- Prologue
- Case: three men start the work simultaneously.
- Case: Three Men Start Working One After Another
- Case: Share in Daily Wage
- Case: Share in money after work finished?
- Case: Time equations: set of two people working
- Case: Time Ratios are given
Prologue
Speed Time work questions routinely appears in almost every aptitude exam
A can complete a piece of work in 10 days, B can do it in 12 days, C can to it in 15 days. Now A starts the work and continues for 2 days, then B joins and then C joins ….etc.etc. how much time will it take?
- Sometimes, Instead of character names, they’ll give your three pipes. Pipe A can fill the tank in X minutes, Pipe B can do in ….But the approach remains the same. (if pipe is emptying then use minus sign in the speed.)
- Such questions take very little time. Just LCM, fill in the table, and use STD formula and done.
Case: three men start the work simultaneously.
Aiyyar can paint Gokuldham society in 10 days, Bhide Master can do it in 12 days and Champaklal can do it in 15 days. If all three of them start working simultaneously from day 1, how many days will it take for them to finish the painting work?
It is similar to the pipes and cisterns problem.so make a table
Aiyyar | Bhide | Champak | Together | |
Speed | ||||
Time | 10 | 12 | 15 | |
Distance |
The “time” for each individual character, is already given in the problem.
Now take the LCM of Time, means that factorize number into its prime numbers (2,3,5,7,11,13,17,19 etc)
10=2×5
12=2^{2}×3
15=3×5
Now take each prime number in its highest power and multiply them together.
LCM=2^{2}×3^{1}×5^{1}=60
- So LCM (10,12,15)=60. This is the total distance covered, and in this case it means total work done. i.e. entire surface area of walls of Gokuldham is 60 square meters.
- Whether these characters paint the society individually or together, the area of the walls will remain one and same 60.
Aiyyar | Bhide | Champak | Together | |
Speed | ||||
Time | 10 | 12 | 15 | |
Distance | 60 | 60 | 60 | 60 |
Not only empty box is of speed.
Simply write the number required to get 60 by multiplication.
Example Aiyyer’s time is 10 days. So to get 60 from ten, you’ve to multiply 10 with 6.
(Alternatively) for Aiyyer’s case
Speed x time = distance
Speed x 10 days = 60 ; because it is already given in the problem that Aiyyar takes 10 days.
Speed = 60/10 = 6.
Write 6, in Aiyyar’s speed box.
Similarly for Bhide and Champak you’ll get 5 and 4 respectively.
Aiyyar | Bhide | Champak | Together | |
Speed | 6 | 5 | 4 | 6+5+4=15 |
Time | 10 | 12 | 15 | |
Distance | 60 | 60 | 60 | 60 |
When these three characters work together, their speed will increase: 6+5+4=15.
Now we have to find out the time required,
Again STD formula
Speed x time = distance
15 x time = 60
Time = 60/15= 4 days.
Final answer: if these three characters work together, they can paint the society in only four days.
Case: Three Men Start Working One After Another
Same problem, but with little modification
Aiyyar can paint Gokuldham society in 10 days, Bhide Master can do it in 12 days and Champaklal can do it in 15 days. Aiyyar starts painting the society and Bhide joins him after 2 days. Then Both of paint the society together for one day and then Champaklal joins the gang. In this manner, how many total days will be required to finish the paint work?
Part I: Aiyaar starts the work
The table remains seem up to this part
Characters | Work | |||
Aiyyar | Bhide | Champak | Aiyyar starts | |
Speed | 6 | 5 | 4 | 6 |
Time | 10 | 12 | 15 | x2 |
Distance | 60 | 60 | 60 | =12 |
It is given that For the first two days, Aiyyar works alone.
So in the last column, fill up Aiyyar’s speed and time accordingly.
Speed x time = distance
6 x 2 =12 ; because Aiyyar’s speed is 6 and he works for 2 days.
Meaning, in the first two days Aiyyar has painted 12 square meters of walls.
How much work remains?
Total 60 minus 12 = 48.
Part II: Bhide joins
it is given that, after two days Bhide master joins and together Aiyyar and Bhide work for one day. (and then Champak joins)
So make a new column in a table, Aiyyar +Bhide
since they’re working together, this bill will increase and we will do addition : 6 + 5 =11
Characters | Work | ||||
Aiyyar | Bhide | Champak | Aiyyar starts | Bhide Joins | |
Speed | 6 | 5 | 4 | 6 | 6+5=11 |
x Time | 10 | 12 | 15 | x2 | x 1 |
=Distance | 60 | 60 | 60 | 12 | =11 |
Again
Speed x time = distance
11 x 1 =11
Meaning, in one-day Aiyyar and Bhide painted another 11 square meters.
How much work is left??
48 minus 11 =37
Part III : Champaklal joins
Now Champaklal also joins, so combine speed is 6+5+4=15
and remaining work is 37 We have to find out the time
Characters | Work Progress report | |||||
Aiyyar | Bhide | Champak | Aiyyarstarts | BhideJoins | ChampakJoins | |
Speed | 6 | 5 | 4 | 6 | 6+5=11 | 6+5+4=15 |
xTime | 10 | 12 | 15 | 2 | 1 | ?? |
Distance | 60 | 60 | 60 | 60 | 11 | 37 remains |
Again STD
Speed x Time =distance
15 x Time = 37
Time = 37/15= approx. 2.5 days
Caution : This “2.5” is not the final answer. Because they’ve asked you to find total number of days required to finish the paint work. So total time is right from the day one when Aiyyar started.
2+1+2.5=5.5
Final Answer: The work will be finished in 5.5 days.
Case: Share in Daily Wage
Abdul can do a piece of work in 10 days; Bhide in 15 days. They work for 5 days. The rest of the work was finished by Champak in 2 days. If they get Rs.1500 for the whole work. Find the daily wages of Bhide and Champak.
They got Rs.1500 for the whole work, meaning the work is worth Rs.1500 so take it as our distance
Repharse: Abdul can do road-construction of 1500 kms in 10 days!.. and so on
Abdul | Bhide | |
Speed | ||
Time | 10 | 15 |
distance | 1500 | 1500 |
Apply STD formula and find out speed value of each column
A | B | |
Speed | 150 | 100 |
xTime | 10 | 15 |
=distance | 1500 | 1500 |
Given: they (A and B) worked for five days
A | B | A+B | |
Speed | 150 | 100 | 150+100=250 |
Time | 10 | 15 | 5 |
distance | 1500 | 1500 | 250 x 5=1250 |
Combined speed of A+B=250 so in two days they constructed road of 1250 kms.
Remaining work = 1500 minus 1250 = 250kms
This was done by C alone, in two days.
A | B | A+B | C | |
Speed | 150 | 100 | 250 | |
Time | 10 | 15 | 5 | 2 |
distance | 1500 | 1500 | 1250 | 250 |
Run STD formula on C’s column and find his speed
A | B | A+B | C | |
Speed | 150 | 100 | 250 | 125 |
Time | 10 | 15 | 5 | 2 |
distance | 1500 | 1500 | 1250 | 250 |
Their daily wage depends on how much work can they do per day?
Which is actually their speed. E.g. B can construct 100 km roads per day.
So daily wage of B=Rs.100
And daily wage of C=Rs.125
Total daily wage of B and C=100+125=225
Case: Share in money after work finished?
Q. Abdul alone can do a piece of work in 6 days and Bhide alone in 8 days. They undertook to do it for Rs.3200. With the help of Champak, they completed the work in 3 days. How much is to be paid to Champak?
Total road construction is 3200 kms this time.
A | B | |
Speed | ||
Time | 6 | 8 |
distance | 3200 | 3200 |
Run STD formula to find speeds
A | B | |
Speed | 3200/6 | 400 |
Time | 6 | 8 |
distance | 3200 | 3200 |
Given: With the help of C, they completed the work in 3 days.
We don’t know the speed of C, assume it is “c” but construct his column and fill remaining data.
A | B | C | A+B+C | |
Speed | 3200/6 | 400 | c | (3200/6)+400+c |
Time | 6 | 8 | ?? | 3 |
distance | 3200 | 3200 | 3200 |
Run STD formula on third column
[ (3200/6)+400+c ] x 3 = 3200Solve this equation you get c=400/3
Meaning speed of c=400/3
He works for 3 days.
Speed x time = distance
(400/3) x 3 =distance
400= distance
Meaning C covered 400kms
Which means he was paid Rs.400
Case: Time equations: set of two people working
Q. A & B can do a piece of work in 12 days , B and C in 15 , C & A in 20 days. How long would each take separately to do the same work?
In the STD table, when people come and go, we can do addition and subtraction in the “SPEED” boxes only and now in the Time boxes.
Now fill up the table.
A+B | B+C | C+A | |
SPEED | |||
TIME | 12 | 15 | 20 |
DISTANCE |
TAKE THE LCM (12, 15, 20) =60. Based on that we can find individual speeds using STD formula.
A+B | B+C | C+A | |
SPEED | 5 | 4 | 3 |
TIME | 12 | 15 | 20 |
DISTANCE | 60 | 60 | 60 |
Assume that these three entities work together
A+B | B+C | C+A | (A+B)+(B+C)+(C+A) | |
SPEED | 5 | 4 | 3 | 5+4+3=12 |
TIME | 12 | 15 | 20 | ?? |
DISTANCE | 60 | 60 | 60 | 60 |
Speed of A+B+C together can be derived using
(A+B)+(B+C)+(C+A)
= 2(A+B+C)=5+4+3=12
Hence speed of A+B+C together =12/2=6
Now STD formula
Speed x time = distance
6 x time = 60
Time = 10
Meaning if A+B+C work together, they can finish work in 10 days. [Although Time of A+B+C, is not asked in the question, but I’ve given it only for explanation]
Now back to the question, we’ve to find individual time. So far our table looks like this-
A+B | B+C | C+A | A+B+C | |
SPEED | 5 | 4 | 3 | 6 |
TIME | 12 | 15 | 20 | 10 |
DISTANCE | 60 | 60 | 60 | 60 |
Now you can solve it three unknown equation.
Because this is ‘speed’, we can do addition, subtraction and solve it as three unknown equations
Speed of A+B+C=6
Speed of A+B=5
Hence speed of C= 6 –(A+B)= 6-5=1
A+B | B+C | C+A | A+B+C | (A+B+C)-(A+B)=C | |
SPEED | 5 | 4 | 3 | 6 | 6-5=1 |
TIME | 12 | 15 | 20 | 10 | ?? |
DISTANCE | 60 | 60 | 60 | 60 | 60 |
Speed of C is 1 and he has to cover 60 kilometers alone,
STD formula,
1 x Time = 60
Hence it’ll take him 60 days.
A+B | B+C | C+A | A+B+C | (A+B+C)-(A+B)=C | |
SPEED | 5 | 4 | 3 | 6 | 6-5=1 |
TIME | 12 | 15 | 20 | 10 | 60 |
DISTANCE | 60 | 60 | 60 | 60 | 60 |
Do similar procedure for B and A.
A+B | B+C | C+A | A+B+C | C | B | A | |
SPEED | 5 | 4 | 3 | 6 | 6-5=1 | 6-3=3 | 6-4=2 |
TIME | 12 | 15 | 20 | 10 | 60 | 20 | 30 |
DISTANCE | 60 | 60 | 60 | 60 | 60 | 60 | 60 |
Final answer, time taken by A,B,C if they work individually= 30,20,60
Run a cross verification to see the answer is correct. Plug the answer values.
A | B | C | A+B | B+C | C+A | |
Speed | 2 | 3 | 1 | 5 | 4 | 3 |
Time | 30 | 20 | 60 | 12 | 15 | 20 |
Dist | 60 | 60 | 60 | 60 | 60 | 60 |
Yes it satisfies the statements given in the Question.
Case: Time Ratios are given
Q. A takes twice as much time as B and thrice as much as C to finish a piece of work. They together finish the work in one day. Find the time taken by each of them to finish the work.
A takes twice as much time as B,
meaning if B takes 1 day, A takes 2 days
So the ratio of time is A:B=2:1
Similarly A:C=3:1
We can write the same thing as
B:A=1:2
A:C=3:1
If we want the three terms ratio, we want mid term (A) to be same in both equation.
In the first eq. A is 2 and in second eq. A is 3. LCM (2,3)=6
Multiply something in both equations so that we get 6 in both equation.
Meaning, Multiply first eq with (2) and second equation with (3)
B:A=3:6
A:C=6:2
Now we can make a three terms ratio
B:A:C=3:6:2
Keep in mind, this is ratio and not ‘absolute number’
So we fill “Time” boxes as
A’s time = 6x
B’s time = 3x
C’s Time = 2x
Run the STD table, fill the data
A | B | C | A+B+C | |
Speed | 1 | 2 | 3 | 6 |
Time | 6x | 3x | 2x | 1 day |
Distance | 6x | 6x | 6x | 6x |
LCM of (6x,3x,2x)= 6x
From the STD fomula, we can get the individual speeds of A,B,C,= 1,2,3
Together their speed A+B+C=1+2+3=6
For the last column
Speed x time = distance
6 multiplied with 1 =6x
Hence x= 1
Put the value of x=1 in the table and you get individual time
A will take 6 days individually, B will take 3 days and C will take 2 days.
Cross verification (plug the answers we got)
A | B | C | A+B+C | |
Speed | 1 | 2 | 3 | 6 |
Time | 6 | 3 | 2 | 1 day |
Distance | 6 | 6 | 6 | 6 |
From the table,
A takes twice as much time as B and thrice as much as C to finish a piece of work
Yes it satisfies the statement given in the question itself.
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