**Additional Questions
Answer the following questions.
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 .**

Are all whole numbers are natural numbers? Justify your answer?

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:

$24+2 \times 8 \div 2-1$

$=24+2 \times 4-1$

$=24+8-1$

$=32-1$

$=31$