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Published on: 25/09/2019
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.
The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ______________
2.
The HCF of 45 and 75 is ___________
3.
The number of distinct prime factors of the smallest 4 digit number is ___________
4.
The number of prime numbers between 11 and 60 is________.
5.
Explain your answer with reason for the following statements
A number is divisible by 6, if it is divisible by 12.
6.
Find the smallest number which is exactly divisible by all the numbers from 1 to 9.
7.
Find the HCF and the LCM of the numbers 154, 198 and 286.
8.
I am a two digit prime number and the sum of my digits is 10. I am also one of the factors of 57. Who am I?
9.
The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pairs exist upto 100.
10.
Write the smallest and the biggest two digit prime number.
11.
Which of the following cannot be the HCF of two numbers whose LCM is 120?
60
40
80
30
12.
The greatest 4 digit number which is exactly divisible by 8, 9 and 12 is
9999
9996
9696
9936
13.
If the number 6354 * 97 is divisible by 9, then the value * is
2
4
6
7
14.
The factors of a number are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80. What is the number?
80
100
128
160
15.
The only even prime number is
4
6
2
0
16.
There are four Mobile Phones in a house. At 5 a.m, all the four Mobile Phones will ring together. Thereafter, the first one rings every 15 minutes, the second one rings every 20 minutes, the third one rings every 25 minutes and the fourth one rings every 30 minutes. At what time, will the four Mobile Phones ring together again?
17.
Find the HCF of the numbers 18, 24 and 30 by factor tree method.
18.
The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example
19.
The product of 2 two digit numbers is 300 and their HCF is 5. What are the numbers?
20.
Find the length of the longest rope that can be used to measure exactly the ropes of length 1m 20cm, 3m 60cm and 4m.
21.
The LCM of two co-prime numbers is 5005. If one of the numbers is 65, then find the other number.
22.
A book seller has 175 English books, 245 Science books and 385 Mathematics books. He wants to sell the books in a box, subject-wise in equal numbers. What will be the greatest number of the boxes required? Also find the number of books for each subject in a box.
23.
The HCF of two numbers is always a factor of their LCM
24.
The LCM of two successive numbers is the product of the numbers
25.
If a number is divisible by 6, then it must be divisible by 3.
26.
The sum of any number of odd numbers is always even
1.
3
2.
15
3.
( )
2
4.
( )
12
5.
True, because 72 is divisible by 72, and also 72 is divisible by 6.
Therefore this statement is true.
6.
Given: numbers are 1 to 9.
Therefore the smallest number
= \(2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 2\)
= 2520
7.
i) LCM
Therefore LCM = \(2 \times 11 \times 7 \times 9 \times 13\) = 18018
ii) HCF
Therefore HCF = 2 \(\times\) 11 = 22
8.
Given Sum of my digits is 10.
Factors of 57 are 1, 3, 19, 57
Here, 19 and 57 are two digit prime numbers.
Therefore 19 = 1 + 9 = 10.
Therefore I am 19.
9.
The given prime number 13 can be reversed
to get another prime number 31.
Similarly, other any such pairs are (17, 71), (37, 73) and (79, 97)
10.
The Smallest two digit prime number = 11
The Biggest two digit prime number = 97
11.
(c)
80
12.
(d)
9936
13.
(a)
2
14.
(a)
80
15.
(c)
2
16.
This is a LCM related sum. So, we need to find the LCM of 15, 20, 25 and 30.
The LCM of 15, 20, 25 and 30 is 2 \(\times\) 2 \(\times\) 3 \(\times\) 5 \(\times\) 5
= 300 minutes = 5 \(\times\) 60 minutes = 5 \(\times\) 1 hour = 5 hours
Thus, the four Mobile Phones will ring together again at 10.00 a.m.
17.
Let us find the factors of 18, 24 and 30 (use of divisibility test rules will also help).
The factors of 18 are 
9 and 18.
The factors of 24 
8, 12 and 24.
The factors of 30 are 
10, 15 and 30.
The factors that are common to all the three given numbers are 1, 2, 3 and 6 of which 6 is the highest.
Hence, HCF (18, 24, 30) = 6.
Note that 1 is a trivial factor of all numbers.
Let us find the factors of 24 by tree method.
Here, 24 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3
Similarly, we can find the factors of 18 and 30.
18.
Given statement
The sum of any two successive odd numbers is always divisible by 4.
Example:
Let any two successive odd numbers are
(i) 13, 15 (ii) 17, 19
(i) Sum = 13 + 15 = 28 is divisible by 4.
(ii) Sum = 17 + 19 = 36 is divisible by 4.
Therefore given statment is true.
19.
Given
The product of 2 two digit numbers = 300, HCF = 5
300 = \(15 \times 20=60 \times 5=100 \times 3=2 \times 150=75 \times 4\)
But, given that, two digit numbers.
Therefore 60 \(\times\) 5, 100 \(\times\) 3, 2 \(\times\) 150, 75 \(\times\) 4 is not possible.
300 = 15 \(\times\) 20
HCF = 5
Therefore numbers are 15, 20.
20.
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.
21.
We know that, the product of the two numbers = LCM \(\times\) HCF
As the HCF of co-primes is 1,
65 \(\times\) (the other number) = 5005 \(\times\) 1
The other number = 5005 \(\div \) 65 = 77
22.
This is a HCF related problem. So, we need to find the HCF of 175, 245 and 385.
175 = 5 \(\times\) 5 \(\times\) 7; 245 = 5 \(\times\) 7 \(\times\) 7; 385 = 5 \(\times\) 7 \(\times\) 11
The HCF of 175, 245 and 385 is the product of the common factors 5 and 7 i.e, 5 \(\times\) 7=35
Since each box contains equal number of books, the greatest possible number of boxes = 35
The number of English books in each box = 175 \(\div \) 35 = 5
The number of Science books in each box = 245 \(\div \) 35 = 7
The number of Maths books in each box = 385 \(\div \) 35 = 11
Hence, the total number of books in each box is 5+7+11 = 23.
23.
(a)
24.
(a)
25.
(a)
26.
(b)
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