NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev

Class 8 Mathematics by Full Circle

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Class 8 : NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev

The document NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by Full Circle.
All you need of Class 8 at this link: Class 8

Question 1. Write the following numbers in generalized form.
(i) 25        (ii) 73       
(iii) 129        (iv) 302

Solution:   
(i) 25 = 20 + 5
    = 10 x 2 + 5 x 1 = 10 x 2 + 5

(ii)  73 = 70 + 3
    = 10 x 7 + 3 x 1 = 10 x 7 + 3

(iii) 129 = 100 + 20 + 9
   = 100 x 1 + 10 x 2 + 1 x 9 = 100 x 1 + 10 x 2 + 9

(iv) 302 = 100 x 3 + 10 x 0 + 1 x 2 = 300 + 0 + 2

Question 2. Write the following in the usual form.
(i) 10 x 5 + 6     (ii) 100 x 7 + 10 x 1 + 8   
(iii) 100 x a + 10 x c + b

Solution: 
(i) 10 x 5 + 6 = 50 + 6 = 56
(ii) 100 x 7 + 10 x 1 + 8 = 700 + 10 + 8 = 718
(iii) 100 x a + 10 x c + b = 100a + 10c + b = NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev


[Adding the number with reversed digits to the chosen numbers.]
Question: Check what the result would have been if sundaram had chosen the numbers shown below.
1. 27     
2. 39     
3. 64     
4. 17

Solution: 
1.                Chosen number = 27
Number with reversed digits = 72
      Sum of the two numbers = 27 + 72 = 99
Now,                                 99 = 11[9] = 11[2 + 7]
                                              = 11[Sum of the digits of the chosen number]

2.                Chosen number = 39
Number with reversed digits = 93
      Sum of the two numbers = 39 + 93 = 132
Now,                       132 ÷ 11 = 12
i.e.                                  132 = 11[12] = 11[3 + 9]
                                             = 11[Sum of the digits of the chosen number]

3.                Chosen number = 64
Number with reversed digits = 46
                                     Sum = 64 + 46 = 110
                              Now, 110 = 11[10] = 11[6 + 4]
                                             = 11[Sum of the digits of the chosen number]

4.                Chosen number = 17
Number with reversed digits = 71
                                     Sum = 17 + 71 = 88
Now,                                88 = 11[8] = 11[1+ 7]
                                             = 11[Sum of the digits of the chosen number]

[Finding the difference of the chosen number and the number obtained by reversing the digits.]
Question: Check what result would have been if Sundaram had chosen the numbers shown.
1. 17         
2. 21         
3. 96         
4. 37

Solution: 
1.                Chosen number = 17
Number with reversed digits = 71
                            Difference = 71 – 17 = 54
                                             = 9 x [6]
                                             = 9 x [Difference of the digits of the chosen number (7 – 1 = 6)]

2.                Chosen number = 21
Number with reversed digits = 12
                            Difference = 21 – 12 = 9
                                             = 9 x [1]
                                             = 9 x [Difference between the digits of the chosen number (2 – 1 = 1)]

3.                 Chosen number = 96
Number with reversed digits = 69
                            Difference = 96 – 69 = 27
                                             = 9 x [3]
                                             = 9 x [Difference between the digits of the chosen number (9 – 6 = 3)]

4.                Chosen number = 37
Number with reversed digits = 73
                            Difference = 73 – 37 = 36
                                             = 9 x [4]
                                            = 9 x [Difference between the digits of the chosen number (7 – 3 = 4)]

Question: Check what the result would have been if Minakshi had chosen the numbers shown below.
In each case keep a record of the quotient obtained at the end.
1. 132     
2. 469     
3. 737     
4. 901

Solution: 
1.   132         Chosen number = 132
                  Reversed number = 231
                               Difference = 231 – 132 = 99
    We have                 99 ÷ 99 = 1, remainder = 0


2.    469         Chosen number = 469
                   Reversed number = 964
                               Difference = 964 – 469 = 495
  We have                 495 ÷ 99 = 5, remainder = 0


3.                 Chosen number = 737
                 Reversed number = 737
              We have Difference = 737 – 737 = 0
                                   0 ÷ 99 = 0, remainder = 0

4.               Chosen number = 901
               Reversed number = 109
                           Difference = 901 – 109 = 792
    We have           792 ÷ 99 = 8, remainder = 0
Forming three-digit number with given three digits

Note: Generalised form of NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Generalised form ofNCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Generalised form of NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Adding NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
                                                     NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev which is divisible by 37


Question: Check what the result would have been if Sudaram had chosen the numbers shown below.
1. 417     
2. 632     
3. 117     
4. 937

Solution:
1.          Chosen number = 417
Two other numbers with the same digits are 741 and 174
Sum of the three numbers
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
We have 1332 ÷ 37 = 36, remainder = 0

2.            Chosen number = 632
Two other numbers are 263 and 326
Sum of the three numbers
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
We have 1221 ÷ 37 = 33, remainder = 0

3.        Chosen number = 117
Other numbers are 711 and 171
Sum of the three numbers
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
We have 999 ÷ 37 = 27, remainder = 0

4.            Chosen number = 937
Other two numbers are 793 and 379
∴ Sum of the three numbers
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
We have 2109 ÷ 37 = 57, remainder = 0

EXERCISE 16.1
 Question: Find the values of the letters in each of the following and give reasons for the steps
 involved.

NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev  

Solution: 
1. ∵                 A + 5 =NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev2 or 22 or 32, etc.
    ∴        A = 12 – 5 = 7 or 22 – 5 = 17
Since                   A = 17 is not possible
∴                          A = 7
                   NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Since    NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev + 3 + 2 = 6
∴                          B = 6
Thus,                   A = 7, B = 6


2. If                   A + 8 = NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev 3
   Then                    A = 13 – 8 = 5
                      NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
∴              NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev + 4 + 9 = B
or                           B = NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev 4
Clearly                   C = 1
Thus                      A = 5, B = 4 and C = 1


3. ∵           A x A = 1
∴                     A = 1, or A = 5 or A = 6
                NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
If                    A = 1, then
∵                  11 ≠ 9A
∴                   A ≠ 1
If                  A = 5, then
                 75 ≠ 9A
∴               A = 5
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
If            A = 6, then
∵           96 = 9A
∴ A = 6 is the required values of A.
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev

4. We need B x 7 = A
∴ Possible value of B can be 5, [∵ 5 + 7 = 12]
∵ 6 – 3 – 1 = 2
∴ A can be equal to 2
Now  NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
∴ A = 2 and B = 5


5. We need B x 3 = 3B
Since 5 x 3 = 15
∴ Possible value of B can be 5.
Also         0 x 3 = 0, i.e. B = 0 can be another possible value
∵             A x 3 = A + 0 = A
∴ Possible value of A = 5 or A = 0
∴ Since C ≠ 0
∴ Possible value of A = 5
30, B must be    

NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Thus, = 0.
∴       A = 5, B = 0 and C = 1


6. B can be either 5 or 0.
∵                   A x 5 = A
∴ B must be 0
Again A can either be 5 or 0.
∴                  C ≠ 0    ∴ A ≠ 0
∴ A must be equal to 5
NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Thus, we have
Therefore          A = 5, B = 0, C = 2

7. B can be 2, 4, 6 or 8
We need product 111 or 222, or 333 or 444 or 888 out of them 111 and 333 are rejected.
Possible products are 222, 444 or 888
To obtain
The possible value is B = 4
∵     NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev      NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev

∴      A can be either 2 or 7
A x 6 means NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
∴                            A x 6 = 7 x 6 is the accepted value

Now,     NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Thus,           A = 7 and B= 4


8. ∵          10 – 1 = 9
∴                      B = 9
Also  NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
∴                 NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
We have    NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Thus, A = 7 and B = 9

 

9. ∵             8 – 1 = 7
∴                       B = 7
∴                 7 + 4 = 11 ∴ A = 4
           NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Now, we have
Thus               A = 4 and B = 7


10. ∵             10 – 2 = 8
∴                          A = 8
Also               9 – 8 = 1 ∴ B = 1
              NCERT Solutions(Part- 1)- Playing with Numbers Class 8 Notes | EduRev
Now, we have
Thus,                A = 8 and B = 1

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