Exercise 1.1
1. Express each number as product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Answer
(i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
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
Ans:
(i) 26 = 2 × 13
91 =7 × 13
HCF = 13
LCM =2 × 7 × 13 =182
Product of two numbers 26 × 91 = 2366
Product of HCF and LCM 13 × 182 = 2366
Hence, product of two numbers = product of HCF × LCM.
(ii) 510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers 510 × 92 = 46920
Product of HCF and LCM 2 × 23460 = 46920
Hence, product of two numbers = product of HCF × LCM.
(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of two numbers 336 × 54 =18144
Product of HCF and LCM 6 × 3024 = 18144
Hence, product of two numbers = product of HCF × LCM.
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Ans:
(i) 12 = 2 × 2 × 3
15 = 3 × 5
21 =3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420
(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339
(iii) 8 =1 × 2 × 2 × 2
9 =1 × 3 × 3
25 =1 × 5 × 5
HCF =1
LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer
We have the formula that,
Product of LCM and HCF = product of number
LCM × 9 = 306 × 657
Divide both side by 9 we get,
⇒ LCM = 34 × 657 = 22338.
5. Check whether 6ncan end with the digit 0 for any natural number n.
Ans:
If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.
So, value 6n should be divisible by 2 and 5 both.
6n is divisible by 2 but not divisible by 5 as the prime factors of 6 are 2 and 3.
So, it can not end with 0.
6. Explain why 7×11×13 + 13 and 7×6×5×4×3×2×1 + 5 are composite numbers.
Ans:
7 × 11 × 13 + 13
Taking 13 common, we get
13 (7×11 +1 )
⇒ 13(77 + 1 )
⇒ 13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
⇒ 5(1008 + 1)
⇒ 5(1009)
It is product of two numbers and both numbers are more than 1 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.
Ans:
They will be meet again after LCM of both values at the starting point.
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.
Exercise 1.2
1. Prove that √5 is irrational.
Ans:
Let us take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get,
b√5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 …(i)
Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.
That means 5 will divide a. So we can write,
a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
⇒ 5b2 = 25c2
Divide by 25 we get,
b2/5 = c2
Similarly, we get that b will divide by 5 and we have already get that a is divide by 5 but a and b are co prime number. So it contradicts.
Hence, √5 is not a rational number, it is irrational.
2. Prove that 3 + 2√5 is irrational.
Ans:
Let take that 3 + 2√5 is a rational number.
So we can write this number as,
3 + 2√5 = a/b
Here, a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
⇒ 2√5 = (a-3b)/b
Now, divide by 2, we get
√5 = (a-3b)/2b
Here, a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number. But √5 is a irrational number, so it contradicts.
Hence, 3 + 2√5 is a irrational number.
Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2
Ans:
(i) Let take that 1/√2 is a rational number.
So, we can write this number as
1/√2 = a/b
Here, a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get,
1 = (a√2)/b
Now, multiply by b
b = a√2
Divide by a we get
b/a = √2
Here, a and b are integer, so b/a is a rational number. So, √2 should be a rational number
But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a irrational number
(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here, a and b are two co prime number and b is not equal to 0
Divide by 7 we get
√5 = a/(7b)
Here, a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a irrational number.
(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here, a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
⇒ √2 = (a – 6b)/b
Here, a and b are integer so (a-6b)/b is a rational number, So, √2 should be a rational number.
But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.