Edudel, Directorate of Education Govt. of NCT of Delhi Class 6 Mental Maths Chapter 3 Playing with Numbers Questions Solution. In this chapter, there is total 50 math problems. We have given each questions solution step by step.
1.) Which is the smallest odd prime number?
Answer: the smallest old prime number is 3
2.) How many prime numbers are there between1 to 50?
Answer: the prime numbers are there between1 to 50 are2,3,5,7,11,13,17,19,23,29,31,37,41,43,47
Therefore total 15 prime numbers are there between 1 to 50.
3.) Find all pairs of prime numbers less than 15, whose difference is 2.
Answer: Pairs of prime numbers less than 15, whose difference is 2 = (3,5), (5,7), (11,13)
4.) How many twin primes are there between 20 and 30?
Answer: 3 and 5 are twin primes.
So Notwin primes are there between 20 and 30
5.) Find the greatest 4 digit number divisible by 4.
Answer:
The greatest 4-digit number = 9999
The greatest 4-digit number divisible by 4 should be = 9999- 4
The greatest 4-digit number divisible by 4 should be =9996
6.) Find the greatest 2 digit prime number.
Answer: the greatest 2 digit prime number is 97
7.) Determine first five multiples of 12.
Answer:
First five multiples of 12= 12×1, 12 ×2, 12 ×3, 12 ×4, 12 × 5
First five multiples of 12 = 12, 24, 36, 48, 60
8.) Find the HCF of 9 and 13.
Answer:
The Highest Common factor (HCF)of 9 and 13is 1
The only common factor we can see is 1
9.) Find the LCM of 12 and 15.
Answer :
multiplesof 12 = 12,24,36,48,60,72
multiples of 15 = 15, 30, 45, 60, 90, 225
The least common multiple (LCM) of 12 and 15 = 60
10.) A vessel has 10 litres 500 ml of milk. In how many glasses each of capacity 150 ml can it be filled?
Answer:
1 vessel has amount of milk = 10 liters 500 ml
1 vessel has amount of milk = 10500 ml
1 glass capacity that can be filled = 150 ml
Total No. Of glass of milk = ?
Total No. Of glass of milk = 1 vessel has amount of milk ÷ 1 glass capacity that can be filled
Total No. Of glass of milk = 10500 ÷ 150
Total No. Of glass of milk = 70
70 glasses each of capacity 150 ml can be filled
11.) Find all numbers less than 60, which are common multiples of 2 and 3.
Answer:Common multiples of 2 : 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58
Common multiples of 3 : 3,6,9,12, 15,18,21,24,27,30,33,36,39,42,45,48,51,54,57
Common multiples of 2 and 3 = 6, 12, 18, 24, 30, 36, 42, 48, 54
12.) Find all prime numbers less than 15.
Answer :
The prime numbers less than 15 = 2,3,5,7,11,13
13.) Determine all the factors of 27.
Answer : Determine all the factors of 27 is 1, 3, 9, 27
14.) Find the missing number :-
Answer:
- 5 – – – ( 3×5=15)
- 2 – – – ( 3×2 =6)
15.) Complete the factor tree:-
Answer:
X = 2 – – – (30 × 2 = 60)
Y= 30 – – – (10 ×3= 30)
Z= 5 – – – (2×5 =10)
16.) Find the least number which when divided by 12, 16, 24 and 36 leaves the remainder 7 in each case.
Answer:
17.) Three persons start walking at the same time. The lengths of Their steps be 80cm, 85cm and 90cm respectively .Find the minimum distance that can be measured in exact number of steps.
Answer : minimum distance that is we need to find LCM
LCM of given distance = LCM of 80, 85 and 90
LCM of 80, 85 and 90 = 12240 cm
The minimum distance that can be measured in exact number of steps is 12240 com
- Find the sum of prime numbers between 1 and 10.
Answer: the Prime numbers between 1 to 10 = 2,3,5,7
The sum of prime numbers between 1 and 10 = 2+3+5+7 = 17
19.) A machine, on an average, manufactures 250 screws a day. How many screws did it manufacture in the month of November?
Answer :
Per day machine manufactures screws = 250
Month of November = 30 days
Month of November machine manufactures screws = No. Of days in November × Per day manufactures screws
= 7500 screws
20.) The Length, breadth and height of a room are 825cm, 675cm and 450 cm respectively. Find the maximum length of tape that can measure all the dimensions exact number of times.
Answer :
Length = 825 cm
Breadth = 675 cm
Height = 450 cm
Factors of 825 =1, 3, 5, 11, 15, 25, 33, 55, 75, 165,
Factors of 675 = 1, 3, 5, 9, 15, 25, 27, 45, 75, 135,
Factors of 450 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90,
Need to calculate the HCF of the given dimensions to find the longest tape required to measure the three dimensions of the room.
The HCF of 825, 675 and 450 =75 cm
21.) Determine first five multiples of 9.
Answer :first five multiples of 9 are, 9, 18, 27, 36, 45
22.) Find the HCF of 24 and 36.
Answer:
Factors of 24= 1, 2, 3, 4, 6, 8, 12,24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18
The HCF of 24 and 36 = 12
23.) Find the common factors of 4, 8 and 12.
Answer : factors of 4 =1, 2, 4
Factors of 8 = 1, 2,4
Factors of 12 = 1, 2, 3, 4,6
The common factors of 4,8 and 12 are 1, 2, 4
24.) Find the smallest 4 -digit number which is divisible by 6, 8 and 9.
Answer:
Let’s first find smallest number divisible by 6,8,9
Smallest number divisible by 6,8,9 = LCM of 6,8,9
LCM of 6,8,9 = 72
Smallest 4-digit number = 1000
The smallest 4 -digit number which is divisible by 6, 8 and 9
1000 ÷ 72 = 13.88
1000 ÷72 =14
1000 = 14 ×72
The smallest 4 -digit number which is divisible by 6, 8 and 9 = 14 ×72
The smallest 4 -digit number which is divisible by 6, 8 and 9 = 1008
25.) Find the LCM of 12, 15 and 45.
Answer :
Multiples of 12=12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192
Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195
Multiples of 45=45, 90, 135, 180, 225,
The LCM of 12, 15 and 45 = 180
26.) Find the HCF of 27 and 63.
Answer:
Factors of 27 = 1,3,9, 27
Factors of 63= 1,3,7,9, 21, 63
The HCF of 27 and 63 = 9
27.) If a number is divisible by 10, then what is its one’s digit?
Answer: if any number is divisible by 10 then the only digit in units place will be 0
28.) Find the largest 3 -digit number which is exactly divisible by 3.
Answer:
The largest 3-digit number = 999
The largest 3-digit number which is exactly divisible by 3 = 999
29.) Find the prime numbers between 90 and 100.
Answer : The prime numbers between 90 and 100 is 97
30.) 5*2 is a 3 -digit number with * as a missing digit. If the number is divisible by 6, find the least value of the missing digit.
Answer :
If the number is divisible by 6 then it should be divisible by 2 and 3
The number 5*2 has 2 in units place so it is divisible by 2
Now to make it divisible by 3 the sum of digits should be divisible by 3
Sum of digits = 5 + * + 2
= 7+ *
9 is the number which is greater and comes just after 7 and is divisible by 3
So 7 + * = 9
*= 9-7
*=2
31.) What least value should be given to * so that the number 6342*1 is divisible by 3?
Answer:
Here we have 6342 * 1
The number will be divisible by 3 if the sum of digits is divisible by 3
Sum of digits = 6 + 3 + 4 + 2 + * + 1
= 16 + *
Now find the number greater than 16 which is divisible by 3
18 is the number which is greater and comes just after 16 and is divisible by 3
So 16 + * = 18
*= 18-16
*= 2
32.) What least value should be given to * so that the number 915*26 is divisible by 9?
Answer :
Here we have 915 * 26
The number will be divisible by 9 if the sum of digits is divisible by 9
Sum of digits = 9 + 1 + 5 + * + 2 + 6
Sum of digits = 23 + *
Now find the number greater than 23 which is divisible by 9
27 is the number which is greater and comes just after 23 and is divisible by 9
So 23 + * = 27
*= 27-23
*= 4
33.) Find the sum of prime numbers between 60 and 75.
Answer:
Prime number between 60 to 75 are 61, 67, 71, 73
The sum of prime numbers between 60 and 75 = 61 + 67 +71+ 73
= 272
34.) If the HCF of two numbers is 16 and their product is 3072. Find their LCM.
Answer : Let two numbers be X and Y
HCF of X and Y = 16
X × Y = 3072
We know
Product = HCF × LCM
LCM = product ÷ HCF
LCM = 3072 ÷ 16
LCM = 192
35.) What is the HCF of two co-prime numbers?
Answer: HCF is the Highest Common factor of two numbers
Co-prime numbers are the one who has only factor = 1
So the HCF of two co-prime numbers will be 1
36.) If the HCF and LCM of two numbers be 4 and 24 respectively. One of the numbers is 8, then find another number.
Answer :
Let the two numbers be X and Y
HCF of X and Y = 4
LCM of X and Y = 24
Product = HCF × LCM
Product = 4 × 24
Product = 96
Now as given in question Y = 8
So X × Y = 96
X × 8 = 96
X = 96 ÷8
X = 12
The value of other number is 12
37.) How many prime numbers are there between 1 and 30?
Answer :
Prime numbers between 1 to 30 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Total Prime numbers between 1 and 30 are 10
38.) Which of the following number is not a multiple of 27:- 1, 3, 9 and 6?
Answer :
Factors of 27 = 1, 3, 9 ,27
Answer is 6
39.) Which of the given numbers 4, 2, 3 and 8 is not a factor of 12?
Answer :
Factor of 12 = 1, 2, 3, 4, 6 , 12
Therefore ,8 is not a factor of 12
40.) Which of the given numbers 21, 12, 17 and 39 is a prime number?
Answer : A prime number is a natural number greater than 1 which has exactly two factors, 1 and itself.
Factor of 21= 1,3,7,21
Factor of 12 = 1, 2, 3, 4, 6, 12
Factor of 17 = 1, 17
Factor of 39 = 1, 3, 13, 39
Number 17 is a prime number
41.) Which of the following pairs of numbers are co-prime? (30, 415), (17, 68), (16, 81) and (15,100). Answer:
Number with only 1 as a common factor are called co-prime
The number (30, 415 ) has common factor = 1 and 5
The number (17, 68) has common factor =1 and 17
The number (16, 81 ) has common factor = 1
The number (15, 100 ) has common factor= 1 and 5
The number (16, 81 ) has common factor = 1 so the the number (16, 81 ) is co-prime
42.) Which of the given pair of numbers (7, 15), (12, 49), (18, 23) and (12, 21) are not co-prime?
Answer:
Number with only 1 as a common factor are called co-prime
The number (12,21) has common factor 1 and 3 so the number (12,21 ) is not a co-prime.
43.) What is the HCF of 75, 60 and 210?
Answer:
Factors of 75 = 1, 3, 5, 15, 25, 75.
Factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Factors of 210 = 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210.
The HCF of 75, 60 and 210 = 15
44.) The numbers which have more than two factors are called_____.
Answer : Composite numbers
45.) If A is the 5th prime number and B is the 7th prime number then what is B ˗ A?
Answer :
46.) Four bells ring at intervals of 6, 7, 8 and 9 seconds respectively. After how many seconds do all the bells ring together?
Answer :
To calculate the seconds at which bells rings together, we need to find the LCM of given intervals
The LCM of 6, 7, 8 and 9 = 504 seconds
Four bells ring together after every 504 seconds
47.) What is the greatest number that divides 37, 50 and 123 leaving remainders 1, 2 and 3 respectively?
Answer:
1st we subtract the given respective reminders from the numbers
37-1 = 36
50-2= 48
123-3 =120
Now on taking HCF of 36, 48 and 120 we will find that the greatest number
Factors of 36= 1,2,3,4, 6,9,12,18, 36
Factors of 48 = 1,2,3,4,6,8,12,16,24
Factors of 120 = 1,2,3,4,5,6 8,10,12,15,20
HCF of 36, 48 and 120 = 12
48.) What is the least value that should be given to? So that the number 653? 47 is divisible by 11?
Answer :
49.) The LCM of 64 and 48 is 192. What is the HCF of these numbers?
Answer : The LCM of 64 and 48 is 192.
Factors of 64 = 1,2,4,8,16,18
Factors of 48 = 1,2,3,4,6 8,12,16,24
HCF of 64 and 48 = 16
50.) Find the sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 +15?
Answer: The sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 +15 = 64
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