Given x and yare inversely proportional
xy = constant
wheny = 35 and constant = 7; x x 35 = 7

\(x=\cfrac { 1 }{ 5 } \)

From the table itself we observe that as x increases y decreases
 \(\therefore\) x and y are inversely proportional
\(\therefore\) xy = 8 x 32 = 16 x 16 = 32 x 8 = 256 x 1 = 256

Using unitary method we can solve this as follows:
The cost of 2 notebooks = Rs. 24
The cost of 1 notebook = \(\frac { 24 }{ 2 } \) = Rs.12
Therefore, the cost of 9 notebooks = 9 x Rs.12
= Rs.108
Hence, Anbu has to pay Rs.108 for 9 notebooks.

Let x be the required number of charts.
As weight increases, the number of charts also increases. So the quantities are in direct proportion.
Hence \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)

\(\frac { 15 }{ 50 } =\frac { x }{ 2500 } \)
15 x 2500 = x \(\times\) 50
x \(\times\) 50 = 15 x 2500
x = \(\frac { 15\times 2500 }{ 50 } \) = 750
Therefore, 750 charts will weigh 2\(\frac {1}{2}\) kilogram.

Let x be the number of pencils required for 18 children. As the number of children increases, number of pencils also increases.
In the case of direct proportion we take \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)

\(\frac { 6 }{ 24 } =\frac { 18 }{ x } \)
\(\frac { 6\times x }{ 24 } \) = 18
x = \(\frac { 18\times 24 }{ 6 } \) = 72
Hence, 72 pencils are required for 18 children.

For Rs 130 distance travelled = 100 km

300 can be travelled with Rs 390

In \(\cfrac { 1 }{ 2 } \) an hour number of words typed = 540
i.e., In 30 min No. of words typed = 540

= 18
In 12 minutes number of words typed = 18 X 12
= 216
216 words can be typed in 12 min

6 cans of juice costs Rs. 210
1 can of juice costs \(=Rs. \frac{210}{6}=Rs. 35 \)
4 cans of juice costs = Rs. 35 x 4 = Rs. 140.

This is an inverse proportion x₁y₁ = x₂ y₂
10 x 60 = 30 ducks x
\(x=\frac{10 \times 60}{30}\)
x = 20
30 machines take 20 days to dig the same hole.

3 months rent = Rs. 7500
1 month rent \(=\frac{7500}{3}=Rs. 2500\)
One year rent \(=\frac{7500}{3} \times 12=Rs. 30000\)
Thamarai has to pay Rs. 30000 for 1 year.

Length of the road = 600m
\(\cfrac { 2 }{ 3 } \) rd of the road = \(\cfrac { 2 }{ 3 } \times 600m=400m\)
In \(2\cfrac { 1 }{ 2 } \) hrs sumathi sweeps 600m
\(\therefore\) In 1 hr sumathi Sweeps \(\cfrac { 600 }{ 2\frac { 1 }{ 2 } } m=\cfrac { 600 }{ \frac { 5 }{ 2 } } m=120\times 2=240m\)
In \(1\frac { 1 }{ 2 } \) hrs Ramani sweeps 400 m.
\(\therefore\) In 1hr Ramani sweeps \(\cfrac { 400 }{ 1\frac { 1 }{ 2 } } m=\cfrac { 400 }{ 3 } m=400\times \cfrac { 2 }{ 3 } m=\cfrac { 800 }{ 3 } =266.66m\)
\(\therefore\) Ramanl sweeps more. in 1 hr
\(\therefore\) Ramani sweeps speedily.

Let x be the required number of days. The decrease in number of workers lead to the increase in number of days. (Therefore, both are in inverse proportion)
For inverse proportion x₁y₁ = x₂ y₂

Hence 60 x 7 = 42 \(\times\) x
42 \(\times\) x = 60 x 7
x = \(\frac { 60\times 7 }{ 42 } \)
x = 10
In 10 days 42 workers can spin a bale of cotton.