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02 Dec 2020, 12:30
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A
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Difficulty:
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Three professors A, B, and C are evaluating a work. A is 40% more efficient than B and B is 20% more efficient than C. A takes 10 days less than B to complete the evaluation work. A starts the evaluation work and works for 10 days and then B takes over. B evaluates for next 15 days and then stops. In how many days, C can complete the remaining evaluation work?
A. 7.2 days
B. 9.5 days
C. 11.5 days
D. 12.5 days
E. 13.5 days
Are You Up For the Challenge: 700 Level Questions
A. 7.2 days
B. 9.5 days
C. 11.5 days
D. 12.5 days
E. 13.5 days
Are You Up For the Challenge: 700 Level Questions
02 Dec 2020, 21:30
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Bunuel
A is 40% more efficient than B, there \(A_E = 1.4 B_E\).
Since efficiency is inversely proportional to Time, \(B_T = 1.4 * A_T\)
Let B take x days to Finish. Then A takes x - 10 days. Substituting in the above equation and solving
x = 1.4(10 - x)
0.4x = 14
x = 35
So A takes 25 days, B takes 35 days
Given that B is 20% more efficient than C. So, \(B_E = 1.2 C_E\)
In terms of time, \(C_T = 1.2 * B_T\)
Time Taken by C = 1.2 * 35 = 42 days
Let C take x days to evaluate the remaining work.
Fraction of work done by A in 10 days = 10 * \(\frac{1}{25} = \frac{2}{5}\)
Fraction of work done by B in 15 days = 15 * \(\frac{1}{35} = \frac{3}{7}\)
Fraction of work done by C in x days = x * \(\frac{1}{42}\)
Sum of fractions of work = 1 (completion of work is always = 1)
\(\frac{2}{5} + \frac{3}{7} + \frac{x}{42} = 1\)
\(\frac{84 + 90 + 5x}{210} = 1\)
5x = 210 - 174 = 36
x = 7.2 days
Option A
Arun Kumar
General Discussion
jcaguirre91
Joined: 20 Oct 2020
Last visit: 26 Feb 2025
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Location: Mexico
Schools: HBS '25 (D) Booth '25 (I) Sloan '25 (WL)
GMAT 1: 690 Q48 V35
GPA: 3.8
02 Dec 2020, 15:05
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Given
Rate A = 1.4 * Rate B
Rate B = 1.2 * Rate C
Time A= Time B – 10 (days)
Get Time A por completing 1 job
(1) rA=1.4 rB→\frac{1}{tA}=\frac{1.4}{tB}→tB=1.4*tA (2) tA=tB-10
From (1) and (2)
tA=(1.4tA)-10→.4tA=10→tA=10/(.4 )=25 days
Get Time B and Time C each for completing 1 job
tB=tA+10=25+10=35 days
rB=1.2Rc→ 1/tB=1.2/tC→tC=1.2*tB→tC=1.2*35=42 days
W=(rA*tA)+(rB*tB)+(rC*tC)
We can then use the time that each professor evaluated the exams multiplying the rates which are the inverse of the times to the times for each professor in the problem.
W=(rA*tA)+(rB*tB)+(rC*tC)
1=(1/25*10 days)+(1/35*15 days)+(1/42*x days)
Simplify and solve for x, x being the number of days that professor C will spend on the evaluation:
1=2/5+3/7+x/42→1= (14+15)/35+ x/42→ (35-29)/35=x/42→6/35=x/42→x=(42)(6)/35=(6)(6)/5
=7.2 days
Rate A = 1.4 * Rate B
Rate B = 1.2 * Rate C
Time A= Time B – 10 (days)
Get Time A por completing 1 job
(1) rA=1.4 rB→\frac{1}{tA}=\frac{1.4}{tB}→tB=1.4*tA (2) tA=tB-10
From (1) and (2)
tA=(1.4tA)-10→.4tA=10→tA=10/(.4 )=25 days
Get Time B and Time C each for completing 1 job
tB=tA+10=25+10=35 days
rB=1.2Rc→ 1/tB=1.2/tC→tC=1.2*tB→tC=1.2*35=42 days
W=(rA*tA)+(rB*tB)+(rC*tC)
We can then use the time that each professor evaluated the exams multiplying the rates which are the inverse of the times to the times for each professor in the problem.
W=(rA*tA)+(rB*tB)+(rC*tC)
1=(1/25*10 days)+(1/35*15 days)+(1/42*x days)
Simplify and solve for x, x being the number of days that professor C will spend on the evaluation:
1=2/5+3/7+x/42→1= (14+15)/35+ x/42→ (35-29)/35=x/42→6/35=x/42→x=(42)(6)/35=(6)(6)/5
=7.2 days
