Real Numbers Ex 1.2 Solutions
Question - 1 : - Write whether every positive integer can be ofthe form 4q + 2, where q is an integer. Justify your answer.
Answer - 1 : -
Solution:
No, every positive integer cannot be of theform 4q + 2, where q is an integer.
Justification:
All the numbers of the form 4q + 2, where ‘q’is an integer, are even numbers which are not divisible by ‘4’.
For example,
When q=1,
4q+2 = 4(1) + 2= 6.
When q=2,
4q+2 = 4(2) + 2= 10
When q=0,
4q+2 = 4(0) + 2= 2 and so on.
So, any number which is of the form 4q+2 willgive only even numbers which are not multiples of 4.
Hence, every positive integer cannot bewritten in the form 4q+2
Question - 2 : - “The product of two consecutive positiveintegers is divisible by 2”. Is this statement true or false? Give reasons.
Answer - 2 : -
Solution:
Yes, the statement “the product of twoconsecutive positive integers is divisible by 2” is true.
Justification:
Let the two consecutive positive integers = a,a + 1
According to Euclid’s division lemma,
We have,
a = bq + r, where 0 ≤ r < b
For b = 2, we have a = 2q + r, where 0 ≤ r< 2 … (i)
Substituting r = 0 in equation (i),
We get,
a = 2q, is divisible by 2.
a + 1 = 2q + 1, is not divisible by 2.
Substituting r = 1 in equation (i),
We get,
a = 2q + 1, is not divisible by 2.
a + 1 = 2q + 1+1 = 2q + 2, is divisible by 2.
Thus, we can conclude that, for 0 ≤ r < 2,one out of every two consecutive integers is divisible by 2. So, the product ofthe two consecutive positive numbers will also be even.
Hence, the statement “product of twoconsecutive positive integers is divisible by 2” is true.
Question - 3 : - “The product of three consecutive positiveintegers is divisible by 6”. Is this statement true or false? Justify youranswer.
Answer - 3 : -
Solution:
Yes, the statement “the product of threeconsecutive positive integers is divisible by 6” is true.
Justification:
Consider the 3 consecutive numbers 2, 3, 4
(2 × 3 × 4)/6 = 24/6 = 4
Now, consider another 3 consecutive numbers 4,5, 6
(4 × 5 × 6)/6 = 120/6 = 20
Now, consider another 3 consecutive numbers 7,8, 9
(7 × 8 × 9)/6 = 504/6 = 84
Hence, the statement “product of threeconsecutive positive integers is divisible by 6” is true.
Question - 4 : - Write whether thesquare of any positive integer can be of the form 3m + 2, where m is a naturalnumber. Justify your answer.
Answer - 4 : -
Solution:
No, the square of any positive integer cannotbe written in the form 3m + 2 where m is a natural number
Justification:
According to Euclid’s division lemma,
A positive integer ‘a’ can be written in theform of bq + r
a = bq + r, where b, q and r are any integers,
For b = 3
a = 3(q) + r, where, r can be an integers,
For r = 0, 1, 2, 3……….
3q + 0, 3q + 1, 3q + 2, 3q + 3……. are positiveintegers,
(3q)2 = 9q² = 3(3q²) = 3m(where 3q² = m)
(3q+1)2 = (3q+1)² = 9q²+1+6q =3(3q²+2q) +1 = 3m + 1 (Where, m = 3q²+2q)
(3q+2)2 = (3q+2)² = 9q²+4+12q= 3(3q²+4q) +4 = 3m + 4 (Where, m = 3q²+2q)
(3q+3)2 = (3q+3)² = 9q²+9+18q= 3(3q²+6q) +9 = 3m + 9 (Where, m = 3q²+2q)
Hence, there is no positive integer whosesquare can be written in the form 3m + 2 where m is a natural number.
Question - 5 : - A positive integer is of the form 3q + 1, qbeing a natural number. Can you write its square in any form other than 3m + 1,i.e., 3m or 3m + 2 for some integer m? Justify your answer.
Answer - 5 : -
Solution:
No.
Justification:
Consider the positive integer 3q + 1, where qis a natural number.
(3q + 1)2 = 9q2 +6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1, (where m is an integer which is equalto 3q2 + 2q.
Thus (3q + 1)2 cannot beexpressed in any other form apart from 3m + 1.
Question - 6 : - Express each number as a product of its prime factors:-
Answer - 6 : -
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solutions:
(i) 140
We will get the product of its prime factor.
140 = 2× 2 × 5 × 7 × 1 = 22×5×7
Q ii 156
Solutions:
(ii)156
We will get the product of its prime factor.
156 = 2× 2 × 13 × 3 × 1 = 22× 13 × 3
Q iii 3825
Solutions:
(iii) 3825
We will get the product of its prime factor.
3825 =3 × 3 × 5 × 5 × 17 × 1 = 32×52×17
Q iv 5005
Solutions:
(iv) 5005
We will get the product of its prime factor.
5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13
Q v 7429
Solutions:
(v)7429
We will get the product of its prime factor.
7429 =17 × 19 × 23 × 1 = 17 × 19 × 23
Question - 7 : - Find the LCM and HCF of the following pairs of integers and verify thatLCM × HCF = product of the two numbers.
Answer - 7 : -
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
(i) 26 and 91
Solutions:
We will get the product of its prime factor
26 = 2 × 13 × 1
91 = 7 × 13 × 1
LCM(26, 91) = 2 × 7 × 13 × 1 = 182
And HCF (26, 91) = 13
Verification
Now, product of 26 and 91 = 26 × 91 = 2366
And Product of LCM and HCF = 182 × 13 = 2366
Hence, LCM × HCF = product of the 26 and 91.
(ii) 510 and 92
Solutions:
We will get the product of its prime factor
510 = 2 × 3 × 17 × 5 × 1
92 = 2 × 2 × 23 × 1
LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
And HCF (510, 92) = 2
Verification
Now, product of 510 and 92 = 510 × 92 = 46920
And Product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92.
(iii) 336 and 54
Solutions:
We will get the product of its prime factor
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1
54 = 2 × 3 × 3 × 3 × 1
LCM(336, 54) = = 3024
And HCF(336, 54) = 2×3 = 6
Verification
Now, product of 336 and 54 = 336 × 54 = 18,144
And Product of LCM and HCF = 3024 × 6 = 18,144
Hence, LCM × HCF = product of the 336 and 54.
Question - 8 : - Find the LCM and HCF of the following integersby applying the prime factorisation method.
Answer - 8 : -
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Q (i) 12,15 and 21
Solutions:
(i) 12, 15 and 21
We will get the product of its prime factor
12=2×2×3
15=5×3
21=7×3
Therefore,
HCF(12,15,21) = 3
LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420
Q (ii) 17,23 and 29
(ii)17, 23 and 29
We will get the product of its prime factor
17=17×1
23=23×1
29=29×1
Therefore,
HCF(17,23,29) = 1
LCM(17,23,29) = 17 × 23 × 29 = 11339
Q (iii) 8, 9 and 25
(iii)8, 9 and 25
We will get the product of its prime factor
8=2×2×2×1
9=3×3×1
25=5×5×1
Therefore,
HCF(8,9,25)=1
LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800
Question - 9 : - Given that HCF (306, 657) = 9, find LCM (306,657).
Answer - 9 : -
Solution:
We know that,
HCF×LCM=Product of the two given numbers
Then,
9 × LCM = 306 × 657
LCM = (306×657)/9 = 22338
Hence, LCM(306,657) = 22338
Question - 10 : - Check whether 6n can end with the digit 0 for anynatural number n.
Answer - 10 : - 