion. $x_{1} y_{1}=x_{2} y_{2}$
$\begin{aligned}
36 \times 54 &=x \times 81 \\
x &=\frac{36 \times 54}{81} \\
x &=24
\end{aligned}$
$\therefore 24$ machines would be required

Objective Type Questions

Question $10 .$
12 cows can graze a field for 10 days. 20 cows can graze the same field for __days
(i) 15
(ii) 18
(iii) 6
(iv) 8
Solution:
(iii) 6

Question $11 .$
4 typists are employed to complete a work in 12 days. If two more typists are added, they will finish the same work in days
(i) 7
(ii) 8
(iii) 9
(iv) 10
Solution:
(ii) 8