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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Question $1 .$
A pair of prime numbers whose difference is 2, is called
Solution:
Twin primes

Question $2 .$
The first 4 digit number divisible by 3
Solution:
1002

Question $3 .$
Prime triplet
Solution:
$(3,5,7)$

Question $4 .$
100 years =
Solution:
100 years $=1$ Century

Question 5 .
Loss =
Solution:
Loss $=$ CP $-$ SP

Question $6 .$
$526 \mathrm{ml}$
Solution:
$526 \mathrm{ml}=0.526 \mathrm{~L}$

Question $7 .$
No sides are equal
Solution:
Scalene Triangle
 

Question $8 .$
What is the total number of primes up to 100 ?
Solution:
25
 

Question $9 .$
Check whether $(37,39)$ is a twin prime?
Solution:
No, because 39 is not a prime number.
 

Question $10 .$
Check the divisibility by 11 of $684398 ?$
Solution:
In 684398
Sum of digits in odd places $=8+3+8=19$
Sum of digits in even places $=6+4+9=19$.
Difference $=19-19=0$.
684398 is divisible by $11 .$
 

Question $11 .$
Is 53249624 is divisible by 8 ? How?
Solution:
In 53249624 , consider the last three digits 624 , which is divisible by $8 .$ 53249624 is divisible by 8 .

Question $12 .$
Factorise 1056
Solution:

$1056=2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 11$

Question $13 .$
Express 42 and 100 as the sum of two consecutive primes?
Solution:
$\begin{aligned}
&42=19+23 \\
&100=47+53
\end{aligned}$
 

Question $14 .$
A heap of stones can be made up into groups of 21 . When made up into groups of $16,20,25$ and 45 there are 3 stones left in each case. How many stones at least can there be in the heap?
Solution:
LCM of $16,20,25,45=2 \times 5 \times 2 \times 5 \times 2 \times 2 \times 3 \times 3=3600$

The heap contain $3600+3=3603$ stones at least.
3603 stones at least can there be in the heap.
 

Question $15 .$
Find the largest number of four digits exactly divisible by $12,15,18$ and 27
Solution:
The largest number of four digits $=9999$
lcm of $12,15,18,27$ is 540
Dividing 9999 by 540
We get 279 as the remainder

$\mathrm{LCM}=2 \times 3 \times 3 \times 2 \times 5 \times 3=540$
Required number $=9999-279=9720$

Question $16 .$
Find the least number which when divided by $6,7,8,9$ and 12 leaves the same remainder 1 in each case.
Solution:
Required number $=[\mathrm{LCM}(6,7,8,9,12)]+1$
$\operatorname{LCM}(6,7,8,9,12)=3 \times 2 \times 2 \times 2 \times 3 \times 7=504$
Required number $=504+1=505$

Question $17 .$
Product of two co-prime numbers is 117 . Then what will be their LCM?
Solution:
We know that $\mathrm{LCM} \times \mathrm{HCF}=$ Product of two numbers
Also, we know that HCF of two co-primes $=1$
LCM $\times 1=117$
LCM $=117$
 

Question $18 .$
Six bells commence tolling together and toll at intervals of $2,4,6,8,10$ and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Solution:
LCM of $2,4,6,8,10$ and 12 is 120
So the bell will toll together after every 120 seconds i.e 2 minutes.
In 30 minutes, they will toll together $\frac{30}{2}+1=16$ times.
$\mathrm{LCM}=2 \times 2 \times 3 \times 2 \times 5=120$