| Non-Rationalised NCERT Books Solution | ||||||
|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
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| Exercise 1.1 | Example 1 to 4 (Before Exercise 1.2) | Exercise 1.2 |
| Example 5 & 6 (Before Exercise 1.3) | Exercise 1.3 | Example 7 |
Chapter 1 Knowing Our Number
The solutions page for Chapter 1, "Knowing Our Numbers," provides comprehensive, step-by-step guidance through the foundational concepts of the number system introduced at the Class 6 level. This chapter primarily deals with understanding large numbers, place value, comparison, estimation, and the use of brackets, along with an introduction to Roman numerals.
The solutions meticulously address exercises covering topics like identifying the greatest and smallest numbers from a given set, arranging numbers in ascending and descending order, and forming the largest and smallest possible numbers using given digits without repetition.
Significant emphasis is placed on understanding the Indian and International systems of numeration, and the solutions clearly demonstrate how to insert commas correctly and write number names according to both systems. Word problems involving large numbers, often related to population, production, or distances, are broken down logically, showing students how to extract relevant information, decide on the operation (addition, subtraction, multiplication, division), and perform calculations accurately.
Estimation is another key area; the solutions explain rounding off numbers to the nearest tens, hundreds, thousands, etc., and applying these estimations to approximate sums, differences, and products, highlighting the practical utility of estimation in daily life.
The use of brackets is explained through examples demonstrating the order of operations (BODMAS/PEMDAS isn't formally introduced but the principle of solving brackets first is).
Finally, the conversion between Hindu-Arabic numerals and Roman numerals is covered, with solutions showing the rules and patterns involved in writing and interpreting Roman numerals up to a reasonable limit relevant for this grade. The solutions aim not just for correct answers but for conceptual clarity, ensuring students build a strong number sense and computational fluency required for subsequent mathematical learning.
Exercise 1.1
Question 1. Fill in the blanks:
(a) 1 lakh = _______ ten thousand.
(b) 1 million = _______ hundred thousand.
(c) 1 crore = _______ ten lakh.
(d) 1 crore = _______ million.
(e) 1 million = _______ lakh.
Answer:
(a) 1 lakh = 1,00,000
Ten thousand = 10,000
Therefore, 1 lakh = $1,00,000 = 10 \times 10,000 = \mathbf{10}$ ten thousand.
(b) 1 million = 1,000,000
Hundred thousand = 100,000
Therefore, 1 million = $1,000,000 = 10 \times 100,000 = \mathbf{10}$ hundred thousand.
(c) 1 crore = 1,00,00,000
Ten lakh = 10,00,000
Therefore, 1 crore = $1,00,00,000 = 10 \times 10,00,000 = \mathbf{10}$ ten lakh.
(d) 1 crore = 1,00,00,000
1 million = 1,000,000
Therefore, 1 crore = $1,00,00,000 = 10 \times 1,000,000 = \mathbf{10}$ million.
(e) 1 million = 1,000,000
1 lakh = 1,00,000
Therefore, 1 million = $1,000,000 = 10 \times 1,00,000 = \mathbf{10}$ lakh.
Question 2. Place commas correctly and write the numerals:
(a) Seventy three lakh seventy five thousand three hundred seven.
(b) Nine crore five lakh forty one.
(c) Seven crore fifty two lakh twenty one thousand three hundred two.
(d) Fifty eight million four hundred twenty three thousand two hundred two.
(e) Twenty three lakh thirty thousand ten.
Answer:
(a) The numeral representation for Seventy three lakh seventy five thousand three hundred seven is 73,75,307.
(b) The numeral representation for Nine crore five lakh forty one is 9,05,00,041.
(c) The numeral representation for Seven crore fifty two lakh twenty one thousand three hundred two is 7,52,21,302.
(d) The numeral representation for Fifty eight million four hundred twenty three thousand two hundred two (using the International System of Numeration) is 58,423,202.
(e) The numeral representation for Twenty three lakh thirty thousand ten is 23,30,010.
Question 3. Insert commas suitably and write the names according to Indian System of Numeration :
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701
Answer:
(a) The number 87595762 is written with commas according to the Indian System of Numeration as 8,75,95,762.
Its name is Eight crore seventy-five lakh ninety-five thousand seven hundred sixty-two.
(b) The number 8546283 is written with commas according to the Indian System of Numeration as 85,46,283.
Its name is Eighty-five lakh forty-six thousand two hundred eighty-three.
(c) The number 99900046 is written with commas according to the Indian System of Numeration as 9,99,00,046.
Its name is Nine crore ninety-nine lakh forty-six.
(d) The number 98432701 is written with commas according to the Indian System of Numeration as 9,84,32,701.
Its name is Nine crore eighty-four lakh thirty-two thousand seven hundred one.
Question 4. Insert commas suitably and write the names according to International System of Numeration :
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831
Answer:
(a) The number 78921092 is written with commas according to the International System of Numeration as 78,921,092.
Its name is Seventy-eight million nine hundred twenty-one thousand ninety-two.
(b) The number 7452283 is written with commas according to the International System of Numeration as 7,452,283.
Its name is Seven million four hundred fifty-two thousand two hundred eighty-three.
(c) The number 99985102 is written with commas according to the International System of Numeration as 99,985,102.
Its name is Ninety-nine million nine hundred eighty-five thousand one hundred two.
(d) The number 48049831 is written with commas according to the International System of Numeration as 48,049,831.
Its name is Forty-eight million forty-nine thousand eight hundred thirty-one.
Example 1 to 4 (Before Exercise 1.2)
Example 1: Population of Sundarnagar was 2,35,471 in the year 1991. In the year 2001 it was found to be increased by 72,958. What was the population of the city in 2001?
Answer:
Given:
Population of Sundarnagar in 1991 = 2,35,471
Increase in population by the year 2001 = 72,958
To Find:
The population of the city in 2001.
Solution:
To find the population in 2001, we need to add the population in 1991 to the increase in population.
Population in 2001 = Population in 1991 + Increase in population
Population in 2001 = 2,35,471 + 72,958
Performing the addition:
$$\begin{array}{cc} & 2 & 3 & 5 & 4 & 7 & 1 \\ + & & 7 & 2 & 9 & 5 & 8 \\ \hline & 3 & 0 & 8 & 4 & 2 & 9 \\ \hline \end{array}$$
Population in 2001 = 3,08,429
Therefore, the population of the city in 2001 was 3,08,429.
Example 2: In one state, the number of bicycles sold in the year 2002-2003 was 7,43,000. In the year 2003-2004, the number of bicycles sold was 8,00,100. In which year were more bicycles sold? and how many more?
Answer:
Given:
Number of bicycles sold in the year 2002-2003 = 7,43,000
Number of bicycles sold in the year 2003-2004 = 8,00,100
To Find:
1. The year in which more bicycles were sold.
2. The difference in the number of bicycles sold between the two years.
Solution:
First, we compare the number of bicycles sold in the two years:
Sales in 2002-2003 = 7,43,000
Sales in 2003-2004 = 8,00,100
Comparing the two numbers, we see that 8,00,100 > 7,43,000.
Therefore, more bicycles were sold in the year 2003-2004.
Now, we find how many more bicycles were sold in 2003-2004 compared to 2002-2003 by subtracting the smaller number from the larger number.
Difference = Number of bicycles sold in 2003-2004 - Number of bicycles sold in 2002-2003
Difference = 8,00,100 - 7,43,000
Performing the subtraction:
$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c} & & \overset{7}{\cancel{8}} & \overset{9}{\cancel{0}} & \overset{10}{\cancel{0}} & 1 & 0 & 0 \\ -& & 7 & 4 & 3 & 0 & 0 & 0 \\ \hline & & 0 & 5 & 7 & 1 & 0 & 0 \\ \hline \end{array}$
Difference = 57,100
Thus, 57,100 more bicycles were sold in the year 2003-2004.
Example 3: The town newspaper is published every day. One copy has 12 pages. Everyday 11,980 copies are printed. How many total pages are printed everyday?
Answer:
Given:
Number of pages in one copy of the newspaper = 12
Number of copies printed every day = 11,980
To Find:
The total number of pages printed every day.
Solution:
To find the total number of pages printed every day, we need to multiply the number of pages per copy by the number of copies printed.
Total pages printed = Number of pages per copy $\times$ Number of copies printed
Total pages printed = $12 \times 11,980$
Performing the multiplication:
$\begin{array}{cc} & & 1 & 1 & 9 & 8 & 0 \\ \times & & & & & 1 & 2 \\ \hline & & 2 & 3 & 9 & 6 & 0 \\ + & 1 & 1 & 9 & 8 & 0 & \times \\ \hline & 1 & 4 & 3 & 7 & 6 & 0 \\ \hline \end{array}$
Total pages printed = 1,43,760
Therefore, 1,43,760 pages are printed every day.
Example 4: The number of sheets of paper available for making notebooks is 75,000. Each sheet makes 8 pages of a notebook. Each notebook contains 200 pages. How many notebooks can be made from the paper available?
Answer:
Given:
Number of sheets of paper available = 75,000
Number of pages made from each sheet = 8
Number of pages in each notebook = 200
To Find:
The total number of notebooks that can be made from the available paper.
Solution:
First, calculate the total number of pages available.
Total pages available = Number of sheets $\times$ Pages per sheet
Total pages available = $75,000 \times 8$
$\begin{array}{cc} & & 7 & 5 & 0 & 0 & 0 \\ \times & & & & & & 8 \\ \hline & 6 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$
Total pages available = 6,00,000
Now, calculate the number of notebooks that can be made using the total available pages.
Number of notebooks = Total pages available $\div$ Pages per notebook
Number of notebooks = $6,00,000 \div 200$
Number of notebooks = $\frac{6,00,000}{200}$
Number of notebooks = $\frac{6000\cancel{00}}{2\cancel{00}}$
Number of notebooks = $\frac{6000}{2}$
Number of notebooks = 3000
Therefore, 3,000 notebooks can be made from the paper available.
Exercise 1.2
Question 1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Answer:
Given:
Number of tickets sold on the first day = 1094
Number of tickets sold on the second day = 1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the final (fourth) day = 2751
To Find:
The total number of tickets sold on all the four days.
Solution:
To find the total number of tickets sold, we need to add the number of tickets sold on each of the four days.
Total tickets sold = (Tickets sold on day 1) + (Tickets sold on day 2) + (Tickets sold on day 3) + (Tickets sold on day 4)
Total tickets sold = $1094 + 1812 + 2050 + 2751$
Let's perform the addition:
$\begin{array}{cc} & 1 & 0 & 9 & 4 \\ & 1 & 8 & 1 & 2 \\ & 2 & 0 & 5 & 0 \\ + & 2 & 7 & 5 & 1 \\ \hline & 7 & 7 & 0 & 7 \\ \hline \end{array}$
The sum is 7707.
Therefore, the total number of tickets sold on all the four days is 7707.
Question 2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Answer:
Given:
Runs Shekhar has scored so far = 6980
Runs Shekhar wishes to complete = 10,000
To Find:
How many more runs Shekhar needs to reach his goal.
Solution:
To find the number of additional runs Shekhar needs, we must subtract the runs he has already scored from the target number of runs.
Runs needed = (Target runs) - (Runs scored so far)
Runs needed = $10000 - 6980$
Let's perform the subtraction:
$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c} & \overset{0}{\cancel{1}} & \overset{9}{\cancel{10}} & \overset{9}{\cancel{10}} & \overset{10}{\cancel{0}} & 0 \\ % Shows borrowing steps & 1 & 0 & 0 & 0 & 0 \\ -& & 6 & 9 & 8 & 0 \\ \hline & & 3 & 0 & 2 & 0 \\ \hline \end{array}$
The difference is 3020.
Therefore, Shekhar needs 3020 more runs to complete 10,000 runs.
Question 3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Answer:
Given:
Number of votes secured by the successful candidate = 5,77,500
Number of votes secured by his nearest rival = 3,48,700
To Find:
The margin by which the successful candidate won the election.
Solution:
To find the margin of victory, we need to calculate the difference between the votes secured by the successful candidate and the votes secured by his nearest rival.
Margin = (Votes of successful candidate) - (Votes of nearest rival)
Margin = 5,77,500 - 3,48,700
Performing the subtraction:
$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c} & 5 & \cancel{7}^{6} & \cancel{7}^{16} & \cancel{5}^{15} & 0 & 0 \\ - & 3 & 4 & 8 & 7 & 0 & 0 \\ \hline & 2 & 2 & 8 & 8 & 0 & 0 \\ \hline \end{array}$
Margin = 2,28,800
Final Answer:
Therefore, the successful candidate won the election by a margin of 2,28,800 votes.
Question 4. Kirti bookstore sold books worth ₹ 2,85,891 in the first week of June and books worth ₹ 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Answer:
Given:
Value of books sold in the first week =
Value of books sold in the second week =
To Find:
1. Total sale for the two weeks together.
2. Which week had greater sales.
3. By how much the sales were greater in that week.
Solution:
1. Total sale for the two weeks together:
Total sale = Sale in first week + Sale in second week
Total sale =
Performing the addition:
$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c} & & 2 & 8 & \overset{1}{5} & \overset{1}{8} & 9 & 1 \\ + & & 4 & 0 & 0 & 7 & 6 & 8 \\ \hline & & 6 & 8 & 6 & 6 & 5 & 9 \\ \hline \end{array}$
Total sale for the two weeks =
2. Comparing the sales in the two weeks:
Sale in first week =
Sale in second week =
Comparing the two values: $4,00,768 > 2,85,891$.
Therefore, the sale was greater in the second week.
3. Difference in sales:
To find by how much the sale was greater in the second week, we subtract the sale of the first week from the sale of the second week.
Difference = Sale in second week - Sale in first week
Difference =
Performing the subtraction:
$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c} & \overset{3}{\cancel{4}} & \overset{9}{\cancel{0}} & \overset{9}{\cancel{0}} & \overset{16}{\cancel{7}} & \overset{16}{\cancel{6}} & 8 \\ - & 2 & 8 & 5 & 8 & 9 & 1 \\ \hline & 1 & 1 & 4 & 8 & 7 & 7 \\ \hline \end{array}$
Difference =
Final Answer:
The total sale for the two weeks together was
The sale was greater in the second week.
The sale in the second week was greater by
Question 5. Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Answer:
Given:
The digits are 6, 2, 7, 4, 3.
Each digit must be used only once to form a 5-digit number.
To Find:
The difference between the greatest and the least 5-digit number formed using these digits.
Solution:
The given digits are 6, 2, 7, 4, 3.
To form the greatest 5-digit number, we arrange the digits in descending order (from largest to smallest):
7, 6, 4, 3, 2
So, the greatest 5-digit number is 76,432.
To form the least 5-digit number, we arrange the digits in ascending order (from smallest to largest):
2, 3, 4, 6, 7
So, the least 5-digit number is 23,467.
Now, we need to find the difference between the greatest and the least number:
Difference = Greatest number - Least number
Difference = 76,432 - 23,467
Performing the subtraction:
$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c} & 7 & \overset{5}{\cancel{6}} & \overset{13}{\cancel{4}} & \overset{12}{\cancel{3}} & \overset{12}{\cancel{2}} \\ - & 2 & 3 & 4 & 6 & 7 \\ \hline & 5 & 2 & 9 & 6 & 5 \\ \hline \end{array}$
The difference is 52,965.
Final Answer:
The difference between the greatest and the least 5-digit number formed using the digits 6, 2, 7, 4, 3 is 52,965.
Question 6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Answer:
Given:
Average number of screws manufactured by the machine per day = 2,825
To Find:
The total number of screws produced by the machine in the month of January 2006.
Solution:
We need to find the total number of screws produced in January 2006.
First, we find the number of days in the month of January.
Number of days in January = 31 days.
(The year 2006 is not a leap year, but the number of days in January is always 31, regardless of whether the year is a leap year or not.)
Total screws produced = (Average number of screws per day) $\times$ (Number of days in the month)
Total screws produced in January 2006 = $2825 \times 31$
Now, we perform the multiplication:
$\begin{array}{cc} & & 2 & 8 & 2 & 5 \\ \times & & & & 3 & 1 \\ \hline & & 2 & 8 & 2 & 5 \\ + & 8 & 4 & 7 & 5 & \times \\ \hline & 8 & 7 & 5 & 7 & 5 \\ \hline \end{array}$
So, the total number of screws produced is 87,575.
Therefore, the machine produced 87,575 screws in the month of January 2006.
Question 7. A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ₹ 1200 each. How much money will remain with her after the purchase?
Answer:
Given:
Initial amount of money with the merchant =
Number of radio sets ordered = 40
Cost of each radio set =
To Find:
The amount of money remaining with the merchant after the purchase.
Solution:
First, we calculate the total cost of the 40 radio sets.
Cost of one radio set =
Number of radio sets = 40
Total cost of radio sets = Cost per set $\times$ Number of sets
Total cost = $\textsf{₹} 1200 \times 40$
We perform the multiplication:
$\begin{array}{cc} & & 1 & 2 & 0 & 0 \\ \times & & & & 4 & 0 \\ \hline & & 0 & 0 & 0 & 0 \\ + & 4 & 8 & 0 & 0 & \times \\ \hline & 4 & 8 & 0 & 0 & 0 \\ \hline \end{array}$
So, the total cost of 40 radio sets is
Now, we find the money remaining with the merchant.
Remaining money = Initial money $-$ Total cost of radio sets
Remaining money = $\textsf{₹} 78,592 - \textsf{₹} 48,000$
We perform the subtraction:
$\begin{array}{cc} & 7 & 8 & 5 & 9 & 2 \\ - & 4 & 8 & 0 & 0 & 0 \\ \hline & 3 & 0 & 5 & 9 & 2 \\ \hline \end{array}$
The remaining money is
Therefore,
Question 8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
(Hint: Do you need to do both the multiplications?)
Answer:
Given:
Number to be multiplied = 7236
Incorrect multiplier = 65
Correct multiplier = 56
To Find:
The difference between the incorrect answer ($7236 \times 65$) and the correct answer ($7236 \times 56$).
Solution (Method 1: Direct Calculation):
First, calculate the incorrect answer (multiplying by 65):
$\begin{array}{cc} && & 7 & 2 & 3 & 6 \\ &\times & & & & 6 & 5 \\ \hline && 3 & 6 & 1 & 8 & 0 \\ + & 4 & 3 & 4 & 1 & 6 & \times \\ \hline & 4 & 7 & 0 & 3 & 4 & 0 \\ \hline \end{array}$
Incorrect answer = $7236 \times 65 = 470340$
Next, calculate the correct answer (multiplying by 56):
$\begin{array}{cc} && & 7 & 2 & 3 & 6 \\ &\times & & & & 5 & 6 \\ \hline && 4 & 3 & 4 & 1 & 6 \\ + & 3 & 6 & 1 & 8 & 0 & \times \\ \hline & 4 & 0 & 5 & 2 & 1 & 6 \\ \hline \end{array}$
Correct answer = $7236 \times 56 = 405216$
Now, find the difference between the incorrect answer and the correct answer:
Difference = Incorrect answer $-$ Correct answer
Difference = $470340 - 405216$
$\begin{array}{cc} & 4 & 7 & 0 & 3 & 4 & 0 \\ - & 4 & 0 & 5 & 2 & 1 & 6 \\ \hline & 0 & 6 & 5 & 1 & 2 & 4 \\ \hline \end{array}$
Difference = 65124
Alternate Solution (Using the Hint):
The student multiplied by 65 instead of 56. The difference in the multipliers is:
$65 - 56 = 9$
The difference between the incorrect answer and the correct answer is the original number multiplied by the difference in the multipliers.
Difference = $7236 \times (65 - 56)$
Difference = $7236 \times 9$
Now, perform the multiplication:
$\begin{array}{cc} & & 7 & 2 & 3 & 6 \\ \times & & & & & 9 \\ \hline & 6 & 5 & 1 & 2 & 4 \\ \hline \end{array}$
Difference = 65124
This method is quicker as it only requires one multiplication.
Therefore, the student's answer was 65,124 greater than the correct answer.
Question 9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
(Hint: convert data in cm.)
Answer:
Given:
Total length of cloth available = 40 m
Length of cloth needed to stitch one shirt = 2 m 15 cm
To Find:
1. The number of shirts that can be stitched from the total cloth.
2. The amount of cloth that will remain.
Solution:
Following the hint, we first convert all measurements to centimeters (cm).
We know that $1 \text{ m} = 100 \text{ cm}$.
Total length of cloth available:
$40 \text{ m} = 40 \times 100 \text{ cm} = 4000 \text{ cm}$
Length of cloth needed for one shirt:
$2 \text{ m } 15 \text{ cm} = (2 \times 100 \text{ cm}) + 15 \text{ cm} = 200 \text{ cm} + 15 \text{ cm} = 215 \text{ cm}$
Now, we need to find how many times 215 cm fits into 4000 cm. This is a division problem.
Number of shirts = Total cloth available (cm) $\div$ Cloth needed per shirt (cm)
Number of shirts = $4000 \div 215$
We perform the long division:
$\begin{array}{r} 18\phantom{00} \\ 215{\overline{\smash{\big)}\,4000\phantom{)}}} \\ \underline{-~\phantom{(}215\phantom{f)}} \\ 1850\phantom{)} \\ \underline{-~\phantom{(}1720 \phantom{)}} \\ 130\phantom{)} \end{array}$
The quotient is 18 and the remainder is 130.
The quotient represents the number of shirts that can be stitched.
Number of shirts = 18
The remainder represents the amount of cloth left over in centimeters.
Remaining cloth = 130 cm
We can convert the remaining cloth back to meters and centimeters:
$130 \text{ cm} = 100 \text{ cm} + 30 \text{ cm} = 1 \text{ m } 30 \text{ cm}$
Therefore, 18 shirts can be stitched and 1 m 30 cm (or 130 cm) of cloth will remain.
Question 10. 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:
Given:
Weight of one medicine box = 4 kg 500 g
Maximum weight the van can carry = 800 kg
To Find:
The maximum number of boxes that can be loaded into the van.
Solution:
To solve this, we need to have both weights in the same unit. Let's convert both weights to grams (g).
We know that $1 \text{ kg} = 1000 \text{ g}$.
Weight of one box in grams:
$4 \text{ kg } 500 \text{ g} = (4 \times 1000 \text{ g}) + 500 \text{ g}$
$= 4000 \text{ g} + 500 \text{ g}$
$= 4500 \text{ g}$
Maximum weight capacity of the van in grams:
$800 \text{ kg} = 800 \times 1000 \text{ g}$
$= 800000 \text{ g}$
Now, we need to find how many boxes, each weighing 4500 g, can fit into a total capacity of 800000 g.
Number of boxes = Total capacity $\div$ Weight per box
Number of boxes = $800000 \div 4500$
We can simplify the division by removing common trailing zeros:
Number of boxes = $8000 \div 45$
Now we perform the long division:
$\begin{array}{r} 177\phantom{0} \\ 45{\overline{\smash{\big)}\,8000\phantom{)}}} \\ \underline{-~\phantom{(}45\phantom{..)}} \\ 350\phantom{)} \\ \underline{-~\phantom{(}315 \phantom{(}} \\ 350\phantom{)} \\ \underline{-~\phantom{(}315\phantom{(}} \\ 35\phantom{)} \end{array}$
The quotient is 177 and the remainder is 35.
The quotient represents the maximum number of full boxes that can be loaded without exceeding the weight limit.
Number of boxes = 177
The remainder (35 g) represents the leftover capacity, which is not enough to load another full box.
Therefore, 177 such boxes can be loaded in the van.
Question 11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Answer:
Given:
Distance between school and house = 1 km 875 m
The student walks both ways (to school and back) each day.
Time period = 6 days
To Find:
The total distance covered by the student in 6 days.
Solution:
First, let's convert the distance between the school and the house into a single unit, meters (m).
We know that $1 \text{ km} = 1000 \text{ m}$.
Distance = $1 \text{ km } 875 \text{ m}$
$= (1 \times 1000 \text{ m}) + 875 \text{ m}$
$= 1000 \text{ m} + 875 \text{ m}$
$= 1875 \text{ m}$
The student walks this distance twice each day (once to school and once back home).
Distance covered in one day = Distance $\times$ 2
$= 1875 \text{ m} \times 2$
$\begin{array}{cc} & 1 & 8 & 7 & 5 \\ \times & & & & 2 \\ \hline & 3 & 7 & 5 & 0 \\ \hline \end{array}$
Distance covered in one day = $3750 \text{ m}$
Now, we need to find the total distance covered in 6 days.
Total distance covered in 6 days = Distance covered in one day $\times$ 6
$= 3750 \text{ m} \times 6$
$\begin{array}{cc} & 3 & 7 & 5 & 0 \\ \times & & & & 6 \\ \hline 2 & 2 & 5 & 0 & 0 \\ \hline \end{array}$
Total distance covered in 6 days = $22500 \text{ m}$
We can convert this distance back to kilometers and meters.
$22500 \text{ m} = 22000 \text{ m} + 500 \text{ m}$
$= (22000 \div 1000) \text{ km} + 500 \text{ m}$
$= 22 \text{ km } 500 \text{ m}$
Therefore, the total distance covered by the student in six days is 22 km 500 m.
Question 12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Answer:
Given:
Total quantity of curd in the vessel = 4 litres 500 ml
Capacity of one glass = 25 ml
To Find:
The number of glasses that can be filled with the curd.
Solution:
First, we need to convert the total quantity of curd into a single unit, milliliters (ml), to match the capacity of the glasses.
We know the conversion factor: $1 \text{ litre} = 1000 \text{ ml}$.
Total quantity of curd = $4 \text{ litres } 500 \text{ ml}$
$= (4 \times 1000 \text{ ml}) + 500 \text{ ml}$
$= 4000 \text{ ml} + 500 \text{ ml}$
$= 4500 \text{ ml}$
The capacity of each glass is given as $25 \text{ ml}$.
To find the number of glasses that can be filled, we divide the total quantity of curd by the capacity of one glass.
Number of glasses = $\frac{\text{Total quantity of curd}}{\text{Capacity of one glass}}$
Number of glasses = $\frac{4500 \text{ ml}}{25 \text{ ml}}$
Number of glasses = $4500 \div 25$
We perform the long division:
$\begin{array}{r} 180\phantom{0} \\ 25{\overline{\smash{\big)}\,4500\phantom{)}}} \\ \underline{-~\phantom{(}25\phantom{..)}} \\ 200\phantom{)} \\ \underline{-~\phantom{(}200\phantom{(}} \\ 00\phantom{)} \end{array}$
The result of the division is 180.
Therefore, 180 glasses, each of 25 ml capacity, can be filled from the vessel.
Example 5 & 6 (Before Exercise 1.3)
Example 5: Estimate: 5,290 + 17,986.
Answer:
To Estimate:
$5,290 + 17,986$
Solution:
To estimate the sum, we round off the numbers to a suitable place value to simplify the calculation. Rounding to the nearest thousand seems appropriate here.
Rounding 5,290 to the nearest thousand:
The digit in the hundreds place is 2. Since $2 < 5$, we round down.
$5,290 \approx 5,000$
Rounding 17,986 to the nearest thousand:
The digit in the hundreds place is 9. Since $9 \ge 5$, we round up.
$17,986 \approx 18,000$
Now, we add the rounded numbers to get the estimate:
Estimated Sum = $5,000 + 18,000$
Estimated Sum = $23,000$
Therefore, the estimated sum of 5,290 and 17,986 is 23,000.
Example 6: Estimate: 5,673 – 436.
Answer:
To Estimate:
$5,673 - 436$
Solution:
To estimate the difference, we round off the numbers. Since the numbers have different place values (thousands and hundreds), a reasonable approach is to round both numbers to the nearest hundred.
Rounding 5,673 to the nearest hundred:
We look at the tens digit, which is 7. Since $7 \ge 5$, we round up the hundreds digit (6 becomes 7).
$5,673 \approx 5,700$
Rounding 436 to the nearest hundred:
We look at the tens digit, which is 3. Since $3 < 5$, we keep the hundreds digit (4) as it is.
$436 \approx 400$
Now, we subtract the rounded numbers to get the estimate:
Estimated Difference = $5,700 - 400$
$\begin{array}{cc} & 5 & 7 & 0 & 0 \\ - & & 4 & 0 & 0 \\ \hline & 5 & 3 & 0 & 0 \\ \hline \end{array}$
Estimated Difference = $5,300$
Therefore, the estimated difference between 5,673 and 436 is 5,300.
Exercise 1.3
Question 1. Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12,904 +2,888
(d) 28,292 – 21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Answer:
The general rule for estimation typically involves rounding off each number to its greatest (highest) place value.
(a) $730 + 998$
Rounding off 730 to its highest place value (hundreds):
The digit in the tens place is 3, which is less than 5. So, round down.
$730 \approx 700$
Rounding off 998 to its highest place value (hundreds):
The digit in the tens place is 9, which is greater than or equal to 5. So, round up.
$998 \approx 1000$
Estimated sum = $700 + 1000 = 1700$
(b) $796 – 314$
Rounding off 796 to its highest place value (hundreds):
The digit in the tens place is 9, which is greater than or equal to 5. So, round up.
$796 \approx 800$
Rounding off 314 to its highest place value (hundreds):
The digit in the tens place is 1, which is less than 5. So, round down.
$314 \approx 300$
Estimated difference = $800 - 300 = 500$
(c) $12,904 + 2,888$
Here, we round off both numbers to the nearest thousand for a reasonable estimate.
Rounding off 12,904 to the nearest thousand:
The digit in the hundreds place is 9, which is greater than or equal to 5. So, we round up.
$12,904 \approx 13,000$
Rounding off 2,888 to the nearest thousand:
The digit in the hundreds place is 8, which is greater than or equal to 5. So, we round up.
$2,888 \approx 3,000$
Estimated sum = $13,000 + 3,000 = 16,000$
(d) $28,292 – 21,496$
Here, we round off both numbers to the nearest thousand for a reasonable estimate.
Rounding off 28,292 to the nearest thousand:
The digit in the hundreds place is 2, which is less than 5. So, we round down.
$28,292 \approx 28,000$
Rounding off 21,496 to the nearest thousand:
The digit in the hundreds place is 4, which is less than 5. So, we round down.
$21,496 \approx 21,000$
Estimated difference = $28,000 - 21,000 = 7,000$
Ten More Examples:
1. Addition: $62 + 134$
$62 \approx 60$ (nearest ten)
$134 \approx 100$ (nearest hundred)
Estimate: $60 + 100 = 160$
2. Subtraction: $87 - 42$
$87 \approx 90$ (nearest ten)
$42 \approx 40$ (nearest ten)
Estimate: $90 - 40 = 50$
3. Addition: $588 + 310$
$588 \approx 600$ (nearest hundred)
$310 \approx 300$ (nearest hundred)
Estimate: $600 + 300 = 900$
4. Subtraction: $975 - 521$
$975 \approx 1000$ (nearest hundred)
$521 \approx 500$ (nearest hundred)
Estimate: $1000 - 500 = 500$
5. Addition: $4321 + 2890$
$4321 \approx 4000$ (nearest thousand)
$2890 \approx 3000$ (nearest thousand)
Estimate: $4000 + 3000 = 7000$
6. Subtraction: $8765 - 3456$
$8765 \approx 9000$ (nearest thousand)
$3456 \approx 3000$ (nearest thousand)
Estimate: $9000 - 3000 = 6000$
7. Addition: $392 + 2550$
$392 \approx 400$ (nearest hundred)
$2550 \approx 3000$ (nearest thousand)
Estimate: $400 + 3000 = 3400$
8. Subtraction: $7123 - 789$
$7123 \approx 7000$ (nearest thousand)
$789 \approx 800$ (nearest hundred)
Estimate: $7000 - 800 = 6200$
9. Addition: $14890 + 920$
$14890 \approx 10000$ (nearest ten thousand)
$920 \approx 900$ (nearest hundred)
Estimate: $10000 + 900 = 10900$
10. Subtraction: $42105 - 2876$
$42105 \approx 40000$ (nearest ten thousand)
$2876 \approx 3000$ (nearest thousand)
Estimate: $40000 - 3000 = 37000$
Question 2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :
(a) 439 + 334 + 4,317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365
Make four more such examples.
Answer:
(a) $439 + 334 + 4,317$
Rough Estimate (Rounding to nearest hundreds):
$439 \approx 400$ (tens digit is 3, round down)
$334 \approx 300$ (tens digit is 3, round down)
$4,317 \approx 4,300$ (tens digit is 1, round down)
Estimated sum = $400 + 300 + 4,300$
$\begin{array}{cc} & & 4 & 0 & 0 \\ & & 3 & 0 & 0 \\ + & 4 & 3 & 0 & 0 \\ \hline & 5 & 0 & 0 & 0 \\ \hline \end{array}$
Rough Estimate = 5,000
Closer Estimate (Rounding to nearest tens):
$439 \approx 440$ (ones digit is 9, round up)
$334 \approx 330$ (ones digit is 4, round down)
$4,317 \approx 4,320$ (ones digit is 7, round up)
Estimated sum = $440 + 330 + 4,320$
$\begin{array}{cc} & & 4 & 4 & 0 \\ & & 3 & 3 & 0 \\ + & 4 & 3 & 2 & 0 \\ \hline & 5 & 0 & 9 & 0 \\ \hline \end{array}$
Closer Estimate = 5,090
(b) $1,08,734 – 47,599$
Rough Estimate (Rounding to nearest hundreds):
$1,08,734 \approx 1,08,700$ (tens digit is 3, round down)
$47,599 \approx 47,600$ (tens digit is 9, round up)
Estimated difference = $1,08,700 - 47,600$
$\begin{array}{cc} & 1 & 0 & 8 & 7 & 0 & 0 \\ - & & 4 & 7 & 6 & 0 & 0 \\ \hline & & 6 & 1 & 1 & 0 & 0 \\ \hline \end{array}$
Rough Estimate = 61,100
Closer Estimate (Rounding to nearest tens):
$1,08,734 \approx 1,08,730$ (ones digit is 4, round down)
$47,599 \approx 47,600$ (ones digit is 9, round up)
Estimated difference = $1,08,730 - 47,600$
$\begin{array}{cc} & 1 & 0 & 8 & 7 & 3 & 0 \\ - & & 4 & 7 & 6 & 0 & 0 \\ \hline & & 6 & 1 & 1 & 3 & 0 \\ \hline \end{array}$
Closer Estimate = 61,130
(c) $8325 – 491$
Rough Estimate (Rounding to nearest hundreds):
$8325 \approx 8300$ (tens digit is 2, round down)
$491 \approx 500$ (tens digit is 9, round up)
Estimated difference = $8300 - 500$
$\begin{array}{cc} & 8 & 3 & 0 & 0 \\ - & & 5 & 0 & 0 \\ \hline & 7 & 8 & 0 & 0 \\ \hline \end{array}$
Rough Estimate = 7,800
Closer Estimate (Rounding to nearest tens):
$8325 \approx 8330$ (ones digit is 5, round up)
$491 \approx 490$ (ones digit is 1, round down)
Estimated difference = $8330 - 490$
$\begin{array}{cc} & 8 & 3 & 3 & 0 \\ - & & 4 & 9 & 0 \\ \hline & 7 & 8 & 4 & 0 \\ \hline \end{array}$
Closer Estimate = 7,840
(d) $4,89,348 – 48,365$
Rough Estimate (Rounding to nearest hundreds):
$4,89,348 \approx 4,89,300$ (tens digit is 4, round down)
$48,365 \approx 48,400$ (tens digit is 6, round up)
Estimated difference = $4,89,300 - 48,400$
$\begin{array}{cc} & 4 & 8 & 9 & 3 & 0 & 0 \\ - & & 4 & 8 & 4 & 0 & 0 \\ \hline & 4 & 4 & 0 & 9 & 0 & 0 \\ \hline \end{array}$
Rough Estimate = 4,40,900
Closer Estimate (Rounding to nearest tens):
$4,89,348 \approx 4,89,350$ (ones digit is 8, round up)
$48,365 \approx 48,370$ (ones digit is 5, round up)
Estimated difference = $4,89,350 - 48,370$
$\begin{array}{cc} & 4 & 8 & 9 & 3 & 5 & 0 \\ - & & 4 & 8 & 3 & 7 & 0 \\ \hline & 4 & 4 & 0 & 9 & 8 & 0 \\ \hline \end{array}$
Closer Estimate = 4,40,980
Four More Examples:
1. $582 + 211 + 6,789$
Rough Estimate (Nearest Hundreds): $600 + 200 + 6,800 = 7,600$
Closer Estimate (Nearest Tens): $580 + 210 + 6,790 = 7,580$
2. $9,876 – 2,345$
Rough Estimate (Nearest Hundreds): $9,900 - 2,300 = 7,600$
Closer Estimate (Nearest Tens): $9,880 - 2,350 = 7,530$
3. $15,405 + 672 + 98$
Rough Estimate (Nearest Hundreds): $15,400 + 700 + 100 = 16,200$
Closer Estimate (Nearest Tens): $15,410 + 670 + 100 = 16,180$
4. $2,05,123 - 88,945$
Rough Estimate (Nearest Hundreds): $2,05,100 - 88,900 = 1,16,200$
Closer Estimate (Nearest Tens): $2,05,120 - 88,950 = 1,16,170$
Question 3. Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples
Answer:
The general rule for estimating products involves rounding off each factor to a convenient place value that makes the calculation easier, such as its greatest place value or nearest hundred/thousand.
(a) $578 \times 161$
Rounding off both numbers to the nearest hundred.
Rounding off 578:
The tens digit is 7 ($\ge 5$), so we round up.
$578 \approx 600$
Rounding off 161:
The tens digit is 6 ($\ge 5$), so we round up.
$161 \approx 200$
Estimated product = $600 \times 200 = 1,20,000$
(b) $5281 \times 3491$
Here, we can get a reasonable estimate by rounding off 5281 to the nearest thousand and 3491 to the nearest hundred.
Rounding off 5281 to the nearest thousand:
The hundreds digit is 2 ($< 5$), so we round down.
$5281 \approx 5,000$
Rounding off 3491 to the nearest hundred:
The tens digit is 9 ($\ge 5$), so we round up.
$3491 \approx 3,500$
Estimated product = $5,000 \times 3,500 = 1,75,00,000$
(c) $1291 \times 592$
Rounding off both numbers to the nearest hundred gives a good estimate.
Rounding off 1291 to the nearest hundred:
The tens digit is 9 ($\ge 5$), so we round up.
$1291 \approx 1,300$
Rounding off 592 to the nearest hundred:
The tens digit is 9 ($\ge 5$), so we round up.
$592 \approx 600$
Estimated product = $1,300 \times 600 = 7,80,000$
(d) $9250 \times 29$
We round off 9250 to the nearest thousand and 29 to the nearest ten.
Rounding off 9250 to the nearest thousand:
The hundreds digit is 2 ($< 5$), so we round down. This gives 9000. However, for easier calculation, we can also round it up to 10,000.
$9250 \approx 10,000$
Rounding off 29 to the nearest ten:
The ones digit is 9 ($\ge 5$), so we round up.
$29 \approx 30$
Estimated product = $10,000 \times 30 = 3,00,000$
Four More Examples:
1. $81 \times 479$
$81 \approx 80$ (rounding to nearest ten)
$479 \approx 500$ (rounding to nearest hundred)
Estimated Product = $80 \times 500 = 40,000$
2. $63 \times 182$
$63 \approx 60$ (rounding to nearest ten)
$182 \approx 200$ (rounding to nearest hundred)
Estimated Product = $60 \times 200 = 12,000$
3. $3947 \times 56$
$3947 \approx 4000$ (rounding to nearest thousand)
$56 \approx 60$ (rounding to nearest ten)
Estimated Product = $4000 \times 60 = 2,40,000$
4. $2105 \times 888$
$2105 \approx 2000$ (rounding to nearest thousand)
$888 \approx 900$ (rounding to nearest hundred)
Estimated Product = $2000 \times 900 = 18,00,000$
Example 7
Example 7: Write in Roman Numerals
(a) 69
(b) 98
Answer:
(a) 69
We break down the number 69 into its components according to Roman numeral values:
$69 = 60 + 9$
We know that $60 = 50 + 10$, which is represented as L (50) followed by X (10), i.e., LX.
We know that $9 = 10 - 1$, which is represented as I (1) before X (10), i.e., IX.
Combining these parts:
$69 = 60 + 9 = \text{LX} + \text{IX} = \text{LXIX}$
So, 69 in Roman Numerals is LXIX.
(b) 98
We break down the number 98 into its components according to Roman numeral values:
$98 = 90 + 8$
We know that $90 = 100 - 10$, which is represented as X (10) before C (100), i.e., XC.
We know that $8 = 5 + 1 + 1 + 1$, which is represented as V (5) followed by I (1) three times, i.e., VIII.
Combining these parts:
$98 = 90 + 8 = \text{XC} + \text{VIII} = \text{XCVIII}$
So, 98 in Roman Numerals is XCVIII.
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