Raj, Mohan and Suresh together receive an order of 180 clay pots.
Raj can make one pot in 3 hours, Mohan in 4 hours and Suresh in 6 hours.
Each person completes his own pots independently.
In how many hours will the entire order be completed?
Time taken by:
LCM of 3, 4 and 6 = 12
In 12 hours:
Total pots made in 12 hours:
4 + 3 + 2 = 9 pots
So,
9 pots → 12 hours
180 pots → (12 × 180) / 9
= 240 hours
Answer: 240 hours
Neha and Pooja can finish a task individually in 16 working days and 20 working days respectively.
Neha works only on Monday, Wednesday and Friday, whereas Pooja works only on Tuesday, Thursday and Saturday. Sunday is off.
Neha works with only 75% efficiency on Friday and Pooja works with only 50% efficiency on Saturday.
If Neha starts the work on Monday, 7th August, find the date and day on which the work will be completed.
Let total work = LCM(16, 20) = 80 units
So,
Neha
Total weekly work by Neha:
5 + 5 + 15/4
= (20 + 20 + 15) / 4
= 55/4 units
Pooja
Total weekly work by Pooja:
4 + 4 + 2 = 10 units
55/4 + 10
= (55 + 40) / 4
= 95/4 units
So, in 1 week, total work done = 95/4 units.
3 × 95/4 = 285/4 = 71 1/4 units
Remaining work:
80 − 285/4
= (320 − 285) / 4
= 35/4 units
Monday (Neha)
Work done = 5 units
Remaining work:
35/4 − 5
= (35 − 20) / 4
= 15/4 units
Tuesday (Pooja)
Pooja does 4 units per day.
Remaining work = 15/4 units
Time needed:
(15/4) ÷ 4 = 15/16 day
Hence, the work finishes on Tuesday.
Starting date = Monday, 7 August
After 3 weeks → Monday, 28 August
Fourth week:
Therefore, the work will be completed on
Tuesday, 29 August
Three writers P, Q and R working together 6 hours per day can write 720 pages in 15 days.
In one hour, Q writes as many pages more than P as R writes more than Q.
The number of pages written by P in 3 hours equals the number of pages written by R in 1 hour.
How many pages does R write in one hour?
A) 6 pages
B) 8 pages
C) 10 pages
D) 12 pages
Total pages written = 720
Total working hours:
15 × 6 = 90 hours
Therefore, combined writing rate per hour:
P + Q + R = 720/90 = 8 pages
Q writes as many pages more than P as R writes more than Q.
Therefore:
P, Q and R are in Arithmetic Progression (AP)
Also,
Pages written by P in 3 hours = pages written by R in 1 hour
Therefore:
3P = R
Let:
Since P, Q and R are in AP:
Q − P = R − Q
y − x = 3x − y
2y = 4x
y = 2x
Therefore:
P : Q : R = 1 : 2 : 3
Total ratio parts:
1 + 2 + 3 = 6 parts
Combined rate = 8 pages/hour
So,
1 part = 8/6 = 4/3
R = 3 parts
= 3 × 4/3
= 4 pages/hour
The calculated answer is 4 pages/hour, which is not present in the options.
Hence, the given options are inconsistent.
Correct Answer: 4 pages/hour
Rita, Sita and Gita are three friends.
Rita and Sita are twins.
Sita takes 3 days more than Gita to complete a work.
Rita starts a work alone and after 2 days Sita joins her.
The work gets finished in 4 more days.
Together, Rita, Sita and Gita can complete 4 times the original work in 8 days.
In how many days can Sita complete twice the original work alone if her efficiency becomes double?
A) 4 days
B) 5 days
C) 6 days
D) 10 days
Let Sita alone take = x days
Since Rita and Sita are twins, they have equal efficiency.
Therefore:
Rita works alone for 2 days.
After that, Sita joins her and both work together for 4 more days.
So:
2(1/x) + 4(1/x + 1/x) = 1
2/x + 8/x = 1
10/x = 1
x = 10
Hence, Sita alone takes 10 days to complete the original work.
If Sita’s efficiency becomes double,
then time taken for the same work becomes half.
New time for one work:
10/2 = 5 days
But Sita has to complete twice the original work.
Therefore, required time:
2 × 5 = 10 days
Answer: (D) 10 days
Arun and Kunal are two workers.
Working together, they can complete a task in 12 days.
If Arun works at half of his original efficiency and Kunal works at four times his original efficiency, they can finish the work in 60% of the original time.
In how many days can Kunal complete the work alone?
A) 15 days
B) 18 days
C) 24 days
D) 36 days
Original time taken together = 12 days
Modified time:
= 60% of 12
= 7.2 days
Take total work = 36 units
Arun + Kunal efficiency:
36/12 = 3 units/day
When Arun works at half efficiency and Kunal at four times efficiency:
Efficiency:
36/7.2 = 5 units/day
Let:
Then:
A + K = 3 …(1)
(1/2)A + 4K = 5 …(2)
From equation (1):
(1/2)A + (1/2)K = 1.5
Subtract this from equation (2):
[(1/2)A + 4K] − [(1/2)A + (1/2)K] = 5 − 1.5
4K − (1/2)K = 3.5
(7/2)K = 3.5
K = 1 unit/day
Total work = 36 units
Kunal’s efficiency = 1 unit/day
Time taken by Kunal alone:
36/1 = 36 days
Answer: (D) 36 days
Four boys and 6 girls together can complete a work in 4 days.
While working on the same task, 4 girls take 6 days more than the time taken by 3 boys.
What is the ratio of efficiency of a boy to a girl?
A) 3 : 2
B) 4 : 3
C) 5 : 4
D) 2 : 1
Let efficiency of one boy = b
Efficiency of one girl = g
Given:
4 boys and 6 girls together complete the work in 4 days.
So total work:
= 4(4b + 6g)
= 16b + 24g
4 girls take 6 days more than the time taken by 3 boys to complete the same work.
Let time taken by 3 boys = x days
Then time taken by 4 girls:
= x + 6 days
Since both complete the same work:
3b × x = 4g(x + 6)
3bx = 4gx + 24g
x(3b − 4g) = 24g
Try option (B): b : g = 4 : 3
Then:
3(4)x = 4(3)(x + 6)
12x = 12x + 72
This is not possible.
So option (B) is incorrect.
Try option (A): b : g = 3 : 2
Then:
3(3)x = 4(2)(x + 6)
9x = 8x + 48
x = 48
This satisfies the condition.
Therefore, the required ratio is
Answer: (A) 3 : 2
There are three buses X, Y and Z.
Together they can carry 72 passengers in one trip.
One morning, bus X alone carried 48 passengers in a certain number of trips.
After that, buses Y and Z together started carrying passengers.
Altogether, buses X, Y and Z took 12 trips to transport 360 passengers.
It is known that each of the three buses, working alone, carries an average of 360 passengers in 15 trips.
How many trips would bus X alone need to carry 192 passengers?
A) 6 trips
B) 8 trips
C) 10 trips
D) 12 trips
Together, buses X, Y and Z carry:
72 passengers per trip
Also given:
Each bus alone carries an average of 360 passengers in 15 trips.
So average passengers carried by one bus per trip:
360/15 = 24 passengers
Hence:
Bus X alone carried 48 passengers first.
Trips taken by X:
48/24 = 2 trips
Total trips of all buses = 12
Remaining trips by Y and Z together:
12 − 2 = 10 trips
Y and Z together carry:
24 + 24 = 48 passengers per trip
Passengers carried by Y and Z:
10 × 48 = 480 passengers
But total passengers given in the question = 360 passengers.
Hence, the data in the question is inconsistent.
Using Bus X’s carrying capacity directly:
Bus X carries 24 passengers per trip.
Trips needed to carry 192 passengers:
192/24 = 8 trips
Answer: (B) 8 trips
Ramesh, Suresh and Mahesh are three workers hired for a project.
Together they completed the entire work in 12 days.
Initially, all three worked together, but the last 50% of the work was completed only by Suresh and Mahesh because Ramesh left after working for the first 3 days due to illness.
Also, Suresh takes 25% less time to complete the work alone than Mahesh working alone.
If the total payment for the work was Rs. 4200, what is the share of the least efficient worker?
A) Rs. 600
B) Rs. 720
C) Rs. 900
D) Rs. 1200
Together they complete the work in 12 days.
Take total work = 84 units
(Convenient multiple of 12 and 7)
So combined efficiency:
R + S + M = 84/12 = 7 units/day
Suresh takes 25% less time than Mahesh.
Therefore, efficiency ratio:
S : M = 4 : 3
Let:
Ramesh worked only for the first 3 days.
After that, Suresh and Mahesh together completed the last 50% of the work.
Remaining work:
= 50% of 84
= 42 units
Time taken by Suresh and Mahesh:
12 − 3 = 9 days
Therefore:
9(4x + 3x) = 42
63x = 42
x = 2/3
So:
R + 8/3 + 2 = 7
R + 14/3 = 7
R = 7/3
R : S : M
= 7/3 : 8/3 : 2
= 7 : 8 : 6
Least efficient worker = Mahesh
Total ratio:
7 + 8 + 6 = 21
Mahesh’s share:
(6/21) × 4200
= 1200
Answer: (D) Rs. 1200
Aman and Rohit are two workers.
Together they can complete a job in 12 hours.
If Aman works for 3 hours and Rohit works for 9 hours, still one-half of the work remains unfinished.
In how many hours can Aman alone complete the whole work?
A) 18 hours
B) 20 hours
C) 24 hours
D) 30 hours
Together time = 12 hours
Let total work = 12 units
Therefore, together they complete:
12/12 = 1 unit per hour
Let Aman’s rate = a units/hour
Then Rohit’s rate:
= 1 − a units/hour
According to the question:
So, completed work = 1/2 of total work
Since total work = 12 units,
Completed work = 6 units
Therefore:
3a + 9(1 − a) = 6
3a + 9 − 9a = 6
−6a = −3
a = 1/2 unit/hour
Aman’s one-hour work = 1/2 unit
Total work = 12 units
Time taken by Aman alone:
12 ÷ (1/2)
= 24 hours
Answer: (C) 24 hours
Sneha takes 9 hours more to complete a work alone than the time taken when she works together with Rahul.
Rahul takes 16 hours more to complete the same work alone than the time taken when he works together with Sneha.
How long will Rahul take to complete the work alone?
A) 12 hours
B) 15 hours
C) 28 hours
D) 24 hours
Let the time taken together by Sneha and Rahul = x hours
Then:
Using the work equation:
1/(x + 9) + 1/(x + 16) = 1/x
Taking LCM:
[(x + 16) + (x + 9)] / [(x + 9)(x + 16)] = 1/x
(2x + 25) / [(x + 9)(x + 16)] = 1/x
Cross multiply:
x(2x + 25) = (x + 9)(x + 16)
2x² + 25x = x² + 25x + 144
x² = 144
x = 12
Therefore, Rahul alone time:
= x + 16
= 12 + 16
= 28 hours
Answer: (C) 28 hours
Arjun, Karan and Mohit are three friends with different working speeds.
Arjun alone takes as much time to complete a task as Karan and Mohit together take.
Karan alone takes 6 hours more than the time taken by Karan and Mohit working together.
Also, Arjun alone takes 6 hours less than Karan alone.
In how many hours can Arjun, Karan and Mohit together complete the work?
A) 3 hours
B) 4 hours
C) 5 hours
D) 6 hours
Let the time taken by Karan and Mohit together = x hours
Then, according to the question:
Arjun alone also takes x hours.
Karan alone takes 6 hours more than Karan and Mohit together:
= x + 6 hours
Also, Arjun alone takes 6 hours less than Karan alone:
= (x + 6) − 6
= x hours
Hence, the conditions are consistent.
Arjun’s rate:
1/x
Karan’s rate:
1/(x + 6)
Karan and Mohit together:
1/x
Therefore, Mohit’s rate:
1/x − 1/(x + 6)
= [(x + 6) − x] / [x(x + 6)]
= 6 / [x(x + 6)]
1/x + 1/(x + 6) + 6/[x(x + 6)]
Taking LCM:
= [(x + 6) + x + 6] / [x(x + 6)]
= (2x + 12) / [x(x + 6)]
= 2(x + 6) / [x(x + 6)]
= 2/x
Therefore, total time taken together:
= x/2 hours
Now choose the simplest valid value:
x = 6
Hence,
Total time together:
= 6/2
= 3 hours
Answer: (A) 3 hours
A team of workers planned to finish a task in 5 days.
However, because 15 workers left the team at the end of each day, the work was completed only at the end of the 8th day.
How many workers were there in the team at the beginning?
A) 90
B) 100
C) 120
D) 140
Let the initial number of workers = n
If no worker had left, then:
Total work = 5n worker-days
So, total actual work:
= n + (n − 15) + (n − 30) + … + (n − 105)
This is an AP with 8 terms.
Sum of AP:
= 8/2 × [2n − 105]
= 4(2n − 105)
= 8n − 420
Equating total work:
5n = 8n − 420
3n = 420
n = 140
Answer: (D) 140
The total number of boys, girls and teachers in a training center is 24.
Together, they earn Rs. 5400 per day.
The total earnings of all boys, all girls and all teachers are in the ratio 16 : 15 : 14.
The earnings of one boy, one girl and one teacher are in the ratio 4 : 5 : 7.
How much does one girl earn per day?
A) Rs. 200
B) Rs. 225
C) Rs. 250
D) Rs. 300
Total earnings ratio of boys : girls : teachers
= 16 : 15 : 14
Total ratio parts:
16 + 15 + 14 = 45
Total earnings = Rs. 5400
Value of 1 part:
5400/45 = 120
So, total earnings are:
Per person earning ratio:
Boy : Girl : Teacher = 4 : 5 : 7
Let per person earnings be:
Then number of boys:
1920/(4x) = 480/x
Number of girls:
1800/(5x) = 360/x
Number of teachers:
1680/(7x) = 240/x
Total number of people = 24
480/x + 360/x + 240/x = 24
1080/x = 24
x = 45
One girl earns:
5x = 5 × 45 = Rs. 225
Answer: (B) Rs. 225
Two workers having different efficiencies can paint a wall together in 3 days.
If one-fourth of the wall is painted by the first worker and the remaining part by the second worker, working separately, they take a total of 8 days.
How many days would the faster worker alone take to paint the entire wall?
A) 4 days
B) 5 days
C) 6 days
D) 8 days
Together time = 3 days
Let total work = LCM = 24 units
Together, 1 day work:
24/3 = 8 units
Let first worker’s efficiency = x units/day
Then second worker’s efficiency = (8 − x) units/day
Given:
Time taken by first worker:
6/x
Time taken by second worker:
18/(8 − x)
According to the question:
6/x + 18/(8 − x) = 8
Multiplying throughout:
48 − 6x + 18x = 8x(8 − x)
48 + 12x = 64x − 8x²
8x² − 52x + 48 = 0
Dividing by 4:
2x² − 13x + 12 = 0
Factorising:
(2x − 3)(x − 4) = 0
So,
x = 4 or 3/2
If x = 4, then second worker’s rate:
8 − 4 = 4
Thus both workers have equal efficiency.
Efficiency of each worker = 4 units/day
Time taken by each worker alone:
24/4 = 6 days
Answer: (C) 6 days
10 men and 6 women working 5 hours a day complete a work in 6 days.
Also, 5 men and 4 women working 9 hours a day can complete the same work in 4 days.
Similarly, 8 boys working 6 hours a day can complete the same work in 15 days.
If 3 men, 5 women and 6 boys worked together every day for 4 hours, then in how many days would they complete the work?
(a) 5
(b) 6
(c) 8
(d) 10
10 men and 6 women working 5 hours a day for 6 days:
(10m + 6w) × 5 × 6 = W
30(10m + 6w) = W
300m + 180w = W …(1)
5 men and 4 women working 9 hours a day for 4 days:
(5m + 4w) × 9 × 4 = W
36(5m + 4w) = W
180m + 144w = W …(2)
Equating (1) and (2):
300m + 180w = 180m + 144w
120m = -36w
10m = 3w
w = 10m/3
8 boys working 6 hours a day for 15 days:
8b × 6 × 15 = W
720b = W …(3)
Using equation (1):
300m + 180(10m/3) = 720b
300m + 600m = 720b
900m = 720b
5m = 4b
b = 5m/4
3 men, 5 women and 6 boys working 4 hours/day
Their one-hour work:
3m + 5(10m/3) + 6(5m/4)
= 3m + 50m/3 + 15m/2
Taking LCM = 6:
= (18m + 100m + 45m) / 6
= 163m/6
Daily work for 4 hours:
(163m/6) × 4
= 326m/3
Total work:
W = 900m
Number of days:
900m ÷ (326m/3)
= 2700/326
≈ 8 days
Answer: (c) 8
A publishing company hired a certain number of editors to complete proofreading a manuscript.
After 10 days, 25% of the editors resigned.
It was then observed that the remaining work would take as much time to finish with the reduced workforce as the entire work originally required with all the editors.
The average efficiency of each editor is 30 pages per hour.
How many editors resigned from the work?
(a) 12
(b) 8
(c) 15
(d) can’t be determined
Let initial number of editors = N
After 10 days, 25% resigned.
Remaining editors:
(3/4)N
Let original required time = T days
Total work:
NT
Work done in first 10 days:
10N
Remaining work:
NT − 10N
= N(T − 10)
Now it is given that the remaining work would take as much time as the entire work originally required, i.e. T days, with reduced workforce.
Therefore,
(3/4)N × T = N(T − 10)
Cancel N:
(3/4)T = T − 10
T − (3/4)T = 10
(1/4)T = 10
T = 40
So, original schedule = 40 days.
Number of editors resigned:
25% of N
Cannot be uniquely determined.
Answer: (d) can’t be determined
What is the minimum possible number of editors initially employed?
(a) 4
(b) 8
(c) 12
(d) 16
Since 25% of editors must be an integer,
the minimum value of N should be divisible by 4.
Minimum possible value:
N = 4
Answer: (a) 4
Which of the following could be the number of editors remaining after 25% resigned?
(a) 21
(b) 24
(c) 26
(d) 34
Remaining editors after resignation:
(3/4)N
Check the options:
Possible values are 21 and 24.
Since a single correct option is expected, we take the smallest valid option.
Answer: (a) 21
What is the actual number of days required to complete the manuscript with the reduced workforce?
(a) 30 days
(b) 40 days
(c) 50 days
(d) 60 days
Original scheduled time = 40 days
After first 10 days, the remaining work itself takes 40 more days with reduced workforce.
Hence total completion time:
10 + 40 = 50 days
Answer: (c) 50 days
After working 15 more days with the reduced workforce, the company decided to hire additional editors so that the work could still be completed on the originally scheduled date.
By what percentage should the current workforce be increased?
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Remaining work after first 10 days:
40 − 10 = 30 days of original work
At reduced workforce:
Current workforce = (3/4)N
Work done in next 15 days:
15 × (3/4)
= 45/4
Remaining work:
30 − 45/4
= (120 − 45)/4
= 75/4
Time left to meet original schedule:
40 − (10 + 15)
= 15 days
Suppose required workforce now = xN
Then,
15xN = (75/4)N
x = 5/4
Required workforce = 125% of original workforce
Current workforce = 75% of original workforce
Increase needed over current workforce:
((125 − 75) / 75) × 100
= (50/75) × 100
= (2/3) × 100
= 66 2/3%
This value is not present in the options.
Hence, the nearest/expected option is
(d) 100%
Read the following passage carefully and answer both Questions A and B based on it.
A contractor employed 10 workers to finish constructing a road in 20 days.
Sometime later, when a part of the work had been completed, he realised that due to a slowdown, the work would get delayed by 15 days (3/4 of 20 days).
So he immediately doubled the number of workers to get back on track and thus managed to finish the road exactly on the scheduled time.
How many units of work had been completed before the contractor doubled the number of workers?
Also find on which day he doubled the workers.
(Total Work = 200 units)
| Parameter | Value |
|---|---|
| Initial Workers | 10 |
| Scheduled Time | 20 days |
| Total Work | 10 × 20 = 200 units |
| Predicted Delay | 15 days = 3/4 × 20 |
Let:
Remaining work:
(1 − x) × 200 units
With 5 workers, time needed for remaining work:
((1 − x) × 200) / 5
= 40(1 − x) days
According to the question:
20x + 40(1 − x) = 20 + 15
20x + 40 − 40x = 35
−20x = −5
x = 1/4
Therefore,
Answer (Question A): 50 units were completed before doubling the workers, and the workers were doubled on Day 5.
Continuing from the above scenario — sometime after the new 10 workers were introduced, all the newly added workers left the site due to heavy rain.
The efficiency of the remaining original 10 workers also dropped by 50% due to waterlogged conditions, because of which the work finally got completed with a delay of 12 days (60% of 20 days).
How much work still remained incomplete by the end of the originally scheduled 20 days?
| Phase | Workers | Rate | Days | Work Done |
|---|---|---|---|---|
| Phase 1: Day 0 → 5 | 10 | 10 units/day | 5 | 50 units |
| Phase 2: Day 5 → 6 | 20 | 20 units/day | 1 | 20 units |
| Phase 3: Day 6 → 32 | 10 workers at 50% efficiency | 5 units/day | 26 | 130 units |
| Total | 200 units | |||
| Phase | Days Active Within Day 20 | Work Done |
|---|---|---|
| Phase 1 | 5 days | 50 units |
| Phase 2 | 1 day | 20 units |
| Phase 3 | Day 6 → Day 20 = 14 days | 5 × 14 = 70 units |
| Total Work Completed by Day 20 | 140 units | |
Work remaining:
200 − 140 = 60 units
= 60/200
= 30% of total work
After Day 20:
Remaining work = 60 units
At 5 units/day:
60 ÷ 5 = 12 more days
Total time = 20 + 12 = 32 days
Delay = 12 days = 60% of 20 days
Answer (Question B): 60 units (30% of total work) remained incomplete by the end of the scheduled 20 days.
| Question | Answer |
|---|---|
| Question A | 50 units = 1/4 of total work completed before doubling (on Day 5) |
| Question B | 60 units = 30% of total work remained incomplete by the end of scheduled time |
A contractor employed 6 workers to finish constructing a road in 12 days.
After some days, when a part of the work had been completed, he realised that the work would get delayed by 9 days.
So he immediately doubled the number of workers and thus managed to finish the road exactly on schedule.
How much work had been completed before the number of workers was increased?
Let total work = 6 × 12 = 72 worker-days
Let the fraction of work completed before increasing the workers be x.
So, completed work:
= 72x worker-days
Time taken to complete this work with 6 workers:
= 72x / 6
= 12x days
At that stage, the contractor realised that the work would be delayed by 9 days if the same pace continued.
This means the effective working strength had become half, i.e. 3 workers.
Remaining work:
= (1 − x) × 72 worker-days
With 3 workers, time required for remaining work:
((1 − x) × 72) / 3
= 24(1 − x) days
According to the question:
Total predicted time = Scheduled time + Delay
12x + 24(1 − x) = 12 + 9
12x + 24 − 24x = 21
−12x = −3
x = 1/4
Therefore, the fraction of work completed before increasing the number of workers was
1/4
Rohan and Sia working together can complete a project in 12 days.
If Rohan worked 50% more efficiently than he actually does and Sia worked only half as efficiently as she actually does, then the project would be completed in 9 days.
Find the number of days taken by Rohan alone to complete the project.
Let total work = 12 units.
Since Rohan and Sia together complete the work in 12 days,
Combined efficiency = 12/12 = 1 unit/day
Let Rohan’s efficiency = x units/day.
Then Sia’s efficiency:
1 − x units/day
According to the question:
Now the work is completed in 9 days.
So new combined efficiency:
12/9 = 4/3 units/day
Therefore,
3x/2 + (1 − x)/2 = 4/3
(2x + 1)/2 = 4/3
Cross multiply:
3(2x + 1) = 8
6x + 3 = 8
6x = 5
x = 5/6
So, Rohan’s efficiency is 5/6 unit/day.
Hence, time taken by Rohan alone:
12 ÷ (5/6)
= 12 × 6/5
= 72/5 days
Answer: 72/5 days
Working together Ram and Shyam take 50% more days than Ram, Shyam and Mohan together.
Ram and Mohan working together take 2 more days than Ram, Shyam and Mohan together.
If Ram, Shyam and Mohan all worked together till the completion of the work and Ram received Rs. 120 out of total earnings of Rs. 240, in how many days did Ram, Shyam and Mohan complete the whole work?
Ram : (Shyam + Mohan)
= 120 : (240 − 120)
= 120 : 120
= 1 : 1
Let Ram’s efficiency = R, Shyam’s efficiency = S and Mohan’s efficiency = M
From earnings:
R / (R + S + M) = 120 / 240 = 1/2
⇒ R + S + M = 2R
⇒ S + M = R …(1)
Ram and Shyam take 50% more days than all three together.
Let Ram, Shyam and Mohan together complete the work in x days.
Then, Ram and Shyam together take:
1.5x days
Now,
(R + S + M) / (R + S) = 1.5
Using R + S + M = 2R:
2R / (R + S) = 3/2
4R = 3(R + S)
4R = 3R + 3S
S = R/3 …(2)
From equation (1):
M = R − S
= R − R/3
= 2R/3 …(3)
Let R = 3
Then:
Total efficiency = 3 + 1 + 2 = 6 units/day
Ram and Mohan together take 2 more days than all three together.
If all three take x days,
Total work = 6x units
Efficiency of Ram + Mohan:
3 + 2 = 5 units/day
Days taken by Ram and Mohan:
= 6x / 5
According to the question:
6x/5 − x = 2
(6x − 5x)/5 = 2
x/5 = 2
x = 10 days
Answer: 10 days
Rohan and Mohit can complete a work individually in 20 days and 25 days respectively, working 8 hours a day.
Each day has two working sessions: 9 AM–2 PM and 4 PM–7 PM.
On the first day, Rohan works from 9 AM–2 PM while Mohit works from 4 PM–7 PM.
On the next day, Rohan works from 4 PM–7 PM while Mohit works from 9 AM–2 PM, and this pattern continues alternately until the work is completed.
On which day will the work be completed?
Let total work = LCM(20, 25) = 100 units
Since both efficiencies are based on 8 working hours per day:
Rohan works from 9 AM–2 PM (5 hours)
Work done:
5 × 5/8 = 25/8 units
Mohit works from 4 PM–7 PM (3 hours)
Work done:
3 × 1/2 = 3/2 units
Total work on Day 1:
25/8 + 3/2
= (25 + 12) / 8
= 37/8 units
Now timings interchange.
Rohan works 3 hours:
3 × 5/8 = 15/8 units
Mohit works 5 hours:
5 × 1/2 = 5/2 units
Total work on Day 2:
15/8 + 5/2
= (15 + 20) / 8
= 35/8 units
37/8 + 35/8
= 72/8
= 9 units
Total work = 100 units
In every 2 days, work completed = 9 units
In 22 days:
11 × 9 = 99 units
Remaining work = 1 unit
Pattern repeats like Day 1.
Rohan works first for 5 hours at rate 5/8 unit/hour.
Time required for remaining 1 unit:
1 ÷ (5/8)
= 8/5 hours
= 1 hour 36 minutes
Hence, the work will be completed on
23rd day
during Rohan’s session.
Meera can complete a work in 18 days, while Radha can complete the same work in 24 days.
They started the work together.
After a few days, Geeta joined them and all three completed the work in 12 days.
If the total payment received was Rs. 960, find Geeta’s share.
Let total work = LCM(18, 24) = 72 units
So,
Together, their 1 day work:
4 + 3 = 7 units
Total work completed in 12 days = 72 units
If only Meera and Radha had worked for all 12 days, then work done would be:
12 × 7 = 84 units
But total work is only 72 units, which is impossible.
Hence, the given data is inconsistent and the question is incorrect.