CBSE Class 6 Maths Advanced Knowing Our Numbers Long Answer Questions Set 01

Long Answer Questions

Teacher's Note:
1. Indian: 3 digits, then 2, 2, ... International: groups of 3.
2. Say the group names crore, lakh, thousand (Indian) or million, thousand (International).

Question 2. Write all possible three digit numbers (without repeating the digits), by using the digits. (i) 6, 7, 5 (ii) 9, 0, 2
Answer:
(i) Using 6, 7, 5: 567, 576, 657, 675, 756, 765 (6 numbers).
(ii) Using 9, 0, 2: 0 cannot be the first digit, so the numbers are 209, 290, 902, 920 (4 numbers).
In simple words: 6 numbers in (i) and 4 numbers in (ii).

Teacher's Note:
1. List by the first digit to avoid missing any.
2. Numbers such as 029 are not 3-digit numbers.

Teacher's Note:
1. Without repetition, start building from the left.
2. Check that each number uses every given digit once.

Teacher's Note:
1. With 0 repeated, the first digit still has to be non-zero.
2. Check that exactly one digit appears twice.

Teacher's Note:
1. Digits must all be different.
2. Do not put 0 in the thousands place.

Question 6. A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0.
Answer:
The first four digits 9, 9, 7, 9 may be arranged in any order, and the smallest arrangement is 7, 9, 9, 9.
The last six digits use 8, 3, 5, 6, 0 with one digit repeated. To make the smallest, repeat the smallest digit 0 and write the digits in ascending order: 0, 0, 3, 5, 6, 8.
The smallest mobile number is 7999003568.
In simple words: Arrange the first four digits as 7999, then 003568.

Teacher's Note:
1. Repeat the smallest digit (0) for the smallest number.
2. Here the first digits can be rearranged; a mobile number may begin with 0 in the later places.

Question 7. In a five digit number, digit at ten's place is four times the digit at unit's place, digit at hundred's place is 0, digit at thousand's place is 5 times of the digit at unit's place and ten thousand's place digit is double the digit at ten's place. Write the number.
Answer:
Let the digit at the unit's place be u.
Ten's digit = 4u, thousand's digit = 5u, hundred's digit = 0, ten thousand's digit = 2 × 4u = 8u.
All digits must be at most 9, so 8u ≤ 9, giving u = 1 (u = 0 would make the number non-five-digit).
Digits: ten thousands 8, thousands 5, hundreds 0, tens 4, ones 1.
The number is 85041.
In simple words: Take the unit digit as 1 and build the rest.

Teacher's Note:
1. Use a variable for the unit digit.
2. The limit of 9 for a digit decides the value.

Question 8. Estimate each of the following by rounding off each number to nearest hundreds: (a) 874 + 478 (b) 793 + 397 (c) 11244 + 3507 (d) 17677 + 13589
Answer:
(a) 874 ≈ 900, 478 ≈ 500: 900 + 500 = 1400.
(b) 793 ≈ 800, 397 ≈ 400: 800 + 400 = 1200.
(c) 11244 ≈ 11200, 3507 ≈ 3500: 11200 + 3500 = 14700.
(d) 17677 ≈ 17700, 13589 ≈ 13600: 17700 + 13600 = 31300.
In simple words: Round each number to hundreds and then add.

Teacher's Note:
1. Look at the tens digit.
2. Compare with the exact sums to see the error.

Question 9. Give the approximate value by estimating. (a) 3228 + 572 (b) 8010 − 2507 (c) 32 × 58 (d) 108 × 47
Answer:
(a) 3228 ≈ 3200, 572 ≈ 600: 3200 + 600 = 3800.
(b) 8010 ≈ 8000, 2507 ≈ 3000: 8000 − 3000 = 5000.
(c) 32 ≈ 30, 58 ≈ 60: 30 × 60 = 1800.
(d) 108 ≈ 100, 47 ≈ 50: 100 × 50 = 5000.
In simple words: Round to the greatest place and then calculate.

Teacher's Note:
1. For products round each factor to its greatest place.
2. For sums and differences, the book rounds to hundreds or thousands.

Question 10. The number of candidates appearing for class 10 board examination conducted by CBSE was 14, 58, 937 in year 2002; 16, 93, 487 in year 2003; 24, 13, 468 in year 2004 and 40, 05,093 in year 2005. Find the total number of candidates who appeared for the examination in these four years.
Answer:
Total = 14,58,937 + 16,93,487 + 24,13,468 + 40,05,093.
14,58,937 + 16,93,487 = 31,52,424.
31,52,424 + 24,13,468 = 55,65,892.
55,65,892 + 40,05,093 = 95,70,985 candidates.
In simple words: Add the four numbers: 95,70,985.

Teacher's Note:
1. Add two numbers at a time.
2. Keep the Indian commas in the answer.

Question 11. The number of scooters produced in a year was 25, 43, 163. Out of these 16,43, 078 were sold. How many were still left ?
Answer:
Produced = 25,43,163 and sold = 16,43,078.
Left = 25,43,163 − 16,43,078 = 9,00,085 scooters.
In simple words: 9,00,085 scooters were left.

Teacher's Note:
1. Subtract the sold number from the produced number.
2. Check: 9,00,085 + 16,43,078 = 25,43,163.

Question 12. A milk depot sells 657 litres of milk every day. How much milk will it sell in 1 year ? (Take 1 year = 365 days)
Answer:
Milk sold in a year = 657 × 365.
657 × 365 = 657 × 300 + 657 × 65 = 1,97,100 + 42,705 = 2,39,805 litres.
In simple words: 2,39,805 litres in a year.

Teacher's Note:
1. Multiply the daily amount by 365.
2. Split 365 as 300 + 65 to make the multiplication easier.

Question 13. The students of class VI of a school collected Rs. 3, 37, 875 for Prime Minister's Relief fund. If each child contributed Rs. 255, how many children are there in the school ?
Answer:
Number of children = 3,37,875 ÷ 255.
255 × 1325 = 3,37,875, so there are 1325 children.
In simple words: There are 1325 children.

Teacher's Note:
1. Divide the total money by the money per child.
2. Check by multiplying 255 × 1325.

Question 14. The diameter of Jupiter is 142800000 meters. Insert commas suitably and write the diameter according to International system of Numeration.
Answer:
Group the digits in threes from the right: 142,800,000 metres.
In words: one hundred forty-two million eight hundred thousand metres.
In simple words: 142,800,000 m.

Teacher's Note:
1. International commas come after every 3 digits.
2. In the Indian system it would be 14,28,00,000.

Question 15. India's population has been steadily increasing from 439 million in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in Indian System of Numeration, using commas suitably.
Answer:
Increase = 1028 million − 439 million = 589 million = 589,000,000.
In the Indian system: 58,90,00,000, which is 5890 lakh.
In simple words: The increase is 58,90,00,000.

Teacher's Note:
1. Subtract first, then re-group the commas.
2. 1 million = 10 lakh.

Question 16. In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and 2,16,813 children in the next month. Find the difference of the number of children getting polio drops in the two months.
Answer:
Difference = 2,16,813 − 2,12,583 = 4,230 children.
In simple words: 4,230 more children got drops the next month.

Teacher's Note:
1. Subtract the smaller number from the bigger.
2. Borrow carefully in the hundreds place.

Question 17. Solve the following: (a) V − I (b) VII − IV (c) IX − III (d) XL − VII
Answer:
(a) 5 − 1 = 4, that is IV.
(b) 7 − 4 = 3, that is III.
(c) 9 − 3 = 6, that is VI.
(d) 40 − 7 = 33, that is XXXIII.
In simple words: Convert, subtract and write the result in Roman numerals: IV, III, VI, XXXIII.

Teacher's Note:
1. Change to numbers, calculate, then change back.
2. 33 is XXX + III.

Question 18. Solve the following: (a) X + V (b) V + III (c) XV + II (d) IX + IX (e) VI + II
Answer:
(a) 10 + 5 = 15 = XV.
(b) 5 + 3 = 8 = VIII.
(c) 15 + 2 = 17 = XVII.
(d) 9 + 9 = 18 = XVIII.
(e) 6 + 2 = 8 = VIII.
In simple words: The sums are XV, VIII, XVII, XVIII and VIII.

Teacher's Note:
1. Convert, add and convert back.
2. 18 is X + VIII, not XIIX.

Question 19. Estimate the following by rounding off each number to its greatest place : (a) 439 + 334 + 4317 (b) 108734 − 47599 (c) 87 × 317 (d) 4489 ÷ 394
Answer:
(a) 439 ≈ 400, 334 ≈ 300, 4317 ≈ 4000: 400 + 300 + 4000 = 4700.
(b) 108734 ≈ 100000, 47599 ≈ 50000: 100000 − 50000 = 50000.
(c) 87 ≈ 90, 317 ≈ 300: 90 × 300 = 27000.
(d) 4489 ≈ 4000, 394 ≈ 400: 4000 ÷ 400 = 10.
In simple words: The estimates are 4700, 50000, 27000 and 10.

Teacher's Note:
1. Keep only the first digit and replace the rest by zeros (after rounding).
2. Compare with the exact values to see how close they are.

Question 20. A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin ?
Answer:
The tin capacity must divide 120, 180 and 240 exactly, and it must be the greatest such number, that is the HCF.
120 = 2³ × 3 × 5, 180 = 2² × 3² × 5, 240 = 2⁴ × 3 × 5.
HCF = 2² × 3 × 5 = 60.
The greatest capacity of a tin is 60 litres.
Check: 120 ÷ 60 = 2, 180 ÷ 60 = 3, 240 ÷ 60 = 4 tins.
In simple words: The tin holds 60 litres.

Teacher's Note:
1. Use the highest common factor for the biggest equal tin.
2. HCF is taught later in Class 6; here a simple check of common factors also works.

Question 21. Express each of the following numbers as a Roman numerals : (a) 26 (b) 63 (c) 72 (d) 99 (e) 556 (f) 769 (g) 26495
Answer:
(a) 26 = XX + VI = XXVI.
(b) 63 = LX + III = LXIII.
(c) 72 = LXX + II = LXXII.
(d) 99 = XC + IX = XCIX.
(e) 556 = D + L + VI = DLVI.
(f) 769 = DCC + LX + IX = DCCLXIX.
(g) 26495 = 26000 + 495 = \( \overline{\mathrm{XXVI}} \) + CD + XC + V = \( \overline{\mathrm{XXVI}} \)CDXCV.
In simple words: Split each number into thousands, hundreds, tens and ones.

Teacher's Note:
1. The bar over a numeral multiplies it by 1000.
2. 99 is XCIX, never IC.

Question 22. Write each of the following in Hindu-Arabic numeral : (a) CLXVI (b) CDXLVI (c) DCCLXVI (d) \( \overline{\mathrm{LIX}} \)CLXIX (e) \( \overline{\mathrm{XXVI}} \)CDXCV
Answer:
(a) CLXVI = 100 + 50 + 10 + 5 + 1 = 166.
(b) CDXLVI = 400 + 40 + 6 = 446.
(c) DCCLXVI = 500 + 200 + 60 + 6 = 766.
(d) \( \overline{\mathrm{LIX}} \)CLXIX = 59 × 1000 + 169 = 59,169.
(e) \( \overline{\mathrm{XXVI}} \)CDXCV = 26 × 1000 + 495 = 26,495.
In simple words: The answers are 166, 446, 766, 59,169 and 26,495.

Teacher's Note:
1. Read the barred part as a number and multiply it by 1000.
2. Break CD, XL, XC into subtractive pairs.