Sunday, April 24, 2016

Chapter 6.5 - Metric weights - Solved examples

In the previous section we learned how the metric weights are expressed as decimals. In this section we will see some solved examples.

Solved example 6.7
The weight of a piece of steel tube was balanced by
• 3 blocks of 1 kg each +
• 5 blocks of 1 hectogram each +
• 7 blocks of 1 decagram each +
• 4 blocks of 1 gram each. 
What is the weight of the tube in kg?
Solution:
• 3 blocks of 1 kg each = 3 kg
• 5 blocks of 1 hectogram each = 5 hectograms.
1 hectogram = 110 kg = 0.1 kg
∴ 5 hectograms = 0.5 kg
• 7 blocks of 1 decagram each = 7 decagrams.
1 decagram = 1100 kg = 0.01 kg
 7 decagrams = 0.07 kg
• 4 blocks of 1 gram each = 4 grams.
1 gram = 11000 kg =  0.001 kg
 4 grams= 0.004 kg

Thus total weight = 3 + 0.5 + 0.07 + 0.004 = 3.574 kg

Second method:We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place. (Details here)
We have:
Number of Kilograms = 3; hectograms = 5; decagrams = 7 and grams = 4
So the wt. in kg = 3.574

Solved example 6.8The weight of a bag of rice was balanced by 5 blocks of 1 kg each plus 3 blocks of 1 hectogram each plus 8 blocks of 1 gram each. What is the weight of the bag of rice in kg? If the wt. of the bag alone is 0.254 kg, what is the net wt. of rice?
Solution:
• 5 blocks of 1 kg each = 5 kg
• 3 blocks of 1 hectogram each = 3 
1 hectogram = 0.1 kg
 3 hectograms = 0.3 kg
• 8 blocks of 1 gram each = 8 grams.
1 gram= 0.001 kg
 8 grams= 0.008 kg

Thus total weight = 5 + 0.3 + 0.008 = 5.308 kg

Second method:
We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place. 
We have:
Number of Kilograms = 5; hectograms = 3; decagrams = 0 and grams = 8
So the wt. in kg = 5.308

Wt. of bag alone = 0.254 kg.
There fore wt. of rice = 5.308 - 0.254 = 5.054 kg

Solved example 6.9
The weight of a parcel brought through courier is known to be 2.839 kg. How many blocks each of kg, hectogram, decagram and gram will have to be used to balance this weight ?
Solution:
Weight = 2.839 kg. 
The ‘whole number part’ is 2. So there are full 2 kilograms.
The decimal portion 0.839 kg gives the quantity between 2 kg and 3 kg
This can be split as: 8/10 + 3/100 + 9/1000  - - - (1)
[proof:
• 8/10 = 800/1000
• 3/100 = 30/1000
• 9/1000 = 9/1000
• Total = 800/1000 + 30/1000 + 9/1000 = 839/1000 = 0.839]
From (1), we can say there are 8 ‘one tenths of a kg’ in 0.839. But 1 one tenth of a kg is 1 hectogram. So there are 8 hectograms.
From (1), we can say there are 3 ‘one hundredths of a kg’ in 0.839. But 1 one hundredth of a kg is one decagram. So there are 3 decagrams.
From (1), we can say there are 9 ‘one thousandths of a kg’ in 0.839. But 1 one thousandth of a kg is one gram. So there are 9 grams.

Thus we can write: 2.839 = 2 kg + 8 hectograms + 3 decagrams + 9 grams.

Second method:
We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place.
We have:
Wt. = 2.839 kg
So, Number of Kilograms = 2; hectograms = 8; decagrams = 3 and grams = 9


Solved example 6.10
When some Tomatoes were weighed on an electronic balance, the reading was 3.402 kg. Split this weight into kg, hectograms, decagrams and grams.
Solution:
Second method:
We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place.
We have:
Wt. = 3.402 kg
So, Number of Kilograms = 3; hectograms = 4; decagrams = 0 and grams = 2

Solved example 6.11
Weight of a certain object is 3.256 kg. Express this in decagrams
Solution:
Weight = 3.256 kg. That is., 3 kg + 0.2 kg + 0.05 kg + 0.006 kg = 3 kg + 2 hectograms + 5 decagrams + 6 grams.

We have to convert each item into decagrams:
• 3 kg = 30 hectograms ( 1 kg = 10 hectogram)
30 hectograms = 300 decagrams ( 1 hectogram = 10 decagrams)
• 2 hectograms = 20 decagrams ( 1 hectogram = 10 decagrams)
• 5 decagrams = 5 decagrams
• 6 grams = 0.6 decagrams ( 1 decagram = 10 grams  1 gram = 0.1 decagram)

So we get 3.256 kg = 300 + 20 + 5 + 0.6 = 325.6 decagrams

Solved example 6.12
Weight of a certain object is 5.029 kg. Express this in grams
Solution:
Weight = 5.029 kg. That is., 5 kg + 0.0 kg + 0.02 kg + 0.009 kg = 5 kg + 0 hectograms + 2 decagrams + 9 grams.

Now we have to convert each item into grams:

• 5 kg = 50 hectograms ( 1 kg = 10 hectogram)
50 hectograms = 500 decagrams ( 1 hectogram = 10 decagrams)
500 decagrams = 5000 grams ( 1 decagram = 10 grams)
[Once we understand the basics, we need not write the detailed steps. We need write only this:
5 kg = 5000 grams ( 1 kg = 1000 grams)]
• 0 hectograms = 0 grams
• 2 decagrams = 20 grams ( 1 decagram = 10 grams)
• 9 grams = 9 grams
So we get 5.029 kg = 5000 + 0 + 20 + 9 = 5029 grams

So we have learned how to 
• split a given kg weight into smaller quantities like hectograms, decagrams and grams. 
• combine given smaller quantities into kg. 
• express the given wt in any one unit. 
Now we will see how we can use the set of ‘standard weights’ in day to day life, for finding the weights using a balance. 
In day to day life, we do not use hectograms and decagrams. We use only kilograms and grams. Hectograms and decagrams are used only for some special purposes like ‘quantitity of agricultural products’ obtained from a certain area of farm land. Centigrams and decigrams are also not used. For small quantities we use grams and milligrams only.

So we have to learn to find the quantities in terms of kilograms and grams only. For example, suppose an object weighs 2 kg and 6 hectograms. We do not have hectograms in the set of ‘standard weights’. But we do have kilograms and grams. After putting a 2 kg weight on the right side of the balance, we must put 600 grams above it. Because 6 hectograms = 600 grams. Then the two sides will balance.

We must be able to do the reverse also. That is., if we are given a weight, we must be able to express it in terms of kilograms and/or grams. Consider an example: The reading in an electronic balance is 4.283 kg. We can use this reading to convey the idea. People will understand it. For those who want finer details, we can split it into kg and grams. 4.283 kg is 4 kg plus 283 grams.

proof:
0.283 kg = 2 hectograms + 8 decagrams + 3 grams
• 2 hectograms = 20 decagrams  ( 1 hectogram = 10 decagrams)
20 decagrams = 200 grams ( 1 decagram = 10 grams)
• 8 decagrams = 80 grams ( 1 decagram = 10 grams)
• 3 grams = 3 grams

Total = 200 + 80 + 3 = 283 grams.

From the above proof, we can note the following points:
• We get a wt. in kg
     ♦ The tenths give us hectograms
     ♦ The hundredths give us decagrams
     ♦ The thousandths give us grams
• We want the 'hectograms' and the 'decagrams' to go. For that:
     ♦ Multiply the digit in the tenths place by 100
     ♦ Multiply the digit in the hundredths place by 10
     ♦ keep the digit in the thousandths place as such
• Add the three items. This will give the 'quantity after the decimal point' in grams

An even easier method is to multiply the decimal part by 1000.

Some examples:
3.041 kg = 3 kg + [.041 × 1000] grams = 3 kg + 41 grams
5.002 kg = 5 kg + [.002 × 1000] grams = 5 kg + 2 grams
9.305 kg = 9 kg + 305 grams
2.3 kg = 2 kg + 300 grams
2.03 kg = 2 kg + 30 grams

Readers are advised to write the proof for each of the above examples in all the 3 methods.

So we have seen how the metric weights are expressed as decimals. In the next section we will see the expression of metric volumes as decimals.

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