1. Real Numbers Mathematics class 10 exercise Exercise 1.4
1. Real Numbers Mathematics class 10 exercise Exercise 1.4 ncert book solution in english-medium
NCERT Books Subjects for class 10th Hindi Medium
Exercise 1.1
Exercise 1.1
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Sol:
(1) 135 and 225
a = 225, b = 135 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
225 = 135 ×1 + 90
135 = 90 ×1 + 45
90 = 45 × 2 + 0 {when we get r=0, our computing get stopped}
b = 45 {b is HCF}
Hence: HCF = 45
Sol:
(ii) 196 and 38220
a = 38220, b = 196 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
Sol:
(iii) 867 and 255
a = 867, b = 255 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Sol:
Let a is the positive odd integer
Where b = 6,
When we divide a by 6 we get reminder 0, 1, 2, 3, 4 and 5, {r < b}
Here a is odd number then reminder will be also odd one.
We get reminders 1, 3, 5
Using Euclid’s division algorithm
So we get
a = 6q + 1, 6q+3 and 6q+5
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Sol:
Maximum number of columns = HCF (616, 32)
a = 616, b = 32 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
616 = 32 ×19 + 8 {when we get r=0, our computing get stopped}
32 = 8 × 4 + 0
b = 8 {b is HCF}
HCF = 8
Hence: Maximum number of columns = 8
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Solution :
To Show :
a2 = 3m or 3m + 1
a = bq + r
Let a be any positive integer, where b = 3 and r = 0, 1, 2 because 0 ≤ r < 3
Then a = 3q + r for some integer q ≥ 0
Therefore, a = 3q + 0 or 3q + 1 or 3q + 2
Now we have;
⇒ a2 = (3q + 0)2 or (3q + 1)2 or (3q +2)2
⇒ a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
⇒ a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 3 + 1
⇒ a2 = 3(3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
Let m = (3q2) or (3q2 + 2q) or (3q2 + 4q + 1)
Then we get;
a2 = 3m or 3m + 1 or 3m + 1
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let , a is any positive integer
By using Euclid’s division lemma;
a = bq + r where; 0 ≤ r < b
Putting b = 9
a = 9q + r where; 0 ≤ r < 9
when r = 0
a = 9q + 0 = 9q
a3 = (9q)3 = 9(81q3) or 9m where m = 81q3
when r = 1
a = 9q + 1
a3 = (9q + 1)3 = 9(81q3 + 27q2 + 3q) + 1
= 9m + 1 where m = 81q3 + 27q2 + 3q
when r = 2
a = 9q + 2
a3 = (9q + 2)3 = 9(81q3 + 54q2 + 12q) + 8
= 9m + 2 where m = 81q3 + 54q2 + 12q
⇒ The EndExercise 1.2
EXERCISE 1.2
Q1. Express each number as a product of its prime factors:
(i) 140
Solution:
= 22 × 5 × 7
(ii) 156
Solution:
= 22 × 3 × 13
(iii) 3825
Solution:
= 32 × 52 × 17
(iv) 5005
Solution:
= 5 × 7 × 11 × 13
(v) 7429
Solution:
Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
Solution:
26 = 2 × 13
91 = 7 × 13
Common factors = 13
∴ HCF = 13
LCM = 2 × 7 × 13 = 182
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
26 × 91 = 13 × 182
2366 = 2366
Hence Varified,
(ii) 510 and 92
Solution:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
Common factors = 2
∴ HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
510 × 92 = 2 × 23460
46920 = 46920
Hence Varified,
(iii) 336 and 54
Solution:
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
Common factors = 2 × 3
∴ HCF = 6
LCM = 2 × 2 × 2× 2 × 3 × 3 × 3 × 7 = 3024
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
336 × 54 = 6 × 3024
18144 = 18144
Hence Varified,
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution:
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Common Factors = 3
HCF = 3
LCM = 3 × 2 × 2 × 5 × 7 = 420
(ii) 17, 23 and 29
Solution:
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
Solution:
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
There is no common factor except 1.
∴ HCF = 1
LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5
= 8 × 9 × 25
= 1800
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9
LCM × HCF = N1 × N2
LCM = 22338
Q5. Check whether 6n can end with the digit 0 for any natural number n.
Solution:
Prime factorisation of 6n = (2 × 3 )n
While, Any natural number which end with digit 0 has
the prime factorisation as form of (2 × 5 )n
Therefore, 6n will not end with digit 0.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Let A = 7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 × 78
Hence this is composite number because It has at least one positive divisor other than one.
Similarily,
Let B = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
Hence this is also a composite number because It has at least one positive divisor other than one.
Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Sonia takes 18 minutes in one round.
Ravi takes 12 minutes in one round
they will meet again at the starting point after LCM(18, 12) minutes
18 = 2 × 3 × 3
12 = 2 × 2 × 3
HCF = 2 × 3 = 6
= 36 minutes
Exercise 1.3
Exercise 1.3
Exercise 1.4
Exercise 1.4
Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
Q2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:
Q3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p , q what can you say about the prime factors of q?
(i) 43.123456789 (ii) 0.120120012000120000. . . (iii) 43.123456789
Solution:
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Mathematics Chapter List
1. Real Numbers
2. Polynomials
3. Pair of Linear Equations in Two Variables
4. Quadratic Equations
5. Arithmetic Progressions
6. Triangles
7. Coordinate Geometry
8. Introduction to Trigonometry
9. Some Applications of Trigonometry
10. Circles
11. Constructions
12. Areas Related to Circles
13. Surface Areas and Volumes
14. Statistics
15. Probability
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