Exercise 1.2 - Chapter 1 Numbers Term 2 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Fill in the blanks.
(i) The HCF of 45 and 75 is
(ii) The HCF of two successive even numbers is
(iii) If the LCM of 3 and 9 is 9 , then their HCF is
(iv) The LCM of 26,39 and 52 is
(v) The least number that should be added to 57 so that the sum is exactly divisible by $2,3,4$ and 5 is
Solution:
(i) 15
(ii) 156
(iii) 2
(iv) 3
(v) 3

Question $2 .$
Say True or False
(i) The numbers 57 and 69 are co-primes.
(ii) The HCF of 17 and 18 is $1 .$
(iii) The LCM of two successive numbers is the product of the numbers.
(iv) The LCM of two co-primes is the sum of the numbers.
(v) The HCF of two numbers is always a factor of their LCM.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True

Question $3 .$
Find the HCF of each set of numbers using prime factorisation method.
(i) 18,24
(ii) 51,85
(iii) 61,76
(iv) 84,120
(v) $27,45,81$
(vi) $45,55,95$
Solution:
(i) 18,24 .
Prime factorisation of $18=2 \times 3 \times 3$
Prime factorisation of $24=2 \times 2 \times 2 \times 3$
Common factors of 18 and $24=2 \times 3=6$
$\operatorname{HCF}(18,24)=6$

(ii) 51,85
Prime factorisation of $51=3 \times 17$
Prime factorisation of $85=5 \times 17$
Common factors of 51 and $85=17$
$\operatorname{HCF}(51,85)=17$

(iii) 61,76
Prime factorisation of $61=1 \times 61$
Prime factorisation of $76=2 \times 2 \times 19 \times 1$
Common factors of 61 and $76=1$
$\operatorname{HCF}(61,76)=1$

(iv) 84,120
Prime factorisation of $84=2 \times 2 \times 3 \times 7$
Prime factorisation of $120=2 \times 2 \times 2 \times 3 \times 5$
Common factors of 84 and $120=2 \times 2 \times 3$
$\operatorname{HCF}(84,120)=12$

(v) $27,45,81$
Prime factorisation of $27=3 \times 3 \times 3$
Prime factorisation of $45=3 \times 3 \times 5$
Prime factorisation of $81=3 \times 3 \times 3 \times 3$
Common factors of $27,45,81=3 \times 3=9$
$\operatorname{HCF}(27,45,81)=9$

(vi) $45,55,95$
Prime factorisation of $45=3 \times 3 \times 5$
Prime factorisation of $55=5 \times 11$
Prime factorisation of $95=5 \times 19$
Common factors of $45,55,95=5$

Question $4 .$
Find the LCM of each set of numbers using prime factorisation method.
(i) 6,9
(ii) 8,12
(iii) 10,15
(iv) 14,42
(v) $30,40,60$
(vi) $15,25,75$
Solution:
(i) 6,9
Prime factorisation of $6=2 \times 3$
Prime factorisation of $9=3 \times 3$
Product of common factors $=3$
Product of other factors $=2 \times 3=6$
LCM $(6,9)=3 \times 6=18$
(ii) 8,12
$\begin{aligned}
&8=2 \times 4=2 \times 2 \times 2 \\
&12=2 \times 6=2 \times 2 \times 3
\end{aligned}$
Product of common factors $=2 \times 2=4$
Product of other factors $=2 \times 3=6$
LCM $=$ Product of common factors $\times$ Product of other factors $=4 \times 6=24$
LCM $(8,12)=24$.
(iii) 10,15 $10=2 \times 5$ $15=3 \times 5$
$\begin{aligned}
&10=2 \times 5 \\
&15=3 \times 5
\end{aligned}$
Product of common factors $=5$
Product of other factors $=2 \times 3=6$
LCM $(10,15)=$ Product of common factors $\times$ Product of other factors $=5 \times 6=30$

(iv) 14,42 $14=2 \times 7$ $42=2 \times 21=2 \times 3 \times 7$
$\begin{aligned}
&14=2 \times 7 \\
&42=2 \times 21=2 \times 3 \times 7
\end{aligned}$
Product of common factors $=2 \times 7$
Product of other factor $=3$
LCM $(14,42)=$ Product of common factors $\times$ Product of other factors $=2 \times 7 \times 3=42$
$\operatorname{LCM}(14,42)=42$
(v) $30,40,60$
$30=3 \times 2 \times 5$
$40=2 \times 2 \times 2 \times 5$
$60=2 \times 3 \times 2 \times 5$
Product of highest powers of the common factors $=3 \times 2^{3} \times 5=120$
$\operatorname{LCM}(30,40,60)=120$
(vi) $15,25,75$
$15=5 \times 3$
$25=5 \times 5$
$75=5 \times 5 \times 3$
Product of the highest powers of the common factors $=3 \times 5^{2}=3 \times 25=75$
$\operatorname{LCM}(15,25,75)=75$

Question $5 .$
Find the HCF and the LCM of the numbers 154,198 and 286 .
Solution:
Prime factorisation of $154=2 \times 7 \times 11$
Prime Factorisation of $198=2 \times 3 \times 3 \times 11$
Prime factorisation of $286=2 \times 11 \times 13$

To find HCF

Product of common factors of 154,198 and $286=2 \times 11=22$
HCF $(154,198,286)=22$

To find LCM
Product of common factors of atleast two numbers $=2 \times 11=22$

Product of other factors $=7 \times 3 \times 3 \times 13=819$
LCM $(154,198,286)=$ Product of common factors $\times$ Product of other factors $=22 \times 819=18,018$ LCM $(154,198,286)=18,018$

Question $6 .$
What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?
Solution:
This Problem is HCF related problem
Prime factorisation of $80=2 \times 2 \times 2 \times 2 \times 5$
Prime factorisation of $100=2 \times 2 \times 5 \times 5$
Prime factorisation of $120=2 \times 2 \times 2 \times 3 \times 5$
Product of common factors 80,100 and $120=2 \times 2 \times 5=20$
$\mathrm{HCF}(80,100,120)=20$
The volume of the vessel $=20$ litres

Question $7 .$
The traffic lights at three different road junctions change after every 40 seconds, 60 seconds and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?
Solution:
This is an LCM related problem
Finding the LCM of 40,60 and 72
60 seconds $=1 \mathrm{~min}$
$360 \mathrm{~min}=6 \mathrm{~min}$

$\operatorname{LCM}(40,60,72)=2 \times 2 \times 3 \times 5 \times 2 \times 3=360$
After 360 seconds they will change again i.e after six minutes they will change again i.e at $8.06$ am they will change again simultaneously.
 

Question $8 .$
The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible?
Solution:
Let the numbers be $14 x$ and $14 y$
$14 x \times 14 y=14 \times 210$
$\Rightarrow x y=\frac{210}{14}=\frac{30}{2}=15$
$\mathrm{x}=1, \mathrm{y}=15 ; \mathrm{x}=3, \mathrm{y}=5$
$(14,210),(42,70)$ Two pairs
 

Question $9 .$
The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36 , then find the other number.
Solution:
$\mathrm{HCF}=12$
Product of two numbers $=\mathrm{LCM} \times \mathrm{HCF}$
$36 \times$ other number $=72 \times 12$
Other number $=\frac{72 \times 12}{36}$
Other number $=24$

Objective Type Questions
Question 10 .
Which of the following pairs is co-prime?
(a) 51,63
(b) 52,91
(c) 71,81
(d) 81,99
Solution:
(c) 71,81

Question $11 .$
The greatest 4 digit number which is exactly divisible by 8,9 and 12 is
(a) 9999
(b) 9996
(c) 9696
(d) 9936
Solution:
(d) 9936

Question $12 .$
The HCF of two numbers is 2 and their LCM is 154 . If the difference between numbers is 8 , then the sum is
(a) 26
(b) 36
(c) 46
(d) 56
Solution:
(b) 36
 

Question $13 .$
Which of the following cannot be the HCF of two numbers whose LCM is $120 ?$
(a) 60
(b) 40
(c) 80
(d) 30
Solution:
(c) 80