Unit Test Questions And Solutions Prime Time - Class 6 Mathematics | Question Answer

Time: 1 hour M.M. 30 

Attempt all questions. 

• Question numbers 1 to 5 carry 1 mark each. 
• Question numbers 6 to 8 carry 2 marks each. 
• Question numbers  9 to 11 carry 3 marks each. 

Q1: Find all multiples of 50 that lie between 500 and 600.

Ans: Here, multiples of 50 are: 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600
Hence, multiples of 50 that lie between 500 and 600 are: 550.

Q2: How many prime numbers are there from 10 to 20?

Ans: In total, there are 4 prime numbers between 10 and 20.
They are 11, 13, 17, and 19.

Q3: How many composite numbers are there from 10 to 20?

Ans: The total number of composite numbers from 10 to 20 is 7.
They are 10, 12, 14, 15, 16, 18 and 20.

Q4: There is no prime number whose units digit is 4. (T/F)

Ans: True
Sol: A prime number must end in 1, 3, 7, or 9 (except for the number 2). Any number ending in 0, 2, 4, 6, or 8 is divisible by 2, making it a non-prime.

Q5: A product of primes can also be prime. (T/F)

Ans: False
Sol: A product of prime numbers is only prime if it involves precisely one prime number. When two or more prime numbers are multiplied, the result is a composite number, not a prime.

Q6: Is the first number divisible by the second? Use prime factorization.
(a) 150 and 25
(b) 84 and 28
(c) 224 and 16
(d) 800 and 80

Ans: 
(a) Prime Factors of 150 and 25:
150 = 2 × 3 × 5 × 5, and 25 = 5 × 5
Since 150 contains sufficient factors of 5, it is divisible by 25.
(b) Prime Factors of 84 and 28:
84 = 2 × 2 × 3 × 7, and 28 = 2 × 2 × 7
Since 84 includes the required factors to match those in 28, it is divisible by 28.
(c) Prime Factors of 224 and 16:
224 = 2 × 2 × 2 × 2 × 2 × 7, and 16 = 2 × 2 × 2 × 2
Since 224 includes the required factors to match those in 16, it is divisible by 16.
(d) Prime Factors of 800 and 80:
800 = 2 × 2 × 2 × 2 × 2 × 5 × 5, and 80 = 2 × 2 × 2 × 2 × 5
Since 800 includes the required factors to match those in 80, it is divisible by 80.

Q7: Observe that 5 is a prime number, and 2 × 5 + 1 = 11 is also a prime. Are there other primes for which doubling and adding 1 gives another prime? Find at least five such examples.

Ans: The five prime numbers for which doubling and adding 1 gives another prime are:

• 2 (since 2 × 2 + 1 = 5)
• 3 (since 2 × 3 + 1 = 7)
• 11 (since 2 × 11 + 1 = 23)
• 23 (since 2 × 23 + 1 = 47)
• 29 (since 2 × 29 + 1 = 59)

Q8: Find the prime factorization of these numbers without multiplying first.
(a) 72 × 36
(b) 120 × 48

Ans:
(a) Prime factors of 72 = 2 × 2 × 2 × 3 × 3
Prime factors of 36 = 2 × 2 × 3 × 3
Combined prime factorization of 72 × 36 = 2 × 2 × 2 × 3 × 3 × 2 × 2 × 3 × 3

(b) Prime factors of 120 = 2 × 2 × 2 × 3 × 5
Prime factors of 48 = 2 × 2 × 2 × 2 × 3
Combined prime factorization of 120 × 48 = 2 × 2 × 2 × 3 × 5 × 2 × 2 × 2 × 2 × 3

Q9: Which of the following pairs of numbers are co-prime?
(a) 24 and 35
(b) 40 and 97
(c) 50 and 225

Ans: 
(a) Here, factors of 24 = 1 × 2 × 2 × 2 × 3, and factors of 35 = 1 × 5 × 7
No common factor other than 1.
Hence, 24 and 35 are co-prime numbers.

(b) We have factors of 40 = 1 × 2 × 2 × 2 × 5, and factors of 97 = 1 × 97
No common factor other than 1.
Hence, 40 and 97 are co-prime numbers.

(c) Given numbers are 50 and 225.
Here, factors of 50 = 1 × 2 × 5 × 5, and factors of 225 = 1 × 3 × 3 × 5 × 5.
Clearly, 5 is a common factor of 50 and 225.
Hence, 50 and 225 are not co-prime numbers.

Q10: Consider these statements:
(a) Only the last two digits matter when deciding if a given number is divisible by 4.
(b) If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.
(c) If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.
Do you agree? Why or why not?

Ans: (a) Yes, correct.
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
Example: $3,2163,216$ → last two digits $1616$. Since $16÷4=416\; \backslash div\; 4\; =\; 4$, the whole number is divisible by 4.

(b) Yes, correct. 
If the last two digits form a number divisible by 4, then the entire number is divisible by 4 (because multiples of 100 are always divisible by 4).
Example: $5,732$ → last two digits $3232$. Since $32÷4=832\; \backslash div\; 4\; =\; 8$,  $5,732$5,732 is divisible by 4.
(c) Yes, correct. 
If the entire number is divisible by 4, then its last two digits must also form a number divisible by 4.
Example: $2,420÷4=60$. The last two digits are $2020$, which is divisible by 4.

Q11: Who am I?
(a) I am a number less than 50. One of my factors is 8. The sum of my digits is 12.
(b) I am a number less than 100. Two of my factors are 4 and 6. One of my digits is twice the other.

Ans: (a) 8 is a common factor of 8, 16, 24, 32, 40, and 48, which are less than 50. Check the digit sum:

• 8 → sum = 8 (not 12)

• 16 → sum = 1+6 = 7 (not 12)

• 24 → sum = 2+4 = 6 (not 12)

• 32 → sum = 3+2 = 5 (not 12)

• 40 → sum = 4+0 = 4 (not 12)

• 48 → sum = 4+8 = 12

Sum of digits of 48 is 12 and 48 is less than 50. 
(b) Common factors of 4 and 6 are 12, 24, 36, 48, 60, 72, 84 and 96 (which are less than 100). The number where one digit is twice the other are 12, 24, 36 & 48.