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Samacheer Kalvi 6th Maths Solutions Term 2 Chapter Chapter 1 Numbers Ex 1.3

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3

Miscellaneous Practice Problems

Question 1.
Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number up to 16.
Solution:
Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13

Question 2.
Is 173 a prime? Why?
Solution:
yes, because it has two factors.

Question 3.
For which of the numbers, from n = 2 to 8, is 2n – 1 a prime?
Solution:
Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3 Q3
For n = 2, 3, 4, 6 and 7 it is prime.

Question 4.
Explain your answer with the reason for the following statements.
(i) A number is divisible by 9, if it is divisible by 3.
(ii) A number is divisible by 6, if it is divisible by 12.
Solution:
(i) False 42 is divisible by 3 but it is not divisible by 9
(ii) True 36 is divisible by 12. Also divisible by 6.

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter Chapter 1 Numbers Ex 1.3

Question 5.
Find A as required
(i) The greatest 2 digit number 9A is divisible by 2.
(ii) The least number 567A is divisible by 3.
(iii) The greatest 3 digit number 9A6 is divisible by 6.
(iv) The number A08 is divisible by 4 and 9.
(v) The number 225A85 is divisible by 11.
Solution:
(i) A number is divisible by 2 if it is an even number.
Greatest 2 digit even number is 98.
∴ A = 8

(ii) A number is divisible by 3 if the sum of its digits is divisible by 3
Sum of digits of 567A = 5 + 6 + 7+ A = 18 + A
∴ 18 is divisible by 3
∴ A maybe 0
The number will be 5670

(iii) A number is divisible by 6 if it is divisible by both 2 and 3
9A6 is even and so divisible by 2
If A = 9 then the sum of digits will be = 24 which is divisible by 3.
The number will be 996 and A = 9

(iv) 08 is divisible by 4, so A08 is divisible by 4.
If A = 1 then the sum of digits will be 9 which is divisible by 9.
The number will be 108 and A = 1

(v) 5 + A + 2 – (8 + 5 + 2) = 7 + A – 15 = -8 + A
∴ A = 8

Question 6.
Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.
Solution:
False 12 is divisible by both 4 and 6. But not divisible by 24

Question 7.
The sum of any two successive odd numbers ir always divisible by 4. Justify this statement with an example.
Solution:
True.
The sum of any two consecutive odd numbers is divisible by 4
For example 11 + 13 = 24, divisible by 4
Also, all the consecutive odd numbers are of the form 4n + 1 or 4n + 3
Their sum = 4x + 4 which is divisible by 4.

Question 8.
Find the length of the longest rope that can be used to measure exactly the ropes of length lm 20 cm, 3m 60 cm and 4 m.
Solution:
Length of ropes are 4 m, 3 m 60 cm and lm 20 cm = 400 cm, 360 cm, 120 cm
Finding HCF (400, 360, 120)
10, 9 and 3 has no common divisor
HCF (400, 360, 120) = 2 × 2 × 2 × 5 = 40
The length of the rope will be 40 cm
Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3 Q8

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter Chapter 1 Numbers Ex 1.3

Challenge Problems

Question 9.
The sum of three prime numbers is 80. The difference of two of them is 4. Find the numbers.
Solution:
Three prime numbers 2, 37, 41
Sum 2 + 37+ 41 = 80
The difference between two of them 41 – 37 = 4

Question 10.
Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.
Solution:
Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

Question 11.
Find the smallest number which is exactly divisible by all the numbers from 1 to 9.
Solution:
To find the smallest number we have to Find the LCM (1, 2, 3, 4, 5, 6, 7, 8, 9)
LCM is 2 × 3 × 2 × 5 × 7 × 2 × 3 = 2520
The required number is 2520
Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3 Q11

Question 12.
The product of any three consecutive number is always divisible by 6. Justify this statement with an example.
Solution:
2 × 3 × 4 = 24 is divisible by 6.

Question 13.
Malarvizhi, Karthiga and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively. Assuming that they met each other on the 1st of October, when will all the three again?
Solution:
Find the LCM (5, 6, 10)
LCM (15, 25, 30) = 5 × 6 = 30
They meet again after 30 days
∴ They met on 1st October
They will meet again on 31st October
Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3 Q13

Question 14.
In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5 th floors respectively. On which floors, will both of them stop together?
Solution:
LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90 and 105.

Samacheer Kalvi 6th Maths Solutions Term 2 Chapter Chapter 1 Numbers Ex 1.3

Question 15.
The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?
Solution:
Given that HCF of 2 numbers is 5
The numbers may like 5x and 5y
Also given their product = 300
5x × 5y = 300
⇒ 25xy = 300
⇒ \(x y=\frac{300}{25}\)
⇒ xy = 12
The possible values of x andy be (1, 12) (2, 6) (3, 4)
The numbers will be (5x, 5y)
⇒ (5 × 1, 5 × 12) = (5, 60)
⇒ (5 × 2, 5 × 6) = (10, 30)
⇒ (5 × 3, 5 × 4) = (15, 20)
(5, 60) is impossible because the given the numbers are two digit numbers.
The remaining numbers are (10, 30) and (15, 20)
But given that HCF is 5
(10, 30) is impossible, because its HCF = 10
The numbers are 15, 20

Question 16.
Find whether the number 564872 is divisible by 88?
Solution:
564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.

Question 17.
Wilson, Mathan and Guna can complete one round of a circular track in 10, 15 and 20 minutes respectively. If they start together at 7 a.m from the starting point, at what time will they meet together again at the starting point?
Solution:
LCM (10, 15, 20) = 60
∴ They will meet at 8 a.m

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