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Sia ? 6 years, 6 months ago
Since, the difference between the savings of two consecutive months is Rs. 20, therefore the series is an A.P.
Here, the savings of the first month is Rs. 50
First term, a = 50, Common difference, d = 20
No. of terms = no. of months
No. of terms, n = 12
{tex}S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}
{tex}= \frac { 12 } { 2 } [ 2 \times 50 + ( 12 - 1 ) 20 ]{/tex}
= 6[100 + 220]
= 6(320)
= 1920
After a year, Ramakali will save Rs. 1920.
Yes, Ramakali will be able to fulfill her dream of sending her daughter to school.
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Beauty Queen? Miss Sweetu? 7 years, 5 months ago
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Sia ? 6 years, 6 months ago
Let height of each candle = {tex}x\ unit.{/tex}
First candle burns off in 6 hours.
Second candle burns off in 8 hours.
Height of 1st candle after burning for 1 hr = {tex}\frac{x}{6}{/tex}{tex}unit{/tex}
and height of 2nd candle after burning for 1 hr = {tex}\frac{x}{8}{/tex}{tex}unit{/tex}
Let the required time {tex}= y\ hrs{/tex}.
Length of 1st candle burnt after y hrs = {tex}\frac{y \times x}{6}{/tex}{tex}unit{/tex}
Height of 1st candle left = {tex}\left(x-\frac{x y}{6}\right){/tex}
Length of 2nd candle burnt after y hrs = {tex}\left(\frac{y \times x}{8}\right){/tex}{tex}unit{/tex}
Height of 2nd candle left = {tex}\left(x-\frac{x y}{8}\right){/tex}
According to the question,
Height of 1st candle = {tex}\frac{1}{2} \times{/tex}Height of 2nd candle
{tex}\Rightarrow \quad x-\frac{x y}{6}=\frac{1}{2}\left(x-\frac{x y}{8}\right){/tex}
{tex}\Rightarrow \quad x\left(1-\frac{y}{6}\right)=\frac{1}{2} x\left(1-\frac{y}{8}\right){/tex}
{tex}1-\frac{y}{6}=\frac{1}{2}\left(1-\frac{y}{8}\right){/tex}
{tex}\Rightarrow \quad 2-\frac{y}{3}=1-\frac{y}{8}{/tex}
{tex}2 -1 ={/tex} {tex}\frac{y}{3}-\frac{y}{8}{/tex}
1 = {tex}\frac{8 y-3 y}{24}{/tex}
{tex}\Rightarrow{/tex} {tex}24 = 5y{/tex}
{tex}\Rightarrow{/tex} {tex}y ={/tex} {tex}\frac{24}{5}{/tex}
{tex}y = 4.8 hours = 4\ hours\ 48\ minutes.{/tex}
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Prachi Chandila 7 years, 5 months ago
3Thank You