Works W1 and W2 are done by two persons A and B.

Question

Works W1 and W2 are done by two persons A and B. A takes 80% more time to do the work W1 alone than he takes to do it together with B. How much percent more time B will take to do the work W2 alone than he takes to do it together with A?

A 100%
B 110%
C 120%
D 125%
E can not be determined

Show Answer

Official Answer

D

A 100%
B 110%
C 120%
D 125%
E can not be determined

yashikaaggarwal · Answer

Suppose W1=W2
Let A+B do 108 work in 1 hour
A takes 1.8 hours to complete (W1) 108 work alone.
That means he does 60 work/hour
So B completed only 48 work that particular hour.
B will take 108/48 time to complete W2 alone
B takes 2.25 hours to complete work w2
The percent more time B will take to do the work W2 alone than he takes to do it together with A = 1*(100+x)/100 = 2.25
=> 100+x = 225
=> x = 125

IMO D

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NikuHBS · Answer

The ratio of time taken by an individual is inversely proportional to the rate of doing work.

\frac{Time(A) }{ Time(A+B)} = \frac{180 }{100} = \frac{9}{5}

Therefore, 
 \frac{Rate(A) }{ Rate(A+B)} = \frac{100 }{180} = \frac{5}{9}

On simplification, we get
 \frac{Rate(A) }{ Rate(B)} = \frac{5}{4}

Thus, 
 \frac{Rate(B) }{ Rate(A+B)} = \frac{4}{9}

Hence,
 \frac{Time(B) }{ Time(A+B)} = \frac{9}{4}

Thus, we can see that Time taken by B alone to finish the work is 9/4th of the total time taken by A and B together to finsih the work W2 (It doesn't matter how much is the work W2. Time required would be in the same proportion.)

Hence,
B takes  125 %more time as compared to the time required by both A and B together.

Option D

KG04 · Answer

Let W1 be 180 units and time taken by A alone to do it be 18min and A+B together 10min
=> A does 10units/min and A+B together do 18 units/min
=> B alone does 8units/min
=> Time taken by B alone to do W1 => 180/8 = 22.5min
=> Percentage of time taken by B more than A+B together = (22.5-10)*100/10 =125%
This percentage will be same for both W1 and W2 since it is a ratio.
=> 125%

urshi · Answer

Answer : E

The amount of effort required to complete W2 is unknown. Hence, We will not be able to provide the answer with the information given.

ashwini2k6jha · Answer

Let a, b be the amounts of work done per day. by A-and B-respectively.

Let A and B complete W, in n days. Then
(a+b)n=W and 1.8 n a=W1
a+b=1.8a
b=0.8a
a=1.25b
Now let A and B complete W2 in m days.Then B will take=(2.25-1/1)100=125% more
Answer D

vtexcelgmat · Answer

MayankSingh · Answer

IMO D

Assuming that work W1 & W2 are of same type*, W2 = y * W1 (y = any number)

For W1:  
Let (A+B) take "X" days
Then, A takes= 1.8X days.
Let time taken by B alone= B days

AB/(A+B) = 1.8X
1.8B = 1.8X + B
B= 2.25X

For W1:  
W2= y* W1
(A+B) will take = y*X days
B alone will take = y* 2.25X days

Req. percentage = (2.25-1)y*x / y*X x 100 = 125%

D 125%

firas92 · Answer

Let a and b be the rates of  A  and  B  respectively

For Work  W1 ,
 (a+b)t=W1  or  \frac{W1}{t}=a+b 
 a(1.8t)=W1  or  \frac{W1}{t}=1.8a

So  a+b=1.8a  or  b=0.8a

We have the ratio of work rates of  A  and  B  so we can calculate the percentage increase in time for a given amount of work.

Eliminate (E)

For Work  W2 ,
 b(T1)=W2  or  0.8a(T1)=W2

T1=\frac{W2}{0.8a}

(a+b)(T2)=W2  or  (a+0.8a)(T2)=W2 , which is  1.8a(T2)=W2

T2=\frac{W2}{1.8a}

We are asked to find  \frac{T1-T2}{T2}*100

\frac{T1-T2}{T2}*100 = \frac{\frac{W2}{0.8a}-\frac{W2}{1.8a}}{\frac{W2}{1.8a}}*100

Factoring out and cancelling  \frac{W2}{a}  from the numerator and denominator,

\frac{T1-T2}{T2}*100 = \frac{\frac{1}{0.8}-\frac{1}{1.8}}{\frac{1}{1.8}}*100 = \frac{\frac{1}{8}-\frac{1}{18}}{\frac{1}{18}}*100 = \frac{\frac{9-4}{72}}{\frac{1}{18}}*100 = \frac{5}{72}*18*100 = 1.25*100 = 125 %

Answer is  (D)

Kinshook · Answer

Given: 
1. Works W1 and W2 are done by two persons A and B. 
2. A takes 80% more time to do the work W1 alone than he takes to do it together with B. 
Asked: How much percent more time B will take to do the work W2 alone than he takes to do it together with A?
A 100%
B 110%
C 120%
D 125%
E can not be determined

Let the time taken by A alone to complete W1 be a & A alone to complete W1 be b.

1/a + 1/b = 1/t 
a = 1.8t = 9t/5
1/1.8t + 1/b = 1/t
1/b = 1/t - 1/1.8t = .8/1.8t = 4/9t
b = 9t/4 = 2.25t

Percent more time B will take to do the work W2 alone than he takes to do it together with A = 125%; Which is independent of work

IMO D

Archit3110 · Answer

work done by A = x & of B = y
Let work W be completed together in n days
(x+y)*n= W
1.8*nx = W1
x+y= 1.8x
y=.8x
x=1.25y

so for W2 total time ; 1.25y+y ; 2.25y
% time taken ; (2.25y-y/y) *100 ; 125%
OPTION D

Works W1 and W2 are done by two persons A and B. A takes 80% more time to do the work W1 alone than he takes to do it together with B. How much percent more time B will take to do the work W2 alone than he takes to do it together with A?
A 100%
B 110%
C 120%
D 125%
E can not be determined

saumyagupta1602 · Answer

Method I
time A takes alone ie t2= (1+80/100) time A and B take together ie t1 ..for W1
time A takes alone= 9/5 time A and B take together 
a*t2/w1= (a+b)*t1/w1 ; t2=t1((180/100)
a*t1((180/100)= (a+b)*t1 
a*9/5=a+b
4/5a=a/b

similarly, 
b*t3/w2= (a+b)*t1 /w2
b*t3 = (a+b)*t1
b*t1(1+x/100) = (a+b)*t1
b*(1+x/100)=a+b
4/5a*(1+x/100)=9/5a
(1+x/100)=9/4
x=125% >>option D

Method II
A B A+B if A and B together take t hours 
9/5t t(1+x/100) t

1/A+1/B=1/(A+B)
5/9t+ 100/t(100+x)= 1/t
4/9t=100/(t*(100+x))
4x=500
x=125>> option D

GMATWhizTeam · Answer

Solution

• Let us assume that  R_A  and  R_B  are the rates ( work per unit time) of A and B, respectively.
• Also, let us assume that  t_1  and  t_2  are times taken by A and B together to complete  W_1  and  W_2  respectively.
 o Time taken by A alone to complete the work  W_1 = 1.8t_1

o Thus,  R_A = \frac{W_1}{1.8t_1} = \frac{W_1}{t_1}*\frac{10}{18} = \frac{W_1}{t_1}*\frac{5}{9}

• Since,  (R_A + R_B)t_1 = W_1

o This means,  R_B = \frac{W_1}{t_1} - R_A= \frac{W_1}{t_1} - \frac{W_1}{t_1}*\frac{5}{9} = \frac{W_1}{t_1}*\frac{4}{9}

• Now,  t_2 = \frac{W_2}{R_A + R_B} = \frac{W_2}{\frac{W_1}{t_1} (\frac{4}{9} + \frac{5}{9})} = \frac{W_2 * t_1}{W_1}

• And time taken by B  = \frac{W_2}{ R_B} = \frac{W_2}{\frac{W_1}{t_1} *(\frac{4}{9})} = \frac{W_2*t_1}{W_1} * \frac{9}{4}

• Thus the required percentage  = \frac{Time \space taken \space by \space B \space – t_2}{t_2}*100 =\frac{( \frac{W_2*t_1}{W_1} * ( \frac{9}{4} – 1) )} {\frac{W_2 * t_1}{W_1}}*100 = (\frac{9}{4} – 1 )*100 = 125  %

Thus, the correct answer is  Option D.

ArijitR · Answer

Lets say A and B together does the work in T hrs.
A&B does 1/T of the work in in 1 hr.
A alone does the complete work in T+80T/100 hrs = 9T/5 hrs 
In 1 hr A does 5/9T of the work
B alone does (1/T - 5/9T)= 4/9T part of the work in 1 hr. 
B alone does the complete work in 9T/4 hrs, B takes (T-9T/4)= 5T/4 hrs extra than A&B combined. 
So its 125% of the combined time.

rebrahimi12 · Answer

Make both works equal to 1 and substitute numbers.

Assume A and B together can do 1 job in 1 hour. If A alone takes 80% longer, it would take A 9/5 of an hour.

A's work rate is 5/9. B's work rate must be 4/9 to maintain 1 hour if they work together.

Thus B can complete a job in 9/4 hours.

Percent greater = (9/4 - 1)/1 x 100 = 1.25/1 x 100 = 125%

Demonicgod · Answer

Let  A’s rate = a  and  B’s rate = b  for the same kind of work.
Given:
A takes  80% more time  to do W1  alone  than  with B .
So,
TA=1.8 TA+BT_A = 1.8 \, T_{A+B}TA=1.8TA+B
For work = 1 unit:
TA=1a,TA+B=1a+bT_A = \frac{1}{a}, \quad T_{A+B} = \frac{1}{a+b}TA=a1,TA+B=a+b1
Substitute:
1a=1.8⋅1a+b\frac{1}{a} = 1.8 \cdot \frac{1}{a+b}a1=1.8⋅a+b1
Cross multiply:
a+b=1.8aa + b = 1.8aa+b=1.8a b=0.8ab = 0.8ab=0.8a
 
Now find % more time B takes alone vs with A for work W2
Assuming same type of work (otherwise impossible to determine):
 
  B alone time :  
TB=1b=10.8a=1.25aT_B = \frac{1}{b} = \frac{1}{0.8a} = \frac{1.25}{a}TB=b1=0.8a1=a1.25
 
  Together time :  
TA+B=1a+b=11.8a=0.555...aT_{A+B} = \frac{1}{a+b} = \frac{1}{1.8a} = \frac{0.555...}{a}TA+B=a+b1=1.8a1=a0.555...
Now find how many % more B takes alone:
TBTA+B=1.25/a0.555.../a=2.25\frac{T_B}{T_{A+B}} = \frac{1.25/a}{0.555.../a} = 2.25TA+BTB=0.555.../a1.25/a=2.25
This means B takes  2.25 times , i.e.  125% more  time.
 
✅  Correct Answer: D) 125%

Works W1 and W2 are done by two persons A and B.

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Works W1 and W2 are done by two persons A and B.

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