### Theory:

1. Nandhini bought $$3 \ m \ 40 \ cm$$ of blue colour ribbon and $$4 \ m \ 50 \ cm$$ of red colour ribbon. What was the total length of the ribbon bought by her?

Solution:

Length of blue colour ribbon $$=$$ $$3 \ m \ 40 \ cm$$

Length of red colour ribbon $$=$$ $$4 \ m \ 50 \ cm$$

To find the total length, we need to add two quantities.

Total length of the ribbon $$=$$ Length of blue colour ribbon $$+$$ Length of red colour ribbon

$\begin{array}{l}m\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{cm}\\ 3\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}40\\ \underset{¯}{4\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}50}\\ \underset{¯}{7\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}90}\end{array}$

Therefore, the total length of the ribbon bought by Nandhini was $$7 \ m \ 90 \ cm$$.

2. Roshini bought $$4 \ m \ 20 \ cm$$ of blue colour ribbon and $$2 \ m \ 50 \ cm$$ of red colour ribbon. What was the difference between the length of two colour ribbons?

Solution:

Length of blue colour ribbon $$=$$ $$4 \ m \ 20 \ cm$$

Length of red colour ribbon $$=$$ $$2 \ m \ 50 \ cm$$

To find the difference between the length of ribbons, we need to subtract two quantities.

Difference between the length of the ribbons $$=$$ Length of blue colour ribbon $$-$$ Length of red colour ribbon

$\begin{array}{l}m\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{cm}\\ 4\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}20\\ \underset{¯}{2\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}50}\\ \underset{¯}{1\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}70}\end{array}$

Therefore, the difference between the length of two ribbons was $$1 \ m \ 70 \ cm$$.

3. A glass can hold $$300 \ ml$$ of juice. How much litres of juice will be there in $$6$$ glasses?

Solution:

Quantity of juice a glass can hold $$=$$ $$300 \ ml$$

Quantity of juices in $$6$$ glasses $$=$$ $$300 \ ml$$ $$\times$$ $$6$$

$\begin{array}{l}\underset{¯}{\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}300×6}\\ \underset{¯}{1800\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}}\end{array}$

Quantity of juices in $$6$$ glasses $$=$$ $$1800 \ ml$$

We need to find the how much litres of juice can be held in $$6$$ glasses, so convert $$ml$$ to $$l$$.

$$1000 \ ml$$ $$=$$ $$1 \ l$$

$$1 \ ml$$ $$=$$ $\frac{1}{1000}$ $$l$$

$$1800 \ ml$$ $$=$$ $\frac{1800}{1000}$ $$=$$ 1.8 $$l$$

Therefore, $$6$$ glasses can hold 1.8 litres of juice.

4. $$5$$ $$kg$$ of flour to be packed in a small packets, each packet can hold $$250 \ g$$ of flour. How many packets are required to fill $$5 \ kg$$ of flour?

Solution:

Total quantity of flour $$=$$ $$5 \ kg$$

Quantity of flour each packet can hold $$=$$ $$250 \ g$$

Both numbers should be in the same units to do fundamental operations.

So, let us first convert $$kg$$ to $$g$$, to find the number of packets.

$$1 \ kg$$ $$=$$ $$1000 \ g$$

$$5 \ kg$$ $$=$$ $$5 \times 1000 = 5000 \ g$$

Number of packets required $$=$$ $$\frac{\text{Total quantity of flour}}{\text{Quantity of flour each packet can hold}}$$

$$=$$ $\frac{5000}{250}$

$$=$$ 20

Therefore, 20 packets are required to pack $$5 \ kg$$ of flour.

Important!
When we write measurement with different units:

The left side should be a higher unit

The right side should be a lower unit

Example: $$5 \ m \ 15 \ cm$$

Here, $$m$$ is a higher unit and $$cm$$ is a lower unit.