Additional Questions - 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

Exercise 4.1
Question $1 .$
The amount of extension in an elastic spring varies directly as the weight hung on it. If a weight of $150 \mathrm{gm}$ produces an extension of $2.9 \mathrm{~cm}$, then what weight would produce an extension of $17.4$ cm?
Solution:
To produce $2.9 \mathrm{~cm}$ extension weight needed $=150 \mathrm{gm}$
To produce $1 \mathrm{~cm}$ extension weight needed $=\frac{150}{2.9} \mathrm{gm}$
$\therefore$ To produce $17.4 \mathrm{~cm}$ extension weight needed $=\frac{150 \times 17.4}{2.9} \mathrm{gm}$

 Question 2 .
Reeta types 540 words during half on hour. How many words would she type in 12 minutes?
Solution:
In $\frac{1}{2}$ an hour number of words typed $=540$
i.e., In $30 \mathrm{~min}$ No. of words typed $=540$
$\therefore \quad$ In $1 \mathrm{~min}$ No. of words typed $=\frac{540}{30}$
$=18$

In 12 minutes number of words typed $=18 \times 12$ $=216$
216 words can be typed in $12 \mathrm{~min}$
Question $3 .$
A call taxi charges ₹ 130 for $100 \mathrm{~km}$. How much would one travel for ₹ 390 ? Solution:
For $₹ 130$ distance travelled $=100 \mathrm{~km}$
For $₹ 1$ distance travelled $=\frac{10 \varnothing}{13 \emptyset} \mathrm{km}=\frac{10}{13} \mathrm{~km}$
$\therefore$ For $₹ 390$ distance travelled $=\frac{10 \times 390}{13} \mathrm{~km}=300 \mathrm{~km}$
$300 \mathrm{~km}$ can be travelled with ₹ 390

Exercise $4.2$
Question 1 .
In the following table find out $x$ and $y$ vary directly or inversely?

Solution:
From the table itself we observe that as $x$ increases $y$ decreases.
$\therefore \mathrm{x}$ and $\mathrm{y}$ are inversely proportional
$\therefore \mathrm{xy}=8 \times 32=16 \times 16=32 \times 8=256 \times 1=256$

Question 2 .
If $x$ and $y$ vary inversely as each other and $x=10$ when $y=6$. Find $y$ when $x=15$.
Solution:
Since $x$ and $y$ vary inversely as each other
$\mathrm{xy}=$ constant
$10 \times 6=15 \mathrm{xy}$
$60=15 \mathrm{y}$
$\begin{aligned}
&y=\frac{60}{15} \\
&y=4
\end{aligned}$

Question $3 .$
If $x$ and $y$ vary inversely and if $y=35$ find $x$ when constant of variation is $7 .$
Solution:
Given $x$ andy are inversely proportional
$\mathrm{xy}=$ constant
when $y=35$ and constant $=7 ; x \times 35=7$
$x=\frac{1}{35}=\frac{1}{5}$
$x=\frac{1}{5}$

Exercise $4.3$
Question $1 .$
Sumathi sweeps $600 \mathrm{~m}$ long road in $2 \frac{1}{2}$ hrs. Ramani sweeps $\frac{2}{3}$ rd of same road in $1 \frac{1}{2}$ hrs. Who sweeps more speedily?
Solution:
Length of the road $=600 \mathrm{~m}$
$\frac{2}{3} \mathrm{rd}$ of the road $=\frac{2}{3} \times 600 \mathrm{~m}=400 \mathrm{~m}$
In $2 \frac{1}{2}$ hrs sumathi sweeps $600 \mathrm{~m}$.
$\therefore$ In $1 \mathrm{hr}$ sumathi Sweeps $=\frac{600}{2 \frac{1}{2}} \mathrm{~m}=\frac{600}{\frac{5}{2}} \mathrm{~m}=120 \times 2=240 \mathrm{~m}$.
In $1 \frac{1}{2}$ hrs Ramani sweeps $400 \mathrm{~m}$.
$\therefore$ In $1 \mathrm{hr}$ Ramani sweeps $\frac{400}{1 \frac{1}{2}} \mathrm{~m}=\frac{400}{\frac{3}{2}} \mathrm{~m}=400 \times \frac{2}{3} \mathrm{~m}=\frac{800}{3}=266.66 \mathrm{~m}$
$\therefore$ Ramani sweeps more in $1 \mathrm{hr}$.
$\therefore$ Ramani sweeps speedily.

Question $2 .$
Suma weaves 25 baskets in 35 days. In how many days will she weave 110 baskets?

Solution:
To weave 25 baskets time taken $=35$ days
$\therefore \quad$ To weave 1 basket time taken $=\frac{35}{25}$ days
$\therefore$ To weave 110 baskets time taken $=\frac{35}{25} \times 110=154$ days