Example problems based on fundamental operations with different units — lesson. Mathematics State Board, Class 6.

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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

$\begin{array}{l} m cm \\ 3 40 \\ \underline{4 50} \\ \underline{7 90} \end{array}$

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

$\begin{array}{l} m cm \\ 4 20 \\ \underline{2 50} \\ \underline{1 70} \end{array}$

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

$=$

$300 \ ml$

$\times$

$6$

$\begin{array}{l} \underline{300 \times 6} \\ \underline{1800} \end{array}$

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.

$5$

$kg$ 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.

$1$

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2. Example problems based on fundamental operations with different units

Theory: