Math, asked by potterhead32, 2 months ago

Can anyone tell me the answer to these questions in the pic with step-by-step explanation.

Don't post irrelevant answers on this question plz.

Attachments:

Answers

Answered by abhi569

(i) 17.56 bar can be expressed as

(a) 1756/99 (b) 12/99 (c) 1739/99 (d) 173/99

Let 17.56 = x ...(1)

As there are two digit under bar, multiply both sides by 10² or 100:

=> 100 × 17.56 = 100 × x

=> 1756.56 = 100x ...(2)

Subtract (1) from (2) :

1756.56 = 100x

- 17.56 = - x

1739 = 99x [bar cancelled]

=> 1739 = 99x

=> 1739/99 = x = 17.56

(ii): 16/27 is equivalent to

(a) 59.2 (b) 0.592 (c) 0.592 (d) 0.595

on dividing,

16/27 = 0.592592592592....

Notice that the repetitive term is '592'

Thus, 16/27 = 0.592

(iii): 0.163 bar can be expressed as

(a) 49/300 (b) 16/18 (c) 16/300 (d) 16/90

Let 0.163 = x

Multiply both sides by 100:

=> 16.3 = 100x ...(1)

As there is only one digit under bar, multiply both sides by 10¹ or 10:

=> 10 × 16.3 = 10 × 100x

=> 163.3= 1000x ...(2)

Subtract (1) from (2) :

163.3 = 1000x

- 16.3 = - 100x

147 = 900x [bar cancelled]

=> 147 = 900x

=> 147/900 = x

=> 49/300 = x = 0.163

(iv) 0.324 bar can be expressed as

(a)107/99 (b)107/330 (c)306/107(d)106/324

Let 0.324= x

=> 3.24 = 10x

As there are two digit under bar, multiply both sides by 10² or 100:

=> 100 × 3.24 = 100 × 10x

=> 324.24 = 1000x ...(2)

Subtract (1) from (2) :

324.24 = 1000x

- 3.24 = - 10x

321 = 990x [bar cancelled]

=> 321 = 990x

=> 321/990 = x

=> 107/330 = x = 0.324

(v): 3/18 is equivalent to

(a) 0.16 (b) 0.16 (c) 1.16 (d) 16.6

On solving 3/18 comes out to be 0.16666...

Notice that the repetitive term is '6'

Thus, 3/18 = 0.16

Answered by BrainlyRish

$\underline {\underline {\bf (1) }}\\$

⠀⠀⠀⠀ $\bf{17.\overline {56} }$ can be expressed as :

$\sf \dfrac{1756}{99}\\$
$\sf \dfrac{12}{99}\\$
$\bf \dfrac{1739}{99}\\$ ✔
$\sf \dfrac{173}{99}\\$

⠀⠀❍ Let's Assume $\sf{17.\overline {56} }$ be a .

$\qquad :\implies \:\: \bf a = \:17.\overline {56} \qquad\longrightarrow \:\:Eq.1\\$

⠀⠀⠀⠀Therefore,

$\qquad:\implies \sf 100a = 100 \times \sf{17.\overline {56} } \\\qquad :\implies \:\: \bf 100a = \:1756.\overline {56} \qquad\longrightarrow \:\:Eq.2\\$

⠀⠀⠀⠀Now Subtract Eq.1 from Eq.2 :

$\qquad :\implies \:\: \sf a = \:17.\overline {56} \qquad\longrightarrow \:\:Eq.1\\\qquad :\implies \:\: \sf 100a = \:1756.\overline {56} \qquad\longrightarrow \:\:Eq.2\\ \qquad :\implies \:\: \sf 100a - a = \:1756.\overline {56} - 17.\overline {56} \\ \qquad :\implies \:\: \sf 99a = \:1756. \cancel { \overline {56}} - 17.\cancel {\overline {56}} \\ \qquad :\implies \:\: \sf 99a = \:1756 - 17 \\ \qquad :\implies \:\: \sf 99a = \:1739 \\\qquad :\implies \:\: \bf a = \:\dfrac{1739}{99} \\$

Therefore,

$\qquad :\implies \:\: \bf 17.\overline {56} \:= a = \:\dfrac{1739}{99} \\$

Therefore,

⠀⠀⠀⠀⠀ $\therefore {\underline{ \sf \:The\:Correct \:AnswEr \:is \:\bf c \:\sf\: or\: \bf \dfrac{1739}{99} }}\\$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\underline {\underline {\bf (2) }}\\$

⠀⠀⠀⠀ $\sf{\dfrac {16}{27} }$ is equivalent to :

$\sf 59.\overline{2}$
$\sf 0.5\overline{92}$
$\bf 0.\overline{592}$ ✔
$\sf 0.595$

⠀⠀⠀⠀

⠀⠀⠀⠀For Explanation refer Attachment 1 .

Therefore,

⠀⠀⠀⠀⠀ $\therefore {\underline{ \sf \:The\:Correct \:AnswEr \:is \:\bf c\: \sf or\: \bf 0.\overline{592} }}\\\\\\$

$\underline {\underline {\bf (3) }}\\$

⠀⠀⠀⠀ $\bf{0.16\overline {3} }$ can be expressed as :

$\bf \dfrac{49}{300}\\$ ✔
$\sf \dfrac{16}{18}\\$
$\sf \dfrac{16}{300}\\$
$\sf \dfrac{16}{90}\\$

⠀⠀❍ Let's Consider $\sf{0.16\overline {3} }$ be a .

⠀⠀⠀⠀Therefore,

$\qquad:\implies \sf 1000a = 1000 \times \sf{0.16\overline {3} } \\$

$\qquad :\implies \:\: \bf 1000a = \:163.\overline {3} \qquad\longrightarrow \:\:Eq.1\\$

$\qquad:\implies \sf 100a = 100 \times \sf{0.16\overline {3} } \\$

$\qquad :\implies \:\: \bf 100a = \:16.\overline {3} \qquad\longrightarrow \:\:Eq.2\\$

⠀⠀⠀⠀Now Subtract Eq.1 from Eq.2 :

$\qquad :\implies \:\: \bf 100a = \:163.\overline {3} \qquad\longrightarrow \:\:Eq.1\\$

$\qquad :\implies \:\: \bf 100a = \:16.\overline {3} \qquad\longrightarrow \:\:Eq.2\\$

$\qquad :\implies \:\: \sf 1000a - 100 a = \:163.\overline {3} - 16.\overline {3} \\\qquad :\implies \:\: \sf 900a = \:163. \cancel { \overline {3}} - 16.\cancel {\overline {3}} \\\qquad :\implies \:\: \sf 900a = \:163 - 16 \\\qquad :\implies \:\: \sf 900a = \:147 \\\qquad :\implies \:\: \sf a = \cancel {\:\dfrac{147}{900}} \\\qquad :\implies \:\: \bf a = \:\dfrac{49}{300} \\$

Therefore,

$\qquad :\implies \:\: \bf 0.16\overline {3} \:= a = \:\dfrac{49}{300} \\$

Therefore,

⠀⠀⠀⠀⠀ $\therefore {\underline{ \sf \:The\:Correct \:AnswEr \:is \:\bf a\:\: \sf or\: \bf \dfrac{49}{300} }}\\\\\\$

$\underline {\underline {\bf (4) }}\\$

⠀⠀⠀⠀ $\bf{0.3\overline {24} }$ can be expressed as :

$\sf \dfrac{107}{99}\\$
$\bf \dfrac{107}{330}\\$ ✔
$\sf \dfrac{306}{107}\\$
$\sf \dfrac{106}{324}\\$

⠀⠀❍ Let's Consider $\sf{0.3\overline {24} }$ be a .

⠀⠀⠀⠀Therefore,

$\qquad:\implies \sf 10a = 10 \times \sf{0.3\overline {24} } \\$

$\qquad :\implies \:\: \bf 10a = \:3.\overline {24} \qquad\longrightarrow \:\:Eq.1\\$

$\qquad:\implies \sf 1000a = 1000 \times \sf{0.3\overline {24} } \\$

$\qquad :\implies \:\: \bf 1000a = \:324.\overline {24} \qquad\longrightarrow \:\:Eq.2\\$

⠀⠀⠀⠀Now Subtract Eq.1 from Eq.2 :

$\qquad :\implies \:\: \sf 1000a - 10 a = \:324.\overline {24} - 3.\overline {24} \\\qquad :\implies \:\: \sf 990a = \:324. \cancel { \overline {24}} - 3.\cancel {\overline {24}} \\\qquad :\implies \:\: \sf 990a = \:324 - 3 \\\qquad :\implies \:\: \sf 990a = \:147 \\\qquad :\implies \:\: \sf a = \cancel {\:\dfrac{321}{990}} \\\qquad :\implies \:\: \bf a = \:\dfrac{107}{330} \\$