Exercise 1.7 - Chapter 1 - Number Systems - 8th Maths Guide Samacheer Kalvi Solutions

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Miscellaneous Practice Problems
Question 1.
If $\frac{3}{4}$ of a box of apples weighs $3 \mathrm{~kg}$ and $225 \mathrm{gm}$, how much does a full box of apples weigh?
Answer:
Let the total weight of a box of apple $=x \mathrm{~kg}$.
Weight of $\frac{3}{4}$ of a box apples $=3 \mathrm{~kg} 225 \mathrm{gm}$.
$=3.225 \mathrm{~kg}$
$\frac{3}{4} \times \mathrm{x}=3225$
$\mathrm{x}=\frac{3.225 \times 4}{3} \mathrm{~kg}$
$=1.075 \times 4 \mathrm{~kg}=4.3 \mathrm{~kg}$
$=4 \mathrm{~kg} 300 \mathrm{gm}$
Weight of the box of apples $=4 \mathrm{~kg} 300 \mathrm{gm}$.

Question $2 .$
Mangalam buys a water jug of capacity $3 \frac{4}{5}$ litre. If she buys another jug which is $2 \frac{2}{3}$ times as large as the smaller jug, how many litre can the larger one hold?
Answer:

Capacity of the small waterug $=3 \frac{4}{5}$ litres.
Capacity of the big jug $=2 \frac{2}{3}$ times the small one.
$=2 \frac{2}{3} \times 3 \frac{4}{5}=\frac{8}{3} \times \frac{19}{5}=\frac{152}{15}$
$=\frac{2}{15}$ litres
Capacity of the large jug $=\frac{2}{15}$ litres.

Question $3 .$
Ravi multiplied $\frac{25}{8}$ and $\frac{16}{5}$ to obtain $\frac{400}{120}$. He says that the simplest form of this product is $\frac{10}{3}$ and Chandru says the answer in the simplest form is $3 \frac{1}{3}$. Who is correct? (or) Are they both correct?
Explain.
Answer:
$\begin{aligned}
\text { Product of } \frac{25}{8} \text { and } \frac{16}{15} &=\frac{25}{8} \times \frac{16}{15} \\
&=\frac{10}{3}=3 \frac{1}{3} \\
\text { Answer obtained } &=\frac{400}{120}
\end{aligned}$

$=\frac{400 \div 40}{120 \div 40}=\frac{10}{3}=3 \frac{1}{3}$
$\therefore$ The product is $\frac{400}{120}$ and its simplest form improper fraction is $\frac{10}{3}$
And mixed fraction is $3 \frac{1}{3}$
$\therefore$ Both are correct
 

Question $4 .$
Find the length of a room whose area is $\frac{153}{10}$ sq.m and whose breadth is $2 \frac{11}{20} \mathrm{~m}$.
Answer:
Length of the room $\times$ Breadth $=$ Area of the room
Breadth of the room $=2 \frac{11}{20} \mathrm{~m}$
Area of the room $=\frac{153}{10} \mathrm{sq} . \mathrm{m}$
Length $x 2 \frac{11}{20}=\frac{153}{10}$
Length $=\frac{153}{10} \div 2 \frac{11}{20}=\frac{153}{10} \div \frac{51}{20}=\frac{153^{3}}{10} \times \frac{20^{2}}{51_{1}}=6 \mathrm{~m}$
Length of the room $=6 \mathrm{~m}$

Question $5 .$
There is a large square portrait of a leader that covers an area of $4489 \mathrm{~cm}^{2}$. If each side has a 2 $\mathrm{cm}$ liner, what would be its area?
Answer:

Area of the square $=4489 \mathrm{~cm}^{2}$
(side) $2=4489 \mathrm{~cm}^{2}$
(side) $2=67 \times 67$
side $=67^{2}$
Length of a side $=67$
Length of a side with liner $=67+2+2 \mathrm{~cm}$ $=71 \mathrm{~cm}$

Area of the larger square $=71 \times 71 \mathrm{~cm}^{2}$
$=5041 \mathrm{~cm}^{2}$
Area of the liner $=$ Area of big square $-$ Area of small square
$\begin{aligned}
&=(5041-4489) \mathrm{cm}^{2} \\
&=552 \mathrm{~cm}^{2}
\end{aligned}$

Question $6 .$
A greeting card has an area $90 \mathrm{~cm}^{2}$. Between what two whole numbers is the length of its side?
Answer:

$\begin{aligned}
&\text { Area of the greeting card }=90 \mathrm{~cm}^{2} \\
&\text { (side) }{ }^{2}=90 \mathrm{~cm}^{2} \\
&(\text { side })^{2}=2 \times 5 \times 3 \times 3=2 \times 5 \times 3^{2} \\
&\sqrt{(\text { side })^{2}}=\sqrt{2 \times 5 \times 3^{2}} \\
&\text { Side }=3 \sqrt{2 \times 5} \\
&\text { side }=3 \sqrt{10} \mathrm{~cm} \\
&\text { side }=3 \times 3.2 \mathrm{~cm} \\
&\text { side }=9.6 \mathrm{~cm}
\end{aligned}$
$\therefore$ Side lies between the whole numbers 9 and 10 .

Question $7 .$
225 square shaped mosaic tiles, each of area 1 square decimetre exactly cover a square shaped verandah. How long is each side of the square shaped verandah?
Answer:
Area of one tile $=1$ sq.decimeter
Area of 225 tiles $=225$ sq.decimeter
225 square tiles exactly covers the square shaped verandah.
$\therefore$ Area of 225 tiles = Area of the verandah
Area of the verandah $=225$ sq.decimeter
side $\times$ side $=15 \times 15$ sq.decimeter
side $=15$ decimeters
Length of each side of verandah $=15$ decimeters.

Question $8 .$
If $\sqrt[3]{1906624} \times \sqrt{x}=3100$, find $x$.
Answer:

Answer:
$\begin{aligned} \sqrt[3]{1906624} \times \sqrt{x} &=3100 \\ \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 31 \times 31 \times 31} \times \sqrt{x}=3100 \\ \sqrt[3]{2^{3} \times 2^{3} \times 31^{3}} \times \sqrt{x} &=3100 \\ 2 \times 2 \times 31 \times \sqrt{x} &=3100 \\ \sqrt{x} &=\frac{3100}{2 \times 2 \times 31} \\ \sqrt{x} &=25 \\ \text { Squaring on both sides } \quad(\sqrt{x})^{2} &=25^{2} \\ x &=625 \end{aligned}$

Question $9 .$
If $2^{m-1}+2^{m+1}=640$, then find ' $m$ '.
Answer:
Given $2^{\mathrm{m}-1}+2^{\mathrm{m}+1}=640$
$\begin{aligned}
&\mathrm{m}-1=7 \\
&\mathrm{~m}=7+1 \\
&\mathrm{~m}=8
\end{aligned}$
$\begin{aligned}
&2^{\mathrm{m}-1}+2^{\mathrm{m}+1}=128+512 \\
&2^{\mathrm{m}-1}+2^{\mathrm{m}+1}-2^{7}+2^{9}
\end{aligned}$
[consecutive powers of 2]
Powers of 2:
$2,4,8,16,32,64,128,256,512, \ldots$
 

Question $10 .$
Give the answer in scientific notation:
A human heart beats at an average of 80 beats per minute. How many times does it beat in
i) an hour?
ii) a day?
iii) a year?
iv) 100 years?
Answer:
Heart beat per minute $=80$ beats
(i) an hour
One hour $=60$ minutes
Heart beat in an hour $=60 \times 80$
$=4800$
$=4.8 \times 10^{3}$
(ii) In a day
One day $=24$ hours $=24 \times 60$ minutes
$\therefore$ Heart beat in one day $=24 \times 60 \times 80=24 \times 4800=115200$
$=1.152 \times 10^{5}$

(iii) a year
One year $=365$ days $=365 \times 24$ hours $=365 \times 24 \times 60$ minutes
$\therefore$ Heart beats in a year $=365 \times 24 \times 60 \times 80$
$=42048000$
$=4.2048 \times 10^{7}$
(iv) 100 years
Heart beats in one year $=4.2048 \times 10^{7}$
heart beats in 100 years $=4.2048 \times 10^{7} \times 100=4.2048 \times 10^{7} \times 10^{2}$ $=4.2048 \times 10^{9}$

Challenging Problems:
Question $11 .$
In a map, if 1 inch refers to $120 \mathrm{~km}$, then find the distance between two cities $\mathrm{B}$ and $\mathrm{C}$ which are $4 \frac{1}{6}$ inches and $3 \frac{1}{3}$ inches from the city A which lies between the cities $B$ and $C$.
Answer:

1 inch $=120 \mathrm{~km}$
Distance between $\mathrm{A}$ and $\mathrm{B}=4 \frac{1}{6}$
Distance between $\mathrm{A}$ and $\mathrm{C}=3 \frac{1}{3}$
$\therefore$ Distance between $\mathrm{B}$ and $\mathrm{C}=4 \frac{1}{6}+3 \frac{1}{3}$ inches $=\frac{25}{6}+\frac{10}{3}=\frac{25}{6}+\frac{20}{6}=\frac{25+20}{6}=\frac{45}{6}$ inches
1 inch $=120 \mathrm{~km}$
$\therefore \frac{45}{6}$ inches $=\frac{45}{6} \times 120 \mathrm{~km}=900 \mathrm{~km}$
Distance between $\mathrm{B}$ and $\mathrm{C}=900 \mathrm{~km}$
 

Question 12 .
Give an example and verify each of the following statements.
(i) The collection of all non-zero rational numbers is closed under division.
Answer:
let $\mathrm{a}=\frac{5}{6}$ and $\mathrm{b}=\frac{-4}{3}$ be two non zero rational numbers. $a \div b=\frac{5}{6} \div \frac{-4}{3}=\frac{5}{6} \times \frac{3}{-4}=\frac{5}{-8}$ is in $\mathrm{Q}$
$\therefore$ Collection of non-zero rational numbers are closed under division.
(ii) Subtraction is not commutative for rational numbers.
Answer:
let $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=-\frac{5}{6}$ be two rational numbers.

$\begin{aligned} a-b &=\frac{1}{2}-\left(-\frac{5}{6}\right)=\frac{1}{2}+\left(+\frac{5}{6}\right)=\frac{3}{6}+\frac{5}{6}=\frac{3+5}{6}=\frac{8}{6} \\ &=1 \frac{2}{6}=1 \frac{1}{3} \end{aligned}$ $b-a=-\frac{5}{6}-\frac{1}{2}=-\frac{5}{6}-\frac{3}{6}=\frac{-5-3}{6}=\frac{-8}{6}=-1 \frac{1}{3}$ $a-b \neq b-a$
(iii) Division is not associative for rational numbers.
Answer:
Let $\mathrm{a}=\frac{2}{5}, \mathrm{~b}=\frac{6}{5}, \mathrm{c}=\frac{3}{5}$ be three rational numbers.
From (1) and $(2) \quad \frac{1}{5} \neq \frac{5}{9}$ $a \div(b \div c) \neq(a \div b) \div c$ $\therefore$ Division is not associative for rational numbers.

(iv) Distributive property of multiplication over subtraction is true for rational numbers. That is, a $(b-c)=a b-a c$.
Answer:
Let $\mathrm{a}=\frac{2}{9}, \mathrm{~b}=\frac{3}{6}, \mathrm{c}=\frac{1}{3}$ be three rational numbers.
To prove $a \times(b-c)=a b-b c$
$\begin{aligned}
a \times(b-c) &=\frac{2}{9} \times\left(\frac{3}{6}-\frac{1}{3}\right) \\
&=\frac{2}{9} \times\left(\frac{3-(1 \times 2)}{6}\right)=\frac{2}{9} \times \frac{(3-2)}{6} \\
&=\frac{2}{9} \times \frac{1}{6}=\frac{1}{27}
\end{aligned}$
$\begin{aligned}
a b-a c &=\left(\frac{2}{9} \times \frac{3}{6}\right)-\left(\frac{2}{9} \times \frac{1}{3}\right)=\frac{1}{9}-\frac{2}{27} \\
&=\frac{(1 \times 3)-2}{27}=\frac{3-2}{27}=\frac{1}{27}
\end{aligned}$

$\therefore$ From (1) and (2)
$a \times(b-c)=a b-b c$
$\therefore$ Distributivity of multiplication over subtraction is true for rational numbers.
(v) The mean of two rational numbers is rational and lies between them.
Answer:
Let $\mathrm{a}=\frac{2}{11}$ and $\mathrm{b}=\frac{5}{6}$ be two rational numbers
Mean of $a$ and $b$ is $c=\frac{1}{2}(a+b)=\frac{1}{2}\left(\frac{2}{11}+\frac{5}{6}\right)=\frac{1}{2}\left(\frac{(2 \times 6)+(5 \times 11)}{66}\right)$ $=\frac{1}{2} \times\left(\frac{12+55}{66}\right)=\frac{1}{2} \times \frac{67}{66}=\frac{67}{132}$ is in $\mathrm{Q}$.
$\begin{aligned}
\text { Also } \frac{2}{11}=& \frac{2 \times 12}{11 \times 12}=\frac{24}{132} \\
\frac{5}{6}=\frac{5 \times 22}{6 \times 22}=\frac{110}{132} \\
\therefore \frac{24}{132}<\frac{67}{132}<\frac{110}{132}
\end{aligned}$
$\therefore$ The mean lies between the given rational numbers $\frac{2}{11}$ and $\frac{5}{6}$
 

Question $13 .$
If $\frac{1}{4}$ of a ragi adai weighs 120 grams, what will be the weight of $\frac{2}{3}$ of the same ragi adai?
Answer:
Let the weight of 1 ragi adai $=x$ grams

given $\frac{1}{4}$ of $x=120 \mathrm{gm}$
$\begin{aligned}
&\frac{1}{4} \times x=120 \\
&x=120 \times 4 \\
&x=480 \mathrm{gm} \\
&\therefore \frac{2}{3} \text { of the adai }=\frac{2}{3} \times 480 \mathrm{gm}=2 \times 160 \mathrm{gm}=320 \mathrm{gm} \\
&\frac{2}{3} \text { of the weight of adai }=320 \mathrm{gm}
\end{aligned}$

Question $14 .$
If $\mathrm{p}+2 \mathrm{q}=18$ and $\mathrm{pq}=40$, find $\frac{2}{p}+\frac{1}{q}$
Answer:
Given $p+2 q=18$.
$\begin{aligned}
&\frac{2}{p}=40 \ldots \cdots \cdots(2) \\
&\frac{2}{q}=\frac{(2 \times q)+(1 \times p)}{p q}=\frac{2 q+p}{p q}=\frac{18}{40}[\because \text { from (1) and (2)] } \\
&\frac{2}{p}+\frac{1}{q}=\frac{9}{20}
\end{aligned}$

Question $15 .$
Find $x$ if $5 \frac{x}{5} \times 3 \frac{3}{4}=21$.
Answer:
$\begin{aligned}
5 \frac{x}{5} \times 3 \frac{3}{4} &=21 \\
\frac{25+x}{5} \times \frac{15}{4} &=21 \\
\frac{25+x}{5} &=21+\frac{15}{4} \\
\frac{25+x}{5} &=21 \times \frac{4}{15}
\end{aligned}$

$\begin{aligned}
\frac{25+x}{5} &=\frac{28}{5} \\
25+x &=\frac{28 \times 5}{5} \\
25+x=28 \\
x=28-25 \\
x=3
\end{aligned}$

Question $16 .$
By how much does $\frac{1}{\frac{10}{11}}$ exceed $\frac{\frac{1}{10}}{11}$ ?
Answer:
The difference $=\frac{1}{\frac{10}{10}}-\frac{\frac{1}{10}}{11}=\left(\frac{1}{1} \times \frac{11}{10}\right)-\left(\frac{1}{10} \times \frac{1}{11}\right)=\frac{11}{10}-\frac{1}{110}=\frac{121-1}{110}=\frac{120}{110}=\frac{12}{11}$ $\frac{1}{\frac{10}{11}}$ exceed $\frac{\frac{1}{10}}{11}$ by $\frac{12}{11}$
 

Question $17 .$
A group of 1536 cadets wanted to have a parade forming a square design. Is it possible? If it is not possible, how many more cadets would be required?
Answer:
Number of cadets to form square design
$1536=\overline{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3}$

The numbers 2 and 3 are unpaired
$\therefore$ It is impossible to have the parade forming square design with 1536 cadets.

$39 \times 39=1521$
Also $40 \times 40=1600$
$\therefore$ We have to add $(1600-1536)=64$ to make 1536 a perfect square.
$\therefore 64$ more cadets would be required to form the square design.

Question $18 .$
Evaluate: $\sqrt{286225}$ and use it to compute $\sqrt{2862.25}+\sqrt{28.6225}$
Answer:

$\therefore \sqrt{286225}=535$ $$ \sqrt{2862.25}=\sqrt{\frac{286225}{100}}=\frac{\sqrt{286225}}{\frac{100}{12865}}=\frac{535}{10}=53.5 $$ $$ \sqrt{28.6225}=\sqrt{\frac{286225}{1000}}=\frac{\sqrt{286225}}{10000}=\frac{535}{100}=5.35 $$ $\therefore \sqrt{2862.25}+\sqrt{28.6225}=53.5+5.35=58.85$
Simplify: $\left(3.769 \times 10^{5}\right)+\left(4.21 \times 10^{5}\right)$
Answer:
$\left(3.769 \times 10^{5}\right)+\left(4.21 \times 10^{5}\right)=3,76,900+4,21,000$
$=7,97,000$
$=7.979 \times 10^{5}$

Question $20 .$
Order the following from the least to the greatest: $16^{25}, 8^{100}, 3^{500}, 4^{400}, 2^{600}$
Answer:
$\begin{aligned}
&16^{25}=\left(2^{4}\right)^{25}=2^{100} \\
&8^{100}=\left(2^{3}\right)^{100}=2^{300} \\
&4^{400}=\left(2^{2}\right)^{400}=2^{800} \\
&2^{600}=2^{600}
\end{aligned}$
Comparing the powers we have.
$2^{100}<2^{300}<2^{600}<2^{800}$
$\therefore$ The required order: $16^{25}, 8^{100}, 3^{500}, 4^{400}, 2^{600}$