Exercise 1.2
Q 1, Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 1
Express each number as product of its prime factors:
(i) 140
Sol :
$=2\times 2\times 5\times 7$
$=2^2\times 5\times7$
(ii) 156
Sol :
$=2\times 2\times 3\times 13$
$=2^2\times 3\times 13$
(iii) 3825
Sol :
$=3\times 3\times 5\times 5\times 17$
$=3^2\times 5^2\times 17$
(iv) 5005
Sol :
$=5\times 7\times 11\times 13$
(v) 7429
Sol :
$=17\times 19\times 23$
Q - 2(ii & iii), Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 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
Sol :
$26=2\times 13$
$91=7\times 13$
HCF = 13
LCM = $2\times 7\times 13 = 182$
Product of the two numbers $=26\times 91=2366$
$HCF\times LCM=13\times182$ = 2366
Hence,Product of the two numbers = HCF × LCM
(ii) 510 and 92
Sol :
$510=2\times 3\times 5\times 17$
$92=2\times 2\times 23$
HCF = 2
LCM = $2\times 2\times 3\times 5\times 17\times 23=23460$
Product of the two numbers $=510\times 92=46920 $
$HCF\times LCM=2\times 23460$ = 46920
Hence,Product of the two numbers = HCF × LCM
(iii) 336 and 54
Sol :
$336=2\times 2\times 2\times 2\times 3\times 7$
$336=2^4\times 3\times 7$
$54=2\times 3\times 3\times 3$
$54=2\times 3^3$
$HCF=2\times 3=6$
$LCM=2^4\times 3\times 7$ = 3024
Product of the two numbers $=336\times 54=18144$
$HCF\times LCM=6\times 3024$ = 18144
Hence,Product of the two numbers = HCF × LCM
Q - 3, Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 3
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12 , 15 and 21
Sol :
$12=2^2\times 3$
$15=3\times 5$
$21=3\times 7$
HCF = 3
LCM $=2^2\times 3\times 5\times 7=420$
(ii) 17 , 23 and 29
Sol :
$17=1\times 17$
$23=1\times 23$
$29=1\times 29$
HCF = 1
LCM $=17\times 23\times 29=11339$
(iii) 8 , 9 and 25
Sol :
$8=2\times 2\times 2$
$9=3\times 3$
$25=5\times 5$
HCF = 1
LCM $=2\times 2\times 2\times 3\times 3\times 5\times 5\times$ = 11339
Q - 4, Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 4
Given that HCF (306, 657) = 9, find LCM (306, 657)
Sol :
We know that , LCM × HCF = Product of the two numbers
$\therefore LCM\times HCF$ $=306\times 657$
LCM$=\dfrac{306\times 657}{HCF}$ $=\dfrac{306\times 657}{9}$
LCM = 22338
Q - 5, Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 5
Check whether $6^n$ can end with the digit 0 for any natural number n.
Sol :
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as $10 = 2 × 5$
Prime factorisation of $6^n=(2\times 3)^n$
It can be observed that 5 is not in the prime factorisation of $6^n$.
Hence, for any value of n , $6^n$ will not be divisible by 5 .
Therefore, $6^n$ cannot end with the digit 0 for any natural number n .
Q - 6, Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 6
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Sol :
Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that
$=7\times 11\times 13\times +13$
$=13\times(7\times 11\times +1)$
$=13\times(77\times +1)$
$=13\times 78$
$=13\times 13\times 6$
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
$7\times 6\times 5\times 4\times 3\times 2\times 1 +5$
$5\times(7\times 6\times 4\times 3\times 2\times 1 +1)$
$5\times(1008+1)$
$5\times(1009)$
1009 cannot be factorised further. therefore, the given expression has 5 and 1009 as its factors. Hence ,it is a composite number .
Q - 7, Ex 1.2 - Real Numbers - Chapter 1 - Maths Class 10th - NCERT
Question 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?
Sol :
It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
$18=2\times 3\times 3$
$12=2\times 2\times 3$
LCM of 12 and 18 $=2\times 2\times 3\times 3=36$
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
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