Real Numbers Question Answers: NCERT Class 10 Mathematics

(i) Here, we have to find H.C.F of 135 and 225

First divide divide the larger integer smaller integer

Since, 225 > 135

Therefore, by Euclid’s Division algorithm

225 = 135 × 1 + 90                                       (i)

Here 90 ≠ 0, so proceed the same procedure further

Again by E.D.L,              (E.D.L = Euclid’s division algorithm)

135 = 90 × 1 + 45                                         (ii)

As we know, 45 ≠ 0 therefore, again by E.D.L

90 = 45 × 2 + 0                                             (iii)

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 45.

From (i), (ii) and (iii)

H.C.F (225, 135) = H.C.F (135, 90) = H.C.F (90, 45) = 45.

(ii) Here, we have to find H.C.F of 38220 and 196

First divide the larger integer smaller integer

Since, 3822 > 196

Therefore by Euclid’s Division Algorithm

38220 = 196 × 195 + 0

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 196.

Hence, H.C.F (38220, 196) = 196.

(iii) Here, we have to find H.C.F of 867 and 255

First divide the larger integer smaller integer

Since, 867 > 255

Therefore, by Euclid’s Division algorithm

867 = 255 × 3 + 102                           (i)

Remainder 102 ≠ 0, so proceed the same procedure further using E.D.L

255 = 102 × 2 + 51                            (ii)

Here, 51 ≠ 0 again using E.D.L = 51 × 2

102 = 51 × 2 + 0                                (iii)

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 51.

From (i), (ii) and (iii)

H.C.F (867, 255) = H.C.F (255, 102) = H.C.F (102, 51) = 51.

Let a be any positive integer b = 6. Then by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0

So, values of r we get, r = 0, 1, 2, 3, 4, 5

When, r = 0, then, a = 6q + 0,

similarly for r = 1, 2, 3, 4, 5 the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5 respectively.

If a = 6q, 6q+2, 6q+4, then a is even and divisible by 2.

Therefore, any positive odd integer is in the form of 6q+1, 6q+3, 6q+5

Where q is some integer.

Total no. of army contingent members = 616

No. of army band members = 32

To find max. numbers of the same columns in which the both groups march. We have to find it.

Their highest common factor. since, 616 > 32 then, by Euclid’s algorithm

616 = 32 × 19 + 8                            (i)

Here 8 ≠ 0 then by again using Euclid algorithm

32  =  8 × 4 + 0                                (ii)

Here, r =0 so we cannot proceed further. The divisor at this Stage is 8.

So the no. of columns is 8.

Let x be any positive integer and b = 3.

Then, by euclid’s algorithm

x = 3q + r, where q ≥ 0 and r = 0, 1, 2          [0 ≤ r ≤ b]

Case (i) :    For r = 0, x = 3q, = x² = 9q², taking 3 as common,

                   x² = 9q²  = 3 (3q²), which is of the form 3m, where m = 3q².

Case (ii) :   For r = 1, x = 3q + 1

                   x² = 9q² + 1 + 6q, taking 3 as common,

                   = 3 (3q² + 2q) + 1, which is of the form 3m + 1, where m = 3q² + 2q

Case (iii) :   For r = 2, 3q + 2

                   x² = 9q² + 4 + 12q = (9q² + 12q + 3) + 1, taking 3 as common,

                   = 3 (3q² + 4q + 1) + 1, which is of the form 3m +1, where m = 3q² + 4q + 1

                   Hence, x² is either of the form 3m, 3m + 1 for some integer m.

Let a be any positive integer and b = 3. Then, by euclid’s algorithm

a = 9q + r, where q ≥ 0 and r = 0, 1, 2                                         [ 0 ≤ r ≤ b ]

For r = 0, x = 3q,   or

For r = 1, x = 3q +1

For r = 2, x = 3q + 2

Now by taking the cube of all the three above terms, we get,

Case (i) :      when r = 0, then,

                     x³ = (3q)³ = 27q³  = 9 (3q³) = 9m; where m = 3q

Case (ii) :     when  r = 1, then,

                     x³ = (3q +1)³ = (3q)³ + 1³ +3 × 3q × 1 (3q + 2 ) = 27q³ + 1 +27q² + 9q

                     Taking 9 as common factor, we get,

                     x³ = 9 (3q³ +3q² + q) +1

                     Putting  (3q³ + 3q² + q) = m, we get,

                     x³ = 9m + 1

Case (iii) :     when r = 2, then,

                     x³ = (3q + 2)³ = (3q)³ + 2³ + 3 × 3q × 2 (3q + 2) = 27q³ + 54q² + 36q + 8

                     Taking 9 as common factor , we get 

                     x = 9 (3q + 6q + 4q) + 8

                     Putting (3q + 6q + 4q) = m, we get,

                     x = 9m + 8,

Therefore, it is proved that the cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.

(i) 140 

By taking the L.C.M of 140, we will get the product of its prime factors.
Therefore, 140 = 2 × 2 × 5 × 7 ×1 = 22 × 5 × 7

(ii) 156

By taking the L.C.M of 156, we will get the product of its prime factors.
Therefore, 156 = 2 ×  2 × 13 × 3 × 1 =  22 × 13 × 3

(iii) 3825

By taking the L.C.M of 3825, we will get the product of its prime factors.
Therefore, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32 × 52 × 17 × 1

(iv) 5005 

By taking the L.C.M of 5005, we will get the product of its prime factors.
5005 = 5 × 5 × 11 × 13 × 1 = 5 × 7 × 11 × 1

(v) 7429 

By taking the L.C.M of 7429, we will get the product of its prime factors.
Therefore, 7429 = 17 × 19 × 13 × 1

(i) 26 and 91

First we have to find the L.C.M and H.C.F of 26, 91 using fundamental theorem of arithmetic,
26 = 13 × 2
91 = 13 × 7
For L.C.M - list all the prime factors (only once) of 26, 91 with their greatest power.
L.C.M (26, 91) = 13 × 2 × 7 = 182
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (26, 91) = 13
Verification :   L.C.M (26, 91) × H.C.F (26, 91) = 26 × 91
                      182 × 13 = 2366,       => 2366 = 2366
                      Hence, proved.

(ii) 510 and 92

First we have to find the L.C.M and H.C.F of 510, 92 using fundamental theorem of arithmetic,
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
For L.C.M - list all the prime factors (only once) of 510, 92 with their greatest power.
L.C.M (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (510, 92) = 2
Verification :  L.C.M (510, 92) × H.C.F (510, 92) = 510 × 92
                     23460 × 2 = 46920,                 =>  46920 = 46920
                     Hence, proved.

(iii) 336 and 54

First we have to find the L.C.M and H.C.F of 336, 54 using fundamental theorem of arithmetic,
336 = 2 × 2 × 2 × 2  × 3 × 7 = 2⁴ × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 3³
For L.C.M - list all the prime factors (only once) of 336, 54 with their greatest power.
L.C.M (336, 54) = 3024
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (336, 54) = 2 × 3 = 6
Verification : L.C.M (336, 54) × H.C.F (336, 54) = 336 × 54
                     3024 × 6 = 18144                    => 18144 = 18144
                     Hence, proved.

(i) 12, 15 and 21
12 = 2 × 3 × 2
15 = 3 × 5
21 = 3 × 7
L.C.M (12, 15, 21) = 2 × 3 × 2 × 5 × 7 = 420
H.C.F (12, 15, 21) = 3

(ii) 17, 23 and 29
17 = 17 × 1
23 = 23 × 1
29 = 29 ×1
L.C.M (17, 23, 29) = 11339
H.C.F (17, 23, 29) = 1

(iii) 8, 9 and 25
8 = 2 × 2 × 2 × 1
9 = 3 × 3 × 1
25 = 5 × 5 × 1
L.C.M (8, 9, 25) = 1800
H.C.F (8, 9, 25) = 1

Given,
H.C.F (306, 657) = 9
First number (a) = 306
Second number (b) = 657
As we know, H.C.F × L.C.M = a × b
9 × L.C.M = 306 × 657
L.C.M = (306 × 657) / 9 = 22338
Hence, L.C.M (306, 657) = 22338

Suppose the number 6ⁿ for any n € N ends with zero. In this case 6ⁿ is divisible by 5.

Prime factorization of 6ⁿ = (2 × 3)ⁿ

Therefore, the prime factorization of 6ⁿ doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, 6ⁿ is not divisible by 5 and thus it proves that 6ⁿ cannot end with the digit 0 for any natural number n.

Composite numbers are factors other than 1 and itself.
From here we can observe that :

(a) 7 × 11 × 13 + 13 = 13 (7 × 11 × 1 + 1) = 13 (77 + 1)
               = 13 × 78
               = 13 × 13  × 6

Therefore, this expression has 6 and 13 as its factors. Hence, it is a composite number.

(b) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 (7 × 6 × 1 × 4 × 3 × 2 × 1 + 1)
                                                  = 5 × (1008 + 1)
                                                  = 5 × 1009

1009 can not be factored further. Therefore, this expression has 5 and 1009 as its factors.
Hence, it is a composite number.

Time taken by Sonia to drive one round = 18 min
Time taken by Ravi to drive one round = 15 min
Let they both meet after time ‘t’. Here, ‘t’ is L.C.M of 18, 12
Now, 18 = 2 × 3 × 3
12 = 2 × 3 × 2
L.C.M (18, 12) = 2² × 3² = 36

Thus, they both meet after 36 minutes at the starting point.

Note: - If x be a rational number, then x can be expressed in the form p/q where p and q are Co- primes. Then, if the prime factorization of denominator (q) is in the form of 2^m × 5ⁿ where n, m are non- negative integers, then x has a decimal expansion which terminates. If the prime factorization is not in this form of 2^m × 5ⁿ then x has a decimal expansion which is non terminating.

(i)                  13 / 3125 

    = 5 × 5 × 5 × 5 × 5 = 5⁵ (on factorization of q)

    Since, its factorization contains only power of 5.

    Therefore, it has a terminating decimal expansion.

(ii)                 17 / 5    

                      = 2 × 2 × 2 = 2³                                        (on factorization of q)

                      Since, its factorization contains only power of 2.

Therefore, it has a terminating decimal expansion.

(iii)                  64 / 455

  = 5 × 7 × 13 = 5¹ × 7 × 13                           (on factorization of q)

                      Since the factorization of not in the 2^m ×5ⁿ

                     Therefore, it has a non -terminating decimal expansion.

(iv)                  15 / 1600

                        = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁶ × 5²          (on factorization of q)

                      Since the factorization of q is in the form 2^m ×5ⁿ

                      Therefore, it has a terminating decimal expansion

  (V)                 29 / 343                              

                           = 7 × 7 × 7 = 7³                                         (on factorization of q)

                       Since the factorization of denominator is not in the form 2^m × 5ⁿ

                       Therefore it has a non -terminating repeating decimal

 (vi)                    23 / 2³ 5²

                         Since factorization of q already given and it is in the form 2^m × 5ⁿ

                     Therefore it has a terminating decimal expansion.

(vii)               129 / 2² 5⁷ 7⁵

                       Since factorization of q already given but it is not in the form 2^m × 5ⁿ

                     Therefore it has a terminating decimal expansion.

(viii)                6 / 15 = 2/5

                       5 = 5 × 1                                                (on factorization of q)

                      Since 5 is the only factor in denominator.                    

                     Therefore it has a non -terminating repeating decimal expansion.

(ix).                35 / 50 = 7 / 10

                       10 = 2 × 5 = 2¹ × 5                                (on factorization of q)

                      Since the factorization of denominator is in the form 2^m × 5ⁿ

                      Therefore it has a terminating decimal expansion.

(x)                 77 / 210 = 11/30

                        30 = 2 × 3 × 5 × 7 = 2¹ × 3 × 5 × 7.            (on factorization of q) 

                    Since the factorization of denominator is not in the form 2^m × 5ⁿ

                    Therefore it has a non -terminating repeating decimal expansion.

      (i)                  13 / 3125    

 (ii)          17 / 8     

(iii)            64/455 is non terminating repeating decimal expansion.

(iv)           15 / 1600 

(v)                 29/343 is non terminating repeating decimal expansion.

(vi)                23 / 2³ 5²

(vii)              120 / 2³ 5⁷ 7⁵ is non terminating repeating decimal expansion.

(viii)               6/15 = 2/5