CBSE Class 6 Mathematics Prime Time MCQs Set 11

Answer: (c) 9

Explanation:
1. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
2. 24 ÷ 9 = 2 remainder 6, so 9 does not divide 24 exactly.
3. So 9 is not a factor of 24.

Teacher's Note:
1. List factors in pairs: 1 × 24, 2 × 12, 3 × 8, 4 × 6.
2. Any number missing from the pairs is not a factor.

Answer: (b) 12

Explanation:
1. ‘Idli-vada’ is said at a number that is a multiple of 4 and of 6 at the same time.
2. Multiples of 4: 4, 8, 12, 16, … Multiples of 6: 6, 12, 18, …
3. The first number in both lists is 12.

Teacher's Note:
1. Write a few multiples of each number and look for the first match.
2. 24 is a common multiple too, but not the first one.

Answer: (d) exactly two factors

Explanation:
1. A prime number has exactly two factors: 1 and the number itself, for example 7 = 1 × 7.
2. A number with more than two factors is called composite.

Teacher's Note:
1. Prime: exactly two factors.
2. Composite: more than two factors.

Answer: (c) neither a prime nor a composite number

Explanation:
1. The only factor of 1 is 1 itself, so it has just one factor.
2. A prime needs exactly two factors and a composite needs more than two.
3. So 1 is neither prime nor composite.

Teacher's Note:
1. Count the factors of 1: there is only one.
2. 1 stays outside both groups.

Answer: (a) 2

Explanation:
1. The factors of 2 are only 1 and 2, so 2 is prime.
2. Every other even number also has 2 as a factor, giving it more than two factors.
3. So 2 is the only even prime.

Teacher's Note:
1. 4, 6 and 8 are all divisible by 2, so they are composite.
2. Remember: 2 is the smallest prime too.

Answer: (b) 1

Explanation:
1. Co-prime numbers have 1 as their only common factor, for example 4 and 9.
2. Every pair of numbers shares the factor 1, so co-prime means no other common factor exists.

Teacher's Note:
1. 0 is never a factor of a number.
2. Co-prime is a property of a pair of numbers.

Answer: (d) 2 × 2 × 5

Explanation:
1. 20 = 4 × 5 and 4 = 2 × 2, so 20 = 2 × 2 × 5.
2. 2 × 10 and 4 × 5 still contain the composite numbers 10 and 4.
3. 2 × 5 × 5 = 50, not 20.

Teacher's Note:
1. In a prime factorisation every factor must be prime.
2. Multiply back to check the product.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The last digit of 8560 is 0, so it is divisible by 10. A is true.
2. R is the correct divisibility rule for 10, so R is true.
3. The rule in R is exactly why 8560 is divisible by 10, so R explains A.

Teacher's Note:
1. Check A and R separately first.
2. Then ask: does R give the reason for A?

Answer: (c) A is true, but R is false.

Explanation:
1. The factors of 15 are 1, 3, 5 and 15. That is four factors, so 15 is composite. A is true.
2. 15 does not have exactly two factors, so R is false.

Teacher's Note:
1. List all factors before deciding.
2. A number with exactly two factors would be prime, not composite.

Answer: (d) A is false, but R is true.

Explanation:
1. 51 = 3 × 17, so 51 is divisible by 3. R is true.
2. Because 3 divides 51, it has the factors 1, 3, 17 and 51, so 51 is composite. A is false.

Teacher's Note:
1. 51 looks prime but is not; add its digits: 5 + 1 = 6.
2. Always test small primes 2, 3, 5, 7 before calling a number prime.

Answer: (a) Only Statement I is true.

Explanation:
1. The multiples of 6 are 6, 12, 18, 24, … and all of them are even, since 6 = 2 × 3. Statement I is true.
2. 8 is even but is not a multiple of 6, so Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. One counter-example is enough to show a statement is false.
2. ‘Every A is B’ does not mean ‘every B is A’.

Answer: (b) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r)

Explanation:
1. 18 = 2 × 3 × 3 (q) and 25 = 5 × 5 (p).
2. 28 = 2 × 2 × 7 (s) and 45 = 3 × 3 × 5 (r).
3. So the match is (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r).

Teacher's Note:
1. Multiply each product in Column II to see which number it makes.
2. Then pick the option that agrees.

Answer: (b) 6

Explanation:
1. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
2. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
3. Common to both: 1, 2, 3, 4, 6, 12, which makes 6 common factors.

Teacher's Note:
1. Write both lists fully before comparing.
2. Do not forget 1.

Answer: (c) 24

Explanation:
1. Multiples of 8: 8, 16, 24, 32, … Multiples of 12: 12, 24, 36, …
2. The first number in both lists is 24.
3. 4 is a common factor, not a multiple, and 96 = 8 × 12 is a common multiple but not the smallest.

Teacher's Note:
1. A multiple is never smaller than the number itself.
2. The product is a common multiple but not always the first one.

Answer: (d) 41 and 43

Explanation:
1. Twin primes are two primes whose difference is 2.
2. 41 and 43 are both prime and 43 − 41 = 2.
3. The other pairs differ by 6, 6 and 4, so the answer is 41 and 43.

Teacher's Note:
1. Check both conditions: both prime, and a gap of 2.
2. Twin primes always have one even number between them.

Answer: (b) 4

Explanation:
1. The middle region is shared by both circles, so it holds the common factors of 12 and 18.
2. Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18.
3. Common factors: 1, 2, 3, 6. So 4 numbers go in the middle.

Teacher's Note:
1. The overlap of two circles means ‘in both lists’.
2. 4 and 12 stay on the left; 9 and 18 stay on the right.

Answer: (a) 15 and 28

Explanation:
1. 15 = 3 × 5 and 28 = 2 × 2 × 7 have no prime factor in common, so their only common factor is 1.
2. 14 and 21 share 7; 16 and 24 share 8; 18 and 27 share 9.
3. So 15 and 28 are co-prime.

Teacher's Note:
1. Break each number into primes and look for a shared prime.
2. Co-prime numbers need not be prime themselves.

Answer: (c) Both the statements are true.

Explanation:
1. 2 × 2 × 2 × 7 = 8 × 7 = 56, and all the factors are prime. Statement I is true.
2. 63 = 3 × 3 × 7, so 7 is a prime factor of both 56 and 63. Statement II is true.
3. So both statements are true.

Teacher's Note:
1. Multiply back to check a factorisation.
2. A common prime factor appears in both factorisations.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The last three digits of 3128 form 128 = 8 × 16, so 128 is divisible by 8. R is true.
2. By this test, 3128 is divisible by 8 (3128 = 8 × 391). A is true.
3. R is the very reason A holds, so R explains A.

Teacher's Note:
1. For 8, use the last three digits.
2. Check: 8 × 391 = 3128.

Answer: (b) 7436

Explanation:
1. For 4 we look at the number made by the last two digits: 54, 36, 18 and 42.
2. Only 36 = 4 × 9 is a multiple of 4.
3. So 7436 is divisible by 4.

Teacher's Note:
1. For 4, the last two digits decide.
2. All four options are even, so the test for 2 is not enough.

Answer: (d) A is false, but R is true.

Explanation:
1. 42 = 2 × 3 × 7 and 12 = 2 × 2 × 3.
2. 12 needs two 2s, but 42 has only one 2, so 42 is not divisible by 12 (42 ÷ 12 = 3 remainder 6). A is false.
3. 2 and 3 really are prime factors of 42, so R is true.

Teacher's Note:
1. Every prime of the divisor must appear in the number, as many times as needed.
2. Having the right primes is not enough; count them.

Answer: (a) Only Statement I is true.

Explanation:
1. 100 is a multiple of 4, so the last two digits decide divisibility by 4. Statement I is true.
2. For 8 we need the last three digits. For example, 324 ends in 24 (a multiple of 8) but 324 is not divisible by 8. Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. Test for 4: last two digits. Test for 8: last three digits.
2. A counter-example settles a false statement.

Answer: (c) 10

Explanation:
1. The empty box splits into 2 and 5, so the number in it is 2 × 5 = 10.
2. Check: 9 × 10 = 90, and 9 = 3 × 3, so 90 = 3 × 3 × 2 × 5.
3. The empty box holds 10.

Teacher's Note:
1. A box in a factor tree is the product of the two boxes below it.
2. Check the top: the two branches must multiply to 90.

Answer: (d) 275

Explanation:
1. 150 = 50 × 3, 200 = 50 × 4 and 300 = 50 × 6 are all multiples of 50.
2. 275 = 25 × 11, so it is a multiple of 25, but 275 ÷ 50 = 5 remainder 25.
3. So the answer is 275.

Teacher's Note:
1. Multiples of 50 end in 00 or 50.
2. Multiples of 25 end in 00, 25, 50 or 75.

Answer: (c) 5

Explanation:
1. A number that is shaded and circled is a multiple of 2 and also of 3, that is, a multiple of 6.
2. In the grid these are 6, 12, 18, 24 and 30.
3. So 5 numbers are both shaded and circled.

Teacher's Note:
1. Common multiples of 2 and 3 are the multiples of 6.
2. Count them straight from the picture to check.

Answer: (b) Statement I is true but Statement II is false.

Explanation:
1. Two different primes have only 1 as a common factor, so Statement I is true.
2. 8 and 9 are both composite, yet their only common factor is 1. So Statement II is false.
3. The answer is: Statement I is true but Statement II is false.

Teacher's Note:
1. 8 and 9 is a handy example of composite co-primes.
2. Note that options (b) and (c) here are worded differently from usual.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 7 is prime and 9 = 3 × 3, so 7 and 9 share no prime factor and are co-prime.
2. So their first common multiple is 7 × 9 = 63. A is true.
3. R is a true general rule, and it is exactly what gives A, so R explains A.

Teacher's Note:
1. First check that the pair is co-prime.
2. Then the product is the first common multiple.

Answer: (d) 7160

Explanation:
1. All four numbers end in 0, so all are divisible by 2, 5 and 10.
2. For 8, look at the last three digits: 540, 550, 420 and 160. Only 160 = 8 × 20 is a multiple of 8.
3. 7160 also passes the test for 4 (60 = 4 × 15). So the answer is 7160.

Teacher's Note:
1. Use the strongest test, the one for 8, first.
2. A number divisible by 8 is automatically divisible by 4 and 2.

Answer: (d) (i)-(q), (ii)-(s), (iii)-(r), (iv)-(p)

Explanation:
1. 17 and 19 are primes differing by 2, so they are twin primes (q).
2. 8 = 2 × 2 × 2 and 15 = 3 × 5 are composite with no common prime, so they are co-prime (s).
3. 6 divides both 12 and 18 (r), and 1 is neither prime nor composite (p). So the match is (i)-(q), (ii)-(s), (iii)-(r), (iv)-(p).

Teacher's Note:
1. Start with the easiest pair: the number 1 goes with (p).
2. Then eliminate options.

Answer: (c) 5

Explanation:
1. Row 1 has product 10 = 2 × 5, so P and Q are 2 and 5 in some order.
2. Column I has product 14 = 2 × 7, so P must be 2 (5 does not divide 14). Hence Q = 5.
3. Then R = 7 and S = 3. Check: 7 × 3 = 21 and 5 × 3 = 15.

Teacher's Note:
1. Write each product as a product of two primes first.
2. A box sits in one row and one column, so it must fit both.