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Ask QuestionPosted by Dhruv Rana 7 years, 5 months ago
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Sia ? 6 years, 6 months ago
Let father's age be x years and the sum of the ages of his two children be y years.
Then,
{tex}x = 2y{/tex}
{tex}\Rightarrow{/tex} {tex}x - 2y = 0{/tex} ....(i)
20 years hence,
Father's age = {tex}(x + 20) years{/tex}
Sum of the ages of two children {tex}= y + 20 + 20 = (y + 40) years{/tex}
Then, we have
{tex}x + 20 = y + 40{/tex}
{tex}\Rightarrow{/tex} {tex}x - y = 20 {/tex}....(ii)
Multiplying (ii) by 2, we get
{tex}2x - 2y = 40{/tex} ...(iii)
Subtracting (i) from (iii), we have
x = 40
Thus, the age of father is 40 years.
Posted by M P 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
By the question,two pipes running together can fill a tank in {tex}11 \frac { 1 } { 9 }{/tex} minutes. If one pipe takes 5 minutes more than the other to fill the tank,we have to find the time in which each pipe would fill the tank separately.
Let time taken by pipe A be x minutes, and time taken by pipe B be x + 5 minutes.
In one minute pipe A will fill {tex}\frac { 1 } { x }{/tex} tank
In one minute pipe B will fill {tex}\frac { 1 } { x + 5 }{/tex} tank
pipes A + B will fill in one minute = {tex}\frac { 1 } { x } + \frac { 1 } { x + 5 }{/tex} tank
Now according to the question.
{tex}\frac { 1 } { x } + \frac { 1 } { x + 5 } = \frac { 9 } { 100 }{/tex}
or, {tex}\frac { x + 5 + x } { x ( x + 5 ) } = \frac { 9 } { 100 }{/tex}
or, 100(2x + 5) = 9x(x + 5)
or, {tex}200 x + 500 = 9 x ^ { 2 } + 45 x{/tex}
or, {tex}9 x ^ { 2 } - 155 x - 500 = 0{/tex}
or, {tex}9 x ^ { 2 } - 180 x + 25 x - 500 = 0{/tex}
or,9x(x - 20) + 25(x - 20) = 0
or, (x-20)(9x + 25) = 0
or, {tex}x = 20 , \frac { - 25 } { 9 }{/tex}
rejecting negative value, x = 20 minutes
and x + 5 = 25 minutes
Hence pipe A will fill the tank in 20 minutes and pipe B will fill it in 25 minutes.
Posted by Roshan Kumar 7 years, 5 months ago
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Posted by Astha Biswas 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
a4 + a8 = 24 (Given)
{tex} \Rightarrow {/tex} a + 3d + a + 7d = 24
{tex} \Rightarrow {/tex} 2a + 10d = 24
{tex} \Rightarrow {/tex} a + 5d = 12 ..... (i)
a6 + a10 = 44
{tex} \Rightarrow {/tex} a + 5d + a + 9d = 44
{tex} \Rightarrow {/tex} 2a + 14d = 44
{tex} \Rightarrow {/tex} a + 7d = 22 ...... (ii)
On solving equation (i) and (ii)
d = 5, a = -13
First three terms are -13, -8, -3.
Posted by Astha Biswas 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
For sin A,
By using identity {tex}cosec ^ { 2 } A - \cot ^ { 2 } A = 1 \Rightarrow \cos e c ^ { 2 } A = 1 + \cot ^ { 2 } A{/tex}
{tex}\Rightarrow \frac { 1 } { \sin ^ { 2 } A } = 1 + \cot ^ { 2 } A{/tex}
{tex}\Rightarrow \sin A = \frac { 1 } { \sqrt { 1 + \cot ^ { 2 } A } }{/tex}
For secA,
By using identity {tex}\sec ^ { 2 } A - \tan ^ { 2 } A = 1 \Rightarrow \sec ^ { 2 } A = 1 + \tan ^ { 2 } A{/tex}
{tex}\Rightarrow \sec ^ { 2 } A = 1 + \frac { 1 } { \cot ^ { 2 } A } = \frac { \cot ^ { 2 } A + 1 } { \cot ^ { 2 } A } \Rightarrow \sec ^ { 2 } A = \frac { 1 + \cot ^ { 2 } A } { \cot ^ { 2 } A }{/tex}
{tex}\Rightarrow \sec A = \frac { \sqrt { 1 + \cot ^ { 2 } A } } { \cot A }{/tex}
For tanA,
{tex}\tan A = \frac { 1 } { \cot A }{/tex}
Posted by Anurag Kumar 7 years, 5 months ago
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Karan Meena 7 years, 5 months ago
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Posted by Aarav Jaiswal 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the two numbers be x and y.
Then, by first Condition, ratio of these two numbers = 5:6.
x : y = 5 : 6
{tex}\Rightarrow\;\;\frac xy=\;\frac56{/tex}
{tex}\Rightarrow\;\; y=\;\frac{6x}5{/tex} ............ (i)
As per second condition if 8 is subtracted from each of the numbers, then ratio becomes 4:5.
{tex}\Rightarrow\;\;\frac{x-8}{y-8}=\;\frac45{/tex}
5x - 40 = 4y - 32
5x - 4y = 8 ........... (ii)
Now Put the value of y from (i) to (ii), we get
{tex}5x-4\left(\frac{6x}5\right)=\;8{/tex}
⇒ 25x - 24x = 40
⇒ x = 40
Put the value of x in equation (i), we get
{tex}\begin{array}{l}y=\;\frac65\times40\\=6\times\;8\\=\;48\end{array}{/tex}
Posted by Bipasha Bakshi 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a and d be the first term and common difference respectively of the given A.P. Then
an = a + (n - 1)d
{tex}\frac { 1 } { n } ={/tex} mth term
{tex}\Rightarrow \frac { 1 } { n } {/tex}= a + ( m - 1 ) d ...(i)
{tex}\frac { 1 } { m }{/tex}= nth term
{tex}\Rightarrow \frac { 1 } { m } {/tex}= a + ( n - 1 ) d ...(ii)
On subtracting equation (ii) from equation (i), we get
{tex}\frac { 1 } { n } - \frac { 1 } { m } = {/tex} [a+ (m-1) d] -[ a+ (n -1)d]
= a + md - d - a - nd + d
{tex}= ( m - n ) d{/tex}
{tex} \Rightarrow \frac { m - n } { m n } = ( m - n ) d {/tex}
{tex}\Rightarrow d = \frac { 1 } { m n }{/tex}
Putting d = {tex}\frac { 1 } { m n }{/tex} in equation (i), we get
{tex}\frac { 1 } { n } = a + \frac { ( m - 1 ) } { m n } {/tex}
{tex}\Rightarrow \frac { 1 } { n } = a + \frac { 1 } { n } - \frac { 1 } { m n } {/tex}
{tex}\Rightarrow a = \frac { 1 } { m n }{/tex}
{tex}\therefore{/tex} (mn)th term = a + (mn - 1) d
= {tex}\frac { 1 } { m n } + ( m n - 1 ) \frac { 1 } { m n } {/tex}{tex}\left[ \because a = \frac { 1 } { m n } = d \right]{/tex}
= {tex}\frac { 1 } { m n } + \frac { mn } { m n } - \frac { 1 } { m n }{/tex}
= 1
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