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## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers 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 ml _____

Solution:

526 ml = 0.526 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 × 2 × 2 × 2 × 2 × 3 × 11

Question 13.

Express 42 and 100 as the sum of two consecutive primes?

Solution:

42 = 19 + 23

100 = 47 + 53

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 × 5 × 2 × 5 × 2 × 2 × 3 × 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

LCM = 2 × 3 × 3 × 2 × 5 × 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 = [LCM (6, 7, 8, 9, 12)] + 1

LCM (6, 7, 8, 9, 12) = 3 × 2 × 2 × 2 × 3 × 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 LCM × HCF = Product of two numbers

Also, we know that HCF of two co-primes = 1

LCM × 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.

LCM = 2 × 2 × 3 × 2 × 5 = 120