Work/Rate – Mini Test
Solutions
Note: The problems in this topic can mostly be solved by using a formula.
1.
If w women can do a job in d days, then how many days will it take (w + n) women to do the job if all the women work at the same pace?
A.
B.
C.
D.
E.
Solution:
Formula used: time taken for x persons to perform a task = Time taken for 1 person to perform the task / x
Details:
Here time taken for 1 woman to perfom the task = dw.
So time taken for n+w women to perform the task = dw/(n+w)
2.
At their respective rates, pump A, B, and C can fulfill an empty tank, or pump-out the full tank in 2, 3, and 6 hours. If A and B are used to pump-out water from the half-full tank, while C is used to fill water into the tank, in how many hours, the tank will be empty?
A. 2/3
B. 1

So from the above time taken to empty= ¾ hours
Hence C.
C. 3/4
D. 3/2
E. 2
Solution:
Formula Used: t1/ta + t2/tb + t3/tc=1
The first and second terms and the RHS are negative because they represent draining.
So the above equation becomes,
-t1/ta - t2/tb + t3/tc = -1
Details: ta =2/2 tb= 3/2 tc= 6/2 t1=t2=t3= time taken to empty half full tank=?
3.
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets.
A. 4
B. 6
C. 8
D. 10
E. 12
Solution:
Formula Used: 1/ ta + 1/tb = 1/tab since ta and tb are related and tab is given
Since the time taken by x is asked for 2W convert the other info for an output of 2W widgets.
Details: ta-tb=4 days for 2w widgets

tab= 4.8 days for 2w widgets
So 1/(tb+4) + 1/tb = ¼.8
We can find tb from above and hence ta. ta=12 days
Hence E.
4.
It takes machine A 'x' hours to manufacture a deck of cards that machine B can manufacture in
'y' hours. If machine A operates alone for 'z' hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?
A. (100xy – z)/(x + y)
B. y(100x – z)/(x + y)
C. 100y(x – z)/(x + y)
D. (x + y)/(100xy – z)
E. (x + y – z)/100x
Solution:
Formula used: t1/ta + t2/tb = 1
Details:
Let both operate together for b hours. t1= z + b t2= b ta = 100x tb=100y
(z+b) / 100x + b/100y = 1
Y(z+b) + xb = 100xy
by + bx +zy= 100xy
b= (100xy –zy) / (x+y)
Hence answer is choice B.
5.
If Dev works alone he will take 20 more hours to complete a task than if he worked with Tina to complete the task. If Tina works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dev to complete the task? What is the ratio of the time taken by Dev to that taken by Tina if each of them worked alone to complete the task?

A. 4 : 1
B. 2 : 1
C. 10 : 1
D. 3 : 1
E. 1 : 2
Solution:
Formula used = 1/ta + 1/tb = 1/ tab
Details: ta = tab + 20 and tb = tab + 5
So, 1/ (tab+20) + 1/(tab+5) = 1/tab
We get tab = 10 and so ta = 30 and tb = 15, ta:tb = 2:1
Hence B.
6.
Machine A and machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produces?
A. 6
B. 6.6
C. 60
D. 100
E. 110
Solution:
Formula Used: t1*ra =p and t2*rb=p
Details:
We have t1=t2+10 and rb = 1.1 *ra and p=660
Substituting the above relations in the formula,
(t2+10)*ra = 660 --- (1) t2* 1.1*ra = 660--- (2)
(1)/ (2) -> t2*ra + 10 * ra= t2*1.1*ra
t2+10 =1.1*t2
t2=100 hrs and so ra= 660/110= 6

Hence A.
7.
Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?
A. 6
B. 7
C. 8
D. 9
E. 10
Solution:
Formula Used: t1/ta + t2/tb + t3/tc=1
Details:
Carla works throughout . So finding the number of days Carla worked, we can find how long the work lasted. Let it be x. t1=2, ta=10 t2=x-3, tb=12 t3=x, tc=15
Substituting the above values in the formula, we get x=7. Hence B.
8.
A furniture manufacturer has two machines, but only one can be used at a time. Machine A is utilized during the first shift and Machine B during the second shift, while both work half of the third shift. If Machine A can do the job in 12 days working two shifts and Machine B can do the job in 15 days working two shifts, how many days will it take to do the job with the current work schedule?
A. 14
B. 13
C. 11
D. 9
E. 7
Solution:
Formula Used: 1/ta + 1/tb = 1/tab
Details: ta = 12 days * 2 shifts is the time it takes to complete the task alone

But A works only 1.5 shifts per day
So, 12 * 2 = x * 1.5. x=16 days. So A has to actually work for 16 days to complete the task
alone. Similarly B has to work for 20 days
So, 1/16 + 1/20 = 1/tab , tab= 80/9 = 9 days (approx). Hence D.
9 . 6 Workers should finish a job in 8 days. After 3 days, 4 workers joined them. How many days do they need to finish the same job?
A. 3
B. 4
C. 5
D. 6
E. 7
Solution:
Formula Used: n* tab = n1*t1+ n2*t2+…
Details:
n= 6, tab=8
n1=6, t1=3
n2=(6+4)=10
t2=?
We have, 48=18+10*t2, t2 = 3.
So totally t1+ t2 = 3+3= 6 days are needed to finish the same job. Hence D.
10 . Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone? a.4/7 b. 1/3 c. 2/3 d.3/4

Solution:
Formula Used: Work done per unit time by 4 pumps working together = 3/ta + 1/tb
Details:
3/ta + 1/tb = 3/ta + 1/ (2/3) ta = 9/ (2* ta). Time taken = (2/9) *ta
Time taken by the larger pump = (2/3)* ta
The ratio is (2/9) * ta / (2/3) * ta = 1/3
Hence B.
11. A water tank has three taps A, B, and C. A fills four buckets in 24 min, B fills 8 buckets in 1 hour and C fills 2 buckets in 20 minutes. If all the taps are opened together a full tank is emptied in 2 hours. If a bucket can hold 5 litres of water, what is the capacity of the tank? a.120 litres b. 240 litres c. 180 litres d. 60 litres
Solution:
Formula Used: (t1/ta) * P + (t2/tb)*P + (t3/tc) * P = P
Details: t1=t2=t3=2 hrs = 120 min
A bucket holds 5 litres of water. ra= P/ta= (4*5)/24 = 5/6 rb=P/tb =(8*5)/60= 2/3 rc=P/tc =(2*5)/20= 1/2
Note we do not know P and ta, tb, tc but we can find ra, rb and rc as above.
So we have, 120*(5/6) + 120* (2/3) + 120 *(1/2) = 240 litres. Hence B.

12. There is leak in the bottom of a tank. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which admits 6 litres per hour and the tank is now emptied in 12 hours. What is the capacity of the tank? a.28.8 litres b. 36 litres c. 144 litres d. Cannot be determined
Solution:
Formula Used: (t1/ta)*P + (t2/tb)*P=P
Since the second term is a leak and the tank eventually drains they are negative.
So the equation becomes,
(t1/ta)*P – (t2/tb)*P = -P
Details:
We have t1=t2=12, tb=8
P/ta=6
So we have, 12*6 – (12/8)*P =-P
P=144 litres. Hence C.
13. Ram starts working on a job and works on it for 12 days and completes 40% of the work. To help him complete the work he employs Ravi and together they work for another 12 days and the work gets completed. How much more efficient is Ram than Ravi? a.50% b. 200% c. 60% d. 100%
Solution:

Formula Used: t1/ta + t2/tb =1
Details: t1=12+12=24, ta=? But we know 40% of the work is done in 12 days. So ta =30 t2=12, tb=?
We have 24/ta + 12/tb= 1
24/30 +12/tb = 1
Tb=60
Ta/tb=1/2 . So ra/rb = 2/1. So Ram is twicw as efficient or 100% more efficient than Ravi.
14. A,B and c can independently do a work in 15 days,20 days and 30 days respectively. They work together for sometime after which C leaves. A total of Rs.18,000 is paid for the work and B gets Rs.6000 more than C. For how many days did A work? a. 2 b. 4 c. 6 d.8
Solution:
Formula Used: (t1/ta) * P + ( t2/tb) * P + (t3/tc) * P =P
Details:
We have t1=t2, ta =15, t2=20 and tc=30, p=18000
The equation becomes,, t1/15 + t1/20 + t3/30 = 1 -- (1)
B was paid 6000 more than C and as t2=t1,
So, ( t1/20) *P – (t3/30) *P =6000
The equation becomes, t1/20 - t3/30 = 1/3 --- (2)
Solving (1) and (2), we get t1=8. Hence D.

15. There's a lot of work in preparing a birthday dinner. Even after the turkey is in oven, there's still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table. Three friends, Asit, Arnold, and Afzal, work together to get all of these chores done. The time it takes them to do the work together is six hours less than Asit would have taken working alone, one hour less than Arnold would have taken, and half the time Afzal would have taken working alone.
How long did it take them to do these chores working together? a.20 minutes b.30 minutes c. 40 minutes d. 50 minutes
Solution:
Formula Used: 1/ta + 1/tb + 1/ tc = 1/tabc
Details:
ta = tabc+6, tb=tabc+1, tc= 2*tabc
So, 1/ (tabc+6) + 1/ (tabc+1) +1/ (2*tabc) = 1/tabc
Solving, we have tabc = 40 minutes. Hence C.