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Two gardeners, Burton and Philip, work at independent
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26 May 2017, 11:43
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68% (02:44) correct 32% (03:01) wrong based on 146 sessions
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Two gardeners, Burton and Philip, work at independent constant rates to prune a garden full of roses. If both gardeners start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Philip were to work at twice Burton’s rate, they would take only 20 minutes. How long would it take Philip, working alone at his normal rate, to tune the garden full of roses? A. 1 hour 20 minutes B. 1 hour 45 minutes C. 2 hours D. 2 hours 20 minutes E. 3 hours
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Two gardeners, Burton and Philip, work at independent
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26 May 2017, 14:16
Let the rate of work of Burton be x and rate of work(Philip) is y 45(x+y) = 900 units(which we are assuming as the total units of work to be done) > eqn1 Also, 20(2x + x) = 900 units of work (If Philip works at twice Burton’s rate, they would take only 20 minutes)20(3x) = 900 x = 15 units/minute(Burton's rate) Substituting x=15 in eqn1, we get 45*(15 + y ) = 900 or y = 5 units/minute, which is the rate of Philip's work. Hence, time take for Burton to complete the work is 900/5 =180 minutes(Option E)
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Re: Two gardeners, Burton and Philip, work at independent
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16 Sep 2018, 19:47
Quote: Two gardeners, Burton and Philip, work at independent constant rates to prune a garden full of roses. If both gardeners start at the same time and work at their normal rates, they will complete the job in 45 minutes working together. However, if Philip were to work at twice Burton’s rate, working together they would take only 20 minutes. How long would it take Philip, working alone at his normal rate, to tune the garden full of roses?
A. 1 hour 20 minutes B. 1 hour 45 minutes C. 2 hours D. 2 hours 20 minutes E. 3 hours
Prune a garden full of roses = 1 job Burton takes (say) 2x minutes to do 1 job alone. If Philip takes x minutes to do 1 job alone (to work at twice Burton’s rate), together they would do the job in 20min, hence: \(\frac{1}{{20}} = \frac{1}{{2x}} + \frac{{1 \cdot \boxed2}}{{x \cdot \boxed2}} = \frac{3}{{2x}}\,\,\,\,\, \Rightarrow \,\,\,x = 30\,\,\,\left[ {\min } \right]\) Conclusion: Burton takes 2x = 60 minutes to do this job alone. If Philip takes y minutes to do 1 job alone (our FOCUS!), from the fact that together they would do it in 45min, we have: \(\frac{1}{{45}} = \frac{1}{{60}} + \frac{1}{y}\,\,\,\,\, \Rightarrow \,\,\,\frac{1}{y} = \frac{{1 \cdot \boxed4}}{{3 \cdot 15 \cdot \boxed4}}  \frac{{1 \cdot \boxed3}}{{4 \cdot 15 \cdot \boxed3}} = \,\,\frac{1}{{3 \cdot 4 \cdot 15}}\,\,\,\,\left[ {\frac{1}{{\min }}} \right]\) \(? = y = 3 \cdot 60\,\,\min = 3{\text{h}}\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Two gardeners, Burton and Philip, work at independent
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17 Sep 2018, 15:27
Kritesh wrote: Two gardeners, Burton and Philip, work at independent constant rates to prune a garden full of roses. If both gardeners start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Philip were to work at twice Burton’s rate, they would take only 20 minutes. How long would it take Philip, working alone at his normal rate, to tune the garden full of roses?
A. 1 hour 20 minutes B. 1 hour 45 minutes C. 2 hours D. 2 hours 20 minutes E. 3 hours let b=Burton's rate 3b*20=1→ b=1/60 if it takes Burton 60 minutes to complete entire job alone, then in 45 minutes he can complete 3/4 of the job, so Philip, in the same 45 minutes, completes 1/4 of the job, and will complete the entire job alone in 4*45=180 minutes 3 hours E



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Re: Two gardeners, Burton and Philip, work at independent
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18 Sep 2018, 18:42
Kritesh wrote: Two gardeners, Burton and Philip, work at independent constant rates to prune a garden full of roses. If both gardeners start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Philip were to work at twice Burton’s rate, they would take only 20 minutes. How long would it take Philip, working alone at his normal rate, to tune the garden full of roses?
A. 1 hour 20 minutes B. 1 hour 45 minutes C. 2 hours D. 2 hours 20 minutes E. 3 hours Let’s let B = the number of minutes for Burton to do the job alone. Thus, Burton’s rate is 1/B. We also let P = the number of minutes for Philip to do the job alone; his rate is 1/P. Since they are working together, we can combine their rates and create the following equation: 1/B + 1/P = 1/45 If Philip were to work at twice Burton’s rate, his new rate would be 2/B, and we would have: 2/B + 1/B = 1/20 3/B = 1/20 60 = B Substituting, we have: 1/60 + 1/P = 1/45 Multiplying by 180P, we have: 3P + 180 = 4P 180 = P Thus, it will take Philip 180 minutes, or 3 hours, to prune the garden, working alone. Answer: E
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Re: Two gardeners, Burton and Philip, work at independent
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09 Oct 2019, 22:54
Hi,
The best way to solve time/work questions is to use numbers instead of fractions. To avoid using fractions, we should not consider the work to be 1 but consider it to be the LCM of the times given to us.
Given
time (B + P) = 45 minutes
Since Philip works at twice Burton's rate
time (B + 2B) = 20 minutes
Now let us consider the work to be the LCM of the times i.e. 45 and 20, which is 180 units.
Work = 180 units
rate (B + P) = 180/45 > 4units/hr
rate (B + 2B) = 180/20 > 9units/hr
Now rates can be treated as algebraic equations,
3B = 9 > B = 3units/hr
B + P = 4 > P = 1unit/hr
Now the question asks for the time that Philip will take to prune the garden.
time (P) = work/rate(P) > 180/1 > 180 minutes > 3 hours




Re: Two gardeners, Burton and Philip, work at independent
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09 Oct 2019, 22:54






