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Published on: 04/09/2019
Numbers
Download Tamil Nadu 6th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Your friend says that every odd number is prime. Give an example to prove him/her wrong.
2.
Write the smallest and the biggest three digit composite number.
3.
Express 42 and 100 as the sum of two consecutive primes.
4.
The greatest 4 digit number which is exactly divisible by 8, 9 and 12 is
9999
9996
9696
9936
5.
If the number 6354 * 97 is divisible by 9, then the value * is
2
4
6
7
6.
The prime factorisation of 60 is 2 \(\times\) 2 \(\times\) 3 \(\times\) 5. Any other number which has the same prime factorisation as 60 is
30
120
90
impossible
7.
The difference between two successive odd numbers is
1
2
3
0
8.
The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ______________
9.
The LCM of 26, 39 and 52 is ____________
10.
If the LCM of 3 and 9 is 9, then their HCF is _______________
11.
Find the LCM of 156 and 124.
12.
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?
13.
Find the length of the longest rope that can be used to measure exactly the ropes of length 1m 20cm, 3m 60cm and 4m.
14.
The LCM of two numbers is 432 and their HCF is 36. If one of the numbers is 108, then find the other number.
15.
The HCF of two numbers is always a factor of their LCM
16.
The LCM of two co-primes is the sum of the numbers
17.
The LCM of two successive numbers is the product of the numbers
1.
Answer : Given statement is false.
Example : Let the odd number be 9.
9 is an odd number, but not prime.
Therefore, every odd number is not prime.
2.
The smallest 3 digit composite number = 100
The biggest three digit composite number = 999
3.
42 = 19+23;
100 = 47+53
4.
(d)
9936
5.
(a)
2
6.
(d)
impossible
7.
(b)
2
8.
3
9.
156
10.
3
11.
Solution: By Division method
Step 1: Start with the smallest prime factor and go on dividing till all the numbers are divided as given below.
Step 2: LCM = product of all prime factors
= 2 \(\times\) 2 \(\times\) 3 \(\times\) 13 \(\times\) 31 = 4836
Thus, the LCM of 156 and 124 is 4836.
By Prime Factorisation method
Step 1: We write the prime factors of 156 and 124 as given below (use of divisibility test rules will also help).
156 = 2 \(\times\) 78 = 2 \(\times\) 2 \(\times\) 39 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 13
124 = 2 \(\times\) 62 = 2 \(\times\) 2 \(\times\) 31
Step 2: The product of common factors is 2 \(\times\)2 and also the product of the factors that are not common is 3 \(\times\) 13 \(\times\) 31.
Step 3: Now, LCM = product of common factors x product of factors that are not common
= (2 \(\times\) 2) \(\times\) (3 \(\times\) 13 \(\times\) 31) = 4 \(\times\) 1209 = 4836
Thus, LCM of 156 and 124 is 4836.
(or)
156 = 2 \(\times\) 78 = 2 \(\times\) 2 \(\times\) 39 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 13;
124 = 2 \(\times\) 62 = 2 \(\times\) 2 \(\times\) 31
The prime factor 2 appears a maximum of 2 times in the prime factorization of 156 and 124, the prime factor 3 appears only 1 time in the prime factorization of 156, the prime factor 13 appears only 1 time in the prime factorization of 156 and the prime factor 31 appears only 1 time in the prime factorization of 124.
Hence, the required LCM = (2 \(\times\) 2) \(\times\) 3 \(\times\) 13 \(\times\) 31 = 4836.
12.
Given
Wilson, Mathan and Guna can complete 1 round of a circular track in 10,15 and 20 minutes respectively.
Starting point at 7 a.m.
LCM = \(5 \times 2 \times 3 \times 2\) = 60 min = 1 hour = 1a.m.
⇒ 7a.m. + 1a.m. = 8a.m.
Therefore after 60 minutes at 8 a.m. will they meet together again at the starting point.
13.
Given statement
The ropes of length 1 m 20 cm, 3 m 60 cm and 4 m.
1 m 20 cm = 120 cm, 3 m 60 cm = 360 cm and 4 m = 400 cm
To find the length of the longest rope.
HCF = \(5 \times 2 \times 2 \times 2\) = 40
Therefore the length of the longest rope = 40 cm.
14.
We know that, the product of the two numbers = LCM \(\times\) HCF
108 \(\times\) (the other number) = 432 \(\times\) 36
The other number = (432 \(\times\) 36) \(\div \) 108 = 144
15.
(a)
16.
(b)
17.
(a)
6th Standard Syllabus & Materials
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