Answer: Option B. -> 90, 91, 92, 93, 94, 95, 96
:
B
All the numbers in the option:90, 91, 92, 93, 94, 95, 96have more than 2 factors, hence they are all composite numbers.

Answer: Option C. -> 1, itself   
:
C
1 is an exact divisor of all the numbers.Also, any number is divisible byitself.
Therefore,1 is a factor of every number and every number is a factor of itself.

Answer: Option D. -> 95
:
D
19× 1 = 19
19× 2 = 38
19× 3= 57
19× 4= 76
19×5 = 95
19× 6= 114
Hence, the largest two-digit multiple of 19 is 95.

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Definition: 1 Mark
Steps: 1 Mark
Solution: 1 Mark
A Common Multiple is a number that is a multiple of two or more numbers.
We can find all the factors of 16, as shown below.
16=1$\times$16
16=2$\times$8
16=4$\times$4
Since the number is divisible by 16, it will also be divisible by its factor i.e. 1, 2, 4, 8, 16.
So, clearly, the number which is divisible by 16, will also be divisible by 1, 2, 4, 8.

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Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
For a minimum number of biscuits, first, we find the LCM of 12, 15 and 21.
The prime factors of the following numbers are:
12=2$\times$2$\times$3
15=3$\times$5
21=3$\times$7
The maximum number of times the prime factor 2 occurs is 2.
The maximum number of times the prime factor 3 occurs is 1.
The maximum number of times the prime factor 5 occurs is 1.
The maximum number of times the prime factor 7 occurs is 1.
$\therefore$ LCM= 3 $\times$ 4 $\times$ 5 $\times$ 7 = 420
$\therefore$ Minimum number of packets of brand A $=\frac{420}{12}=35$
Minimum number of packets of brand B $=\frac{420}{15}=28$
And minimum number of packets of brand C $=\frac{420}{21}=20$

:
Smallest Number: 1 Mark
Highest Common factor: 2 Marks
The first four prime numbers are 2, 3, 5, 7.
1 is neither a prime nor a composite number so we will not include that.
Now, all these numbers are prime numbers,so the smallest number possible which is a factor of these numbers is the lowest common multiple of these numbers.
Since these are all prime numbers, their lowest common multiple is the product of these numbers, which is:
2 $\times$ 3 $\times$ 5 $\times$ 7 = 210
Hence, 210 is the required number.
The given numbers are prime numbers, so all of them will have the highest common factor equal to 1.

:
A factor of a number is an exact divisor of that number.

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Steps: 1 Mark
Solution: 1 Mark
The dimensions of the garden can be found by finding the factors of the number 256.
The various factors of 256 are as follows:
256=1$\times$256
256=2$\times$128
256=4$\times$64
256=8$\times$32
256=16$\times$16
We will stop at (16,16), because they are equal.
$\therefore$ The various dimensions of the garden can be, (1,256),(2,128),(4,64),(8,32),(16,16).

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Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
The greatest capacity of a tin is the HCF of 120, 180 and 240 liters.
The prime factorisation of the given numbers is :
120=2$\times$2$\times$2$\times$3$\times$5
180=2$\times$2$\times$3$\times$3$\times$5
240=2$\times$2$\times$2$\times$2$\times$3$\times$5
Hence the highest common factor of 120,180,240=2$\times$2$\times$3$\times$5
$\because$ Highest common factor of 120, 180 and 240 is 60.
$\therefore$ Required HCF = 60 litres.
$\therefore$ Greatest capacity of a tin = 60 litres.

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Concept: 1 Mark
Steps: 2 Marks
Solution: 1 Mark
The given numbers are 2,4,6,8,10 and 12.
The factors of the given numbers are:
2=1$\times$2
2=2$\times$2
6=2$\times$3
8=2$\times$2$\times$2
10=2$\times$5
12=2$\times$2$\times$3
Now ,the maximum number of times prime factor 2 occurs is 3,
the maximum number of times prime factor 3 occurs is 1,
the maximum number of times prime factor 5 occurs is 1.
So, LCM=2$\times$2$\times$2$\times$3$\times$5 = 120
L.C.M. of 2, 4, 6, 8, 10 and 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 30 minutes, they will toll together $\frac{30}{2}$+ 1 = 16 times.