On this page, you will read important Class 10 Mathematics Chapter 1 Real Number Exercise 1.1.
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- Class : 10th
- Chapter : 1
- Subject : Mathematics
- Topic : Real Number Exercise 1.1
- Resource : Notes, Important Question & Practice Paper
1. Express Each Number as a Product of its Prime Factors :
(i). 140
(ii). 156
(iii). 3825
(iv). 5005
(v). 7429
Solution : (i).
Hence,
140 = 2 × 2 × 5 × 7
= 2² × 5 × 7
= 2 × 5 × 7
Solution : (ii).
Hence,
156 = 2 × 2 × 3 × 13
= 2² × 3 × 13
Solution : (iii).
Hence,
3825 = 3 × 3 × 5 × 5 × 17
= 3² × 5 × 17
Solution : (iv).
Hence,
3825 = 5 × 7 × 11 × 13
(v). 7429
Hence,
7429 = 17 × 19 × 23
2. 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
(ii). 510 and 92
(iii). 336 and 54
(i). Solution : Finding L.C.M
L.C.M. = 13 × 2 × 7
L.C.M. = 182
(i). Solution : Finding H.C.F
H.C.F = 13
Now, we have to verify that
Since, L.H.S = R.H.S
Hence Verified.
(ii). Solution : Finding L.C.M
L.C.M. = 2 × 2 × 3 × 5 × 17 × 23
(ii). Solution : Finding H.C.F
H.C.F = 2
Now, we have to verify that
Since, L.H.S = R.H.S
Hence Verified.
(ii). Solution : Finding L.C.M
L.C.M. = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7
L.C.M. = 3024
(iii). Solution : Finding H.C.F
H.C.F = 6
Now, we have to verify that
Since, L.H.S = R.H.S
Hence Verified.
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i). 12, 15 and 21
(ii). 17, 23 and 29
(iii). 8, 9 and 25
(i). Solution :- Finding H.C.F
H.C.F = 3
(i). Solution :- Finding L.C.M
L.C.M. = 2 × 2 × 3 × 5 × 7
L.C.M. = 420
(ii). Solution : Finding H.C.F
17 = 17
23 = 23
29 = 29
Since no Factor is Common so, 1 must be common
H.C.F = 1
(ii). Solution : Finding L.C.M
L.C.M = 17 × 23 × 19
L.C.M = 11339
(iii). Solution :- Finding H.C.F
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
Since no Factor is Common so, 1 must be common
H.C.F = 1
(iii). Solution :- Finding L.C.M
L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 × 5
L.C.M. = 8 × 9 × 25
L.C.M. = 1800
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution :- We Known That
H.C.F × L.C.M = Product of Number
9 × L.C.M = 306 × 657
L.C.M = (306 × 657)/9
L.C.M = 22338
L.C.M of 306 $ 657 is 22338
5. Check whether 6″ can end with the digit 0 for any natural number n.
Solution :- Let us tAKE The Example of a Number which ends with the digit 0
so,
10 = 2 × 5
100 = 2 × 2 × 5 × 5
Here we note number ending with 0 has both 2 and 5 their prime factor whereas
6ⁿ = (2 × 3)ⁿ
Does note have 5 as a prime factor so, it does not end with zero
∴ 6ⁿ cannot end with zero for any natural number n.
6. Explain why 7x11x13+13 and 7x6x5x4x3x2x1+5 are composite numbers.
Solution:-
Checking 7 × 11 × 13 + 13
7 × 11 × 13 + 13
= 13 × ( 7 × 11 + 1)
= 13 × ( 77 + 1 )
= 13 × 78
= 13 × 13 × 3 × 2
Since it has more than two factors
(13, 3, 2, 1, 13 × 78)
It is a Composite Number
Checking 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Now,
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 × (7 × 6 × 4 × 3 ×2 × 1 + 1)
= 5 × (1009 + 1)
= 5 ×1009
It is a composite Number
7. 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:-
Time taken by Sonia to drive one round
= 18 minutes
Time taken by Ravi to drive one round
= 12 minutes
Time taken by both to meet again =18 &12
We find L.C.M 18 and 12 using Prime factorization
L.CM = 2 × 2 × 3 × 3
= 36
Hence, taken by both to meet again= L.CM of 18 & 12
= 36 minutes
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