Alternate Based MCQ Quiz - Objective Question with Answer for Alternate Based - Download Free PDF
Last updated on Jun 29, 2025
Latest Alternate Based MCQ Objective Questions
Alternate Based Question 1:
A and B alone can do a piece of work in 4 and 9 days, respectively. In how many days will the work be completed, if they both work on alternate days, starting with B?
Answer (Detailed Solution Below)
Alternate Based Question 1 Detailed Solution
Given:
A can do a piece of work = 4 days
B can do a piece of work = 9 days
Formula used:
Total work = efficiency × time
Calculation:
| Efficiency | Person | Time | Total work |
| 9 | A | 4 | 36 |
| 4 | B | 9 |
Now,
(B + A) = 4 + 9 = 13 units = 2 days
⇒ (13 × 2) = 26 units = (2 × 2) = 4 days
⇒ 30 units = 5 days
⇒ 36 units = 5 + (6/9) = 5\(2\over3\) days
∴ The correct answer is 5\(2\over3\) days.
Alternate Based Question 2:
P and Q can complete a piece of work in 28 days and 35 days, respectively. P begins to do the work and they work alternatively one at a time for one day each. In how many days will the whole work be completed?
Answer (Detailed Solution Below)
Alternate Based Question 2 Detailed Solution
Given:
P can complete the work in 28 days.
Q can complete the work in 35 days.
P and Q work alternatively, starting with P.
Formula Used:
1. Work done in 1 day = 1 / Time taken to complete the work.
2. Total work = LCM of individual times.
3. Work completed in 2 days (P and Q working alternatively) = Work done by P in 1 day + Work done by Q in 1 day.
4. Remaining work = Total work - Work completed in full cycles.
Calculation:
LCM(28, 35) = 140 units (Total work).
Work done by P in 1 day = 140/28 = 5 units.
Work done by Q in 1 day = 140/35 = 4 units.
Work done in 2 days (1 cycle of P and Q working alternatively).
Work done in 2 days = 5 + 4 = 9 units.
Number of full cycles = Total work / Work done in 2 days = 140 / 9.
Full cycles = 15 (remaining work will be calculated next).
Work completed in 15 cycles = 15 × 9 = 135 units.
Remaining work = Total work - Work completed = 140 - 135 = 5 units.
On the 31st day, it is P's turn. P can complete 5 units of work in 1 day.
Total days = Days for full cycles + 1 (remaining work).
Total days = (15 × 2) + 1 = 31 days.
The total time required to complete the work is 31 days.
Alternate Based Question 3:
A tank can be filled by pipe A in 4 hours and pipe B in 6 hours. At 8:00 a.m., pipe A was opened. At what time will the tank be filled if pipe B is opened at 9:00 a.m.?
Answer (Detailed Solution Below)
Alternate Based Question 3 Detailed Solution
Given:
Pipe A fills the tank in 4 hours
Pipe B fills the tank in 6 hours
Pipe A is opened at 8:00 a.m.
Pipe B is opened at 9:00 a.m.
Formula used:
Work done by A in 1 hour = 1/4
Work done by B in 1 hour = 1/6
Calculation:
From 8:00 a.m. to 9:00 a.m., only pipe A is working:
⇒ Work done by A in 1 hour = 1/4
From 9:00 a.m. onwards, both pipes A and B are working:
Combined work done by A and B in 1 hour = 1/4 + 1/6 = (3 + 2)/12 = 5/12
Remaining work after 9:00 a.m. = 1 - 1/4 = 3/4
Time taken to complete remaining work = (3/4) / (5/12)
⇒ Time = (3/4) × (12/5) = 9/5 hours = 1.8 hours
1.8 hours = 1 hour and 0.8 × 60 minutes = 1 hour and 48 minutes
Therefore, the tank will be filled at 9:00 a.m. + 1 hour 48 minutes = 10:48 a.m.
∴ The correct answer is option (3).
Alternate Based Question 4:
Pipe X can fill a tank in 9 hours and Pipe Y can fill it in 21 hours. If they are opened on alternate hours and Pipe X is opened first, in how many hours shall the tank be full?
Answer (Detailed Solution Below)
Alternate Based Question 4 Detailed Solution
Given:
Pipe X fills the tank in 9 hours.
Pipe Y fills the tank in 21 hours.
Pipes are opened alternately, with Pipe X opened first.
Formula used:
Rate of Pipe X = \(\frac{1}{9} \)
Rate of Pipe Y = \(\frac{1}{21} \)
Calculations:
In 2 hours (1 hour by each pipe), the total amount of the tank filled is:
\(\frac{1}{9} + \frac{1}{21}\)
To add these fractions, find the LCM of 9 and 21, which is 63:
\(\frac{1}{9} = \frac{7}{63} and \frac{1}{21} = \frac{3}{63}\)
\(\frac{7}{63} + \frac{3}{63} = \frac{10}{63} \)
In every 2-hour period, \(\frac{10}{63}\) of the tank is filled. Let the total time taken be t hours.
After \(t = 2n \) hours, the total fill will be:
\(\frac{10}{63} \times n \)
We want the tank to be full, i.e., 1 unit of the tank:
\(\frac{10}{63} \times n = 1\)
⇒ \(n \) = 63/10 = 6.3
Therefore, \(t \) = 2n = 2(6.3) = 12.6 hours.
The tank will be full in approximately 12.6 hours, or 12 hours and 36 minutes.
The correct answer is option 2.
Alternate Based Question 5:
Pipes A and B can fill a tank in 6 hours and 8 hours, respectively. Both pipes are opened together for 3 hours. After that pipe A is closed, and B continues to fill the tank. In how many hours will the tank be filled?
Answer (Detailed Solution Below)
Alternate Based Question 5 Detailed Solution
Given:
Pipe A can fill the tank in 6 hours.
Pipe B can fill the tank in 8 hours.
Both pipes are opened together for 3 hours, then A is closed and B continues.
Formula Used:
Work done = Time × Rate of work
Calculation:
Rate of work of pipe A = 1/6 (tank/hour)
Rate of work of pipe B = 1/8 (tank/hour)
Work done by both pipes in 3 hours = 3 × (1/6 + 1/8)
\( \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}\)
Work done by both pipes in 3 hours = 3 × (7/24) = 21/24 = 7/8 (of the tank)
Remaining work = 1 - 7/8 = 1/8 (of the tank)
Time taken by pipe B to fill the remaining 1/8 of the tank = (1/8) / (1/8) = 1 hour
Total time to fill the tank = 3 hours (both pipes) + 1 hour (pipe B) = 4 hours
The tank will be filled in 4 hours.
Shortcut Trick
Alternate Method
| Pipe | Time to Fill Tank (hours) | Efficiency (Tank/hour) | Work Done in 3 hours |
|---|---|---|---|
| Pipe A | 6 | 1/6 | (1/6) × 3 = 1/2 |
| Pipe B | 8 | 1/8 | (1/8) × 3 = 3/8 |
Total work done by both pipes in 3 hours = 1/2 + 3/8 = 7/8 of the tank.
Remaining work = 1 - 7/8 = 1/8 of the tank.
Time taken by pipe B to fill the remaining 1/8 of the tank = (1/8) ÷ (1/8) = 1 hour.
Total time to fill the tank = 3 hours (both pipes working) + 1 hour (pipe B alone) = 4 hours.
Top Alternate Based MCQ Objective Questions
A,B and C can do a piece of work in 30 days, 40 days and 50 days, respectively. Beginning with A, if A, B and C do the work alternatively then in how many days will the work be finished?
Answer (Detailed Solution Below)
Alternate Based Question 6 Detailed Solution
Download Solution PDFGiven:
A can do a piece of work = 30 days
B can do a piece of work = 40 days
C can do a piece of work = 50 days
Formula used:
Total work = efficiency × time
Calculation:
| Efficiency | Person | Time | Total work |
| 20 | A | 30 | 600 |
| 15 | B | 40 | |
| 12 | C | 50 |
According to the question:
⇒ (20 + 15 + 12) = 47 units = 3 days
⇒ 47 × 12 = 564 units = 3 × 12 = 36 days
⇒ (564 + 20 + 15) = 599 units = 38 days
Total work = 600 units = 38 + (1/12) = 38\(1\over12\) days.
∴ The correct answer is 38\(1\over12\) days.
An inlet pipe can fill an empty tank in \(4\frac{1}{2}\) hours while an outlet pipe drains a completely filled tank in \(7\frac{1}{5}\) hours. The tank is initially empty. and the two pipes are alternately opened for an hour each, till the tank is completely filled, starting with the inlet pipe. In how many hours will the tank be completely filled?
Answer (Detailed Solution Below)
Alternate Based Question 7 Detailed Solution
Download Solution PDFGiven:
An inlet pipe can fill an empty tank in \(4\frac{1}{2}\) hours while an outlet pipe drains a completely filled tank in \(7\frac{1}{5}\) hours.
Concept used:
Efficiency = (Total work / Total time taken)
Efficiency = work done in a single day
Calculation:Time taken by A = 9/2 hours
Please note that after 20 hours, the remaining capacity = 6 units
Now in the 21st hour, pipe A will work and fill the tank so no need to add time after that.
Time taken by pipe A to fill 6 units = 6/8 = 3/4 hours
So,
To do a certain work, Ajay and Bharat work on alternate days, with Bharat starting the work on the first day. Ajay can finish the work alone in 32 days. If the work gets completed in exactly 8 days, then Bharat alone can finish 7 times the same work in ____________ days.
Answer (Detailed Solution Below)
Alternate Based Question 8 Detailed Solution
Download Solution PDFFormula Used:
Total work = Efficiency × Time taken
Calculation
Ajay can finish the work alone in 32 days
A’s one day work = 1/32
A and B complete the whole work in = 8 days
Ajay and Bharat work on alternate days, with Bharat starting the work on the first day so, we can say B will work only 4 days A will work only:
= 8 - 4 = 4 days
If A’s 4-day work = 4/32 = 1/8
Remaining work = 1 – [1/8] = 7/8
B complete 7/8 work in = 4 days
B complete whole work in = 4 × [8/7] = 32/7 days
B alone can finish 7 times the same work in = [32/7] × 7 = 32 days
B alone can finish 7 times the same work in 32 days.
A and B alone can do a piece of work in 4 and 9 days, respectively. In how many days will the work be completed, if they both work on alternate days, starting with B?
Answer (Detailed Solution Below)
Alternate Based Question 9 Detailed Solution
Download Solution PDFGiven:
A can do a piece of work = 4 days
B can do a piece of work = 9 days
Formula used:
Total work = efficiency × time
Calculation:
| Efficiency | Person | Time | Total work |
| 9 | A | 4 | 36 |
| 4 | B | 9 |
Now,
(B + A) = 4 + 9 = 13 units = 2 days
⇒ (13 × 2) = 26 units = (2 × 2) = 4 days
⇒ 30 units = 5 days
⇒ 36 units = 5 + (6/9) = 5\(2\over3\) days
∴ The correct answer is 5\(2\over3\) days.
R, S and T can finish a work in 20, 15 and 10 days, respectively. R works on all days and S and T work on alternate days with T starting the work on the first day. In how many days is the work finished?
Answer (Detailed Solution Below)
Alternate Based Question 10 Detailed Solution
Download Solution PDFGiven:
R, S and T can finish a work in 20, 15 and 10 days, respectively
R works on all days and S and T work on alternate days with T starting the work on the first day.
Formula Used:
Efficiency = Total work/Time taken
Calculation:
Let the total work be 60 units(LCM of 20, 15, and 10).
Efficiency of R = 60/20 = 3 units/day
The efficiency of S = 60/15 = 4 units/day
Efficiency of T = 60/10 = 6 units/day
On the first day, R and T work = 9 units
On the second day, R and S work = 7 units
2 day's work → 16 units
6 day's work → 48 units
7 day's work → 57 units
The remaining work of 3 unit is done by R and S in = 3/7 days
Total Time = 7 + 3/7 = 52/7 days
∴ In 52/7 days the work was finished.
Ravi can do a piece of work in 40 days and Sudha can do the same piece of work in 60 days. If they work on alternative days starting with Sudha on the first day, then in how many days will the work be completed?
Answer (Detailed Solution Below)
Alternate Based Question 11 Detailed Solution
Download Solution PDFGiven:
Ravi can do a piece of work = 40 days
Sudha can do the same work = 60 days
Formula used:
Total work = efficiency × time
Calculation:
| Efficiency | Person | Time | Total work |
| 3 | Ravi | 40 | 120 |
| 2 | Sudha | 60 |
According to the question:
⇒ (Sudha + Ravi) = 2 days
⇒ (2 + 3) = 5 units = 2 days
⇒ 5 × 24 = 120 units = 2 × 24 = 48 days.
∴ The correct answer is 48 days.
X and Y can complete a work in 9 days and 36 days, respectively. X begins to do the work and they work alternately one at a time for one day each. The whole work will be complete in:
Answer (Detailed Solution Below)
Alternate Based Question 12 Detailed Solution
Download Solution PDFGiven:
X and Y can complete a work in 9 days and 36 days, respectively.
Formula used:
Work = Efficiency × Time
Calculation:
Total work = LCM of (9, 36)
Total work = 36 unit
The efficiency of X = 36/9 = 4 unit/day
The efficiency of Y= 36/36 = 1unit/day
A and B work alternatively, starting with A, then
5 units of work is done in 2 days by both of them.
= (5 × 7)unit → (2 × 7)days
In 14 days the work done by both = 35 units
The remaining work is 1 unit, which will be done by X in = 1/4
Total time taken = 14 + 1/4
The whole work will be completed in 14(1/4) days.
Shortcut Trick
A, B and C can finish a work in 15 days working alternately in the same order. The efficiency of A is the same as that of B, and the efficiency of C is equal to that of A. In how many days will the work be finished if they work alternately in the order C, A and B?
Answer (Detailed Solution Below)
Alternate Based Question 13 Detailed Solution
Download Solution PDFGiven:
Efficiency of A = Efficiency of B
Efficiency of C = Efficiency of A
Formula used:
Total work = efficiency × time
Calculation:
According to the question:
Efficiency of A = efficiency of B = efficiency of C
A, B and C can finish a work in 15 days working alternately.
⇒ 3 days = (1 + 1 + 1) = 3 units of work
⇒ 3 × 5 = 15 days = 3 × 5 = 15 units of work
Now, C,A and B can do the work alternately then,
⇒ (1 + 1 + 1) = 3 units of work = 3 days
⇒ 3 × 5 = 15 units of work = 3 × 5 = 15 days
∴ The correct answer is 15 days.
A can do a piece of work in 8 days, while B can do it in 18 days. In how many days will the work be completed if they both work on alternate days starting with B?
Answer (Detailed Solution Below)
Alternate Based Question 14 Detailed Solution
Download Solution PDFGiven:
A can do a piece of work in 8 days, while B can do it in 18 days.
Concept used:
Efficiency = (Total work / Total time taken)
Efficiency = work done in a single day
Calculation:
Let the total work is the 72 units (LCM of 8 and 18)
The efficiency of A is 72 / 8 = 9 units
The efficiency of Bi is 72 / 18 = 4 units
In 2 days they do 9 + 4 = 13 units
They do 65 units of work in 5× 2 = 10 days
Then B does 4 units in 1 day
Then A does 3 units in 1/3 day
Total number of days will be 10 + 1 + 1/3 = 11\(\frac{1}{3}\)
Raju, Shobh and Mohan can do a work in 15 days, 20 days and 25 days respectively. In how many days,will the work be finished, if they do it on alternate days, Starting from Raju?
Answer (Detailed Solution Below)
Alternate Based Question 15 Detailed Solution
Download Solution PDFGiven:
Raju, Shobh and Mohan can do a work in 15 days, 20 days and 25 days respectively.
Concept used:
Total work = Efficiency (Work done per day) × Total time taken
Calculation:
Let the total work be 300 units.
Efficiency of A (Raju) = 300/15 = 20 units/day
Efficiency of B (Shobh) = 300/20 = 15 units/day
Efficiency of C (Mohan) = 300/25 = 12 units/day
In three days, working on alternate days, they complete = (20 + 15 + 12) = 47 units of work
In such a manner, total work in 6 such sets of 3 days = 47 × 6 = 282 units
Remaining work = 300 - 282 = 18 units
Time taken to complete 18 units of work by A = \(\frac {18}{20} = \frac {9}{10}\ days\)
Total time taken = 6 × 3 + \(\frac {9}{10}\) = \(18\frac{9}{{10}}\) days
∴ The total time taken to complete the work is \(18\frac{9}{{10}}\) days.