Friday, July 29, 2016

Chapter 6.11 - Recurring decimals - Solved examples

In the previous section we wrote 56  in the decimal form. In this section, we will solve the rest of the problems.

(ii) 311 : 1. We know that 311 = (3×10) (11×10)
2. Let us rearrange the right side: 311 = 310 × 1011 
3. In the above result, we cannot write 1011 as a mixed fraction. So in this problem, we cannot use ten. We will start with 100

First cycle:
1. We know that 311 = (3×100) (11×100)
2. Let us rearrange the right side: 311 = 3100 × 10011 
3. In the above result, we can write 10011 as (9 + 111 )
4. So (2) becomes 311 = 3100 × (9 + 111). So we get:
5. 311 = 27100 + 31100
Second cycle:
1. We know that 311 = (3×1000) (11×1000)
2. Let us rearrange the right side: 311 = 31000 × 100011 
3. In the above result, we can write 100011 as (90 + 1011)
4. So (2) becomes 311 = 31000 × (90 + 1011). So we get:
5. 311 = 2701000 + 3011000
Third cycle:
1. We know that 311 = (3×10000) (11×10000)
2. Let us rearrange the right side: 311 = 310000 × 1000011 
3. In the above result, we can write 1000011 as (909 + 111 )
4. So (2) becomes 311 = 310000 × (909 + 111). So we get:
5. 311 = 272710000 + 3110000

Based on the above results we can write:
 311  27100 + 31100  =   0.27 +  31100
 311  =  2701000   +  3011000      =   0.27 +  3011000
 311    =  272710000   +  3110000       =   0.2727 + 3110000

The fractions 271002701000272710000, etc., gets closer and closer to 311 . Based on this, we can write the decimal form of  311 as:
In method 2, dots are given for both 2 and 7 because, they repeat as a pattern. For the same reason, the line is placed over both 2 and 7 in method 3. The result obtained when we divide 3 by 11 using a calculator is shown below:



(iii) 2311 : This is an improper fraction. We must first convert it into the sum of a natural number and a proper fraction. It is cone as follows:
2311 2211 + 111 = 2 + 111
Now we will work with 111 and add the final result to 2

1. We know that 111 = (1×10) (11×10)
2. Let us rearrange the right side: 111 = 110 × 1011 
3. In the above result, we cannot write 1011 as a mixed fraction. So in this problem, we cannot use ten. We will start with 100

First cycle:
1. We know that 111 = (1×100) (11×100)
2. Let us rearrange the right side: 111 = 1100 × 10011 
3. In the above result, we can write 10011 as (9 + 111 )
4. So (2) becomes 111 = 1100 × (9 + 111). So we get:
5. 111 = 9100 + 11100
Second cycle:
1. We know that 111 = (1×1000) (11×1000)
2. Let us rearrange the right side: 111 = 11000 × 100011 
3. In the above result, we can write 100011 as (90 + 1011)
4. So (2) becomes 111 = 11000 × (90 + 1011). So we get:
5. 111 = 901000 + 1011000
Third cycle:
1. We know that 111 = (1×10000) (11×10000)
2. Let us rearrange the right side: 111 = 110000 × 1000011 
3. In the above result, we can write 1000011 as (909 + 111 )
4. So (2) becomes 111 = 110000 × (909 + 111). So we get:
5. 111 = 90910000 + 1110000

Based on the above results we can write:
 111  9100 + 11100  =   0.09 +  11100
 111  =  901000   +  1011000      =   0.09 +  1011000
 111    =  90910000   +  1110000       =   0.0909 + 1110000

The fractions 910090100090910000, etc., gets closer and closer to 111 . Based on this, we can write the decimal form of  111 as: 111 = 0.0909...

But the problem was to convert 2111. So we can write 2111 = 2.0909...



(iv) 113 : 1. We know that 113 = (1×10) (13×10)
2. Let us rearrange the right side: 113 = 110 × 1013 
3. In the above result, we cannot write 1013 as a mixed fraction. So in this problem, we cannot use ten. We will start with 100

First cycle:
1. We know that 113 = (1×100) (13×100)
2. Let us rearrange the right side: 113 = 1100 × 10013 
3. In the above result, we can write 10013 as (7 + 913 )
4. So (2) becomes 113 = 1100 × (7 + 913). So we get:
5. 113 = 7100 + 91300
Second cycle:
1. We know that 113 = (1×1000) (13×1000)
2. Let us rearrange the right side: 113 = 11000 × 100013 
3. In the above result, we can write 100013 as (76 + 1213)
4. So (2) becomes 113 = 11000 × (76 + 1211). So we get:
5. 113 = 761000 + 1213000
Third cycle:
1. We know that 113 = (1×10000) (13×10000)
2. Let us rearrange the right side: 113 = 110000 × 1000013 
3. In the above result, we can write 1000013 as (769 + 313 )
4. So (2) becomes 113 = 110000 × (769 + 313). So we get:
5. 113 = 76910000 + 3130000
Fourth cycle:
1. We know that 113 = (1×100000) (13×100000)
2. Let us rearrange the right side: 113 = 1100000 × 10000013 
3. In the above result, we can write 10000013 as (7692 + 413 )
4. So (2) becomes 113 = 1100000 × (7692 + 413). So we get:
5. 113 = 7692100000 + 41300000
Fifth cycle:
1. We know that 113 = (1×1000000) (13×1000000)
2. Let us rearrange the right side: 113 = 11000000 × 100000013 
3. In the above result, we can write 100000013 as (76923 + 113 )
4. So (2) becomes 113 = 1100000 × (76923 + 113). So we get:
5. 113 = 769231000000 + 113000000

Based on the above results we can write:
 113  7100 + 91300  =   0.07 +  31100
 113  =  761000   +  1213000      =   0.076 +  1213000
 113    =  76910000   +  3130000       =   0.0769 + 3130000
 113    =  7692100000   +  41300000       =   0.07692 + 41300000

 113    =  769231000000   +  113000000       =   0.076923 + 113000000

The fractions 7100761000769100007692100000769231000000  etc., gets closer and closer to 113 . Based on this, we can write the decimal form of  113 as: 113 = 0.076923...

In the next section we will see some more solved examples.

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