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Ask QuestionPosted by Updesh Kumar 7 years, 9 months ago
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Posted by Sankalp Kumar 7 years, 9 months ago
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Posted by Pushpendar Verma 7 years, 9 months ago
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Nyayir Riba 7 years, 9 months ago
Posted by Yash Jain 7 years, 9 months ago
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Posted by Pushpendar Verma 7 years, 9 months ago
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Posted by Srbh Vishwakarma 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let a be the first term and d be the common difference of the given A.P. Then,
Sm = Sn
{tex}\Rightarrow \quad \frac { m } { 2 } \{ 2 a + ( m - 1 ) d \} = \frac { n } { 2 } \{ 2 a + ( n - 1 ) d \}{/tex}
{tex} \Rightarrow{/tex} 2a (m - n) + {m (m - 1) - n (n - 1)} d = 0
{tex} \Rightarrow{/tex} 2a (m - n) + {m2 - m - n2 + n}d = 0
{tex} \Rightarrow{/tex} 2a (m - n) + {(m2 - n2) - (m - n)} d = 0
{tex} \Rightarrow{/tex}2a (m - n) + {(m - n) (m +n) - (m - n)} d = 0
{tex} \Rightarrow{/tex} (m - n) {2a + (m + n - 1)d} = 0
{tex} \Rightarrow{/tex} 2a + (m + n - 1)d = 0 {tex} [ \because m - n \neq 0 ]{/tex} ...(i)
Now, {tex} S _ { m + n } = \frac { m + n } { 2 } \{ 2 a + ( m + n - 1 ) d \} = \frac { m + n } { 2 } \times {/tex}0 = 0 [Using (i)]
Posted by Sagar Arya 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let the original average speed of the train be x km/hr.
Time taken to cover 63 km {tex} = \frac{{63}}{x}{/tex} hours
Time taken to cover 72 km when the speed is increased by 6 km/hr {tex} = \frac{{72}}{{x + 6}}{/tex} hours
By the question,we have,
{tex}\frac{{63}}{x} + \frac{{72}}{{x + 6}} = 3{/tex}
{tex} \Rightarrow \frac{{21}}{x} + \frac{{24}}{{x + 6}} = 1{/tex}
{tex} \Rightarrow \frac{{21x + 126 + 24x}}{{{x^2} + 6x}} = 1{/tex}
{tex} \Rightarrow{/tex} 45x + 126 = x2 + 6x
{tex} \Rightarrow{/tex} x2 - 39x - 126 = 0
{tex} \Rightarrow{/tex} x2 - 42x + 3x - 126 = 0
{tex} \Rightarrow{/tex} x(x - 42) + 3(x - 42) = 0
{tex} \Rightarrow{/tex} (x - 42)(x + 3) = 0
{tex} \Rightarrow{/tex} x - 42 = 0 or x + 3 = 0
{tex} \Rightarrow{/tex} x = 42 or x = -3
Since the speed cannot be negative, {tex}x \ne -3{/tex}.
Thus, the original average speed of the train is 42 km/hr.
Posted by Yash Jain 7 years, 9 months ago
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Posted by Vaibhav Mittal 7 years, 9 months ago
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Nyayir Riba 7 years, 9 months ago
Posted by Suneha Suneha 7 years, 9 months ago
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Sankalp Kumar 7 years, 9 months ago
Posted by Sakshi Yadav 7 years, 9 months ago
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Posted by Kushal Kaushik 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Total number of possible outcomes = 36
- Doublet are { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}.
Total number of doublets = 6
{tex}\therefore{/tex} Prob (getting a doublet) = {tex}\frac{6}{36}{/tex}={tex}\frac{1}{6}{/tex} - Outcome whose sum of digits is 10 ={(4, 6) (5, 5) (6, 4)}.
Number of favorable outcomes =3
{tex}\therefore{/tex} Prob (getting a sum 10) = {tex}\frac{3}{36}{/tex} ={tex}\frac{1}{12}{/tex}
Posted by Yash Jain 7 years, 9 months ago
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Posted by Manish Anand 7 years, 9 months ago
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Ritik Chauhan 7 years, 9 months ago
Posted by Khushbu . 7 years, 9 months ago
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Khushbu . 7 years, 9 months ago
Posted by Sankalp Kumar 7 years, 9 months ago
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Aisha Singh 7 years, 9 months ago
Posted by Teerth Raj Chaudhary 7 years, 9 months ago
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Posted by Aakash Kumar 7 years, 9 months ago
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Chandrashekhar Wankhede 7 years, 9 months ago
Ritik Chauhan 7 years, 9 months ago
Posted by Ritik Chauhan 7 years, 9 months ago
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Posted by Kunal Gautam 7 years, 9 months ago
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Posted by Sanatan Chauhan 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Suppose the time taken by the pipes of larger and smaller diameters alone to fill the pool be x hours and y hours respectively.
Let the total volume of the pool be V cubic units.
{tex}\therefore{/tex}In 1 hour volume of the water that comes out of the pipe of larger diameter is {tex}\frac { V } { x }{/tex} cubic units.
In 4 hours, the volume of the water that comes out of the pipe of larger diameter is {tex}\frac { 4 V } { x }{/tex} cubic units.
The volume of the water that comes out of the pipe of smaller diameter in {tex}9 \text { hours is } \frac { 9 V } { y }.{/tex}
According to the first condition,
{tex}\therefore \quad \frac { 4 V } { x } + \frac { 9 V } { y } = \frac { 1 } { 2 } V \Rightarrow \frac { 4 } { x } + \frac { 9 } { y } = \frac { 1 } { 2 }{/tex} .............(i)
According to the second condition,
{tex}\therefore \quad \frac { 12 V } { x } + \frac { 12 V } { y } = V \Rightarrow \frac { 12 } { x } + \frac { 12 } { y } = 1{/tex} ............................(ii)
Putting {tex}\frac { 1 } { x } = u \text { and } \frac { 1 } { y } = v{/tex} in (i) and (ii), we obtain
{tex}4 u + 9 v = \frac { 1 } { 2 }{/tex} ..........................(iii)
{tex}12 u + 12 v = 1{/tex} ........................(iv)
Multiplying (iii) by 3 and subtracting from (iv), we get
{tex}- 15 v = - \frac { 1 } { 2 } \Rightarrow v = \frac { 1 } { 30 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 30 } \Rightarrow y = 30{/tex}
Substituting {tex}v = \frac { 1 } { 30 }{/tex} in (iii), we get
{tex}4 u + \frac { 9 } { 30 } = \frac { 1 } { 2 } \Rightarrow 4 u = \frac { 1 } { 2 } - \frac { 9 } { 30 } = \frac { 1 } { 5 } \Rightarrow u = \frac { 1 } { 20 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 20 } \Rightarrow x = 20{/tex}
Thus, the pipes of larger and smaller diameters fill the swimming pool alone in {tex}20\ hours\ and\ 30 \ hours{/tex} respectively.
Posted by Sarthak Bhalla 7 years, 9 months ago
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Chandrashekhar Wankhede 7 years, 9 months ago
Posted by Manjit Singh 7 years, 9 months ago
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Posted by ♥️Bhavika Naik?? 7 years, 9 months ago
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Posted by Ritik Chauhan 7 years, 9 months ago
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♥️Bhavika Naik?? 7 years, 9 months ago
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Shivansh Chouhan 7 years, 9 months ago
1Thank You