CBSE Class 8 Mathematics Algebra Play MCQs Set 11

Answer: (c) the sum of the two numbers directly below it

Explanation:
1. The rule of a number pyramid: every box holds the sum of the two boxes directly below it.
2. Options (a) and (b) use the wrong operation, and (d) adds the wrong boxes.

Teacher's Note:
1. Upwards we add; downwards we subtract.
2. Only the two boxes touching from below are added.

Answer: (b) explain why number tricks and puzzles always work

Explanation:
1. In this chapter algebra is used to study number tricks and puzzles, to explain why they always work, and then to invent new ones.
2. Measuring, drawing graphs and writing large numbers are not what the chapter is about.

Teacher's Note:
1. A letter-number stands for every possible number at once.
2. So one piece of algebra covers all cases of a trick.

Answer: (d) 9

Explanation:
1. Write the number as 10a + b; its reverse is 10b + a.
2. Difference = (10b + a) − (10a + b) = 9b − 9a = 9(b − a).
3. So the difference is always a multiple of 9 (for example 62 − 26 = 36 = 9 × 4).

Teacher's Note:
1. A multiple of 9 is also a multiple of 3, but '3 only' is wrong.
2. The quotient after dividing by 9 is the difference of the digits.

Answer: (a) 11

Explanation:
1. (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
2. So the sum is always a multiple of 11. Examples: 31 + 13 = 44, 28 + 82 = 110.

Teacher's Note:
1. Collect the a terms and the b terms separately.
2. Difference gives 9, sum gives 11.

Answer: (c) a + 7

Explanation:
1. Going one row down on a calendar means going forward one full week, that is, 7 days.
2. So the number directly below a is a + 7.
3. (a + 1 is to the right of a, and a + 8 is diagonally below right.)

Teacher's Note:
1. Right means +1; down means +7.
2. Check on a real calendar: below 5 is 12.

Answer: (d) 5

Explanation:
1. Start with x: double → 2x; add 10 → 2x + 10; divide by 2 → x + 5.
2. Subtract the starting number: (x + 5) − x = 5.
3. The letter-number cancels, so the answer is 5 for every starting number.

Teacher's Note:
1. Adding 2k before halving makes the answer k.
2. Here 10 = 2 × 5, so the answer is 5.

Answer: (b) 19

Explanation:
1. Middle row: 4 + 6 = 10 and 6 + 3 = 9.
2. Top: 10 + 9 = 19.
3. Check with the rule top = a + 2b + c: 4 + 12 + 3 = 19.

Teacher's Note:
1. The middle number of the bottom row is counted twice.
2. Use the formula as a quick check.

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

Explanation:
1. The steps are x → 2x → 2x + 4 → x + 2 → (x + 2) − x = 2. So R is true.
2. Since x cancels, the answer is 2 whatever the starting number, so A is true.
3. The cancelling of x is exactly why everybody gets 2, so R explains A.

Teacher's Note:
1. A trick works for everyone when the letter-number cancels out.
2. Here R gives the reason, not just a related fact.

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

Explanation:
1. A is true: the top box is the sum of the two boxes below, a + b.
2. R is false: a pyramid can also be filled downwards by subtraction, and letter-numbers help when neither direction works at first.

Teacher's Note:
1. If a box and one box below it are known, the other box below is their difference.
2. So pyramids can be filled in both directions.

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

Explanation:
1. 7 × 11 = 77 and 77 × 13 = 1001, so Statement I is true.
2. abcabc = abc × 1000 + abc = abc × 1001, so Statement II is true (e.g. 538 × 1001 = 538538).
3. So both statements are true.

Teacher's Note:
1. Writing a number twice shifts one copy three places to the left.
2. 1001 = 7 × 11 × 13 is worth remembering.

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

Explanation:
1. (i) ba − ab = (10b + a) − (10a + b) = 9b − 9a = 9(b − a) → (s).
2. (ii) ab + ba = 11a + 11b = 11(a + b) → (p).
3. (iii) abcabc = abc × 1001 → (q).
4. (iv) abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) → (r).
5. So the matching is (i)-(s), (ii)-(p), (iii)-(q), (iv)-(r).

Teacher's Note:
1. Expand each number by place value before simplifying.
2. Match the easiest pairs first and then check the options.

Answer: (b) 16

Explanation:
1. Middle row: 5 + 2 = 7 and 2 + 7 = 9.
2. Top box: 7 + 9 = 16.

Teacher's Note:
1. Check: 5 + 2 × 2 + 7 = 16.
2. Add neighbouring pairs, one row at a time.

Answer: (a) a + 3b + 3c + d

Explanation:
1. Second row: a + b, b + c, c + d.
2. Third row: a + 2b + c and b + 2c + d.
3. Top: a + 3b + 3c + d; the coefficients 1, 3, 3, 1 are symmetric.

Teacher's Note:
1. Inner bricks are counted more often than end bricks.
2. Coefficients 1, 2, 1 then 1, 3, 3, 1 follow a pattern.

Answer: (d) 100M + 165 + D

Explanation:
1. Multiply the month by 5: 5M; add 6: 5M + 6; multiply by 4: 20M + 24.
2. Add 9: 20M + 33; multiply by 5: 100M + 165; add the day: 100M + 165 + D.
3. That is why 165 is subtracted to get 100M + D.

Teacher's Note:
1. Track the expression step by step, multiplying every term.
2. The fixed part 165 comes only from the added numbers 6 and 9.

Answer: (c) qp × r

Explanation:
1. The largest digit r should be the multiplier.
2. The other two digits are written in decreasing order, so the two-digit number is qp.
3. So the largest product is qp × r (e.g. 3, 5, 9 give 53 × 9 = 477).

Teacher's Note:
1. qp × r = 10qr + pr and rp × q = 10qr + pq; pr > pq.
2. Biggest digit multiplies; the rest go in decreasing order.

Answer: (b) 477

Explanation:
1. Use the largest digit 9 as the multiplier and write 5, 3 in decreasing order: 53 × 9 = 477.
2. The next best arrangement, 93 × 5 = 465, is smaller.

Teacher's Note:
1. Check the two strongest candidates only.
2. 495 cannot be made from these digits.

Answer: (a) 12

Explanation:
1. The block is a, a + 1, a + 7, a + 8, with sum 4a + 16.
2. 4a + 16 = 64 gives 4a = 48 and a = 12.
3. The block is 12, 13, 19, 20, so the smallest number is 12.

Teacher's Note:
1. The top-left number is always the smallest.
2. Check: 12 + 13 + 19 + 20 = 64.

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

Explanation:
1. R is true: (10b + a) − (10a + b) = 9(b − a).
2. So the difference between a two-digit number and its reverse is always a multiple of 9.
3. 25 is not a multiple of 9, so no such number exists. A is false.

Teacher's Note:
1. A true general rule can prove a claim false.
2. The difference can only be 9, 18, 27, …, 72.

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

Explanation:
1. A is true: the sum is 111a + 111b + 111c = 111(a + b + c), and 111 = 3 × 37.
2. R is true: it is just the place-value form of abc.
3. But R alone does not explain divisibility by 37; the reason is the factor 111 = 3 × 37. So R is not the correct explanation.

Teacher's Note:
1. Place value is a tool used in the proof, not the reason itself.
2. The key fact is 111 = 3 × 37.

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

Explanation:
1. A is true: the steps turn the date into 100M + 165 + D, so 165 is a fixed extra.
2. R is false: after subtracting 165, the last two digits give the day D (at most 31), and the digits before them give the month.

Teacher's Note:
1. 100M pushes the month into the hundreds and thousands places.
2. The day fits in the last two digits.

Answer: (b) Only Statement II is true

Explanation:
1. Statement I is false: the largest digit should be the multiplier, not the tens digit.
2. Statement II: 23 × 5 = 115, 32 × 5 = 160, 52 × 3 = 156, so 32 × 5 is the greatest. True.
3. So only Statement II is true.

Teacher's Note:
1. Statement II itself shows that putting 5 in the tens place is not best.
2. Always compute before choosing.

Answer: (a) Only Statement I is true

Explanation:
1. The sum is 4a + 16 = 4(a + 4), always a multiple of 4. 50 is not a multiple of 4, so Statement I is true.
2. 4a + 16 = 80 gives 4a = 64 and a = 16, not 20. Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. Every correct sum must be a multiple of 4.
2. Solve the equation instead of dividing the sum by 4.

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

Explanation:
1. Subtract 165 to get 100M + D.
2. 296 − 165 = 131 → 31 January (s); 1269 − 165 = 1104 → 4 November (p).
3. 394 − 165 = 229 → 29 February (q); 1390 − 165 = 1225 → 25 December (r).
4. So the answer is (i)-(s), (ii)-(p), (iii)-(q), (iv)-(r).

Teacher's Note:
1. Last two digits give the day; the digits before them give the month.
2. 1104 means month 11, day 04.

Answer: (d) 37

Explanation:
1. Let the middle box of the bottom row be c. The middle row is 15 + c and c + 11.
2. (15 + c) + (c + 11) = 70 gives 2c + 26 = 70, so 2c = 44 and c = 22.
3. The marked box is 15 + 22 = 37 (the other middle box is 33, and 37 + 33 = 70).

Teacher's Note:
1. When neither upward nor downward filling works, use a letter-number.
2. Always check that the two middle boxes add to the top.

Answer: (a) Every number appearing in the pyramid, including the top number, is again a Virahanka-Fibonacci number

Explanation:
1. The pyramid rule (add the two numbers below) is the same as the sequence rule (add the two previous terms).
2. So every new box again holds a Virahanka-Fibonacci number. For 1, 2, 3: the pyramid is 3, 5 and then 8.
3. This example rules out (b), (c) (1 + 2 + 3 = 6 ≠ 8) and (d) (for 1, 2 the top is 3, which is odd).

Teacher's Note:
1. Two neighbouring terms always add to the next term.
2. One example can rule out several wrong options.

Answer: (c) 21

Explanation:
1. Let the fixed number be k. The steps give 3x, then 3x + k, then x + \( \frac{k}{3} \), and finally \( \frac{k}{3} \).
2. \( \frac{k}{3} \) = 7 gives k = 21.

Teacher's Note:
1. Whatever we multiply by, we must add that multiple of the answer.
2. Check with 4: 12, 33, 11, 7.

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

Explanation:
1. R is true: writing the digits twice multiplies the number by 1001, and 1001 = 7 × 11 × 13.
2. Dividing by 7, 11 and 13 in turn divides by 1001 in all, which gives back abc. So A is true.
3. R gives exactly this reason, so R explains A. Example: 407407 ÷ 7 ÷ 11 ÷ 13 = 407.

Teacher's Note:
1. Dividing by 7, 11 and 13 one after another is the same as dividing by 1001.
2. Try it on any three-digit number.

Answer: (d) Both Statement I and Statement II are true

Explanation:
1. The top is a + 2b + c, so b is counted twice. Statement I is true.
2. Reversing gives c + 2b + a, the same value. Statement II is true.
3. So both statements are true.

Teacher's Note:
1. Note that options (c) and (d) are in a different order here.
2. The coefficients 1, 2, 1 are symmetric.

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

Explanation:
1. Rule: largest digit as multiplier, other two in decreasing order.
2. 31 × 7 = 217 (r); 53 × 9 = 477 (s); 42 × 6 = 252 (q); 54 × 8 = 432 (p).
3. So the answer is (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p).

Teacher's Note:
1. Apply the same rule to every set.
2. A quick check: 64 × 2 = 128 is much smaller than 42 × 6.

Answer: (a) 6

Explanation:
1. Let the centre number be m. The nine dates are m − 8, m − 7, m − 6, m − 1, m, m + 1, m + 6, m + 7, m + 8.
2. The added and subtracted parts cancel, so the sum is 9m. Then 9m = 126 gives m = 14.
3. The smallest number is the top-left one: 14 − 8 = 6. The block runs from 6 to 22.

Teacher's Note:
1. The sum of a 3 × 3 block is 9 times the centre.
2. The top-left number is the centre minus 8.