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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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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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

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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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
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