Class 10 Real Numbers Exercise 1.2
SSC Real Numbers :
- Positive or negative, Large or small, whole numbers or decimal numbers are all Real Numbers.
They are called “Real Numbers” because they are not Imaginary Numbers.
Class 10 Real Numbers Exercise 1.2
EXERCISE – 1.2
1Q. Express each of the following number as a product of its prime factors.
i) 140 ii) 156 iii) 3825 iv) 5005 v) 7429
Sol : 140
140 = 2 X 2 X 5 X 7 = 22 X 5 X 7
Sol : ii) 156
156 = 2 X 2 X 3 X 13 = 22 X 3 X 13
Sol : iii) 3825
3825 = 3 X 3 X 5 X 5 X 17
= 32 X 52 X 17
Sol : iv) 5005
👉 5005 = 5 X 7 X 11 X 13
Sol : v) 7429
👉 7429 = 17 X 19 X 23
👉 CLICK HERE FOR REAL NUMBER EXERCISE 1.1
👉 CLICK HERE FOR REAL NUMBER EXERCISE 1.3
👉 CLICK HERE FOR REAL NUMBER EXERCISE 1.4
SSC Real Numbers
2Q. Find the L.C.M and H.C.F of the following integers by the prime factorization method.
Sol. i) 12, 15 and 21 ii) 17, 23 and 29
iii) 8, 9 and 25 iv) 72 and 108
Sol : i) 12, 15 and 21
12 = 2 X 2 X 3 = 22 X 3
15 = 3 X 5
21 = 3 X 7
L.C.M = 22 X 3 X 5 X 7 = 420
H.C.F = 3
Sol : ii) 17, 23 and 29
The given numbers 17, 23 and 29 are all primes.
L.C.M = their product
= 17 X 23 X 29 = 11339
H.C.F = 1
iii) 8, 9 and 25
8 = 2 X 2 X 2 = 23
9 = 3 X 3 = 32
25 = 5 X 5 = 52
L.C.M = 23 X 32 X 52 = 1800
( OR )
8, 9 and 25 are relatively prime, there fore L.C.M is equal to their product. (i.e.) L.C.M = 8 X 9 X 25 = 1800
H.C.F = 1
iv) 72 and 108
72 = 2 X 2 X 2 X 3 X 3
72 = 23 X 32
108 = 22 X 33
L.C.M = 23 X 33 = 8 X 27 = 216
H.C.F = 22 X 32 = 4 X 9 = 316
v) 306 and 657
306 = 2 X 32 X 17
657 = 32 X 73
L.C.M = 2 X 32 X 17 X 73 = 22338
H.C.F = 32 = 9.
SSC Real Numbers
Class 10 Real Numbers Exercise 1.23Q. Check whether 6n can end with the digit ‘0’ for any natural number n.
Sol : Given number = 6n = ( 2 X 3 )n
The prime factors here are 2 and 3 only.
To be end with 0; 6n should have a prime factor 5 and also 2. So,
6n can’t end with zero.
Class 10 Real Numbers exercise 1.2
SSC Real Numbers
4Q. Explain why 7 X 11 X 13 + 13 and 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5 are composite numbers.
Sol : Given numbers are
7 X 11 X 13 + 13 and
7 X 6 X 5 X 4 X 3 X 2 X 1 + 5
13 ( 7 X 11 + 1 ) and 5 ( 7 X 6 X 4 X 3 X 2 X 1 + 1 )
13 K and 5 L, where K = 78 and
L = 7 X 6 X 4 X 3 X 2 X 1 + 1 = 1009
As the given numbers can be written as product of two numbers, they are composite.
Class 10 Real Numbers exercise 1.2
5Q. How will you show that ( 17 X 11 X 2 ) + ( 17 X 11 X 5 ) is a composite number? Explain.
Sol : ( 17 X 11 X 2 ) + ( 17 X 11 X 5 )
= ( 17 X 11 ) ( 2 + 5 )
= ( 17 X 11 ) ( 7 ) = 187 X 7
Now the given expression is written as a product of two integers and hence it is a composite number.
👉 CLICK HERE FOR REAL NUMBER EXERCISE 1.1
👉 CLICK HERE FOR REAL NUMBER EXERCISE 1.3
👉 CLICK HERE FOR REAL NUMBER EXERCISE 1.4
Class 10 Real Numbers Exercise 1.2
For Real Number chapter of SSC Examination
