Alternate Based MCQ Quiz in मल्याळम - Objective Question with Answer for Alternate Based - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 3, 2025
Latest Alternate Based MCQ Objective Questions
Top Alternate Based MCQ Objective Questions
Alternate Based Question 1:
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 1 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 2:
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 2 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.
Alternate Based Question 3:
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 3 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 4:
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 4 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 5:
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 5 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.