Real Numbers Exercise – 1.2 NCRT Download pdf || UP Board
Real Numbers
Exercise – 1.2
1. Express
each number as product of its prime factors:
(i) 140
Solution:
(ii)
156
Solution:
(Solve as first question)
The prime factor of 156 = 2 × 2 × 3 × 13
= 22 × 3 × 7
(iii)
3825
Solution:
The prime factor of 3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17
(iv)
5005
Solution:
The prime factor of 5005 = 5 × 7 × 11 × 13
(v)
7429
Solution:
The prime factor of 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
Solution:
The prime factor of 26 = 2 × 13
The prime factor of 91 = 7 × 13
HCF = 13
LCM = 2×7×13 = 182
product of the two numbers = 26×91 = 2366
HCF×LCM = 13×182 = 2366
product of the two numbers = LCM
× HCF
(ii)
510 and 92
Solution:
510 = 2×3×5×17
92 = 2×2×23
HCF = 2
LCM = 2×2×3×5×17×23 = 23460
product of the two numbers = 510×92 = 46920
HCF×LCM = 2×23460 = 46920
product of the two numbers = LCM × HCF
(iii)
336 and 54
हल:
336 = 2×2×2×2×3×7
54 = 2×3×3×3
LCM = 24×33×7 = 3024
HCM = 2×3 = 6
product of the two numbers = 336×54 = 18144
HCF×LCM = 6×3024 = 18144
product of the two numbers = LCM × HCF
3. 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 = 3×5
21 = 3×7
HCF = 3
LCM = 22×3×5×7 = 420
(ii)
17, 23 and 29
Solution:
17 = 17×1
23 = 23×1
29 = 29×1
HCF = 1
LCM = 17×23×29 = 11339
(iii)
8, 9 and 25
Solution:
8 = 2×2×2×1
9 = 3×3×1
25= 5×5×1
HCF = 1
LCM = 2×2×2×3×3×5×5 = 1800
4. Given that HCF(306, 657) = 9, find LCM(306, 657).
Solution:
HCF(306, 657) = 9
We know that,
LCM × HCF = product of the two numbers
LCM(306, 657)
= 22338
5. Check whether 6n can end with the
digit 0 for any natural number n.
Solution:
If a number ends on 0, it is divisible by 2 and 5 (eg- 10)
The prime factor of 6n = (2×3)n
There is no 5 in the prime factorization of 6n. So 6n will not be divisible by 5.
Hence, for any natural number n, the number 6n cannot end with the digit 0.
6. Explain
why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers.
Solution:
First Number = 7×11×13+13
= 13×(7×11×1+1)
= 13×(77+1)
= 13×78
= 13×13×6
This
number has more than two factors (1, 6, 13)
so it is a composite
number, because composite numbers have more than two factors.
Second
Number =
7×6×5×4×3×2×1+5
= 5×(7×6×4×3×2×1+1)
=
5×(1008+1)
=
5×1009
This
number has more than two factors (1, 5, 1009)
so 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:
Sonia takes 18 minutes while Ravi takes 12 minutes, after a time both will be at their starting
positions that time will be the LCM of 12 and 18.
18 = 2×3×3
12 = 2×2×3
LCM(18,
12) = 2×2×3×3 = 36
Hence,
Ravi and Sonia will meet again at the starting place after 36
minutes.
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