# Additional Questions - Chapter 1 Numbers Term 1 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024]

Question 1 .

How many thousands are there in 1 lakhs?
Solution:
$\frac{1,00,0000}{1000}=100$ Thousands

Question 2 .
The difference between successor and predecessor of any number is 2 . Is it true? Justify your answer.
Solution:
It is true that the difference between successor and predecessor of any number is 2 .
Because the difference between any number and its predecessor is 1 .
Also the difference between the number and its successor is 1 .
The total difference is 2 .

Question 3 .
The expanded form of the number $6,00,001$ is given as $6 \times 100000+1 \times 1$. Can you write like this
Comment.
Solution:
Yes. We can write the expansion of the number 600001 as $6 \times 100000+1 \times 1$.
Because $6 \times 100000+1 \times 1=600000+1=600001$

Question 4 .
Write the greatest five digit number using the digits $2,3,4,0$ and 7 .
Solution:
Greatest five digit number $=74320$

Question 5 .
Can you write the least five digit number using the digits $2,3,4,0$ and 7 as 02347 . Why? What will be the
correct number?
Solution:
No, we cannot write the least five digit number using the digits $2,3,4,0$ and 7 as 02347 . If it is 02347 ,

the left most zero has no value. It becomes 4 digit number 2347 .
The correct number will be 20347 .

Question 6 .
Write the relation between Largest two digit number and Smallest three digit number.
Solution:
Largest two digit number $+1=$ Smallest three digit number.
$99+1=100$

Question 7 .
Name the property being illustrated in each of the cases.
$1 .(30+20)+10=30+(20+10)$
$2.10 \times 35=(10 \times 30)+(10 \times 5)$
Solution:
$1 .$ Associativity
2 . Distribution of multiplication over addition.

Question 8 .
$10 \mathrm{crore}=$
Solution:
100 million

Question 9 .
The heights of five boys in class VI are $135,141,129,132,145$ (in centimetres) in height. Arrange their
heights as how they stand in the assembly?
Solution:
$129 \mathrm{~cm}<132$ cm $<135$ cm $<141$ cm $<145$ cm

Question 10 .
The number lock has the password number with 3 digits. The number is the least even number and less
Solution:
102

Question 11 .
Arrange in ascending order. $123456,123546,123623,123511$
Solution:
$123456<123511<123546<123623$ Question 12. Arrange in descending order. $8461,7535,2943,6214$
Solution: $8461>7535>6214>2943$

Question $13 .$
Find the numbers between 572634 and 562634 which is approximated to ten thousand place.
Solution:
$562634<570000<572634$

Question $14 .$
Evaluate the following:
(a) $44 \div 2+(7+80 \div 10)-14+23$
(b) $17 \times 6-4-2+20-(22+18)$
(c) $16 \times 144 \div 16 \div 9+16+15-20$
(d) $12 \times 36 \div 12 \div 3+5+6-2$
(e) $15-[17+30 \div 6-(6+6)+7]$
Solution:

(a) $44 \div 2+(7+80 \div 10)-14+23$ (Given)
$=44 \div 2+(7+8)-14+23$ (To complete the bracket $\div$ done first)
$=44 \div 2+15-14+23$ (Bracket completed second)
$=22+15-14+23(\div$ completed third $)$
$=37-37$ (+ completed fourth)
$=0(-$ completed last $)$
$\therefore 44 \div 2+(7+80 \div 10)-14+23=0 .$
(b) $17 \times 6-4-2+20-(22+18)$ (Given)
$=17 \times 6-4-2+20-40$ (Bracket completed first)
$=102-4-2+20-40$ ( $\times$ completed second)
$=102-4-22-40(+$ completed third)
$=98-22-40(\div$ completed one by one)
$=76-40$
$=36$
$\therefore 17 \times 6-4-2+20-(22+18)=36$

(c) $16 \times 144 \div 16 \div 9+16+15-20$ (Given)
$=16 \times 9 \div 9+16+15-20$ (\div completed first)
$=16 \times 1+16+15-20(\div$ completed second $)$
$=16+16+15-20$ ( $\times$ completed third)
$=32+15-20$ (+ completed fourth)
$=47-20$ (+ completed fifth)
$=27(-$ completed last $)$
$\therefore 16 \times 144 \div 16 \div 9+16+15-20=27$
(d) $12 \times 36 \div 12 \div 3+5+6-2$ (Given)
$=12 \times 3 \div 3+5+6-2$ (\div completed first)
$=12 \times 1+5+6-2$ (\div completed second)
$=12+5+6-2$ ( completed third )
$=17+6-2$ (+ completed forth)
$=23-2$ (+ completed fifth)

$\therefore 12 \times 36 \div 12 \div 3+5+6-2=21$
(e) $15-[17+30 \div 6-(6+6)+7]$ (Given)
$=15-[17+30 \div 6-12+7]$ (Inner bracket completed first)
$=15-[17+5-12+7](\div$ completed second)
$=15-[22-19](+$ completed third $)$
$=15-3$ (bracket completed forth)
$=12(-$ completed last $)$
$\therefore 15-[17+30 \div 6-(6+6)+7]=12$.

Question 15 .
An export company produced 235219 shirts, 158342 trousers and 11704 jackets in a year. What is the
total production of all the three items in that year?
Solution:
Number of shirts produced $=235219$
Number of trousers produced $=158342$
Number of jackets produced $=11704$
Total production of all items $=405265$
Total production of all items in that year $=4,05,265$

Question 16 .
India's population has been steadily increasing from 439 million in 1961 to 1028 million in 2001 . Find
the total increase in population from 1961 to 2001 . Write the increase in population in the Indian system
of Numeration using commas suitably.
Solution:
Population of India in $1961=439$ millions $=439,000,000$
Population of India in $2001=1028$ millions $=1,028,000,000$
Increase in population from 1961 to $2001=$ Population in $2001-$ Population in 1961
$=1028000000-439000000$
$=589000000$
$=589$ million.
Increase in population in Indian System $=58,90,00,000$

Question 17 .
A person had ₹ $10,00,000$ with him. He purchased a flat for ₹ $8,70,000$. With the remaining money, he
has to buy a T.V. for 1 lakh. How much money was left with him to buy a T. V?
Solution:
Total money the person had = ₹ $10,00,000$
Cost of flat = ₹ $8,70,000$
Remaining money = ₹ $1,30,000$
Now he has ₹ $1,30,000$. So it is enough to buy a TV for ₹ $1,00,000$.

Question 18 .
A box contains 50 packets of biscuits, each weighing $120 \mathrm{~g}$. How many such boxes can be loaded in a
van, which cannot carry more than $900 \mathrm{~kg}$ ?
Solution:
Given: Total number of packets $=50$.
Weight of each packet $=120 \mathrm{~g}$
Weight of a box $=50 \times 120 \mathrm{~g}=6000 \mathrm{~g}=6 \mathrm{~kg}[\because 1000 \mathrm{~g}=1 \mathrm{~kg}]$
Required number of boxes $=\frac{900}{6}=150$.
150 boxes are required.

Question 19 .
How much money was collected from 5342 students for a charity show, if each student contributed ₹
670 ?
Solution:
Total number of students $=5342$
Contribution of each student $=₹ 670$
Total money collected $=5342 \times 670=₹ 35,79,140$
Total money collected $=₹ 35,79,140$

Question 20 .
Estimate the following to the nearest hundreds
(a) $439+334+4317$
(b) $1,08,734-47,599$
(c) $8325-491$
(d) $4,89,348-48,365$

Solution:
(a) $439+334+4317$
$439 \Rightarrow 400$
$334 \Rightarrow 300$
$4317 \Rightarrow 4300$
Sum $=5,000$
(b) $1,08,734-47,599$
$1,08,734 \Rightarrow 1,08,700$
$47,599 \Rightarrow 47,600$
Difference $=61,100$
(c) $8325-491$
$8325 \Rightarrow 8300$
$491 \Rightarrow 500$
Differences $=7,800$
(d) $4,89,348-48,365$
$4,89,348 \Rightarrow 4,89,300$

Question $21 .$
Estimate the following products:
(a) $578 \times 161$
(b) $5281 \times 3491$
(c) $1291 \times 592$
(d) $9250 \times 29$
Solution:
(a) $578 \times 161$
$578 \Rightarrow 600$
$161 \Rightarrow 200$
Estimated product is $600 \times 200=1,20,000$
(b) $5281 \times 3491$
$5281 \Rightarrow 5000$
$3491 \Rightarrow 3500$
Estimated Product $=5000 \times 3500=1,75,00,000$
(c) $1291 \times 592$
$1291 \Rightarrow 1300$
$592 \Rightarrow 600$
Estimated Product is $=1300 \times 600=7,80,000$
(d) $9250 \times 29$
$9250 \Rightarrow 9000$
$29 \Rightarrow 30$
Estimated Product is $9000 \times 30=2,70,000$

Question 22 .
Solution:
No, all whole numbers are not natural numbers.
Because ' 0 ' belongs to the whole number system. But it is not in a natural number system.
All whole numbers except ' 0 ' are natural numbers.

Question $23 .$
Use associative property of addition to add $847+306+453$
Solution:
\begin{aligned} &847+306+453 \\ &=(847+453)+306 \\ &=1300+306 \end{aligned}

\begin{aligned} &=1606 \\ &\therefore 847+306+453=1606 \end{aligned}

Question 24 .
Find the value of $(1063 \times 127)-(1063 \times 27)$
Solution:
$(1063 \times 127)-(1063 \times 27)$
$=1063(127-27)[$ Taking 1063 as common $]$
$=1063 \times 100$
$=106300$.
i.e $(1063 \times 127)-(1063 \times 27)=106300$

Question 25 .
Find the product using suitable properties
(a) $738 \times 103$
(b) $1005 \times 168$
Solution:
(a) We have $738 \times 103$
$=738 \times(100+3)$
$=738 \times 100+738 \times 3[$ By distributive property of multiplication over addition $]$
$=73800+2214$
$=76014$
(b) $1005 \times 168$
$=(1000+5) \times 168$
$=(168 \times(1000+5)($ By commutative property $)$
$=(168 \times 1000)+(168 \times 5)$
$=1,68,000+840$
$=1,68,840$

Question 26.

Write the largest six-digit number and write the number names in words using the Indian and International system.

Solution:

The largest six-digit number is 999999

Question $27 .$
In a mobile store, the number of mobiles sold during a month is 1250 , Assuming that the same number of mobiles are sold every month, find the number of mobiles sold in 2 years.
Solution:
Number of mobiles sold in 1 month $=1250$
1 year $=12$ months
2 years $=2 \times 12=24$ months
Number of mobiles sold in 24 months $=1250 \times 24=30,000$
Number of mobiles sold in 2 years $=30,000$

Question 28 .
Simplify $24+2 \times 8 \div 2 \quad 1$
Solution:
$24+2 \times 8 \div 2-1$
$=24+2 \times 4-1$
$=24+8-1$
$=32-1$
$=31$