CBSE Class 6 Maths Advanced Knowing Our Numbers MCQs Set 02

Answer: (b) 10

Explanation:
1. Numbers from 195 to 204 round to 200 (nearest ten).
2. These are 195, 196, 197, 198, 199, 200, 201, 202, 203, 204.
3. There are 10 such numbers.

Teacher's Note:
1. Find the smallest number that rounds up to 200 (195) and the greatest number that rounds down to 200 (204).
2. Count both ends.

Answer: (c) 7499

Explanation:
1. Numbers from 6500 to 7499 round to 7000.
2. The greatest of them is 7499.

Teacher's Note:
1. The hundreds digit must be 4 or less for the top limit.
2. 7500 would round up to 8000.

Answer: (a) 550

Explanation:
1. Numbers from 550 to 649 round to 600.
2. The smallest is 550.

Teacher's Note:
1. A tens digit of 5 rounds up.
2. 549 would round down to 500.

Answer: (d) 21

Explanation:
1. Ones place: 1, 11, 21, 31, 41, 51, 61, 71, 81, 91 gives 10 times.
2. Tens place: 10 to 19 gives 10 times.
3. The number 100 has a 1 in the hundreds place, which is 1 more time.
4. Total = 10 + 10 + 1 = 21.

Teacher's Note:
1. Count place by place so that no 1 is missed.
2. Remember that 11 contains two 1s and that 100 also contains one.

Answer: (c) A is the largest of the 4 numbers

Explanation:
1. A has 9 in the thousands place, so A is at least 9000.
2. B, C and D have 9 only once and not in the thousands place, so their thousands digit is at most 8; each is less than 9000.
3. So A is the largest of the four numbers.
4. B versus C and D versus the others cannot be decided because the other digits are unknown.

Teacher's Note:
1. Compare the thousands digit first.
2. If the other digits are unknown, only a statement that is true for every possible case is correct.

Answer: (c) ten lakhs

Explanation:
1. Indian places from the right: ones, tens, hundreds, thousands, ten thousands, lakhs, ten lakhs, crores.
2. The 7th digit from the right is in the ten lakhs place.

Teacher's Note:
1. Count 7 places from the right.
2. The 8th digit is in the crores place.

Answer: (c) 10

Explanation:
1. Ones → tens → hundreds → thousands: each step to the left multiplies the value by 10.
2. So the place value increases 10 times.

Teacher's Note:
1. Each place is ten times the place on its right.
2. Moving to the right divides the value by 10.

Answer: (b) 9

Explanation:
1. 39 = XXXIX = X + X + X + IX.
2. Each X needs 2 sticks, so 4 X's need 8 sticks, and I needs 1 stick.
3. Total = 8 + 1 = 9 sticks.

Teacher's Note:
1. X is made with 2 crossing sticks.
2. Write the Roman numeral first, then count the sticks.

Answer: (c) 90000

Explanation:
1. The smallest 5-digit number is 10000 and the greatest is 99999.
2. Count = 99999 − 10000 + 1 = 90000.

Teacher's Note:
1. Count = last − first + 1.
2. There are 90000 five-digit numbers, 9000 four-digit numbers.

Answer: (c) 12

Explanation:
1. The largest 5-digit number is 99999.
2. The largest with three distinct digits is 99987.
3. Difference = 99999 − 99987 = 12.

Teacher's Note:
1. Find each number separately.
2. Subtract the smaller from the larger.

Answer: (b) 9

Explanation:
1. Numbers from 535 to 544 round to 540.
2. Greatest − smallest = 544 − 535 = 9.
3. Options (A) and (D) are both 10, so only 9 can be correct.

Teacher's Note:
1. The range of a rounded ten covers 10 numbers, but the difference between its ends is 9.
2. Count the ends, not the number of values.

Answer: (b) 3

Explanation:
1. The abacus shows 9 beads in H, 4 in T and 8 in O, so the number is 900 + 40 + 8 = 948.
2. With 4 tens and 8 ones, the number is 100 × (beads in H) + 48. For it to lie between 550 and 650 the hundreds must be 6 (648).
3. So 9 − 6 = 3 beads must be removed.

Teacher's Note:
1. Read the abacus rod by rod to get the number.
2. A number between 550 and 650 with 48 at the end is 648.

Answer: (a) Two hundred fifty six trillion two billion one hundred seventeen million one hundred sixteen thousand two hundred twenty three

Explanation:
1. Group in threes from the right: 256,002,117,116,223.
2. The groups are 256 trillion, 002 billion, 117 million, 116 thousand and 223.
3. In words: two hundred fifty-six trillion two billion one hundred seventeen million one hundred sixteen thousand two hundred twenty-three.

Teacher's Note:
1. Groups of three, from the right: ones, thousand, million, billion, trillion.
2. A group 002 still counts, so billions are 2.

Answer: (b) One hundred trillion five hundred sixty seven

Explanation:
1. Group in threes from the right: 100,000,000,000,567.
2. The groups are 100 trillion, 000 billion, 000 million, 000 thousand and 567.
3. In words: one hundred trillion five hundred sixty-seven.

Teacher's Note:
1. The number has 15 digits, so the first group is trillions.
2. Empty groups are not read.

Answer: (d) 13

Explanation:
1. 385 = 5 × 7 × 11, so the first three primes are 5, 7, 11.
2. 1001 = 7 × 11 × 13, so the last three primes are 7, 11, 13.
3. The four primes are 5, 7, 11, 13 and the last is 13.

Teacher's Note:
1. Factorise each product into primes.
2. The two products share two primes.

Answer: (c) 81

Explanation:
1. There are 90 two-digit numbers (10 to 99).
2. Those with equal digits are 11, 22, ..., 99, which are 9 numbers.
3. Required count = 90 − 9 = 81.

Teacher's Note:
1. Count all, then remove the unwanted ones.
2. The tens digit has 9 choices and the ones digit has 9 choices: 9 × 9 = 81.

Answer: (a) 67

Explanation:
1. The sum 100 is even, so one of the three primes must be 2 (the only even prime). The other two add to 98.
2. Let them be x and x + 36: x + x + 36 = 98, so x = 31 and the other is 67.
3. The primes are 2, 31 and 67, and one of them is 67.

Teacher's Note:
1. Three odd numbers would give an odd sum, so one prime is 2.
2. Then use the difference 36.

Answer: (c) 75

Explanation:
1. Dividend = divisor × quotient + remainder.
2. 4150 = d × 55 + 25, so d × 55 = 4125.
3. d = 4125 ÷ 55 = 75.

Teacher's Note:
1. Use the division algorithm.
2. Check: 75 × 55 + 25 = 4125 + 25 = 4150.

Answer: (a) 5,400,000

Explanation:
1. In 467,285 the digit 6 is in the ten-thousands place, so its place value is 60,000.
2. In 30,792 the digit 9 is in the tens place, so its place value is 90.
3. Product = 60,000 × 90 = 5,400,000.

Teacher's Note:
1. Read each number separately from its own place chart.
2. 6 × 9 = 54, then add the zeros: 4 + 1 = 5 zeros.

Answer: (b) 10

Explanation:
1. 1 million = 10,00,000 (Indian system).
2. 1 lakh = 1,00,000, so 10,00,000 ÷ 1,00,000 = 10 lakhs.

Teacher's Note:
1. Learn: 1 million = 10 lakh, 10 million = 1 crore.
2. Divide the larger number by 1 lakh.

Answer: (b) (2 × 100,000) + (5 × 1,000)

Explanation:
1. (A) 200,000 + 600 = 200,600.
2. (B) 200,000 + 5,000 = 205,000.
3. (C) 30,000 + 600 + 70 = 30,670.
4. (D) 30,000 + 5,000 + 70 = 35,070.
5. The greatest is (B) 205,000.

Teacher's Note:
1. Expand each option into an ordinary number.
2. The two options with 2 × 100,000 are far larger than the others.

Answer: (b) 99899999

Explanation:
1. Largest 8-digit number = 99999999.
2. Smallest 6-digit number = 100000.
3. Difference = 99999999 − 100000 = 99899999.

Teacher's Note:
1. The smallest n-digit number is 1 followed by n − 1 zeros.
2. Subtract carefully, place by place.

Answer: (d) 456, 465, 564, 546, 645, 654

Explanation:
1. With the digits 4, 5, 6 used once each there are 3 × 2 × 1 = 6 numbers.
2. With 4 first: 456, 465. With 5 first: 546, 564. With 6 first: 645, 654.
3. The group with exactly these six is (d).
4. The other options contain numbers like 565, 544 or 444 that repeat a digit, or leave some out.

Teacher's Note:
1. List by the first digit so none is missed.
2. Reject any option with a repeated digit.

Answer: (d) 5

Explanation:
1. Batteries needed = 20 × 4 = 80.
2. Packages = 80 ÷ 16 = 5.

Teacher's Note:
1. First find the total, then divide by the pack size.
2. 80 ÷ 16 = 5 exactly, so no part-pack is needed.

Answer: (d) Rs. 3080

Explanation:
1. Earnings = 110 × 28 = 3080.
2. Rounding gives about 100 × 30 = 3000, which is closest to Rs. 3080 among the options.
3. Rs. 1000, Rs. 500 and Rs. 9000 are far from 3000.

Teacher's Note:
1. Estimate with round numbers (100 × 30) and pick the nearest option.
2. Note that 3080 happens to be the exact product too.

Answer: (b) Ten

Explanation:
1. 1 million = 1,000,000.
2. 1 hundred thousand = 100,000.
3. 1,000,000 ÷ 100,000 = 10.

Teacher's Note:
1. A hundred thousand is the same as a lakh.
2. So 1 million = 10 lakh.

Answer: (a) 64000

Explanation:
1. 21397 rounds to 21000 (hundreds digit 3).
2. 42505 rounds to 43000 (hundreds digit 5).
3. Sum ≈ 21000 + 43000 = 64000.

Teacher's Note:
1. Round each number to the nearest thousand, then add.
2. A hundreds digit of 5 rounds up.

Answer: (b) 3,813

Explanation:
1. Total sweets = 19 × 228 = 4332.
2. Left after giving 519 = 4332 − 519 = 3,813.

Teacher's Note:
1. Multiply first and then subtract.
2. Check 19 × 228 = 20 × 228 − 228 = 4560 − 228.

Answer: (d) 7875

Explanation:
1. Swap the tens digit with the hundreds digit and the ones digit with the thousands digit (the number abcd becomes dcba).
2. 8865 → 5688; 8765 → 5678; 7865 → 5687; 7875 → 5787.
3. The largest is 5787, which comes from 7875.

Teacher's Note:
1. Do the swap for every option before comparing.
2. Compare the new numbers, but answer with the original option.

Answer: (a) 91710

Explanation:
1. 21396 rounds to 21400, 27808 rounds to 27810 and 42504 rounds to 42500.
2. Sum = 21400 + 27810 + 42500 = 91710.

Teacher's Note:
1. Look at the ones digit to round to tens.
2. Add the rounded numbers, not the originals.

Answer: (b) C can be subtracted from X, M , D and L.

Explanation:
1. (A), (C) and (D) are correct rules.
2. The symbol C can be subtracted only from D and M, not from X or L.
3. So (B) is not satisfied.

Teacher's Note:
1. I is subtracted from V and X; X from L and C; C from D and M.
2. Check each option against the rules.

Answer: (a) I and II

Explanation:
1. (I) 38 + 49 + 65: ones add to 22, so 2 is written and 2 is carried, but only 1 is carried. The true sum is 152 and the student wrote 142.
2. (II) 45 + 37 + 29: ones add to 21, so 1 is written and 2 should be carried, but only 1 is carried. The true sum is 111 and the student wrote 101.
3. (III) 32 + 14 + 19: the true sum is 65. The student's ones digit 4 is a different slip (2 + 4 + 9 = 15 should give 5), not a carry mistake.
4. The same kind of error (carrying 1 instead of 2) occurs in I and II.

Teacher's Note:
1. Find the correct sum for each and see where it differs.
2. Look at what was carried from the ones column.

Answer: (d) CXXXIV

Explanation:
1. LIX = 59, XXXVI = 36, IX = 9, XC = 90, LX = 60.
2. 59 + 36 + 9 + 90 − 60 = 134.
3. 134 = 100 + 30 + 4 = CXXXIV.
4. CXXXXVI is not a valid numeral (X repeated four times).

Teacher's Note:
1. Convert to numbers, calculate, then convert back.
2. Remember that a symbol is not repeated more than 3 times.

Answer: (b) 684

Explanation:
1. Words per page = 9,95,904 ÷ 1,456.
2. 1456 × 684 = 995,904, so the answer is 684.

Teacher's Note:
1. Divide the total words by the number of pages.
2. Verify by multiplying back.

Answer: (b) (a) – (q), (b) – (s), (c) – (r), (d) – (p)

Explanation:
1. MDCLXX = 1000 + 500 + 100 + 50 + 20 = 1670, so (a) – (q).
2. MMDCCII = 2000 + 500 + 200 + 2 = 2702, so (b) – (s).
3. MCCLXVI = 1000 + 200 + 50 + 10 + 5 + 1 = 1266, so (c) – (r).
4. MCCXXII = 1000 + 200 + 20 + 2 = 1222, so (d) – (p).
5. This is option (B).

Teacher's Note:
1. Convert each Roman numeral and match it with the number in Column-II.
2. Add the symbols from left to right.

Answer: (d) 15899

Explanation:
1. Numbers from 15850 to 15949 round to 15900.
2. 14955 rounds to 15000, 14999 to 15000, 15086 to 15100 and 15899 to 15900.
3. So 15899 is the answer.

Teacher's Note:
1. Round each option to the nearest hundred.
2. Look at the tens digit.

Answer: (b) 90003

Explanation:
1. Smallest 6-digit odd number = 100001.
2. Largest 4-digit even number = 9998.
3. Difference = 100001 − 9998 = 90003.

Teacher's Note:
1. Odd numbers end in 1, 3, 5, 7, 9; even numbers end in 0, 2, 4, 6, 8.
2. Do not use 100000 (it is even) or 9999 (it is odd).

Answer: (a) 1.62

Explanation:
1. 108 thousandths = 108 ÷ 1000 = 0.108.
2. 15 ones = 15.
3. 0.108 × 15 = 1.62.

Teacher's Note:
1. Thousandths means divide by 1000.
2. This question uses decimals, which come later in the Class 6 course.

Answer: (c) 5660 × 2230

Explanation:
1. 5663 ≈ 5660 and 2234 ≈ 2230 are the closest roundings.
2. (D) uses 2330, which is far from 2234.
3. The best estimation is 5660 × 2230.

Teacher's Note:
1. The best estimate stays closest to the original numbers.
2. Compare each rounded value with the original.

Answer: (d) Predecessor, 1, successor

Explanation:
1. Every natural number has a predecessor except 1 (its predecessor 0 is not a natural number).
2. Every natural number has a successor.
3. So the blanks are predecessor, 1, successor.

Teacher's Note:
1. Natural numbers start from 1.
2. Whole numbers start from 0, where 0 has no predecessor.