CBSE Class 6 Mathematics Knowing Our Numbers Worksheet Set 06

Question 1. Given below are a few digits, you have to make numbers following the instructions
Given: 0, 4, 5, 6, 9, 7, 1
(a) Write a number between 5 and 6 million.
(b) Write a number greater than 80 lakhs
(c) Write the largest and the smallest 7-digit numbers without repeating the digits.
(d) Write the largest 7-digit number with 0 at the thousands place.
(e) Write the smallest 7-digit number with 9 at the hundreds place.
Answer:
(a) 5,609,741 (Using all the given digits 0, 4, 5, 6, 9, 7, 1 to form a number between 5 and 6 million)
(b) 9,765,410 (Since 80 lakhs equals 8,000,000, any 7-digit number starting with 9 is greater)
(c) Largest: 9,765,410, Smallest: 1,045,679
(d) 9,765,041
(e) 1,046,957
In simple words: Numbers are formed by arranging the given digits according to place value rules and specific constraints.

Exam Tip: When forming the smallest number, do not place 0 at the extreme left as it becomes a 6-digit number instead of 7.

Question 2. Write the following in Roman numerals:
(i) 89 (ii) 762 (iii) 657 (iv) 596 (v) 888 (vi) 944 (vii) 33
Answer:
(i) \(89 = 80 + 9 = \text{LXXX} + \text{IX} = \text{LXXXIX}\)
(ii) \(762 = 700 + 60 + 2 = \text{DCC} + \text{LX} + \text{II} = \text{DCCLXII}\)
(iii) \(657 = 600 + 50 + 7 = \text{DC} + \text{L} + \text{VII} = \text{DCLVII}\)
(iv) \(596 = 500 + 90 + 6 = \text{D} + \text{XC} + \text{VI} = \text{DXCVI}\)
(v) \(888 = 800 + 80 + 8 = \text{DCCC} + \text{LXXX} + \text{VIII} = \text{DCCCLXXXVIII}\)
(vi) \(944 = 900 + 40 + 4 = \text{CM} + \text{XL} + \text{IV} = \text{CMXLIV}\)
(vii) \(33 = 30 + 3 = \text{XXX} + \text{III} = \text{XXXIII}\)
In simple words: Break down each number into hundreds, tens, and units, then write their Roman equivalents.

Exam Tip: Remember the primary symbols: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000.

Question 3. Write each of the following in Hindu Arabic numerals:
(i) DCCXXIV (ii) CCCLXV (iii) LXIII (iv) CMXCIX (v) MCMVII
Answer:
(i) DCCXXIV = \(500 + 200 + 20 + 4 = 724\)
(ii) CCCLXV = \(300 + 50 + 10 + 5 = 365\)
(iii) LXIII = \(50 + 10 + 3 = 63\)
(iv) CMXCIX = \((1000 - 100) + (100 - 10) + (10 - 1) = 900 + 90 + 9 = 999\)
(v) MCMVII = \(1000 + (1000 - 100) + 5 + 2 = 1000 + 900 + 7 = 1907\)
In simple words: Add the values of Roman symbols, keeping in mind that a smaller symbol before a larger one means subtraction.

Exam Tip: Group symbols from left to right and apply the subtraction rule for combinations like IV, IX, XL, XC, CD, and CM.

Question 4. Fill in the boxes with \(>\), \(<\) or \(=\) sign.
(a) XC \(\square\) LXXIV
(b) 76 \(\square\) LXXVI
(c) LXXXV \(\square\) CV
(d) XCIX \(\square\) 89
(e) LXVI \(\square\) 66
Answer:
(a) XC (\(90\)) \(>\) LXXIV (\(74\))
(b) 76 (\(76\)) \(=\) LXXVI (\(76\))
(c) LXXXV (\(85\)) \(<\) CV (\(105\))
(d) XCIX (\(99\)) \(>\) 89 (\(89\))
(e) LXVI (\(66\)) \(=\) 66 (\(66\))
In simple words: Convert Roman numerals to Hindu-Arabic numbers first to easily compare their values.

Exam Tip: Always evaluate both sides into standard numbers before deciding on the inequality symbol.

Question 5. The number 392492 contains two 9s and two 2s.
(i) How many times larger is the first 9 compared to the second 9?
(ii) How many times is the second 2 is smaller compared to the first 2? (compare the place values)
Answer:
(i) In 392492, the first 9 is at the ten-thousands place, so its place value is \(90,000\).
The second 9 is at the tens place, so its place value is \(90\).
Ratio = \(\frac{90,000}{90} = 1,000\)
The first 9 is \(1,000\) times larger than the second 9.
(ii) In 392492, the first 2 is at the thousands place, so its place value is \(2,000\).
The second 2 is at the ones place, so its place value is \(2\).
Ratio = \(\frac{2}{2,000} = \frac{1}{1,000}\)
The second 2 is \(1,000\) times smaller than the first 2.
In simple words: Place value increases by a factor of 10 for each step to the left in the place value chart.

Exam Tip: Identify the exact position of each digit from right to left (ones, tens, hundreds, thousands, ten-thousands, lakhs) before calculating place values.

Question 6. Name five large cities in India. Find their population. Also find the distance in Km between each pair of these cities.
Answer:
Five large cities in India and their approximate populations (as per recent census estimates):
1. Mumbai - 12,500,000
2. Delhi - 11,000,000
3. Bengaluru - 8,400,000
4. Hyderabad - 6,800,000
5. Chennai - 7,100,000

Distances between pairs of cities (approximate road/rail distances in km):
- Mumbai to Delhi: 1,400 km
- Mumbai to Bengaluru: 980 km
- Mumbai to Hyderabad: 710 km
- Mumbai to Chennai: 1,330 km
- Delhi to Bengaluru: 2,150 km
- Delhi to Hyderabad: 1,580 km
- Delhi to Chennai: 2,180 km
- Bengaluru to Hyderabad: 570 km
- Bengaluru to Chennai: 350 km
- Hyderabad to Chennai: 630 km
In simple words: Large cities have high populations, and distances are measured across states in kilometers.

Exam Tip: Use standard geographical data sources when looking up real-world statistics for projects.

Question 7. Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Answer:
Weight of one box = \(4\text{ kg } 500\text{ g} = 4,500\text{ g}\)
Maximum weight the van can carry = \(800\text{ kg} = 800,000\text{ g}\)
Number of boxes = \(\frac{800,000}{4,500} = \frac{8000}{45} = \frac{1600}{9} = 177.77\)
Since fractional boxes cannot be loaded, the maximum number of full boxes is 177.
In simple words: Convert both weights into grams and divide the total capacity by the weight of one box, then take the whole number part.

Exam Tip: Always convert all quantities into the same unit (grams in this case) before performing division.

Question 8. The numbers 3, 6, 10 can be written in triangular form. Represent the next 5 triangular numbers by drawing dots.
Answer:
Triangular numbers are formed by the sum of consecutive natural numbers: \(1, 3, 6, 10, 15, 21, 28, 36, 45, \dots\)
The given numbers are the 2nd (\(3\)), 3rd (\(6\)), and 4th (\(10\)) triangular numbers.
The next 5 triangular numbers are:
- 5th triangular number = \(1 + 2 + 3 + 4 + 5 = 15\)
- 6th triangular number = \(1 + 2 + 3 + 4 + 5 + 6 = 21\)
- 7th triangular number = \(1 + 2 + 3 + 4 + 5 + 6 + 7 = 28\)
- 8th triangular number = \(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\)
- 9th triangular number = \(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45\)
In simple words: Each triangular number adds one more row of dots than the previous number, forming an equilateral triangle.

Exam Tip: The \(n\)-th triangular number is given by the formula \(\frac{n(n+1)}{2}\).

Question 9. Study the following pattern and write 4 more steps:
\(11 \times 11 = 121\)
\(111 \times 111 = 12321\)
\(1111 \times 1111 = 1234321\)
Answer:
The next 4 steps in the pattern are:
\(11111 \times 11111 = 123454321\)
\(111111 \times 111111 = 12345654321\)
\(1111111 \times 1111111 = 1234567654321\)
\(11111111 \times 11111111 = 123456787654321\)
In simple words: The product numbers mirror the count of ones being multiplied, counting up to that number and back down to 1.

Exam Tip: Count the number of ones in each multiplier to know the peak digit in the center of the product.

Question 10. A bus started its journey and reached different places with a speed of \(60\text{ km/hr}\). The journey is shown below.
(i) Find the total distance covered by the bus from A and D.
(ii) Find the total distance covered by the bus from D to G.
(iii) Find the total distance covered by the bus, if it starts from A and returns back to A.
(iv) Can you find the difference of distance from C to D and D to E.
(v) Find out the time taken by the bus to reach
(a) A to B (b) C to D (c) E to G (d) total journey
Answer:
Based on the route diagram with distances: A to B = \(4130\text{ km}\), B to C = \(3140\text{ km}\), C to D = \(2160\text{ km}\), D to E = \(8140\text{ km}\), E to F = \(4830\text{ km}\), F to G = \(2550\text{ km}\), G to A = \(1290\text{ km}\).
(i) Distance from A to D (via B and C) = \(4130 + 3140 + 2160 = 9,430\text{ km}\)
(ii) Distance from D to G (via E and F) = \(8140 + 4830 + 2550 = 15,520\text{ km}\)
(iii) Total distance for a complete round trip starting and returning to A = \(4130 + 3140 + 2160 + 8140 + 4830 + 2550 + 1290 = 26,240\text{ km}\)
(iv) Distance from C to D = \(2160\text{ km}\). Distance from D to E = \(8140\text{ km}\).
Difference = \(8140 - 2160 = 5,980\text{ km}\)
(v) Time = \(\frac{\text{Distance}}{\text{Speed}}\), where speed = \(60\text{ km/hr}\):
(a) A to B time = \(\frac{4130}{60} = 68.83\text{ hours}\) (or \(68\text{ hrs } 50\text{ mins}\)
(b) C to D time = \(\frac{2160}{60} = 36\text{ hours}\)
(c) E to G time (via F) = \(\frac{4830 + 2550}{60} = \frac{7380}{60} = 123\text{ hours}\)
(d) Total journey time = \(\frac{26240}{60} = 437.33\text{ hours}\) (or \(437\text{ hrs } 20\text{ mins}\)
In simple words: Total distance is the sum of path segments, and time is calculated by dividing distance by speed.

Exam Tip: Double-check all intermediate segment additions before calculating time.

CROSS-NUMBER PUZZLE

Fit all of the numbers written in words as clues. Using each number once. There is only one way in which all of the numbers will fit.
Across:
4. Four hundred forty nine thousand, seven hundred and sixty eight. (\(449,768\))
7. Twenty million, two hundred eighty three thousand, six hundred and forty two. (\(20,283,642\))
8. One crore, ninety three lakh, eighty four thousand and three. (\(1,93,84,003\))
10. Three hundred forty five million, six hundred ninety eight thousand and fifty seven. (\(345,698,057\))
Down:
1. Eight hundred and nine. (\(809\))
2. Five hundred and eighty six. (\(586\))
3. Two hundred ninety three thousand and forty two. (\(293,042\))
5. Thirty eight lakh, two thousand seven hundred and forty eight. (\(38,02,748\))
6. Two million six hundred eighty three thousand seven hundred and forty eight. (\(2,683,748\))
9. Five hundred eighty three million, seventy nine thousand, six hundred and forty six. (\(583,079,646\))

Answer:
The puzzle grid is filled by placing each number into its respective Across and Down intersecting grid cells based on digit lengths and overlapping digit values.
In simple words: Cross-number puzzles work just like crossword puzzles, but with numbers instead of words.

Exam Tip: Verify intersecting digits between Across and Down numbers to ensure correct placement.

Rounding Robots

Round the numbers on the robots to the nearest hundred. Draw a line from each robot to the correct battery.
Answer:
- Robot 746 rounded to the nearest hundred = \(700\)
- Robot 489 rounded to the nearest hundred = \(500\)
- Robot 757 rounded to the nearest hundred = \(800\)
- Robot 550 rounded to the nearest hundred = \(600\)
- Robot 969 rounded to the nearest hundred = \(1,000\)
- Robot 74 rounded to the nearest hundred = \(100\)
- Robot 27 rounded to the nearest hundred = \(0\)
- Robot 872 rounded to the nearest hundred = \(900\)
- Robot 252 rounded to the nearest hundred = \(300\)
- Robot 220 rounded to the nearest hundred = \(200\)
- Robot 361 rounded to the nearest hundred = \(400\)
In simple words: Look at the tens digit; if it is 5 or more, round up to the next hundred. If it is less than 5, round down.

Exam Tip: When the tens digit is 5, always round up to the higher hundred.