⭐A and B together can do a work in 72 minutes. but B takes 1 hour more than A takes to do the same work, when each works alone. Find the time taken by A and B each working alone do the same work.⭐
⚒️CLASS 10 CHAPTER :QUADRATIC EQUATIONS⚒️
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Answers
Note: 1 hour = 60 minutes.
According to the given condition,
Given that A and B can do work in 72 minutes.
⇒ 1/A + 1/B = 1/72
Let the time taken by A to finish the work is 'x' hours.
⇒ 1/A = 1/x.
Given that B takes 1 hour more than A takes to do the same work
⇒ 1/B = 1/x + 60.
Now,
Frame the equations and solve, we get
⇒ (1/x) + (1/x + 60) = (1/72)
LCM = 72x(x + 60)
⇒ 72(x + 60) + 72x = x(x + 60)
⇒ 72x + 4320 + 72x = x^2 + 60
⇒ 144x + 4320 = x^2 + 60
⇒ x^2 - 84x - 4320 = 0
⇒ x^2 - 120x + 36x - 4320 = 0
⇒ x(x - 12) + 36(x - 12) = 0
⇒ (x - 12)(x + 36) = 0
⇒ x = 120,-36{Neglect -ve values}
⇒ x = 120 minutes
⇒ x = 2 hours.
Then:
⇒ 1/B = 1/x + 60
= 1/120 + 60
= 1/180
= 180 minutes.
= 3 hours.
Therefore:
⇒ Time taken by A to do the same work = 2 hours.
⇒ Time taken by B to do the same work = 3 hours.
Hope this helps!
Let A alone finish the work = x mi
part of the work finished by
A in a day = 1/x ----( 1 )
B alone finish the work = ( x+60)mi
part of the work finished by
B in a day = 1/(x+60) ---( 2 )
Given ,
A and B together can do work = 72mi
A and B together can do
work in a day = 1/72 ---( 3 )
=> 1/x + 1/(x+60) = 1/72
=> (x+60+x)/[x(x+60)] = 1/72
=> (2x+60)/(x²+60x) = 1/72
=> 72(2x+60) = x²+60x
=> 144x + 72 × 60 = x² + 60x
=> x² + 60x - 144x - 72 × 60 = 0
=> x² - 84x - 72 × 60 = 0
=> x² - 120x + 36x - 72 × 60 = 0
=> x( x - 120 )+ 36( x - 120 ) = 0
=> ( x - 120 )( x + 36 ) = 0
Therefore ,
x - 120 = 0 or x + 36 = 0
x = 120 or x = -36
Now ,
A alone can do the work = x = 120mi
= 2 hr
B alone can do the work = ( x + 60 )
= ( 120 + 60 )
= 180 mi
= 3 hr
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