PDF chapter test TRY NOW

**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

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

**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

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

**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 \(=\) \(300 \ ml\) \(\times\) \(6\)

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

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.