Class 6 Maths Chapter 3 Playing With Numbers | Expected Questions For Annual Examination

EXPECTED QUESTIONS FOR ANNUAL EXAMINATION | CLASS 6 MATHEMATICS | PLAYING WITH NUMBERS –Chapter 3

Answer the following (2 Marks):

1. Find the LCM of 12, 30 and 60.

2. Find the HCF of 48 and 64.

3. Manu wants to place 48 chairs and 32 tables in a hall. What is the highest number of rows possible if each row has the same number of chairs and same number of tables?

4. Using divisibility test, determine which of the following numbers are divisible by 11.

a) 42801
b) 5698317

5. Write down separately the prime and composite numbers less than 20.

6. What is the smallest number, which when divided separately by 15, 20 and 30; will in each case leave 7 as a remainder?

7. Express the following as the sum of two odd primes.

a) 34
b) 24

8. Which of the following numbers are co prime?

a) 18 and 35
b) 17 and 68
c) 30 and 60

9. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case?

10. Determine the smallest three digit number which is exactly divisible by 6, 8 and 12.

ANSWERS:

1. Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120 ……………………….

Multiples of 30 are 30, 60, 90, 120, 150……………………….
Multiples of 60 are 60, 120, 180, 240 ……………….
Common Multiples are 60, 120 …………………..
LCM = 60

2. Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 64 are 1, 2, 4, 8, 16, 32 and 64.
Common Factors are 1, 2, 4, 8 and 16.
HCF = 16

3. Here we need to find the HCF of 48 and 32.

Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors 32 are1, 2, 4,8,16 and 32.
Common Factors are 1, 2, 4, 8 and 16.
HCF = 16

4. a) 42801

Sum at odd places = 1 + 8 + 4 = 13
Sum at even places = 0 + 2 = 2
Difference = 13 – 2 = 11, which is a multiple of 11.
So 42801 is divisible by 11.
b) 5698317
Sum at odd places = 7 + 3+ 9+ 5 = 24
Sum at even places = 1 + 8+ 6 = 15
Difference = 24 – 15 = 9, which is not a multiple of 11.
So 5698317 is not divisible by 11.

5. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.

Composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.

6. To find the smallest number, we need to find the LCM of 15, 20 and 30.

Multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120 ……………………
Multiples of 20 are 20, 40, 60, 80,100, 120 …………………….
Multiples of 30 are 30, 60, 90, 120………………..
Common Multiples are 60, 120 …………………..
LCM = 60
When leave 7 as a remainder, the smallest number is 60 + 7 = 67.

7. a) 34 = 31 + 3

b) 24 = 19 + 5

8. a) 18 and 35.
Factors of 18 are 1, 2, 3, 6, 9 and 18.
Factors of 35 are 1, 5, 7 and 35.
Common factors are 1.
So 18 and 35 are co-prime.
b) 17 and 68
Factors of 17 are 1 and 17.
Factors of 68 are 1, 2, 4, 17, 34 and 68.
Common factors are 1 and 17.
So 17 and 68 are not co-prime.
c) Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
Common factors are 1, 2, 3, 5, 10, 15 and 30.
So 30 and 60 are not co-prime.

 

9. Here we need to find the LCM of 6, 15 and 18.

Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90…………………
Multiples of 15 are 15, 30, 45, 60, 75, 90………………
Multiples of 18 are 18, 36, 54, 72, 90…………….
Common Multiples are 90.
LCM = 90
When leaves a remainder 5 in each case = 90 +5 = 95

10. Here we need to find the LCM of 6, 8 and 12.

Multiples of 6 are 6, 12, 18, 24, 30 36………………..
Multiples of 8 are 8, 16, 24, 32, 40………………………
Multiples of 12 are 12, 24, 36, 48……………
Common Multiples are 24, 48………………
LCM = 24
Now multiples of 24 are 24, 48, 72, 96, 120…………………..
The smallest three digit number is 120.

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