CBSE Class 10 maths chapter 1 Ex 1.2 || NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2

\( 140=2 \times 2 \times 5 \times 7=2^{2} \times 5 \times 7 \)
\(\begin{array}{|l|l|}
\hline 2 & 140 \\
\hline 2 & 70 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1 \\
\hline
\end{array}\)

\( 156=2 \times 2 \times 3 \times 13=2^{2} \times 3 \times 13 \)
\(\begin{array}{|l|l|}
\hline 2 & 156 \\
\hline 2 & 78 \\
\hline 3 & 39 \\
\hline 13 & 13 \\
\hline & 1 \\
\hline
\end{array}\)

\( 3825=3 \times 3 \times 5 \times 5 \times 17=3^{2} \times 5^{2} \times 17 \)
\(\begin{array}{|l|l|}
\hline 3 & 3825 \\
\hline 3 & 1275 \\
\hline 5 & 425 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1 \\
\hline
\end{array}\)

\( 5005=5 \times 7 \times 11 \times 13 \)
\(\begin{array}{|l|l|}
\hline 5 & 5005 \\
\hline 7 & 1001 \\
\hline 11 & 143 \\
\hline 13 & 13 \\
\hline & 1 \\
\hline
\end{array}\)

\( 7429=17 \times 19 \times 23 \)
\(\begin{array}{|l|l|}
\hline 17 & 7429 \\
\hline 19 & 437 \\
\hline 23 & 23 \\
\hline & 1 \\
\hline
\end{array}\)

\( 26=2 \times 13 \)
\( 91=7 \times 13 \)
\( \mathrm{HCF}=13 \)
\( \mathrm{LCM}=2 \times 7 \times 13=182 \)
Product of the two numbers \( =26 \times 91=2366 \)
\( \mathrm{HCF} \times \mathrm{LCM}=13 \times 182=2366 \)
Hence, product of two numbers \( =\mathrm{HCF} \times \mathrm{LCM} \)

\( 510=2 \times 3 \times 5 \times 17 \)
\( 92=2 \times 2 \times 23 \)
\( \mathrm{HCF}=2 \)
\( \mathrm{LCM}=2 \times 2 \times 3 \times 5 \times 17 \times 23=23460 \)
Product of the two numbers \( =510 \times 92=46920 \)
\( \mathrm{HCF} \times \mathrm{LCM}=2 \times 23460\)
\(\mathrm{HCF} \times \mathrm{LCM}=46920
\)
Hence, product of two numbers \( =\mathrm{HCF} \times \mathrm{LCM} \)

\( 336=2 \times 2 \times 2 \times 2 \times 3 \times 7\)
\(336=24 \times 3 \times 7\)
\(54=2 \times 3 \times 3 \times 3\)
\(54=2 \times 3^{3}\)
\(\text {HCF }=2 \times 3=6\)
\(\text {LCM }=24 \times 3^{3} \times 7=3024\)
Product of the two numbers \( =336 \times 54=18144 \)
\(\mathrm{HCF} \times \mathrm{LCM}=6 \times 3024=18144\)
Hence, product of two numbers \( =\mathrm{HCF} \times \mathrm{LCM} \)

Prime factors of any number is the representation of a number as a product of prime numbers it is composed of
for example: Prime factors of \( 20=2 \times 2 \times 5 \mathrm{HCF}= \) Highest common factor \( = \) The product of the factors that are common to the numbers LCF \( = \) Least Common Factor \( = \) Product of all the factors of numbers without duplicating the factor
let us write prime factors of the given numbers \( 12=2 \times 2 \times 3=22 \times 3 \)
\(
15=3 \times 5\)
\(21=3 \times 7
\)
Only 3 is common in all the three numbers, therefore
\(\mathrm{HCF}=3\)
As 3 is common in all three numbers, it will be taken as 1 time in the product of calculating LCM
\( \text {LCM }=2^{2} \times 3 \times 5 \times 7=420 \)

Prime factors of any number is the representation of a number as a product of prime numbers it is composed of
for example: Prime factors of \( 20=2 \times 2 \times 5 \mathrm{HCF}= \) Highest common factor \( = \) The product of the factors that are common to the numbers LCF \( = \) Least Common Factor \( = \) Product of all the factors of numbers without duplicating the factor
Let us write the prime factors of the given numbers
\(17=1 \times 17\)
\(23=1 \times 23\)
\(29=1 \times 29
\)
As only 1 is common from all the three factors
\(\mathrm{HCF}=1\)
As nothing except 1 is common from all three numbers, simply multiplying them will give LCM of numbers.
\(\mathrm{LCM}=17 \times 23 \times 29=11339\)

Prime factors of any number is the representation of a number as a product of prime numbers it is composed of
for example: Prime factors of \( 20=2 \times 2 \times 5 \mathrm{HCF}= \) Highest common factor \( = \) The product of the factors that are common to the numbers LCF \( = \) Least Common Factor \( = \) Product of all the factors of numbers without duplicating the factor
Let us write the prime factors of the given numbers
\(8=2 \times 2 \times 2\)
\(9=3 \times 3\)
\(25=5 \times 5\)
As nothing is common in the numbers
\(
\mathrm{HCF}=1
\)
As nothing is common in factors of numbers, numbers are simply multiplied to obtain LCM
\(\text {LCM }=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5=1800
\)

Given: \( \operatorname{HCF} \) of \( (306,657)=9 \)
We know that,
\( \mathrm{LCM} \times \mathrm{HCF}= \) product of two numbers
\( \mathrm{LCM} \times \mathrm{HCF}=306 \times 657 \)
\(\mathrm{LCM}=\frac{306 \times 657}{\mathrm {H C F}}=\frac{306 \times 657}{9}\)
\(\mathrm{LCM}=22338\)

We need to find can \( 6^{\mathrm{n}} \) end with zero
If any number has last digit 0,
Then, it should be divisible by 10
Factors of \( 10=2 \times 5 \)
So, Value \( 6^{\mathrm{n}} \) should be divisible by 2 and 5
Prime factorisation of \( 6^{\mathrm{n}}=(2 \times 3)^{\mathrm{n}} \)
Hence,
\( 6^{\mathrm{n}} \) is divisible by 2 but not by 5.
It can not end with 0.

Composite numbers are those numbers, which can be written in the form of the product of two or more integers, and at least one of them should not be 1
(i) \(7 \times 11 \times 13+13\)
\(=(7 \times 11 \times 13)+(13 \times 1)\)
Taking 13 as common, we get,
\(=13 \times(7 \times 11+1)\)
\(=13 \times(77+1)\)
\(=13 \times 78\)
\(=13 \times 13 \times 6\)
As the given no is a multiple of two or more integers, one of them being other than 1.
Hence, it is a composite number.
Therefore, it is a composite number.
(ii) \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1+5\)
\(=(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)+(5 \times 1)\)
Taking 5 as common, we get,
\(=5 \times(7 \times 6 \times 4 \times 3 \times 2 \times 1+1)\)
\(=5 \times(1008+1)\)
\(= 5 \times 1009 \)
As the given no is a multiple of two integers, one of them being other than 1.
Hence, it is a composite number.

Both of them start from the same point and start moving in same direction
Sonia takes 18 minutes and Ravi takes 12 minutes to complete the circle. After 12 minutes Ravi will be back to the starting point and Sonia must have covered \( (\frac{12 }{ 18})=\frac{2 }{ 3} \) of the rounds. After 12 more minutes Ravi has completed 2 rounds and Sonia must have covered \( (\frac{24 }{ 18})=\frac{4 }{ 3} \) of the rounds. After 12 more minutes Ravi has completed 3 rounds and Sonia must have completed \( (\frac{36 }{ 18})=2 \) rounds. Hence after 36 minutes both will again meet at the starting point.
Alternate Method:
They will meet again after LCM of both values at starting point.
\(18=2 \times 3 \times 3\)
And
\(12=2 \times 2 \times 3\)
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.