CBSE Class 8 Mathematics Number Play MCQs Set 11

Answer: (d) 2u − 4v

Explanation:
1. 2u − 4v = 2(u − 2v), so 2 is a factor for every integer u and v; it is always even.
2. The others are only sometimes even: 3g + 5h = 8 when g = h = 1, but 11 when g = 2, h = 1.
3. b² + 1 = 2 when b = 1 but 5 when b = 2; 6m − 3n = 3 when m = n = 1.

Teacher's Note:
1. An expression is always even when 2 can be taken out as a common factor.
2. One odd value is enough to show that an expression is not always even.

Answer: (b) the sum of its digits is divisible by 9

Explanation:
1. Each place value is 1 more than a multiple of 9: 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, and so on.
2. So a number and the sum of its digits leave the same remainder on division by 9.
3. Hence a number is divisible by 9 exactly when the sum of its digits is divisible by 9.

Teacher's Note:
1. The last-digit test works for 2, 5 and 10, not for 9.
2. For 9 (and 3) always look at the digit sum.

Answer: (c) 9

Explanation:
1. The digital root is found by adding the digits again and again until one digit is left.
2. For a multiple of 9 the digit sum is again a multiple of 9, so the process ends at 9, never at 0.
3. Examples: 63 → 6 + 3 = 9; 999 → 27 → 9.

Teacher's Note:
1. The digital root matches the remainder on division by 9, except that remainder 0 shows up as 9.
2. A digital root can never be 0 for a non-zero number.

Answer: (a) 3

Explanation:
1. 5 + 8 + 4 + 7 + 6 = 30.
2. 3 + 0 = 3, so the digital root is 3.

Teacher's Note:
1. Keep adding until a single digit is left.
2. You can also drop 9s at once: 5 + 4 = 9 and 8 + 7 + 6 = 21 → 3.

Answer: (b) 3

Explanation:
1. In 1000d + 100c + 10b + a, the letter c multiplies 100, so it is the hundreds digit.
2. 5307 = 1000 × 5 + 100 × 3 + 10 × 0 + 7, so c = 3.

Teacher's Note:
1. a is the units digit and the letters go up with the place value.
2. Write the number in expanded form to read off each letter.

Answer: (a) A + B

Explanation:
1. Write A = 13p and B = 13q.
2. A + B = 13p + 13q = 13(p + q), which is a multiple of 13.
3. The other options add 1, 5 or 7, which are not multiples of 13, so they are not always divisible by 13.

Teacher's Note:
1. If a number divides two numbers, it divides their sum and their difference.
2. Adding a non-multiple spoils the divisibility.

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

Explanation:
1. R is true: 8 − 2 + 5 = 11, a multiple of 11.
2. By the shortcut for 11, a number whose alternating sum is 0 or a multiple of 11 is divisible by 11, so A is true (528 = 11 × 48).
3. The alternating sum is exactly the reason, so R correctly explains A.

Teacher's Note:
1. Start the alternating signs with ‘+’ at the units digit.
2. Always confirm with a quick multiplication: 11 × 48 = 528.

Answer: (c) Both Statement I and Statement II are true.

Explanation:
1. Every place value above the units place is a multiple of 10, so a units digit 0 makes the whole number a multiple of 10. Statement I is true.
2. 10 = 2 × 5, so every multiple of 10 is also a multiple of 2 and of 5. Statement II is true.
3. So both statements are true.

Teacher's Note:
1. A number divisible by k is divisible by every factor of k.
2. 2 and 5 are factors of 10.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: 5 + 3 + 4 + 6 = 18, a multiple of 9, so 5346 is divisible by 9 (5346 = 9 × 594).
2. R is true: 5346 ends in 6, so it is even.
3. Being even has nothing to do with divisibility by 9, so R does not explain A.

Teacher's Note:
1. Divisibility by 9 depends only on the digit sum.
2. Two true facts need not be linked as cause and reason.

Answer: (d) 5k + 3

Explanation:
1. Numbers leaving remainder 3 on division by 5 are 3 more than a multiple of 5.
2. A multiple of 5 is 5k, so the numbers are 5k + 3: 3, 8, 13, 18, and so on.

Teacher's Note:
1. Divisor times k, plus the remainder.
2. 5k − 3 gives remainder 2, not 3.

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

Explanation:
1. 45 → 4 + 5 = 9, so (i)-(s).
2. 61 → 6 + 1 = 7, so (ii)-(r).
3. 123 → 1 + 2 + 3 = 6, so (iii)-(q).
4. 2048 → 2 + 0 + 4 + 8 = 14 → 1 + 4 = 5, so (iv)-(p).

Teacher's Note:
1. Find all four roots first, then pick the option.
2. Two-step sums like 14 → 5 are easy to forget.

Answer: (a) 2970

Explanation:
1. Divisible by 9: digit sum a multiple of 9, so 2970, 4509 and 3456 pass; 6160 (sum 13) fails.
2. Divisible by 11: alternating sum 0 or a multiple of 11, so 2970 (0) and 6160 (−11) pass.
3. Only 2970 passes both tests (2970 = 9 × 330 = 11 × 270).

Teacher's Note:
1. Read each column with its own test.
2. A negative multiple of 11, like −11, also passes the test for 11.

Answer: (c) 3 and 8

Explanation:
1. 24 = 3 × 8, and 3 and 8 have no common factor, so LCM(3, 8) = 24.
2. A number divisible by both 3 and 8 is therefore divisible by 24.
3. For the other pairs the LCM is only 12 (LCM(4, 6) = LCM(2, 12) = LCM(4, 12) = 12); for example, 12 passes them but is not a multiple of 24.

Teacher's Note:
1. Choose two factors of 24 with no common factor.
2. Divisibility by both k and m guarantees only the LCM.

Answer: (b) 3

Explanation:
1. Digit sum = 5 + z + 1 + 4 = 10 + z.
2. 10 + z must be 12, 15 or 18, so z = 2, 5 or 8.
3. That is three digits.

Teacher's Note:
1. z can be any digit from 0 to 9.
2. Values of z that work differ by 3.

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

Explanation:
1. R is the correct rule, so R is true.
2. LCM(9, 6) = 18, not 54, so only divisibility by 18 is guaranteed.
3. 18 is divisible by both 9 and 6 but not by 54, so A is false.

Teacher's Note:
1. Multiply the two numbers only when they have no common factor.
2. One counter-example is enough to show A is false.

Answer: (a) Only Statement I is true.

Explanation:
1. n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1), always a multiple of 3. Statement I is true.
2. n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 4(n + 1) + 2, which leaves remainder 2 on division by 4. Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. Check with 1 + 2 + 3 + 4 = 10, which is not a multiple of 4.
2. Algebra shows why the pattern for 3 does not carry over to 4.

Answer: (c) 7

Explanation:
1. 5 + 4 + 7 + 3 + 6 = 25, and 2 + 5 = 7.
2. The digital root is the remainder on division by 9, so the remainder is 7.

Teacher's Note:
1. A remainder on division by 9 can never be 9.
2. Check: 9 × 6081 = 54729 and 54736 − 54729 = 7.

Answer: (b) If two even numbers each leave a remainder of 2 on division by 4, their sum is a multiple of 4.

Explanation:
1. A = 14 (3 rows of 4 and 2 left over) and B = 10 (2 rows of 4 and 2 left over); both are of the form 4p + 2.
2. The two left-over groups of 2 join to make one more row of 4: (4p + 2) + (4q + 2) = 4(p + q + 1).
3. Here 14 + 10 = 24, a multiple of 4. This is option (b).

Teacher's Note:
1. Count the full rows and the left-over dots separately.
2. Two remainders of 2 add to make a complete group of 4.

Answer: (c) (i)-(r), (ii)-(q), (iii)-(p)

Explanation:
1. (i) 8a + 8b = 8(a + b): always a multiple of 8, so (r).
2. (ii) 3m + 3n = 3(m + n) is a multiple of 6 only when m + n is even (3 + 3 = 6 yes, 3 + 6 = 9 no), so (q).
3. (iii) Even + odd is odd, and an odd number is never a multiple of 4, so (p).

Teacher's Note:
1. Use algebra for 'always' and an example plus a non-example for 'sometimes'.
2. An odd number cannot be a multiple of any even number.

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

Explanation:
1. A is true: (6a + 2) + (6b + 2) + (6c + 2) = 6(a + b + c + 1).
2. R is false: the sum of three even numbers need not be a multiple of 6, for example 2 + 4 + 8 = 14.
3. So A is true but R is false.

Teacher's Note:
1. The three remainders 2 + 2 + 2 = 6 make one more group of 6.
2. Test a general claim like R with small numbers.

Answer: (c) 14

Explanation:
1. The product has three digits, so FG is at least 34, and 3 × G must end in G; this happens only for G = 0 or 5.
2. G = 0 would give 30F = HH0, that is 3F = 11H, which no single digits satisfy.
3. G = 5: checking 35, 45, …, 95, only 75 × 3 = 225 has the form HHG.
4. So F = 7, G = 5, H = 2 and F + G + H = 14.

Teacher's Note:
1. Start with the units digit in every cryptarithm.
2. A digit times 3 ends in itself only for 0 and 5.

Answer: (a) Only Statement I is true.

Explanation:
1. 5k + 3 is 3 more than a multiple of 5, so the remainder is 3. Statement I is true.
2. 5k − 2 = 5(k − 1) + 3, so it also leaves remainder 3, not 2 (for example 8 = 5 × 2 − 2 leaves 3). Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. 5k − 2 and 5k + 3 describe the same numbers.
2. Rewrite with a plus sign to read the remainder.

Answer: (d) 2 and 3

Explanation:
1. 6 = 2 × 3, and 2 and 3 have no common factor, so LCM(2, 3) = 6.
2. A number divisible by both 2 and 3 is divisible by 6.
3. Divisibility by 2 and 4 says nothing about 3 (4 passes), and 3 and 4 is too strong (6 fails the test for 4).

Teacher's Note:
1. Even and digit sum a multiple of 3 means divisible by 6.
2. Pick factors with no common factor whose product is the divisor.

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

Explanation:
1. R is true: switching +b to −b changes the value by 2b, which is even.
2. Numbers that differ by an even number have the same parity.
3. Every one of the eight expressions can be reached from any other by such switches, so A is true and R explains it.

Teacher's Note:
1. Parity means being odd or even.
2. An even change never turns odd into even or even into odd.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. 8(7b − 3) − 4(11b + 1) = 56b − 24 − 44b − 4 = 12b − 28 = 12(b − 3) + 8.
2. It always leaves remainder 8 on division by 12, so it is never a multiple of 12. A is true.
3. 12b − 28 = 2(6b − 14) is even, so R is true; but being even does not stop a number from being a multiple of 12, so R does not explain A.

Teacher's Note:
1. Simplify first, then rewrite as a multiple of 12 plus a remainder.
2. Many even numbers, like 24, are multiples of 12.

Answer: (b) Only Statement II is true.

Explanation:
1. LCM(6, 4) = 12, so only divisibility by 12 is certain; 12 is divisible by 6 and 4 but not by 24. Statement I is false.
2. LCM(9, 4) = 36, as 9 and 4 have no common factor. Statement II is true.
3. So only Statement II is true.

Teacher's Note:
1. Use the LCM, not the product, when the numbers share a factor.
2. 9 and 4 share no factor, so their LCM equals their product.

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

Explanation:
1. 407: 7 − 0 + 4 = 11, a multiple of 11, remainder 0, so (i)-(p).
2. 916: 6 − 1 + 9 = 14 = 11 + 3, remainder 3, so (ii)-(q).
3. 743: 3 − 4 + 7 = 6, remainder 6, so (iii)-(s).
4. 258: 8 − 5 + 2 = 5, remainder 5, so (iv)-(r).

Teacher's Note:
1. A positive alternating sum less than 11 is the remainder itself.
2. Subtract 11 when the alternating sum is 11 or more.

Answer: (a) 18

Explanation:
1. Region B holds numbers that are multiples of 9 but not multiples of 27.
2. 27, 54 and 81 are multiples of 27, so they lie in region C.
3. 18 = 9 × 2 is a multiple of 9 but not of 27, so 18 lies in region B.

Teacher's Note:
1. Every multiple of 27 is a multiple of 9, and every multiple of 9 is a multiple of 3.
2. That is why the ovals sit one inside another.

Answer: (c) 4

Explanation:
1. 36 = 4 × 9 and 4, 9 have no common factor, so the number must pass the tests for 4 and for 9.
2. Test for 4: the last two digits 4q must be a multiple of 4, so q = 0, 4 or 8.
3. Test for 9: 6 + p + 4 + q = 10 + p + q must be a multiple of 9. q = 0 gives p = 8; q = 4 gives p = 4; q = 8 gives p = 0 or 9.
4. The numbers are 6840, 6444, 6048 and 6948: four pairs.

Teacher's Note:
1. Do not miss the second value of p when q = 8.
2. 10 + p + q can be 18 or 27.

Answer: (d) 4

Explanation:
1. 18m + 27n + 22 = 9(2m + 3n) + 22, which is 22 more than a multiple of 9.
2. 22 = 9 × 2 + 4, so the number leaves remainder 4 on division by 9.
3. The digital root equals this remainder, so it is 4.

Teacher's Note:
1. Multiples of 9 do not change the digital root.
2. Only the leftover 22 matters: 2 + 2 = 4.